__________________________________________________________

Guidelines

For design of concrete and reinforced concrete structures made of heavy-weight and light-weight concrete without

reinforcement prestress

(Addition to SNiP 2.03.01-84)

__________________________________________________________________

Moscow Central Project Institute of Standard Designs

1989

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1. GENERAL RECOMMENDATIONS

Basic Positions

Recommendations of the present Guidelines are applied to design of concrete and reinforced concrete structures produced without reinforcement priestess made of heavy-weight, fine and light-weight concrete and used by temperature no more than 50 Celsius degree above zero and no less than 70 Celsius degree below zero.

Notes: 1. Recommendations of the Guidelines are not applied to design of concrete and reinforced concrete structures for water development facilities, bridges, transport tunnels, pipes under filling dams, highways and aerodromes covering. 2. Definitions “heavy-weight concrete”, “fine concrete” and “light-weight concrete” are used in accordance with GOST 25192-82.

Light-weight concretes may have compact and porous structure that’s why in the present Guidelines there are used definitions “light-weight concrete” for light-weight concrete of compact structure and “porous concrete” for light-weight concrete of porous structure with inter-granular openings more than 6 percent.

Types of light-weight and porous concretes as well as their application fields are given in Annex 1.

Concrete and reinforced concrete members of buildings and structures for corrosive

atmosphere and high humidity conditions should be designed considering requirements of SNiP 1.03.11-85.

(1.4) Prefabricated members structures must conform to requirements of mechanized

production at specialized plants. It is wise to enlarge elements of prefabricated structures as big as it is possible according to weight-lift ability of installing mechanisms, producing and transportation conditions.

(1.5) For monolithic structures it is necessary to use dimensions applicable for inventory

form work as well as enlarged three-dimensional reinforcement cages.

(1.6) It is necessary to pay more attention to rigidity and working life of connections.

Joints and connection structures of members must provide reliable transferring of forces, durability of members in connection zones as well as connection of additional concrete in joints with concrete of structure by means of different structural and technological measures.

(1.7) Concrete members are used:

a) in structures being pressed by little eccentricities of longitudinal forces, not exceeding the values given in Item 3.4;

b) in specific cases in structures being pressed by larger eccentricities as well as in bending structures if their failure does note constitute a danger for human life and equipment safety (members base on solid base etc).

Note: Structures are considered as concrete ones if their durability during the use period is provided only by concrete

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(1.8) Design winter temperature of outside air is taken as average air temperature of the coldest five-days week depending on the construction region according to SNiP 2.01.01-82. Design technological temperatures are settled in the project statement.

Environment air humidity is determined as average relative humidity of outside air of the hottest month according to the construction region in compliance with SNiP 2.01.01-82 or as relative air humidity of rooms of heated buildings.

Numerical values of given in the present document concrete and reinforcement design characteristics, limit values of crack openings and deflections are used only during design. For construction quality estimation it is necessary to follow the requirements of correspondent state standards and technical specifications.

Basic Calculation Requirements

(1.10) Concrete and reinforced concrete structures must meet the requirements of the load-carrying capacity calculation (first class limit states) and according to serviceability limit state (second class limit states). a) Calculation according to the first class limit states must protect structures against:

- Unstable, elastic or other failure (rigidity calculation considering deflection of the structure before failure);

- Structure stable form failure or position failure. - Endurance rupture (endurance limit calculation of the structure which is under

effect of repeated load – moving and pulsating); - Failures under influence of stresses and adverse environmental impacts (periodic

or permanent aggressive influences, freezing and melting etc);

b) Calculation according to the first class limit states must protect structures against: - Exceeding crack opening (calculation of the crack opening); - Exceeding displacements – deflections, rotation angles, vibration (deformation

calculation).

It is possible not to make calculation of concrete structures according to second class limit states as well as regarding the endurance limit. Notes: 1. Calculations of repeated loads are made in compliance with the recommendations of “Guidelines for design of prestressed reinforced concrete structures made of heavy-weight and light-weight concrete” (Moscow, 1986). 2. Calculations of the form stability or position stability as well as calculations of influence of stresses and adverse environmental impacts are made according to normative documents or Guidelines.

(1.11) Design to limit state of the structure in general as well as of members of structure must be made as a rule for all stages – manufacturing, transportation, installing and use; at the same time design schemes must meet the accepted construction solutions.

(1.12) Loads and effects values, values of safety factors as regards the load fγ , combinations

coefficients as well as dividing of loads into dead loads and live loads must be taken according to requirements of SNiP 2.01.07-85.

Loads values must be multiplied by safety factors as regards the purpose taken according to “Registration rules of responsibility degree of buildings and structures during design” approved by Gosstroy of the USSR.

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Loads considered during calculations of first class limit states (exploitative) must be taken according to Items 1.15 and 1.17. At the same time to long-term loads belong also a part of total value of short-term loads settled in SNiP 2.01.07-85 and short-term load inserted into the calculation must be taken as reduced by the value considered in long-term load (for example if snow load for the IIIrd region is s = 1000 H/m2 so snow long-term load will be

30010003.0 =×=s H/m2 and snow short-term load 7003001000 =−=s H/m2).

Combinations coefficients belong to total value of short-term loads. It is necessary to consider temperature climatic effects for structures not protected against solar irradiation for climatic sub-regions IVA according to SNiP 2.01.01-82.

(1.13) During calculation of members of prefabricated structures as regards the forces growing during their lifting, transportation and installation it is necessary to insert the load of the member weight with dynamic factor equal to: 1.60 – during transportation 1.40 – during lifting and installing

In this case it is also necessary to consider the load safety factor.

(1.15) Forces in statically indefinable reinforced concrete structures caused by loads and forced displacements (as result of changes of temperature, concrete moisture, supports displacements etc) as well as forces in statically indefinable reinforced concrete structures during their calculation as regards the deformation scheme must be determined considering inelastic concrete and reinforcement deformations and cracks presence.

It is possible to determine forces in statically indefinable reinforced concrete structures on the basis of their linear elasticity for structures whose calculation methods considering inelastic characteristics of reinforced concrete are not worked out as well as for intermediate stage of the calculation considering inelastic characteristics.

(1.16) Width of long-lived and short-lived crack openings for members used in non-aggressive conditions must not exceed values mentioned in Table 1.

Members mentioned in Position 1a of Table 1 can be designed without prestressing only by special justification Table 1 (1, 2)

Limit width of crack opening, mm Work conditions of the structure

Short-lived acrc1 Long-lived acrc2

1. members carrying the load of liquids or gases if the section is a) fully stretched b) partly compressed

2. members carrying the load of granular materials 3. members used in the ground with variable ground-

water elevations 4. other members

0.2 0.3 0.3 0.3

0.4

0.1 0.2 0.2 0.2

0.3

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Note. By short-lived crack opening we shall basically understand opening under effect of dead loads, long-term and short-term loads; by long-lived crack opening we shall understand – only under effect of dead loads and long-term loads. At the same time safety factor is equal to 1.

(1.19) For under-reinforced concrete members whose load-carrying capacity becomes exhausted concurrent with crack opening in the stretched concrete zone, sectional area of longitudinal stretched reinforcement must be increased by no less than 15 percent in comparison with calculations requirements.

Such increase is to be fulfilled upon the following condition

ucrc MM ≥ ,

where crcM is crack opening moment determined according to Item 4.2 replacing value

serbtR , by serbtR ,2.1 ;

uM is moment corresponding to load-carrying capacity exhaust and determined

according to Items 3.15-3.80; for eccentric compressed and stretched members values are determined relating to the axis going through core point the most distant from the stretched zone (see Item 4.2).

This requirement can be applied to elements which rest on a solid base.

(1.20) Deflections of members of reinforced concrete structures must not exceed limit values

settled considering the following requirements:

a) technological requirements (normal running conditions of cranes, process units, machines, etc);

b) structural requirements (neighbor elements influence; given grade of slope, etc); c) esthetic requirements (a person’s impression of structure workability).

Deflection limits values are given in Table 2.

Deformation calculation must be made by technological or constructive requirements as regards dead loads, short-term and long-term loads; by esthetic loads as regards dead

loads and long-term loads. At the same time it is taken 0.1=fγ

By dead loads, short-term and long-term loads beams and slabs deflections must not exceed 1/150 of a span and 1/75 of an overhanging length in all cases.

Limit deflections values can be increased by the height of a camber if it is not restricted by technological or constructive requirements.

If in the lower room with plain ceiling there are partition walls (not supporting) located across the span of member l and if the distance between these partition walls is lp so the deflection of the member within the distance lp (counted from the line connecting top points of partition walls axes) can be taken up to 1/200lp, at the same time limit deflection must be no more than 1/500l.

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Table 2 (4)

Structure members Deflection limits

1. Crane beams

For manually operated cranes For electric cranes

500

l

600

l

2. Floors with a plane ceiling and roof members (except the ones mentioned in position 4) if the span is:

l < 6

6 ≤ l ≤ 7.5

l > 7.5

200

l

3 cm

250

l

3. Floors with ribbed ceiling and stairs members if the span is:

l < 5

5 ≤ l ≤ 10

l > 10

200

l

2.5 cm

400

l

4. Roof elements of agricultural building for production purpose if the span is:

l < 6

6 ≤ l ≤ 10

l > 10

150

l

4 cm

250

l

5. Suspended wall panels if the span is:

l < 6

6 ≤ l ≤ 7.5

l > 7.5

200

l

3 cm

250

l

Symbols: l is beams or slabs span; for consoles it is necessary to take l equal to double overhanging length.

(1.20) For not connected with neighbor members structures of floor slabs, flights of stairs,

platforms etc it is necessary to run additional check as regards the instability: additional deflection caused by short-term center-point load 1000 H by the worst loading scheme must be no more than 0.7 mm.

(1.22) The distance between contraction joints must be settled according to the calculation. It is possible not to make the calculation if the distance between contraction joints by design if temperature of outside air 40 Celsius degrees below zero and higher doesn’t exceed values given in Table 3. For framework buildings and structures without top-running bridge crane if in the considered direction there are bracings (stiffening diaphragms) the values given in Table 3 can be multiplied by the coefficient equal to:

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ftt δδδδ ∆= ,

but no less than one,

Where t∆δ is the coefficient taken equal to ε

δ+∆

⋅=

−

−

∆

w

tt

5

5

10

1050 for heated buildings and

c

tt∆

=∆

60δ for not heated buildings and structures (here cw tt ∆∆ , are design

temperature changes in Celsius degrees determined in compliance with SNiP 2.01.07-85, ε – is coefficient of strain of longitudinal elements caused by vertical loads. For reinforced concrete elements it is possible to

take 4101 −⋅=ε , for other members 4103 −⋅=ε );

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/ hll =δ (Here l is the length of the column between fixing points, h is the height

of the column section in the direction under consideration);

1100/4.0 ≤+= extϕδϕ (Here extϕ is external air humidity in percents during the

hottest month of the year, taken in accordance with SNiP 20.1.01-82).

When considering the coefficient δ the distance between contraction joints must be no more than 150 m for heated buildings made of prefabricated structures, 90 m – for heated buildings made of prefabricated-monolithic and monolithic structures; for not heated building and structures the mentioned above values must be reduced by 20 percent.

Table 3

Maximum distances in meters between contraction joints allowable without calculation for structures

located

Structures

inside of heated buildings or in the ground

inside of not heated buildings

in the open

1. Concrete structures a) prefabricated b) monolithic:

by constructive reinforcement without constructive reinforcement

40

30 20

35

25 15

30

20 10

2. reinforced concrete structures: a) prefabricated-frame structure:

one-storey multi-storey

b) prefabricated-monolithic and monolithic structures: frame structures solid structures

72 60

50 40

60 50

40 30

48 40

30 25

Note: For reinforced concrete frame structures (pos. 2) the distances between contraction joints are determined without bracings or if bracings are located in the middle of the temperature block.

During calculation of the floor as regards all limit states the weight of partition walls located

along the slabs span is considered in the following manner: a) The load from weight of blind rigid partition wall (for example reinforced concrete

prefabricated wall made of horizontal members, reinforced concrete monolithic wall, stone wall, etc) is applied concentrated at the distance of 1/12 of the partition wall length from its edges;

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b) If there is an opening in the rigid partition wall and the opening is located within one half of the partition wall so the load from the smaller pier (including the load of the half part above the opening) is applied concentrated at the distance 1/3 of the pier length and the load of weight of another part of the partition wall is applied at the distance 1/12 of the length of this part from the opening edge and from the partition wall edge; if the opening is arranged differently so the load is applied at the distance 1/18 of corresponding parts of a partition wall and of their edges;

c) If there are two and more openings in a partition wall so the load of the weight of this partition wall is applied concentrated on the centers of parts supported on the floor;

d) For other partition walls 60 percent of their weight is distributed along the partition wall length (on the parts between openings) and 40 percent of the weight is applied in compliance with sub-items “a” – “b”.

Local load among members of prefabricated floors made of hollow-cored or solid slabs is

spread in the following manner if the joints between slabs are grouted well: a) By calculations as regards all limit states it is taken the following spread of load from

the weight of partition walls located along the span of slabs with the same width: - If the partition wall is located within one plate so this plate carries 50 percent of

the partition wall weight and two neighbor plates carry 25 percent of its weight; - If the partition wall is supported on two neighbor plates so the weight of the

partition wall is spread among them. b) By calculations of the second class limit states local concentrated loads located within

a center third of the slab span are applied on the width no more than a length of the span; by the durability calculation such spread of concentrated loads can be applied only if neighbor plates are doweled (see Item 3.115).

Note. If the floor is formed of two slabs supported at three sides and the partition wall is located within one slab so this slab carries 75 percent of the partition wall weight; in this case the load from the partition wall is transferred according to Item 1.20 if the partition wall is located both along and across the slab.

2. MATERIALS FOR CONCRETE AND REINFORCED CONCRETE

STRUCTURES

Concrete

(2.3) For concrete and reinforced concrete structures it is necessary to use the following materials:

a) concrete class as regards the resistance against compression

- heavy-weight concrete – B3.5; B5; B7.5; B10; B12.5; B15; B20; B25; B30; B35; B40; B45; B50; B55; B60;

- fine concrete groups: A – aging concrete or concrete tempered by pressure of air on the sand with fineness modulus more than 2.0 – B3.5; B5; B7.5; B10; B12.5; B15; B20; B25; B30; B (Rus. – Б) – the same with fineness modulus 2.0 and less – B3.5; B5; B7.5; B10; B12.5; B15; B20; B25; B30; C (Rus. – В) – autoclaved concrete – B15; B20; B25; B30; B40; B45; B50; B55; B60;

- light-weight concrete if average density grades are the following: D800, D900 – B2.5; B3.5; B5; B7.5* D1000, D1100 – B2.5; B3.5; B5; B7.5; B10; B12.5*; D1200, D1300 – B2.5; B3.5; B5; B7.5; B10; B12.5; B15*;

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D1400, D1500 – B3.5; B5; B7.5; B10; B12.5; B15; B20*; B25*; B30*; D1600, D1700 – B5; B7.5; B10; B12.5; B15; B20; B25*; B30*; B35*; D1800, D1900 – B10; B12.5; B15; B20; B25*; B30*; B35*; B40*; D2000 – B20; B25; B30; B35*; B40*;

- porous concrete if average density grades are: D800, D900, D1000 – B2.5; B3.5; B5; B7.5;

D1100, D1200, D1300, D1400 – B3.5; B5; B7.5 b) concrete class as regards the resistance to frost:

heavy-weight and fine concrete – F50; F75; F100; F150; F200; F300; F400; F500; light-weight concrete – F25; F35; F50; F75; F100; F150; F200; F300; F400; F500; porous concrete – F15; F25; F35; F50; F75; F100;

c) concrete class as regards the water permeability – W2; W4; W6; W8; W10; W12; d) concrete class as regards the average density:

light-weight concrete – D800; D900; D1000; D1100; D1200; D1300; D1400; D1500; D1600; D1700; D1800; D1900; D2000;

porous concrete – D800; D900; D1000; D1100; D1200; D1300; D1400 ____________

* The present grade of light-weight concrete based on natural aggregate, foamed slag and fly ash aggregate can be used only if it is approved by the manufacturing plant.

Notes: 1. It is necessary to take concrete grade according to resistance to axis tension for structures whose resistance to tension is the main characteristic in compliance with SNiP 2.03.01-84.

2. Definitions “concrete grade” and “concrete class” see in GOST 25192-82. 3. According the present Guidelines porous concrete can be used only for eccentric compressed

concrete and reinforced concrete members.

(2.4) Concrete age conforming to its grade according to resistance to compression is taken in compliance with possible terms of structure loading by design loads, mode of building, concrete hardening conditions. In case if there is no this data concrete age is taken 28 days.

Concrete strength of members of prefabricated structures is taken according to GOST 13015.0-83.

(2.5) For reinforced concrete structures it is impossible to use: - heavy-weight and fine concrete less than B7.5 concrete grade according to

resistance to compression; - light-weight concrete of grade B2.5 as regards the resistance to compression – for

one-layer structures; - concrete of grade no less than B25 – for heavily loaded reinforced concrete axial

element (for example for columns carrying heavy crane loads and for columns of lower storeys of multistory buildings);

- concrete of grade no less than B15 for thin-walled reinforced concrete structures as well as for walls of buildings and structures built up in slip or traveling forms.

For concrete compressed members it is not recommended to use more than B30 concrete grade.

(2.8) For building-in of members joints of prefabricated reinforced concrete structures concrete grade must be taken according to work conditions of joined members but it must be no less than B7.5.

(2.9) Concrete grades as regards resistance to frost and to water of concrete and reinforced concrete structures (according to their use mode and design winter temperatures of outside air in the construction region) must be the following:

- no less than the ones shown in Table 4 – for buildings structures (except external walls of heated buildings);

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- no less than the ones sown in Table 5 – for external walls of heated buildings.

Table 4 (9)

Structure work conditions Concrete grade no less than

according to resistance to frost

according to resistance to water

for building structures (except external walls of heated buildings) of responsibility degree

mode characteristics design winter temperature of external air, in Celsius

degrees

I II III I II III

W4

W2

W2

Not regulat

ed

W6

W4

W2 Not regulated

Lower than 40 degrees below zero Lower than 20 degrees below zero up to 40 degrees below zero Lower than 5 degrees below zero up to 20 degrees below zero 5 degrees below zero and more

F300

F200

F150

F100

F200

F150

F100

F75

F150

F100

F75

F50 Not regulated

W2 Not regulat

ed

W4

W2 Not regulated

Lower than 40 degrees below zero Lower than 20 degrees below zero up to 40 degrees below zero Lower than 5 degrees below zero up to 20 degrees below zero 5 degrees below zero and more

F200

F100

F75

F50

F150

F75

F50

F35*

F100

F50

F35*

F25*

Not regulated

Not regulated

W4 W2 Not regulat

ed

1. Alternate freezing and melting a) in waterlogged state

(for example structures located in the ground layer seasonally melting in the permafrost region)

b) in conditions of

occasional water saturation (for example overland structures exposed to atmospheric effects)

c) in air humidity conditions without occasional water saturation (for example structures exposed to atmospheric effects but protected against atmospheric precipitation)

Lower than 40 degrees below zero Lower than 20 degrees below zero up to 40 degrees below zero Lower than 5 degrees below zero up to 20 degrees below zero 5 degrees below zero and more

F150

F75

F50

F35*

F100

F50

F35*

F25*

F75

F35*

F25*

F15**

Not regulated

Not regulated

Not regulated

2. Possible occasional temperature influence lower than 0 degree below zero

a) in waterlogged state (for example structures located in the ground or under water)

Lower than 40 degrees below zero Lower than 20 degrees below zero up to 40 degrees below zero Lower than 5 degrees below zero up to 20 degrees below zero 5 degrees below zero and more

F150

F75

F50

F35*

F100

F50

F35*

F25*

F75

F35

F25

Not regulat

ed

Not regulated

Not regulated

Not regulated

Not regulated

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b) in air humidity conditions (for example internal structures of heated buildings during construction and assembling)

Lower than 40 degrees below zero Lower than 20 degrees below zero up to 40 degrees below zero Lower than 5 degrees below zero up to 20 degrees below zero 5 degrees below zero and more

F75

F50

F35*

F25*

F50

F35*

F25*

F15**

F35*

F25*

F15**

Not regulat

ed

Not regulated

Not regulated

Not regulated

Not regulated

* For heavy-weight and fine concrete the grades as regards resistance to frost are not regulated. ** For heavy-weight, fine and light-weight concrete the grades as regards resistance to frost are not regulated. Notes: 1. Concrete grades as regards resistance to frost and to water for water supply and sewer systems buildings as well as for piles and pile shells must be taken in compliance with requirements of corresponding normative documents. 2. Design winter temperatures of external air are taken according to instructions of Item 1.8.

Table 5 (10)

Structure work conditions Minimum concrete grade according to resistance to frost of external walls of heated buildings made of

light-weight, porous concrete

heavy-weight, fine concrete

for building structures (except external walls of heated buildings) of responsibility degree

relative degree of humidity of internal air inside of

rooms intϕ , in percents

design winter temperature of external air, in Celsius

degrees

I II III I II III

F150

F75

F50

F100

F50

Not regulat

ed

1. intϕ > 75

Lower than 40 degrees below zero Lower than 20 degrees below zero up to 40 degrees below zero Lower than 5 degrees below zero up to 20 degrees below zero 5 degrees below zero and more

F100

F75

F50

F35

F75

F50

F35

F25

F50

F35

F25

F15*

F200

F100

F75

F50

Not regulated

F75 F50 F100

F50

Not regulated

F35

F25

F15* Not regulated

2. 60 < intϕ ≤ 75 Lower than 40 degrees below zero Lower than 20 degrees below zero up to 40 degrees below zero Lower than 5 degrees below zero up to 20 degrees below zero 5 degrees below zero and more

F75

F50

F35

F25

F50

F35

F25

F15*

Not regulated

F75 F50 Not regulat

ed

F25

F15* Not regulated

F35

F25

F15* Not regulated

3. intϕ ≤ 60 Lower than 40 degrees below zero Lower than 20 degrees below zero up to 40 degrees below zero Lower than 5 degrees below zero up to 20 degrees below zero 5 degrees below zero and more

F50

F35

F25

F15* Not regulated

* For light-weight concretes the grades as regards resistance to frost are not regulated.

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Notes: 1. If structures made of heavy-weight, fine and light-weight concretes have vapor- and hydro-insulation so their grades as regards resistance to frost shown in the present table must be decreased by one degree. 2. Design winter temperatures of external air are taken according to instructions of Item 1.8.

(2.10) For building-in of members joints of prefabricated reinforced concrete structures exposed to freezing temperature of external air during use period or assembling it is necessary to use concretes of design grades as regards resistance to frost and water no less than grades of concrete of joined members.

For light-weight concretes it is necessary to take concrete grades as regards average density in compliance with Table 6. Table 6

Grades regarding average density for Light-weight concrete grade as regards the resistance to compression

expanded-clay concrete shungite concrete

slag-pumeconcrete

slag-concrete

perlite concrete concrete of natural expanded aggregate

agloporite concrete

B2.5 B3.5 B5 B7.5 B10 B12.5 B15 B20 B25 B27.5* B30 B35 B40

D800–D1000 D800–D1100 D800–D1200 D900–D1300

D1000–D1400 D1000–D1400 D1200–D1700 D1300–D1800 D1300–D1800 D1400–D1800 D1500–D1800 D1600–D1900 D1700–D1900

D1000–D1400 D1100–D1500 D1200–D1600 D1300–D1700 D1400–D1800 D1400–D1800 D1600–D1800 D1700–D1900 D1800–D1900 D1900–D2000

– – –

D800–D900 D800–D1000 D800–D1100 D900–D1200

D1000–D1300 D1000–D1400 D1300–D1600

– – – – – –

D800–D1200 D900–D1300

D1000–D1400 D1100–D1500 D1200–D1600 D1200–D1600 D1500–D1700 D1600–D1800 D1700–D1900 D1800–D2000 D1900–D2000

– –

D1000–D1200 D1100–D1300 D1200–D1400 D1300–D1500 D1400–D1600 D1400–D1600 D1600–D1800 D1700–D1900 D1700–D1900 D1800–D2000 D1900–D2000

– –

* Is used with a view to economize cement in comparison with use of concrete of grade B30 and to save other technical-economical characteristics of the structure

Standard and Design Characteristics of Concrete

(2.11) Standard resistance of concrete is also resistance to centric compression of prism

(prism strength) bnR and resistance to centric tension btnR .

Design resistances of concrete bnR and btnR according to concrete class B are given in

Table 7.

(2.11, 2.13) Design resistances of concrete for first class limit states bR and btR are

determined by means of dividing of standard resistances into safety factors for concrete equal to:

by tension 3.1=bcγ ; by compression 5.1=btγ .

Design concrete resistances bR and btR are to be decreased (or increased) by means of

multiplying by concrete work conditions coefficients biγ considering work conditions of

the structure, process of manufacturing, sections dimensions etc.

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Table 7 (12)

Standard resistances of concrete bnR and btnR and design resistances for second

class limit states serbR , and serbtR , , in Mega Pascal (kilogram-force per cm2) if

concrete grade as regards resistance to compression is

Resistance type Concrete

B2.5 B3.5 B5 B7.5 B10 B12.5 B15 B20

Axial compression (prism strength)

bnR and serbR ,

heavy-weight, fine, light-weight

1.9 (19.4)

2.7 (27.5)

3.5 (35.7)

5.5 (56.1)

7.5 (76.5)

9.5 (96.9)

11.0 (112)

15.0 (153)

heavy-weight, fine1, light-weight with dense aggregate

0.29 (2.96)

0.39 (4.00)

0.55 (5.61)

0.70 (7.14)

0.85 (8.67)

1.00 (10.2)

1.15 (11.7)

1.40 (14.3)

Axial tension

btnR and serbtR ,

Light-weight concrete with porous aggregate2

0.29 (2.96)

0.39 (4.00)

0.55 (5.61)

0.70 (7.14)

0.85 (8.67)

1.00 (10.2)

1.10 (11.2)

1.20 (12.2)

Standard resistances of concrete bnR and btnR and design resistances for second

class limit states serbR , and serbtR , , in Mega Pascal (kilogram-force per cm2) if

concrete grade as regards resistance to compression is

Resistance type Concrete

B25 B30 B35 B40 B45 B50 B55 B60

Axial compression (prism strength)

bnR and serbR ,

heavy-weight, fine, light-weight

18.5 (189)

22.0 (224)

25.5 (260)

29.0 (296)

32.0 (326)

36.0 (367)

39.5 (403)

43.0 (438)

heavy-weight, fine1, light-weight with dense aggregate

1.60 (16.3)

1.80 (18.4)

1.95 (19.9)

2.10 (21.4)

2.20 (22.4)

2.30 (23.5)

2.40 (24.5)

2.50 (25.5)

Axial tension

btnR and serbtR ,

Light-weight concrete with porous aggregate2

1.35 (13.8)

1.50 (15.3)

1.65 (16.8)

1.80 (18.4)

– – – –

1 For fine concrete of groups Б (see Item 2.1) values btnR and serbtR , are decreased by 15 percent.

2 For expanded-clay perlite concrete on expanded perlite sand values btnR and serbtR , are decreased by 15 percent.

Note. For porous concrete values bnR and serbR , are taken the same as for light-weight concrete and values btnR and

serbtR , are multiplied by coefficient 0.7.

Design resistances of concrete for second class limit states serbR , and serbtR , are taken

equal to standard resistances and are inserted into the calculation with the concrete

work condition coefficient 0.1=biγ .

Design resistances of concrete according to concrete resistance to compression are given: in Table 8 – for the first class limit states; in Table 7 – for the second class limit states.

Design resistances given in Table 8 include work condition coefficient 2bγ considering

duration of loads action influence and strength gain of concrete; coefficient 2bγ usage

order is given in Item 3.1.

14

In case of need design resistances of concrete given in Table 8 must be multiplied by work conditions coefficients according to Table 9.

(2.14) Concrete tangent modulus of elasticity values Eb by tension and compression are taken according to Table 11.

For concretes being permanently frozen and melted (see pos. 1 of Table 4) values Eb

given in Table 11 must be multiplied by work condition coefficient 6bγ taken according

to Table 10.

(2.15) Linear temperature deformation coefficient btα by temperature variation from 40

degree below zero up to 50 degree above zero is taken equal to:

- 5101 −× ˚C-1 – for heavy-weight, fine and light-weight concrete with fine dense aggregate;

- 5107.0 −× ˚C-1 – for light-weight concrete with fine porous aggregate.

- 5108.0 −× ˚C-1 – for porous concrete.

(2.16) Prime coefficient of concrete deformation v (Poisson number) is taken equal to 0.2 for all concrete types and modulus of shear of concrete G is taken equal to 0.4, corresponding values Eb given in Table 11.

For determination of weight of reinforced concrete or concrete structures concrete density is taken equal to: 2400 kg/m3 – for heavy-weight concrete; 2200 kg/m3 – for fine concrete; for light-weight and porous concrete it is necessary to multiply concrete grade as regards average density D by 1.05 – for concrete grade B12.5 and more, and by

100/1 w+ (where w is gravimetric humidity of concrete during its use determined according to SNiP II-3-79**, it is possible to take w equal to 10 percent) – for concrete grade B10 and less. During calculation of structures at stage of manufacturing and transportation light-weight and porous concrete density is determined considering

transport volume humidity ω by formula 1000100

ω+D where ω = 15 and 20 percent

correspondingly for light-weight and porous concrete grade B10 and less and ω = 10 percent for light-weight concrete of class B12.5 and more.

Reinforced concrete density by reinforcement content 3 percent and less can be taken more than concrete density by 100 kg/m3; if reinforcement content is more than 3 percent so density is determined as a sum of concrete and reinforcement weight per unit of volume of reinforced concrete structure. At the same time weight of 1 m of reinforcement steel is taken according to Annex 4 and weight of strip iron, angle steel and section steel – according to state standards. During determination of external walling structures weight made of light-weight concrete of grade B100 and less it is necessary to consider high density of textured layers. For determination of loads of dead weight of the structure it is possible to take its specific weight kN/m3 equal to 0.01 of density kg/m3.

Reinforcement

(2.19) As non-prestressed reinforcement of reinforced concrete structures (except the ones

mentioned in Item 2.15):

15

it is necessary to use:

a) ribbed rod reinforcement A-III, and At-IIIC; b) ribbed regular reinforcement wire of class Bp-I in welded meshes and frameworks

it is possible to use:

c) ribbed rod reinforcement A-II and plain reinforcement A-I for cross reinforcement as well as for working longitudinal reinforcement if other kinds of reinforcement can’t be used;

d) regular reinforcement wire of class Bp-I for bound stirrups of beams up to 400 mm high and columns.

Reinforcement grade A-III, At-IIIC, A-II and A-I must be used in form of welded frameworks and welded meshes. Under economical justification it is possible to use non-prestressed reinforcement A-IV, A-V and A-VI as pressed reinforcement, and reinforcement A-IV as stretched reinforcement. It is also possible to use reinforcement A-IIIв as stretched reinforcement. The elements with mentioned above reinforcement must be designed in compliance with “Guidelines for design of prestressed reinforced concrete structures made of heavy-weight and light-weight concrete” (Gosstroy USSR, 1986) As constructive reinforcement of reinforced concrete structures it is also possible to use regular plain bars B-I. Notes: 1. In the present document there is used the definition “bar” for reinforcement of any diameter, type and section. 2. Special purpose rod reinforcement A-II is lettered as Ac-II with the letter “c”.

(2.20) In structures with non-prestressed reinforcement which are under gas or liquid

pressure: it is necessary to use:

a) rod reinforcement A-II and A-I; it is possible to use:

b) rod reinforcement A-III and At-IIIC; c) reinforcement wire Bp-I.

(2.23) When choosing type and grade of steel for reinforcement as well as rolled iron for embedded elements it is necessary to consider temperature conditions of use of the structure and loading schemes according to Table 12 and 13.

During installation works performed during cold seasons in climatic regions with design winter temperature less than 40 Celsius degrees below zero load-carrying capacity of structures with reinforcement which can be used only in heated buildings must be provided reasoning from design resistance of reinforcement with reduction

factor 0.7 and from design load with safety factor 0.1=fγ

(2.24) For lifting loops of members of prefabricated reinforced concrete and concrete

structures it is necessary to use hot-rolled reinforcement steel Ac-II of grade 10ГТ and A-I of grade ВСт3сп2 and ВСт3пс2.

If the installation of structures is possible by design winter temperature lower than 40 Celsius degrees below zero so it is possible to use steel of grade ВСт3пс2 for lifting loops.

Table 8

Design resistance of concrete for the first class limit states bR and btR , Mega Pascal (kilogram-force/cm2) if class of concrete as regards the

resistance to compression is

Resistance type Concrete Work condition coefficient

2bγ B2.5 B3.5 B5 B7.5 B10 B12.5 B15 B20 B25 B30 B35 B40 B45 B50 B55 B60

0.9 1.3 (13.3)

1.9 (19.4)

2.5 (25.5)

4.0 (4.08)

5.4 (55)

6.7 (68.5)

7.7 (78.5)

10.5 (107)

13.0 (133)

15.5 (158)

17.5 (178)

20.0 (204)

22.5 (230)

25.0 (230)

27.0 (275)

29.5 (300)

1.0 1.5 (15.3)

2.1 (21.4)

2.8 (28.6)

4.5 (45.9)

6.0 (61.2)

7.5 (76.5)

8.5 (86.7)

11.5 (117)

14.5 (148)

17.0 (173)

19.5 (199)

22.0 (224)

25.0 (255)

27.5 (280)

30.0 (306)

33.0 (336)

Axial compression (prism strength) Rb

Heavy-weight, fine and light-weight

1.1 1.6 (16.3)

2.3 (23.4)

3.1 (32.6)

4.9 (50)

6.6 (67.3)

8.2 (83.5)

9.4 (96)

12.5 (128)

16.0 (163)

19.0 (194)

21.5 (219)

24.0 (245)

27.5 (280)

30.5 (310)

33.0 (334)

36.5 (370)

0.9 0.18 (1.84)

0.23 (2.34)

0.33 (3.33)

0.43 (4.39)

0.51 (5.20)

0.59 (6.01)

0.67 (6.83)

0.80 (8.16)

0.95 (9.7)

1.10 (11.2)

1.15 (11.7)

1.25 (12.7)

1.30 (13.3)

1.40 (14.3)

1.45 (14.8)

1.50 (15.3)

1.0 0.20 (2.04)

0.26 (2.65)

0.37 (3.77)

0.48 (4.89)

0.57 (5.81)

0.66 (6.73)

0.75 (7.65)

0.90 (9.18)

1.05 (10.7)

1.20 (12.2)

1.30 (13.3)

1.40 (14.3)

1.45 (14.8)

1.55 (15.8)

1.60 (16.3)

1.65 (16.8)

Heavy-weight, fine1 and light-weight concrete with fine dense aggregate

1.1 0.22 (2.24)

0.29 (2.96)

0.41 (4.18)

0.53 (5.40)

0.63 (6.43)

0.73 (7.45)

0.82 (8.36)

1.00 (10.2)

1.15 (11.7)

1.30 (13.3)

1.45 (14.8)

1.55 (15.8)

1.60 (16.3)

1.70 (17.3)

1.75 (17.8)

1.80 (18.4)

0.9 0.18 (1.84)

0.23 (2.34)

0.33 (3.33)

0.43 (4.39)

0.51 (5.20)

0.59 (6.01)

0.66 (6.73)

0.72 (7.34)

0.81 (8.26)

0.90 (9.18)

1.00 (10.2)

1.10 (11.2)

– – – –

1.0 0.20 (2.04)

0.26 (2.65)

0.37 (3.77)

0.48 (4.89)

0.57 (5.81)

0.66 (6.73)

0.74 (7.55)

0.80 (8.16)

0.90 (9.18)

1.00 (10.2)

1.10 (11.2)

1.20 (12.2)

– – – –

Axial tension Rbt

Light-weight concrete with fine porous aggregate2

1.1 0.22 (2.24)

0.29 (2.96)

0.41 (4.18)

0.53 (5.40)

0.63 (6.43)

0.73 (7.45)

0.81 (8.26)

0.90 (9.18)

1.00 (10.2)

1.10 (11.2)

1.2 (12.2)

1.30 (13.3)

– – – –

1 For fine concrete of group Б (see Item 2.1) values Rbt are decreased by 15 percent. 2 For expanded-clay perlite concrete on expanded perlite sand values Rbt are decreased by 15 percent. Notes: 1. For porous concrete values Rb are taken the same like for light-weight concrete and values Rbt are multiplied by the coefficient 0.7.

2. Application conditions of work condition coefficient 2bγ are given in Item 3.1.

3. Design concrete resistance with the work condition coefficient 0.12 =bγ are taken in compliance with Table 13 of SNiP 2.03.01-84.

Table 9 (15)

Work condition coefficient of concrete Factors providing work condition coefficient insertion

graphical symbol number identification

1. Concreting in vertical position (concreting layer height is more than 1.5 m)

3bγ 0.85*

2. Concreting of monolithic poles and reinforced concrete columns with maximum section dimension less than 30 cm

5bγ 0.85

3. Alternate freezing and melting 6bγ See Table 10

4. Use of not protected against solar radiation structures in climatic sub-region IVA according to SNiP 2.01.01-82

7bγ 0.85

5. Concrete structures 9bγ 0.90

6. Concrete structures of heavy-weight concrete B35 and higher or of fine concrete B25 and higher

10bγ 0.3+ω≤1 (value ω see in

Item 3.14) 7. Concrete for joints filling of prefabricated elements

if thickness of the joint is less than 1/5 of the least dimension of the member section and less than 10 cm.

12bγ 1.15

*For members of porous concrete 3bγ = 0.80

Notes: 1. Work condition coefficients according pos. 3-5 must be considered during determination of design resistances Rb and Rbt, according other positions only during determination of Rb. 2. Work conditions coefficients of concrete are inserted independently on each other but at the same time their

product [including 2bγ (see Item 3.1)] must be no less than 0.45.

Table 10 (17)

Work conditions coefficient of

concrete 6bγ by alternate freezing and

melting of the structure

Structure application conditions

Design winter temperature of external air, Celsius degrees

for heavy-weight and fine concrete

for light-weight and porous

concrete Lower than 40 degrees below zero Lower than 40 degrees below zero up to 40 degrees below zero Lower than 5 degrees below zero up to 20 degrees below zero 5 degrees below zero and higher

0.70

0.85

0.90

0.95

0.80

0.90

1.00

1.00

Alternate freezing and melting

a) in water saturated state (see pos. 1a of Table 4);

b) in conditions of occasional water saturation (see pos. 1b of Table 4)

Lower than 40 degrees below zero 40 degrees below zero and higher

0.90

1.00

1.00

1.00

Notes: 1. Design winter temperature of external air is taken according to Item 1.8. 2. If concrete grade as regards resistance to frost in comparison with a required one according to Table 4 the coefficient of the present table can be decreased by 0.05 according to each decrease step but they cannot be more than 1.

Table 11 (18)

Prime concrete modulus of elasticity 310−⋅bE Mega Pascal (kilogram-force/cm2) if concrete class as regards resistance to compression is Concrete

B2.5 B3.5 B5 B7.5 B10 B12.5 B15 B20 B25 B30 B35 B40 B45 B50 B55 B60

Heavy-weigh: - of air hardening; - exposed to thermal treatment by air pressure

–

–

0.95

(96.9) 8.5

(86.7)

13.0 (133) 11.5 (117)

16.0 (163) 14.5 (148)

18.0 (184) 16.0 (163)

21.0 (214) 19.0 (194)

23.0 (235) 20.5 (209)

27.0 (275) 24.0 (245)

30.0 (306) 27.0 (275)

32.5 (331) 29.0 (296)

34.5 (352) 31.0 (316)

36.0 (367) 32.5 (332)

37.5 (382) 34.0 (347)

39.8 (398) 35.0 (357)

39.5 (403) 35.5 (362)

40.0 (408) 36.0 (367)

Fine concrete of groups: A–of air hardening;

exposed to thermal treatment by air pressure

Б– of air hardening;

exposed to thermal treatment by air pressure

В–of autoclave hardening

–

–

–

–

–

7.0

(71.4) 6.5

(66.3) 6.5

(66.3) 5.5

(56.1) –

10.0 (102) 9.0 (92) 9.0

(91.8) 8.0

(81.6) –

13.5 (138) 12.5 (127) 12.5 (127) 11.5 (117)

–

15.5 (158) 14.0 (143) 14.0 (143) 13.0 (133)

–

17.5 (178) 15.5 (158) 15.5 (158) 14.5 (148)

–

19.5 (199) 17.0 (173) 17.0 (173) 15.5 (158) 16.5 (168)

22.0 (224) 20.0 (204) 20.0 (204) 17.5 (178) 18.0 (184)

24.0 (245) 21.5 (219) 21.5 (219) 19.0 (194) 19.5 (199)

26.0 (265) 23.0 (235) 23.0 (235) 20.5 (209) 21.0 (214)

27.5 (280) 24.0 (245)

–

– 22.0 (224)

28.5 (291) 24.5 (250)

–

– 23.0 (235)

–

–

–

– 23.5 (240)

–

–

–

– 24.0 (245)

–

–

–

– 24.5 (250)

–

–

–

– 25.0 (255)

Light-weight and porous of grade as regards average density D:

800 1000 1200 1400 1600 1800 2000

4.0 (40.8)

5.0 (51.0)

6.0 (61.2)

7.0 (71.4)

–

–

–

4.5 (45.9)

5.5 (56.1)

6.7 (68.3)

7.8 (79.5)

9.0 (91.8)

–

–

5.0 (51.0)

6.3 (62.4)

7.6 (77.5)

8.8 (89.7) 10.0 (102) 11.2 (114)

–

5.5 (56.1)

7.2 (73.4)

8.7 (88.7) 10.0 (102) 11.5 (117) 13.0 (133) 14.5 (148)

–

8.0 (81.6)

9.5 (96.9) 11.0 (112) 12.5 (127) 14.0 (143) 16.0 (163)

–

8.4 (85.7) 10.0 (102) 11.7 (119) 13.2 (135) 14.7 (150) 17.0 (173)

–

–

10.5 (107) 12.5 (127) 14.0 (143) 15.5 (158) 18.0 (184)

–

–

–

13.5 (138) 15.5 (158) 17.0 (173) 19.5 (199)

–

–

–

14.5 (148) 16.5 (168) 18.5 (189) 21.0 (214)

–

–

–

15.5 (158) 17.5 (178) 19.5 (199) 22.0 (224)

–

–

–

–

18.0 (184) 20.5 (209) 23.0 (235)

–

–

–

–

–

21.0 (214) 23.5 (240)

–

–

–

–

–

–

–

–

–

–

–

–

–

–

–

–

–

–

–

–

–

–

–

–

–

–

–

–

Notes: 1. Fine concrete groups are given in Item 2.1. 2. For light-weight and porous concrete by intermediate values of concrete grade as regards average density initial elasticity modulus is taken according to linear interpolation. 3. For light-weight and porous concrete values Eb are given by use gravimetric humidity w which is 5 percent for concrete B12.5 and higher and 10 percent – for concrete B10 and lower. If

for concrete B10 and lower gravimetric humidity w determined in compliance with SNiP II-3-79** is more than 10 percent so values Eb can be increased according to Table 11 if relative grade as regards average density D (100+w)/110 (where D is concrete grade as regards average density).

4. For heavy-weight concrete exposed to autoclave treatment values Eb given in Table 11 for natural hardening concrete must be multiplied by the coefficient 0.75. 5. For not protected against solar radiation structures designed for use in climatic sub-region IVA according to SNiP 2.01.01-82 Eb given in Table 11 must be multiplied by the coefficient

0.85

19

Table 12 (Annex 1) Use conditions of the structure by

static load dynamic and repeated load

in open air and in not heated buildings by design temperature in Celsius degrees

in open air and in not heated buildings by design temperature in Celsius degrees

Reinforcement types and documents regulating its quality

Reinforce-ment class

Steel grade Reinforcement diameter, mm

in heated buildings

up to 30 degrees below zero

lower than 30 degrees below

zero up to 40

degrees below zero

lower than 40 degrees below

zero up to 55

degrees below zero

lower than 55 degrees below

zero up to 70

degrees below zero

in heated buildings

up to 30 degrees below zero

lower than 30 degrees below

zero up to 40

degrees below zero

lower than 40 degrees below

zero up to 55

degrees below zero

lower than 55 degrees below

zero up to 70

degrees below zero

Hot-rolled plain rod reinforcement, GOST 5781-82 and GOST 380-71

A-I Ст3сп3 Ст3пс3 Ст3кп3 ВСт3сп2 ВСт3пс2 ВСт3кп2 ВСт3Гпс2

6–40 6–40 6–40 6–40 6–40 6–40 6–18

+ + + + + + +

+ + + + + + +

+ + – + + – +

+ – – + + – +

+1 – – + – – +1

+ + + + + + +

+ + + + + + +

– – – + + – +

– – – + – – +

– – – + – – +1

A-II ВСт5сп2 ВСт5пс2 18Г2С

10–40 10–16 18–40 40–80

+ + + +

+ + + +

+ + – +

+1

+1 – +

+1

– – +1

+ + + +

+ + +1

+

+1

+1

– +

– – – +

– – – +1

Ас-II 10ГТ 10–32 + + + + + + + + + +

Hot-rolled ribbed rod reinforcement

A-III 35ГС 25Г2С 32Г2Рпс

6–40 6–8

10–40 6–22

+ + + +

+ + + +

+ + + +

+1

+ + +1

– + +1 –

+ + + +

+ + + +

+1

+ + +1

– + +1 –

– – – –

Ausform robbed rod reinforcement

Ат-IIIC БСт5пс БСт5сп

10–22 + + + +1 – + + +1 – –

Regular ribbed reinforcement wire

Bp-I – 3–5 + + + + + + + + + +

1 Can be used only in bound framework meshes Notes: 1. In the present table sign “+” means – allowable, sign “–” means not allowable. 2. Design temperature is taken according to instructions of Item 1.8. 3. In the present table the loads must be considered to be dynamic if quantity of these loads during calculation of the structure as regards the rigidity is more than 0.1 of static load;

repeated loads are the loads which require calculation of the structure as regards robustness.

Table 13 (Annex 2)

Design temperature, Celsius degrees

up to 30 degrees below zero lower than 30 degrees below zero up to 40 degrees below zero

Embedded elements characteristics

Steel grade according to GOST 380-71

sheet steel thickness, mm

Steel grade according to GOST 380-71

sheet steel thickness, mm

1. Calculated as regards the loads

a) static; b) dynamic and

repeated

ВСт3кп2 ВСт3пс6 ВСт3Гпс5 ВСт3сп5

4–30 4–10

11–30 11–25

ВСт3пс6 ВСт3пс6 ВСт3Гпс5 ВСт3сп5

4–25 4–10

11–30 11–25

2. Constructive (not calculated as regards any forces)

БСт3кп2 ВСт3кп2

4–10 4–30

БСт3кп2 БСт3кп2

4–10 4–30

Notes: 1. Design temperature is taken according to Item 1.8 instructions. 2. When using low-alloyed steel for example steel grade 10ГС2С1, 09ГС2С, 15 ХСНД as well as by design

temperature lower than 40 Celsius degrees below zero choosing of steel grade and electrodes must be performed as for steel welded structures in compliance with requirements of SNiP II-23-81.

3. Design resistances of steel are taken according to SNiP II-23-81.

Table 14 (19, 20)

Type and class of reinforcement

Standard resistances against tension Rsn and

design resistances against tension for the second class

limit states Rs,ser, mega Pascal (kilogram-

force/cm2)

Type and class of reinforcement

Standard resistances against tension Rsn and

design resistances against tension for the second class

limit states Rs,ser, mega Pascal (kilogram-

force/cm2)

Rod reinforcement A-I A-II A-III and Ат-IIIC

235 (2400) 295 (3000) 390 (4000)

Reinforcement wire of class Bp-I, diameter:

3 mm 4 mm 5 mm

410 (4200) 405 (4150) 395 (4050)

Table 15 (22, 23)

Design resistances of reinforcement for the first classes limit states, mega Pascal (kilogram-force/cm2)

against tension

Type and class of reinforcement

of longitudinal reinforcement Rs

Of cross reinforcement (stirrups and bend-up bars)

Rsw

against compression Rsc

Rod reinforcement of classes: A-I A-II A-III with diameter:

6–8 mm 10–40 mm

Ат-IIIC Reinforcement wire of class Bp-II with diameter:

3 mm 4 mm 5 mm

225 (2300) 280 (2850)

355 (3600) 365 (3750) 365 (3750)

375 (3850) 356 (3750) 360 (3700)

175 (1800) 225 (2300)

285 (5900)* 290 (3000)* 390 (3000)*

270 (2750); 300 (3050)** 265 (2700); 295 (3000)** 260 (2650); 290 (2950)**

225 (2300) 280 (2850)

355 (3600) 365 (3750) 365 (3750)

375 (3850) 365 (3750) 360 (3700)

* In welded frameworks for stirrups made of reinforcement A-III and Ат-IIIC with diameter less than 1/3 of diameter of longitudinal bars values Rsw are taken equal to 255 Mega Pascal (2600 kilogram-force/cm2).

** For bound frameworks.

21

Standard and design characteristics of reinforcement

(2.25) For characteristic strength of reinforcement Rsn it is necessary to take the least controlled values: - for rod reinforcement – physical yield limit; - for regular reinforcement wire – stress equal to 0.75 of rapture strength. Standard resistances Rsn for main types of non-prestressed reinforcement are given in Table 14.

(2.26) Design strength of reinforcement against tension and compression Rs and Rsc for the first class limit states are determined by means of dividing of characteristic strength into safety factor γs taken equal to:

a) 1.05 – for rod reinforcement A-I and A-II; 1.07 – for rod reinforcement Ат-IIIC and A-III with diameter 10–40 mm 1.10 – for rod reinforcement A-III with diameter 6–8 mm;

b) 1.10 – for reinforcement wire Bp-I.

Design extension strength of reinforcement for the second group limit states is taken equal to characteristic strength. Design extension and compression strength of reinforcement used during calculation according to the first class limit states are given in the Table 15 and by calculations according to the second class limit states – in Table 14.

(2.28) Design strength of cross reinforcement (stirrups and bend-up bars) Rsw get decreased in

comparison with Rs by means of multiplying by the work conditions coefficients 1sγ and

2sγ :

a) independently on type and class of reinforcement – by the coefficient 8.01 =sγ

considering unevenness of forces spread in reinforcement in the length dimension of the section;

b) for rod reinforcement of class A-III and Aт-IIIC with diameter no less than 1/3 of diameter of longitudinal bars and for reinforcement wire of class Bp-I in welded

frameworks – by the coefficient 9.02 =sγ considering the welded joint brittle failure

possibility.

Design strengths Rsw with consideration of the mentioned above work conditions

coefficients 1sγ and 2sγ are given in Table 15.

Besides if the considered section is locates in anchor zone of reinforcement so design

strengths Rs and Rsc are multiplied by work conditions coefficient 5sγ considering

incomplete anchorage of reinforcement and determined according to Item 3.44. For elements made of light-weight concrete B7.5 and less design resistances Rsw of cross

reinforcement A-I and Bp-I are to be multiplied by work conditions coefficient 8.07 =sγ .

(2.30) Values of reinforcement elasticity modulus Es are taken equal to:

210 000 mega Pascal (2 100 000 kilogram-force/cm2) – for reinforcement A-I and A-II 200 000 mega Pascal (2 000 000 kilogram-force/cm2) – for reinforcement A-III and Aт-IIIC 170 000 mega Pascal (1 700 000 kilogram-force/cm2) – for reinforcement Bp-I

22

3. CALCULATION OF CONCRETE AND REINFORCED CONCRETE MEMBERS

AS REGARDS THE FIRST CLASS LIMIT STATES.

3.1. For registration of loads influence on the concrete strength it is necessary to calculate concrete and reinforced concrete members as regards their strength:

a) regarding dead loads, long-term and short-term loads except loads of short duration (wind

loads, crane loads and other during production, transportation, installation, etc) as well as regarding special loads caused by deformation of collapsible, swelling, permanently frozen soils and soil of that kind; in that case design tension and compression strength of concrete

Rb and Rbt are taken according to Table 8 if 9.02 =bγ :

b) regarding all loads action including loads of short duration; in that case design strength of

concrete Rb and Rbt are taken according to Table 8 by 1.12 =bγ *

* If by consideration of special loads in compliance with instructions of norms it is necessary to insert a work

conditions coefficient (for example when consideration of earthquake loads) so it is taken 0.12 =bγ

If the structure is used in conditions favorable for concrete strength developing [hardening under the water, in humid soil or if surrounding air humidity is more than 75 percent (see

Item 1.8)] so calculation according to case “a” is made by 0.12 =bγ .

Strength conditions must be fulfilled as according to case “a” as according to case “b”. In case of absence of loads of short duration or emergency calculation is made only as according to case “b” if the following condition is met:

III FF 82.0< (1)

where FI is the force (moment MI, cross force QI or longitudinal force NI) from the loads

used by the calculation according to case “a”; at the same time in the calculations of sections normal to longitudinal axis of eccentric loaded members moment MI is taken relating to the axis going through the most stretched (or the least pressed) reinforcement rod, and for concrete members – relating to stretched or the leased compressed surface;

FII is the force from the loads used by calculation according to case “b”.

It is possible to make the calculation only according to case “b” if the condition (1) is not

fulfilled, taking design resistances Rb and Rbt (by 0.12 =bγ ) with the

coefficient 1.1/9.0 ≤= IIIbl FFγ .

For eccentric pressed members calculation according to un-deformed scheme values FI and FII can be determined without considering member deflection. For structures used in conditions favorable for concrete strength developing, condition (1)

becomes III FF 9.0< and the coefficient IIIbl FF /=γ .

23

CALCULATION OF CONCRETE MEMBERS STRENGTH

3.2. (3.1) Calculation of strength of concrete members must be made for sections normal to their longitudinal axis. According to work conditions of members they are calculated considering as well as without considering resistance of tensile zone of concrete.

Without consideration of resistance of tensile zone of concrete the calculation of eccentric pressed members mentioned in Item 1.7a considering that limit state is characterized by failure of compressed concrete. With consideration of resistance of tensile zone of concrete the calculation of members mentioned in Item 1.7b as well as members for which the presence of cracks is not allowed according to use conditions of the structure (members under the pressure of water, cornices, parapets, etc). At the same time it is considered that limit state is characterized by failure of tensile concrete (crack formation). In case if appearance of diagonal cracks is possible (for example members of T- or double T-section under lateral forces) it is necessary to make the calculation of concrete members according to condition (13). Besides it is necessary to make the calculation as regards local compression in compliance with Item 3.93.

Eccentric Pressed Members

3.3. (3.2, 1.21) During calculation of eccentric pressed concrete members it is necessary to take into account the occasional eccentricity of longitudinal force ea caused by not considered in the calculation factors. In any eccentricity ea is taken no less than

- 1/600 of the member length or of distance between its sections fixed against displacement;

- 1/30 of the member height; - 10 mm (for prefabricated members if there are no any other justified values ea)

For members of statically non-definable structures the value of eccentricity of longitudinal

force relating to center of gravity of the given section 0e is taken equal to eccentricity of

static calculation of the structure but no less than ea.

In members of statically non-definable structures eccentricity 0e is determined as a sum of

eccentricities according to static calculation of the structure and occasional one.

3.4. (3.3) By elasticity of members 14/0 >il (for rectangular sections by 4/0 >hl ) it is

necessary to consider the influence of deflections in the eccentricity plane of longitudinal force and in the plane normal to it on the load-carrying capacity of members by means of

multiplying of values 0e by coefficient η (see Item 3.7). In the calculation from eccentricity

plane of longitudinal force value 0e is taken equal to occasional eccentricity.

24

Use of eccentric pressed concrete members (except the cases provided in Item 1.7b) is not

allowed by eccentricities of longitudinal force considering deflections 0e η which are more

than: a) according to the loads combinations

- 0.9y……….by basic combination; - 0.95y….......by special load combination;

b) according to concrete class: - y –10……..by B10 and higher; - y –20……..by B7.5 an lower (here y is the distance from the center of gravity of the section to the most compressed concrete fiber). 3.5. (3.4) In eccentric compressed concrete members it is necessary to design constructive reinforcement in cases mentioned in Item 5.122.

3.6. (3.5) Calculation of eccentric compressed concrete members must be made without considering tensile concrete according to the following condition:

bb ARN ≤ (2)

where Ab area of compressed zone of concrete determined according to the condition that its center of gravity is congruent with point of external resultant forces (Draft 1). Draft 1. Forces scheme and stress distribution across the cross-section of compressed concrete member

without considering the tensile concrete resistance

1 – center of gravity of compressed zone area; 2 – the same of the whole section area.

For members of rectangular section Ab is determined by the following formula:

−=

h

ebhAb

η021 (3)

Eccentric compressed concrete elements which can not have any cracks according to use conditions (see Item 3.2) must be checked independently on calculation according to condition (2) but in compliance with the following condition:

re

WRN

plbt

−≤

η0

(4)

For members of rectangular section condition (4) has the following view:

ϕη

−

≤

h

e

bhRN bt

06

75.1 (5)

Calculation of eccentric pressed members mentioned in Item 1.7b must be made according to the condition (2) or (4). In formulas (3)–(5):

η is the coefficient determined by the formula (8);

25

r is the distance from the center of gravity of the section to the heart point most distant from the tensile zone determined by the following formula:

A

Wr ϕ= (6)

serb

b

R ,

6.1σ

ϕ −= But is taken no more than 1.0;

bσ – Maximum compression stress determined as for elastic body;

Wpl – is sectional modulus for end tensile fiber considering non-elastic deformations of tensile concrete determined by the following formula:

002

b

b

pl Sxh

IW +

−= (7)

where Ib0 is moment of inertia of concrete pressed zone section area relating to zero line;

Sb0 is static moment of concrete pressed zone section area relating to zero line; h – x is the distance from the zero line to the tensile surface:

1

12

b

b

AA

Sxh

+=− ;

Ab1 is area of compressed zone of concrete supplemented in tensile zone with the rectangle with width b equal to the width of section along the zero line and with height h –x (Draft 2);

Sb1 is static moment of area Ab1 relating to stretched surface.

Draft 2. To definition Ab1.

It is possible to determine Wpl by the following formula:

0WWpl γ=

where γ – see in Table 29. 3.7. (3.6) Coefficient η considering deflection influence on the eccentricity of longitudinal

force 0e must be determined by the following formula:

crN

N−

=

1

1η (8)

where Ncr is relative critical force determined by the following formula:

+

+= 1.0

1.0

11.0

)/(

4.62

0 el

b

crhl

IEN

δϕ (9)

(here I is moment of inertia of concrete section). For elements of rectangular section formula (9) has the following view:

+

+= 1.0

1.0

11.0

)/(

533.02

0 el

b

crhl

AEN

δϕ (9a)

In formulas (9) and (9a):

lϕ – Coefficient considering influence of long duration of the load on the member

deflection:

26

1

11M

M l

l βϕ += (10)

but no more than 1+β here β is coefficient taken by Table 16;

M1 is the moment relating to tensile or the least compressed surface of the section caused by influence of dead loads, short-term and long-term loads;

M1l is the same but caused by dead loads and long-term loads; l0 is determined according to Table 17;

eδ – The coefficient taken equal to he /0 but no less than

be Rh

l01.001.05.0 0

min, −−=δ

(Here bR is in Mega Pascals).

Note. During calculation of the section according to cases “a” and “b” (see Item 3.1) it is possible to

determine min,eδ only once taking Rb by 1.02 =bγ .

Table 16 (30)

Concrete Coefficient β in formula (10)

1. Heavy-weight concrete 1.0

2. Fine concrete: group A group Б group В

1.3 1.5 1.0

3. Light-weight concrete - with artificial coarse and fine aggregate:

dense porous

- with natural coarse aggregate

1.0 1.5 2.5

4. Porous concrete 2.0 Note: Fine concrete groups are given in Item 2.1.

Table 17 (31)

Walls and columns support character Design length 0l of eccentric pressed concrete

members

1. with supports above and below: a) with hinges on both ends independently

on displacement of supports; b) by one end restraint and possible

displacement of supports for - multi-span buildings - one-span buildings

H

1.25H

1.50H

2. free supported 2.00H Symbols in Table 17: H – the height of the column (wall) within the first storey except the thickness of the floor slab or the height of free supported structure.

3.8. The calculation considering deflection of eccentric pressed concrete members of rectangular section made of heavy-weight concrete of class no higher than B20 can be made due to the diagram (Draft 3). At the same time the following condition must be met:

27

bhRN bnα≤

Where nα is determined according to the diagram (Draft 3) in compliance with values

he /0 and hl /0=λ .

Draft 3. Diagram of load carrying capacity of eccentric compressed concrete elements.

Explanation: ––––– by 0.1/ 11 =MM l

-------- By 5.0/ 11 =MM l

Bending Elements

3.9. (3.8) Calculation of bending concrete elements must be made according to the following condition:

plbtWRM ≤ (11)

where Wpl is determined by Formula (7); for members of rectangular section Wpl is taken equal to:

5.3

2bh

Wpl = (12)

Besides for members of T- and double T-section the following condition must be met:

btxy R≤τ (13)

Where xyτ – shear stresses determined as for elastic material at the level of center of

gravity of the section.

Examples of Calculation

Example 1. Given: a concrete panel of the wall between apartments, thickness h = 200 mm, height H = 2.7 mm manufactured vertically (in the mounting) of expanded-clay concrete with glass sand of class B15, concrete grade as regards average density is D1600 (Eb = 14 000 Mega Pascal) total load per 1 m of the wall is N = 900 kN, including dead load and long-term loads Nl = 540 kN; no load of short duration.

It is required to test the strength of the wall panel. Calculation is made according to Item 3.6 as regards the longitudinal force N =

900 kN applied with occasional eccentricity ae determined according to Item 3.3.

As 67.630

200

30==

h mm < 10 mm occasional eccentricity is taken equal to 10

mm, which means 100 =e mm. The connection of the panel above and below is considered to

be hinge connection, so design length 0l in compliance with Table 17 is 7.20 == Hl m.

As panel elasticity 45.132.0

7.20 >==h

l so the calculation is made with

consideration of deflection in compliance with Item 3.7.

Coefficient lϕ is determined according to formula (10) by 0.1=β (see Table

16). As eccentricity of longitudinal force doesn’t depend on load characteristics so here it is

possible to take 6.0900

5401 ===N

N

M

M ll ,

So 6.16.0111

1 =+=+=M

M l

l βϕ

As there are no loads of short duration so design concrete strength Rb in

compliance with Item 3.1 is taken considering the coefficient 90.02 =bγ that is bR = 7.7 mega

28

Pascal and in compliance with Table 9 considering work conditions coefficients 85.03 =bγ and

90.09 =bγ we get 89.590.085.07.7 =××=bR mega Pascal.

As 200

10306.089.501.05.1301.05.001.001.05.0 00

min, =>=⋅−⋅−=−−=h

eR

h

lbeδ so we

take 306.0min, == ee δδ .

Critical force Ncr is determined by formula (9a) taking section area A for 1 m of the wall length, that is A = 200×1000 = 200 000 mm2:

3

2

3

2

0

1018981.0306.01.0

11.0

5.136.1

2000001014533.01.0

1.0

11.0

)/(

533.0⋅=

+

+⋅

⋅⋅⋅=

+

+=

el

b

crhl

AEN

δϕN = 1898kN

from this 902.1

1898

9001

1

1

1=

−

=

−

=

crN

Nη

If we check condition (2) using formula (3):

954000200

902.1102120000089.5

21 0 =

⋅⋅−⋅=

−=

h

ebhRAR bbb

ηN = 954 kN > N = 900 kN,

that is the strength of the panel is provided. CALCULATION OF REINFORCED CONCRETE MEMBERS STRENGTH

3.10.(3.9) Calculation of reinforced concrete members as regards their strength must be made for the sections normal to their longitudinal axis as well as for inclined sections of the most dangerous direction. By torque moments it is necessary to check the strength of spatial sections in stretched zone bounded by torsion fracture of the most dangerous of all possible directions. Besides it is necessary to make the calculation of members as regards local loads (bearing stress, punching force, cleavage).

Bending Elements

3.11.(3.11) Calculation of sections normal to longitudinal axis of the member when bending moment acts in the plane of section symmetry axis and reinforcement is concentrated at surfaces perpendicular to the mentioned plane must be made in compliance with Items 3.15-3.23 according to the ratio between the value of relative

height of concrete compressed zone 0/ hx=ξ determined according to requirements

for equilibrium and the value of relative height of compressed concrete zone Rξ (see

Item 3.14) whereby limit state of limit state of the member comes at the same time with the stress equal to design strength Rs in the stretched reinforcement.

3.12. (3.18) Calculation of ring cross section bending elements if the ration of internal and

external radii is 5.0/ 21 ≥rr with reinforcement evenly spread in a circumferential

direction (if there are no less than 6 longitudinal bars) must be made as for eccentric compressed members in compliance with Items 3.69 and 3.70 by N = 0 and by bending

moment value instead of 0Ne .

3.13. Calculation of normal sections not mentioned in Items 3.11, 3.12 and 3.24 is made by formulas of general case of normal section calculation in compliance with Item 3.76

taking N = 0 in formula (154) and replacing eN by M (projection of bending moment on the plane perpendicular to the straight line which bounds compression zone) in condition (153). If symmetry axis of the section is not congruent with the moment plane or is absent

29

at all so location of the compressed zone bounding must conform to the additional condition of parallelism of moments planes of internal and external forces.

3.14. (3.12) Value Rξ is determined by the following formula:

−+

=

1.111

,

ω

σ

ωξ

usc

s

RR

(14)

where ω is characteristic of concrete compressed zone determined by the following formula:

bR008.0−= αω (15)

here α is the coefficient equal to:

0.85……for heavy-weight concrete 0.80……for fine concrete (see Item 2.1) of group A 0.75……for fine concrete of groups Б and В 0.80……for light-weight and porous concrete

500, =uscσ Mega Pascal by coefficient 9.02 =bγ (see Item 3.1);

400, =uscσ Mega Pascal by coefficient 0.12 =bγ or 1.12 =bγ ;

Rs, Rb are in mega Pascals.

Values ω and Rξ are given in table 18 – for members of heavy-weight concrete; in

Table 19 – for members of fine concrete of group A, light-weight and fine concrete

Table 18

Values ω, Rξ , αR and ψc for members of heavy-weight concrete of classes Concrete work

conditions coefficient

2bγ

Class of tensile

reinforcement

Symbol

B12.5 B15 B20 B25 B30 B35 B40 B45 B50 B55 B60

Any ω 0.796 0.788 0.766 0.746 0.726 0.710 0.690 0.670 0.650 0.634 0.614

A-III (Ø10–40) and BP-I (Ø4; 5)

Rξ αR ψc

0.662 0.443 4.96

0.652 0.440 4.82

0.627 0.430 4.51

0.604 0.422 4.26

0.582 0.413 4.03

0.564 0.405 3.86

0.542 0.395 3.68

0.521 0.381 3.50

0.500 0.376 3.36

0.484 0.367 3.23

0.464 0.355 3.09

A-II Rξ αR ψc

0.689 0.452 6.46

0.680 0.449 6.29

0.650 0.439 5.88

0.632 0.432 5.55

0.610 0.424 5.25

0.592 0.417 5.04

0.571 0.408 4.79

0.550 0.399 4.57

0.531 0.390 4.38

0.512 0.381 4.22

0.490 0.370 4.03

0.9

A-I Rξ αR ψc

0.708 0.457 8.04

0.698 0.455 7.82

0.674 0.447 7.32

0.652 0.439 6.91

0.630 0.432 6.54

0.612 0.425 6.27

0.591 0.416 5.96

0.570 0.407 5.68

0.551 0.399 5.46

0.533 0.391 5.25

0.510 0.380 5.01

Any ω 0.790 0.782 0.758 0.734 0.714 0.694 0.674 0.650 0.630 0.610 0.586

A-III (Ø10–40) and BP-I (Ø4; 5)

Rξ αR ψc

0.628 0.431 3.89

0.619 0.427 3.79

0.591 0.416 3.52

0.563 0.405 3.29

0.541 0.395 3.12

0.519 0.384 2.97

0.498 0.374 2.83

0.473 0.361 2.68

0.453 0.350 2.56

0.434 0.340 2.46

0.411 0.327 2.35

A-II Rξ αR ψc

0.660 0.442 5.07

0.650 0.439 4.94

0.623 0.429 4.6

0.593 0.417 4.29

0.573 0.409 4.07

0.551 0.399 3.87

0.530 0.390 3.69

0.505 0.378 3.49

0.485 0.367 3.34

0.465 0.357 3.21

0.442 0.344 3.06

1.0

A-I Rξ 0.681 0.673 0.645 0.618 0.596 0.575 0.553 0.528 0.508 0.488 0.464

30

αR ψc

0.449 6.31

0.447 6.15

0.437 5.72

0.427 5.34

0.419 5.07

0.410 4.82

0.400 4.59

0.389 4.35

0.379 4.16

0.369 3.99

0.356 3.80

Any ω 0.784 0.775 0.750 0.722 0.698 0.678 0.653 0.630 0.606 0.586 0.558

A-III (Ø10–40) and BP-I (Ø4; 5)

Rξ αR ψc

0.621 0.428 3.81

0.610 0.424 3.71

0.581 0.412 3.44

0.550 0.399 3.19

0.523 0.386 3.00

0.502 0.376 2.86

0.481 0.365 2.73

0.459 0.351 2.65

0.429 0.346 5.52

0.411 0.327 2.35

0.385 0.312 2.23

1.1

A-II Rξ αR ψc

0.650 0.439 4.97

0.642 0.436 4.84

0.613 0.425 4.49

0.582 0.413 4.16

0.556 0.401 3.91

0.534 0.391 3.72

0.514 0.382 3.53

0.485 0.361 3.34

0.477 0.363 3.29

0.442 0.344 3.06

0.417 0.330 2.91

A-I Rξ αR ψc

0.657 0.447 6.19

0.665 0.444 6.02

0.636 0.434 5.59

0.605 0.422 5.17

0.579 0.411 4.86

0.558 0.402 4.63

0.537 0.393 4.42

0.509 0.379 4.16

0.500 0.375 4.09

0.464 0.356 3.80

0.439 0.343 3.62

bR008.085.0 −=ω ;

−+

1.111

,

ω

σ

ωξ

usc

s

RR

; ( )RRR ξξα 5.01−= ;

−

=

1.11

,

ω

σψ

s

usc

c

R

Note: Values ω, Rξ αR and ψc given in Table 18 are calculated without considering coefficients biγ according to Table 9.

Table 19

Values ω, Rξ , αR and ψc for members of fine concrete of group A, light-weight

and porous concrete of classes

Concrete work

conditions coefficient

2bγ

Class of tensile

reinforcement

Symbol

B5 B7.5 B10 B12.5 B15 B20 B25 B30 B35 B40

Any ω 0.780 0.768 0.757 0.746 0.738 0.716 0.696 0.676 0.660 0.640

A-III (Ø10–40) and BP-I (Ø4; 5)

Rξ αR ψc

0.643 0.436 4.71

0.629 0.431 4.54

0.617 0.427 4.39

0.604 0.422 4.26

0.595 0.418 4.16

0.571 0.408 3.92

0.551 0.399 3.75

0.528 0.388 3.55

0.510 0.380 3.42

0.490 0.370 3.28

A-II Rξ αR ψc

0.671 0.446 6.14

0.657 0.441 5.92

0.644 0.437 5.73

0.632 0.432 5.55

0.623 0.429 5.43

0.599 0.420 5.12

0.577 0.411 4.86

0.556 0.401 4.63

0.539 0.394 4.46

0.519 0.384 4.27

0.9

A-I Rξ αR ψc

0.690 0.452 7.64

0.676 0.488 7.36

0.664 0.444 7.13

0.652 0.439 6.91

0.643 0.436 6.75

0.619 0.427 6.37

0.597 0.419 6.05

0.576 0.410 5.76

0.559 0.403 5.56

0.539 0.394 5.31

Any ω 0.774 0.761 0.747 0.734 0.725 0.700 0.672 0.648 0.628 0.608

A-III (Ø10–40) and BP-I (Ø4; 5)

Rξ αR ψc

0.609 0.424 3.70

0.594 0.418 3.56

0.578 0.411 3.42

0.563 0.405 3.29

0.553 0.400 3.22

0.526 0.388 3.01

0.496 0.373 2.82

0.471 0.360 2.67

0.451 0.349 2.55

0.432 0.339 2.45

A-II Rξ αR ψc

0.641 0.436 4.82

0.626 0.430 4.64

0.610 0.424 4.45

0.595 0.418 4.29

0.585 0.414 4.19

0.558 0.402 3.67

0.528 0.389 3.48

0.503 0.377 3.30

0.482 0.366 3.33

0.463 0.356 3.19

1.1

A-I Rξ αR ψc

0.663 0.443 6.00

0.648 0.438 5.71

0.633 0.433 5.54

0.618 0.427 5.34

0.608 0.423 5.21

0.581 0.412 4.89

0.551 0.399 4.57

0.526 0.388 4.33

0.506 0.378 4.14

0.486 0.368 3.97

bR008.080.0 −=ω ;

−+

1.111

,

ω

σ

ωξ

usc

s

RR

; ( )RRR ξξα 5.01−= ;

−

=

1.11

,

ω

σψ

s

usc

c

R

Note: Values ω, Rξ , αR and ψc given un Table 19are calculated without considering coefficients according to Table 9.

RECTANGULAR SECTIONS

31

3.15. Calculation of rectangular sections with reinforcement concentrated at compressed and tensile surface of the member (Draft 4), is made in the following manner according to the height of compressed zone:

bR

ARARx

b

sscss

'−= (16)

a) by Rh

xξξ ≤=

0

– for the condition

( ) ( )'5.0 0

'

0 ahARxhbxRM sscb −+−≤ (17)

b) by Rξξ > – for the condition

)'( 0

'2

0 ahARbhRM sscbR ++≤ α (18)

Where ( )RRR ξξα 5.01−=

At the same time design load-carrying capacity of the section can be increased by means

of replacing of value Rα by mR αα 2.08.0 + in the condition (18) where by 1≤ξ

( )ξξα 5.01−=m or according to table 20. Values Rξ and αR are determined according to

table 18 and 19. If 0≤x so the strength is checked according to the following condition

( )'0 ahARM ss −≤ (19)

Note. If the height of compressed zone determined considering of a half of compressed reinforcement,

'5.0 '

abR

ARARx

b

sscss ≤−

= so design load carrying capacity of the section can be increased if the

calculation will be made by formulas (16) and (17) without considering compressed reinforcement '

sA .

Draft 4. Loads scheme in rectangular cross section of bending reinforced concrete element.

3.16. It is recommended to design bending elements so that to provide the fulfillment of the

condition Rξξ < . It is possible not to meet this condition only in case when the section

area of stretched reinforcement is determined according to the calculation as regards the second class limit states or if it’s taken on the grounds of constructive solutions.

3.17. Checking of rectangular sections strength with single reinforcement is made

- by 0hx Rξ< in compliance with the condition:

( )xhARM ss 5.00 −≤ (20)

Where height of compressed zone is bR

ARx

b

ss=

- by 0hx Rξ≥ in compliance with the condition: 2

0bhRM bRα≤ (21)

at the same time design load carrying capacity of the section can be increased using

recommendations of Item 3.15b [ Rξ , Rα - see formula (4) or Table 18 and 19].

3.18. Choosing of longitudinal reinforcement is made in the following manner. It is necessary

to calculate the following value:

2

0bhR

M

b

m =α (22)

If Rm αα ≤ (see Table 18 and 19) so that means that compressed reinforcement is not

required.

32

If there is no compressed reinforcement so section area of tensile reinforcement is determined by the following formula:

0hR

MA

b

sζ

= (23)

Where ζ is determined according to Table 20 according to value mα .

If Rm αα > so it is necessary to enlarge the section or to increase the concrete grade, or to

fix compressed reinforcement in compliance with Item 3.19.

By consideration of the concrete work conditions coefficient 9.02 =bγ (see Item 3.1)

tensile reinforcement can be chosen according to Annex 2.

Table 20

ξ ζ mα ξ ζ

mα ξ ζ mα

0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10 0.11 0.12 0.13 0.14 0.15 0.16 0.17 0.18 0.19 0.20 0.21 0.22 0.23 0.24 0.25

0.995 0.990 0.985 0.980 0.975 0.970 0.965 0.960 0.955 0.950 0.945 0.940 0.935 0.930 0.925 0.920 0.915 0.910 0.905 0.900 0.895 0.890 0.885 0.880 0.875

0.010 0.020 0.030 0.039 0.049 0.058 0.068 0.077 0.086 0.095 0.104 0.113 0.122 0.130 0.139 0.147 0.156 0.164 0.172 0.180 0.188 0.196 0.204 0.211 0.219

0.26 0.27 0.28 0.29 0.30 0.31 0.32 0.33 0.34 0.35 0.36 0.37 0.38 0.39 0.40 0.41 0.42 0.43 0.44 0.45 0.46 0.47 0.48 0.49 0.50

0.870 0.865 0.860 0.855 0.850 0.845 0.840 0.835 0.830 0.825 0.820 0.815 0.810 0.805 0.800 0.795 0.790 0.785 0.780 0.775 0.770 0.765 0.760 0.755 0.750

0.226 0.234 0.241 0.243 0.255 0.262 0.269 0.276 0.282 0.289 0.295 0.302 0.308 0.314 0.320 0.326 0.332 0.338 0.343 0.349 0.354 0.360 0.365 0.370 0.375

0.51 0.52 0.53 0.54 0.55 0.56 0.57 0.58 0.59 0.60 0.62 0.64 0.66 0.68 0.70 0.72 0.74 0.76 0.78 0.80 0.85 0.90 0.95 1.00

–

0.745 0.740 0.735 0.730 0.725 0.720 0.715 0.710 0.705 0.700 0.690 0.680 0.670 0.660 0.650 0.640 0.630 0.620 0.610 0.600 0.575 0.550 0.525 0.500

–

0.380 0.385 0.390 0.394 0.399 0.403 0.407 0.412 0.416 0.420 0.428 0.435 0.442 0.449 0.455 0.461 0.466 0.471 0.476 0.480 0.489 0.495 0.499 0.500

– For bending moments of rectangular section:

0

'

bhR

ARAR

b

sscss −=ξ ;

( )2

0

0

' '

bhR

ahARM

b

ssc

m

−−α ; ( )ξξα 5.01−=m ; ξζ 5.01−= .

3.19. Cross sections areas of tensile As and compressed '

sA reinforcement corresponding to

minimum of their sum for members of concrete of class B30 and lower should be determined if compressed reinforcement is required according to the calculation (see Item 3.18), by the following formulas:

( )'4.0

0

2

0'

ahR

bhRMA

sc

b

s−

−= (24)

'055.0s

s

b

s AR

bhRA += (25)

33

If taken section area of compressed reinforcement '

sA far exceeds the value calculated by

formula (24) so section area of tensile reinforcement is determined according to actual

value of area '

sA by the following formula:

'

0 s

s

b

s AR

RbhA += ξ (26)

Where ξ is determined according to Table 20 depending on the value

( )0

'2

0

0

'

≥−−

=bhR

ahARM

b

ssc

mα which must conform to requirement Rm αα ≤

(see table 18 and 19).

T- AND DOUBLE T-SECTIONS 3.20. Calculation of sections which have a flange in compressed zone (T-sections and double

T-sections, etc) must be made depending on the compressed zone bounding position: a) if the bounding of compressed zone goes in the flange (Draft 5a) that is the following

condition is met: '''

sscffbss ARhbRAR +≤ (27)

The calculation is made as for rectangular section which is '

fb wide in compliance with

Items 3.15 and 3.17;

b) if the bounding of compressed zone goes in the rib (Draft 5b) that is condition (27) is not met, so the calculation is made according to the following condition

( ) ( ) ( ) ( )'5.05.0 0

''

0

''

0 ahARhhhbbRxhbxRM sscfffbb −+−−+−≤ (28)

At the same time concrete compressed zone height x is determined by the following formula:

( )bR

hbbRARARx

b

ffbsscss

''' −−−= (29)

And it is taken no more than 0hRξ (see Table 18 and 19).

If 0hx Rξ≥ so condition (28) can be written in the following form:

( ) ( ) ( )'5.0 0

''

0

''2

0 ahARhhhbbRbhRM sscfffbbR −+−−+≤ α (30)

Where Rα – see in Table 18 and 19.

At the same time it is necessary to consider the recommendations of Item 3.16.

Notes: 1. by variable height of a flange overhang it is possible to take the value '

fh equal to average height of

overhangs.

2. Compressed flange width '

fb inserted into the calculation must not exceed the values given in Item 3.23.

Draft 5. Compressed zone bounding position in T-section of bending reinforced concrete element.

a – in a flange; b – in a rib

3.21. Required section area of compressed reinforcement is determined by the following formula:

( ) ( )( )'

5.0

0

'0

''20'

ahR

hhhbbRbhRMA

sc

fffbbR

s−

−−−−=

α (31)

34

where Rα – see in Table 18 and 19.

3.22. Required section area of tensile reinforcement is determined in the following manner:

a) if compressed zone border goes in a flange, that is the following condition is met:

( ) ( )'5.0 0

''

0

''ahARhhhbRM sscfffb −+−≤ (32)

so section area of tensile reinforcement is determined as for rectangular cross section '

fb

wide in compliance with Items 3.18 and 3.19; b) if compressed zone border goes in a rib that is condition (32) is not met so cross section

of tensile section is determined by the following formula:

( )[ ]s

sscffb

sR

ARhbbbhRA

'''0 +−+ξ

(33)

where ξ is determined according to Table 20 depending on the value

( ) ( ) ( )2

0

0''

0'' '5.0

bhR

ahARhhhbbRM

b

sscfffb

m

−−−−−=α (34)

At the same time the condition Rm αα ≤ must be met (see Table 18 and 19).

3.23. (3.16) Value '

fb inserted into the calculation is taken according to the condition that

the width of an overhang to each side of the rib must be no less than 1/6 of the span of the member and no more than:

a) by cross ribs or by 2/11.0' −≥ hh f of the clear distance between longitudinal ribs;

b) without cross ribs (or if the distance between them is more than the distance between

longitudinal ribs) and '' 61.0 ff hhh −< ;

c) by console overhangs of a flange:

By '' 61.0 ff hhh −≥ ;

By '' 31.005.0 ff hhhh −<≤ ;

By hh f 05.0' < - overhangs are not taken into account.

Examples of Calculation

Rectangular section

Example 2. Given: the section with dimensions b = 300 mm, h = 600 mm, a = 40 mm;

9.02 =bγ (no loads of short duration); bending moment M = 200 KN·m; heavy-weight

concrete B15 ( 7.7=bR Mega Pascal); reinforcement A-II (Rs = 280 Mega Pascal).

It is required to determine the cross section area of longitudinal reinforcement.

Calculation. 560406000 =−=h mm. Longitudinal reinforcement is chosen according to

Item 3.18. Value mα is determined by Formula (22):

276.05603007.7

102002

6

2

0

=⋅⋅

⋅==

bhR

M

b

mα

According to Table 18 for a member of concrete B15 with reinforcement A-II by 9.02 =bγ , we

find that 449.0=Rα .

As 449.0276.0 =<= Rm αα so that means that compressed reinforcement is not required.

According to table 20 by 276.0=mα we find 835.0=ζ

Required cross section of tensile reinforcement is to be determined by formula (23):

35

1528560835.0280

10200 6

0

=⋅⋅

⋅==

hR

MA

s

sζ

mm2.

It is taken 2Ø28 + 1Ø25 ( 1598=sA mm2)

Example 3. Given: a section with dimensions b = 300 mm; h = 800 mm; a = 70 mm; tensile

reinforcement A-III (Rs = 365 Mega Pascal); its section area 2945=sA mm2 (6Ø25);

9.02 =bγ (no loads of short duration); heavy-weight concrete B25 (Rb = 13 Mega Pascal);

bending moment M = 550 kN·m. It is required to check the section strength.

Calculation. 730708000 =−=h mm. Section strength is calculated according to Item 3.17.

Value x is determined in the following manner:

27630013

2945365=

⋅

⋅==

bR

ARx

b

ss mm

According to table 18 for concrete B25 with reinforcement A-III by 9.02 =bγ , we

find 604.0=Rξ .

As 604.038.0730

276

0

=<=== Rh

xξξ

so the strength is to be checked according to condition (20):

( ) ( ) 6

0 104.6362765.073029453655.0 ⋅=⋅−⋅=− xhAR ss N· mm 4.636= kN·m > M = 550 kN ·m,

that means the strength is corresponding to norms. Example 4. Given: a section with dimensions b = 300 mm; h = 800 mm; a = 50 mm;

reinforcement A-III ( 365== scs RR Mega Pascal); bending moment with consideration of

crane load 780=IIM kN·m; moment without consideration crane load 670=IM kN·m;

heavy-weight concrete B15 ( 5.8=bR Mega Pascal by 0.12 =bγ ).

It is required to determine the section area of longitudinal reinforcement. Calculation is made as regards the total load correcting design resistance of concrete according to Item 3.1.

As 1.105.1670

7809.09.0 <===

I

II

blM

Mγ so we take 93.805.15.8 =⋅=bR Mega Pascal.

We calculate 750508000 =−h mm.

We determined required area of longitudinal reinforcement according to Item 3.18. Value mα

is determined according to Formula (22):

518.075030093.8

107802

6

2

0

=⋅⋅

⋅==

bhR

M

b

mα

As 42.0518.0 =>= Rm αα (see Table 18 by 0.12 =bγ ) by given dimensions of the section and

concrete class it is required compressed reinforcement. The following calculation is made according to Item 3.19. Taking 30'=a mm we determine required section of compressed and tensile reinforcement by formulas (24) and (25):

( )674

30750365

75030093.84.010780

)'(

4.0 216

0

2

0' =−

⋅⋅⋅−⋅=

−

−=

ahR

bhRMA

sc

b

s mm2

3702674365

93.875030055.055.0 '0 =+⋅⋅⋅

=+= s

s

b

s AR

RbhA mm2.

We take 763' =sA mm2 (3Ø18); As = 4021 mm2 (5Ø32).

36

Example 5. Given: a section with dimensions b = 300 mm; h = 700 mm; a = 50 mm; a’ = 30

mm; heavy-weight concrete B30 (Rb = 15.5 MPa by 9.02 =bγ ); reinforcement A-III

( 365=sR MPa); section area of compressed reinforcement 942' =sA mm2 (3Ø20); bending

moment M = 580 kN·m. It is required to determined section area of tensile reinforcement.

Calculation: 650507000 =−=h mm. The calculation is made considering the area of

compressed reinforcement according to Item 3.19.

Value mα is determined in the following manner:

( )187.0

6503005.15

)30650(94236510580'2

6

2

0

0

'

=⋅⋅

−⋅−⋅=

−−=

bhR

ahARM

b

ssc

mα ;

413.0187 =<= Rm αα (See Table 18)

According to Table 20 by 187=mα we find 21.0=ξ . Required area of tensile reinforcement is

determined by Formula (26):

2680942365

5.1565030021.0'0 =+⋅⋅⋅

=+= s

s

b

s AR

RbhA

ξ mm2

We take 3Ø36 (Rs = 3054 mm2). Example 6. Given: a section with dimensions b = 300 mm; h = 700 mm; a = 70 mm; a’ = 30

mm; heavy-weight concrete B25 (Rb = 13 MPa by 9.02 =bγ ); reinforcement A-III

( 365== scs RR MPa); section area of stretched reinforcement 4862=sA mm2 (6Ø32), of

tensile reinforcement 339' =sA mm2 (3Ø12); bending moment M = 600 kN·m.

It is required to check the section strength.

Calculation: 630707000 =−=h mm. The section strength is checked in compliance with

Item 3.15. The height of compressed zone x is determined by Formula (16):

( )420

30013

3394826365'

=⋅

−=

−=

bR

ARARx

b

sscss mm

We find 604.0=Rξ and 422.0=Rα according to table 18.

As 420=x mm 380630604.00 =⋅=> hRξ mm so section strength is to be checked according

to condition (18):

( ) ( ) 62

0

'2

0 104.7273063033936563030013422.0' ⋅=−⋅+⋅⋅⋅=−+ ahARbhR sscbRα N·mm =

4.727 kN·m > M = 600 kN·m, that is section strength is provided. T-SECTIONS AND DOUBLE T-SECTIONS

Example 7. Given: a section with dimensions 1500' =fb mm, 50' =fh mm, and b = 200 mm,

h = 400 mm, a = 40 mm; heavy-weight concrete B25 (Rb = 13 MPa by 9.02 =bγ );

reinforcement A-III (Rs = 365 MPa); bending moment M = 300 kN·m. It is required to determine the section area of longitudinal reinforcement.

Calculation: 360404000 =−=h mm. The calculation is made according to Item 3.22 on the

hypothesis that compressed reinforcement is not required according to the calculation.

37

We check the condition (32) taking 0' =sA ;

( ) 6'

0

'' 106.326505.036050150013)5.0( ⋅=⋅−⋅⋅=− fffb hhhbR N·mm = 326.6 kN·m, that is the

border of compressed zone goes in the flange and the calculation is made as for rectangular section with the width b = b’f = 1500 mm in compliance with Item 3.18.

We determine the value mα :

422.0119.0360150013

103002

6

2

0

=<=⋅⋅

⋅== R

b

mbhR

Mαα (See Table 18),

that is compressed reinforcement is not required. The section area of stretched reinforcement is calculated by formula (23). For that according to

Table 20 by 119.0=mα we find 938.0=ζ and

2434360938.0365

10300 6

0

=⋅⋅

⋅==

hR

MA

s

sζ

mm2

We take 4Ø28 (As = 2463 mm2).

Example 8. Given: a section with dimensions 400' =fb mm, 120' =fh mm, and b = 200 mm,

h = 600 mm, a = 60 mm; heavy-weight concrete B15 (Rb = 7.7 MPa by 9.02 =bγ );

reinforcement A-III (Rs = 365 MPa); bending moment M = 270 kN·m. It is required to determined section area of tensile reinforcement.

Calculation: 540606000 =−=h mm. The calculation is made in compliance with Item 3.22

on the hypothesis that compressed reinforcement is not required.

As ( ) 6'

0

'' 104.1771205.05401204007.7)5.0( ⋅=⋅−⋅⋅=− fffb hhhbR N·mm = 177.4 kN·m < M =

= 270 kN·m, that is the border of compressed zone goes in the rib, section area of stretched reinforcement is calculated by formula (33).

For that we determine the value mα :

( ) ( ) ( )44.0404.0

5402007.7

1205.05401202004007.710270)5.0(2

6

2

0

'0

''

=<=⋅⋅

⋅−⋅−−⋅=

−−−= R

b

fffb

mbhR

hhhbbRMαα

(see Table 18), so compressed reinforcement is not required.

According to Table 20 by 404.0=mα we find 563.0=ξ , then

( )[ ] ( )[ ] 1789365

7.7120200400540200563.0''

0 =−+⋅⋅=−+=s

b

ffsR

RhbbbhA ξ mm2.

We take 4Ø25 (As = 1964 mm2).

Example 9. Given: a section with dimensions 400' =fb mm, 100' =fh mm, b = 200 mm,

h = 600 mm, a = 70 mm; heavy-weight concrete B25 (Rb = 13 MPa by 9.02 =bγ ); tensile

reinforcement A-III (Rs = 365 MPa), its section area As = 1964 mm2 (4Ø25); 0' =sA ; bending

moment M = 300 kN·m. It is required to check the strength of the section.

Calculation: 530706000 =−=h mm. The section strength is checked in compliance with

Item 3.20, taking 0' =sA . As 7168601964365 =⋅=ss AR N >

52000010040013'' =⋅⋅=ffb hbR N, the border of compressed zone goes in the rib. The section

strength is checked according to condition (28). For that we determine the height of compressed zone x by Formula (29):

38

( )20013

100200400131964365)( ''

⋅

−−⋅=

−−=

bR

hbbRARx

b

ffbss =

= 176 mm 320530604.00 =⋅=< hRξ mm ( Rξ is found according to table 18);

( ) ( )+⋅−⋅⋅=−−+− 1765.053017620013)5.0()(5.0 '

0

''

0 ffhbb hhhbbRxhbxR

( ) 6101.327)1005.0530(10020040013 ⋅=⋅−−+ N·mm = 327.1 kN·m > M =300 kN·m,

that is the strength of the section is provided. MEMBERS WORKING IN SKEW BENDING 3.24. Calculation of rectangular sections, T-sections, double T- and L-sections of members

working in skew bending can be made taking the form of compressed zone according to Draft 6, at the same time the following condition must be met:

( )[ ] sxscxovwebbx SRSxhARM ++−≤ ,10 3/ (35)

where Mx is a component of a bending moment in plane of axes x (for axes x and y we take to perpendicular axes going trough the center of gravity of section of tensile reinforcement parallel to the section sides; for a section with a flange axis x is taken parallel to the rib plane);

ovbweb AAA −= (36)

Ab compressed concrete zone area equal to:

b

sscss

bR

ARARA

'−= (37)

Aov is the area of the most compressed overhang of a flange; x1 the measurements of compressed zone along the most compressed lateral side

of the section determined by the following formula:

( )ysyscyovwebb

webb

MSRSAbR

ARx

−++=

,0

2

13

2 (38)

b0 is the distance from the center of gravity of the section of tensile reinforcement to the most compressed lateral face of the rib (side);

Sov,y is static moment of the area Aov in the plane of axis y relating to axis x;

Ssy is static moment of the area '

sA in the plane of axis y relating to axis x;

My is a component of bending moment in the plane of axis y; Sov,x is static moment of the area Aov in the plane of axis x relating to axis x;

Ssx is static moment of the area '

sA in the plane of axis x relating to axis y.

Draft 6. Form of compressed zone in cross section of reinforced concrete element working in biaxial

bending

a – T-section; b – rectangular section; 1 – plane of bending moment; 2 – center of gravity of tensile reinforcement section.

If considered in the calculation tensile reinforcement rods are located in plane of axis x (Draft 7) value x1 is determined by the following formula:

βctgAttx web22

1 ++−= (39)

Where

−+

−= 00

,,5.1 hctgb

A

SctgSt

web

xovyovβ

β;

β is angle of dip of the bending moment plane to axis x that is ctgβ = Mx/My.

39

Drafts 7. Section with tensile reinforcement rods in the plane of axis x.

Formula (39) must be also used independently on reinforcement location if it is necessary to determine limit value of bending moment by given angle β. During calculation of rectangular sections values Aov, Sov,x and Sov,y in formulas (35), (36), (38) and (39) are taken equal to zero.

If ovb aA < or '

1 2.0 fhx < so that means that the calculation is made as for rectangular

section '

fbb = wide.

If the following condition is met:

ov

web

bb

Ax

=<

5.11 (40)

(where ovb the width of the least compressed overhang of the flange), so the calculation is

made without considering skew bending that is according to formulas of Items 3.15 and 3.20 as regards moment M = Mx at the same time it is necessary to check the condition (41) taking x1 as by skew bending. During determination of the value Ab by formula (37) the stress in the closest to compressed zone border tensile bar must be no less than Rs that corresponds to the following condition:

( ) R

iovi

ov

ihtgbb

xtgbξ

θ

θξ ≤

+=

==

0

'

0

1

'

(41)

Where Rξ – see Tables 18 and 19

ii hb 00 , are the distances from the rod under consideration to the most compressed and

lateral surface of the rib (side) and to the most compressed surface normal to axis x (see Draft 4);

'

ovb – The width of the most compressed overhang;

θ – Angle of slope of the line bounding the compressed zone to axis y; value of tgθ is determined by the following formula:

webA

xtg

2

21=θ .

If condition (41) is not met so the calculation of the section is made by means of step-by-step approximation and replacing in formula (37) value Rs for each tensile rod by stress values equal to:

( ) sicsi R1/ −ξωψσ But no more than Rs,

Where ωψ ,c are taken according to Table 18 and 19, at the same time axes x and y must

be drawn through resultant of forces in tensile rods.

During design of structures value iξ must not exceed value Rξ more than by 20 percent,

at the same time it is possible to make only one repeated calculation with replacement of

values Rs in formula (37) for tensile rods for which Ri ξξ > by stresses equal to :

( )s

ic

si R3

21/ +−ξωψσ (42)

By repeated calculation value x1 is determined by formula (39) independently on location of tensile rods.

40

The calculation as regards skew bending is made according to Item 3.27 if the following conditions are met: - for rectangular sections, T- and L-sections with a flange in compressed zone

hx >1 (43)

- for double T-, T- and L-sections with a flange in tensile zone

θtgbhhx tovf ,1 −−> (44)

where hf, bov,ttgθ is the height and the width of the least tensile overhang of a flange (Draft 8).

Draft 8. T-section with compressed zone going into the least tensile overhang of a flange.

When using formula (37) it is recommended to take reinforcement located near tensile surface which s parallel to axis y for tensile reinforcement with the area As, and to take reinforcement located near tensile surface which s parallel to axis y but on one the most

compressed side of axis x (see Draft 6) for compressed reinforcement with the area '

sA .

3.25. It is recommended to determine required quantity of tensile reinforcement by skew

bending for rectangular section, T- and L-section elements with a flange in compressed zone by means of Draft 9. For that αs is determined by means of the diagram depending on the following values:

2

00

,

hbR

SRSRM

b

sxscxovbx

mx

−−α ;

0

2

0

,

hbR

SRSRM

b

syscyovby

my

−−=α

[symbols see in Formulas (35)–(38)].

If 0<mxα so the calculation is made as for rectangular section taking '

fbb = .

If value αs on the diagram is located on the left side of the curve corresponding to

parameter 0b

bbov +, choosing of reinforcement is made without considering skew bending

that is according to Items 3.18, 3.19 and 3.22 as regards the moment xMM = .

Draft 9. Diagram of bearing capacity of rectangular, T- and L-sections for members working in skew

bending

Required area of tensile reinforcement by work condition of its total design resistance is determined by the following formula:

( ) '

00 s

s

b

ovss AR

RAhbA ++= α (45)

where Aov – see Formula (36). Center of gravity of the section of actual tensile reinforcement must be distant from tensile surfaces no more than the taken in the calculation center of gravity. Otherwise the calculation is made one more time taking the new center of gravity of tensile reinforcement.

41

Work condition of tensile reinforcement with total design resistance is the fulfillment of condition (41). For members made of concrete B25 and lower condition (41) is always met if value αs on Draft 9 is located inside the area bounded by coordinate axes and a curve corresponding

to parameter 0

' / bbov .

If condition (41) is not met so it is necessary to put (increase) compressed reinforcement or increase class of concrete or to increase dimensions of the section (especially dimensions of the most compressed overhang).

Values αs on the diagram must not be located between axis myα and a curve

corresponding to parameter hh /0 . Otherwise x is more than h and the calculation is made

according to Item 3.27. 3.26. The calculation of skew bending of rectangular and double T-shaped symmetric sections

with symmetrically located reinforcement can by made according to Item 3.76 taking N = 0.

3.27. For sections not settled in Items 3.24–3.26 as well as if conditions (43) and (44) are met and if reinforcement is spread on the section that disturbs the determination of values As

and '

sA and location of centers of gravity of tensile and compressed reinforcement,

calculation as regards skew bending must be made using the formula of general case of normal section calculation (see Item 3.76) considering instructions of Item 3.13. Order of use of formulas of general case is the following: 1) two perpendicular to each other axes x and y are drawn through the center of gravity

of the section of the most tensile rod if possible parallel to section sides; 2) By means of step-by-step approximation it is chosen the location of a line bounding

compressed zone so that the equation (154) by N = 0 was met after insertion into it

the value siσ determined by formula (155). At the same time angle of slope of this

line θ is taken as a permanent one and equal to angle of slope of neutral axis determined as for elastic body;

3) there are determined moments of internal stresses relating to axes x and y – Myu and Mxu.

If both these moments are more or less then correspondent components of external moment (My or Mx) so the strength of the section is correspondingly provided (if more) and not provided (if less). If one of moments (for example Myu) is less than the correspondent component of external moment My and the other moment is more than the component of external moment (that is Mxu < Mx) so it is taken a larger angle of slope θ (more than the one taken earlier) and the same calculation is made one more time. Examples of Calculation

Example 10. Given: reinforced concrete collar beam of the roof with the slope 1:4 (ctgβ

= 4); section and location of reinforcement is corresponding to Draft 10; 9.02 =bγ (no

loads of short duration); heavy-weight concrete B25 (Rb = 13 MPa); stretched

42

reinforcement class A-III (Rs = 365 MPa); its section area As = 763 mm2 (3Ø18); 0' =sA ;

bending moment in a vertical plane M = 82.6 kN·m. It is required to check the strength of the section. Calculation. In compliance with Draft 10 it is:

3603

301304000 =

⋅−−=h mm;

903

30112020 =

⋅+⋅=b mm;

752

150300' =−

== ovov bb mm;

902

2080' =+=fh mm

Draft 10. To calculation example 10.

1 – bending moment plane; 2 – center of gravity of tensile reinforcement section

According to Formula (37) we determine the area of compressed zone of concrete Ab:

2142013

763365=

⋅==

b

ss

bR

ARA mm2.

Area of the most compressed overhang of a flange and static moments of its area relating to axes x and y are correspondingly equal to:

67509075'' =⋅=fovov hbA mm2;

( ) ( ) 4'

0, 1006.862/759067502/ ⋅=+=+= ovovyov bbAS mm3;

( ) ( ) 4'

0, 106.2122/9036067502/ ⋅=+=+= fovxov hhAS mm3.

As ovb AA > so the calculation is continued as for T-section.

14670675021420 =−== ovbweb AAA mm2.

The components of bending moment in the plane of axes y and x are correspondingly equal to (by ctgβ = 4):

2041

6.82

1sin

22=

++==

ββ

ctg

MMM y kN·m;

80420 =⋅== βctgMM yx kN·m

According to formula (38) we determine dimensions of compressed zone of concrete x1

relating to the most compressed side of the section, taking 0=syS :

( ) 2232000000)8606001467690(13

1467013

3

2

3

2 2

,0

2

1 =−+⋅

⋅=

−+=

yyovwebb

webb

MSAbR

ARx mm

Let’s check condition (40):

8.9775150

146705.15.1 2

=+

⋅=

+ ov

web

bb

A mm < x1 = 233 mm

So the calculation is to be continued by formulas foe skew bending.

Let’s check condition (41) for the least tensile rod. According to Draft 10 300 =ib mm,

370304000 =−=ih mm:

695.1146702

223

2

212 =

⋅==

webA

xtgθ ;

43

( ) ( )604.064.0

370695.17530

223695.175

00

1 =>=++

+⋅=

++

== R

iovi

ov

ihtgbb

xtgbξ

θ

θξ

(see Table 18). Condition (41) is not met. Let’s make one more calculation; in formula (37) we replace

value Rs for the least tensile rod by stress sσ determined by formula (42), correcting

values h0 and b0.

From Table 18 we have 746.0=ω and 26.4=cψ .

( ) ( )sss

ic

s RRR 902.03

2164.0/746.026.4

3

21/=

+−=

+−=

ξωψσ

As all rods have the same diameter so new values bA , 0b and 0h will be equal to:

207203

902.0221420 =

+=bA mm2;

0.92902.02

30902.012020 =

+

⋅+⋅=b mm;

7.359902.02

301304000 =

+

⋅−−=h mm

Then let’s determine values yovS , , xovS , and Aweb:

4

, 1041.87)2/7592(6750 ⋅=+=yovS mm3;

4

, 104.212)2/907.359(6750 ⋅=+=xovS mm3;

13970675020720 =−=webA mm2.

Value x1 we determine by formula (39):

8.1597.35949213970

212400048741005.15.1 00

,,=

−⋅+

−⋅=

−+

−= hctgb

A

SctgSt

web

xovyovβ

βmm

7.21041397028.1598.1592 22

1 =⋅⋅++−=++−= βctgAttx web mm

Let’s check the section strength according to condition (35) taking 0=sxS :

( )[ ] 64

,10 102.80104.2123

7.2107.35913970133/ ⋅=

⋅+

−=+− xovwebb SxhAR N·mm >

> Mx = 80·106 N·mm, that is the section strength is provided. Example 11. According to data of Example 10 it is necessary to choose the area of tensile reinforcement by moment in vertical plane M = 64 kN·m. Calculation. Components of bending moment in the plane of axes y and x are equal to:

6245.1541

1064

1sin

2

6

2=⋅=

+

⋅=

+==

ββ

ctg

MMM y kN·m

We determine required quantity of reinforcement according to Item 3.25.

Taking values 0,,,, '

,0,000 == sysyyvxv SSSShb according to Example 10 we find values

mxα and myα :

227.03609013

104.2121310622

46

2

00

,=

⋅⋅

⋅⋅−⋅=

−=

hbR

SRM

b

xovbx

mxα ;

44

114.03609013

1006.8613105.152

46

0

2

0

,=

⋅⋅

⋅⋅−⋅=

−=

hbR

SRM

b

yovby

myα .

As 0>mxα so the calculation is continued as for T-section.

As the point with coordinates 227.0=mxα and 114.0=myα on Drawing 9 is located on

the right side of the curve corresponding to parameter 5.290

75150

0

=+

=+

b

bb ov and on

the left side of the curve corresponding to parameter 83.090

75

0

==b

bov so the

reinforcement will work with total design resistance that is condition (41) is met. Required area of tensile reinforcement is determined according to formula (45).

According to Draft 9 by 227.0=mxα and 114.0=myα we find 25.0=sα

Then by 0' =sA we have

( ) ( ) 529365

1367503609025.000 =+⋅⋅=+=

s

b

ovssR

RAhbA α mm2.

We take rods 3Ø16 ( 603=sA mm2) and arrange them as it is shown on Draft 10.

Example 12. Given: hinged wall plate of a public building with a span 5.8 m with the cross section according to Draft 11; light-weight concrete B3.5, average density grade D1100; reinforcement A-III; load on the panel at stage of use – dead weight and weight of located above glass (including the pier) 3 m height 3.93 kN/m2, from plane of the panel – wind load 0.912 kN/m2. It is required to check the strength of the panel at stage of use. Calculation. First we determine bending moments in the middle section of the panel in the plane and out of the plane of the panel. According to Item 2.13 we determine the load of dead weight of the panel. As class of light weight concrete is lower than B12.5 so density of concrete is

121011001.11.1 =⋅== Dγ kg/m3. So the load of dead weight of the panel will be:

94.401.012102.134.001.0 =⋅⋅⋅=⋅= γbhqw kN/m

And considering safety factor 2.1=fγ (as 1800<γ kg/m3)

92.594.42.1 =⋅=wq kN/m

Load of located above glazing weight is 8.11393.3 =⋅=gq kN/m.

Total load in the panel plane is:

72.178.1192.5 =+=+= gwx qqq kN/m,

and moment of this load in the middle of the panel is:

5.748

8.572.17

8

2

=⋅

==lq

M x

x kN·m

Wind load per 1 m of the panel length considering the load of located above and below glazing is:

( ) 83.332.1912.0 =+=yq kN/m

and moment of this load is:

1.168

8.583.3

8

22

=⋅

==lq

My

y kN·m

45

As reinforcement is spread unevenly on the section so the strength is to be checked by formulas of a general case in compliance with Item 3.76 (considering Item 3.13). We’ll give numbers to all rods as it is shown on Draft 11. Through the center of the most stretched rod 1 we draw axis x parallel to dimension h = 1195 mm and axis y parallel to dimension b = 340 mm.

Draft 11. To example of calculation 12

! – 8 – rods

Angle θ between axis y and the straight line bounding the compressed zone is taken as for the calculation of elastic body as regards biaxial bending:

67.234.0

195.1

5.74

1.1622

=

=

==

b

h

M

M

I

I

M

Mtg

x

y

y

x

x

yθ .

In the first approximation we determine the area of concrete compressed zone by formula (37) that is taking all rods with total design strength, at the same time rod 8 is taken as a compressed one and other ones are taken as tensile ones. For rods 1, 2, 7, 8 (Ø10) we have Rs = Rsc = 365 MPa, and for rods 3 – 6 (Ø6) – Rs = 355 MPa, then:

126250113355236365 =⋅+⋅=ss RR N;

286505.78365' =⋅=ssc AR N.

As there is a wind load so value Rb is taken considering the coefficient 1.12 =bγ that is

3.2=bR MPa.

424403.2

28650126250'

=−

=−

=b

sscss

bR

ARARA mm2

Area of compressed zone on the assumption that it’s of triangular form is determined by

formula θtg

xAb

2

2

1= where x1 is dimension of compressed zone by the section side h so x1

is:

476424407.26221 =⋅⋅== bAtgx θ mm < h = 1195 mm.

Dimension y1 of compressed zone by section side b is:

17276.2

47611 ===

θtg

xy mm < b = 340 mm,

that is compressed zone is of triangular form in fact. If we put these dimension on Draft 11 so we can see that rod 8 is located in compressed

zone and all other rods are in tensile zone. Let’s check the stress siσ in the closest to

tensile zone border rods that is in rods 6 – 8 by formula (155), determining the ratios

ih

x

0

=ξ by formula xiyi

iatga

x

==

θξ 1 where xia and yia are the distances from i-rod to

the most compressed side of the section in direction of axes x and y.

Taking 400, =uscσ MPa, 782.03.2008.08.0008.08.0 =⋅−=−= bRω we get

13841082

1782.0

1.1

782.01

4001

1.11 1

,−=

−

−

=

−

−

=ii

usc

siξξξ

ωω

σσ (MPa)

The calculations we summarize in the following Table: Rod number

siA , mm2 yia , mm xia , mm xiyi atga +θ iξ ssi R<>σ

46

, mm MPa

6 7 8

28.3 78.5 78.5

40 300 40

555 80 80

662 881 187

0.719 0.54

2.545

120.9<355 620>365

-959<-365

During determination of Ab it was taken incorrect stress only for rod 6: 355 MPa instead

of 120.9 MPa. In this rod we take stress larger than the calculated one, – 1606 =sσ MPa.

From equation (150) by N = 0 we determine value Ab:

400803.2

286503.2816085355263365=

−⋅+⋅+⋅==

∑b

sisi

bR

AA

σ mm2.

By a similar way we determine 4634008067.221 =⋅⋅=x mm.

So for rod 6 we have:

699.0662

46316 ==

+=

xiyi atga

x

θξ ;

1641384699.0

10826 =−=sσ MPa,

That is value 6sσ is close to the excepted one and so it is not necessary to calculate

values Ab and x1 one more time. Let’s determine moments of internal forces relating to axes y and x – Mxu and Myu.

( ) −

−⋅=−Σ−

−=

3

4631115400803.2

31

11 xixsisixbbxu aaA

xaARM σ

( ) ( ) ( )−−⋅+−−⋅⋅− 55511153.28160355101511153.282355

( )( ) 6104.788011155.783655.78365 ⋅=−⋅−⋅− N·mm =

+ 78.4 kN·m > 5.74=xM kN·m;

17367.2

46311 ===

θtg

xy mm;

( ) −

−⋅=−Σ−

−=

3

173300400803.2

31

11 yiysisiybbyu aaA

yaARM σ

( )( ) =−⋅−⋅+⋅+⋅− 403005.783653.281603.283555.78365 61055.18 ⋅= N·mm = 18.5 kN·m > 1.16=yM kN·m.

As both internal moments exceed both components of external moment so the section strength is provided.

CALCULATION OF SECTIONS INCLINED TO LONGITUDINAL AXIS OF THE MEMBER

3.28. (3.29) It is necessary to calculate inclined sections of reinforced concrete members to

provide the strength: - against lateral force along inclined band between inclined cracks in compliance with

Item 3.30; - against lateral force along inclined crack for members with cross reinforcement in

compliance with Items 3.31-3.39, for members without cross reinforcement - in compliance with Items 3.40 and 3.41;

- against bending moment along inclined crack in compliance with Items 3.42-3.47.

47

Short consoles of columns are calculated as regards the lateral forces along inclined compressed band between the load and the support in compliance with Item 3.99. Beams loaded by one or two point forces located no farther than h0 from the support as

well as short beams with a span 02hl ≤ are to be calculated as regards lateral force

considering the strength of inclined compressed band between the load and the support considering proper recommendations. It is possible to calculate such beams as members without cross reinforcement according to Item 3.40. Note. In the present document under cross reinforcement we understand stirrups and bend-up bars. Definition “stirrups” includes cross rods of welded frameworks and stirrups of bound frameworks.

3.29. Distances between stirrups s, between a support and a bend-up bar end s1 as well as between the end of a previous and beginning of the next bending s2 (Draft 12) must be no more than value smax:

Q

bhRs btb

2

04max

ϕ= (46)

Where 4bϕ – see table 21.

Besides, these distances must correspond to constructive requirements of Items 5.69 and 5.71. Draft 12. Distances between stirrups, support and bendings.

By linear width b variation along the height it is necessary to insert [into the formula (46) and the following ones] the width of the member at the level of the middle of the section height (without considering flanges).

CALCULATION OF MEMBERS AS REGARDS LATERAL FORCE ALONG INCLINED COMPRESSED BAND

3.30. (3.30) Calculation of reinforced concrete members as regards lateral force to provide the

strength along inclined stripe between inclined cracks must be made according to the following condition:

0113.0 bhRQ bbw ϕϕ≤ (47)

where Q is lateral force in a normal section taken at the distance from the support no less than h0;

1wϕ – Coefficient considering influence of stirrups normal to the member axis

and determined by the following formula:

ww αµϕ 511 += (48)

but no more than 1.3;

Herebs

Asw

w =µ ;

1bϕ – Coefficient determined by the following formula:

bb Rβϕ −= 11 (49)

here β is the coefficient taken equal to 0.01 – for heavy-weight and fine concrete; 0.02 – for light-weight concrete;

Rb is in MPa.

48

CALCULATION OF INCLINED SECTION AS REGARDS LATERAL FORCE ALONG INCLINED CRACK Members with constant height reinforced by stirrups without bend-up bars

3.31. The strength of inclined section as regards lateral force along inclined crack (Draft 13) is

made according to the following condition:

swb QQQ +≤ (50)

where Q is lateral force of external load located on one side of inclined section under review; by vertical load on the top surface of the member value Q is taken in the normal section going through the most distant from the support end of inclined section; by the load on the bottom surface of the member or within the height of its section it is also possible to take value Q as the most distant from the support end of inclined section if stirrups installed according to Item 3.97 are not considered in the present calculation; at the same time it is necessary to consider absence of live load within the inclined section;

Qb is cross force taken by concrete and equal to:

c

MQ b= , (51)

( ) 2

02 1 bhRM btfbb ϕϕ += ; (52)

2bϕ – Coefficient considering concrete type and determined by Table 21;

fϕ – Coefficient considering influence of compressed flanges of T- and double-T-

section elements and determined by the following formula:

( )0

''

75.0bh

hbb ff

f

−=ϕ (53)

but no more than 0.5,

At the same time value )( 'bb f − is taken no more than '3 fh ;

reinforcement in the rib is anchored in the flange where there is cross reinforcement connecting flange overhangs with a rib;

c is projection length of inclined axis upon member determined according to Item 3.32.

Draft 13. Forces model in inclined section of elements with stirrups during its calculation as regards lateral

force

Table 21

Coefficients Concrete

2bϕ 3bϕ 4bϕ

Heavy-weight concrete Fine concrete Light-weight concrete

- D1900 - D1800 and lower by fine aggregate

dense porous

2.00 1.70

1.90

1.75 1.50

0.6 0.5

0.5

0.4 0.4

1.5 1.2

1.2

1.0 1.0

Value Qb is taken no less than ( ) 03min, 1 bhRQ btfbb ϕϕ += ( 3bϕ - see Table 21);

Qsw is cross force carried by stirrups equal to:

49

0cqQ swsw = (54)

here qsw is force in stirrups per length unit of the member within inclined section determined by the following formula:

s

ARq swsw

sw = ; (55)

c0 is projection length of inclined crack upon longitudinal axis of the member taken equal to:

sw

b

q

Mc =0 (56)

But no more than c and no more than 02h as well as no less than 0h if 0hc >

At the same time for stirrups installed according to the calculation (that is when requirements of Items 3.40 and 3.41 are not met) it is necessary to follow the following requirement:

0

min,

2h

Qq

b

sw ≥ (57)

It is possible not to follow the requirement (57) if in formula (52) it is considered such

reduced value bRbt when condition (57) becomes an equation that is if3

22

02b

b

swb qhMϕ

ϕ= ;

in that case it is always 00 2hc = but no more than c.

3.32. By checking of condition (50) in general case it is taken a row of inclined sections by

different values c not exceeding the distance from the support to the section with

maximum bending moment and no more than ( ) 032 / hbb ϕϕ .

If point loads act on the element so values c are taken equal to the distances from the support to the point of the force impact (Draft 14).

Draft 14. Location of design inclined sections by point loads

1 – inclined section checked as regards the lateral force Q1; 2 – the same of force Q2

By calculation of the member as regards the distributed load q value c is taken equal to

sw

b

q

M and if swqq 65.01 > so it is necessary to take

sw

b

qq

Mc

+=

1

where q1 is

determined in the following manner:

a) if there is actual distributed load qq =1 ;

b) If load q includes live load which reduces to equivalent distributed load v (when diagram of moment M of accepted in the calculation load v always bends round the

diagram M of any actual live load), 2/1 vgq += (where g is dead continuous load).

At the same time value Q is taken equal to cqQ 1max − where maxQ is lateral force in

support section.

3.33.Required density of stirrups shown by means of swq (see Item 3.31) is determined in the

following manner: a) by point forces located at the distances ci of the support acting on the element, for

each inclined section with the projection length ci not exceeding the distance to the

50

section with maximum bending moment value qsw is determined according to the

coefficient bi

bii

iQ

QQ −=χ by one of the following formulas:

If0

0min,

02

ææh

c

Q

Q

bi

b

ii =< ,

1æ

æ

0

0

0

)(+

=i

ii

iswc

Qq ; (58)

If0

0 ææc

ci

ii ≤≤ ,

0

)(c

QQq bii

isw

−= ; (59)

If00

æh

c

c

c i

i

i ≤< ,

( )

b

bii

iswM

QQq

2

)(

−= ; (60)

If0

æh

ci

i > ,

0

)(h

QQq bii

isw

−= ; (61)

(here h0 is taken no more than ci).

Finally it is taken maximum value )(iswq .

In formulas of Item 3.33: Q is lateral force in the normal section located at the distance ci from the

support; Qbi is determined by formula (51) by c = ci;

bb MQ ,min, See Item 3.31;

c0 is taken equal to ci but no more than 2h0;

b) by distributed load q required density of stirrups is determined by the following formulas:

By 6.0

maxbiQ

Q ≤

b

b

swM

QQq

4

2

1

2

max −= ; (62)

By 6.0

1max1

0

b

b

b QQQ

h

M>>+

( )

b

b

swM

QQq

2

1max −= ; (63)

51

In both cases swq is taken no less than

−

0

1max

2h

QQ b ;

By 1

0

max b

b Qh

MQ +≥

0

1max

h

QQq b

sw

−= (64)

In case if calculated value swq doesn’t correspond to condition (57) so it must be

calculated by the following formula: 2

0

max

2

1

3

2

0

max1

3

2

0

max

222

−

+−+=

h

Qq

h

Qq

h

Qq

b

b

b

b

swϕ

ϕ

ϕ

ϕ,

Here 11 2 qMQ bb =

maxQ – Lateral force in the support section;

32 ,, bbbM ϕϕ – see in Item 3.31;

q1 see in Item 3.32.

3.34. If from the support to the span density of stirrups reduces from 1swq to 2swq (for example

by stirrups spacing increase) it is necessary to check condition (50) by values c increasing

l1 the length of the member part with stirrups quantity 1swq (Draft 15). At the same time

value swQ is taken equal to:

By 011 clc <−

( )( )121011 lcqqcqQ swswswsw −−−= ;

By 01102 clcc >−>

( )12 lcqQ swsw −= ;

By 021 clc >−

022cqQ swsw = ,

Where 0201 ,cc are determined by formula (56) by swq equal to 1swq and 2swq .

Draft 15. to the calculation of inclined sections by density of stirrups alterations.

By distributed load acting on the element the length of the part with density 1swq is taken

no less than value l determined in the following manner:

If 211 swsw qqq −> ,

21

1max0111

/

swsw

swb

qq

cqQcqcMcl

−

+−+−= ,

Where ( )211 swsw

b

qqq

Mc

−−= but no more than 0

3

2 hb

b

ϕ

ϕ;

At the same time if 211 56.1 swsw qqq −>

So21 sw

b

qq

Mc

+= ;

52

If 211 swsw qqq −≤

( )01

1

012min,max

1 cq

cqQQl

swb−

+−=

Here 1q - see Item 3.32.

If condition (57) is not met for the value 2swq so the length l1 is calculated by values

322

2

0 /2 bbswb qhM ϕϕ= and 20min, 2 swb qhQ = corrected according to Item 3.31, at the same

time the sum 012min, cqQ swb + is taken no less than the uncorrected value min,bQ .

Members with a constant height reinforced by bend-up bars

3.35. Strength test of inclined section against lateral force for members with bend-up bars is made according to condition (50) with addition to the right part of the condition (50) the following value:

θsin,, incsswimcs ARQ = (65)

Where incsA , – section area of bend-up bars crossing dangerous inclined crack with the

projection length 0c ;

θ is angle of slope of bend-up bars to the longitudinal axis of the member.

Value 0c is taken equal to the length of the member part within the inclined section under

review for which the expression 0

,0,c

MQcqQQQ b

incsswbincssw ++=++ gets minimum

value. For that it is necessary to consider the parts from the end of inclined section or from the end of bend-up bar within the length c to the beginning of the bend-up bar which is close to the support (Draft 16) at the same time the length of the part is taken no

more than 0c determined by formula (56) and inclined cracks not crossing bend-up bars

by values 0c less than the ones determined by formula (56) are not considered in the

calculation.

Draft 16. For determination of the most dangerous inclined section for members with bend-up bars by

calculation as regards lateral force.

1 – 4 – possible inclined sections; 5 – considered inclined section

On Draft 16 the most dangerous inclined crack corresponds to minimum value of the following expressions:

1 – 011,01 /sin cMARcq bincsswsw ++ θ ;

2 – 022,02 /sin cMARcq bincsswsw ++ θ ;

3 – 033,03 /sin cMARcq bincsswsw ++ θ ;

4 – 022,1,0 /sin)( cMAARcq bincsincsswsw +++ θ

[here 0c - see formula (56)].

Values c are taken equal to the distances from the support to the ends of bend-up bars as well as to points of concentrated forces; besides it is necessary to examine inclined

sections which cross the last plane of bend-up bars and end at the distance 0c determined

by formula (56) from the beginning of the last and next to the last planes of bend-up bars (Draft 17).

53

Location of bend-up bars must meet the requirements of Items 3.29, 5.71 and 5.72.

Draft 17. Location of inclined sections of the member with bend-up bars

1 – 4 – calculated inclined sections.

Variable height elements with cross reinforcement

3.36. (3.33) Calculation of members with inclined compressed surfaces as regards lateral forces is made according to Items 3.31, 3.32, 3.34 and 3.35 considering recommendations

of Items 3.37 and 3.38 taking maximum value 0h as working height (Draft 18, a).

It is also recommended to make calculation of members with inclined stretched sections as regards lateral forces in compliance with Items 3.31, 3.32, 3.34 and 3.35 taking

maximum value 0h within inclined section (Draft 18, b).

Draft 18. Beams with variable height and inclined surface

a – compressed; b – tensile

Angle β between compressed and tensile surfaces of member must meet the requirement 4.0<βtg .

3.37. For beams without bend-up bars with height evenly increasing from the support to the

span (see Draft 18) calculated as regards distributed load q inclined section is tested according to condition (50) by the most disadvantageous value c determined in the following manner:

if the following equation is met:

swincsw qqqq 5.256.01 −< (66)

so value c is determined in by the following formula

1

1

qqqq

Mc

swincinc

b

++= (67)

if condition (66) is not met so value c is calculated by the following formula:

1

1

qqq

Mc

swinc

b

++= (At the same time cc =0 ) (68)

As well as if )4/( 2

011 hMq bsw < ,

1

1

2 qtgqq

Mc

swinc

b

++=

β(At the same time 00 2hc = ) (69)

Here βϕ 2

2 btgRq btbinc = ;

1bM – Value bM determined by formula (52) as for support section of beam with

working height 01h without considering enlarged footing b;

β – Angle between compressed and tensile surfaces of the beam;

1q – See Item 3.32.

Working height h0 is taken equal to βctghh += 010 .

54

By decrease of density of stirrups from 1swq at the support to 2swq in the span it is necessary

to examine condition (50) by values c exceeding l1 the length of the part of the member

with density of stirrups 1swq , at the same time value swQ is determined according to Item

3.34.

Parts of beams with permanent decrease of working height 0h must not be less than the

accepted value c. By point loads acting on the beam, it is necessary to examine inclined sections by values c taken in compliance with Item 3.32 as well as if 01>βtg determined by formula (68)

by 01 =q .

3.38.For consoles without bend-up bars with height evenly exceeding from the free supported

beam to the support (Draft 19) in general case it is necessary to examine condition (50)

taking inclined sections with values c, determined by formula (68) by 01 =q and taken

no more than the distances from the beginning of the inclined section in tensile zone to

the support. At the same time for 01h and Q there are taken working height and shear at

the beginning of the inclined section in tensile zone. Besides it is necessary to examine

inclined sections carried on to the support if cc <2 .

Draft 19. Console with the height decreasing from the support to the free supported end

By distributed loads acting on the console the inclined section is located in tensile zone of normal sections going through the points of application of these loads (see Draft 19). By distributed load or linear increasing to the support the console is calculated as an element with the constant height according to Items 3.31 and 3.32 taking working height

0h in support section.

Members with cross reinforcement by biaxial bending

3.39.Calculation as regards shear force of members with rectangular section exposed to biaxial bending is made according to the following condition:

( ) ( )

1

22

≤

+

ybw

y

xbw

x

Q

Q

Q

Q (70)

Where yx QQ , are components of shear force acting in the plane of symmetry x and in

the normal to it plane y at the most distant from the support end of inclined section;

( ) ( )ybwxbw QQ , Are limit shear forces on inclined section by acting of these forces in

planes x and y and taken equal to the right part of the condition (50). By distributed load acting on the member it is possible to determine value c according to

Item 3.32 for each plane x and y. Note. Bend-up bars are not considered by calculation as regards the shear force by biaxial bending.

Members without cross reinforcement

3.40. (3.32) Calculation of members without cross reinforcement as regards the shear force is made according to the following conditions:

55

a) 0max 5.2 bhRQ bt≤ (71)

Where maxQ – maximum shear force at the support surface;

b) c

bhRQ btb

2

04ϕ≤ (72)

where Q is shear force at the end of inclined section;

4bϕ – Coefficient determined by Table 21;

c – Is projection length of inclined section starting from the support; value c is

taken no more than 0max 5.2 hc = .

In continuous flat slabs with constrained edges (connected with other elements or having

supports) it is possible to divide the mentioned value maxc by the coefficient α:

hb /05.01+=α (73) but no more than 1.25.

By testing of condition (72) in general case it is taken value c no more than maxc .

By point loads acting on the element values c are taken equal to the distances from the support to the application points of these loads (Draft 20).

Draft 20. Location of the most disadvantageous sections in elements without cross reinforcement.

1 – inclined section tested as regards shear force action Q1; 2 – the same for force Q2

During calculation of the element as regards distributed load if the following condition is met:

( )2

0max

41

/ hc

bRq btbϕ

≤ (74)

so value c in condition (72) is taken equal to maxc and if condition (74) is not met so

1

40

q

bRhc btbϕ

= (75)

here q1 is taken by distributed loads in compliance with Item 3.32 and by continuous load with linear variable density it is taken equal to average density of the support part with the length equal to a quarter of the beam (slab) span or to a half of the console overhang

but no more than maxc .

3.41.For elements with variable height of the section during testing of condition (71) value 0h

is taken in support section and during testing of condition (72) as average value 0h within

inclined section.

For elements with the section height increasing by increase of shear force value maxc is

taken equal to βtg

hc

25.11

5.2 01max

+= at the same time for continuous flat plates mentioned in

Item 3.40 βα tg

hc

25.1

5.2 01max

+= ,

Where 01h – working height in support section;

β is angle between tensile and compressed surfaces of the element; α see formula (73) where h can be taken according to support section.

56

By distribution load acting on such element value c in condition (72) is taken equal to:

( )bRqtghc

btb21

201/4/

1

ϕβ += (76)

but no more than maxc where q1 – see Item 3.40.

CALCULATION OF INCLINED SECTIONS AS REGARDS BENDING MOMENT 3.42. (3.35) Calculation of elements as regards bending moment to provide the strength along

inclined crack (Draft 21) must be made according to the following condition:

incsincsswswswswsss zARzARzARM ,,Σ+Σ+≤ (77)

where M is moment of external load located on one side of considered inclined section relating to the axis which is perpendicular to the moment action plane and going through the point of application of resultant force Nb in compressed zone (Draft 22);

sss zAR – Moments sum relating to the same axis from forces in longitudinal

swswsw zARΣ Reinforcement, stirrups and bend-up bars crossing stretched zone of

incsincssw zAR ,,Σ inclined section;

incssws zzz ,,, – The distances from the planes of longitudinal reinforcement, stirrups

and bend-up bars to the mentioned axis.

Draft 21. Forces scheme in inclined section by its calculation as regards bending moment.

Draft 22. Determination of design value of the moment by calculation of inclined section a – for a free supported beam; b – for a console

The height of compressed zone measured by a line normal to longitudinal axis of the member is determined according to the requirements of equilibrium of forces projections in concrete of compressed zone and in reinforcement crossing the inclined section onto a longitudinal axis of the member according to Items 3.15 and 3.20. If there are bend-up bars in the member so in the numerator for value x it is necessary to add value

θcos,incssw ARΣ (where θ is angle of inclination of bend-up bars to the longitudinal axis

of the member).

Value zs can be taken equal to xh 5.00 − but considering compressed reinforcement no

more than 'ah − .

Value swswsw zARΣ by constant stirrups quantity is determined by the following formula: 25.0 cqzAR swswswsw =Σ (78)

Where swq is force in the stirrups per unit of length (see Item 3.31);

c is the length of the projection of incline section onto the longitudinal axis of the member measured between points of application of resultant forces in tensile reinforcement and compressed zone (see Item 3.45).

Values incsz , for each plane of bend-up bars are determined by the following formula:

57

( ) θθ sincos 1, aczz sincs −+= (79)

where a1 is the distance from the beginning of the inclined section to the beginning of a bend-up bar in tensile zone (see Draft 21).

3.43. (3.35) Calculation of inclined sections as regards the moment is made in points of break

or bending of longitudinal reinforcement as well as at surfaces of end free support of a beams and at free end of consoles if there is no special anchors of longitudinal reinforcement. Besides, calculation of inclined sections as regards the moment is made in points of abrupt change of the element configuration. It is possible not to make calculation of inclined sections as regards the moment if conditions (71) and (72) are met by multiplying their right parts by 0.8 and by values c no

more than max8.0 c .

3.44. If inclined section crosses longitudinal inclined reinforcement without anchors so design

strength of this reinforcement Rs within anchorage zone must be decreased by means of

multiplying it by the work condition coefficient 5sγ equal to:

an

x

sl

l=5γ (80)

where lx is the distance from the end of reinforcement to the cross point of inclined section with longitudinal reinforcement;

lan anchorage zone length determined by the following formula:

dR

Rl an

b

s

anan

∆+= λω (81)

Here anan λω ∆, are coefficients taken equal to:

For end free supports of beams 5.0=anω , 8=∆ anλ

For free ends of consoles 7.0=anω ; 11=∆ anλ

In case of use of plain rods coefficient anω is taken equal to 0.8 for supports of beams

and to 1.2 for ends of consoles. If end free supports have confinement or cross reinforcement bounding longitudinal

reinforcement without welding coefficient anω must be divided by value vµ121+ and

coefficient anλ∆ is decreased by the value bb R/10σ here vµ is volume coefficient of

reinforcement determined as for welded meshes by formula (99), for stirrups – by

formula as

Asw

v2

=µ (where Asw and s – are correspondingly section area of a bounding

stirrup and its spacing), in any case value µv is taken no more than 0.06.

Concrete compression stress on the support bσ is determined by means of dividing of

support reaction by support area of the element and is taken no more than bR5.0 .

The length anl is taken no less than 20d or 250 mm for free ends of consoles, at the same

time the length of anchorage anl can be determined considering the data of Table 45

(Pos.1).

58

In case if cross reinforcement or distribution bars are welded to longitudinal tensile bars so considered in the calculation strengthening of longitudinal reinforcement RsAz is decreased by the following value:

btwwww RdnN27.0 ϕ= (82)

Taken no less than wws ndR28.0

In formula (82):

wn – Number of welded rods along the length lx;

wϕ – Coefficient taken by Table 22;

wd – Diameter of welded bars.

Table 22

wd 6 8 10 12 14

wϕ 200 150 120 100 80

Finally value RsAz is taken no more than value RsAz without considering 5sγ and Nw.

3.45. For free supported beams the most disadvantageous inclined section begins from the

surface support and has the projection length c for beams with permanent section height equal to:

qq

ARFQc

sw

incsswi

+

−−=

θsin, (83)

but no more than maximum length of support part beyond which condition (72) is met multiplying the right part by 0.8 and by c no more than 0.8cmax. In formula (83):

Q – shear force in support section; Fi,q – point load and distributed load within inclined section;

As,inc – section area of bend-up bars crossing the inclined section; θ – angle of slope of bend-up bars to longitudinal axis of the element;

qsw – the same like in formula (55).

If value c determined considering point load Fi will be less than the distance to the surface of the support to force Fi and value c determined without considering force Fi – more than this distance so it is necessary to take the distance to force Fi for value c. If within the length c the stirrups change their density from qsw1 at the beginning of the

inclined section to qsw2 so value c is determined by formula (83) by 2swsw qq = and if

numerator is decreased by the value ( ) 121 lqq swsw − (where l1 is the length of the part with

stirrups quantity qsw1). For beams loaded by distributed load q with constant density of stirrups without bends condition (77) can be replaced by the following condition:

( )( )qqMzARQ swsss +−≤ 02 (84)

where Q – is shear force in support sectionж

59

M0 – is the moment in the section along the surface of the support.

For consoles loaded by point forces (Draft 22, b) the most disadvantageous inclined section begins from the point of application of point forces near free end and has the projection length c for consoles with a constant height equal to:

sw

incssw

q

ARQc

θsin,1 −= (85)

but no more than the distance from the beginning of the inclined section to the support (here Q1 is shear force in the beginning of the inclined section). For consoles loaded by only by distributed load q the most disadvantageous inclined section ends at the support section and has the projection length c equal to:

( )qql

zARc

swan

sss

+= (86)

At the same time if anllc −< so it is possible not to make the calculation of inclined

section. In formula (86):

As – is section area of reinforcement going to the free end; zs – see Item 3.42; value zs is determined for support section; lan – length of the anchorage zone (see Item 3.44).

For members with the section height increasing at the same time with the increase of bending moment by determination of the projection length of the most disadvantageous section by formulas (83) and (85) numerators of these formulas are to be decreased by value RsAstgβ – by inclined compressed surface, and by value RsAssinβ – by inclined tensile section (where β is angle of slope of the surface to the horizontal line).

3.46.To provide the strength of inclined sections to bending moment in members of constant height with stirrups longitudinal tensile bars break in the span must be get to the point of break in theory (that is behind the normal section where external moment becomes equal to the bearing capacity of the section without considering broken bars, Draft 23) by length no less than value w determined by the following formula:

dq

ARQw

sw

incssw5

2

sin,+

−=

θ (87)

where Q is shear force in the normal section going through the theoretical break point;

θ,,incsA The same like in formula (83);

b is diameter of a broken bar; qsw see in Item 3.31.

For beams with inclined compressed surface, numerator of formula (87) is decreased by RsAstgβ and for beams with inclined tensile surface – by RsAssinβ (where β is angle of slope of the surface to the horizontal line). Besides, it is necessary to consider requirements of Item 5.44. For members without cross reinforcement value w is taken equal to 10d, at the same time point of theoretical break must be located on the part of the element on which condition (72) is met multiplying the right part by 0.8 and by value c no more than 0.8cmax.

Draft 23. Break of tensile bars in the span

60

1 – point of theoretical break; 2 – diagram M

3.47.To provide strength of inclined sections against bending moment the beginning of bar

bending in tensile zone must be distant from the normal section where bend-up bar is used according to the moment no less than by h0/2 and the end of the bar bending must be located no closer than that normal section where bar bending is not required according to the calculation.

CALCULATION OF INCLINED SECTIONS IN UNDERCUTS

3.48.For members with sharply-changing section height (for example for beams and consoles

with undercuts) it is necessary to make calculation as regards shear force for inclined sections going at the console support created by a undercut (Draft 24) according to Items 3.31-3.39, at the same time it is necessary to insert working height h01 of the short console formed by an undercut into the calculation formulas.

Stirrups necessary for inclined section strength must be fixed behind the end of undercut on the part no less than w0 determined by formula (88).

Draft 24. The most disadvantageous inclined sections n members with undercut

1 – inclined compressed stripe; 2 – by calculation as regards shear force; 3 – the same as regards bending moment; 4 – the same as regards bending moment beyond undercut

3.49. For free supported beams with undercuts it is necessary to make the calculation as

regards bending moment in inclined section going through re-entrant angle of undercut (see Draft 24) according to Items 3.42-3.45. At the same time longitudinal tensile reinforcement in the short reinforcement formed by the undercut must be get behind the end of the undercut at the length no less than lan (see Item 5.44) and no less than w0 equal to:

( )da

q

ARARQw

sw

incsswswsw10

sin20

,11

0 ++−−

=θ

(88)

where Q1 is shearing force in the normal section at the end of undercut; Asw1 is section area of additional stirrups located at the end of the undercut on the

part no more than 4/01h long and which are not considered by

determination of stirrups quantity qsw at the undercut; As,inc is section area of bend-up bars going through the re-entrant angle of

undercut; a0 is the distance from the console support to the end of the undercut;

d is diameter of a broken bar.

Stirrups and bend-up bars fixed at the end of the undercut must conform to the following requirement:

( )0011,1 /1sin hhQARAR incsswswsw −≥+ θ (89)

Where 001 , hh – working height in the short console of the undercut and in the beam

beyond the undercut. If bottom reinforcement of the element has no anchorage so according to Items 3.42-3.45 it is also necessary to check the strength of inclined section located beyond the undercut

and beginning behind the mentioned stirrups at the distance no less than 010 hh − from the

end face (see Draft 24). At the same time in the calculation it is not considered longitudinal reinforcement of short console and projection length c is taken no less than

61

the distance from the inclined section to the end of mentioned reinforcement. Besides anchorage length lan for bottom reinforcement is taken as for free ends of consoles. Calculation of the short console of undercut is made according to Items 3.99 and 3.100 taking direction of compressed strip from external edge of area of bearing to resultant force in additional stirrups with the section area Asw1 at the level of compressed

reinforcement of beams that is by ( )

( ) ( )2

sup

2

01

2

012

'

'sin

xalah

ah

++−

−=θ (where lsup – see Item

3.99, ax – see Draft 24); at the same time in Formula (207) coefficient 0.8 is replaced by 1.0.

EXAMPLES OF CALCULATION Calculation of inclined sections as regards lateral force

Example 13. Given: reinforced concrete floor slab with dimensions of cross section according

to Draft 25; heavy-weight concrete B15 ( 7.7=bR MPa 67.0=btR MPa by 9.02 =bγ ; 3105.20 ⋅=bE MPa); rib of the slab is reinforced by plane welded framework with cross

reinforcement rods A-III, diameter 8 mm ( 3.50=swA mm2; 285=swR MPa; 5102 ⋅=sE MPa), with spacing 100=s mm; equivalent live load v = 18 kN/m; gravity load of

the slab and floor g = 3.9 kN/m; cross force on the support 62max =Q kN.

It is required to test the strength of inclined strip of the rib between inclined cracks as well as strength of inclined sections to shear force. Draft 25. For example of calculation 13

Calculation. 292583500 =−=h mm. Strength of inclined strip is to be calculated

according to condition (47).

We determine coefficients 1wϕ and 1bϕ :

0059.010085

3.50=

⋅==

bs

Asw

wµ ;

76.9105.20

1023

5

=⋅

⋅==

b

s

E

Eα ;

So 3.129.10059.076.951511 <=⋅⋅+=+= ww αµϕ ;

For heavy-weight concrete 01.0=β ;

923.07.701.0111 =⋅−=−= bb Rβϕ ,

So 68300292857.7923.029.13.03.0 11 =⋅⋅⋅⋅⋅=bhRbbw ϕϕ N > Qmax = 62 kN,

that is the strength of inclined section is provided. Strength of inclined section to shear force is to be tested according to condition (50). We determine values Mb and qsw:

0.22 =bϕ (See Table 21);

As 39085475' =−=− bb f mm > 1505033 ' =⋅=fh mm so we take 150' =− bb f , and

62

( )5.0227.0

29285

5015075.075.0

0

''

<=⋅

⋅=

−=

bh

hbb ff

fϕ ;

( ) ( ) 622

02 1092.112928567.0227.0121 ⋅=⋅⋅+=+= bhRM btfbb ϕϕ N·mm = 11.92 kN·m;

143100

3.50285=

⋅=

s

ARq swsw

sw N/mm (kN/m)

We determine value min,bQ taking 6.03 =bϕ :

( ) 122402928567.0227.013min, =⋅⋅+= bbQ ϕ N = 12.24 kN.

As 21292.02

24.12

2 0

min,=

⋅=

h

Qb kN/m < qsw = 143 kN/m,

so condition (57) is met that means it is not necessary to correct value Mb. According to Item 3.32 we determine projection length of the most disadvantageous inclined section c:

9.122/189.32/1 =+=+= vgq kN/m (N/mm),

As 8014356.056.0 =⋅=swq kN/m > 9.121 =q kN/m so value c is to be determined only by the

following formula:

962.09.12

92.11

1

===q

Mc b m;

So 4.12962.0

92.11===

c

MQ b

b kN > 2.12min, =bQ kN;

3.49962.9.12621max =⋅−=−= cqQQ kN

Projection length of inclined crack is equal to:

288.0143

9.110 ===

sw

b

q

Mc m < 02h

As 292.0288.0 00 =<= hc m so we take 292.000 == hc m, so 8.41292.01430 =⋅== cqQ swsw

kN. Let’s check condition (50):

2.548.414.12 =+== swb QQ kN > 3.49=Q kN,

that is strength of inclined section to shear force is provided. Besides it is necessary to meet a requirement of Item 3.29:

5.1171062

2928567.05.13

2

max

2

04max =

⋅

⋅⋅⋅==

Q

bhRs btbϕ

mm > s = 100mm.

Conditions of Item 5.69 1752/3502/ ==< hs mm and s < 150 mm are also met. Example 14. Given: free supported reinforced concrete beam of the floor with the span l = 5.5 m; equivalent distributed live load on the beam v = 36 kN/m; dead load g = 14 kN/m;

dimensions of the cross section b = 200 mm, h = 400 mm, 3700 =h mm; heavy-weight

concrete B15 ( 7.7=bR MPa 67.0=btR MPa by 9.02 =bγ ); stirrups of reinforcement A-I

( 175=swR MPa).

It is required to determine diameter and spacing of stirrups at the support as well as to find out how it is possible to increase spacing of stirrups. Calculation. The largest shear force in the support section is equal to:

63

5.1372

5.550

2max =

⋅==

qlQ kN,

Where 501436 =+=+= gvq kN/m

We determine required density of stirrups of support part according to Item 3.33b.

By means of formula (52) by 0=fϕ and 0.22 =bϕ (see table 21) we get:

622

02 107.3637020067.02 ⋅=⋅⋅⋅== bhRM btbb ϕ N·mm = 36.7 kN·m.

According to Item 3.32

322/36142/1 =+=+= vgq kN/m (N/mm);

4.68327.3622 11 =⋅== qMQ bb kN

As 11406

4.68

6.0

1 ==bQ kN < Qmax = 137.5 kN, and 1674.68

37.0

7.361

0

max =+=+< b

b Qh

MQ kN,

Stirrups quantity is to be determined by formula (63):

( )130

7.36

4.685.137)(22

1max =−

=−

=b

b

swM

QQq kN/m (N/mm)

At the same time as ( )

44.933702

104.685.137

2

3

0

1max =⋅

−=

−

h

QQ b N/mm < 130 N/mm,

So 130=swq N/mm

According to Item 5.69 spacing s1 at the support must be no more than h/2 = 200 and 150 mm

and in the span – 3004

3=h and 500 mm. Maximum allowable spacing at the support

according to Item 3.29 is equal to:

200105.137

37020067.05.13

2

max

2

04max =

⋅

⋅⋅⋅==

Q

bhRs btbϕ

mm

We take stirrups spacing at the support 1501 =s mm and in the span – 3002 1 =s mm so:

111175

15013011 =

⋅==

sw

sw

swR

sqA mm2.

We take two stirrups with diameter 10 mm in the cross section (Asw = 157 mm2). So excepted density of stirrups at he support and in the span will be equal to:

2.183150

157175

1

1 =⋅

==s

ARq swsw

sw N/mm;

6.912.1835.05.0 12 =⋅== swsw qq N/mm

Let’s check condition (57) by means of calculation of min,bQ :

( ) 2975037020067.06.01 03min, =⋅⋅⋅=+= bhRQ btfbb ϕϕ N.

Then 2.403702

29750

2 0

min,=

⋅=

h

Qb N/mm < 2.1831 =swq N/mm;

2.402 0

min,=

h

Qb N/mm < 1.962 =swq N/mm

So values 1swq and 2swq are not to be corrected.

We determine the length of part l1 with stirrups quantity 1swq according to Item 3.34. As

6.91221 ==− swswsw qqq N/mm > 321 =q N/mm so value l1 is to be calculated according to

the following formula:

64

=−⋅−−⋅

=−−−

= 44832

4486.9129750105.137 3

01

1

012min,max

1 cq

cqQQl

swb

= 1637 mm > 375.14

5.5

4==

l m

(Here 4482.183

107.36 6

1

01 =⋅

==sw

b

q

Mc mm)

We take the length of the part with the stirrups spacing 1501 =s mm equal to 1.64 m.

Example 15. Given: reinforced concrete beam of the floor loaded by point forces as it’s shown on Draft 26a; dimensions of cross section – according to Draft 26b; heavy-weight concrete B15

( 67.0=btR MPa by 9.02 =bγ ); stirrups of reinforcement A-I ( 175=swR MPa).

It is required to determine diameter and spacing of stirrups as well as to find out how it is possible to increase spacing of stirrups. Draft 26. For example of calculation 15

Calculation. First we determine value Mb according to Item 3.31:

22 =bϕ (See Table 21);

1302/60100' =+=fh mm (see Draft 26b);

14080220' =−=− bb f mm < '3 fh ;

810808900 =−=h mm;

( )5.0211.0

81080

13014075.075.0

0

''

<=⋅

⋅=

−=

bh

hbb ff

fϕ ;

( ) ( ) 622

02 102.858108067.0211.0121 ⋅=⋅⋅+=+= bhRM btfbb ϕϕ N·mm = 85.2 kN·m.

Let’s determine required density of stirrups according to Item 3.33a, taking projection length of inclined section c equal to the distance from the support to the first weight – c1 = 1.35 m.

Shear force at the distance c1 from the support is equal to 2.1051 =Q kN (see Draft 26).

From formula (51) we have:

11.6335.1

2.85

1

1 ===c

MQ b

b kN > ( ) =+= 03min, 1 bhRQ btfbb ϕϕ

( ) 55.318108067.0211.016.0 =⋅⋅+= kN

Then æ1 667.011.63

11.632.105

1

11 =−

=−

=b

b

Q

QQ

As 35.11 =c m 62.181.022 0 =⋅=< h m so we take 35.110 == cc m;

æ01 417.081.02

35.1

11.63

55.31

2 0

0

1

min,=

⋅==

h

c

Q

Q

b

b

As æ01 = 0.417 < æ1 = 0.667 < c1/c0 = 1 so value )1(swq is to be determined by the following

formula:

18.3135.1

11.632.105

0

11)1( =

−=

−=

c

QQq b

sw kN/m

Let’s determine qsw by value c equal to the distance from the support to the second weight – c2 = 2.85 m.

9.2985.2

2.85

2

2 ===c

MQ b

b kN 55.31min, =< bQ kN

65

We take 55.31min,2 == bb QQ kN.

Corresponding shear force is equal to 1.582 =Q kN. As c2 = 2.85 m > 62.12 0 =h m we take

62.12 00 == hc m.

æ2 <=−

=−

= 842.055.31

55.311.58

2

22

b

b

Q

QQ æ02 1

2 0

0

2

min,==

h

c

Q

Q

b

b

Therefore value )2(swq is to be determined by formula (58):

93.172

1

62.1

1.58

1æ

æ

02

02

0

2)2( ==

+=

c

Qqsw kN/m

So value )2(swq is determined by formula (58):

( ) 93.172

1

62.1

1.58

1æ

æ

02

02

0

22 ==

+=

c

Qqsw kN/m

We take maximum value 18.31)1( == swsw qq kN/m.

According to welding conditions (see Item 5.13) we take diameter of stirrups 6 mm (Asw = 28.3 mm2) then spacing of stirrups in support part is:

15918.31

3.281751 =

⋅==

sw

swsw

q

ARs mm

We take 1501 =s mm. Spacing of stirrups in the span we take equal to 30015022 12 =⋅== ss .

The length of the part with spacing 1s is determined according to the condition of strength in

compliance with Item 3.34; at the same time

33150

3.28175

1

1 =⋅

==s

ARq swsw

sw N/mm;

5.165.0 12 == swsw qq N/mm;

5.16221 ==− swswsw qqq N/mm

Let’s take the length of the part with stirrups spacing 1s equal to the distance from the support

to the first weight l1 = 1.35 m; let’s check condition (50) by value c equal to the distance from the support to the second weight c = 2.85 m > l1. Value c01 is to be determined by formula (56)

by 331 =swq kN/m:

6.133

1.85

1

01 ===sw

b

q

Mc m < 62.12 0 =h m

As 5.135.185.21 =−=− lc m 6.101 =c m so value Qsw in condition (50) is to be taken equal to:

( )( ) 05.285.15.166.133121011 =⋅−⋅=−−−= lcqqcqQ swswswsw kN;

55.31min, == bb QQ kN;

6.5905.2855.31 =+=+ swb QQ kN > Q2 = 58.1 kN,

that is strength of this inclined section is provided. The larger value c is not taken into consideration as by this value shear force sharply reduces.

So the length of the part with stirrups spacing 1501 =s mm is to be taken equal to 35.11 =l m.

Example 16. Given: reinforced concrete beam of monolithic floor with cross section dimensions according to Draft 27a; location of bend-up bars – according to Draft 27b; equivalent live load on the beam is v = 96 kN/m, dead load is g = 45 kN/m; shear force at the

support Qmax = 380 kN/m; heavy-weight concrete B15 ( 67.0=btR MPa by 9.02 =bγ ); two-leg

stirrups with diameter 6 mm (Asw = 57 mm2) of reinforcement A-I ( 175=swR MPa), spacing

66

150=s mm; bend-up bars A-II ( 225=swR MPa), with the section area: of the first plane –

6281,, =incsA mm2 (2Ø20), of the second plane – 4022,, =incsA mm2 (2Ø16).

It is required to examine strength of inclined sections as regards shear force.

Calculation: 560406000 =−=h mm. According to Item 3.31 we find values Mb and qsw:

22 =bϕ (See Table 21);

30010033 '' =⋅==− ff hbb mm;

( )5.0134.0

560300

10030075.075.0

0

''

<=⋅

⋅=

−=

bh

hbb ff

fϕ ;

( ) ( ) 622

0 1014356030067.0134.01212 ⋅=⋅⋅+=+= bhRM btfb ϕ kN·m;

5.66150

57175=

⋅==

s

ARq swsw

sw N/mm

According to tem 3.32 we find 932/96452/1 =+=+= vgq kN/m.

According to condition (50) considering formula (65) let’s check inclined section with the projection length equal to the distance from the support to the end of the second plane of bend-up bars, that is by c = 50 + 520 + 300 = 870 mm = 0.87 m. Shear force at the distance 87.0=c m from the support is:

1.29987.0933801max =⋅−=−= cqQQ kN

Let’s determine projection of dangerous inclined crack c0 in compliance with Item 3.35. First we determine maximum value c0 by formula (56):

466.15.66/143/max,0 === swb qMc m 12.156.022 0 =⋅=> h m;

We take 12.1max,0 =c m. As 87.0=c m < 12.1max,0 =c m so we take 87.00 == cc for this

inclined section. Inclined crack located between the end of the second and the beginning of the first bending plane (that is not crossing bend-up bars) is not considered in the calculation as

30.00 =c m max,0c< .

For the first plane of bend-up bars: 3

,1, 109.99707.0225628sin1 ⋅=⋅⋅== θswincsincs RAQ N = 99.9 kN.

So 1.3229.9987.05.6687.0

1431,0 =+⋅+=++ incssw Qcq

c

M kN > Q =299.1 kN

that is strength of the present section is provided.

Let’s check inclined section that ends at the distance 12.10 =c m from the beginning of the

first plane of bend-up bars that is by 69.112.152.005.0 =++=c m. Shear force at the distance 69.1=c m from the support is 8.22269.193380 =⋅−=Q kN.

For the second plane of bend-up bars:

3

2,2, 109.63707.0225402sin ⋅=⋅⋅== θswincsincs RAQ N = 63.9 kN.

For this section we take inclined crack going from the end of inclined section to the beginning

of the first plane of bend-up bars that is =0c 12.1max,0 =c m. Inclined cracks going from the

end of inclined section to the support and to the beginning of the second plane of bend-up bars

are not considered as in the first case 69.10 == cc m 12.1max,0 => c m and in the second case

the crack crosses bend-up bars by max,00 cc < .

67

So 2239.6312.15.6669.1

1432,0 =+⋅+=++ incssw

b Qcqc

M kN > Q =222.8 kN,

that is strength of the present section is provided.

Let’s check inclined section which ends at the distance 12.1max,0 =c m from the beginning of

the second plane of bend-up bars that is by 51.212.152.030.052.005.0 =++++=c m. Shear force at the distance c = 2.51 m from the support is equal to Q = 380 – 93·2.51 = 146.6 kN.

For this section it is obvious that =0c 12.1max,0 =c m and the inclined crack doesn’t cross bend-

up bars, that is 0, =incsQ . As c = 2.51 m 87.156.06.0

20

3

2 ==> hb

b

ϕ

ϕm so we take

5.7687.1

143

0

3

2min, ====

h

MQQ

b

b

b

bb

ϕ

ϕkN.

So 151012.15.665.76,0 =+⋅+=++ incsswb QcqQ kN > Q = 146.6 kN that is strength of any

inclined sections is provided. In compliance with Item 3.29 let’s check the distance between the beginning of the first plane of bend-up bars and the end of the second plane taking shear force at the end of the second

plane of bend-up bars Q = 299.1 kN and 5.14 =bϕ :

1.316101.299

56030067.05.13

22

04 =⋅

⋅⋅⋅=

Q

bhRbtbϕmm < 300mm,

that is the requirement of Item 3.29 is met. Example 17. Given: a reinforced concrete corner beam with a span 8.8 m; continuous distributed load q = 46 kN/m (Draft 28a); dimensions of the cross section – according to Draft

28b; heavy-weight concrete B25 ( 95.0=btR MPa by 9.02 =bγ ); stirrups made of

reinforcement A-I ( 175=swR MPa) diameter 8 mm ( 3.50=swA mm2), spacing s = 150 mm.

It is required to examine strength of inclined sections as regards shear force. Draft 28. For calculation example 17 Calculation. The calculation is made according to Item 3.37.

Working height of support section is 5208060001 =−=h mm (Draft 28b).

Let’s determine values 1fϕ and 1bM by formulas (53) and (52) as for a support section:

2002/100150' =+=fh mm;

200100300' =−=− bb f mm < '3 fh ;

( )5.0577.0

520100

20020075.075.0

01

''

1 >=⋅

⋅=

−=

bh

hbb ff

fϕ ,

We take 5.01 =fϕ ; 22 =bϕ (see Table 21);

( ) ( ) 622

01121 1006.7752010095.05.0121 ⋅=⋅⋅+=+= bhRM btbbb ϕϕ N·mm = 77.06 kN·mm.

68

Let’s determine value qinc taking tgβ = 1/12:

32.112/10095.02 22

2 =⋅⋅== βϕ btgRq btbinc N/mm (kN/m)

As there is continuous load so we take 461 == qq kN/m.

Let’s check condition (66):

9.1032.17.585.27.5856.05.256.0 =⋅−⋅=− incswsw qqq kN/m 461 =< q kN/m

Condition (66) is not met so value c is to be determined by formula (68):

853.0467.5832.1

06.77

1

1 =++

=++

=qqq

Mc

swinc

b m,

At the same time 853.00 == cc m

Working height of the cross section 0h at the distance 853.0=c m from the support is:

591.012/853.052.0010 =+=+= βctghh m.

Let’s determine value bM by 5910 =h mm:

5.0508.0591100

20020075.0

)(75.0

0

''

>=⋅

⋅=

−=

bh

hbb ff

fϕ ;

We take 5.0=fϕ ;

( ) 622

02 1055.9959110095.05.012)1( ⋅=⋅⋅+=+= bhRM btfbb ϕϕ N·mm = 99.55 kN·m.

Let’s check condition (50) taking shear force at the end of inclined section equal to:

2.163853.0462

8.846

211max =⋅−

⋅=−=−= cq

qlcqQQ kN;

8.166853.07.58853.0

55.990 =⋅+=+=+ cq

c

MQQ sw

b

swb kN > Q = 163.2 kN,

that is strength of inclined sections as regards shear force is provided. Example 18. Given: a console with dimensions according to Draft 29; concentrated force F = 300 kN located at the distance 0.8 m from the support; heavy-weight concrete B15

( 67.0=btR MPa by 9.02 =bγ ); two-leg stirrups with diameter 8 mm ( 101=swA mm2) of

reinforcement A-I ( 175=swR MPa), spacing s = 200 mm.

It is required to check the strength of inclined sections as regards shear force. Draft 29. For the calculation example 18

Calculation. In compliance with Item 3.38 according to condition (50) let’s check the inclined section going from the point of application of concentrated force by value c

determined by formula (68). Working height at the point of application of concentrated force is equal to

30550950

800)300650(65001 =−−−=h mm (see Draft 29).

By formula (52) let’s determine value 1bM taking 22 =bϕ (see Table 21) and 0=fϕ :

( ) 622

0121 109.4930540067.0121 ⋅=⋅⋅⋅⋅=+= bhRM btfbb ϕϕ N·mm

69

Value swq is: 4.88200

101175=

⋅==

s

ARq swsw

sw N/mm (kN/m).

Taking tgβ = 369.0950

300650=

− (see Draft 29) we determine qinc:

73369.040067.02 22

2 =⋅⋅⋅== βϕ btgRq btbinc N/mm,

therefore taking q1 = 0 we have

5564.8873

109.49 61 =

+

⋅=

+=

swinc

b

qq

Mc kN

As value c does not exceed the distance from the load to the support so we take c = 556 mm and determine working height h0 at the end of inclined section:

510369.0556305010 =⋅+=+= βctghh mm

As 51022 0 ⋅=h mm 5580 => c mm so we take 5560 =c mm.

Value Mb is 622

02 104.13951040067.02 ⋅=⋅⋅⋅== bhRM btbb ϕ N·mm = 139.4 kN·m

Therefore 9.299556.04.88556.0

4.1390 =⋅+=+=+ cq

c

MQQ sw

b

swb kN ≈ Q = 300 kN,

that is strength of the present section is provided. For inclined section located from the load to the support we determine value c0 by formula (56)

taking 600506500 =−=h mm: 62 1019360040067.02 ⋅=⋅⋅⋅=bM N·mm;

14784.88

10193 6

0 =⋅

==sw

b

q

Mc mm 120060022 0 =⋅=> h mm,

We take 12002 00 == hc mm.

As 12000 =c mm 800=> c mm so it is possible not to examine the mentioned above inclined

section. So section of any inclined section is provided. Example 19. Given: continuous floor slab without cross reinforcement 3x6 m, h = 160 mm thick, monolithically connected with beams along the perimeter; equivalent distributed live load v = 50 kN/m2; load of dead weight and floor 9=g kN/m2; a = 20 mm; heavy-weight

concrete B25 ( 95.0=btR MPa by 9.02 =bγ ).

It is required to examine the slab strength as regards shear force.

Calculation: 140201600 =−=−= ahh mm. The calculation is made for the strip b = 1m =

= 1000 mm with a span l = 3m; total load on the slab is 59950 =+=+= gvq kN/m.

Shear force on the support is

5.882

359

2max =

⋅==

qlQ kN

Let’s check condition (71): 3

0 10333140100095.05.25.2 ⋅=⋅⋅⋅=bhRbt N 5.88max => Q kN

Let’s check condition (72). As lateral edges of the slab are connected with beams so value cmax is to be determined considering coefficient 25.116.0/605.01/05.01 >⋅+=+= hbα (here b = 6 m – distance between lateral edges of the slab) that is 25.1=α :

2801402225.1

5.25.2000max =⋅==== hhhc

αmm

According to Item 3.32 we have :

70

342/5092/1 =+=+= vgq kN/m = 34 N/mm;

5.14 =bϕ (See Table 21)

As ( )

3562

100095.05.1

/22

0max

4 =⋅⋅

=hc

bRbtbϕN/mm > q1 = 34 N/mm so we take 280max == cc

mm 28.0= m.

Shear force at the end of inclined section is 7928.0345.881max =⋅−=−= cqQQ kN.

322

04 1075.99280

140100095.05.1⋅=

⋅⋅⋅=

c

bhRbtbϕN = 99.75 kN > Q = 79 kN,

that is the strength of the slab as regards shear force is provided. Example 20. Given: a panel of a tank with variable thickness from 262 (point of embedment) to 120 mm (at free supported end), overhang length is 4.25 m; lateral earth pressure is

considering loads of vehicle on the ground surface decreases linearly from 690 =q kN/m2 at

the point of embedment to q = 7 kN/m2 at the free supported end; a = 22 mm; heavy-weight

concrete B15 ( 82.0=btR MPa by 1.12 =bγ ).

It is required to examine the strength of the panel as regards shear force.

Calculation. Working height of the panel at the point of embedment is 2402226201 =−=h

mm. Let’s determine tgβ (β is the angle between the tensile and the compressed surfaces):

0334.04250

120262=

−=βtg .

The calculation is made for the strip of the panel b = 1 m = 1000 mm wide. Let’s check conditions of Item 3.40. Lateral force at the point of embedment is:

5.16125.42

769max =

+=Q kN

Let’s check condition (71) taking 240010 == hh mm:

492240100082.05.25.2 0 =⋅⋅⋅=bhRbt kN > Qmax = 117 kN,

that is the condition is met. As panels are connected with each other and the width of the tank side is more than 5h so we determine value cmax considering the coefficient 25.1=α :

4640334.025.125.1

2405.2

25.1

5.2 01max =

⋅+

⋅=

+=

βα tg

hc mm

Average load density at the support part 464max =c mm long is

( ) 6.6524250

464769691 =

⋅−−=q N/mm. From table 21 5.14 =bϕ

As ( ) ( )

1037100082.05.1/4.654/0334.0

1240

/4/

12

41

201 =⋅⋅+

=+

=bRqtg

hcbtbϕβ

mm

464max => c mm so we take 464max == cc mm.

Let’s determine working height of the section at the distance 2

c from the support (that is

average value 0h within the length c):

2320334.02

464240

2010 =−=−= βtg

chh mm

Shear force at the distance c = 464 mm from the support is:

71

1.131464.06.655.1611max =⋅−=−= cqQQ kN

Let’s check condition (72):

322

04 107.142464

232100082.05.1⋅=

⋅⋅⋅=

c

bhRbtbϕN > Q = 131.1 kN,

that is the strength of the panel as regards lateral force is provided. Calculation of inclined sections as regards bending moment

Example 21. Given: a free supported reinforced concrete beam with a span l = 5.5 m with the distributed load q = 29 kN/m; the structure of the support part of the beams is taken according

to Draft 30; heavy-weight concrete of class B15 ( 7.7=bR MPa; 67.0=btR MPa by

9.02 =bγ ); longitudinal reinforcement A-III ( 365=sR MPa) without anchors, section area is

982=sA mm2 (2Ø25) and 226' =sA mm2 (2Ø12); stirrups of reinforcement A-I ( 175=swR

MPa) with diameter 6 mm, spacing s = 150 mm are welded to longitudinal rods. It is required to test the strength of inclined sections as regards bending moment. Draft 30. For the calculation of example 21

Calculation: 360404000 =−=−= ahh mm. As tensile reinforcement has no anchors so

the calculation of inclined sections as regards the moment is necessary. Let’s take the beginning of inclined section at the surface of the support. Therefore

10sup −= llx mm = 280 – 10 = 270 mm (see Draft 30).

By formula (81) we determine the anchorage length anl taking 5.0=anω and 8=∆ anλ :

7932587.7

3655.0 =

+=

∆+= d

R

Rl an

b

s

anan λω mm

As anx ll < so design resistance of tensile reinforcement is decreased by means of its

multiplying by the coefficient 340.0793

2705 ===

an

x

sl

lγ , therefore 1.124340.0365 =⋅=sR MPa.

As within the length lx four vertical and two horizontal cross rods are welded to tensile rods

(see Draft 30) so we increase force 3109.1219821.124 ⋅=⋅=ss AR N by value wN .

Taking 6=wd mm, 6=wn , 200=wϕ (see Table 22) we get 322 1026.2067.0620067.07.0 ⋅=⋅⋅⋅⋅== btwwww RdnN ϕ N.

Therefore 2.14226.209.121 =+=ss AR kN

As this value does not exceed value RsAs determined without considering 5sγ and wN that is

it’s equal to 310358982365 ⋅=⋅ N so we take 2.142=ss AR kN.

The height of compressed zone is determined by formula (16):

392007.7

226365102.142 3'

=⋅

⋅−⋅=

−=

bR

ARARx

b

sscss mm 352'2 ⋅=< a mm

In compliance with Item 3.42 we take 32535360'0 =−=−= ahz s mm.

Let’s determine value qsw by formula (55):

4.68150

57175=

⋅==

s

ARq swsw

sw N/mm

72

Let’s determine the projection length of the most disadvantageous section by formula (83)

taking value Q equal to support reaction of the beam, that is 802

5.529

2=

⋅==

qlQ kN as well

as Fi = 0 and As,inc = 0:

821294.68

1080 3

=+

⋅=

+=

qq

Qc

sw

mm

Let’s determine maximum length ls of the support part behind which condition (72) is met with

multiplying of the right part by 0.8 and by 0max1 28.0 hccc =≤= ; that is according to the

following equation:

1

2

04max /8.0 cbhRqlQQ btbs ϕ=== .

Supposing that 02hls > we take maximum value 01 2hc =

Then by 5.14 =bϕ we get:

176029

3602

36020067.05.18.01080

2

8.0 23

0

2

04max

=⋅

⋅⋅⋅⋅−⋅

=

−

=q

h

bhRQ

l

btb

s

ϕ

mm >

72036022 0 =⋅=> h mm

As 1760=sl mm > c =821 mm so we take c = 821 mm.

Moment of external forces relating to the axis located in the middle of the height of inclined section here is equal to the bending moment in the normal section going through the mentioned

axis; that is at the distance 9148213/2803/sup1 =+=+=+ clcl mm from the point of

application of support reaction:

( )( )

612

914.029914.080

2

22

11 =

⋅−⋅=

+−+=

clqclQM kN·m

Let’s check the strength according to condition (77) considering formula (78):

4.6905.236.468214.685.0325102.1425.0 232 =+=⋅⋅+⋅⋅=+ cqzAR swsss kN·m >

> M = 61 kN·m, that is the strength of inclined sections as regards bending moment is provided. As the beam has no bend-up bars and is loaded by distributed load so the strength of inclined

section can be also checked according to the more simple formula (84) taking ×== 8010 QlM 63 104.79310 ⋅=⋅× N·mm:

( )( ) ( )( ) 366

0 101.87294.68104.7103.4622 ⋅=+⋅−⋅=+− qqMzAR swsss N = 87 kN >

> Q = 80 kN.

Example 22. Given: a collar beam of a multi-storey frame with diagrams of bending moments and shear forces of distributed load q = 228 kN/m according to Draft 31; heavy-weight

concrete B25; cross reinforcement and longitudinal reinforcement A-III ( 365=sR MPa;

290=swR MPa); cross section of the support part – according to Draft 31; stirrups with

diameter 10 mm, spacing s = 150 mm ( 236=swA mm2).

It is required to determine the distance from the left support to the break point of the first rod of top reinforcement. Draft 31. For the calculation example 32

73

Calculation. Let’s determine limit bending moment which stretches support reinforcement

without considering a broken rod according to the condition (19) as 1609=sA mm2 < '

sA , that

is x < 0:

( ) 405507401609365)'( 0 =−⋅=−= ahARM ssu kN·m

According to the moment diagram we determine distance x from the support to the point of theoretical break of the first rod in compliance with the following equation:

uMxq

xql

xl

MMMM =+−

−−= 2

'supsup

sup22

,

So ( )

−

⋅

−+=

−−

−+−

−+=

9.4228

300600

2

9.42

22

sup

2'supsup

'supsup

q

MM

ql

MMl

ql

MMlx

u

( )334.0

228

4056002

2.4228

300600

2

9.42

=−

−

⋅

−+− m.

Shear force at the o\point of theoretical break is:

554334.0228620max =⋅−=−= qxQQ kN

Let’s determine value qsw :

456150

236290=

⋅==

s

ARq swsw

sw N/mm

By formula (87) let’s determine length w:

7563254562

105445

2

3

=⋅+⋅

⋅=+= d

q

Qw

sw

mm

So the distance from the support to the point of rod break can be taken equal to 1090756334 =+=+ wx mm.

Let’s determine required distance lan from the point of rod break to the vertical section where it is fully used, according to Table 45:

930322929 =⋅== dlan mm < 1090 mm

That is the rod is to be broken at the distance 1090 mm from the support. Example 23. Given: adjoining of a prefabricated reinforced concrete floor beam to the collar

beam by means of cutting as it’s shown on Draft 32a; heavy-weight concrete B25 ( 13=bR

MPa; 95.0=btR by 9.02 =bγ ); stirrups and bend-up bars of reinforcement A-III with diameter

12 and 16 mm ( 452=swA mm2; 804, =incsA mm2); section area of additional stirrups of

cuttings 4021 =swA mm2 (2Ø16); longitudinal reinforcement A-III according to Draft 32b;

shear force at the support Q = 640 kN. It is required to examine the strength of inclined sections. Draft 32. To the calculation example 23 Calculation. Let’s determine the strength of inclined section of the cutting as regards shear

force according to Item 3.31 taking 3700 =h mm, b = 730 mm (see Draft 32), 22 =bϕ (see

Table 21): 622

02 1019037073095.02 ⋅=⋅⋅⋅== bhRM btbb ϕ N·mm

By value c equal to the distance from the support to the first load – c = 1.5 m we have

74

36

107.1261500

10190⋅=

⋅==

c

MQ b

b N < =min,bQ

3

03 1015437073095.06.0 ⋅=⋅⋅⋅== bhRbtbϕ N

( 6.03 =bϕ – see Table 21),

So we take 310154 ⋅=bQ N;

1152100

452255=

⋅==

s

ARq swsw

sw T/mm

4061152

10190 6

0 =⋅

==sw

b

q

Mc mm 02h<

At the same time 5.10 =< cc m and 00 hc > .

Thereafter 33

10 10738402290406115210154 ⋅=⋅+⋅+⋅=++ swswswb ARcqQ N > Q = 640 kN,

that is even without considering bend-up bars the strength of cutting as regards the shear force is provided. Let’s check if there are enough additional stirrups and bend-up bars according to condition

(89). According to Draft 32 45=θ degrees; 6002/80607000 =−−=h mm; 37001 =h mm;

3

,1 10281707.080429040229045sin ⋅=⋅⋅+⋅=°+ incsswswsw ARAR N ×=

−> 4601

0

01

h

h

245600

3701 =

−× kN

Let’s check the strength of the inclined section going through the reentrant angle of cutting as regards bending moment. The most disadvantageous value c is determined by formula (83) considering bend-up bars and

additional stirrups in the numerator and taking 0=iF and 0=q :

( )312

1152

1028110640sin 33,1

=⋅−⋅

=+−

=sw

incsswswsw

q

ARARQc

θ mm

As longitudinal reinforcement of the short console is anchored in the support so we consider

this reinforcement with total design strength; that is with 365=sR MPa.

According to Draft 32 we have 1256' == ss AA mm2 (4Ø20). As 0,' == xAA ss so

−=−= 370'01 ahz s

32050 =− mm

According to (79) taking 301 =a mm we get

( ) ( ) 425707.030312707.0320sincos 1, =−+⋅=−+= θθ aczz sincs mm

Let’s check condition (77) taking:

( ) ( ) 62

11

2 108.883031240229031211525.05.0 ⋅=−⋅+⋅⋅=−+=Σ acARcqzAR swswswswswsw N·mm;

( ) ( ) 63

0 1028331213010640 ⋅=+⋅=+= caQM N·mm;

=⋅⋅+⋅+⋅⋅=Σ+Σ+ 425804290108.883201256365 6

,, incsincsswswswswsss zARzARzAR

6106.334 ⋅ N·mm > M = 283 kN·m, that is the strength of the inclined section is provided. Let’s determine required length of longitudinal tensile reinforcement going behind the cutting end by the following formula:

( )=++

−−= da

q

ARARQw

sw

incsswswsw10

sin20

,11

0

θ

75

( )9532010130

1152

10281106402 33

=⋅++⋅−⋅

= mm 6002030 =⋅=> anl mm

Let’s find out if it is necessary to install anchors for bottom reinforcement of the beam. For that let’s check inclined section located beyond the cutting and beginning at the distance

230370600010 =−=− hh mm from the end of the beam. Then 22010230 =−=xl mm

The length of anchorage for bottom reinforcement is determined according to pos. 1 of Table

45, where for concrete B25 and reinforcement A-III there is 29=anλ thereafter

11604029 =⋅=anl mm 220=> xl .

Design strength of bottom reinforcement is to be decreased by means of multiplying by the

coefficient 19.01160

2205 ===

an

x

sl

lγ ; that is 2.6919.0365 =⋅=sR MPa.

According to Draft 32 5027=sA mm2 (4Ø40)

Taking the fact into account that within the length 220=xl mm two top rods have two welded

vertical rods and two bottom rods have two vertical and one horizontal welded rod, we increase

the force ARs by value wN determined by formula (82) taking 10=wn , 12=wd mm,

100=wϕ (see Table 22):

9576095.012100107.07.0 22 =⋅⋅⋅⋅== btwwww RdnN ϕ N =⋅⋅⋅=< 10123658.08.0 22

wws ndR

420000= N

Thereafter 4436009576050272.69 =+⋅=ss AR N 31018355027365 ⋅=⋅< N

Taking 730' == fbb mm we determine the height of the compressed zone x:

=⋅

⋅⋅+⋅−=

+−=

73013

707.08042901256365443600cos,

'

bR

ARARARx

b

incsswsscss θ

8.15= mm 100502'2 =⋅=< a mm

And so 55050600'0 =−=−= ahz s mm

The most disadvantageous value c is equal to:

5551152

640000===

swq

Qc mm ( ) 7232309530100 =−=−−< hhw mm

that is by such value c the inclined section crosses longitudinal reinforcement of the short console. We take the end of inclined section at the end of the mentioned above reinforcement

that is at the distance 9530 =w mm from the cutting; at the same time c = 723 mm. Design

moment M in the section going through the end of inclined section is:

( ) ( ) 693953.013.046000 =+=+= waQM kN·m;

( ) ( ) 851707.070723707.0550sincos 1, =−+⋅=−+= θθ aczz sincs mm

[Where 702303001 =−=a mm (see Draft 32)]

Let’s check condition (77):

=⋅⋅+⋅

+⋅=++ 8518042902

7231152550443600

2

2

,,

2

incsincssw

sw

sss zARcq

zAR

6105.743 ⋅= N·mm 693=> M kN·m, that is the strength of the inclined section is provided and anchors for bottom reinforcement are not required. Let’s check the strength of the short console of the cutting according to Items 3.99 and 3.100 considering Item 3.31.

Let’s check condition (207) taking 130sup =l mm, 90=xa mm, 32050370'01 =−=− ah mm

(see Draft 32) thereafter

76

( )( ) ( ) ( )

679.090130320

320

'

'sin

22

2

2

sup

2

01

2

012 =++

=++−

−=

xalah

ahθ .

Taking 0=wµ and replacing 0.8 by 1.0 we have =⋅⋅⋅= 679.013073013sin 2

sup θblRb

310838 ⋅= N 3

01 1089837073095.05.35.3 ⋅=⋅⋅⋅=< bhRbt N; that is the right part of condition

(207) is equal to 838 kN. As Q = 640 kN < 838 kN so the strength of the compressed zone is provided.

Let’s check condition (208) taking 220sup1 =+= xall mm, 3200 =h mm, 1256=sA mm2

(4Ø20):

33

0

1 10440320

22010640 ⋅=⋅=

h

lQ N 3104581256365 ⋅=⋅=< ss AR N,

that is there is enough longitudinal reinforcement in the short console. Eccentric Compressed Members

GENERAL POSITIONS

3.50. (1.21) During calculation of eccentric pressed reinforced concrete members it is

necessary to consider accidental eccentricity ae resulting from not considered in the

calculation factors. Anyway eccentricity ae is taken no less than:

- 1/600 of the member length or of the distance between its fixed sections; - 1/30 of the section height; - 10 mm (for structures formed of prefabricated members if there are no any other

experiment justified values ae ).

For members of statically undeterminable structures (including columns of frame work buildings) longitudinal force eccentricity value relating to the center of gravity of the

given section 0e is taken equal to the eccentricity calculated according to static

calculation of the structure, but no less than ae .

In members of statically determinable structures (for example formwork poles, electric

power line supports) eccentricity 0e is calculated as a sum of eccentricities – determined

according to static calculation and accidental one.

3.51.Calculation of eccentric compressed members is to be made considering the deflection influence in the longitudinal force eccentricity plane (in the bending plane) and in the normal to it plane. In the last case it is assumed that longitudinal force is applied with the

eccentricity 0e equal to the accidental eccentricity ae (see Item 3.50).

The deflection influence is considered according to Items 3.54 and 3.55. It is possible not to make calculation as regards the bending plane if the member

elasticity il /0 (for rectangular section – hl /0 ) in the bending plane is more than the

elasticity in the plane normal to the bending plane.

77

If there are design eccentricities in two directions exceeding accidental eccentricities ae

so it is necessary to make calculation as regards the skew eccentric compression (see Items 3.73 – 3.75).

3.52. For frequent types of compressed members (rectangular section; double-T section with

symmetrical reinforcement; round and ring section with reinforcement distributed along the perimeter) the normal sections strength calculation is made according to Items 3.61-3.75). For other kinds of sections and by unspecified location of longitudinal reinforcement the calculation of normal sections is made by formulas of the general case of the normal section calculation of eccentric compressed member according to Item 3.76. During calculation by means of computers it is recommended to follow the instructions of Item 3.76.

If condition bs AA 02.0' > is met, so it is necessary to consider decrease of actual concrete

area by value '

sA in formulas of Items 3.61-3.76.

3.53.Strength of inclined sections of eccentric compressed members is calculated similar to

calculation of bending moments in compliance with Items 3.28-3.49. At the same time

value bM is determined by the following formula:

( ) 2

02 1 bhRM btnfbb ϕϕϕ ++= (90)

Where 0

1.0bhR

N

bt

n =ϕ but no more than 0.5; value min,bQ is taken equal to

( ) 03 1 bhRbtnfb ϕϕϕ ++ and in formulas (72)–(76) coefficient 4bϕ is replaced by

( )nb ϕϕ +14

Total coefficient nf ϕϕ ++1 is taken no more than 1.5.

Longitudinal forces influence is not considered if they cause bending moments which have the same signs like moments caused by lateral load. For eccentric compressed members of statically undeterminable structures for which longitudinal force is located in the center of gravity of the section it is possible always to consider longitudinal forces influence. If there is no lateral load within the span of eccentric compressed member so it is possible not to calculate the strength of inclined sections if there are no normal cracks [that is if

condition (233) is met with replacing of serbtR , by btR ].

Member deflection influence

3.54.(3.24, 3.6) During calculation of eccentric compressed members it is necessary to consider the deflection influence on the bearing capacity as a rule by means of calculation of the structure according to deformed scheme considering non-elastic concrete and reinforcement deformations as well as crack formation.

78

It is possible to make calculation of the structure according to non-deformed scheme considering the member deflection influence by means of multiplying of the eccentricity

0e by the coefficient η determined by the following formula:

crN

N−

=

1

1η (91)

Where crN – relative critical force determined by the following formulas:

for members of any section:

+

+

+= s

el

b

cr II

l

EN α

δϕ1.0

1.0

11.04.620

(92)

for members of rectangular section:

( )

−+

++

=

2

0

10

'

3

1.01.0

11.0

/

6.1

h

ah

hl

bhEN eb

cr µαϕ

δ (93)

In formulas (92) and (93):

sII , – Inertia moments of concrete section and section of all reinforcement relating

to the center of gravity of concrete section;

lϕ – Coefficient considering long-term action of the load on the member deflection

in the limit state and equal to:

1

11M

M l

l βϕ += , (94)

But no more than β+1 (here β – see Table 16);

lMM 11 , – are moments of external forces relating to the axis parallel to the line bounding

the compressed zone and going through the center of the most stretched or the least compressed reinforcement rod (by the whole compressed zone) and caused by the total load and load and by dead loads and long-term loads. For members calculated according to Items 3.61, 3.62, 3.65–3.68 it is possible to

determine 1M and lM 1 relating to the axis going through the center of gravity

of all reinforcement S. If bending moments (or eccentricities) caused by total loads or by the sum of dead loads and long-term loads have different sighs so

by absolute value of the total load eccentricity he 1.00 > it is taken 0.1=lϕ ; if

this condition is met so value lϕ is to be taken equal to ( )×−+= 11 110 lll ϕϕϕ

he /0× where 1lϕ is determined by formula (94), taking 1M equal to the

product of the longitudinal force N caused by the total load by the distance from the center of gravity of the section to the axis going through the center of the most stretched (the least compressed) by dead loads and by long-term loads reinforcement rod;

eδ – Coefficient taken equal to he /0 but no less than

be Rh

l01.001.05.0 0

min, −−=δ (95)

(Here bR is given in mega-Pascal, it is possible to take 0.12 =bγ ; for round and ring

sections value h is replaced by D);

0l – is taken according to Item 3.55;

79

b

sss

E

E

bh

AA'+

=µα (96)

During calculation of rectangular sections with reinforcement located along the height of

the section according to Item 3.63 in value '

ss AA + it is not considered 2/3 of

reinforcement located at the surfaces parallel to the bending plane ( slA2 ), and value

h

ah '0 − in formula (93) is taken equal to 121 δ− .

For members made of fine concrete of group Б it is necessary to insert numbers 5, 6 and 1, 4 instead of 6, 4 and 1, 6 in formulas (92) and (93).

Eccentricity 0e used in the present Item can be determined relating to the center of

gravity of concrete section.

By elasticity of the element 14/0 <il (for rectangular sections 4/0 <hl ) it is

taken 1=η .

By of the element 35/14 0 <≤ il ( 10/4 0 <≤ hl ) and by 025.0'

≤+

=A

AA ssµ it is

possible to take: For rectangular sections:

( )2

0 /15.0

hl

AEN b

cr = ,

For other sections:

2

0

2

l

IEN b

cr = .

By crNN > it is necessary to increase the section dimensions.

By design eccentricities in two directions coefficient η can be determined for each direction and multiplied by the corresponding eccentricity.

3.55.(3.25) It is recommended to determine design length 0l of eccentric compressed

reinforced concrete members as for frame structure members considering its deformed state by the most disadvantageous for the present element load distribution considering inelastic deformations of materials and cracks.

For members of frequent structures it is possible to take 0l equal to:

a) for columns of multistory buildings by no less than two spans and connections of

collar-beams to columns calculated as fixed connections by:

- Prefabricated floor structures…….. Hl =0

- Cast-in-situ floor structures ……… Hl 7.00 =

[where H is the height of the storey (the distance between the centers of joints)];

b) for columns of one-storey buildings with hinge connection of bearing structures of the roof hard in their plane (able to transfer horizontal forces) as well as for trestles – according to Table 23;

80

c) for trusses and arches elements – according to table 24. Table 23 (32) Design length 0l for columns of one-storey

buildings if they are calculated in

the plane perpendicular to the cross frame or parallel

to the trestle axis

if there are if there are no

Buildings and columns characteristics

the plane of the cross frame or

plane perpendicular to the trestle

axis bracings in the plane of

longitudinal row of columns or anchor

supports

Semi-infinite

1.5 H1 0.8 H1 1.2 H1 Bottom part of columns by crane runaway beams

simply supported

1.2 H1 0.8 H1 0.8 H1

Semi-infinite beam

2.0 H2 1.5 H2 2.0 H2

Considering the load of canes

Top part of columns by crane runaway beams

simply supported beam

2.0 H2 1.5 H2 1.5 H2

one-span beam

1.5 H 0.8 H1 1.2 H Bottom part of building beams multi-span

beam 1.2 H 0.8 H1 1.2 H

Semi-infinite beam

2.5 H2 1.5 H2 2.0 H2

With crane bridges

Without considering the load of cranes

Top part of columns by crane runaway beams

simply supported beam

2.0 H2 1.5 H2 1.5 H2

one-span beam

1.5 H 0.8 H 1.2 H Bottom part of building beams multi-span

beam 1.2 H 0.8 H 1.2 H

Tapered columns

Top part of columns 2.5 H2 2.0 H2 2.5 H2

one-span beam

1.5 H 0.8 H 1.2 H

Buildings

Without crane

bridges Columns of constant section of buildings

multi-span beam

1.2 H 0.8 H 1.2 H

Semi-infinite

2.0 H1 0.8 H1 1.5 H1 Trestles Crane trestle by crane runaway beams simply

supported 1.5 H1 0.8 H1 H1

81

Hinge connection

2.0 H H 2.0 H Trestle for pipelines By connection of columns to a span structure

Fixed connection

1.5 H 0.7 H 1.5 H

Symbols of Table 23

H – total height of the column from the top of the foundation to the horizontal structure in the corresponding plane; H1 – the height of the bottom part of the column from the top of the foundation to the bottom of the crane runaway

beam; H2 – the height of the top part of the column from the step of the column to a horizontal structure of the corresponding

plane. Note. If there are bracings up to the top of columns in buildings with bridge cranes so design length of the bottom part

of columns in the plane of the axis of longitudinal column row is taken H2.

Table 24 (33)

Elements Design length 0l of trusses and arches members

1. Trusses elements: a) top chord of truss during calculation: - in the truss plane:

by 10 8/1 he <

by 10 8/1 he ≥

- out of the truss plane for the part under the skylight (if the width of the skylight is 12 m and more); in other cases

b) inclined braces and poles during calculation:

- in the truss plane - out of the truss plane

by 5.1/ 21 <bb

by 5.1/ 21 ≥bb

0.9l 0.8l

0.8l

0.9l

0.8l

0.9l 0.8l

2. Arches a) during calculation in the arch plane: - triple-hinged - double-hinged - hingeless b) during calculation out of the arch plane

(any)

0.580L 0.540L 0.365L

L

Symbols of Table 24: l – the length of the element between centers of adjoining connections; for top chord of the truss during calculation out

of the arch plane this is the distance between its fixing points; L – the length of the arch along its geometrical axis; during calculation out of the arch plane this is the length of the

arch between its fixing points according to the arch plane; h1 – section height of the top chord; b1, b2 – the width of the section of the top chord and of the pole (inclined brace) of the truss.

3.56.Columns deflection influence of multistory framework buildings are to be considered taking moments M in support sections of columns equal to:

thhvv MMMM ++= ηη , (97)

Where vM – moment caused by vertical loads on the floors;

82

vη – Coefficient equal to one and in embedment into foundation it is determined by

formula (91) by ??.00 =l (H is the height of the storey) and by considering only

vertical loads;

hM – Moment caused by horizontal loads (wind loads and seismic loads);

hη – is coefficient η determined according to Items 3.54 and 3.55 considering all

loads;

tM – Moments caused by forced horizontal displacements (for example temperature

deformation of floors, displacement of hard bracing diaphragms).

Moments caused by all loads for sections in the middle third of the column length are multiplied by the coefficient determined according to Items 3.54 and 3.55 and moments in other sections are determine by linear interpolation. Values of moments in support columns sections determined by formula (97) must be considered by determination of moments in adjoining to the column elements (foundations, collar-beams with fixed connections).

Confinement reinforcement influence

3.57.(3.22) Calculation of solid-section elements of heavy-weight and fine concrete with confinement reinforcement in the shape of welded meshes, spiral or ring reinforcement (Draft 33) must be determined according to Items 3.61-3.68, 3.71-3.76 inserting into the

calculation only a part of concrete section efA bounded by axes of end rods of a mesh or

of a spiral and replacing bR by changed strength of concrete redbR , and calculating the

compressed concrete zone characteristics ω considering the influence of confinement by formula (104).

Draft 33. Compressed elements with confinement reinforcement.

a – in the shape of welded meshes; b – in the shape of spiral reinforcement.

The deflection influence of the element with confinement reinforcement on the eccentricity of longitudinal force is considered according to Item 3.58.

Elasticity efil /0 of elements with confinement reinforcement must be no more than:

- 55 – by confinement reinforcement by means of meshes (for rectangular sections

16/0 ≤efhl );

- 35 – By confinement reinforcement by means of spirals (for round

sections 9/0 ≤efdl ) where efi , efh , efd are correspondingly inertia radius, height and

diameter of the section part inserted into the calculation.

Value redbR , is determined by the following formulas:

a) by reinforcement by welded cross meshes

xysxybredb RRR ,, ϕµ+= , (98)

Where xysR , – design strength of reinforcement meshes

sA

lAnlAn

ef

ysyyxsxx

xy

+=µ (99)

Here xsxx lAn ,, are: quantity of rods, cross section area and length of a mesh rod

(calculating in axes of end rods) in one direction;

83

ysyy lAn ,, – The same in another direction;

efA is section area of reinforcement within the meshes contours;

s is distance between meshes; φ is efficiency factor of confinement reinforcement determined by the

following formula:

ψϕ

+=

23.0

1 (100)

10

,

+=

b

xysxy

R

Rµψ (101)

bxys RR ,, Are given in MPa

For members made of fine concrete efficiency factor φ must be taken no more than 1.

b) By reinforcement of spiral or ring reinforcement

−+=

ef

cirscirbredbd

eRRR 0

,,

5.712µ (102)

Where cirsR , – design strength of the spiral;

cirµ –reinforcement factor equal to:

sd

A

ef

cirs

cir

,4=µ (103)

Here cirsA , – cross-section area of spiral reinforcement;

efd – Section diameter inside of the spiral;

s – Is spiral spacing

0e – Eccentricity of longitudinal force application (without considering

deflection influence). Reinforcement coefficients values which are determined by formulas (99) and (103) for members of fine concrete must be no more than 0.04. During determination of limit value of relative height of compressed zone for sections with confinement reinforcement it is necessary to insert the following value into formula (14):

9.0008.0 2 ≤+−= δαω bR (104)

Where α – coefficient taken according to Item 3.14 instructions;

2δ – Coefficient equal to µ10 but taken no more than 0.15 [here µ is

reinforcement coefficient xyµ or cirµ determined by formulas (99) and (103)

correspondingly for meshes and spirals]. Confinement reinforcement is considered in the calculation on conditions that bearing capacity of the element determined according to the instructions of the present Item

(using efA and redbR , ) is more than its bearing capacity determined according to its total

section A and to the value of concrete design strength bR without considering

confinement reinforcement. Besides confinement reinforcement must meet the constructive requirements of Items 5.78-5.80.

3.58.(3.22).During calculation of elements with confinement reinforcement according to undeformed scheme member deflection influence on the longitudinal force eccentricity is

84

considered according to Items 3.54-3.56. At the same time value crN determined by

formula (92) or (93) is to be multiplied by coefficient 0.1/05.025.0 01 ≤+= efclϕ and

value min,eδ is determined by formula ( )befefe Rclcl 01.0/1.00.1/01.05.0 00min, −−+=δ

where efc is the height or diameter of considered part of the section.

Besides during determination of crN dimensions of the section are taken according to the

considered part of the section.

3.59.(3.22). In elements made of heavy-weight concrete with confinement reinforcement at the shape of meshes it is recommended to use longitudinal high-strength reinforcement A-V and A-VI using its increased strength equal to:

sscredsc RRR ≤+

+=

23

13,

1

1

λδ

λδ (105)

Where 21 ,λλ – see Table 25;

ssc RR ,

θψδ 6.13 = ;

Here

−+=

1001258.0

, b

ef

tots R

A

Aθ

(106) but no more than 1.0 and no more than 1.6;

efA,ψ – see Item 3.57;

totsA , – Section area of all longitudinal high-strength reinforcement;

bR – In MPa.

Table 25

21 ,λλ and scR in MPa by coefficient 2bγ (see Item 3.1) equal to

0.9 1.0 or 1.1

Reinforcement

class

1λ 2λ scR 1λ 2λ scR

sR

MPa

sersR ,

MPa

A-V A-VI

1.25 2.04

0.53 0.77

500 500

2.78 3.88

1.03 1.25

400 400

680 815

785 980

ssc

s

RR

R 10001

2

1

−

=λ ;

ssc

s

RR

R 100012

−=λ .

Value usc,σ in formulas (14) and (155) is taken equal to 3, 1000380 δσ +=usc but no

more than 1200 MPa. The mentioned elements of rectangular section with reinforcement concentrated at the most or the least compressed surfaces are calculated according to Items 3.65 and 3.61 if the height of compressed zone x determined by formula (107a) or (110a) is more than the

limit value 0hRξ by replacement sR by sR8.0 in the calculation formula. Otherwise the

calculation is made according to Item 3.41 of the “Manual for design of prestressed

reinforced concrete structures of heavy-weight and light-weight concrete” taking 0=spσ .

In this case use of confinement reinforcement and high-strength reinforcement is inefficient.

85

3.60.(3.23).During calculation of eccentric compressed elements with confinement reinforcement along with calculation as regards the strength according to Item 3.57 instructions it is necessary to make calculations providing crack resistance of protection cover of concrete.

The calculation is made in compliance with instructions of Items 3.61-3.68 and 3.71-3.76

according to performance values of design loads ( 0.1=fγ ) considering total area of

concrete section and taking design resistances serbR , and sersR , for limit states of the

second group and reinforcement design resistance against compression equal to value

sersR , but no more than 400 MPa.

During determination of value Rξ in formulas (14) and (155) it is taken 400, =uscσ MPa

and in formula (15) coefficient 0.008 is replaced by 0.006. When considering the elasticity influence it is necessary to use Item 3.54 instructions

determining value min,eδ by formula (95) replacing bR010.0 by serbR ,008.0 .

Calculation of members of symmetrical section by location of the longitudinal force in the

symmetry plane.

RECTANGULAR SECTIONS WITH SYMMETRICAL REINFORCEMENT 3.61.Strength examination of rectangular sections with symmetrical reinforcement

concentrated at the most compressed and tensile (the least compressed) surfaces of the element is made in the following manner according to the compressed zone height x:

bR

Nx

b

= : (107)

a) by 0hx Rξ≤ (Draft 34) – according to the following condition

( ) ( )'5.0 0

'

0 ahARxhbxRN sscbe −+−≤ ; (108)

Draft 34. Forces scheme in rectangular cross-section of eccentric compressed element.

b) By 0hx Rξ> – according to condition (108) taking the height of compressed zone

equal to 0hx ξ= where ξ is determined by following formulas:

- For elements of concrete of class B30 and lower:

( )

sR

RsRn

αξ

ξαξαξ

21

21

+−

+−= (109)

- For members of concrete of class more than B30:

ωαψααψαααψα

ξ sc

nscsnscs +

+++

−+−=

2

22 (110)

In formulas (109) and (110):

0bhR

N

b

n =α ; 0bhR

AR

b

ss

s =α ;

−

=

1.11

,

ω

σψ

s

usc

c

R

:

ωψξ ,, sR – See Table 18 and 19.

86

Value e is determined by the following formula:

2

'00

ahee

−+= (111)

At the same time eccentricity of longitudinal force 0e relating to the center of gravity of

the section is determined considering the deflection of the element according to Items 3.54-3.56. Notes. 1. If the height of compressed zone which is determined considering a half of compressed

reinforcement is '2/

abR

ARNx

b

ss <+

= , so design bearing capacity of the section can be increased

using condition (108) by 0' =sA and bR

ARNx

b

ss+= .

2. Formula (110) can be also used during determination of elements of concrete B30 and lower class.

3.62.Required quantity of symmetrical reinforcement is determined in the following manner

according to relative value of longitudinal force 0bhR

N

b

n =α :

a) by Rn ξα ≤

( )δ

ααα

−

−−==

1

2/110' nnm

s

b

sR

bhRAA (112)

b) by Rn ξα >

( )δ

ξξα

−

−−==

1

2/110' m

s

b

ssR

bhRAA (113)

Where ξ – relative height of compressed zone determined by formula (109) or (110).

Value sα in formula (109) can be determined by the following formula:

( )δ

αααα

−

−−=

1

2/11 nnm

s (114)

And in formula (110) it’s determined by formula (114) replacing nα by ( ) 2/Rn ξα + .

In formulas (112)–(114):

2

0

1bhR

N

b

e

m =α ; 0

'

h

a=δ .

Value e is determined by formula (111).

If value 'a does not exceed 015.0 h so required quantity of reinforcement can be

determined by the diagram of Draft 35 using the following formula:

s

b

sssR

bhRAA 0' α== ,

Where sα is determined according to the diagram of Draft 35 in relation to the following

values:

2

0bhR

M

b

m =α ; 0bhR

N

b

n =α

At the same time value of moment M relating to the center of gravity is determined considering the element deflection in compliance with Items 3.54-3.56.

87

Draft 35. Bearing capacity diagrams of eccentric compressed elements of rectangular section with

symmetrical reinforcement

0bhR

N

b

n =α ; 2

0bhR

M

b

m =α ; 0bhR

AR

b

ss

s =α

By static calculation as regards the undeformed scheme and by using the coefficient

1>η reinforcement is chosen according to the mentioned formulas and to the diagram of

Draft 35 in general case by means of step-by-step approximation. For members of heavy-weight concrete of class B15-B50 as well as if light-weight

concrete B10-B40 by average density grade no less than D1800 by 25/0 ≤= hlλ and by

'a no more than 015.0 h reinforcement can be chosen without step-by-step approximation

by means of diagrams of Annex 3, at the same time there are used values M without considering coefficient η.

3.63.If there is reinforcement located along the height of the section so calculation of eccentric compressed elements can be made by formulas (117) and (118) considering all reinforcement as evenly distributed along the rods centers of gravity lines (Draft 36). At

the same time section area of reinforcement slA , located at one of the surfaces parallel to

the bending plane taken equal to:

( )1,1 += llssl nAA (115)

Where lsA ,1 is section area of one intermediate rod; by different diameters it is taken

average section area of the rod;

ln – The number of intermediate rods.

Draft 36. The scheme taken by calculation of the eccentric compressed element of rectangular section with

reinforcement located along the height of the section

Section area of reinforcement stA located at one of the surfaces perpendicular to the plane

of bending is:

sl

tots

st AA

A −=2

,, (116)

Where totsA , – section of total reinforcement in the section of element.

The section strength is to be examined according to the relative height of compressed

zoneωα

ααξ

/21

1

sl

sln

h

x

+

+== :

a) By Rξξ ≤ section strength is checked according to the following condition:

( ) ( )( ) ( )[ ]1

2

11111

2

0 2105.0115.0 δαξαδξδξαξξ −+−−−−+−≤ stslslbbhRNe , (117)

Where ω

ξξ =1 ;

bhR

N

b

n =1α ;

( )15.0 δα

−=

bhR

AR

b

sls

sl ;bhR

AR

b

sts

st =α ; h

a11 =δ (see Draft 36)

88

b) By Rξξ > section strength is checked according to the following condition:

nRna

nna

mbt RbhRNeαα

ααα

−

−≤ 12

0 , (118)

Where bhR

AR

b

totss

na

,1+=α – relative value of longitudinal force by even compression of

the whole section;

nRmR αα , – Relative values of a bending moment and of longitudinal

force by compression zone height hRξ :

( ) ( )( ) ( )1

2

11111 2105.0115.0 δαξαδξδξαξξα −+−−−−+−= stRslRRslRRmR ;

( )12 1 −+= RslRnR ξαξα ;

ω

ξξ R

R =1 ; ωξ ,R – see table 18 and 19.

Eccentricity of longitudinal force 0e is determined considering the element deflection

according to Items 3.54-3.56.

Note. By location of reinforcement within end quarters of height 12ah − (see Draft 36) the calculation is

made in compliance with Items 3.61 and 3.62 considering reinforcement S and S’ as concentrated along their centers of gravity.

3.64. Calculation of compressed elements made of heavy-weight concrete B15-B40 or of

light-weight concrete B12.5-B30 and by average density grade no less than D1800 as regards longitudinal force applied with the eccentricity taken according to item 3.50

equal to accidental eccentricity 30/hea = by hl 200 ≤ can be made according to the

following condition:

( )totsscb ARARN ,+≤ ϕ , (119)

Where φ is the coefficient determined by the following formula:

( ) sbsbb αϕϕϕϕ −+= 2 (120)

But it’s taken no more than sbϕ

Here sbb ϕϕ , are coefficients taken according to Tables 26 and 27;

AR

AR

b

totss

s

,=α ;

totsA , – See Item 3.63;

By 5.0>sα it is possible to take sbϕϕ = without using formula (120).

Table 26

Coefficient bϕ by hl /0 Concrete

N

N l

6 8 10 12 14 16 18 20

Heavy-weight 0 0.5 1.0

0.93 0.92 0.92

0.92 0.91 0.91

0.91 0.90 0.89

0.90 0.89 0.86

0.89 0.86 0.82

0.88 0.82 0.76

0.86 0.78 0.69

0.84 0.72 0.61

89

Light-weight 0 0.5 1.0

0.92 0.92 0.91

0.91 0.90 0.90

0.90 0.88 0.86

0.88 0.84 0.80

0.86 0.79 0.71

0.82 0.72 0.62

0.77 0.64 0.54

0.72 0.55 0.45

Table 27

Coefficient bϕ by hl /0 Concrete

N

N l

6 8 10 12 14 16 18 20

A. By haa 15.0'<= and if there are no intermediate rods (see a sketch) or by section area of these rods less

than 3/,totsA

Heavy-weight 0 0.5 1.0

0.93 0.92 0.92

0.92 0.92 0.91

0.91 0.91 0.90

0.90 0.89 0.89

0.89 0.88 0.87

0.88 0.86 0.84

0.86 0.83 0.79

0.84 0.79 0.74

Light-weight 0. 0.5 1.0

0.92 0.92 0.92

0.92 0.91 0.91

0.91 0.90 0.90

0.89 0.88 0.88

0.88 0.86 0.85

0.85 0.83 0.80

0.82 0.77 0.74

0.77 0.71 0.67

B. By haah 15.0'25.0 ≥=> or by section are of no intermediate rods (see a sketch) is equal or more than

3/,totsA independently on value a

Heavy-weight 0 0.5 1.0

0.92 0.92 0.92

0.92 0.91 0.91

0.91 0.90 0.89

0.89 0.88 0.86

0.87 0.85 0.82

0.85 0.81 0.77

0.82 0.76 0.70

0.79 0.71 0.63

Light-weight 0. 0.5 1.0

0.92 0.92 0.91

0.91 0.91 0.90

0.90 0.89 0.88

0.88 0.86 0.84

0.85 0.81 0.76

0.81 0.73 0.68

0.76 0.65 0.60

0.69 0.57 0.52

Symbols in Table 26 and 27:

lN – Longitudinal force of dead loads and long-term loads:

N – Longitudinal force of all loads

Sketch

1 – Considered plane;

2 – Intermediate rods.

RECTANGULAR SECTIONS WITN ASYMMETRICAL REINFORCEMENT 3.65.The strength of rectangular sections with asymmetrical reinforcement concentrated at the

most compressed and tensile (the least compressed) surfaces of the element is calculated according to Item 3.61, at the same time formulas (107), (109) and (110) have the following form:

bR

ARARNx

b

sscss

'−+= ; (107a)

( ) ( ) ( )sR

ssRssRn

αξ

ααξααξαξ

21

1 ''

+−

−+++−= ; (109a)

ωαψααψαααψα

ξ sc

nscsnscs +

−++

−+−=

2''

22; (110a)

Where0

''

bhR

AR

b

ssc

s =α

3.66. Areas of sections of compressed and tensile reinforcement corresponding to minimum of their sum is determined by following formulas: For elements made of concrete B30 and lower:

90

( )0

'

4.0

0

2

0' ≥−

−=

ahR

bhRNeA

sc

b

s ; (121)

'055.0s

s

b

s AR

NbhRA +

−= ; (122)

For elements of concrete more than B30:

( )0

'0

2

0' ≥−

−=

ahR

bhRNeA

sc

bR

s

α; (123)

'0s

s

bR

s AR

NbhRA +

−=

ξ (124)

Where RR ξα , are determined according to Tables 18 and 19 and are taken no more than

0.4 and 0.55.

By negative value sA determined by formula (122) and (124) it is taken minimum section

area of reinforcement S according to constructive requirements but no less than the following value:

( ) ( )( )'

'2/'

0

0min,

ahR

ahbhReahNA

sc

b

s−

−−−−= ; (125)

And the section area of reinforcement S’ is determined in the following manner:

- by negative value of min,sA – by the following formula:

( ) ( ) ( )

sc

bbbb

sR

beRbhRNNbaRNbaRNA

22'' 0

2

'+−−−−−

= , (126)

- by positive value of min,sA – by the following formula:

min,

'

s

sc

b

s AR

bhRNA −

−= ; (127)

If accepted section area of compressed reinforcement '

, factsA is more than the value

calculated by formula (121) or (123) (for example by negative value '

sA ) so section area

of tensile reinforcement can be decreased according to the following formula:

s

factsscb

sR

ARNbhRA

',0 +−

=ξ

; (128)

Where ξ is determined according to table 20 in compliance with the following value:

( )2

0

0', '

bhR

ahARNe

b

factssc

m

−−=α ; (129)

If there is no compressed reinforcement or it’s not considered in the calculation so section area of tensile reinforcement is always determined by formula (128) but at the same time

it is necessary to meet the requirement Rm αα < .

I-SECTIONS WITH SYMMETRICAL REINFORCEMENT

3.67. The strength of I-sections with symmetrical reinforcement concentrated in flanges (see

Draft 37) is made in the following manner. Draft 37. Loads scheme in the cross-I-section of eccentric compressed element

91

If the following condition is met (that is the border of compressed zone lies in the flange) so

the calculation is made as for a rectangular section '

fb wide in compliance with Item 3.61:

''

ffb hbRN ≤ (130)

If condition (130) is not met (that is the border of compressed zone goes in the rib) so the

calculation is made according to the compressed zone heightbR

ARNx

b

ovb−= :

a) by 0hx Rξ≤ section area is examined according to the following condition:

( ) ( ) ( )'2/2/ 0

''

00 ahARhhARxhbxRNe sscfovbb −+−+−≤ (131)

b) by 0hx Rξ> section area is examined according to condition (131) determining the height

of compressed zone by the following formula:

+

−+++

−++−= ωαψ

αααψααααψαsc

novscsnovscshx

2

022

(132)

Where0bhR

AR

b

ss

s =α ; 0bhR

N

b

n =α ; 0bh

Aov

ov =α

ωξψ ,, Rc – See table 18 and 19

ovA – Area of compressed flange overhangs equal to ( ) ffov hbbA −= '

If value x determined by formula (132) is more than fhh − (that is the border of compressed

zone goes along the least compressed flange) it is possible to consider the increase of the

bearing capacity due to the less compressed flange. In that case (if ff bb =' ) the calculation is

made by formulas (131) and (132) replacing fb by '

fb and '

fh by )( '

ff hhh −+ taking

( )( )fffov hhhbbA −−−−= ' .

Note. By variable height of the flange overhangs values fh and '

fh are taken equal to the average height of

overhangs.

3.68.Required quantity of symmetrical reinforcement of I-sections is determined in the following manner.

If condition (130) is met so reinforcement is chosen as for rectangular sections '

fb wide

according to Item 3.62. If condition (130) is not met so reinforcement is chosen according to relative height of compressed zone ξ:

ovn ααξ −= (133)

a) by Rξξ ≤

( )δ

αξξα

−

−−−==

1

2/1 ,10' ovmm

s

b

ssR

bhRAA (134)

b) by Rξξ >

( )δ

αξξα

−

−−−==

1

2/1 ,1110' ovmm

s

b

ssR

bhRAA (135)

92

Where relative height of compressed zone 01 / hx=ξ is determine according to formula

(132) by:

( )δ

αξξαα

−

−−−=

1

1 ,1 ovmm

s (136)

In formulas (133)–(136):

ovn αα , – See Item 3.67;

2

0

1,bhR

Ne

b

m =α ; 0/' ha=δ ;

( )0

'

, /5.01 hh fovovm −= αα .

RING CROSS-SECTIONS 3.69.The strength of ring cross-sections (Draft 38) by the ratio of inner and outer radius

5.0/ 21 ≥rr and reinforcement distributed along the circle (by no less than 6 longitudinal

rods) is calculated in the following manner according to relative area of compressed zone

of concrete cirξ :

totssb

totss

cirARAR

ARN

,

,

7.2+

+=ξ (137)

a) by 6.015 << cirξ – according to the following condition:

( ) ( )( )circirstotss

cir

stotssmb rARrARArRNe ξξπ

πξ3.12.07.11

sin,,0 +−++≤ (138)

b) by 15.0≤cirξ – according to the following condition

( ) stotss

cir

stotssmb rARrARArRNe ,1

,0 29.0sin

++≤π

πξ (139)

Where totssb

totss

cirARAR

ARN

,

,

1

75.0

+

+=ξ (140)

c) By 6.0≥cirξ – according to the following condition

( )π

πξ 2,0

sin cir

stotssmb rARArRNe +≤ (141)

Where totssb

cirARAR

N

,

2+

=ξ (142)

In formulas (137)–(142):

totsA , – Section area of total longitudinal reinforcement;

221 rr

rm

+= ;

sr – Circle radius going through the center of gravity of considered reinforcement.

Eccentricity of longitudinal reinforcement 0e is determined considering the element

deflection according to Items 3.54–3.56.

Draft 38. The scheme taken by calculation of the ring cross-section of eccentric compressed element

93

3.70.It is possible to check the strength and to determine required quantity of reinforcement of

ring sections mentioned in Item 3.69 by ms rr ≈ by means of diagrams of Draft 39 using

the following formulas:

ArRNe mbmα≤0 (143)

s

b

stotsR

ARA α=, (144)

AR

N

b

n =α ; mb

mArR

Ne0=α ;AR

AR

b

totss

s

,=α

Where values mα and sα are determined in compliance with the diagram according to

values AR

AR

b

totss

s

,=α and

mb

mArR

Ne0=α as well asAR

N

b

n =α . At the same time eccentricity

0e is determined considering the deflection of the element according to Items 3.54–3.56.

Draft 39. Bearing capacity diagrams of eccentric compressed elements of ring-section

ROUND SECTIONS 3.71.The strength of round sections (Draft 40) with reinforcement distributed along the circle

(no less than 60 longitudinal rods) by concrete class no less than B30 is checked according to the following condition:

s

cir

totss

cir

b rARArRNe

++≤ ϕ

π

πξ

π

πξ sinsin

3

2,

3

0 (145)

Where r – radius of the cross section;

cirξ – Relative area of the compressed concrete zone determined in the following

manner: If the following condition is met:

totssb ARARN ,645.077.0 +≤ (146)

According to the following equation:

totssb

cir

btotss

cirARAR

ARARN

,

,2

2sin

+

++= π

πξ

ξ ; (147)

If equation (146) is not met so it’s determined according to the following equation:

totssb

cir

b

cirARAR

ARN

,

2

2sin

+

+= π

πξ

ξ ; (148)

ϕ – Coefficient considering the work of tensile reinforcement and taken equal to; if

condition (146) is met ( ) circir ξξϕ 55.116.1 −= but no more than one; if condition

(146) is not met so 0=ϕ ;

totsA , – Section area of the total longitudinal reinforcement;

sr – Circle radius going through the center of gravity of longitudinal reinforcing rods.

Eccentricity of longitudinal force 0e is determined considering the deflection of the

element according to Items 3.54–3.56.

Draft 40. The scheme taken by the calculation of the round section of the eccentric compressed element

94

3.72.It is possible to check the strength and to determine required quantity of reinforcement of round sections mentioned in Item 3.71 by means of diagrams of Draft 41 using the following formulas:

ArRNe bmα≤0 ; (149)

s

b

stotsR

ARA α=, , (150)

Where values mα and sα are determined by Draft 41 according to values AR

AR

b

totss

s

,=α

andArR

Ne

b

m

0=α as well asAR

N

b

n =α . At the same time eccentricity 0e is determined

considering the deflection of the element according to Items 3.54–3.56.

Draft 41. Bearing capacity diagrams of eccentric compressed elements of round section

CALCULATION OF ELEMENTS WORKING IN BIAXIAL ECCENTRIC COMPRESSION 3.73.Calculation of normal sections of elements working in biaxial eccentric compression is

made in general case according to Item 3.76 determining location of the straight line which bounds the compressed zone by means of step-by-step approximation.

3.74.Calculation of elements of rectangular section with symmetrical reinforcement as regards

biaxial compression can be made by means of diagrams of Draft 42. Draft 42. Diagrams of bearing capacity of rectangular section elements with symmetrical reinforcement

working in biaxial eccentric compression

a – by 2.0=sα ; b – by 4.0=sα ; c – by 6.0=sα ; d – by 0.1=sα (wherebhR

AR

b

totss

s

,=α )

Section strength is considered to be provided if points with coordinates 0/ xx MM and 0/ yy MM on the diagram corresponding to parameter sα are located inside of the part

bounded by a curve corresponding to parameter 1nα and by axes of coordinates.

Values xM and yM are represented by bending moments caused by external loads

relating to the center of gravity of the section and acting in symmetry planes x and y. Influence of the element deflection is considered by means of multiplying of moments

xM and yM by coefficients xη and yη determined for planes x and y according to Item

3.54 by the longitudinal force N.

Values 0

xM and 0

yM are represented by limit bending moments which can be taken by

the section in symmetry planes x and y considering longitudinal force N applied in the center of gravity of the section.

Values of limit moments 0

xM and 0

yM are represented by right parts of equations (117)

and (118). At the same time discretely located reinforcement rods are replaced by distributed reinforcement.

95

( ) ( )β

β

+−−++=

121 ,1,10,1 ysxssxxssx AAAnAA ; (151)

sx

tots

sy AA

A −=2

, (152)

Where sysx AA , – area of reinforcement located at surfaces normal to symmetry axes x and

y (Draft 43);

ysxs AA ,1,1 , – Area of each intermediate rod located at surfaces normal to symmetry

axes x and y;

xn – Number of intermediate rods with the area xsA ,1 located along one side

of the section;

0sA – Angle rod area;

x

y

y

x

h

h

M

M=β ;

yx hh , – Section height by eccentric compression in planes x and y;

totsA , – Section area of total longitudinal reinforcement.

Parameters sα and 1nα are determined by following formulas:

bhR

AR

b

totss

s

,=α ;

bhR

N

b

n =1α

Draft 43. Location of reinforcement in rectangular section by calculation as regards biaxial eccentric

compression

a – actual; b – design

3.75.Calculation of elements of symmetrical I-section by 53/ −=bb f and 25.015.0/ −=hh f

with symmetrical reinforcement located in flanges of the section as regards biaxial compression can be made by means of diagrams of bearing capacity given on Draft 44.

Draft 44. Bearing capacity of elements of symmetrical I-section working in biaxial eccentric compression

a – by 2.0=sα ; b – by 4.0=sα ; c – by 0.1=sα ; d – by 4.1=sα (wherebhR

AR

b

totss

s

,=α ); e – by

8.1=sα ; f – by 8.2=sα (wherebhR

AR

b

totss

s

,=α )

The calculation is made similar to the calculation given in Item 3.74 for elements of rectangular section.

Taken by the section in the symmetry axis x going in the rib limit moment 0

xM

represented by the right part of condition (131) decreased by ( ) 2/'0 ahN − ; limit moment 0

yM can be determined as for rectangular section made up of two flanges according to

Item 3.63.

GENERAL CASE OF CALCULATION OF NORMAL SECTIONS OF ECCENTRIC COMPRESSED ELEMENTS (BY ANY SECTIONS, EXTERNAL FORCES AND BY ANY REINFORCEMENT)

96

3.76.(3.28)Calculation of the eccentric compressed element in general case (Draft 45) must be made according to the following equation:

sisibb SSReN σΣ−≤ (153)

Where e – the distance of longitudinal force to the axis parallel to the line bounding compressed zone and going through the center of gravity of a tensile rod which is most distant from the mentioned line;

bS – Static moment of the section area of compressed zone of concrete relating to

the mentioned axis;

siS – Static moment of the section area of i-rod of longitudinal reinforcement

relating to the mentioned axis;

siσ – Stress in the i-rod of longitudinal reinforcement determined according to the

present item instructions.

The height of compressed zone x and stresses siσ are determined according to the

following equations:

0=−Σ− NAAR sisibb σ (154)

−

−

= 1

1.11

,

i

usc

siξ

ωω

σσ (155)

In formulas (154) and (155):

siA – Section area of i-rod of longitudinal reinforcement;

iξ – Relative height of compressed concrete zone equal toi

ih

x

0

=ξ where ih0 is the

distance from the axis going through the center of gravity of the i-rod section and parallel to the line bounding the compressed zone to the most distant point of compressed zone of the section (see draft 45);

ω – Characteristics of concrete compressed zone determined by formulas (15) or (104);

usc,σ – See Items 3.14 and 3.59.

Draft 45. Forces scheme and stresses diagram in the section normal to the longitudinal axis of the

reinforced concrete element, in general case according to the calculation as regards the strength

I – I the plane parallel to the bending moment plane or the plane going through the point of application of the

longitudinal force and resultant force of internal compression and tension forces; A – pint of application of

resultant forces in compressed reinforcement and in concrete of compressed zone; Б – the same in tensile

reinforcement; 1 – 8 – rods.

Stress siσ is inserted into the calculation with the sign determined by formula (155), at the

same time stresses with sign “plus” symbolize tension stress and are taken no more than

siR and stresses with sign “minus” symbolize compression stresses and are taken no

more than scR .

To determine the location of the compressed zone by biaxial eccentric compression except formulas (154) and (155) it is necessary to meet the additional requirement: points of application of external longitudinal force, resultant force of compression forces in concrete and reinforcement and resultant force in tensile reinforcement must belong to one straight line (see Draft 45). If it is possible to identify the specific axis (for example symmetry axis or axis of the rib of a L-section) by biaxial eccentric compression it is necessary to make the calculation

97

according to two conditions: (153) determining values e , bS and siS relating to axis x

going through the center of gravity of the most tensile rod parallel to the mentioned above specific axis and according to condition (153) determining values e ,

bS and siS relating to axis y which crosses axis x at right angle in the center of the most

tensile rod. At the same time the location of the straight line bounding the compressed zone is chosen by means of step-by-step approximation according to equations (154) and (155) taking the angle of slope of this line θ constant and equal to the angle of slope of the neutral axis determined as for elastic material. Section strength will be provided only if condition (153) is met relating to both axes (x and y). If equation (153) is not met during all examinations so the strength of the section is not provided and it is necessary to increase reinforcement, dimensions of the section or to increase concrete class. If the condition is met only relating to one axis so it is necessary to determine the shapes of compressed zone one more time by different angle θ and to make similar calculation one more time.

EXAMPLES OF CALCULATION

RECTANGULAR SECTIONS WITH SYMMETRICAL REINFORCEMENT Example 24. Given: a column with the frame work, section b = 400 mm, h = 500 mm,

40'== aa mm; heavy-weight concrete B25 ( 4107.2 ⋅=bE MPa); reinforcement A-III

( 365== scs RR MPa; 5102 ⋅=sE MPa); its section area is 1232' == ss AA mm2 (2Ø28);

longitudinal forces and bending moments: from dead loads and long-term loads 650=lN kN,

140=lM kN·m; from wind load 50=shN kN, 73=shM kN·m; design length of the

column 60 =l m.

It is required to check the strength of the column section.

Calculation: 460405000 =−=h mm. As there are forces from the short-term loads (wind

load) so the calculation is made according to case “a” in compliance with Item 3.1. Forces from wind load are equal to:

70050650 =+=N kN; 21373140 =+=M kN·m

Let’s determine moments of external forces relating to tensile reinforcement IM and

IIM calculated considering and without consideration a short-term load (wind load):

3602

04.046.0700213

2

'01 =

−+=

−+==

ahNMMM II kN·m

5.2762

04.046.0560140

2

'01 =

−+=

−+==

ahNMMM lllI kN·m

As 29536082.082.0 =⋅=IIM kN·m so the calculation is made only according to case “b” (see

Item 3.1) that is as regards all loads taking 16=bR MPa (by 1.12 =bγ ).

As 10125.0/6/0 >==hl so the calculation is made considering the column deflection

according to Item 3.54, calculating crN by formula (93).

For that we determine:

98

77.1360

5.276111

1

1 =+=+=M

M l

l βϕ

[Here 0.1=β for heavy-weight concrete (see Table 16)];

0913.0107.2500400

102123224

5'

=⋅⋅⋅

⋅⋅⋅=

+=

b

sss

E

E

bh

AAµα ;

30410700

102133

6

0 =⋅

⋅==

M

Ne mm > 30/hea =

So that means accidental eccentricity is not to be taken into account.

As 22.0161201.05.001.001.05.0608.0500

304 0min,

0 =⋅⋅−=−−=>== be Rh

l

h

eδ so we

take 608.00

h

ee =δ .

Coefficient η is to be determined by formula (91): Value е is:

.55,05492

40460115,1304

2

00 мmm

ahee ≅=

−+⋅=

′−+= η

Let’s determine the height of compressed zone х by formula (107): mm.

ξR = 0.55 (See Table 18).

As х = 109,4 mm < ξRh0 = 0,55 · 460 = 253 mm so section strength is to be checked according to condition (108):

( ) ( ) ( )

( ) ,38555,07006,472106,47240460

12323654,1095,04604,109400165,0

6

00

mkNNemkNmmN

ahARxhbxR sscb

⋅=⋅=>⋅=⋅⋅=−×

×⋅+⋅−⋅⋅=′−′+−

т. е. прочность сечения обеспечена. Example 25. Given: section of the element with dimensions b = 400 mm, h = 500 mm; a = a' =

= 40 mm; heavy-weight concrete В25 (Eb = 2,7 · 104 MPa); symmetrical reinforcement A-III (Rs =

= Rsc = 365 MPa; Es = 2 · 105 MPa); longitudinal forces and bending moments: from dead loads and long-term loads Nl = 600 kN, Ml = 170 kN·m; from wind load Nsh = 200 kN, Мsh = 110 kN·m; design length l0 = 8 m.

It is required to determine section area of reinforcement. Calculation: h0 = 500 – 40 = 460 mm. As there are wind load forces so let’s check condition

(1). For that we determine: kN·m; kN·m; kN; kN·m. As 0,82 MII = 0,82 · 448 = 368 kN·m > MI = 296 kN·m so the calculation is made only

according to case „b", that is as regards all loads, taking Rb = 16 MPa (by γb2 = 1,1). As l0/h = 8000/500 = 16 > 10 so the calculation is made considering the element deflection

according to Item 3.54, calculating Ncr by formula (93). For that we determine:

[β = 1,0 – See Table 16];

30/35010800

102803

6

0 hemmN

Me a =>=

⋅

⋅==

(See Item 3.50)

99

As е0/h = 350/500 = 0,7 > δe,min = 0,5 – 0,01 – 0,01Rb, so we take δe = = 0,7.

On the first approximation we take µ = 0,01, = 7,4, So

( )

( ) kNH

h

ah

hl

bhEN

l

eb

cr

33721033720522,00477,01075,33500

404604,701,0

66,13

1,07,01,0

11,0

16

500400107,26,1

3

1,01,0

11,0

/

6,1

36

2

2

42

0

2

0

=⋅=+⋅=

−⋅+

⋅

++

×

×⋅⋅⋅⋅

=

′−+

+

+= αµ

ϕ

δ

Coefficient η is: Considering the element deflection value е is: mm. Required reinforcement we determine according to Item 3.62. Let’s determine the following values:

By Table 18 we find ξR = 0,55.

As αn < ξR , so value Аs = is to be determined by Formula (112): ( )

( ),1413

087,01

2/272,01272,0395,0

365

46040016

1

2/1

2

10

mm

R

bhRAA nnm

s

b

ss

=−

−−⋅⋅=

=−

−−=′=

δ

ααα

From which

As the present reinforcement is more than reinforcement taken during determination of Ncr (µ =

= 0,01) so value Аs = 1413 mm2 is determined „on the safe side" and it can be decreased if value µ. is specified more exact

We take µ = (0,01 + 0,014)/2 = 0,012 and determine value Аs = :

[

;3724103724500

40460

4,7012,00477,01075,33

3

2

6

kNН

N cr

=⋅=

−×

×⋅+⋅=

mm; mm2.

Finally we take As = = 1362 mm2 (2 ∅ 25 + 1 ∅ 22).

Example 26. According to Example 25 data it is necessary to determine reinforcement area,

using diagrams of Annex 3. Calculation. In compliance with Example 25: N = 800 kN; М = 280 kN·m; = 16; = 0,66.

Let’s determine values αn и αm:

By diagram б of Annex 3 by αn = 0,272, αm = 0,207 and λ = 15 we find αs = 0,16.

By diagram в of Annex 3 by αn = 0,272, αm = 0,207 and λ = 20 we find αs = 0,2.

100

Value αs corresponding to λ = 16 is to be determined by means of linear interpolation: So reinforcement area is: mm2.

We take Аs = = 1362 mm2 (2 ∅ 25 + 1 ∅ 22). Example 27. Given: a column with the multistory framework with the section dimensions b = = 400 mm, h = 500 mm; a = а’ = 40 mm; heavy-weight concrete В25 (Eb = 2,7 · 104 MPa); symmetrical reinforcement А-III (Rs = Rsc = 365 MPa; Еs = 2 · 105 MPa); longitudinal forces and bending moments in the support section of the column: from dead loads and long-term loads on the floors Nl = 2200 kN, Ml = 259 kN·m; from wind loads Nsh = 0, Msh = 53,4 kN·m; no short-term loads on the floor; design length of the column is l0 = 6 м. It is required to determine reinforcement area. Calculation. h0 = h – а = 500 – 40 = 460 mm. As there is the wind load force we determine condition (1). We determine: kN·m; kN; kN·m; kN·m. As 0,82 MII = 0,82 · 784,4 = 643 кН·м < MI = 721 kN·m so condition (1) is not met and we make the calculation twice: according to case „а" — as regards dead loads and long-term loads

by Rb = 13 MPa (that is by γb2 = 0,9) and according to case „b" — as regards all loads by Rb =

= 16 МПа (that is by γb2 = 1,1). The calculation is made for the support section. Calculation according to case „а". As l0/h = 6000/500 = 12 > 4 according to Item 3.54 so it is

necessary to consider the column deflection. But in compliance with Item 3.56 coefficient ηv for columns and multistorey frameworks taken for the moment Mv caused by loads on the floors is taken equal to 1,0 and moment Мh = Msh from wind loads is not considered in the

present calculation, that’s why design moment is М = Мv ηv = 259 kN·m. Design longitudinal force is N = Nl = 2200 kN, so = 118 mm > = 16,7 mm. We take e0 = 118 mm. By formula (111) we determine е = e0 + (h0 – a’)/2 = 118 + (460 – 40)/2 = 328 mm. Required reinforcement is to be determined according to Item 3.62. Let’s determine the following values:

By Table 18 we find ξR = 0,604.

As αn = 0.92 > ξR = 0,604 so value Аs = is to be determined by formula (113). For that we

calculate values αs and ξ by formulas (114) and (109):

( ) ( ).1304

087,01

2/772,01772,0656,0

365

46040013

1

2/1 210 mmR

bhRAA m

s

b

ss =−

−−⋅⋅=

−

−−=′=

δ

ξξα

Calculation by case „b". In compliance with Item 3.54 we determine coefficient η, taking reinforcement, calculated according to case „а", that is:

[β = 1,0 – See Table 16]; mm.

As e0/h = = 0,293 > δe,min = 0,5 – 0,01 l0/h – 0,01 Rb = 0,5 –0,01 · 12 – 0,01 · 16 = 0,22, we

take δe = е0/h = 0,293;

101

By formula (93) we determine Ncr:

( )

,108021500

40460096,0

92,13

1,0293,01,0

11,0

12

500400107,26,1

3

1,01,0

11,0

/

6,1

3

2

2

42

0

2

0

N

h

ah

hl

bhEN

l

eb

cr

⋅=

−×

+⋅

++

×

×⋅⋅⋅⋅

=

′−+

+

+= αµ

ϕ

δ

So coefficient η is:

According to Item 3.56 coefficient η = ηh = 1,38 is multiplied by the wind loads moment Мsh =

= M and coefficient ηv = 1,0, that’s why considering the element deflection the moment is equal to: kN·m. Required reinforcement we determine by Item 3.62 similar to the calculation according to case „а" taking Rb = 16 MPa: mm;

According to Table 18 we find ξR = 0,55.

As αn > ξR so we determine value Аs = by formula (113): So

( ) ( )

.13041228

087,01

2/675,01675,0586,0

365

46040016

1

2/1

22

10

mmмм

R

bhRAA m

s

b

ss

<=

=−

−−⋅⋅=

−

−−=′=

δ

ξξα

Finally we take As = = 1362 mm2 (2 ∅ 25 + 1 ∅ 22) >1304 mm2. Example 28. Given: the element section with dimensions b = 400 mm, h = 600 mm; heavy-

weight concrete В25 (Rb = 16 MPa by γb2 = 1,1; Eb = 2,7 · 104 MPa); reinforcement A-III (Rs

= Rsc = 365 MPa; Еs = 2 · 105 MPa) located in the section as it’s shown on Draft 46; longitudinal forces and bending moments: from all loads N = 500 kN, М = 500 kN·m; from dead loads and long-term loads Nl = 350 kN, Ml = 350 kN·m; design length l0 = 10 м. It is required to examine the section strength.

Черт. 46. For the calculation example 28

The calculation is to be made according to Item 3.63. Taking As1,l = 491 mm2 (∅ 25), ηl = 2

and As,tot = 6890 mm2 (8 ∅ 28 + 4 ∅ 25) we find reinforcement area Asl и Аst: mm2; mm2. According to Draft 46 we have a1 = 45 mm, so As l0/h = 10/0,6 = 16,7 > 10 the calculation is to be made considering the element deflection according to Item 3.54 determining value Ncr by formula (93). For that we determine:

[β = 1.0 (See Table 16)];

102

m.

As e0/h = = 1,67 > δe,min = 0,5 – 0,01 l0/h – 0,01 Rb, so we take δe = е0/h = 1,67.

Value µα we determine as for the section with reinforcement located along the height of the section in compliance with Item 3.54: So

( )

.527110527111,07,13

1,067,11,0

11,0

7,16

600400107,26,1

3

1,01,0

11,0

/

6,1

3

2

42

0

2

0

kNH

h

ah

hl

bhEN

l

eb

cr

=⋅=

+⋅

++

×

×⋅⋅⋅⋅

=

′−+

+

+== αµ

ϕ

δ

Coefficient η is: Let’s determine the following values:

According to Table 18 we find ω = 0,722 and ξR = 0,55.

As 0.24 < ξR = 0.55 so section strength is to be determined by formula (117):

( ) ( )( )[ ( )]( ) ( )[ ( )

( )] mkNNeмкНммН

bhR stslslb

⋅=⋅=>⋅=⋅⋅=⋅−+

+⋅⋅−−−×−+−⋅×

×⋅⋅=−+−−−−+−

552105,150068810688075,021187,0

332,0329,005,0075,0332,01075,0332,0329,024,0124,05,0

600400162105,0115,0

0

6

2

2

1

2

11111

2

η

δαξαδξδξαξξ

That is the section strength is provided. Example 29. Given: column section with dimensions b = 600 mm, h = 1500 mm; heavy-

weight concrete В30 (Rb = 19 MPa by γb2 = 1,1); reinforcement А-III (Rs = 365 MPa) located as it’s shown on Draft 47; longitudinal forces and bending moments determined according to the frame calculation as regards deformed scheme: from all loads N = 12 000 kN, М = 5000 kN·m; from dead loads and long-term loads Nl = 8500 kN, Мl = 2800 kN·m; design length of the column in the bending plane l0 = 18м, out of the bending plane l0 = 12 м; actual column length l = 12 m. It is required to examine the section strength.

Черт. 47. For the calculation example 29

The calculation in the bending plane is made according to Item 3.63.

Taking As1,l = 615, 8 mm2 (∅ 28), ηl = 5 and As,tot = 17 417 mm2 (14 ∅ 32 + 10 ∅ 28), we

find reinforcement area Аsl and Аst: Аsl = Аs1,l (ηl + 1) = 615, 8 (5 + 1) = 3695 mm2, mm2.

Center of gravity of reinforcement located at tensile surface (7 ∅ 32) is distant from this surface mm, So Let’s determine the following values:

103

From Table 18 we find ω = 0,698 and ξR = 0,523. As 0,584 > ξR = 0,523so section strength is to be checked by formula (118). For that we determine:

,5000541310541361,0372,1

702,0372,1

24,0150060019

6

212

mkNMмкНммН

bhRnRna

nna

mRb

⋅=>⋅=⋅⋅=−

−×

×⋅⋅⋅=−

−

αα

ααα

That is the section strength in the bending plane is provided. The calculation out of the bending plane. As design length out of the bending plane l0 = 12 m and ratio l0/b = 12/0,6 = 20 is more than ration l0/h = 18/1,5 = 12, corresponding to the column calculation in the bending plane according to 3.51, so it is necessary to calculate the column out of the bending plane taking eccentricity е0 equal to occasional eccentricity еa. At the same time we replace symbols h and b by b and h, that is we take the dimension of the section out of the bending plane h = 600 мм instead of the section height. As according to Item 3.50 the accidential eccentricity is equal to

600

00012

60020

30

600

30=≥===

lmm

hea

and l0 = 12 m ≤ 20h, so the calculation is made according

to Item 3.64.

Section area of intermediate rods located along the short sides is equal to As,int = 4826 мм2 (6 ∅ 32). As = 5800 mm2

> As,int = 4876 mm2 и а = 50 mm < 0,15h = 0,15 · 600 = 90 mm so we

use Table 27 in the calculation (Part А). According to Tables 26 and 27 by and we find ϕb =

0,674 and ϕsb = 0,77. Value

By formula (120) we determine coefficient ϕ: Let’s check condition (119):

( ) ( ) ,000125001741717365150060019746,0, kNNкНARAR totssb =>=⋅+⋅⋅=+ϕ

That is the section strength out of the bending plane is provided. Example 30. Given: a column with the section 400Х400 mm; design length is equal to the

actual length l = l0 = 6 m; heavy-weight concrete В25 (Rb = 13 MPa by γb2 = 0,9); longitudinal reinforcement A-III (Rsc = 365 MPa); centric applied forces: from dead loads and long-term loads Nl = 1800 kN; from short-term loads Nsh = 200 kN. It is required to determine section area of longitudinal reinforcement. Calculation: in compliance with Item 3.50the calculation is made considering occasional eccentricuty ea. As h/30 = 400/30 = 13,3 mm > = 10 mm so occasional eccentricity is taken equal to ea = h/30, so the calculation can be made according to Item 3.64, taking

20002001800 =+=+= shl NNN kN.

According to Tables 26 and 27 for heavy-weight concrete by Nl/N = 1800/2000 = 0,9, l0/h = =6000/400 = 15, supposing that there are no intermediate rods by а = а' < 0,15 h, we find

ϕb = 0,8 и ϕsb = 0,858.

Taking in the first approximation ϕ = ϕsb = 0,858 according to condition (119) we find:

104

.1025110208010233140040013858,0

102000 3333

, NARN

AR btotss ⋅=⋅−⋅=⋅⋅−⋅

=−=ϕ

So

As αs < 0,5 we specify value ϕ more exact calculating it by formula (120): In a similar manner we determine

.10377102080814,0

102000 333

, NAR totss ⋅=⋅−⋅

=

Calculated value RsAs,tot is more than the accepted in the first approximation value so we determine this value one more time:

.10360102080821,0

102000 333

, NAR totss ⋅=⋅−⋅

=

As determined value RsAs,tot is close to the value accepted in the second approximation so total area of reinforcement section is taken equal to: mm2.

Finally we take As,tot = 1018 mm2 (4 ∅ 18).

RECTANGULAR SECTIONS WITH ASYMETRICAL REINFORCEMENT Example 31. Given: element section with dimensions b = 400 mm, h = 500 mm; a = a' = 40

mm; heavy-weight concrete B25 (Rb = 13 MPa by γb2 = 0,9; Eb = 2,7 · 104); reinforcement A-III (Rs = Rsc = 365 MPa); longitudinal force N = 800 kN; its eccentricity relating to the center of gravity of concrete section е0 = 500 мм; design length l0 = 4,8 m. It is required to determine areas of reinforcement section S and S’. Calculation. h0 = 500 – 40 = 460 mm. As 4 < l0/h = 4,8/0,5 = 9,6 < 10so the calculaiton is made considering the element deflection according to Item 3.54. At the same time supposing

that µ ≤ 0,025we determine value Ncr by a simplified formula:

( ).9110109110

6,9

500400107,215,0

/15,0 3

2

4

2

0

kNНhl

AEN

b

cr =⋅=⋅⋅⋅

==

Coefficient η is to be determined by formula (91): Value e considering the element deflection: mm. Required section area of reinforcement S’ and S we determine by formulas (121) and (122):

( ) ( );01085

40460365

460400134,0758108004,0 223

0

20 >=

−

⋅⋅⋅−⋅⋅=

′−

−=′ mm

ahR

bhRNeA

sc

b

s

.24981085365

108004604001355,055,0 23

0 mmAR

NbhRA s

s

b

s =+⋅−⋅⋅⋅

=′+−

=

As 0,018 < 0,025 so values Аs and are not to be specified more exact.

We take = 1232 mm2 (2 ∅ 28), Аs = 2627 mm2 (2 ∅ 32 + 1 ∅ 36). ELEMENTS WITH CONFINEMENT REINFORCEMENT Example 32. Given: a column of a bracing framework with the section dimensions and location of reinforcement according to Draft 48; heavy-weight concrete В40 (Rb = 20 MPa by

γb2 = 0,9; Rb,ser = 29 MPa; Eb = 3,25 · 104 MPa); longitudinal reinforcement A-VI; confinement reinforcement meshes of rods A-III, diameter 10 mm (Rs,xy = 365 MPa), located with spacing

s = 130 mm along the whole length of the column; longitudinal force by γf > 1,0: from all loads

105

N = 6600 kN, from dead loads and long-term loads Nl = 4620 kN; the same by γf = 1,0: N = 5500 kN and Nl = 3850 kN; primary eccentricity of longitudinal force e0 = ea = 13,3 mm; design length of the column l0 = 3,6 m. It is required to examine the strength of the column. Draft. 48. For the calculation example 32

Calculation. Let’s check the section strength within the meshes contours considering confinement reinforcement according to Item 3.57. Design dimensions of the section

350== efef bh mm. As l0/hef = 3600/350 = 10,3 < 16 so confinement reinforcement can be

considered in the calculation; at the same time it is necessary to consider the column deflection in compliance with Items 3.54 and 3.58 as l0/hef > 4. Taking l0/cef = l0/hef = 10,3 and h = hef = 350 mm we get

Therefore we take δe = δe,min = 0,297. As intermediate rods of confinement reinforcement are located in end quarters of the distance between end rods equal to h –2a1 = 350 – 2 · 22 = 306 mm [58 mm < = 76,5 mm (see Draft 48)] so according to the note to Item 3.63 we take reinforcement S and S’ as concentrated along the lines of their center of gravity. Then considering that all rods have the same diameter we have: mm; mm.

Coefficient ϕl is to be determined by formula (94) taking β = 1,0 (see Table 16) and Critical force Ncr is to be determined by formula (93), taking

mm2 (6 ∅ 25),

and multiplying the calculated value by coefficient ϕ1 = 0,25 + 0,05 = 0,25 + 0,05 · 10,3 = =0,764:

( )

.900101090010764,0350

41309281,0

7,13

1,0297,01,0

11,0

3,10

3503501025,36,1

3

1,01,0

11,0

/

6,1

3

2

2

4

1

2

0

2

0

kNН

h

ah

hl

bhEN

l

eb

cr

=⋅=×

−+

⋅

++

×

×⋅⋅⋅⋅

=

′−+

+

+= ϕαµ

ϕ

δ

Coefficient η is equal to: So in compliance with formula (111), mm. Let’s determine prism strength Rb,red according to Item 3.57.

Taking Аsx = Аsy = 78,5 mm2 (∅ 10), nx = ny = 5, lx = ly = 350 mm and Aef = hefbef = 350 · 350= = 122 500 mm2 (see Draft 48) we determine coefficient so MPa As there is used high-strength reinforcement A-VI so design resistance of reinforcement against compression is to be determined according to Item 3.59: mm2;

106

We take θ = 1,6.

From Table 25 λ1 = 2,04, λ2 = 0,77, Rsc = 500 MPa, Rs = 815 MPa, so

MPaRМПаRR sscredsc 81574277,054,01

04,254,01500

1

1

23

13, =<=

⋅+

⋅+=

+

+=

λδ

λδ

Section strength is to be examined according to condition (108), determining the height of

compressed zone х = ξh0 by formula(110а).

For that we determine value ω by formula (104). As 10µxy = 10 · 0,0173 = 0,173 > 0,15, we

take δ2 = 0,15, then ω = 0,85 – 0,008 Rb + δ2 = 0,85 – 0,008 · 20 + 0,15 = 0,84 ≤ 0,9.

According to Items 3.61 and 3.65 we determine required coefficients αn, αs, and ψc, taking Rb

= Rb,red = 34,3 MPa; σsc,u = 380 + 1000 δ3 = 380 + 1000 · 0,54 = 920 MPa < 1200 MPa иand Rsc = Rsc,red = 742 MPa:

;1037103093503,34 3

0 NbhRb ⋅=⋅⋅=

So

Value ξR with replacing Rs by 0,8Rs is:

that means it was necessary to use formula (110a); mm;

( ) ( ) ( )

( ) ,1109168,06600105,1154413092945742

2845,03092843503,345,0

6

00

mkNNeмН

ahARxhbxR sscb

⋅=⋅=>⋅⋅=−⋅+

+⋅−⋅⋅=′−′+−

that is section strength is provided. Let’s examine resistance to strength of the column protection layer by means of similar

calculation as regards the force N = 5500 kN (by γf = 1,0) taking Rb = Rb,ser = 29 MPa, Rs =

=Rs,ser = 980 МПа, Rsc = 400 МПа, σsc,u = 400 МПа, ω = 0,85 – 0,006 Rb,ser = 0,85 – 0,006 · 29 = 0,679 according to Item 3.60 and considering total section of the column, that means b = h =

=400 mm, a = a′ = 41 + 25 = 66 mm, h0 = 400 – 66 = 334 mm. Critical force Ncr is to be determined by formula (93) taking l0/h = 3600/400 = 9, e0/h =

=13,3/400 = 0,033, δe,min = 0,5 – 0,01 – 0,008 Rb,ser = 0,5 – 0,01 · 9 – 0,008 · 29 = 0,178 >

e0/h, that means δe = δe,min = 0,178.

During determination of coefficient ϕl we consider longitudinal forces N and Nl by γf = 1.0, that means

so ϕl = 1 + 0,7 = 1,7;

107

( )

.900191090019

400

66334215,0

7,13

1,0178,01,0

11,0

9

4004001025,36,1

3

1,01,0

11,0

/

6,1

3

2

2

4

2

0

2

0

ђнH

h

ah

hl

bhEN

l

eb

cr

=⋅⋅=

=

−+

⋅

++⋅⋅⋅⋅

=

=

′−+

++

= µαϕ

δ

Coefficient η is: mm. Let’s make a calculation similar to the calculation as regards the strength:

;103880340029 3

0 NbhRb ⋅=⋅⋅=

mm;

( ) ( )( ) ( )

,8381524,05500

8,957108,957

663342945400305,05,0334

305400295,0

6

00

mkN

NemkNmmN

ahARxhbxR sscb

⋅=⋅=

=>⋅=⋅⋅=

=−⋅+⋅−×

×⋅⋅=′−′+−

That means resistance to cracks of the column protection layer is provided. I-SECTIONS

Example 33. Given: section dimensions and location of reinforcement according to Draft

49; heavy-weight concrete В30 (Eb = 2,9 · 104 MPa; Rb = 19 MPa by γb2 = 1,1); reinforcement

А-III (Rs = Rsc = 365 MPa); its cross-section area As = A′s = 5630 mm2 (7 ∅ 32); longitudinal forces and bending moments: from dead loads and long-term loads Nl = 2000 kN, Ml = 2460 kN·m; from all loads N = 2500 kN, М = 3700 kN·m; design length of the element: in the bending plane l0 = 16,2 m, out of the bending plane l0 = 10,8 m; actual length of the element l = 10,8 m. It is required to check the section strength.

Draft 49. For the calculation examples 33, 34 and 39

The calculation in the bending plane. We tae design thickness of the flange equal to the

average height of overhangs h′f = hf = 200 + 30/2 = 215 mm. Let’s determine the area and inertial moment of concrete section: mm2;

.1012792

215

2

1500

215400212

2154002

12

1500200

48

2

33

mm

I

⋅=

+×

×⋅⋅+⋅

+⋅

=

Radius of inertia of the section is mm.

108

As l0/i = 16 200/520 = 31,1 < 35 and l0/i > 14 so the calculation is made considering the element deflection in compliance with Item 3.54 taking value Ncr equal to:

.270281027028

20016

101279109,222

3

2

84

2

0

kNH

l

IEN b

cr

=⋅=

=⋅⋅⋅⋅

==

Coefficient η is to be determined by formula (91):

Center of gravity of the reinforcement area As and A′s is distant from the nearest surface at а =

=а′ = mm, and h0 = h – a = 1500 – 79 = 1421 mm. Value е considering the element deflection is:

.22932

791421

102500

096,1103700

2 3

60

0 mmah

ee =−

+⋅

⋅⋅=

′−+= η

Let’s check condition (130):

,2500245110245121560019 3kNNkNHhbR ffb =<=⋅=⋅⋅=′′

That means that the calculation is made as for the I-section. Area of compressed flange overhangs is equal to: mm2. Let’s determine the area of compressed zone: mm.

From Table 18 we find ξR = 0,523. As х = 228 mm < ξR h0 = 0,523 · 1421 = 743 mm, section strength is to be checked according to condition (131):

( )

( )

,5725293,22500

584710584779142156303652

21514210008619

2

228142122820019

22

6

000

mkNNe

mkNmmN

ahARh

hARx

hbxR ssc

f

ovbb

⋅=⋅=>

>⋅=⋅⋅=−⋅+

−⋅+

+

−⋅⋅=′−′+

′−+

−

That means the section strength in the bending plane is provided. Calculation out of the bending plane. Let’s determine radius of the section out of the bending plane: mm4; mm. As elasticity out of the bending plane l0/i = 10 800/134 = 80 is more than elasticity in the bending plane l0/i = 31.1 so according to Item 3.51 we check the section strength out of the bending plane taking eccentricity е0 equal to occasional eccentricity еа. At the same time the section height is h = 600 мм. As in compliance with Item 3.50 occasional eccentricity is еа = mm> mm, we take еа = that allows to make the calculation by in compliance with Item 3.64 as for the rectangular section without considering the „reserve” of the rib section, that means taking b = 2 · 215 = 430 mm.

Section area of intermediate rods located along both flanges is As,int = 4826 mm2 (6 ∅ 32), and

section area of all rods is As,tot = 11 260 mm2 (14 ∅ 32). As As,tot/3 = 11 260/3 = 3750 mm2 < <As,int = 4826 mm2 so we use Table 27 (Part Б) in the calculation. From Table 27 for heavy-

weight concrete by Nl/N = 2000/2500 = 0,8 и l0/h = 10,8/0,6 = 18 we find ϕsb = 0,724.

Value So, ϕ = ϕsb = 0,724. Let’s check condition (119):

( ) ( ) 3

, 1065251126036560043019724.0 ⋅=⋅+⋅⋅=+ totsscb ARARϕ N 2500=> N kN

That means the section strength out of the bending plane is provided.

109

Example 34. Given: section dimensions and location of reinforcement according to Draft

49; heavy-weight concrete В30 (Rb = 19 MPa by γb2 = 1,1; Eb = 2,9 · 104 MPa); symmetrical reinforcement A-III (Rs = Rsc = 365 MPa); longitudinal force N = 6000 kN; bending moment М = 3100 kN·m; design length of the element: in the bending plane l0 = 16,2 m, out of the bending planel0 = 10,8 m. It is required to determine the section strength of the element.

The calculation in the bending plane. From example 33 we have: h′f = 15 mm; h0=

=1421 mm; а′ = 79 mm; Ncr = 28 270 kN.

By formula (91) we determine coefficient η: Considering the element deflection value е is equal to:

.13272

79142127,1

106000

103100

22 3

600

0 mmah

N

Mahee =

−+

⋅

⋅=

′−+=

′−+= ηη

Let’s check condition (130):

,6000245110245121560019 3kNNkNHhbR ffb =<=⋅=⋅⋅=′′

That means the calculation is to be made as for the I-section. Area of compressed flange overhangs is equal to: mm2.

Let’s determine values αn, αm1, αov, αm,ov, δ:

From Table 18 we find ξR = 0,523.

As ξ = αn – αov = 1,111 – 0,302 = 0,809 > ξR = 0,523 so reinforcement area is to be determined

by formula (135). For that we determine values αs and 0

1h

x=ξ by formulas (136) and (132)

( ) ( ).292,0

055,01

279,02/809,01809,0037,1

1

2/1 ,1=

−

−−−=

−

−−−=

δ

αξξαα

ovmm

s

From Table 18 we find ψс = 3,0 and ω = 0,698.

;18,02

111,1302,0292,03292,0

2=

−+⋅+=

−++ novscs αααψα

,602,0698,03292,018,0

18,022

2

2

1

=⋅⋅++

+−=+

−+++

−++−= ωψα

αααψααααψαξ cs

novscsnovscs

So

( ) ( ).5278

055,01

279,02/602,01602,0037,1

365

142120019

1

2/12,1110 mm

R

bhRAA

ovmm

s

b

ss =−

−−−⋅⋅=

−

−−−=′=

δ

αξξα

We take As = A′s = 5630 mm2 (7 ∅ 32). The calculation out of the bending plane is to be made similar to Example 33. RING SECTIONS

110

Example 35. Given: section with internal radius r1 = 150 mm, external radius r2 = 250 mm;

heavy-weight concrete В25 (Rb = 16 MPa by γb2 = 1,1); longitudinal reinforcement A-III (Rs =

=Rsc = 365 MPa); its section area As,tot = 1470 mm2 (13 ∅ 12); longitudinal force from total load N = 1200 kN, its eccentricity relating to the center of gravity of the section considering the element deflection is е0 = 120 mm. It is required to check the section strength. Calculation. Let’s calculate the area of the ring section: mm2 Relative area of concrete compressed zone:

;502,014703657,260012516

1470365101200

7,2

3

,

,=

⋅⋅+⋅

⋅+⋅=

+

+=

totssb

totss

cirARAR

ARNξ

mm.

As 0,15 < ξcir = 0,502 < 0,6 so section strength is to be checked according to condition (138):

( ) ( )( )

( )( )

( )( )

,14412,012004,175104,175

502,03,12,0502,07,11200147036514,3

502,0180sin200147036520060012516

3,12,07,11sin

0

6

,,

mkNNemkNmmN

rARrARArR circirstotss

cir

stotssmb

⋅=⋅=>⋅=⋅⋅=

=⋅+⋅−⋅⋅+⋅

⋅⋅+⋅⋅=

=+−++

o

ξξπ

ξπ

that is section strength is provided. ROUND SECTIONS Example 36. Given: a section with diameter D = 400 mm; а = 35 mm; heavy-weight concrete

В25 (Rb = 13 MPa by γb2 = 0,9; Eb = 2,7 · 104 MPa); longitudinal reinforcement A-III (Rs =Rsc=

= 365 MPa; Es = 2 · 105 MPa); its section area As,tot = 3140 mm2 (10 ∅ 20); longitudinal forces and bending moments: from dead loads and long-term loads Nl = 400 kN·m; from all loads N = =600 kN, М = 140 kN·m; design length of the element l0 = 4 m. It is required to examine the section strength. Calculation. Let’s calculate:

Area of the round section AD

= =⋅

=π 2 2

4

3 14 400

4125 600

, mm2;

Radius of the section inertia iD

= = =4

400

4100 mm;

Element elasticity .i/l 1440100

40000 >==

So the calculation is made considering deflection of the element according to 3.54 and value Ncr is to be determined by formula (92). For that we determine:

rD

as = − = − =2

400

235 165 mm;

695,1165,0600140

165,04001001111

1

1 =⋅+

⋅++=

+

++=+=

s

slll

lNrM

rNM

M

Mββϕ

[здесь β = 1,0 (see Table 16)];

233233,0600/140/0 ==== “NMe mm

As e D0

233

400/ = = 0,583 > δe,min = 0,5 – 0,01 l0/D – 0,01 Rb so we take δe = e0/D = 0,583.

Inertia moments of the concrete section and all reinforcement are:

ID

= =⋅

= ⋅π 4 4

6

64

3 14 400

641256 10

, mm4;

IA r

s

s tot s= =

⋅= ⋅

,,

2 2

6

2

3140 165

242 74 10 mm4;

111

α = =⋅

⋅=

E

E

s

b

2 10

2 7 107 4

5

4,, .

Then

] .55051055051074,424,7

1,0583,01,0

11,0

695,1

101256

4000

4,107,24,61,0

1,0

11,04,6

36

6

22

0

kNH

II

l

EN s

el

b

cr

=⋅=⋅⋅+

+

+

+

⋅⋅⋅=

+

+

+= α

δϕ

Coefficient η we determine by formula (91):

η =

−

=

−

=1

1

1

1600

5505

112N

Ncr

, .

Section strength is to be examined by means of the diagram of Draft 41.

According to values α n

b

N

R A= =

⋅

⋅=

600 10

13 125 6000 367

3

, ,

α s

s s tot

b

R A

R A= =

⋅

⋅=

, 365 3140

13 125 6000,702 и

a

D= =

35

4000 0875, we find at the diagram αm = 0,51.

As αmRbAr = 0,51 · 13 · 125 600 · 200 = 167 · 106 N·mm = 167 kN·m ×⋅=> 233.06000ηNe

6.15612.1 =× kN·m, the section strength is provided. Example 37. Due to the data of example 36 it is necessary to choose reinforcement using the diagram of Draft 41. Calculation. From example 36 i = 100 mm, А = 125 600 mm2, rs = 165 mm. As l0/i = =4000/100 40 > 35 so we choose reinforcement considering the element deflection determining value Ncr by formula (92). In the first approximation we have As,tot = 0,01 A = 1256 m2, thereafter

IA r

s

s tot s= =

⋅= ⋅

,,

2 2

6

2

1256 165

217 1 10 mm4.

From example 36 ϕl = 1,695, δe = 0,583, I = 1256 · 106 mm4. Then

( ) .3455103455105,126104,1930108,0

101,174,71,0583,01,0

11,0

695,1

101256

4000

107,24,61,0

1,0

11,04,6

366

66

2

4

2

0

kNH

II

l

EN s

el

b

cr

=⋅=⋅+⋅=

=

⋅⋅+

+

+

⋅⋅⋅=

+

+

+= α

δϕ

Coefficient is η =

−

=

−

=1

1

1

1600

3455

1 21N

Ncr

, .

Due to values α n

b

N

R A= =

⋅

⋅=

600 10

13 125 6000 367

3

, ,

αη

m

b

Ne

R Ar= =

⋅ ⋅ ⋅

⋅ ⋅=0

3600 10 233 1 21

13 125 600 2000 518

,, we find αs = 0,74, thereafter

AR A

Rs tot s

b

s

, ,= =⋅

=α 0 7413 125 600

3653310 mm2.

As determined reinforcement is more than the one accepted in the first approximation (As,tot = =1256 mm2) so value As,tot = 3310 mm2 is determined with „reserve” and it can be a bit decreased after value Ncr is specified more exact.

We take As tot, =+

=1256 3310

22283 mm2 and make similar calculation:

I s =⋅

= ⋅2283 165

231 08 10

2

6, mm4;

112

( )N cr = ⋅ + ⋅ ⋅ =0 0108 193 4 10 7 4 31 08 10 45736 6, , , , kN;

η =

−

=1

1600

4573

1151, .

Due to values αm = =0 5181151

1 210 493,

,

,, , αn = 0,367 and

a

D= 0 1, at the diagram of Draft 41 we find

αs = 0,68.

As tot, ,=⋅

=0 6813 125 600

3653042 mm2.

We take As,tot = 3142 mm2 (10 ∅ 20). ELEMENTS WORKING IN SKEW BENDING Example 38. Given: rectangular section of the column with dimensions b = 400 mm, h = 600

mm; heavy-weight concrete В25 (Rb = 16 MPa by γb2 = 1,1); longitudinal reinforcement A-III (Rs = Rsc = 365 MPa) located in the section due to Draft 50; in the section act both longitudinal force N = 2600 kN and bending moments at the same time: in the plane parallel to the dimension h, – Mx = 240 kN·m and in the plane parallel to dimension b, – My = 182,5 kN·m; moments Мх and Мy are given considering the column deflection. It is required to examine the section strength.

Черт. 50. For the calculation examples 38 and 40

I borders of the compressed zone in the first approximation; II final border of the compressed zone

Calculation. The strength is to be checked according to Item 3.74. Symmetry axes which are

parallel to dimensions h and b we symbolize x and y. Let’s determine limit moments M x

0 and

M y

0 . For that we determine distributed reinforcement Asx и Asy. Due to Draft 50 As1, x = 0, nx =

=0, As0 = 804,3 mm2 (∅ 32), As1, y = 314,2 mm2 (∅ 20),

β = = =M

M

h

h

x

y

x

y

240

182 5

400

6000 877

,, .

( ) ( ) ( ) ;605877,01

877,02,3243,80420

121 2

,1,10,1 mmAAAnAA ysxssxxssx =+

−⋅+=+

−−++=β

β

As tot, = 3845 mm2 (4 ∅ 32 + 2 ∅ 20);

AA

Asy

s tot

sx= − = − =,

2

3845

2605 1318 mm2.

When determining moment M x

0 which acts in the plane of axis х due to Item 3.63 we take:

Asl = Asy = 1318 mm2; Ast = Asx = 605 mm2; h = 600 mm; b = 400 mm. δ1 1 50 600 0 083= = =a h/ / , ;

R bhb = ⋅ ⋅ = ⋅16 400 600 3840 103 N;

( ) ( )α

δsl

s sl

b

R A

R bh=

−=

⋅

⋅ −=

0 5

365 1318

3840 10 0 5 0 0830 3

1

3, , ,

, ;

α st

s st

b

R A

R bh= =

⋅

⋅=

365 605

3840 100 063 , ;

α n

b

N

R bh1

3

3

2600 10

3840 100 677= =

⋅

⋅= , .

From Table 18 we find ω = 0,722, ξR = 0,55.

113

As ξα α

α ω=

+

+=

+

+ ⋅=n sl

sl

1

1 2

0 677 0 3

1 2 0 3 0 722/

, ,

, / ,0,534 < ξR = 0,55 so value M x

0 is to be determined by

formula (117) due toξξ

ω1

0 534

0 722= =

,

,= 0,74:

( ) ( )[ ( ) ( )]( )[ ( )( )

( )] .7,464107,464083,02106,074,03,005,0

083,074,01083,074,03,0534,05,01534,05,0600103840

2105,015,015,0

62

3

1

2

11111

20

mkNmmN

bhRM stslslbx

⋅=⋅⋅=⋅−+⋅⋅−

−−−−+⋅−⋅⋅⋅=

=−+−−−×−+−= δαξαδξδξαξξ

When determining moment M y

0 which acts in the plane of axis y, we take: Asl = Asx = 605

mm2; Ast = Asy = 1318 mm2; h = 400 mm; b = 600 mm; δ1

1 50

4000 125= = =

a

h, .

( ) ( );,

,,,bhR

AR

b

slssl 1530

125050103840

605365

50 31

=−⋅

⋅=

δ−=α

α st

s st

b

R A

R bh= =

⋅

⋅=

365 1318

3840 100 1253 , .

As ξα α

α ω=

+

+=

+

+ ⋅=n sl

sl

1

1 2

0 677 0 153

1 2 0 153 0 722/

, ,

, / ,0,583 > ξR = 0,55 so value M y

0 is to be determined by

formula (118) calculating:

α na

s s tot

b

R A

R bh= + = +

⋅

⋅=1 1

365 3845

3840 101 3653

,, ;

ξξ

ω1

0 55

0 7220 762R

R= = =,

,, ;

( ) ( ) ( ) ( )

( ) ( ) ( )( ) ;,,,

,,,,,,,,,,,

,, stRslRRslRRmR

22401250211250

76201530050125076201125076201530550155050

210501150

2

1211111

=⋅−+

+⋅⋅−−−×−+−⋅=

=δ−α+ξα−δ−ξ−×δ−ξα+ξ−ξ=α

( ) ( )α ξ α ξnR R sl R= + − = + ⋅ − =2 1 0 55 0 153 2 0 762 1 0 631 , , , , .

.3221032263,0365,1

677,0365,1224,0400103840 63120 mkNmmNbhRM

nRna

nna

mRby ⋅=⋅⋅=−

−⋅⋅⋅=

−

−=

αα

ααα

As 365,0,

==bhR

AR

b

totss

sα so section strength is to be checked according to the diagram of Draft

42, а, б corresponding to αs = 0,2 and αs = 0,4. At both diagrams the point with coordinates

M Mx x/ 0 = 240/464,7 = 0,516 and M My y/ 0 = 182,5/322 = 0,566 lies inside of the area bounded

by means of the curve corresponding to parameter αn1 = 0,677 and by coordinate axes. That means the section strength is provided. Example 39. Given: the column section, materials characteristics and values of longitudinal

forces from all loads see Example 33; in the section at the same time there are bending moments in the plane parallel to the dimension h, – Mx = 3330 kN·m and in the plane parallel to dimension b, – My = 396 kN·m; moments Мх and Мy are given considering the column deflection. It is required to check the section strength. Calculation. The strength is to be checked according to Item 3.75. Let’s determine limit

moment M x

0 acting in the plane of the symmetry axis х going in the rib. Due to Example 33 the

right part of condition (131) is 5847 kN·m, and so ( )

.41702

079,0421,125005847

25847 00

mkNahN

M x ⋅=−

−=′−

−=

114

Limit moment M y

0 acting in the plane of the symmetry axis y normal to the rib we determine

as for a rectangular section which consists of two flanges due to Item 3.63. So according to Draft 49 we have: h = 600 mm; b = 2 · 215 = 430 mm. Let’s determine distributed reinforcement Asl и Ast:

As l1 804 3, ,= mm2 (∅ 32); ηl = 3;

As tot, = 11 260 mm2 (14 ∅ 32);

Asl = As1, l (nl + 1) = 804,3 (3 + 1) = 3220 mm2; Ast = As, tot/2 – Asl = 11 260/2 – 3220 = 2410 mm2.

From Table 18 we find ω = 0,698 and ξR = 0,523. Rbbh = 19 · 430 · 600 = 4902 · 103 N;

δ1 = a1/h = 0,083;

( ) ( )α

δsl

s sl

b

R A

R bh=

−=

⋅

⋅ −=

0 5

365 3220

4902 10 0 5 0 0830 576

1

3, , ,

, ;

α n

b

N

R bh1

3

3

2500 10

4902 100 51= =

⋅

⋅= , ;

α st

s st

b

R A

R bh= =

⋅

⋅=

365 2410

4902 100 1793 , ;

ξα α

ωξ=

+

+=

+

+ ⋅= < =n sl

sl

R

1

1 2

0 51 0 576

1 2 0 576 0 6980 41 0 523

/

, ,

, / ,, , .

Value M y

0 is to be determine by formula (117) after calculatingξξ

ω1

0 41

0 6980 59= = =

,

,, :

( ) ( )[ ( ) ( )]( )[ ( )( )

( )] mkNммН

bhRM stslslby

⋅=⋅⋅=⋅−+⋅⋅−

−−−−+−⋅⋅⋅=

=−+−−−×−+−=

1029101029083,021179,059,0576,005,0

083,059,01083,059,0576,041,0141,05,060043019

2105,0115,0

62

2

12

1111120 δαξαδξδξαξξ

Let’s check the section strength taking b = 200 mm, h = 1500 mm.

As α s

s s tot

b

R A

R bh= =

⋅

⋅ ⋅=

,, ,

365 11 260

19 200 15000 721 so section strength is to be checked according to

diagrams of Draft 44, б, в, corresponding to αs = 0,6 и αs = 1,0.

At both diagrams the point with coordinates M Mx x/ 0 = 3330/4170 = 0,8 and M My y/ 0 =

=396/1029 = 0,385 lies within the area which is bounded by a curve corresponding to the

parameter αn1 = N/(Rbbh) = 2500 · 103/(19 · 200 · 1500) = 0,44 and by coordinate axes. That means the section strength is provided. Example 40. Given: rectangular section of the column with dimensions b = 400 mm, h = 600

mm; heavy-weight concrete В25 (Rb = 16 MPa by γb2 = 1,1); longitudinal reinforcement A-III (Rs = 365 MPa) due to Draft 50; in the section at the same time there is the longitudinal force N = 2600 kN and bending moments acting in the plane parallel to dimension h, – Мх = 250 kN·m and in the plane parallel to dimension b, My = 200 kN·m; bending moments Мх and My are given considering the column deflection. It is required to examine the section strength using formulas of Item 3.76 for the general calculation case. Calculation. All rods symbolized by numbers as it’s shown on Draft 50. Through the center of gravity of rod 5 we draw an axis х parallel to dimension h = 600 mm and axis y parallel to dimension b.

Angle θ between axis y and the line bounding the compressed zone we take as by the calculation of elastic body as regards the; that means:

115

tgM

M

I

I

M

M

h

b

y

x

x

y

y

x

θ = = =

=

2 2200

250

600

40018, .

Taking value х1 with dimensions of the compressed zone along the section side h for each

rod it is possible to determine the ration ξi = x/h0i by formula ξθi

yi xi

x

a tg a=

+1

, where axi and ayi

are distances from the i-rod to the most compressed side of the section in the lines of axes х and у.

Due to values ξi we determine stress σsi, taking σsc,u = 400 MPa, ω = 0,722 (see Table 18):

−

ξ=

−

ξ×

−

=

−

ξ

ω

ω−

σ=σ 1

722011601

7220

11

72201

4001

111 iii

u,scsi

,,

,

,

,

(MPa)

At the same time if σsi > Rs = 365 MPa, what is equivalent to condition ξi < ξR = 0,55 (see

Table 18) so we take σsi < Rs = 365 MPa.

If σsi < –Rsc = –365 MPa we take σsi = –365 MPa.

The last condition after we insert the equations for σsi into it looks like:

ξi >

−

=0 722

1365

1160

1 054,

, .

Then we determine the sum of forces in all rods ∑Asiσsi. We take equation x1 = h = 600 mm in the first approximation and make mentioned calculations; the results of the calculations are given in the following table:

Number of the rod

Asi, ayi, axi, ayitgθ + axi, х1 = 600 mm х1 = 660 mm

mm2 mm mm mm

(tgθ = 1,8) ξi σsi,

MPa Asiσsi, H ξi σsi,

MPa Asiσsi, N

1 804,3 350 50 680 0,882 –210 –168 900 0,971 –297 –238 877

2 804,3 50 50 140 4,29 –365 –293 570 4,714 –365 –293 570

3 314,2 350 300 930 0,645 138 43 360 0,71 20 6284

4 314,2 50 300 390 1,54 –365 –114 683 1,692 –365 –114 683

5 804,3 350 550 1180 0,508 365 293 570 0,56 339 272 658

6 804,3 50 550 640 0,937 266 213 944

∑Asiσsi = –26 280 N

1,031 –348 –279 896

∑Asiσsi = –648 080 N

As x

tg

1 600

18333

θ= =

, mm < b = 400 mm so compressed zone has triangle form and its area is:

Ax

tgb = =⋅

=12 2

2

600

2 1 8100 000

θ , mm2.

Let’s check condition (154):

,260016261016262802600010016 3kNNkNNAAR sisibb =<=⋅=+⋅=Σ− σ

That means area of compressed zone is decreased.

Let’s increase value х1 up to 660 mm and determine ∑Asiσsi (see the table of the present example).

116

By х1 > h and х1/tgθ = 660/1,8 = 367 mm < b = 400 mm the compressed zone has a trapezoid form and its area is:

( ) ( ).1001201000100121

8,12

600660

2

367660

22

222

1

2

1 mmtg

hx

tg

xAb =−=

⋅

−−

⋅=

−−=

θθ

As RbAb – ∑Asiσsi = 16 · 120 100 + 648 080 = 2570 · 103 N = 2570 кН ≈ N = 2600 kN, so condition (154) is met. Let’s determine moments of internal forces relating to axes у and х. For that let’s determine static moments of the section area of compressed zone relating to these axes:

( )(

;000036406003

6006605501000

3

660550100121

3

232

31

5

2

115

2

1

mmhhx

atg

hxxa

tg

xS xxbx

=

+

−+−+

−=

+

−+

+−−

+

−=

θθ

( )

.000912278,13

600660350100

3

367350100121

323

/

2

3

15

2

115

2

1

mm

tg

hxa

tg

hxtgxa

tg

xS yyby

=

⋅

−−−

−=

=

−−

−−

−=

θθ

θ

θ

Then ( ) ( ) ×−−−−⋅=−Σ−= 29357050550238877[40036000165 xixsisibxbxu aaASRM σ

( ) ( ) ( ) 6109.933]300550114683300550628450550 ⋅=−−−+−× N · mm = 934 kN · m;

( ) ( )[

( ) ( )] .653106535035089627950350683114

503505702930009122716

6

5

mkNmmN

aaASRM yiysisibybyu

⋅=⋅⋅=−−−−

−−−−⋅=−Σ−= σ

Moments of external forces relating to axes у and х are:

;10900502

6001026001025050

2

636

1 mmNh

NMM xx ⋅⋅=

−⋅+⋅=

−+=

.10590502

4001026001020050

2

636

1 mmNb

NMM yy ⋅⋅=

−⋅+⋅=

−+=

As Mxu >Mx1, а Myu > My1 so section strength is provided. CACLULATION OF INCLINED SECTIONS Example 41. Given: a column of a multistory frame work with the section dimensions b = 400

mm, h = 600 mm; а = а′ = 50 mm; heavy-weight concrete В25 (Rbt = 0,95 MPa by γb2 = 0,9); stirrups located along the column surfaces made of reinforcement A-III, diameter 10 mm (Rsw

= 255 MPa; Asw = 157 mm2), spacing s = 400 mm; bending moments in the top and the bottom support sections are Msup = 350 kN·m, Minf = 250 kN·m and stretch the left and the right surface of the column; longitudinal force N = 572 kN; column length (distance between support sections) l = 2,8 m. It is required the strength of inclined sections of the column as regards the shear force. Calculation. h0 = h – a = 600 – 50 = 550 mm. The calculation is made according to Item 3.31 considering recommendations of Item 3.53. Shear force in the in the column is:

QM M

l=

+=

+=

sup inf

,

350 250

2 8214 kN.

As shear force is constant along the column length so the projection length of inclined section is taken maximum possible; tat is equal to

8,2833,155,06,0

20

3

2max =<=== l“hc

b

b

ϕ

ϕm

117

Let’s determine coefficient ϕn:

ϕ ϕn

bt

f

N

R bh= =

⋅ ⋅= < =0 1 0 1

572 000

0 95 400 5500 27 0 5 0

0

, ,,

, , ; .

As с = сmax, Qb = Qb,min = ϕb3 (1 + ϕn)Rbtbh0 = 0,6 (1 + 0,27)0,95 · 400 · 550 = 159,2 · 103 H < <Q = 214 kN so the stirrups are required due to the calculation. Value qsw is to be determined by formula (55):

qR A

ssw

sw sw= =⋅

=255 157

400100 1, N/mm.

Let’s check condition (57):

1,100/7,1445502

102,159

2

3

0

min,=>=

⋅

⋅= sw

bq““н

h

QN·mm

As condition (57) is not met so we determine value by the following formula

M h qb sw b b= 2 02

2 3ϕ ϕ/ ,

So

3

0

320

32

2

0 101,1101,10055022/

/2⋅=⋅⋅==== sw

bb

bbswb

b qhh

qh

c

MQ

ϕϕ

ϕϕN

с0 is taken equal to с0 = 2h0 = 2 · 550 = 1100 mm, and Qsw = qswc0 = 100,1 · 1100 = 110,1 · 103 N. Let’s check condition (50):

214102,220101,110101,110 333 =>⋅=⋅+⋅=+ QHQQ swbkN

That means the section strength as regards the shear force is provided. Центрально- и внецентренно растянутые элементы

CENTRALLY TENSILE ELEMENTS

- (3.26). During calculation of eccentric tensile reinforced concrete elements the following condition must be met:

N R As s tot≤ , , (156)

where As,tot section area of total longitudinal reinforcement.

ECCENTRICTENSILE ELEMENTS

CALCULATION OF RECTANGULAR SECTIONS NORMAL TO THE LONGITUDINAL AXIS OF THE ELEMENT, IF LONGITUDINAL FORCE IS LOCATED IN THE SYMMETRY AXIS - (3.27). Calculation of rectangular sections of eccentric compressed elements with

reinforcement concentrated at the most tensile and at compressed (the lest tensile) surfaces must be made according to the location of longitudinal force N:

a) if longitudinal force N is applied between resultant forces in reinforcement S and S′ (Draft 51, а), that means by е′ ≤ h0 – a′, so the calculation is made due to the following conditions:

( )Ne R A h as s′ ≤ − ′0 ; (157)

( )Ne R A h as s≤ ′ − ′0 ; (158)

118

b) if longitudinal force N is applied beyond the area between resultant forces in

reinforcement S and S′ (Draft51, б), that means by е′ > h0 – a′, so the calculation is made due to the following condition

( ) ( )Ne R bx h x R A h ab sc s≤ − + ′ − ′0 00 5, , (159)

at the same time the height of compressed zone х is determined by formula

xR A R A N

R b

s s sc s

b

=− ′ −

. (160)

Draft 51. Forces scheme and diagram of stresses in the section normal to longitudinal axis of eccentric

tensile reinforced concrete element by its calculation as regards the strength

а longitudinal force N applied between resultant forces in reinforcement S and S′; б the same, beyond the area between resultant forces in reinforcement S and S′

If calculated by formula (160) value х > ξRh0, so it is necessary to insert value х = ξRh0

where ξR is determined by Tables 18 and 19 into the formula (159). If х < 0so section strength is to be checked according to condition (157). By symmetrical reinforcement the strength is examined according to condition (157)

independently of value е′ . Note . If by e′ > h0 – a′ the height of compressed zone determined without considering compressed

reinforcement xR A N

R b

s s

b

=−

, is less than 2а′ so it is possible to increase design bearing capacity after

calculations by formulas (159) and (160) without considering compressed reinforcement.

- Required quantity of longitudinal reinforcement is determined in the following manner:

a) by e′ ≤ h0 – a′ it’s determined section area of reinforcement S and S′ by the following formulas:

( )A

Ne

R h as

=′

− ′0

; (161)

( )′ =

− ′A

Ne

R h as

s 0

; (162)

б) by e′ > h0 – a′ it’s determined section area of stretched reinforcement As by formula

Abh R N

RA

R

Rs

b

s

s

sc

s

=+

+ ′ξ 0

, (163)

where ξ is taken by Table 20 due to value

( )αm

sc s

b

Ne R A h a

R bh=

− ′ − ′0

0

2 . (164)

At the same time condition αm ≤ αR must be met (see Table 18 and 19). Otherwise it is necessary to increase the section of compressed reinforcement ′As

, to increase concrete

class or to increase the section dimensions.

If αm < 0 so section area of tensile reinforcement As is determined by formula (161).

Symmetrical reinforcement area independently of value е′ is chosen due to formula (161). Note . By е′ > h0 – a′ required quantity of reinforcement determined by formula(161) can be decreased if

value ξ, determined due to Table 20 without considering compressed reinforcement, that means due to

119

value αm

b

Ne

R bh=

02 , is less than 2а′/h0. In that case section area of stretched reinforcement As is determined

by the following formula

( )A

N e h

R hs

s

=+ζ

ζ0

0

, (165)

where ζ is determined by formula 20 according to value αm

b

Ne

R bh=

0

.

GENERAL CASE OF CALCULATION OF NORMAL SECTIONS OF ECCENTRIC COMPRESSED ELEMENT (BY ANY SECTIONS, EXTERNAL FORCES AND ANY REINFORCEMENT) - Calculation of sections of the eccentric tensile element in general case (see Draft 45)

must be made due to the following condition

Ne S R Ssi si b b′ ≤ −Σσ , (166)

where ′e distance from the longitudinal force N to the axis parallel to the line which bounds the compressed zone and going through the most distant from the mentioned line point of compressed zone;

Sb static moment of the concrete compressed zone area relating to the mentioned axis;

Ssi static moment of the section area of the longitudinal reinforcement i-rod relating to the mentioned axis;

σsi stress in the i-rod of longitudinal reinforcement.

Compressed zone height х and stresses σsi are determined due to the combined solution of equations (154) and (155) replacing sign „minus” by sign „plus” in front of N. Except formulas (154) and (155) by skew eccentric tension it is necessary to meet additional requirement to determine the location the concrete zone borders: points of application of external longitudinal force, the resultant of compression forces in concrete and reinforcement and the resultant of forces in tensile reinforcement must belong to the straight line (see Draft 45).

CALCULATION OF SECTIONS INCLINED TO THE LONGITUDINAL AXIS OF THE ELEMENT - Calculation of inclined sections of eccentric tensile elements as regards the shear force is

made as for bending moments according to Items 3.283.41. At the same time value Mb in Item 3.31 is determined by the following formula

( )M R bhb b f n bt= + −ϕ ϕ ϕ2 021 , (167)

where ϕn

bt

N

R bh= 0 2

0

, , but no more than 0,8;

value Qb,min is taken equal to ϕb3 (1 + ϕf – ϕn)Rbtbh0. Besides in all formulas of Items 3.29,

3.40 and 3.41 coefficient ϕb4 is replaced by ϕb4 (1 – ϕn). Calculation of inclined sections of eccentric tensile elements as regards bending moment

is made as for bending elements in compliance with Items 3.423.45. At the same time the height of compressed zone in inclined section is determined considering tensional force N by formula (160) or due to Item 3.80.

120

In case if condition e′ < h0 – a′ is met design moment in inclined section can be determined as the moment of external forces located on one side of the inclined section

under review relating to the axis going through the center of gravity of reinforcement S′.

EXAMPLES OF CALCULATION Example 42. Given: a stretched leg of a two-leg column with the cross section dimensions

b = 500 mm, h = 200 mm; а = а′ = 40 mm; longitudinal reinforcement A-III (Rs = Rsc = 365

MPa); its section area As = A′s = 982 mm2 (2 ∅ 25); heavy-weight concrete В25 (Rb = 16 MPa

by γb2 = 1,1); longitudinal force N = 44 kN; maximum bending moment М = 43 kN · m. It is required to examine the strength of the normal section. Calculation. h0 = 200 – 40 = 160 mm;

eM

N0

6

3

43 10

44 10977= =

⋅

⋅= mm;

′ = + − ′ = + − =e eh

a0 2977

200

240 1037 mm;

e eh

a= − + = − + =0 2977

200

240 917 mm.

As there is symmetrical reinforcement so the strength is examined due to condition (157):

( ) ( ) mmNeNmmNahAR ss ⋅⋅=⋅⋅=′<⋅⋅=−⋅=′− 636

0 106,4510371044101,4040160982365

That means condition (157) is not met. As е′ = 1037 mm > h0 – a′ = 120 mm, and the height of compressed zone х, determined by formula (160) without considering compressed reinforcement is

mma““bR

NARx

b

ss 8040224050016

1044982365 3

=⋅=′<=⋅

⋅−⋅=

−=

according to the note to Item 3.78 we check the strength according to Formula (159), taking

х = 40 mm and A′s = 0:

( ) ( ) mmNNemmNxhbxRb ⋅⋅=⋅⋅=>⋅⋅=⋅−⋅⋅=− 636

0 104,409171044106,40405,016040500165,0

that is the strength the normal section is provided.

Example 43. Given: a rectangular section with dimensions b = 1000 mm, h = 200 мм; а = а′ =

=35 mm; heavy-weight concrete В15 (Rb = 7,7 MPa by γb2 = 0,9); longitudinal reinforcement

A-III (Rs = Rsc = 365 MPa); reinforcement section area S′ A′s = 1005 mm2; tension force N = =160 kN; bending moment М = 116 kN·m. It is required to determine section area of reinforcement S. Calculation. h0 = 200 – 35 = 165 mm;

eM

N0

6

3

116 10

160 10725= =

⋅

⋅= mm;

e eh

a= − + = − + =0 2725

200

235 660 mm;

′ = + − ′ = + − =e eh

a0 2725

200

235 790 mm.

As е′ = 790 mm h0 – а′ = 165 – 35 = 130 mm so let’s determine required section area of stretched reinforcement according to Item 3.796. Let’s determine the following value:

( ) ( ).276,0

16510007,7

351651005365660101602

3

2

0

0 =⋅⋅

−⋅−⋅⋅=

′−′−=

bhR

ahARNe

b

ssc

mα

As 0 < αm < αR = 0,44 (see Table 18) so value As is to be determined by formula (163). For that

we find ξ = 0,33 by αm = 0,276 according to Table 20.

121

.25921005365

101607,7165100033,0 23

0 mmR

RA

R

NRbhA

s

sc

s

s

b

s =+⋅+⋅⋅⋅

=′++

=ξ

We take As = 3079 mm2 (5 ∅ 28).

Example 44. Given: a rectangular section with dimensions b = 1000 mm, h = 200 мм; а = а′ =

=40 mm; heavy-weight concrete В15 (Rb = 7,7 MPa by γb2 = 0,9); longitudinal reinforcement A-III (Rs = Rsc = 365 MPa); tension force N = 532 kN; bending moment М = 74 kN·m. It is required to determine section area of symmetrical longitudinal reinforcement. Calculation. h0 = h – a = 200 – 40 = 160 mm;

eM

N0

6

3

74 10

532 10139= =

⋅

⋅= mm;

e eh

a= − + = − + =0 2139

200

240 79 mm;

′ = + − ′ = + − =e eh

a0 2139

200

240 199 mm.

As it is symmetrical reinforcement so section area is to be determined by formula (161):

( ) ( )A A

Ne

R h as s

s

= ′ =′

− ′=

⋅ ⋅

−=

0

3532 10 199

365 160 402417 mm2.

As е′ = 199 mm > h0 – а′ = 120 mm so according to note of Item 3.79 it is possible to decrease value As.

Let’s determine value ξ without considering compressed reinforcement. For that we determine

value αm:

α m

b

Ne

R bh= =

⋅ ⋅

⋅ ⋅=

0

2

3

2

532 10 79

7 7 1000 1600 213

,, .

Due to Table 20 by αm = 0,213 we find ξ = 0,24 and ζ = 0,88. As ξ = <′

=⋅

=0 242 2 40

1600 5

0

, , ,a

h we

determine value As by formula (165):

( ) ( )A A

N e h

R hs s

s

= ′ =+

=⋅ + ⋅

⋅ ⋅=

ζ

ζ0

0

3532 10 79 0 88 160

365 0 88 1602275

,

, mm2.

We take As = A′s = 2281 mm2 (6 ∅ 22). Example 45. Given: a stretched leg of the two-leg column with the section dimensions b = 500

mm, h = 200 mm; а = а′ = 40 mm; heavy-weight concrete В25 (Rbt = 1,15 MPa by γb2 = 1,1); stirrups located along the leg surfaces made of reinforcement A-III (Rsw = 285 MPa); longitudinal tension force N = 44 kN; shear force Q = 143 kN; the distance between the connecting strips of the two-leg column is l = 600 mm. It is required to determine diameter and spacing of stirrups. Calculation. h0 = h – а = 200 – 40 = 160 mm. The calculation is made according to Item 3.33а considering recommendations of Item 3.81.

Value Mb is to be determined by formula (167) by ϕb2 = 2 (see Table 21), ϕf = 0 and

ϕn

bt

N

R bh= =

⋅ ⋅=0 2 0 2

44 000

115 500 1600

, ,,

0,096 < 0,8:

( ) ( )M R bhb b n bt= − = − ⋅ ⋅ = ⋅ϕ ϕ2 02 2 61 2 1 0 096 115 500 160 26 6 10, , , N·mm

As it is constant shear force between the connection strips of the two-leg column so the projection length of inclined section is taken maximum as it’s possible, that is:

c c hb

b

= = = =max ,

ϕ

ϕ2

30

2

0 6160 533 мм < l = 600 mm.

Then

122

QM

cQb

b

b= =⋅

= ⋅ =26 6 10

53349 9 10

6

3,, ,minΗ

As 2h0 = 2 · 160 = 320 mm < с = 533 mm so we take с0 = 2h0 = 320 mm. Определим коэффициент æ :

æ =−

=−

=Q Q

Q

b

b

143 49 9

49 91 866

,

,, .

As c

c0

533

320= = 1,667 < æ = 1,866 <

c

h0

533

160= = 3,33 so stirrups quantity is determined by

formula (63):

( ) ( )q

Q Q

Msw

b

b

=−

=−

=

2 2143 49 9

26 6325 9

,

,, kN/m.

Maximum allowable stirrups quantity according to Item 3.30 is:

( ) ( )6,139

10143

16050015,1096,015,113

22

04max =

⋅

⋅⋅−=

−=

Q

bhRs btnb ϕϕ

mm

Besides in compliance with Item 5.58 the stirrups quantity must be no more than 2h = 2 · 200 = =400 mm. We take stirrups quantity s = 100 mm < smax, so

Aq s

Rsw

sw

sw

= =⋅

=325 9 100

290112 4

,, mm2.

We take two stirrups with diameter 10 mm (Asw = 157 mm2). Elements working in torsion with bending (spatial sections calculation)

ELEMENTS OF RECTANGULAR SECTION

- (3.37) During calculation of elements working in torsion with bending the following requirement must be met:

T R b hb≤ 0 1 2, , (168)

where b, h are the larger and the less dimensions of the element surfaces. At the same time value Rb for concrete more than В30 is taken as for concrete class В30.

- Spatial sections are calculated as regards combined action of torsion and bending moments by location of compressed zone at the element surface perpendicular to the plane of acting of bending moment (Scheme 1 of Draft 52).

Besides, spatial sections are calculated as regards combined action of torsion moments and shear forces by location of compressed zone at the element surface parallel to the plane of acting of bending moment (Scheme 2 of Draft 53).

Draft 52. Forces scheme in spatial section of the 1

st scheme

Draft 53. Forces scheme in spatial section of the 2

nd scheme

- Calculation of the spatial section according to the 1st scheme is made due to the following

condition

( )T Mb

cR A

b

cq c h xs s sw+ ≤ +

−

1

1

1

1 1 1 0 10 5δ , , (169)

123

At the same time value RsAs1 is taken no more than 20 51

0 1

q bM

h xsw +− ,

, and value qsw1 no

more than 1 5

0 51

0 1

,

,.

bR A

M

h xs s −

−

In condition (169):

с1 The length of the projection onto the longitudinal axis of the element of the line bounding the compressed zone of the spatial section; the most disadvantageous value с1 in general case is determined by means of step-by-step approximations and

is taken no more than 2h + b and no more than b2

1δ, at the same time the spatial

section must not be beyond the element borders and its part with one-valued and zero value Т;

As1 section area of all longitudinal rods located at the stretched surface with width b;

qsw1 the force in cross rods located at the stretched surface with the width b per a unit length of the element equal to:

qR A

ssw

sw sw

1

1

1

= , (170)

where Asw1 is section area of one cross rod;

s1 distance between cross rods;

δ1 2=

+

b

h b. (171)

Torsion moment Т and bending moment М are taken in the cross section going through the center of the spatial section (Draft 54, а). The height of compressed zone х1 is determined by formula

xR A R A

R b

s s sc s

b

1

1 1=− ′

, (172)

where ′As1 section area of all compressed rods located at the surface with the width b.

If х1 < 2а′ in condition (169) so it’s taken х1 = 2а′. If х1 > ξRh0 (where ξR see item 3.14) so it is necessary to examine the strength of the normal section according to Item 3.15. Condition (169) must be also met if in the quality of values As1 and Asw1 we take section areas of longitudinal and cross reinforcement located in the compressed by bending zone; in that case value М is taken with sign „minus”. Note . Limitation for value RsAs1 in condition (169) can be used in formula (172) that can cause the increase of design bearing capacity.

Draft 54. Determination of the bending and torsion moments of the shear force acting in the spatial section

а 1st scheme; б 2nd scheme

- The strength as regards longitudinal reinforcement located at stretched by bending

surface (1st Scheme) should be examined: а) for continuous beams and consoles by location of the spatial section at the support as well as for any elements loaded by concentrated forces and torsion moments by location

124

of the spatial section at points of application of these forces and moments on the side of

the part with larger bending moments (Draft 55) according to the following condition

( )( )

( )R A h x M

T Qb

q b h xs s

sw

1 0 1

2

1 1 0 1

0 50 5

4 0 5− ≥ +

−

−,

,

,,max

δ (173)

where Mmax is maximum bending moment at the beginning of the spatial section;

T, Q torsion moment and shear force in the section with the larger bending moment.

At the same time qsw1 b(h0 – 0,5x1) is taken no more than 0 6

1

,;

T

δ

Draft 55. Location of design spatial sections of the 1st scheme in the beam loaded by concentrated forces

1, 2 design spatial sections;

M1, T1, Q1 design forces for the spatial section 1;

М2, Т2, Q2 the same for the spatial section 2

б) for elements loaded only by distributed load q if in the span section with maximum

bending moment Mmax there is a bending moment Т0, due to the following condition

( )( )

,/25,04

5,02

1011

2

0max101

qtxhbq

TMxhAR

sw

ss−−

+≥−δ

(174)

where t distributed torsion moment per a unit length of the element. The strength as regards longitudinal reinforcement located at the compressed by bending surface should be examined for free supported beams according to the condition (173), taking forces Т and Q in the support section by Mmax = 0. If on the considered parts the following condition is not met

T Qb< 0 5, , (175)

so longitudinal reinforcement can be examined only according to the condition of clear bending (see Item 3.15). The strength as regards cross reinforcement located at any surface with the width b should be examined due to the following condition

( )q b h xT

sw1 0 1

1

0 52 2

− ≥, .δ

(176)

Note . Longitudinal reinforcement determined due to the condition (173) can be decreased if the most disadvantageous spatial section with the projection length с1, equal to:

( )c b

R A h x M

T Qb

s s

1

1 0 12

0 5

0 5=

− −

−

,

,,

max (177)

goes beyond the length of the element or its part with one-valued or zero values Т. In that case the calculation is made by means of general method in compliance with Item 3.84 by decreased projection length с1.

- Calculation of the spatial section as regards the 2nd scheme (see Draft 53) is made due to

the following condition

( )T Qb R Ah

cq c b as s sw+ ≤ +

−0 5 22

2

2 2 2 2, ,δ (178)

At the same time RsAs2 value is taken no more than 2qsw2h and value qsw2 no more than 1 5 2,

.R A

h

s s

In condition (178):

125

As2 section area of all tensile longitudinal rods located at the surface wit the width h, parallel to the bending plane;

с2 projection length on the longitudinal axis of the element of the line bounding compressed zone of the spatial section; the most disadvantageous value с2 is determined by the following formula

( )c h

R A b a

T Qb

s s

2

2 22

2

0 5=

−

+ , (179)

and it’s taken no more than h2

2δand no more than 2b + h; at the same time spatial

section must not go beyond the element and its part with one-valued and zero value Т;

qR A

ssw

sw sw

2

2

2

= ; (180)

where Asw2 section area of one cross rod located at the surface with the width h;

s2 the distance between cross rods located at the surface with the width h;

δ 2 2=

+

h

b h; (181)

а2 the distance from the surface with the width h to the axis of longitudinal rods located at this surface.

Torsion moment Т and cross force Q are taken in the cross section going through the center of gravity of the spatial section (see Draft 54, б). In case if condition (175) is met calculation of the spatial section due to the 2nd scheme is not made. Instead of this calculation it is necessary to make a calculation of inclined sections according to Items 3.31–3.38 without considering bend-up bars. At the same

time in corresponding formulas to shear force Q it is added value 3T

b (where Т is

torsion moment in the same cross section like Q) and value q1 is multiplied by coefficient

1 3+e

b

q (where eq is eccentricity of lateral distributed load q which cause the element

torsion). In case if Т < 0,25Qb so it is possible to consider bend-up bars during calculation of inclined sections.

- Required by the calculation as regards the 2nd scheme density of stirrups A

s

sw2

2

can be

determined by the following formulas:

by ( )

122

50

222

≤δ−

+=ϕ

abAR

Qb,T

ss

t

;hR

AR,

s

At

sw

sssw ϕ= 2

2

2 50 (182)

by 1,75 ≥ ϕt > 1

,hR

AR,

s

At

sw

sssw 22

2

2 50 ϕ= (183)

where T, Q is maximum value of corresponding torsion moment and shear force on the part under review.

By ϕt > 1б75 it is necessary to increase the section area of reinforcement As2 or

dimension of the section b so that condition ϕ t ≤ 1,75 was met.

126

If lateral load is applied within the height of the section and acts towards the stretched zone so vertical stirrups quantity must be increased in comparison with the value determined by formulas (182) and (183) in compliance with the calculation as regards the break according to Item 3.97.

T-, I- AND OTHER SECTIONS WITH RE-ENTRANT CORNER

- Cross section of the element must be divided into several rectangles (Draft 56) at the

same time if the height of flange overhangs or the width of the rib are variable so it is necessary to take their average values.

Draft 56. Dividing into rectangles of sections with re-entrant angles during calculation as regards the

torsion with bending

Dimensions of the cross section must meet the following requirement

T R b hb i i≤ 0 1 2, ,Σ (184)

where hi, bi are the larger and the less dimensions of each rectangle. Besides it is necessary to meet requirement of Item 3.30. If within the section height there are flanges whose top or bottom surfaces are not the prolongations of corresponding surfaces of the element so the calculation of spatial sections is made without considering these flanges as for the element of rectangular section in compliance with Items 3.83–3.87.

- Calculation of a spatial section as regards combined action of torsion and bending

moments (the 1st scheme of Draft 57) is made due to the following condition

( ) ( ),5,05,0 10110

1

1

1

xhbqxhc

bAR

c

bMT wfsw

f

ss

f−+−

′≤

′+ (185)

at the same time value RsAs1 is taken no more than 20 51

0 1

q bM

h xsw f +− ,

.

In condition (185):

b′f, bf The width of compressed and stretched surfaces normal to the bending plane;

с1 the length of projection onto the longitudinal axis of the element of the line bounding compressed zone of the spatial section; value с1 is taken corresponding to the value of the slope angle of a spatial crack to the element axis 45° on all surfaces of the element (without considering х1) by formula

( ) ( )c h b b b b b b h b b bf f f f f1 2 2 2 2= + + + − + ′ − = + + ′ − ,

At the same length с1 must not go beyond the element and its part with one-valued or zero values Т;

As1 section area of all longitudinal rods located in the stretched by the bending zone;

х1 the height of compressed zone determined as for the flat cross section of bending moment (see Item 3.20);

qR A

ssw

sw sw

1

1

1

= ; (186)

Asw1, s1 area of cross rods located in one plane in the stretched by bending, and a spacing of these rods;

h0w the distance from the compressed zone to the resultant of forces in cross rods of the stretched zone.

127

Draft 57. Location scheme of the compressed zone in the spatial section of the 1st scheme of the reinforced

concrete element of T- and I-sections working in torsion with bending

С center of gravity of longitudinal stretched reinforcement

Torsion moment Т and bending moment М in condition (185) are taken in the cross section going through the center of the spatial section. In case of changing of cross rods spacing s1 within the length с1 it is necessary to consider average spacing on the part with the length bf located symmetrically relating to the cross section going through the spatial section. Besides it is necessary to check the strength of the normal section in compliance with Item 3.20. Note . Limitation for value RsAs1 by using of condition (185), can be taken into account during calculation of the compressed zone height х1 that will cause the decrease of design bearing capacity.

- Calculation of the spatial section as regards combined action of the torsion moment and

shear force (2nd scheme, Draft 58) is made due to the condition

( ) ( ),5,05,05,0 20220

2

2min, xbhqxbc

hARQbT wswssf −+−≤+ (187)

at the same time condition RsAs2 is taken no more than 2qsw2h. In condition (187):

bf,min the less width of the element flange or if there is one flange – the width of the rib

As2 area of all longitudinal rods located in the stretched zone by the present scheme;

с2 the length of the projection onto the longitudinal axis of the element of the line bounding compressed zone of the spatial section determined by formula:

c b h bf ov2 2 2= + +,min ,

where bov width of the flange overhang located in the stretched zone, at the same time length с2 must not go beyond the element or its part with one-valued or with the zero values of Т;

х2 the height of compressed zone determined as for the flat cross section of bending element by the present scheme of compressed zone location, at the same time compressed overhang of the flange is not taken into account if it sticks out beyond the surface of flange which has less width or beyond the surface of the rib if there is one flange;

qR A

ssw

sw sw

2

2

2

= ; (188)

Asw2, s2 section area of one cross rod located in the stretched zone by the present scheme along the total height h, and its spacing;

b0, b0w distance from the lateral compressed surface of the flange with the width bf,min to the resultant of forces in longitudinal rods with the area As2 and in the cross rods with the area Asw2.

Draft 58. Location schemes of the compressed zone in the spatial section of the 2nd

scheme of reinforced

concrete element of T-, I- and L-sections working in torsion with bending

С center of gravity of longitudinal stretched reinforcement

Torsion moment Т and shear force Q in condition (187) are taken in the cross section going through the center of the spatial section.

128

In case of changing of the cross rods spacing s2 within the length с2 it is necessary to consider average spacing on the part with the length h located symmetrically relating to the cross section going through the center of the spatial section. Besides it is necessary to check the strength of inclined section in compliance with Item 3.31.

RING SECTION ELEMENTS WITH LONGITUDINAL REINFORCEMENT DISTRIBUTED

ALONG THE CIRCLE

- Dimensions of the ring cross section of the element must meet the following requirement

( )T R r rb≤ −0 08 2

3

1

3, ,π (189)

where r1, r2 are internal and external radius of the ring section. The calculation of the spatial section (Draft 59) is made due to the following condition

T Mb

cM

b

cq r cu sr s+ ≤ + β , (190)

where b, c - is the projection length of the line bounding compressed zone onto the cross section of the element and on its longitudinal axis (see Draft 59). Value b is taken equal to

( )b r rs cir= −2 2

2 2cos ,π ξ (191)

value с is determined due to Item 3.91;

Draft 59 Spatial section of the reinforced concrete element of the ring cross section working in torsion with

bending

ξcir - relative area of the concrete compressed zone determined by formula (137),

or by ξcir < 0,15 – by formula (140) by N = 0;

Мu - limit bending moment by clear bending taken equal to the right part of condition (138) or (139);

qR A

ssr

sw sr= ; (192)

Аsr,s – section area of the rod of spiral (ring) reinforcement and spacing of the spiral lapping (rings spacing);

β - coefficient determined by formula

( ) ( )β

π ξ

πξ

π ξπξ= −

−×

−+

12 1 1

b

rs cir

cir

cir

cir

sincos (193)

or according to Draft 60.

Draft 60. Diagram for determination of coefficient β during calculation of elements of ring cross-section as

regards torsion with bending

Torsion moment Т and bending moment М in condition (190) are taken in the cross section going through the center of the spatial section. Besides, it is necessary to check condition (190) as for clear torsion multiplying value Мu

by the ration 4πr q

R A

s sr

s s tot,

, where As, tot is section area of total longitudinal reinforcement.

Value qsr in condition (190) is taken no more than1 5

21

, ,R A

r

M

M

s s tot

s uπ−

.

129

- Condition (190) is checked for spatial sections where projection length с doesn’t go beyond the part with one-valued and zero value Т and is no more than

( )c r cirmax = −2 12π ξ .

For elements with the constant section as regards the length it is recommended to check spatial sections beginning from the normal section with maximum value Т, and by constant values Т – from the section with maximum value М=Ммах. In the last case the most disadvantageous value с is:

c bM M

T Qb

u=−

−2

2max

/.

For elements with variable section as regards the length it is recommended to check several spatial sections located in different places on the length and by values с, equal to:

c bM M

T

u=−

2 ,

At the same time the projection length must not go beyond the length of the element. Dimensions of the cross section are taken corresponding to the center of the spatial section.

CALCULATION EXAMPLES

Example 46. Дано: a collar beam of the end frame floor of the multistory production building loaded by distributed load q=154,4 kN/m and distributed torsion moments t=34,28 kN·m/m; cross section of the collar beam at the support – due to Draft 61, а; diagram of torsion moments caused by vertical dead loads and long-term loads – due to Draft 61, б; diagram of bending moments and cross forces caused by the most disadvantageous for the support section combination of vertical loads and the wind load – due to Draft 61, в, г; diagram of bending moments caused by the most disadvantageous combination of vertical loads – due to Draft 61, д; heavy-weight concrete В25; longitudinal and cross reinforcement А-III (Rs=Rsc=365 MPa; Rsw=290 MPa).

Draft 61. For the calculation example 46

It is required to choose vertical and horizontal cross rods and to check the strength of the collar-beam as regards combined action of torsion and bending. Calculation. As the section has re-entrant angles so we check condition (184) dividing the section into two rectangles with dimensions 800х320 and 155х250 mm and taking Rb = 13 MPa (that is by γ b2 = 0,9);

( )0 1 0 1 13 320 800 155 2502 2 2, ,R b hb i i = ⋅ ⋅ + ⋅ =∑ 6103114 ⋅, Н·мм > Т = 84 кН·м,

that means condition (184) is met. The calculation of spatial sections is made as for a rectangle sections with dimensions b = 300 mm и h = 800 mm, as bottom surface of the collar-beam and the flange form an angle. As for the support section 0,5Qb = 0,5·460·0,3 = 69 kN·m < Т = 84 kN·m according to Items 3.85 and 3.86 so the calculation of the support section according to the 1st and the 2nd scheme is required. Required quantity of vertical rods according to the calculation as regards the 2nd scheme is to be determined due to Item 3.87. First of all we determine coefficients δ 2 и ϕ t :

δ 2 2

800

2 300 8000 571=

+=

⋅ +=

h

b h, ;

130

( ) 222 22

5,0

δϕ

abAR

QbT

ss

t−

+=

( )=

⋅ + ⋅ ⋅ ⋅

⋅ − ⋅ ⋅=

84 10 0 5 460 10 300

365 2304 300 2 50 2 0 5710 851

6 3,

,, ,

where Аs2 = 1609 + 314 + 380 = 2304 mm2 (2∅32 + ∅20 + ∅22). As ϕ t < 1 so stirrups quantity is to be determined by formula (182):

A

s

R A

R h

sw s s

sw

t

2

2

20 5 0 5

365 2304

290 8000 851 154= =

⋅

⋅=, , , ,ϕ mm.

Taking the spacing of vertical stirrups s2 = 100 mm, we find the area of one stirrup: Asw2 = 1,54 · 100 = 154 mm2.

We take stirrups with diameter 14 mm (Asw2 = 154 mm2). Let’s check the strength as regards the longitudinal reinforcement installed at the top stretched surface of the support part of the collar-beam according to Item 3.85а (the 1st scheme).

Due to Draft 61, а we find As1 = 3217 mm2 (4∅32) and As1/ = 1388 mm2 (2∅20 + 2∅22), а' =

=68 mm. By formula (172) we determine the height of compressed zone х1 taking Rb = 16 MPa (that is by γ b2 = 1.1 as wind load is taken into account):

хR A R A

R b

s s sc s

b

11 1 365 3217 1388

16 300139=

−=

−

⋅=

/ ( )мм > 2а/ = 2 · 68 = 136 мм.

Spacing and diameter of horizontal cross rods of the support part we take the same as for vertical stirrups, that means s1=100 mm, Asw1=154 mm2 so

qR A

ssw

sw sw

11

1

290 154

100446 6= =

⋅= , N/m;

δ1 2

300

2 800 3000158=

+=

⋅ +=

b

h b, ;

h0 = 800 - 80 = 720 mm.

Let’s check the equation qsw1b(h0-0,5x1) = 446,6·300(720-0,5·139)=

= 87,2·106 N·mm < 0 6 0 6 84 10

0158126 8 10

1

66, ,

,,

Т

δ=

⋅ ⋅= ⋅ N·mm. So qsw is not to be changed.

Let’s check condition (173):

( )( )

( )=

⋅⋅⋅

⋅⋅⋅−⋅+⋅=

−

−+

6

2366

1011

2

max102.87158.04

300104605.0108410490

5,04

5,0

xhbq

QbTМ

swδ

610494 ⋅ N·mm ( ) ( ) 6

10 108.7631395.072032173655.0 ⋅=⋅−⋅=−< xhAR ss N·mm

that means there is installed enough longitudinal reinforcement according to the strength condition. Due to condition (176) let’s check the strength as regards horizontal cross reinforcement located on the support part:

qsw1b(h0 - 0,5x1) = 446,6·300(720-0,5·139)=87,2·106 Н·мм > Т

2 2

84

2 2 015874 7

1δ=

⋅=

,, kN·m

That means there is installed enough horizontal cross reinforcement on the support part. As it’s shown on Draft 61, б, д, in the section with maximum span bending moment there is a torsion moment that’s why it is necessary to check the strength as regards longitudinal reinforcement installed at the bottom stretched surface in the middle part of the collar-beam span according to condition (174).

For this part of the collar-beam two top rods ∅ 32 are broken that’s why due to Draft 61, а, we

have A/sw = 1609 mm2 (2 ∅ 32); а/ = 62 mm; As1 = 1388 mm2 (2 ∅ 20 + 2 ∅22); а = 68 mm.

131

Let’s determine the height of compressed zone х1, taking Rb = 13 MPa (that is by γ b2 =0,9, as

wind load is not taken into account):

( )x

R A R A

R b

s s sc s

b

11 1 365 1388 1609

13 300=

−=

−

⋅

/

< 0.

We take х1 = 2а', so h0 - 0,5 x1 = h - a - a' = 800 - 68 - 62 = 670 mm. Horizontal cross rods in the middle part of the span are taken with diameter 14 mm (Аsw1 = 154 mm2) and spacing s1 = 200 mm, so

3.223200

154290

1

11 =

⋅==

s

ARq swsw

sw N/mm

From Draft 61, б, д we have:

Т0

2 71 2 45

2 4584 8 9=

−=

, ,

,, kN·m;

Мmax = 321 kN·m Let’s check condition (174):

( )M

T

q b h xt

qsw

max

,

+

− −

=02

1 1 0 1

2

4 0 52

δ

321·106+

+( )

mmN109,326

104,154

28,3426703003,223158,04

109,8 6

62

26

⋅⋅=⋅

−⋅⋅⋅⋅

⋅ <

< RsAs1 (h0 - 0,5x1) = 365·1388·670=339,4·106 N·mm

That means that according to the strength conditions there is installed enough bottom longitudinal reinforcement. Let’s determine distance lх from the zero point of the diagram Т possible for spacing of horizontal cross rods 200 mm using condition (176). Taking Т=tlx we have qsw1b(h0 - 0,5 x1) =

= tlx

2 2 1δ, so

( )m

t

xhbql

sw

x 47,128,34

67,03,223158,0225,022 1011=

⋅⋅⋅=

−=

δ

Therefore spacing of horizontal rods 100 mm on support parts can be 2,45-1,47 ≈ 1 m long. Example 47. Given: a floor beam with the cross section – due to Draft 62, а; location of the loads, diagrams of torsion and bending moments as well as diagram of cross forces – due to Draft 62, б; heavy-weight concrete В25 (Rb = 13 MPa by γ b2

= 0,9); longitudinal and cross

reinforcement А-III (Rs = Rsc = 365 MPa; Rsw = 290 MPa). It is required to check the strength of the beam as regards combined action of torsion and bending. Calculation. Let’s divide the section into two rectangles 200х400 and 350х400 mm and check condition (184):

( )0 1 0 1 13 200 400 350 4002 2 2, ,R b hb i i = ⋅ ⋅ + ⋅ =∑ 84,5·106 Н·мм > Т = 40 kN·m.

Черт. 62. For the calculation example 47

Due to Draft 62, а we have h0 = 800 – 50 = 750 mm. First let’s check the strength of the spatial section according to the 2nd scheme due to Item 3.90. At the same time as point loads applied in the middle of the section height cause the break of

132

the stretched zone of the beam it is necessary to take into account that some vertical stirrups bear the tearing force which is equal (according to Item 3.97).:

kN3,149750

35012801

0

1 =

−=

−=

h

hFF s

(where hs = 400 – 50 = 350 mm). The force per a unit length of the beam in vertical stirrups located at the right surface caused by tearing force F is to be determined by distributing the tearing force on two branches of stirrups and taking the width of the support platform of force F b = 300 mm, then

а = 2hs + b = 2·350 + 300 = 1000 мм = 1 м, that means

N/mm.6,74kN/m6,741

2/3,1492/1 ====a

Fqswa

So considered by the calculation of the spatial section value qsw2 by Аsw2 = 154 mm2 (1∅14) and s2 = 100 mm (see Draft 62, а) will be:

mmqs

ARq swa

swsw

sw N/3726,74100

154290

2

22 =−

⋅=−=

Due to Draft 58, в and 62, а, we take bf,min = 200 mm, h = 800 mm, bov=0, As2 = 1071 mm2

(1∅32 + 1∅12+1∅14). Then value с2 will be:

mmbhbc ovf 1200800200222 min,2 =+⋅=++=

Spatial section is located at the support of the beam. As с2 < 1,94 m that means spatial section doesn’t go beyond the borders of a part with the zero values Т so we live с2 = 1,2 m.

Design values Q and T are taken at the distance с2

2from the support that means

Q Q qc

= − = − =max ,,

,2

2297 5 7

1 2

2293 5кН; Т = 40 kN·m

As RsAs2 = 365·1071 = 391·103 N < 2qsw2h = 2·372·800=595·103 N so we live RsAs2 = 391 kN. The height of compressed zone х2 is determined as for the rectangular section according to Item 3.20, taking for the present scheme h0 = b0 = 200 – 50 = 150 mm and b = h = 800 mm (compressed overhang of a flange is not taken into account). As а' = 50 mm is a great part of h0 = 150 mm so value х2 is determined without considering compressed zone:

xR A

R b

s s

b

22 391 000

13 80037 6= =

⋅= , мм < а' = 50 mm.

Let’s check condition (187) taking bow = bo = 150 mm:

( ) ( )2220

2

2 5,05,0 xbhqxbc

hAR owswss −+− ( )= − ⋅ + ⋅ ×391 000

800

1200150 0 5 37 6 372 800, ,

( ) mmN102,736,375,0150 6 ⋅⋅=⋅−× > Т + 0,5 Qbf,min = 40 + 0,5 ⋅ 293,5 ⋅ 0,2 = 69,35 kN⋅m,

That means the strength as regards the 2nd scheme is provided. Let’s check the strength of the spatial section as regards the 1st scheme according to по 3.89.

Let’s take b'f = b = 200mm; bf= 350mm; Аs1 = 2526 mm2 (3∅32 + 1∅12); А′s1 = 308 mm2

(2∅14); Аsw1 = 154 mm2 (1∅14); s1 = 200 mm. Let’s determine the projection length c1:

c1 = 2h + 2bf + b'f - 2b = 2 ⋅ 800 + 2 ⋅ 350 + 200 - 2 ⋅ 200 = 2100 mm. Spatial section is located on the part between the support and the first load near the point of application of this load. As c1 > 1,94 m that means that the spatial section goes beyond the

133

beam and we take c1 = 1,94 m. Design values М and Т are taken at the distance c1

2 from the

support that means M = ⋅ −⋅

=297 0 975 7 0 97

2285 4

2

,, ,

, kN⋅m; Т = 40 kN⋅m.

The height of compressed zone is determined as for the rectangular section:

mm 31120013

)3082526(365111 =

⋅

−=

′−=

bR

ARARx

b

sscss

At the same time х1 = 311 mm < ξR ho = 0,604 ⋅ 750 = 453 mm (where ξR – see Table. 18);

N/mm 223200

154290

1

11 =

⋅==

s

ARq swsw

sw .

As 2qsw1 bf + M

h x0 10 5− , = 2 ⋅ 223 ⋅ 350 ⋅ + 285 4 10

750 0 5 311

6,

,

⋅

− ⋅ = 636,2 ⋅ 103 N < Rs As1 = 365 ⋅ 2526 =

=922 ⋅ 103 N, we take Rs As1 = 636,9 ⋅ 103 N. Let’s check condition (185) taking how = ho = 750 mm:

m,kN 4,6994,1

2,04,28540N104,85

)3115,0750(350223)3115,0750(1940

200102,636

1

6

3

1110

1

1

mm

)5,0()5,0(

⋅=+=⋅⋅=

=⋅−×⋅+⋅−⋅

′+>

=−+−′

c

bMT

xhbqxhc

bAR

f

owfsw

f

ss

That means the strength as regards the 1st scheme is provided. Calculation of reinforced concrete elements as regards local loads

LOCAL COMPRESSION CALCULATION

- (3.39). During the calculation as regards the local compression of elements without cross

reinforcement the following condition must be met: N R Ab loc loc≤ ψ , 1 , (194)

where N — longitudinal compression force caused by local; Aloc1 — compression area (see Draft 63);

ψ — coefficient пequal to: 1.0 – by local load distributed on the compression area; 0.75 by local load uneven distributed on the compression area (under

ends of beams, girders, and connection beams);

Rb, loc design resistance of concrete against compression determined by formula

Rb, loc = αϕb Rb, (195)

here αϕb ≥ 1.0;

α = 1,0 for concrete grade less than В25;

α = 13,5 Rbt /Rb for concrete grade В25 and more;

ϕb = A Aloc loc2 13 / ,

but no more than the following values: by the load application scheme due to Draft 63, а, в, г, е, and for concrete:

heavy-weight, fine and light-weight concrete of class: more than В7,5 ....................... 2,5 В3,5; В5; В7,5 ........................ 1,5

Light weight concrete of class В2,5 ......... 1,2

by the load application scheme due to Draft 63, б, д, ж independently on concrete class 1,0;

Rb, Rbt — taken as for (see Position 5 Table 9); Aloc2 — расчетная площадь смятия, определяемая в соответствии с п. 3.94.

134

If condition (194) is not met so it is recommended to use confinement reinforcement in form of welded meshes and to calculate the element in compliance with Item 3.95.

- A part which is symmetrical as regards the compression area is included into the design

area Aloc2 (Draft 63). At the same time the following rules must be followed:

By local loads into design area along the whole width b it is necessary to include a part with the length no more than b to each side of the local load border Draft 63, а);

By local edge load along the whole width of the element design area Aloc2 is equal to the compression area Aloc1 (Draft 63, б);

By local loads in the support points of collar-beams and beams ends into the design area it is included the with the width equal to the depth of setting of the collar beam or and with the length no more than the distance between the centers of spans adjoining to the beam (Draft 63, в);

If the distance between beams is more than a double width of the element so the length of design area is determined as a sum of the beam width and of the double width of the element (Draft 63, г);

By local edge load on the element angle (Draft 63, д) design area Aloc2 is equal to the compression area Aloc1;

Draft 63. Determination of design area Aloc2 by the calculation as regards local compression by a local load

а along the whole width of the element; б edge load along the whole width of the element; в, г in support points of of the ends of beams and collar-beams; д — edge load on the element angle; е — on the part of the length and the width of the element; ж — edge load в within the wall pier; и — on the section of irregular shape; I — minimum zone reinforced by meshes by which confinement reinforcement is considered in the calculation

By local load applied on the part of the length and the width of the element design

length is taken due to Draft 63, е. If there are several loads of the mentioned type design areas are bounded by the lines going through the center of distances between points of application of two neighbor loads;

By local edge load located within the wall pier or an I-section separation wall design area Aloc2 is equal to the compression area Aloc1 (черт. 63, ж);

By determination of design area for intricate shape sections it is not necessary to consider the parts which are not connected with the loaded parts and whose safety is not provided (Draft 63, и).

N o t e . By local loads from beams, collar-beams and connection beams working in bending the considered in the calculation depth of the support during determination of Aloc1 и Aloc2 is taken no more than 20 cm.

- (3.41). By determination as regards local compression of the elements made of heavy-

weight concrete with confinement reinforcement in form of welded cross meshes the following condition must be met:

N ≤ R*b,loc Aloc1, (196)

where Aloc1 compression area; R

*b,loc —prism strength of concrete to local compression determined by the

following formula

R*

b,loc = Rb ϕb + ϕµxy Rs,xy ϕs, (197)

here Rs,xy , ϕ, µxy are the same symbols like in Item 3.57;

ϕb = A Aloc loc2 13 / , but no more than 3,5;

ϕs — coefficient considering the influence of confinement reinforcement in the local compression zone; for diagrams of Drafts 63, б, д, ж it is taken

135

ϕs = 1.0, at the same time confinement reinforcement is considered in the calculation by the condition that cross meshes are installed on the area no less than the area bounded by a dotted line on the corresponding schemes

of Draft 63; for schemes of Draft 63, а, в, г, е, и coefficient ϕs is determined by formula

ϕs

loc

ef

A

A= −4 5 3 5 1, , ,

Aef — concrete area within the contours of confinement reinforcing meshes if they are calculated according to end rods for which the following condition

must be met Aloc1 < Aef ≤ Aloc2.

If the compression area border goes beyond the contours of confinement reinforcing meshes (for example see Draft 63, а - д, ж, и) by determination of values Aloc1 and Aloc2

the area occupied by a protection layer is not taken into account. The least depth of confinement reinforcement meshes installation must be determined by formulas:

by loading schemes due to Draft 63, в - е

hN

RAd d

b

loc= −

ϕ 1 ; (198)

by loading schemes due to Draft 63, а, б, ж, и

hb

N

RAd

d

b

loc= −

ϕ1 . (199)

In formulas (198) and (199):

ϕd = 0,5 — by loading schemes due to Draft 63, а, е, и;

ϕd = 0,75 by loading schemes due to Draft 63, в, г;

ϕd = 1,0 — by loading schemes due to Draft 63, б, д, ж.

The number of meshes is taken no less than two. Besides it is necessary to meet constructive requirements of Item 5.79. At the same time if dimensions of a mesh cell are more than 100 mm or more than 1/4 of the less side of the section, so rods of this mesh of

this direction are not taken into account during determination of µху. CALCULATION EXAMPLES

Example 48. Given: a steel pole, supported on the reinforced concrete foundation and centrally loaded by force N = 1000 kN (Draft 64); foundation of heavy-weight concrete В 12,5 (Rb = 6,7

MPa by γb2 = 0,9). It is required to examine the strength of concrete under the pole as regards local compression.

Draft 64. For the calculation 48

T h e c a l c u l a t i o n is made according to instructions of Items 3.93 and 3.94. Design area Аloc2 is to be determined in compliance with Draft 63, е.

Due to Draft 64, we have c1 = 200 mm < b = 800 mm; a1 = 200 ⋅ 2+300 = 700 mm;

60020022001 =+⋅=b мм; Аloc2 = a1 b1 = 700 ⋅ 600 = 420000 mm2.

Compression area is Аloc1 = 300 ⋅ 200 = 60000 mm2. As concrete class is less than В25,

α = 1,0.

Coefficient ϕb is:

ϕb

loc

loc

A

A= = = <2

1

3 3420000

600001 9 2 5, , .

136

Let’s determine design resistance of concrete against compression by formula (195), taking Rb

considering γb9 = 0,9 (see Table 9) as for the concrete structure: Rb = 6,7 ⋅ 0,9 = 6,03 MPa:

Rb, loc = αϕb Rb = 1 ⋅ 9 ⋅ 6,03 = 11,5 MPa

(where αϕb = 1 ⋅ 1,9 = 1,9 > 1,0).

Let’s check condition (194), taking ψ = 1,0 as by even distribution of the local load

ψR Ab loc loc, 1 = 1 ⋅ 11,5 ⋅ 60000 = 690000 N = 690 кН < N = 1000 kN,

that means concrete strength as regards local compression is not provided and that means that it is necessary to use confinement reinforcement. We take confinement reinforcement in form of

meshes made of reinforcing wire Вр-1, diameter 3 mm, dimensions of a cell 100×100 mm and spacing along the height s = 100 mm (Rs,xy = 375 MPa).

Let’s check the strength according to Item 3.95. As ϕb = 1.9 < 3.5, so ϕb = 1.9 is inserted into the calculation.

Coefficient of confinement reinforcement by meshes µxу is determined by formula (99).

Due to Draft 64 we have: пx = 5; lx = 300 mm; пy = 4; ly = 400 mm; Аsx = Аsy = 7,1 mm2 (∅3);

Аef = lx lу = 300 ⋅ 400 = 120000 mm2 > Aloc1 = 60 000 mm2, so

µxy

x sx x y sy y

ef

n A l n A l

A s=

+=

⋅ ⋅ + ⋅ ⋅

⋅=

5 71 300 4 71 400

120000 1000 00183

, ,, .

By formulas (101) and (100) we determine ψ and ϕ:

ψµ

=+

=⋅

+=

xy s xy

b

R

R

, ,

,, ;

10

0 00183 375

6 7 100 041

ϕψ

= =+ +

=1

0 23

1

0 23 0 0413 69

, , ,, .

Coefficient ϕs is:

ϕs = 4,5 - 3,5 Aloc1/Aef = 4,5 - 3,5 ⋅ 60000/120000 = 2,75. Specified concrete strength R*

b,loc is determined by formula (197):

R*

b,loc = Rb ϕb + ϕµxy Rs,xy ϕs = 6,7 ⋅ 1,9+ 3,69 ⋅ 0,00183 ⋅ 375 ⋅ 2,75 = 19,7 МПа. Let’s check condition (196):

R*

b,loc Aloc1 = 19,7 ⋅ 60000 = 1182 ⋅ 103 H > N = 1000 kN, That means concrete strength is provided.

Let’s determine the least depth of meshes setting by formula (198), taking ϕd = 0,5:

mm 100 mm 7,70600007,6

1010005,0

3

1 =<=−⋅

=

−= sloc

b

dd AR

Nh ϕ

that means it’s enough to install two meshes.

CALCULATION AS REGARDS THE PRESSING THROUGH

- (3.42). Calculation of slab structures (without cross reinforcement) as regards pressing

through by forces evenly distributed on the restricted area must be made according to the following condition

F ≤ αRbt um ho, (200)

where F pressing through force;

α — coefficient taken equal to: for heavy-weight concrete ........................ 1,00 for fine concrete ....................................... 0,85

137

for light-weight concrete .......................... 0,80

um arithmetic mean value of perimeters of upper base and lower base of the pyramid which is formed by pressing through within the working height of the section.

By determination of um and F it’s supposed that the punching through takes place on the lateral surface of the pyramid whose less base is the area of application of pressing force and lateral surfaces are inclined at the angle 45° to the horizontal line (Draft 65, а). Pressing through force F is taken equal to the force acting on the pressing pyramid except the loads applied on the larger base of the pressing pyramid (calculation as regards the plane where stretched reinforcement is located) and resisting against the pressing. If supporting scheme is so that pressing can take place only on the surface of the pyramid with the angle of lateral surfaces inclination more than 45° [for example in foundation pile caps (Draft 65, б)] so the right part of condition (200) is determined for actual punching pyramid and is multiplied by ho/с (where с — is the length of horizontal projection of lateral surface of the pressing pyramid). At the same time value of the bearing capacity is taken no less than the value corresponding to the pyramid by с = 0,4hо.

Draft 65. Scheme of the pressing pyramid by angle of inclination of its lateral surfaces to the horizontal line

а 45°; б more than 45°

by installation of stirrups normal to the slab plane within the punching pyramid the calculation must be made due to the following pyramid

F ≤≤≤≤ Fb + 0,8 FSW, (201) But no more than 2Fb,

where Fb — is the right part of condition (200);

Fsw =175ΣAsw the sum of all cross forces, taken by stirrups which cross lateral surfaces

of pressing pyramid (175 MPа limit pressure in stirrups). When considering cross reinforcement value Fsw must be no less than 0,5 Fb.

It is possible to consider the least value Fsw in the calculation by replacement of the right part of condition (201) by 2,8Fsw, but no less than Fb.

By location of stirrups on the restricted area close to the point load it is necessary to make additional calculation as regards the pressing of the pyramid with upper base located along the part contours with cross reinforcement according to condition (200) without considering cross reinforcement. Cross reinforcement must meet requirements of Item 5.75.

CALCULATION AS REGARDS BREAK

- (3.43). Calculation of reinforced concrete elements as regards the break caused by the

load applied on its bottom surface or within its section height (Draft 66) must be made due to the following condition

Fh

hR A

s

sw sw10

−

≤ Σ , (202)

where F – break force; hs – расстояние от уровня передачи отрывающей силы на элемент до центра

тяжести сечения продольной арматуры S; при передаче нагрузки через

138

монолитно связанные балки или консоли принимается, что нагрузка передается на уровне центра тяжести сжатой зоны элемента, вызывающего отрыв;

ΣRswАsw– sum of cross forces taken by stirrups which are installed in addition to the stirrups required by the calculation of inclined or spatial section in compliance

with Items 3.313.39, 3.86, 3.87 and 3.90; these stirrups are located along the break zone length equal to:

а = 2 hs + b, (203) here b – width of the break force силы F transfer area.

By evenly distributed load q, applied within the section height required stirrups intensity is increased by value q(1 - hs/ho)/Rsw.

- Re-entrant angles in the stretched zone of elements reinforced by intersecting

longitudinal rods (Draft 67) must have cross reinforcement enough to take:

а) resultant of forces in longitudinal stretched rods not going into the stretched zone equal to:

F R As s1 122

= cosβ

; (204)

б) 35 percent of resultant forces in all longitudinal stretched rods equal to:

F R As s2 10 72

= , cosβ

. (205)

Required by these calculations stretched reinforcement must be located along the length

s = h tg 3

8β.

Sum of forces projections in cross rods (stirrups) located along this length onto the bisectrix of the angle must be no less than sum F1 + F2,

That means ΣRsw Asw cosθ ≥ F1 + F2. (206)

In formulas (204) (206):

As section area of all longitudinal stretched rods;

Аs1 section area of longitudinal stretched rods not anchored in the compressed zone;

β re-entrant angle in the stretched zone of the element;

ΣAsw cross section of longitudinal reinforcement within the length s;

θ — angle of slope of cross rods onto the bisectrix of angle β. Draft 66. Scheme for determination of the break zone length

а by adjoining of beams; б adjoining of consoles; I center of gravity of compressed zone of the section of adjoining element

Draft 67. Reinforcement of re-entrant angle located in the stretched zone of reinforced concrete element

Calculation of short consoles

- (3.34). Calculation of short consoles of columns [l1 ≤ 0,9 h0; (Draft 68)] as regards the cross force to provide the strength of inclined compressed strip between the load and the support must be made due to the following condition

Q ≤ 0,8 Rb b lsup sin2 θ (1 + 5 αµw), (207) Where right part is taken no less than 3,5Rbtbh0 and no less than 2,5Rbtbh0.

In condition (207):

139

lsup the length of the area of the load supporting along the console overhanging length;

θ — angle of slope of design compressed strip onto the horizontal line расчетной

sin2 02

02

12θ =

+

h

h l;

µw

sw

w

A

bs= — reinforcement coefficient by stirrups located along the console height;

here sw — the distance between stirrups measured along the normal line to them. In the calculation it is necessary to consider the stirrups horizontal and inclined at the angle no more than 45 degrees to the horizontal line. Compression stress in the points of application of the load on the console must be no more than Rb,loc (see Item 3.93). For short consoles included into the hard joint of frame structure value lsup in equation (207) is taken equal to the console overhanging length l1 if the following conditions are

met М/Q ≥ 0,3 m and lsup/l1 ≥ 2/3 (where М and Q — are the moment which stretches the top surface of the collar-beam and the in the normal section of the collar beam along the edge of the console). In this case the right part of condition (207) is taken no more than 5Rbtbh0.

Draft 68. Desigh scheme for the short console by cross force action

By hinge support of prefabricated beam going along the console overhanging length on the short console, if there are no special embedded details fixing the support area (Draft 69) value lsup in condition (207) is taken equal to 2/3 of the length of actual support area. Cross reinforcement of short consoles must meet the requirements of Item 5.77.

Draft 69. Design scheme for the short console by hinge connection of the prefabricated beam going along

the console overhanging length

- By hinge support of the beam on the column console longitudinal reinforcement of the

console is examined due to the following condition

Ql

hR A

o

s s

1 ≤ , (208)

where l1, ho see Draft 68. At the same time longitudinal reinforcement of the console must be installed up to the free supported end of the console and have anchorage (see Items 5.44 and 5.45). By fixed connection of the collar-beam and the column ригеля with monolithing of the joint and welding of lower reinforcement of the collar-beam to reinforcement of the console by means of embedded elements longitudinal reinforcement of the console is examined due to the following condition:

Ql

hN R A

o

s s s

1 − ≤ , (209)

where l1, h0 — overhanging length and working height of the short console; Ns — horizontal force acting on the top of the console from the collar-beam

equal to:

NM Ql

hs

ob

=+ sup / 2

(210)

140

and taken no more than 1,4 kflwRwf + 0,3 Q (where kf and lw — the height and the length

of the angle joint of welding of embedded details of the collar-beam and console; Rf design resistance of angle joints against the cutting of the joint metal determined in compliance with SNiP II-23-81, with electrodes Э42 Rwf = 180 MPa; 0,3 — steel to steel friction coefficient), as well as no more than Rsw Аsw (where Rsw and Аsw — are design resistance and section area of the top reinforcement of the collar-beam). In formulas (209) and (210): M, Q — bending moment and cross force in the normal section of the collar beam along

the edge of the console; if moment М stretches lower surface of the collar-beam so value М is considered in formula (210) with the sign "minus";

lsup — actual length of the load support area along the console overhanging length; hob — working height of the collar beam.

CALCULATION EXAMPLES

Example 49. Given: a free supported prefabricated beam lies on the short console of the column (Draft 70); the length of supporting area lsup,f = 300 mm; console length b = 400 mm; height of the column and overhanging length of the column h = 700 mm, l1 = 350 mm; heavy

weight concrete of the column В25 (Rb = 13 MPа, Rbt = 0,95 MPа by γb2 = 0,9; Еb = 27 ⋅ 103 MPа); longitudinal reinforcement А-III (Rs = 365 МПа); load on the console Q = 700 kN. It is required to check the strength of the column as regards the cross force and to determine the cross section of longitudinal reinforcement and stirrups.

Draft 70. For the calculation example 49

Calculation. H0 = h – а = 700 - 30 = 670 mm. As =⋅⋅⋅= 67040095.05.35.3 0bhRbt 3101.891 ⋅= N 1.891= kN > Q = 700 kN and at the same time

5.63667040095.05.25.2 0 =⋅⋅⋅=bhRbt кН <Q = 700 кН so the console strength is to be

checked according to condition (207). Due to Item 3.99 design length of the support area is to be taken equal to:

lsup = 2/3 lsup, f = 2/3 ⋅ 300 = 200 mm Due to Item 5.77 we take the stirrups spacing equal to

sw = 150 мм < h

4

700

4= = 175 mm

By two-leg stirrups with diameter 10 mm we have Аsw = 157 mm2, so

µw

sw

w

A

bs= =

⋅= ⋅

−157

400 1502 62 10

3, ;

α = =⋅

⋅=

E

E

s

b

20 10

2 7 107 4

4

4,

, ;

sin , ;2

2

212

2

2 2

670

670 3500 786θ =

+=

+=

h

h l

o

o

0,8 Rb blsup sin2θ (1 + 5 αµw) = 0,8 ⋅ 13 ⋅ 400 ⋅ 200 · 0,786 (1+5⋅7,4 ⋅ 2,62 ⋅ 10-3) =

= 717 ⋅ 103 H > Q = 700 kN, That means the strength of the console as regards the cross force is provided.

141

Due to condition (208) let’s determine required section area of the console longitudinal reinforcement:

.23

1mm 1002

365670

35010700=

⋅

⋅⋅==

so

sRh

QlA

We take 3 ∅ 22 (As = 1140 mm2). Calculation of embedded elements and connection details

CALCULATION OF EMBEDDED ELEMENTS

- (3.44). Calculation of normal anchors welded to the flat details of steel embedded

elements as regards bending moments, normal and shearing force caused by static load located in one plane of symmetry of the embedded detail (Draft 71), must be made due to the following condition

A

NQ

Ran

an

an

s

=

+

11 2

2

,λδ

, (211)

where Aan — total area of the anchors cross section of the most stressed row;

Nan maximum stretching force in one row of anchors equal to:

NM

z

N

nan

an

= + ; (212)

Qan shearing force on one anchors row equal to:

QQ N

nan

an

an

=− ′0 3,

; (213)

N′an — maximum compression force in one row of anchors determined by formula

′ = −NM

z

N

nan

an

; (214)

In formulas (211) (214): М, N, Q — moment, normal and shearing forces acting on the embedded element; the

moment is determined relating to the axis located in the area of the external surface of the plate and going through the center of gravity of all;

z distance between the end rows of anchors; nan — number of anchors rows along the shearing force direction; if even transfer of

shearing force Q on all anchor rows is not provided, so by determination of shearing force Qan it is necessary to consider no more than four rows;

λ — coefficient determined for anchor rods with diameter 8 – 25 mm for heavy-weight and fine concrete В12,5 – В50 and light-weight concrete В12,5 – В30 by the following formula

λ β=+

4 75

1 015

3

1

,

( , ),

R

A R

b

an s

(215)

but taken no more than 0.7; for heavy-weight and fine concrete of class more

than В50 coefficient λ is taken as for class В50, and for light-weight concrete of class more than В30 — as for class В30. For heavy-weight concrete

coefficient λ can be determined by Table. 28.

In formula (215): Rb, Rs, – in MPа;

By determination of Rb coefficient γb2 (see Item 3.1) is taken equal to 1.0;

142

Aan1 – section area of the anchor rod of the most stressed row наиболее, cm2;

β – coefficient taken equal to: for heavy-weight concrete ................................ 1,0 for fine concrete of group: А ........................................................................ 0,8 Б and В ............................................................... 0,7

For light-weight concrete .................................. ρm/2300

(ρm average concrete density, kg/m3);

δ – coefficient determined by formula

δω

=+

1

1, (216)

but taken no less than 0,15;

here ω = 0,3 N

Q

an

an

by N′an > 0 (with pressure);

ω = 0,6 N

Q by N′an ≤ 0 (no pressure);

if there are no tension forces in anchors so coefficient в δ is taken equal to 1,0. Section area of anchors of other rows must be taken equal to the section area of anchors of the most stressed row. In formulas (212) and (214) normal force N is considered to be positive if it’s directed from the embedded element (see Draft 71), and negative — if it’s directed towards to it.

In case if normal forces Nan and N′an, as well as shearing force Qan during calculation by formulas (212) – (214) get negative value (211), (213) and (216) they are taken equal to

zero. Besides if Nan gets negative value so in formula (213) it’s taken N′an =N.

By location of the embedded element on the top surface of the detail (during concreting)

coefficient λ is decreased by 20 percent and value N′an in formula (213) is taken equal to zero.

Draft 71. Scheme of forces acting on the embedded element

- Calculation of normal anchors of embedded elements as regards bending moments and

shear forces located in two symmetry planes of the embedded elements as well as normal force and torsion moments is made in compliance with "Recommendations for design of steel embedded elements for reinforced concrete structures" (Moscow, Stroyizdat, 1984).

- (3.45). In the embedded element with anchors welded with overlapping at the angle from

15 to 30 degrees (see Item 5.111) inclined anchors located symmetrically relating to the plane of the shearing force is calculated as regards this shearing force (by Q > N, where

N break force) by formula

AQ N

Ran inc

an

s

,

,,=

− ′0 3 (217)

where Aan,inc total area of the cross section of inclined anchors;

Nan see Item 3.101. At the same time it is necessary to install normal anchors calculated by formula (211) by

δ = 1.0 and by values Qan, equal to 0.1 of shearing force determined by formula (213). It is possible to decrease the section area due to transfer of shearing force equal to

incans ARQ ,9.0− on normal anchors. In that case δ is determined by formula (216).

Таблица 28

143

Anchor Coefficient λ for calculation of normal anchors of embedded elements according to class of heavy-weight

concrete and reinforcement

diameter В15 B20 B25 B30 B40 ≥ B50

мм А-I A-II A-III А-I A-II A-III А-I A-II A-III А-I A-II A-III А-I A-II А-III А-I A-II A-III

8 0,60 0,48 0,66 0,53 0,70 0,57 0,70 0,60 0,70 0,66 0,70 0,70

10 0,58 0,52 0,45 0,64 0,57 0,50 0,69 0,62 0,54 0,70 0,65 0,57 0,70 0,70 0,63 0,70 0,70 0,66 12 0,55 0,50 0,43 0,61 0,55 0,48 0,66 0,59 0,52 0,70 0,62 0,55 0,70 0,69 0,60 0,70 0,70 0,63 14 0,53 0,47 0,41 0,58 0,52 0,46 0,63 0,56 0,49 0,66 0,59 0,52 0,70 0,65 0,57 0,70 0,69 0,60 16 0,50 0,45 0,39 0,55 0,49 0,43 0,59 0,53 0,47 0,63 0,56 0,49 0,69 0,62 0,54 0,70 0,65 0,57 18 0,47 0,42 0,37 0,52 0,46 0,41 0,56 0,50 0,44 0,59 0,53 0,46 0,65 0,58 0,51 0,68 0,61 0,54 20 0,44 0,39 0,34 0,49 0,44 0,38 0,52 0,47 0,41 0,55 0,50 0,43 0,61 0,54 0,48 0,64 0,58 0,50 22 0,41 0,37 0,32 0,46 0,41 0,36 0,49 0,44 0,39 0,52 0,46 0,41 0,57 0,51 0,45 0,60 0,54 0,47 25 0,37 0,33 0,29 0,41 0,37 0,32 0,44 0,40 0,35 0,47 0,42 0,37 0,51 0,46 0,40 0,54 0,49 0,43

Notes: 1. For concrete class В 12,5 coefficient λ is to be decreased by 0,02 in comparison with coefficient λ for concrete class В15.

2. Values of coefficient λ are given by γbi = 1,00.

- On welded to the plate supports made of strip steel or reinforcement lugs (see Item 5.114)

it is possible to transfer no more than 30 percent of shearing force acting on the detail by stresses equal to Rb in concrete under supports. At the same time values of shearing force transferred on the anchors of the embedded element is decreased.

- (3.46). The structure of welded embedded elements with welded to them details which

transfer the load on the embedded elements must provide including into work of anchor rods in compliance with the accepted design. External elements of the embedded details and their welded connections are calculated due to SNiP II-23-81. Durinf calculation of plates and corrugated steel as regards break force it is recommended to take that they have hinge connection with normal anchor rods. If the element which transfers the load is welded to the plate along the line of location of one of the anchor rows so during calculation it is recommended to decrease break force by value пaАan1Rs (where na — number of anchors in the present row).

Besides the thickness of plate t of design embedded element must be examined due to the condition

t ≥ 0,25 dR

Ran

s

sq

, (218)

where dan — diameter of the anchor rod required due to the calculation;

Rsq — Design resistance of rolled-steel against the shearing equal to 0.58 Ry (where Ry see SNiP II-23-81). For welded connections types which provide the larger zone of including the plate into work by pulling out of anchor rods out of it (see position 6 of Table 52) the correction of condition (218) is possible for decrease of the plate thickness. If shearing force Q acts on the embedded element with decreased plate thickness total section area (perpendicular to this force) of the section with welded to it elements in the location zone of anchor rods along the force Q is taken no less than the section area of the plate determined by formula (218).

- If the following condition is met

N′an ≤ 0, (219) -

144

-

Where N′an — see Item 3.101, that means if all normal anchors are stretched so the calculation as regards concrete chipping is to be made in the following manner: а) for normal anchors with reinforcement at the ends (see Item 5.113) – due to the condition

NAR

e

a

e

a

bt≤

+ +

δ δ1 2

1

1

2

2

1 3 5 3 5, ,

, (220)

where А — a projection area on the plane normal to anchors, to chipping surface going from the reinforced ends of anchors at the angle 45 degrees to the anchor axes; by force eccentricity N relating to the anchors center of gravity e0 = M/N dimensions of the projection of the chipping surface in the direction of this eccentricity is decreased by the value equal to 2e0 by corresponding displacement of the inclined surface of the chipping surface (Draft 72); areas of anchor plates are not considered;

δ1 coefficient taken equal: to 0.5 – for heavy-weight and fine concrete; to 0.4 – for light-weight concrete;

δ2 — coefficient taken equal to:

by σbc

bR< 0 25, or

σbc

bR> 0 75, δ2 = 1.0;

by 0 25 0 75, ,≤ ≤σbc

bR δ2 = 1.2.

At the same time if a part of the rod with the length а is located in the concrete zone by 0.25 ≤ σbc/Rb ≤ 0.75 so δ2 is determined by formula

δ2 = 1 + 0.2 a

la, (221)

Here la — anchor rod length; Compression stresses in concrete σbc, perpendicular to the normal anchor and distributed along the whole length are determined as for the elastic material by reliability coefficient 1.0;

a1, a2 — dimensions of the projection of the chipping surface; e1, e2 — eccentricity of force N relating to the center of gravity of area А in the

direction of dimensions а1 and a2;

Draft 72. Scheme of concrete chipping by means of anchors of the embedded element with

reinforcement at the ends by N′′′′an ≤≤≤≤ 0

1 point of application of the normal force N; 2 chipping surface; 3 — projection of the chipping surface on the plane normal to anchors

b) for anchors without reinforcement at the ends the calculation is made due to the condition

NA R

e

a

e

a

R Al h

l

h bt

h

h

h

h

s an a

a

an

≤ +−

+ +

δδ1 2

1

1

2

2

1 3 5 3 5, ,, , (222)

Where Ah – the same like А, if the chipping surface goes at the distance h from the embedded element plane (Draft 73);

ah1, ah2 – dimensions of the chipping surface projection; eh1, eh2 – eccentricity of force N relating to the center of gravity Ah, in the direction

of dimensions ah1 и ah2;

Trang 145

Аan,a – section area of all anchors crossing the chipping surface; lan – anchorage zone length (see Item 5.44). Condition (222) is checked by different values h less than anchorage length or equal to it.

Draft 73. Scheme of concrete chipping by anchors of the embedded element without reinforcement at

the ends by N'an ≤≤≤≤ 0

1 normal force application point N; 2 chipping surface; 3 projection of the chipping surface on the plane normal to anchors

If number of anchors in the direction of the eccentricity is more than two so in

conditions (220) and (222) it is possible to decrease force N by value 12

−

n

M

zan

.

If anchor ends are located close to the concrete surface opposite to the embedded element plate it is necessary to check condition (222) without considering the last member of the right part of the condition by h, equal to the distance from the plate to the opposite surface of the element, at the same time the part of the area Ah, located between end rows of anchors is not considered.

3.107.If condition N′an > 0 is met and if there are reinforcements at the ends of anchors so calculation of concrete chipping (Draft 74) is made due to the following condition

NA R

e

a

an

bt≤

+

δδ1 2 1

1 3 5,

, (223)

Where Nan – see Formula (212); A1 – the same that А in formula (220), if chipping surface begins from the place of

reinforcement of anchors of the most stretched row (see Draft 74); е – eccentricity of force N relating to the center of gravity of area А1 in the direction

of dimension а. It is possible not to make the chipping calculation if ends of anchors are fixed behind the longitudinal reinforcement located at opposite to the embedded element surface of the column, and reinforcement of anchors in form of plates or cross lugs are fixed to the rods of longitudinal reinforcement with diameter: by symmetrical fixing – no less than 20 mm, by asymmetrical fixing – no less than 25 mm (Draft 75). In that case the part of the column between end rows of anchors is checked as regards the cross force equal to (due to Items 3.31 and 3.53):

Q = Nan m Qcol,

WherecolQ – cross force on the part of the column adjoining to the most stretched row

of anchors of the embedded element determined considering the forces acting on the embedded element.

Draft 74. Scheme of concrete chipping by stretched anchors of the embedded element by N′′′′an > 0 1 — projection of the chipping surface on the plane normal to anchors; 2 — anchor plate; 3 — point of application of force Nan

Draft 75. The structure of the embedded element for which chipping calculation is not required

а — embedded element with lugs symmetrically fixed to the longitudinal reinforcement of the column; б diagram Q of the column part with the embedded element; в anchors of the embedded elements with anchor plates asymmetrically fixed to the longitudinal reinforcement of the column; 1 — cross lugs welded to anchors by means of contact welding; 2 anchors; 3 anchor plates

Trang 146

3.108. If shearing force Q acts on the embedded element in the direction towards the element edge (Draft 76), and if there are no inclined anchors so calculation as regards the concrete chipping is made due to the following condition

QR bh

e

b

bt≤

+

δ1

1 3 5,

, (224)

Where δ1 – see Item 3.106; by embedded element located on the top surface of details made of light-weight concrete coefficient δ1 is decreased by 20 percent,

b – element width equal to b = c1 + c2 + s (где c1 и c1 – distances from end rows of anchors to the nearest edges of the element in the direction normal to the shear force taken no more than h, s – distances between end rows of anchors in the same direction);

h – the distance from the outermost anchor row to the edge of the element in the direction of shearing force Q, taken no more than the element width b1 (see Draft 76);

е – eccentricity of force Q relating to the middle of the element width b.

In case if break force N is applied to the embedded element except shearing force Q

the right part of condition (224) is multiplied by the coefficient δn

out bt

N

A R= −1

0 3,, taken

no less than 0,2 (where Aout is the area of the projection on the plane perpendicular to the break force N, and to the chipping surface).

Draft 76. The scheme for the calculation of the concrete chipping by normal anchors of the embedded

element

In case if shearing force is applied to the embedded element with inclined anchors welded with overlapping and which have reinforcement at the ends (see Item 5.113), calculation of concrete chipping is made in compliance with the Recommendations mentioned in Item 3.102.

3.109. If at the ends of anchors of the embedded element there are reinforcements in form of anchor plates (see Item 5.113) so concrete under these reinforcements are checked as regards compression due to the following condition

N R Aloc b b loc≤ αϕ 1 , (225)

Where α, ϕb – coefficients determined due to Item 3.93; Аloc1 – area of concrete plate with the exception of the anchor section area; Nloc – compression force determined in the following manner: а) for anchors by la ≥ 15d:

If crack formation along the anchor is possible as result of concrete tension or in case of use of plane anchor rods by formula

Nloc = Nan1; (226) If crack formation is impossible — by formula

Nloc = Nan1 l l

l

an a

an

−; (227)

Trang 147

б) for anchors by la < 15d value Nloc determined by formulas (226) and (227), is

decreased by Qd l

lan

a

an

1

15 −;

в) for anchors welded with overlapping Nloc is determined by formula Nloc = Qinc. (228)

In formulas (226) (228): Nan1, Qan1 – maximum tension and shearing forces per one normal anchor (see Item

3.101); Qinc – force in the inclined anchor Formula (225) can be used if thickness of anchor plate is less than 0.2 of its length.

3.110. Determination of welded embedded elements details, calculation of inclined anchors welded under the flax material layer to the plate at the angle more than 45 degrees and calculation of pressed embedded details is made according to Recommendations mentioned in Item 3.102.

CALCULATION EXAMPLES

Example 50. Given: embedded element of the column with the welded table for support of the framing beam as well as location and values of loads and framing beams – due to Draft 77; anchors of reinforcement А-III (Rs = 365 MPa); heavy-weight concrete of the column В20; plate of steel grade ВСт3кп2 (Ry = 215 MPa). It is required to design normal anchors of the embedded element and to determine the plate thickness.

Draft 77. For the calculation example 50

C a l c u l a t i o n . Let’s take location of anchors as it’s shown on Draft 77. As all loads act in one direction and don’t cause the torsion, so let’s determine total area of the anchors cross section of the most compressed upper row by formula (211). For that we determine external forces moment:

М = Ql = 150 ⋅ 0,15 = 22,5 kN⋅m Taking z = 0,3м and N = 0, let’s determine maximum tension force in one anchor row by formula (212):

NM

zan = = =22 5

0 375

,

, кН .

On Draft 77 shearing force is Q = 150 kN, anchor rows number is nan = 3. Shearing force per one anchor row is to be determined by formula (213), taking N′an = Nan = =75 kN:

QQ N

nan

an

an

=′−

=− ⋅

=0 3 150 0 3 75

342 5

, ,, кН .

Coefficient δ is to be determined by formula (216).

As N′an > 0, ω = 0,3 N

Q

an

an

= =0 375

42 50 529,

,, ,

So δω

=+

=+

= >1

1

1

1 0 5290 808 015

,, , .

Trang 148

Taking diameter of anchors 16 mm according to Table 28 by concrete class В20 and reinforcement class А-III we find λ = 0,43, so

.mm 432365

808,043,0

42500750001,1

2

2

2

2

21,1

=⋅

+

=

+

=s

an

an

anR

QN

Aλδ

We take two anchors with diameter 18 mm in each row (Aan = 509 mm2). Let’s check condition Aan by coefficient corresponding to the accepted diameter 18 mm, that means by λ = 0,41:

Aan =

+⋅

= <

11 75000

42500

0 41 0 808

365448 509

2

2

,, ,

. мм мм2 2

We take 2∅18. Let’s determine minimum allowable anchors length without forces lan due to Item 5.112. For that we determine coefficient δ3:

δ3

0 3

10 7

0 3

1 42 5 750 7 0 89=

++ =

++ =

,

/.,

,

, /, ,

Q Nan an

Value Rb is taken considering γb2 = 0,9 (no short-term loads), that means Rb = 10,5 MPa. Let’s determine lan, taking σbc < 0,25 Rb, that means ωan = 0,7, ∆λan = 11:

lR

Rdan

an s

b

an= +

= + =δ

ωλ3 0 89 0 7

365

10 511 18 567∆ , ,

, мм.

Considering the fact that area Aan is taken with the reserve let’s specify value lan more exact:

lan = =567448

509500 мм > 400 мм.

As location of anchors in the column by such length is impossible so it is necessary to decrease the length of anchors and reinforce their ends. Due to Item 5.113 anchor ends are to be reinforced by button-heads with diameter dh = 54 m ≥ 3d and concrete is to be checked as regards compression and chipping taking anchors length la = 250 mm > 10d = 10 ⋅ 18 = 180 mm. Compression calculation is to be made due to Item 3.109. Compression area Аloc1 under a button-head of one anchor is:

22

111 mm 20362544

5414,3=−

⋅=−= anhloc AAA .

Let’s suppose that crack formation in the column on the side of the embedded element is possible. So in compliance with Item 3.109 by la = 250 mm < 15 d = 15 ⋅ 18 = 270 mm compression force will be:

.15

kN 452

5,42

567

250270

2

7511 =

−+=

−+= an

an

a

anloc Ql

ldNN

Let’s take maximum value ϕb = 2.5 as design area of concrete Аloc2 here is large; α = 1.0. Let’s check condition (225):

,N 450005345020365,105,211 ==⋅⋅⋅ >= loclocbb NAR Hαϕ

That means the strength as regards compression is provided. As N′an > 0 so chipping calculation is made according to Item 3.107. Anchor ends with reinforcement are not fixed to the longitudinal reinforcement of the column located at the

Trang 149

surface opposite to the embedded element of the column that’s why calculation is made due to the condition (223). Let’s determine value A1 (see Draft 77):

A1 = (2 ⋅ 250 + 54) 400 - 2 314 54

4

2, ⋅ = 217000 mm2.

Force Nan = 75 kN is applied in the center of gravity of area А1, so е = 0. For heavy-weight concrete δ1 = 0.5. Let’s check condition (223) without considering compression stress of concrete (that means δ2 = 1.0) and taking γb2 = 0.9 (that means Rbt = 0.8 MPa): δ1δ2A1Rbt = 0.5 ⋅ 1 ⋅ 217000 ⋅ 0.80 = 86800 N > Nan = 75000 N, That means concrete strength as regards chipping is provided. Accepted distances between anchors in the direction across and along the shearing force equal to 260 mm > 5d = 5 ⋅ 18 = 90 mm and 150 mm > 7d = 7 ⋅ 18 = 126 mm meet the requirements of Item 5.111. The distance from the anchor axis to the column surface equal to 70 mm > 3.5d = 3.5 ⋅ 18 = 63 mm also meet requirements of Item 5.111. Structure of the table welded to the embedded element provides even distribution of forces on the stretched anchors and even transfer of compression stressed on concrete without causing bending of the embedded element plane. That’s why thickness of this plate is to be determined due to condition (218) taking Rsq = 0.58 Ry = 0.58 ⋅ 215 = 125 MPa, and anchor

diameter required by the calculation is dan = 18448

509 = 16.9 mm:

t = 0.25 dR

Ran

s

sq

= 0.25 ⋅ 16.9 365

125 = 12.3 mm.

Due to conditions of mechanized arc welding under flux (see Table 52, position 1) plate thickness must be no more than 0,65d = 0.65 ⋅ 18 = 11.7 mm. We take the plate thickness t = 14 mm.

Example 51. Given: embedded element of the column with welded diagonal member of steel bracings — according to Draft 78, а; tension force in the diagonal member caused by wind loads 270 kN; embedded element anchors of reinforcement A-III (Rs = 365 MPa); embedded element plate of steel ВСт3сп2 (Ry = 215 MPa); heavy-weight concrete of the column В30; column reinforcement — due to draft 78, б, minimum longitudinal force in the column 1100 kN; bending moment in the column at the level of the embedded element in the plane of anchors 40 kN ⋅ m. It is required to design anchors of the embedded elements, to determine the plate thickness and to check the strength of surrounding concrete as regards chipping. C a l c u l a t i o n . We take location of anchor rows along the vertical line as it’s shown on Draft 78, в. The force in the diagonal member is resolved into the normal force N, applied to the embedded element with eccentricity e0 = 100 mm, and shearing force Q:

N = 270 cos 56°20′ = 270 ⋅ 0,555 = 150 kN; Q = 270 sin 56°20′ = 270 ⋅ 0,832 = 225 kN.

By z = 0.42 m and M = Ne0 = 150 ⋅ 0.1 = 15 kN⋅m let's determine maximum tension force in one row of anchors by formula (212):

kN 2,734

150

42,0

15=+=+=

an

ann

N

z

MN

Maximum compression force in row of anchors we determine by formula (214):

Trang 150

′ = − = − = −NM

z

N

nan

an

15

0 42

150

418

,, кН < 0,

Shearing force Qan, applied on one row of anchors is to be determined by formula (213), by N′an = 0:

QQ

nan

an

= =225

4= 56,25 kN.

As N'an = 0,

ω = 0,6 N

Q = 0,6

150

225 = 0 4,

so δω

=+

=+

1

1

1

1 0 4, = 0,845 > 0,15.

Draft 78. For the example calculation 51

Due to Table 28, taking anchor diameter 16 mm by concrete class В30 and anchors of reinforcement А-III we find λ = 0,49, тогда

.mm 465365

845,049,0

562500732001,11,1

2

2

22

2

=

⋅+

=

+

=s

an

an

anR

QN

Aλδ

In each row we take two anchors with diameter 18 mm (Aan = 509 mm2). Let’s check required value Aan by coefficient λ, corresponding to the accepted diameter 18 mm, that means by λ = 0,46:

.mm 509mm 448365

845,046,0

56250732001,1

22

2

2

<=

⋅+

=anA

Let’s take two anchors with diameter 18 mm in each row. Let’s arrange the anchors with the minimum direction between them in the horizontal direction equal to 5d = 5 ⋅ 18 = 90 mm (see Item 5.111). Distances between anchors in the vertical direction (that means in the direction of the shearing force Q) are equal to 140 mm > 7d = 7 ⋅ 18 = 126 mm also corresponds to requirements of Item 5.111. Let’s determine the thickness of the embedded element plate. As the gusset plate which transfers the break force to the embedded element is located in the middle of the distance between vertical rows of anchors so the thickness of the plate is to be determined according to the calculation of the strength of the plate as of the console beam with the overhanging length 35 mm (see Draft 78) as regards tension force in one anchor equal to:

NN

an

an

1 273 2= = , = 36.6 kN.

The width of the console beam is b = 80 mm. Calculation is made due tot he condition

М ≤ RyW, where М = 36600 ⋅ 35 = 1⋅280000 N⋅mm, Wbt

=2

6,

therefore tM

R by

= =⋅ ⋅

⋅

6 6 128 10215 80

4

= 21.2 mm.

We take the plate made of strip steel 22 mm thick, at the same time condition (218) is met:

Trang 151

0,25 dR

Ran

s

sq

= 0,25 18365

130= 12,6 mm < 22 mm and requirements for any kind of T-joint

welding of rods (see Table 52): 0,75d = 0,75 ⋅ 18 = 13,5 mm < 22 mm. Let’s determine minimum allowable length of anchors without reinforcement by formula (316) considering Item 5.112. For that we determine coefficient δ3:

δ3

0 3

10 7

0 3

1 56 25 73 20 7 0 87=

++ =

++ =

,

/,

,

, / ,, , .

Q Nan an

Value Rb is taken considering γb2 = 1.1 as the load on the embedded element is caused only by wind load, that means Rb = 19 МПа. For determination of coefficients ωan and ∆λ we determine maximum and minimum concrete stress within the anchor length. For that we determine area Ared and inertia moment Ired of the column section taking due to Draft 78, б Аs = А′s = 1232 mm (2∅28):

Аred = bh + 2Аs (α - 1) = 400 ⋅ 400 + 2 ⋅ 1232 (6,9 - 1) = 174,5 ⋅ 103 mm2;

Ired = bh

3

12 + 2Аs (α - 1)(0,5h - a)2 =

= 400 400

12

3⋅ + 2 ⋅ 1232 (6,9 - 1) (0,5 ⋅ 400 - 50)2 = 2460 ⋅ 106 mm4

here aE

E

s

b

= =⋅

⋅

2 10

2 9 10

5

4, = 6,9.

Maximum concrete stress at the end of anchor la = 300 mm (that means at the distance у = 300 + 22 - 400/2 = 122 mm from the center of gravity of the section):

σb

red red

N

A

M y

I,max ,

= − =⋅

⋅+

⋅ ⋅

⋅

1100 10

174 5 10

40 10 122

2460 10

3

3

6

6 =

= 6,31 + 1,98 = 8,3 МПа < 0,75 Rb = 14,3 MPa.

Minimum stress of concrete at the beginning of concrete; that means by у = 400

2 - 22 =

178 mm:

σb

red red

N

A

M y

I,min ,= − =

⋅ ⋅

⋅6 31

40 10 178

2460 10

6

6 = 3,42 МПа < 0,25 Rb = 4,75 MPa.

As the anchor is not located in the zone with stress from 0.25Rb до 0.757Rb, so we determine the length of the part of the anchor а, located in this zone:

mm 21842,33,8

75,43,8300

25,0

min,max,

max,=

−

−=

−

−=

bb

bb

a

Rla

σσ

σ

So due to Formula (317),

ωan

a

a

l a a

l=

− +=

− + ⋅=

0 7 0 5 0 7 300 218 0 5 218

3000 555

, ( ) , , ( ) ,, .

∆λan is determined similar to ωan with replacement coefficients 0.7 and 0.5 by 11 and 8 (see Table 44):

∆λan

a

a

l a a

l=

− +=

− + ⋅=

11 8 11 300 218 8 218

3008 82

( ) ( ), .

Allowable anchor length is:

mm. 3051882,819

365555,087,03 =

+=

∆+= d

R

Rl an

b

s

anan λωδ

Trang 152

Considering that area Aan is taken with the reserve we take that lan : lan = 305488

509 = 292 мм.

We take anchor length la = 300 mm. Let’s check the chipping concrete. As all anchors are stretched and have no reinforcement so the calculation is made due to condition (222). Let’s determine the projection area of the chipping surface Аh considering displacement of the inclined surface by 2e0 = 2 ⋅ 100 = 200 mm. By h = la = 300 mm

Ah = (420 - 200 + 2 ⋅ 300) 400 = 32.8 ⋅ 104 mm2. As force N is applied in the center of gravity of area Ah, eh1 = еh2 = 0, δ1 = 0,5 (as for heavy-weight concrete). By formula (221) we get

δ2 = 1 + 0,2 a

la = 1 + 0,2

218

300 = 1.145.

As la = h, RsАап,а (la - h)/lan = 0. Considering that by γb2 = 1.1, Rbt = 1,3 MPa. δ1δ2AhRbt = 0,5 ⋅ 1,145 ⋅ 32,8 ⋅ 104 ⋅ 1,3 = 244300 Н > N = 150 kN.

Let’s check condition (222) by h = 200 mm < la. As at the distance h from the plate chipping surface crosses only two pairs of anchors so

Aan1 = 1018 мм2 (4∅18); Аh = (420 - 200 + 2 ⋅ 200) 400 = 24,2 ⋅ 104 mm2,

δ1δ2AhRbt + RsАап,а

l h

l

a

an

− = 0,5 ⋅ 1,145 ⋅ 24,2 ⋅ 104 ⋅ 1,3 + 365 ⋅ 1018

300 200

300

− =

= 304 ⋅ 103 Н > N = 150 kN. As after decreasing of h concrete bearing capacity is increased so calculation by less values h is not required. Let’s check condition (222) by value h, equal to the column section height; that means h = =400 mm, without considering the area between anchors [(420 - 200) 90 = 19800 mm2]:

Аh = (420 - 200 + 2 ⋅ 400) 400 - 19800 = 388 000 мм2 > 328000 mm2, That means Аh is more than the area determined by h = 300 mm. So, concrete strength against chipping is provided. CALCULATION OF PREFABRICATED COLUMNS JOINTS

3.111. Column joints made by means of welding of reinforcement connecting rods (see

Item 5.90) are calculated for two work stages:

1st stage – before concreting of the joint the calculation is made as regards the loads acting during this construction phase; by determination of forces the joints are accepted as hinged ones; 2nd – after concreting of the joint the calculation is made as regards the loads acting during this construction phase and during use of the building; by determination of forces the joints are accepted as fixed ones.

3.112. Calculation of not concreted joints of the columns mentioned in Item 3.111 (Draft 79) is made as regards local compression of the column concrete by means of centering filler plate due to condition (196) adding to its right part a part of force acting on the reinforcement connecting rods and equal to:

Nout = 0.5 ϕRsc As (229)

Trang 153

Where ϕ – coefficient of longitudinal bending for connecting rods determined in compliance with SNiP II-23-81 (Table 72) by design length l0, equal to actual length of welded connection rods; Аs Section area of all connection rods. At the same time value R

*b,loc is multiplied by coefficient ψloc = 0.75 considering

unevenness of the load spread under centering filler plate and a part of the section area of the column end Aef within the meshes of the confinement reinforcement with dimensions no more than dimensions of the compression area Аloc1 is taken as design area Аloc2.

As area Аloc1 it is taken the area of the centering filler plate or if the centering filler plate is welded to the spread plate (see Draft 79) so it’s taken the area of this plate. At the same time its dimensions must not exceed the dimensions of area Aef, and the plate thickness must be no more than 1/3 of maximum distance from the plate edge to the centering plate.

Draft 79. Not concreted joint of the column

1 centering filler plate; 2 spread plate; 3 welding of reinforcing connecting rods; 4 — confinement reinforcement meshes at the column end

3.113. Calculation of concreted connections of columns mentioned in Item 3.111 is made

as for the column section on the area with cuttings due to Items 3.50 – 3.76 considering the following conditions: а) if there is confinement reinforcement meshes in the column concrete and in the concrete for joints so the calculation is made due to Items 3.57 and 3.60 at the same time it is considered solid section bounded by means of meshes rods located at the surfaces of the concreted part of the column (Draft 80);

Draft 80. Design section of the concreted part of the column with confinement reinforcement meshes in

concrete of the column and concrete of the joint

1 concrete of the column; 2 concrete of the joint; 3 confinement reinforcement meshes

b) if there is confinement reinforcement only in the concrete of the column so the calculation is made either only with consideration of this confinement reinforcement but without considering the concrete of the joint or with consideration of the joint concrete but without considering confinement reinforcement of the column; the strength of the joint is considered to be provided if the strength conditions due to at least one of these calculations are met; c) design resistance of the column concrete and of the joints concrete (Rb or Rb,red) are multiplied by work conditions coefficient equal to γbc = 0.9 and γbs = 0.8; d) in the calculation with consideration of the concreting value ω is determined by formulas (15) or (104) as regards concrete class if it’s located on the whole width of the most compressed surface, and as regards maximum concrete class if along the compressed zone there is located partly concrete of the joints and partly concrete of the column; in formula (104) it is always considered minimum value µxy.

In the calculation of the joint considering concrete of the joint the section area of concreting Аbs should be reduced to the area of the column section by means of multiplying it by the ratio between design resistances of the joint concrete and the column concrete by constant height of the joint concrete section.

Trang 154

For symmetrically reinforced columns of rectangular section calculation of the concreted joint can be made due to formulas of Items 3.67 and 3.68, taking for h′f = hf the height of cutting sections, and for b′f = bf – width of the section reduced to concrete of the column, along the most compressed side of the section. Coefficientη, considering deflection of the column (see Item 3.54), is determined due to geometrical characteristics of the column section beyond the joint zone.

3.114. Joints of columns made by means of connection of ends by means of the cement layer or polymer solution with break of longitudinal reinforcement (see Item 5.91, joints of the 1st and the 2nd type) at the use stage are calculated as eccentric compressed concrete elements due to Item 3.6 considering confinement reinforcement by meshes due to Items 3.57 and 3.60. At the same time design resistance of concrete Rb,red is multiplied by the work condition coefficient γb, equal to 0.9 or 1.0 by filling of the joint by cement or polymer solution. If there is no solution between the ends of the columns (for example in spherical joints, in joints with connected surfaces) the mentioned above work condition coefficient is taken equal to γb = 0,65.

CALCULATION EXAMPLES Example 52. Given: a column joint – due to Draft 81; concrete of the column В30 (Rbc = 15,5 MPa by γb2 = 0,9; Rb,ser = 22 MPa); concrete for joints B20 (Rbs = 10,5 MPa by γb2 = 0,9; Rb,ser = 15 MPa); reinforcement connecting rods А-III (Rs = Rsc = 365 MPa; Rs,ser = 390 MPa), their section area Аs = А's = 4070 mm2 (4∅36); confinement reinforcement meshes made of rods А-III, with diameter 8 mm (Rs,xy = 355 MPa) with spacing s = 70 mm both in concrete of the column; longitudinal force during the use stage N = 3900 kN by γf > 1,0 and N = 3300 kN by γf = 1.0, its eccentricity in the direction perpendicular to the cuttings considering the column bending e0 = 55 mm. It is required to examine the strength of the joint during the use stage and to determine maximum allowable longitudinal force in the joint during the construction stage.

Draft 81. For the calculation example 52

1 reinforcement connection joints; 2 — spread plate; 3 centering filler plate

Calculation during the use stage. In compliance with Item 3.113a we take dimensions of the section along the axes of end rods of meshes, that means b = h = 360 mm, h0 = 330 mm (see Draft 81). Let’s determine design resistance of concrete of the column and of the joint considering confinement reinforcement meshes according to Item 3.57. For the concrete of columns:

Aef = 360 ⋅ 200 = 72 000 mm2 (see Draft 81); nx = 5; lx = 170 mm; пy = 3; ly = 360 mm; Asx = Asy = 50,3 mm2 (∅8);

;,)(,

sA

lAnlAn

ef

ysyyxsxxxy 01930

7072000

36031705350=

⋅

⋅+⋅=

+=µ

;,,

,

R

R

bc

xy,sxy2690

10515

35501930

10=

+

⋅=

+

µ=ψ

.,,,,

022690230

1

230

1=

+=

ψ+=ϕ

Therefore value Rbc,red considering work condition coefficient γbc = 0.9 (see Item 3.113c) is: Rbc,red = γbc (Rbc + ϕµхуRs,ху) = 0,9(15,5 + 2,0 ⋅ 0,0193 ⋅ 355) = 26,3 МПа.

Trang 155

For concrete of the joint in one of the cuttings Aef = 360 ⋅ 80 = 28 800 мм2 (см. черт. 81);

Asx = Asy = 50,3 мм2 (∅8); lx = 65 мм; ly = 360 мм;

µxy

x sx x y sy y

ef

n A l n A l

A s=

+=

⋅ + ⋅

⋅=

50 3 5 65 3 360

28800 700 026

, ( ), ;

ψµ

=+

=⋅

+=

xy s xy

bs

R

R

, ,

,, ;

10

0 026 355

10 5 100 450

ϕψ

=+

=+

=1

0 23

1

0 23 0 450147

, , ,, .

Value Rbs,red considering work condition coefficient γbs = 0.8 is: Rbs,red = γbs (Rbs + ϕµхуRs,ху) = 0,8(10,5 + 1,47 ⋅ 0,026 ⋅ 355) = 19,3 МПа.

Let’s determine value ω by formula (104) according to class of concrete of joints, as the cutting is located along the whole width of the most compressed surface of the column, at the same time we take minimum value µxy = 0.0193: δ2 = 10µxy = 10 ⋅ 0.0193 = 0.19 > 0.15, we take δ2 = 0.15; ω = 0.85 - 0,008Rbs + δ2 = 0.85 – 0.008 ⋅ 10.5 + 0.15 = 0.916 > 0.9, we take ω = 0.9. We reduce the section of the joint to the concrete of the column, at the same time the width of the cutting is equal:

′ = =b bR

Rf

bs red

bc red

,

,

,

,360

19 3

26 3= 264 mm;

The height of the cutting h′f = 80 mm (see Item 81). The joint strength is to be checked according to Item 3.67. For that we determine value ξR by formula (14) taking σsc,и = 500 MPa:

ξω

σ

ωR

s

sc и

R=

+ −

=

+ −

=

1 111

0 9

1365

5001

0 9

11

0 794

, ,

,

,

,

, ;

Aov = (b'f - b) h′f = (264 - 360)80 = - 7680 мм2.

The height of compressed zone is:

mm. 80> mm 4333603,26

76803,263103900

,

,==

⋅

⋅+⋅=

−= f

redbc

ovredbch

bR

ARNx

As х = 433 mm > ξR ho = 0,794 ⋅ 330 = 260 mm, so the height of compressed zone is to be determined by formula (132). For that we determine:

αs

s s

bc red o

R A

R bh= =

⋅

⋅ ⋅=

, ,, ;

365 4070

26 3 360 3300 475

αn

bc red o

N

R bh= =

⋅ ⋅=

, ,, ;

3900000

26 3 360 3301 248

αov

ov

o

A

bh= =

−

⋅= −

7680

360 3300 0646, ;

ψσ

ωc

sc и

sR

=

−

=

−

=,

,

,

,

, ;

111

500

365 10 9

11

7 53

Trang 156

α ψ α α αs c s ov n+ + −=

+ ⋅ − −=

2

0 475 7 53 0 475 0 0646 1 248

21 37

, , , , ,, ;

]

mm. 293)9,053,7475,037,137,1(330 2

2

22

=⋅⋅++−=

=+

−+++

−++−= ωψα

ααψααααψαα τ

cs

novcssnovcss

ohx

Value е is e = eo + h ao − ′

2= 55 +

330 30

2

− = 205 mm.

The joint strength is to be checked according to condition (131): Rbc,redbx (ho - x/2) + Rbc,redAov (ho - h′f/2) + Rsc A′s (ho - a′) = = 26,3 ⋅ 360 ⋅ 293 (330 - 293/2) - 26,3 ⋅ 7680 (330 - 80/2) +

+ 365 ⋅ 4070 (330 - 30) = 896.1 ⋅ 106 H ⋅ мм > Ne = 3900 ⋅ 0,205 = 800 kN ⋅ m, That means the strength joint during the use stage is provided. Let’s check crack resistance of the concreted part of the column according to Item 3.60 по similar to the calculation of the joint strength during the use stage:

ho = h - a = 400 - 50 = 350 mm; ω = 0,85 - 0,006Rbs,ser = 0,85 - 0,006 ⋅ 15 = 0,76;

′ = =b bR

Rf

bs red

bc red

,

,

40015

22 = 273 mm; h′f = 100 mm;

Aov = (b'f - b) h′f = (273 - 400) 100 = -12700 mm2; Rs = Rsc = Rs,ser = 390 MPa;

αs

s s

bc red o

R A

R bh= =

⋅

⋅ ⋅=

,

, ;390 4070

22 400 3500 515

αn

bc red o

N

R bh= =

⋅ ⋅=

,

, ;3300000

22 400 3501 07

091.0350400

12700

0

−=⋅

−==

bh

Aov

ovα ;

32.3

1.1

76.01390

400

1.11

,=

−

=

−

=ω

σψ

s

usc

c

R

;

532.02

07.1091.0515.032.3515.0

2=

−−⋅+=

−++ novacs αααψα;

=

+

−+++

−++−= ωψα

αααψααααψαcs

novscsnovscshx

2

0 22

( ) 25476.032.3515.0532.0532.0350 2 =⋅+++− mm;

2052

5035055

2

'00 =

−+=

−+=

ahee mm

Rbc,serbx (ho - x/2) + Rbc,serAov (ho - h′f/2) + RscA′s (ho - a′) = = 22 ⋅ 400 ⋅ 254 (350 - 254/2) - 22 ⋅ 12700 (350 - 100/2) + 390 ⋅ 4070 (350 - 50) =

= 890,8 ⋅ 106 H⋅мм > Ne = 3300 ⋅ 0,205 = 677 kN⋅m.

Trang 157

Calculation of not concreted joint during the construction stage. Let’s determine design resistance of concrete against compression considering confinement reinforcement due to Items 3.93 and 3.112. Area of a part of the column end section bounded by meshes contours is:

Aef = 170 ⋅ 360 = 61200 mm2. Area of the distribution plate is taken as the compression area as its thickness 20 mm is more than 1/3 of the distance from the plate edge to the centering plate (50 ⋅ 1/3 = 17 mm), at the same time the width of the compression area is taken equal to the width of the mesh 170 mm.

Aloc1 = 200 ⋅ 170 = 34 000 mm2. As 360 mm < 3 ⋅ 200 mm, we take Aloc2 = Aef = 61200 mm2, Therefore:

5.322.134000

6120033

1

2 <===loc

loc

bA

Aϕ ;

56.261200

340005.35.45.35.4 1 =−=−=

ef

loc

sA

Aϕ ;

( )0226.0

7061200

360317053.50=

⋅

⋅+⋅=

+=

sA

lAnlAn

ef

ysyyxsxx

xyµ

As the calculation is made as regards the loads during the construction stage so we take Rbc

= 19 MPa (that means γb2 = 1.1):

277.01019

3550226.0

10,

=+

⋅=

+=

bc

xysxy

R

Rµψ ;

97.1277.023.0

1

23.0

1=

+=

++=

ψϕ .

Value R*

b,loc is determined by formula (197) considering coefficient ψloc = 0,75: R

*b,loc = ψloc (Rbс ϕb + ϕµxy Rs,xy ϕs) = 0,75 (19 ⋅ 1,22 +

+ 1,97 ⋅ 0,0226 ⋅ 355 ⋅2,56) = 47,7 MPa. By formula (229) we determine the force in reinforcement connecting rods. Radius of inertia of reinforcement rod ∅36 is:

4

36

4==

di = 9 мм.

Welded connection rods length is l = lo = 400 мм.

Due to table 72 of SNiP II-23-81 by λ = i

l0 =9

400 = 44,4 and Ry = Rs = 365 МPа we find

ϕ = 0,838, therefore Nout = 0,5 ϕRsАs = 0,5 ⋅ 0,838 × 365 ⋅ 8140 = 1245 ⋅ 103 N. Maximum longitudinal force acting on the not concreted joint is:

N = R*b,loc Aloc1 + Nоut = 47,7 ⋅ 34000 + 1245 ⋅ 103 = 2867 ⋅ 103 N

CALCULATION OF CONCRETE KEYS 3.115. Dimensions of concrete keys which transfer shearing forces between a prefabricated

element and additional concrete (Draft 82), should be determined by formulas:

Trang 158

kkb

knlR

Qt ≥ (230)

kkbt

knlR

Qh

2≥ (231)

Where Q – shearing force transferred by concrete keys; tk, hk, lk – depth, height and length of the concrete key; пk – number of concrete keys inserted into the calculation and taken no more

than tree.

Draft 82. Scheme for the calculation of concrete keys transferring shearing force from the

prefabricated element to monolithic concrete

1 prefabricated element; 2 monolith concrete

By compression force N it is possible to determine the height of concrete keys by formula

kkbt

knlR

NQh

2

7.0−= (231)

and to take it decreased in comparison with the height determined by formula (231), but no more than one half as much. If deck elements are connected by means of concrete keys so the length of keys inserted into the calculation must be no more than a half of a span, at the same time value Q is taken equal to the sum of shearing forces along the whole length of the element. Due to conditions (230) – (232) it is necessary to check the keys of a reinforced concrete element and keys of additional concrete, taking design resistance of concrete keys Rb and Rbt as for concrete structures. Note. By the calculation of a stretched leg of a two-leg column as regards the pulling out of a column pocket it is possible to take into account five keys. 4. CALCULATION OF CONCRETE AND REINFORCED CONCRETE

ELEMENTS AS REGARDS LIMIT STATES OF THE SECOND GROUP

CALCULATION OF CRACK FORMATION OF REINFORCED CONCRETE ELEMENTS 4.1.(4.1). It is necessary to calculate crack formation of reinforced concrete elements as

regards crack formation: - Normal to the longitudinal axis of the element; - Inclined to the longitudinal axis of the element.

Calculation of crack formation is made: a) to find out if it is necessary to calculate the crack growth; b) to determine the deformation calculation case.

In the reinforced element or its part there is no cracks if forces caused by total load (or its part when load cause forces with different signs) and inserted into the calculation with safety factor γf = 1.0, are less than forces acting on the section during crack formation. Total load include dead loads, long-term and short-term loads.

Trang 159

It is possible to take without calculation that bending moments of rectangular and T-sections with compressed flanges have normal to the longitudinal axis cracks on the most compressed parts if required by the calculation reinforcement coefficient µ > 0,005.

4.2.(4.5). Calculation of reinforced concrete elements элементов as regards formation of

normal cracks is made due to the following condition Мr < Мcrc, (233)

where Мr – moment of external forces, located on one side on the considered section, relating to the axis which is parallel to zero line and which goes through the heart point, most distant from the stretched zone where crack formation is checked;

Мcrc – moment acting on the section normal to the longitudinal axis of the element

during crack formation and determined by formula Mcrc = Rbt,serWpl Mshr, (234)

here Mshr – moment of force Nshr caused by concrete settlement relating to the same axis like for determination of Мr; sign of the moment is determined by spinning direction ("plus" – if directions are opposed, "minus" – if direction of moments Мshr and Мr are the same).

For free supported beans and slabs moment Мcrc is determined by formula

Mcrc = Rbt,serWpl - Nshr (eop + r). (235) Force Nshr is considered as external tension force; its value and eccentricity relating to the center of gravity of the section are determined by formulas:

Nshr = σshr (As + A′s); (236)

'

''

0ss

ssss

pAA

yAyAe

−

−= , (237)

where σshr – stress caused by concrete settlement equal to: 40 MPa – for heavy-weight concrete В35 and less by natural hardening and 35 MPa – by heat treating; for other kinds and classes of concrete value σshr is taken due to SNiP 2.03.01-84 (Table5, position 8);

уs, у′s – distance from the center of gravity of the section to the centers if gravity of

sections of reinforcement S and S′.

If reinforcement coefficient µ < 0.01 is possible in formulas (234) and (235) so values Wpl and r are to be determined as for concrete section taking Nshr = 0 and As = A's = 0.

Value Mr is determined by formulas: - For bending element(Draft 83, а)

Мr = М; - For eccentric compressed element (Draft 83, б)

Mr = N(eo - r), (238) - For centrally- and eccentric элементов (черт. 83, в)

Mr = N(eo + r), (239) In formulas (234), (235), (238) and (239): r – the distance from the center of gravity of the section to the heart point which is

most distant from the stretched zone whose crack formation is checked.

Trang 160

Value r is determined by formulas: - For bending elements – by formula

red

red

A

Wr = ; (240)

- For eccentric compressed elements – by formula

red

red

A

Wr ϕ= (241)

here serb

b

R ,

6.1σ

ϕ −=

but it’s taken no less than 0,7 and no more than 1,0; σb – maximum stress in compressed concrete, determined as for elastic body

- For centrally- and eccentric stretched elements – by formula

( )'2 ss

pl

AAA

Wr

++=

α, (242)

Wpl – resistance moment of the transformed section for end stretched fibre considering non-elastic deformations of stretched concrete determined according to Item 4.3.

N o t e . Transformed section includes concrete section as well as section of all longitudinal reinforcement multiplied by the ratio between correspondent modulus of elasticity of reinforcement and concrete.

4.3.(4.7). Resistance moment of the transformed sections for end stretched fibre Wpl

(considering non-elastic deformations of stretched reinforcement) is determined with the assumption that there is no longitudinal force N by formula

( )0

'0002

bssb

pl Sxh

IIIW +

−

++=

αα , (243)

where Ibo, Iso, I′so – inertia moments of sections areas of compressed concrete zone, reinforcement S and S′ relating to the zero line;

Sbo – static moment of the section area of stretched concrete zone relating to the zero line.

Location of the zero line in the general case is determine due to the following condition

( )20

'0

'0

bt

ssb

AxhSSS

−=−+ αα ,

(244) where S′bo, Sso, S′so – static moments of the section area of the compressed concrete zone, reinforcement S and S′ relating to the zero line;

Abt – section area of stretched concrete zone.

For rectangular sections, I- and T-sections condition (244) has the following form:

red

red

A

Sxh =− (245)

where redS – static moment of the area of transformed section calculated without

considering the area of stretched overhangs relating to the end stretched fibre;

Trang 161

redA – area of the transformed section calculated without considering a half of the

area of stretched overhangs.

Condition (245) can be used if calculated according to it zero line crosses the rib of a T- or I-section.

Draft 83. Forces schemes and stresses diagrams in the cross section of the element during calculation of

crack formation, normal to the longitudinal axis of the element

а – by bending; б – by eccentric compression; в – by central and eccentric compression; 1 – heart point; 2 – center of gravity of the transformed section

Value Wpl can be determined by formula

Wpl = [0,292 + 0,75 (γ1 + 2µ1α) + 0,075(γ′1 + 2µ′1α)] bh2;

(246)

Where( )

bh

hbb ff −=1γ ;

( )bh

hbb ff

'''1

−=γ

bh

As=1µ ; bh

As

''1 =µ

b

s

E

E=α

By determined value Wred (see Item 4.2) value Wpl can be also determined by formula

Wpl = γ Wred, (247) where γ – see table 29.

T a b l e 29

Section Coefficient γ Cross section form 1. Rectangular

1,75

2. T-section with a flange located in the compressed zone

1,75

3. T-section with a flange located in the stretched zone: а) by bf/b ≤ 2 independently on value hf/h

б) bf/b > 2 and hf/h ≥ 0,2 в) bf/b > 2 and hf/h < 0,2

1,75

1,75 1,50

4. I-section, symmetrical: а) by b′f/b = bf/b ≤ 2 independently on the ratio h′f/h = hf/h

b) by 2 < b′f/b = bf/b ≤ 6 independently on the ratio h′f/h = hf/h

c) by b'f/b = b′f/b > 6 and h′f/h = h′f/h > 0,2 d) by 6 < b′f/b = bf/b ≤ 15 and h′f/h = hf/h < 0,2 e) by b′f/b = bf/b > 15 and

1,75

1,50

1,50

1,25

1,10

Trang 162

h′f/h = hf/h < 0,2 5. I-section, asymmetrical meeting the requirement b′f/b ≤ 3: а) by bf/b ≤ 2 independently on the ratio hf/h

b) by 2 < bf/b ≤ 6 independently on the ratio hf/h

c) by bf/b > 6 and hf/h > 0,1

1,75

1,50

1,50

6. I-section, asymmetrical meeting the requirement 3 < b′f/b < 8: a) by b′f/b ≤ 4 independently on the ratio hf/h

b) by bf/b > 4 and hf/h ≥ 0,2 c) by bf/b > 4 and hf/h < 0,2

1,50

1,50 1,25

7. I-section, asymmetrical meeting the requirement b′f/b ≥ 8: а) by hf/h > 0,3 б) by hf/h ≤ 0,3

1,50 1,25

8. Ring- and round section

2-0,4D1/D

9. X-section: а) by b′f/b ≥ 2 and 0,9 ≥ h′f/h > 0,2 б) in other cases

2,00

1,75

N o t e s : 1. In Table 29 symbols bf and hf correspond to dimensions of a flange which is stretched by the crack formation calculation, and b′f and h′f – dimensions of a flange which is compressed for that case.

2. Wpl = γ Wred, where Wred – resistance moment for the stretched surface of the transformed section determined according to rules of resistance of elastic materials.

4.4. The parts along the elements where there are no inclined cracks are determined according to the following condition

Q ≤ ϕb3 Rbt.serbho, (248) Where ϕb3 – see Table 21. CALCULATION OF REINFORCED CONCRETE ELEMENTS AS REGARDS THE CRACK FORMATION

4.5. (4.13). Reinforced concrete elements are calculated as regards formation of cracks: - Normal to the longitudinal axis of the element; - Inclined to the longitudinal axis of the element.

Examination of the width of the crack opening is not required if according to the calculation due to Items 4.1 – 4.4 they are not caused by dead loads, long-term loads and short-term loads inserted into the calculation with the safety factor γf = 1.0. For bending and eccentric compressed elements of statically undeterminable systems by one-row reinforcement mentioned in Table 1, position 4, it is not necessary to check the width of the opening of normal cracks in the following cases:

Trang 163

а) for reinforcement А-I and А-II: by any reinforcement coefficient µ, if diameter d ≤ 20 mm; by µ ≥ 0,01, if diameter d = 22 – 40 mm; б) for reinforcement А-III: by any reinforcement coefficient µ, if diameter d ≤ 8 mm; by µ ≥ 0,01, if diameter d = 10 — 25 мм; by µ ≥ 0,015, if diameter d = 28 40 мм; в) for reinforcement Вр-1 – by µ ≥ 0,006 by any diameters. By calculation of crack formation the force caused by concrete settlement Nshr is taken equal to zero.

4.6. In the general case calculation of crack opening is made two times: short-term and long-term crack opening (see Item 1.15).

For elements mentioned in Table 1 position 4 and made of heavy-weight and light-weight concrete it is possible to make only one calculation during examination of opening of cracks normal to the longitudinal axis of the element:

If 3

2≥

r

rl

M

Mso long-term crack formation is checked;

If 3

2<

r

rl

M

M so short-term crack formation is checked,

here Mrl, Mr – is moment Mr (see Item 4.2) caused by the sum of dead loads and long-term loads and by all loads. Calculation of crack opening of normal to the longitudinal axis of the element

cracks

4.7(4.14). The width of the crack opening of normal to the longitudinal axis of the element

cracks, acrc mm, must be determined by formula

( )31005.320 daE

as

s

lcrc −=σ

ηδϕ

(249) where δ – coefficient taken equal to:

1.0............. for bending and eccentric compressed elements 1.2............................................ for stretched elements; ϕl – coefficient taken equal to: 1.00 ............. when considering short-term loads and dead loads and long-term loads of short duration; when considering loads of long duration and dead loads and long-term loads for structures made of: heavy-weight concrete: natural humidity...................... µϕ 1560.1 −=l

Trang 164

In water saturated state (elements taking liquids pressure, as well as elements used in the ground below ground waters level) ...............................................1,20 By changing water saturation and drying ...…………………….. 1,75 Fine concrete of groups: А .................................................................................1,75 Б ................................................................................. 2,00 В ................................................................................. 1,50 Light-weight concrete of class В12.5 and more ..................................... 1.50 Porous concrete ............................................................ 2.00

Values ϕl for fine, light-weight and porous concrete in water saturated state are multiplied by coefficient 0.8, and by changing water saturation and drying – by coefficient 1.2; η – coefficient taken equal to: for reinforcement А-II and А-III – 1.0; А-I – 1.3; Вр-1 – 1.2; σs – stress in the rods of end row of reinforcement S, determined according to Item 4.9; µ – coefficient of the section reinforcement taken equal to the ratio between the reinforcement section S to the concrete section area (by working height ho and without considering compressed overhangs of flanges), but no more than 0.02, at the same time for I-sections, rectangular and T-sections

( )( ) 02.00

≤−−+

=ahbbbh

A

ff

sµ

(250) if hf < а, so stretched overhangs are not considered during calculation; if in eccentric stretched elements force N is located between centers of gravity of reinforcement S and S′, by determination of µ working height ho is taken from the point of application of force N to the least stretched surface, at the same time for the central stretching µ where As,tot – is the area of all longitudinal reinforcement; d – diameter of stretched reinforcement, mm; by different diameters of rods value d is taken equal to:

kk

kk

dndn

dndnd

++

++=

...

...

11

2211

(251) here d1, ..., dk – diameter of rods of stretched reinforcement;

п1, ..., nk – quantity of rods with diameter d1, ..., dk..

Besides it is necessary to consider instructions of Item 4.8.

4.8 (4.14). Width of the crack opening асrc, determined due to Item 4.7 is to be revised in the following cases:

а) if center of gravity of the section of the end row of reinforcement S of bending, eccentric compressed, eccentric stretched by eо ≥ 0,8ho elements is distant from the most stretched fibre of concrete at a2 > 0,2h, value acrc must be increased by multiplying it by coefficient δa, which is:

Trang 165

3

120 2 −= h

a

aδ (252)

and taken no more than 3; b) for under-reinforced bending and eccentric compressed elements made of heavy-weight and light-weight concrete (for example foundations) by 008.0≤µ ( µ – see Item 4.7) value асrc, determined by formula (249), can be decreased if necessary by means of multiplying it by coefficient ϕb, considering work of stretched concrete above the cracks and which is determined by the following formula

ϕb = ϕf1 ϕl1, (253) but no more than 1.0; where ϕf1 – coefficient considering loading level and equal to:

rcrc

crcr

fM

M

MM

MM 0

01

−

−=ϕ (254)

ϕl1 – coefficient considering duration of the load action and equal to: 1.0 – by short-term loads and dead loads and long-term loads of short

duration;

r

crc

lM

M8.11 =ϕ (255)

by dead loads and long-term loads of long duration

but no less than 1.0; Мo – moment by which stretched concrete above the cracks is excluded out of

the work and which is equal to: Мо = Мсrс + ψbh

2Rbt,ser, (256)

Whereη

αµψ 15= , but no more than 0.6;

if Mо < Mr, so coefficient ϕb is not to be determined; Mr – moment determined due to Item 4.2, caused by total load including dead

load, long-term and short-term load; Mcrc – see Item 4.2; ηµ , – see Item 4.7.

By using coefficient ϕb and by3

2<

r

rl

M

M it is not necessary to use recommendations of

Item 4.6; c) for statically undeterminable systems as well as for free supported beams by l/h < 7, near the points of application of point loads and support reactions by µ ≤ 0,02 the width of the crack opening acrc determined by formula (249) can be decreased by means of multiplying by the coefficient ϕloc, considering local features of stressed state in reinforced concrete structures and determined by the following formula

( )( )( )hh

hhah

M

Floc

−

−−−=

0

0

28.2

233.01ϕ ,

(257)

Trang 166

but no less than 0.8 and no more than 1.0, where F – absolute value of the point load or of the support reaction;

М – absolute value of bending moment in the normal section going through the point of application of the point load and of the support reaction (Draft 84);

а – the distance from the point of application of the point load or of the support reaction to the considered section taken in compliance with the Draft 84, but no more than 0,3h;

h – distance from the surface of the element at which it is applied force F to the stretched surface;

ho – the same to the stretched reinforcement (Draft 85);

d) for elements made of light-weight concrete В7,5 and less value acrc must be increased by 20 percent.

Draft 84. Location of support reactions in fixed joints taken for determination of the coefficient ϕϕϕϕloc

а - г prefabricated elements joints; д - и monolithic parts

Draft 85. Design schemes for determination of coefficient ϕϕϕϕloc

а application of load to the compressed surface of the element; б the same to the flanges of the element; в the same along the length of the statically undeterminable beam

4.9 (4.15). Stresses in the stretched reinforcement σs are determined by the following formulas: – for eccentric stretched elements

s

sA

N=σ (258)

– for bending elements

s

sA

N=σ ; (259)

– in eccentric compressed and eccentric stretched elements ( )

zA

zeN

s

s

±=σ (260)

In formula (260) sign «plus» is taken by eccentric stretching, sign «minus» – by eccentric compression. By location of stretching longitudinal force N between centers of gravity of reinforcement S and S′ value еs is taken with the sign «minus». In formulas (259) and (260): z – distance from the center of gravity of the section area of reinforcement S to the

point of application of the resultant of forces in the compressed zone of the section above the crack determined due to Item 4.16, at the same time for eccentric stretched elements by ео < 0,8ho z is taken equal to zs – distance between centers of gravity of reinforcement S and S′, coefficient v in formula (277) is always taken equal to vsh = 0.45 (as by short duration of the load); it is possible to take z the same like by calculation of deformation caused by the same

loads if 01.00

<bh

As .

In case when Mr < Mcrc (see Item 4.2), value σs, is determined by formula

Trang 167

crc

r

crcssM

M,σσ = (261)

where σs,crc – stress in reinforcement by action of the load corresponding to the crack formation determined by formulas (259) and (260) replacing М by Mcrc and N by

r

crc

crcM

MNN = .

By determination of Ncrc moments Mcrc and Mr can be calculated by r = 0,8Wred /Ared.

By location of the stretched reinforcement in several rows along the section height in bending, eccentric compressed and eccentric stretched elements by eo > 0,8ho stresses σs must be multiplied by the coefficient δn, equal to:

1

2

axh

axhn

−−

−−=δ (262)

where х = ξho; value ξ is determined by formula (274); for bending elements it is possible to take value х the same like by calculation of the strength calculation;

а1, a2 — distance from the center of gravity of total reinforcement S and of the end row of rods to the most stretched fibre of concrete.

Value of stress σs caused by the total load determined considering coefficient δn, must be no more than Rs,ser. It is not necessary to check this condition for statically undeterminable structures with reinforcement of one class by its location in one row. Simplified way of determination of σs. For bending moments it is possible to determine σs by the following formula:

u

ssM

MR=σ (263)

where Ми – limit moment of the strength equal to: - By determination of the section strength – equal to the right part of equations

(17) – (21), (28), (30) - By choosing of the reinforcement section

sd

facts

dtotuA

AMM

,,=

here Mtot,d – moment caused by total load with the safety factor as regards the load γf > 1.0;

Аs,fact – actual area of accepted reinforcement; Asd – area of reinforcement required by the strength calculation.

By using of reinforcement of different classed it is necessary to insert design resistance of reinforcement for limit states of the second group Rs for the most hard reinforcement in formula (263). For eccentric compressed elements of heavy-weight and light-weight concrete by Mr ≥ Mcrc it is possible to use σs by formula

crc

s

s

shA

Neϕσ

0

= (264)

Trang 168

where ϕcrc – coefficient determined due to Table 30.

T a b l e 30

ϕf 0h

es

Coefficients ϕcrc by values µα, equal to

0,01 0,02 0,03 0,05 0,07 0,10 0,15 0,20 0,25 0,30 0,40 0,50

0

≤0,8 1,0 1,2 1,5 2,0 3,0 4,0

0,04 0,18 0,31 0,44 0,59 0,74 0,81

0,07 0,22 0,34 0,48 0,62 0,77 0,84

0,10 0,25 0,37 0,50 0,64 0,79 0,86

0,15 0,29 0,40 0,53 0,67 0,82 0,89

0,18 0,31 0,43 0,56 0,70 0,84 0,91

0,22 0,34 0,46 0,58 0,72 0,86 0,94

0,26 0,38 0,49 0,62 0,75 0,89 0,97

0,31 0,42 0,55 0,65 0,78 0,92 1,00

0,34 0,45 0,55 0,67 0,81 0,95 1,02

0,37 0,47 0,57 0,69 0,82 0,96 1,03

0,41 0,50 0,60 0,72 0,85 0,99 1,06

0,45 0,52 0,62 0,74 0,87 1,01 1,08

0,05

≤0,8 1,0 1,2 1,5 2,0 3,0 4,0

0,04 0,17 0,30 0,44 0,59 0,74 0,82

0,04 0,20 0,33 0,46 0,61 0,76 0,84

0,07 0,22 0,35 0,48 0,63 0,78 0,85

0,11 0,26 0,38 0,51 0,65 0,80 0,86

0,14 0,28 0,40 0,53 0,67 0,82 0,90

0,18 0,31 0,43 0,56 0,70 0,84 0,92

0,22 0,34 0,46 0,59 0,72 0,87 0,94

0,26 0,38 0,49 0,61 0,75 0,89 0,97

0,29 0,40 0,51 0,64 0,77 0,91 0,99

0,32 0,42 0,53 0,66 0,79 0,93 1,00

0,36 0,46 0,56 0,68 0,82 0,95 1,03

0,38 0,48 0,58 0,70 0,83 0,97 1,04

0,10

≤0,8 1,0 1,2 1,5 2,0 3,0 4,0

0,03 0,16 0,30 0,44 0,59 0,75 0,83

0,04 0,19 0,32 0,46 0,61 0,76 0,84

0,05 0,21 0,33 0,47 0,62 0,77 0,85

0,09 0,24 0,36 0,50 0,64 0,79 0,87

0,11 0,26 0,38 0,52 0,66 0,81 0,88

0,14 0,28 0,40 0,54 0,68 0,83 0,90

0,16 0,31 0,43 0,56 0,70 0,85 0,92

0,22 0,34 0,46 0,59 0,73 0,87 0,94

0,25 0,37 0,48 0,61 0,75 0,89 0,96

0,28 0,39 0,50 0,63 0,76 0,90 0,98

0,31 0,42 0,53 0,65 0,79 0,93 1,00

0,34 0,44 0,55 0,67 0,80 0,94 1,02

0,20

≤0,8 1,0 1,2 1,5 2,0 3,0 4,0

0,03 0,15 0,29 0,44 0,59 0,75 0,83

0,04 0,17 0,31 0,45 0,60 0,76 0,84

0,05 0,19 0,32 0,46 0,61 0,77 0,85

0,06 0,21 0,34 0,48 0,63 0,78 0,86

0,07 0,23 0,35 0,49 0,64 0,79 0,87

0,10 0,25 0,37 0,51 0,65 0,81 0,88

0,13 0,27 0,40 0,53 0,67 0,82 0,90

0,16 0,30 0,42 0,55 0,69 0,84 0,91

0,19 0,32 0,44 0,57 0,71 0,85 0,93

0,21 0,34 0,45 0,58 0,72 0,87 0,94

0,25 0,37 0,48 0,61 0,75 0,89 0,96

0,28 0,39 0,50 0,63 0,76 0,90 0,98

0,30

≤0,8 1,0 1,2 1,5 2,0 3,0 4,0

0,03 0,15 0,29 0,44 0,59 0,75 0,84

0,04 0,16 0,30 0,45 0,60 0,76 0,85

0,05 0,17 0,31 0,46 0,61 0,77 0,85

0,05 0,19 0,33 0,47 0,62 0,78 0,86

0,06 0,21 0,34 0,48 0,63 0,78 0,86

0,07 0,23 0,35 0,49 0,64 0,79 0,87

0,10 0,25 0,37 0,51 0,66 0,81 0,89

0,12 0,27 0,39 0,53 0,67 0,82 0,90

0,15 0,29 0,41 0,54 0,68 0,83 0,91

0,17 0,30 0,42 0,55 0,70 0,84 0,92

0,20 0,33 0,45 0,58 0,72 0,86 0,94

0,23 0,35 0,46 0,59 0,73 0,88 0,95

0,50

≤0,8 1,0 1,2 1,5 2,0 3,0 4,0

0,04 0,15 0,29 0,44 0,60 0,77 0,85

0,04 0,16 0,30 0,45 0,61 0,77 0,85

0,04 0,16 0,30 0,45 0,61 0,77 0,85

0,04 0,18 0,31 0,46 0,62 0,77 0,86

0,04 0,19 0,32 0,47 0,62 0,78 0,86

0,04 0,20 0,33 0,48 0,63 0,79 0,87

0,06 0,22 0,35 0,49 0,64 0,79 0,87

0,08 0,23 0,36 0,50 0,65 0,80 0,88

0,10 0,24 0,37 0,51 0,66 0,81 0,89

0,12 0,26 0,38 0,52 0,67 0,82 0,89

0,15 0,28 0,40 0,54 0,68 0,83 0,91

0,17 0,30 0,42 0,55 0,69 0,84 0,92

0,70

≤0,8 1,0 1,2 1,5 2,0 3,0 4,0

0,04 0,15 0,29 0,45 0,61 0,77 0,85

0,04 0,15 0,30 0,45 0,61 0,77 0,85

0,04 0,16 0,30 0,45 0,61 0,77 0,85

0,04 0,17 0,31 0,46 0,61 0,77 0,86

0,04 0,18 0,32 0,46 0,62 0,78 0,86

0,04 0,19 0,32 0,47 0,62 0,78 0,86

0,04 0,20 0,34 0,48 0,63 0,79 0,87

0,06 0,21 0,34 0,49 0,64 0,79 0,87

0,07 0,22 0,35 0,49 0,64 0,80 0,88

0,08 0,23 0,36 0,50 0,65 0,80 0,88

0,11 0,25 0,38 0,52 0,66 0,81 0,89

0,13 0,27 0,39 0,53 0,67 0,82 0,90

Trang 169

( )

0

'''

9.0bh

Ahbb s

ff

f

α

ϕ+−

= ; N

Mye ss += ;

b

ss

E

E

bh

A

0

=µα

Calculated by formulas (263) and (264) values σs in case when reinforcement is located in several rows along the section height are multiplied by the coefficient δn.

4.10 (4.14). The width of crack opening caused by all loads is determined as a sum of the width of crack opening caused by dead loads and long-term loads (by ϕl > 1.0) and expansion of the width of the opening caused by short-term loads (by ϕl = 1.0). So the width of the opening is determined by the following formula

−+=

lsl

s

lcrccrc aaϕσ

σ 111,

(265) where acrс,l – the width of crack opening caused by dead loads and long-term loads;

ϕl > 1,0 – see Item 4.7; if value acrс,l is determined considering formula (253), so coefficient ϕl in formula (265) is replaced by the product ϕl ϕl1 (where ϕl1

– see Item 4.8б); σsl, σs – are determined due to Item 4.9, caused by the sum of dad loads and long-term loads as well as by all loads.

Calculation of opening of cracks inclined to the longitudinal axis of the element

4.11(4.17). The width of the crack opening inclined to the longitudinal axis of the element by reinforcement by means of stirrups which are normal to the longitudinal axis must be determined by the following formula

( )wb

w

s

wcw

lcrc

Eh

dE

da

αµ

ησϕ

2115.0

06

0

++

=

(266) where ϕl – coefficient taken equal to:

considering short-term loads and dead loads and long term loads of short duration........ 1,00 considering dead loads and long term loads of long duration for structures made of: heavy-weight concrete: by natural humidity............................................................................................................ 1,50 in water saturated state...................................................................................................... 1,20 by changing water saturation and drying……………………………………………...… 1,75 fine and light-weight concrete – the same like in formula (249);

η – the same like in formula (249); dw – diameter of stirrups;

bs

Asw

w =µ ;

σsw – stress in stirrups determined by the following formula

0

1

hA

QQ

sw

b

sw

−=σ ; (267)

stress σsw must be no more than Rs,ser;

Trang 170

Qb1 – the right part of condition (72) with the coefficient ϕb4, multiplied by 0.8, at the same time Rbt is replaced by Rbt,ser, taken no more than the value corresponding to concrete class В30;

Q – shear force at the end of inclined section with the projection length с.

Value с is taken no more than 2hо. If by the calculation of the element as regard distributed load the following condition is met

bRq serbtb ,41 2.0 ϕ≤ (268)

(where q1 –see Item 3.32), so value с can be taken equal to only 2hо. For elements made of light-weight concrete В7.5 and lower value acrc, determined by formula (266), must be increased by 30 percent. By determination of the width of inclined cracks caused by all loads it is necessary to consider instructions of Item 4.10. At the same time in formula (265) coefficient ϕl is taken according to the present Item and the ratio σsl/σs is replaced by the ration between the stresses σswl/σsw, which are calculated by formula (267) of the sum of dead loads and long-term loads and all loads. It is possible to decrease value асrc by 1.5 times in comparison with the value determined by formula (266), if the element is reinforced by means of longitudinal rods with the same diameter like diameter of stirrups and with the distances along the section height equal to the stirrups spacing s.

EXAMPLES OF CALCULATION

Example 53. Given: a reinforced concrete floor slab with the cross section dimensions (for a half of the slab section) due to Draft 86: b = 85 mm, h = 400 mm, b′f = 725 mm, h′f= 50 mm; heavy-weight concrete В25; main reinforcement A-III (Rs = 365 MPa; Еs = 2⋅105 MPa), located in two rows (a1 = 58 mm; a2 = 33 mm); its section area Аs = 760 mm2 (2∅22) ; total moment in the middle of the span Мtot = 69 kN⋅m; all loads are dead loads and long-term loads; due to the strength calculation Ми = 92,3 kN⋅m and х = 30 mm. It is required to calculate the opening of normal to the longitudinal axis of the element cracks.

Draft 86. For the calculation example 53

C a l c u l a t i o n . ho = h – а = 400 – 58 = 342 mm. So according to Item 4.1 we take that the element works with the cracks in the stretched zone. For determination of the long-duration crack opening we determine the stress in reinforcement σs. Due to formula (263), value σs at the level of the center of gravity of reinforcement is:

2733.92

69365 ===

u

ssM

MRσ MPa

As reinforcement is located in two rows so we calculate coefficient δn is to be determined by formula (262):

Trang 171

08.15830400

3330400

1

2 =−−

−−=

−−

−−=

axh

axhnδ

The stress in the lower reinforcement rod: σs = 273⋅1.08 = 294 MPa.

The width of the crack opening we determine by formula (249). As 02.00262.0 >== µµ

so value µ is taken equal to 0.02. Due to Item 4.7, δ =

1.0; 3.102.0156.1156.1 =⋅−=−= µϕ l ;

η = 1.0; d = 22 mm.

( ) ( ) 16.02202.01005.320102

2940.13.10.11005.320 3

53 =⋅−

⋅⋅⋅=−= d

Ea

s

s

lcrc µσ

ηδϕ mm

Which is less than the maximum allowable width of the crack opening acrc2 = 0,3 mm.

Example 54. Given: reinforced concrete foundation slab with the cross section dimensions h = 300 mm, b = 1150 mm; a = 42 mm; heavy-weight concrete В15 (Rbt,ser = 1.15 MPa; Eb = 2.05⋅104 MPa); main reinforcement A-III (Rs = 365 MPa; Еs = 2⋅105 MPa); its section area Аs = 923 mm2 (6 ∅14); moment in the design section caused by dead loads and long-term loads Ml = 63 kN⋅m, by short-term loads Msh = 4 kN⋅m; limit moment of strength Мu = 80,5 kN⋅m; the foundation is located above the ground waters level. It is required to calculate the opening of normal cracks. C a l c u l a t i o n . ho = h - а = 300 - 42 = 258 mm. Let’s determine if it’s necessary to calculate the width of the crack opening due to Item 4.5. For that we determine the crack formation moment Mcrc.

As 67463 =+=+== shltotr MMMM kN⋅m 8.34=> crcM kN⋅m so moment Mcrc is

determined for the concrete section using formula (246): Mcrc = Rbt,serWpl = 0,292bh

2Rbt,ser = 0.292⋅1150⋅3002⋅1.15 = 34.75⋅106 M⋅mm = 34,8 kN⋅m.

As Mr = Мtot = Ml + Мsh = 63 + 4 = 67 кН⋅м > Mcrc = 34.8 kN⋅m, so it is necessary to check the crack opening. As foundation is located below the ground waters level so maximum allowable crack opening due to Table 1, position 4 is acrc2 = 0.3 mm, that’s why by

3

294.0

67

63>===

tot

l

r

rl

M

M

M

M according to Item 4.6, the calculation is made only as

regards long-duration crack opening caused by moment Ml.

The width of the crack opening is to be determined by formula (249). The stress in the reinforcement σs is to be determined by the simplified formula (263):

2865.80

63365 ===

u

l

ssM

MRσ MPa

Coefficients inserted into formula (249) are taken equal to: δ = 1.0; η = 1.0; d = 14 mm, so

( ) ( ) 34.0140031.01005.320102

2860.155.10.11005.320 3

53 =⋅−

⋅⋅⋅=−= d

Ea

s

s

lcrc µσ

ηδϕ m

m

Trang 172

Which is more than maximum allowable value acrc2 = 0.3 mm, so it is recommended to correct value acrc due to Item 4.8б. As 008.0<µ so such correction is allowable. As

a2 = а = 42 mm < 0.2h = 0.2 ⋅ 300 = 60 mm, so correction of value acrc, due to Item 4.8а is not to be made. Due to formula (256) we determine value Мо after determination of:

76.91005.2

1024

5

=⋅

⋅==

b

s

E

Eα ;

6.0454.076.90031.01515

<=⋅⋅==η

µαψ ;

Mo = Mcrс + ψbh

2Rbt,ser = 34,8⋅106 + 0,454⋅1150⋅3002⋅1.15 = 88.8⋅106 N⋅mm = 88.8 kN⋅m.

As Мо = 88.8 kN⋅m > Mr = 67 kN⋅m, so we determine coefficient ϕb by formula (253):

79.067

8.88

8.348.88

8.34670

01 =

−

−=

−

−=

rcrc

crcr

fM

M

MM

MMϕ

1935.067

8.348.18.11 <===

r

crc

lM

Mϕ ,

so we take ϕl1 = 1;

ϕb = ϕf1ϕl1 = 0.79 < 1.

Considering coefficient ϕb, the width of the crack opening acrc = 0.34⋅0.79 = 0.269 mm < 0.3 mm, that means less than maximum allowable value. Example 55. Given: a reinforced concrete column of the industrial building with the cross section dimensions h = 500 mm, b = 400 mm; a = a′ = 50 mm; heavy-weight concrete В15 (Rb,ser = 11 MPa; Rbt,ser = 1.15 MPa; Eb = 2.05⋅104 MPa); main reinforcement A-III (Еs = 2⋅105 MPa); its section area As = A′s = 1232 mm2 (2 ∅28); longitudinal compression force N = Nl = 500 kN; moment caused by total load Mtot = 240 kN⋅m, including moment of dead loads and long-term loads Ml = 150 kN⋅m. It is required to calculate the column as regards the crack opening. C a l c u l a t i o n . h o = h - a = 500 - 50 = 450 mm. Let’s determine if it’s necessary to calculate the crack opening. For that end we check condition (233). As

0069.0450400

1232

0

=⋅

==bh

Asµ so due to Item 4.2 the resistance moment Wpl is to be

calculated as for the concrete section. Using formula (246), we find: Mcrc = 0,292bh

2Rbt,ser = 0.292⋅400⋅5002⋅1.15 = 33.6⋅106 N⋅mm.

Heart distance r we determine by formula (241). For that we determine σb as for elastic body (without considering reinforcement influence):

9.166/500400

10240

500400

105002

63

=⋅

⋅+

⋅

⋅=+=

W

M

A

N tot

bσ MPa

7.006.011

9.166.16.1

,

<=−=−=serb

b

R

σϕ

We take ϕ = 0,7;

058.04.586

5007.0

6===== mm

h

A

Wr ϕϕ .m

Due to formula (238) we determine moment Mr:

Trang 173

( ) 211058.05002400 =⋅−=−=−= NrMreNM totr kN⋅m 6.33=> crcM kN⋅m

that means condition (233) is not met. So examination of the crack opening is required.

As 3

2573.0

211

058.0500150<=

⋅−=

−=

r

ll

r

rl

M

rNM

M

M due to Item 4.6 we check short-

duration crack opening. For that in compliance with item 4.10 let’s preliminary determine the width of the long-duration cracks caused by forces Ml and Nl by formula (249). At the same time we use simplified formula (264) for σs.

50010500

1015050

2

500

2 3

6

=⋅

⋅+−=+−=

l

l

sN

Ma

he mm;

067.01005.2

1020069.0

4

5

=⋅

⋅==

b

s

E

Eµµα

074.09.0

067.0

9.0

'

9.0 0

'

====αµα

ϕbh

As

f

Due to calculated values ϕf = 0,074, µα = 0,067 and 11.1450

500

0

==h

es we find in Table 30

coefficient ϕcrc = 0,33.

In compliance with Item 4.7, δ = 1.0; η = 1.00; 02.00069.0 <== µµ ;

5.10069.0156.1156.1 =⋅−=−= µϕ l

( ) ( ) 191.0280069.01005.320102

14900.15.10.11005.320 3

53

, =⋅−⋅

⋅⋅=−= dF

as

s

llcrc µσ

ηδϕ m

m The stress in reinforcement σs caused by all loads is to be also determined by formula (264).

68010500

1024050

2

500

2 3

6

=⋅

⋅+−=+−=

N

Ma

he tot

s mm

51.1450

680

0

==h

es

By ϕf = 0,074,µα = 0,067 and 51.10

=h

es coefficient ϕcrc due to Table 30 is 0,522.

320522.04501232

68010500 3

0

=⋅

⋅⋅== crc

s

s

shA

Neϕσ MPa

So due to formula (265),

34.05.1

0.10.1

149

3200.1191.0

0.10.10.1, =

−+=

−+=

lsl

s

lcrccrc aaϕσ

σmm

which is less than maximum allowable value acrcl = 0.4 mm (see Table 1, position 4).

MPaMPahA

Necrc

s

s

sls 1498.14833.04501232

50010500 3

0

≈=⋅

⋅⋅=== ϕσσ

Trang 174

Example 56. Given: free supported bam of the floor with the span l = 5.5 m, loaded by the distributed loads: temporary long-duration equivalent load v = 30 kN/m and dead load g = 12.5 kN/m; dimensions of the cross section b = 200 mm, h = 400 mm, hо = 370 mm; heavy-weight concrete В 15 (Rbt,ser = 1.15 MPa; Eb = 2,05⋅104 MPa); two-legs stirrups made of reinforcement А-I (Еs = 2.1⋅105 MPa) with the spacing s = 150 mm, diameter 8 mm (Аsw = =101 mm2). It is required to make the calculation of inclined cracks. C a l c u l a t i o n . Let’s check if the calculation of cracks is necessary according to condition (248). Maximum shear force in the cross section is:

( ) ( )117

2

5.55.1230

22max =+

=+

==lgvql

Q kN

Due to Table 21, ϕb3 = 0.6. ϕb3Rbt,serbho = 0.6⋅1.15⋅200⋅370 = 51060 N < Qmax = 117 kN, that means inclined cracks are formed and the calculation is required. The calculation is made due to Item 4.11. Let’s determine values Q and Qb1. q1 = g + v/2 = 12.5 + 30/2 = 27.5 kN/n; ϕb4 = 1.5 (see Table 21). As 0.2ϕb4Rbt,serb = 0.2⋅1.5⋅1.15⋅200 = 56.9 N/mm > q1 = 27.5 N/mm, during determination of Qb1 and Q value с is taken equal to с = 2hо = 2⋅370 = 740 mm. So

51060740

37020015.15.15.08.0 220,4

1

⋅⋅⋅⋅==

c

bhRbQ

serbt

b

ϕN;

Q = Qmax – q1 с = 117 – 27.5 · 0.740 = 96.65 kN. Let’s determine the stress in the stirrups by formula (267):

235183150370101

5106096650,

0

1 =<=⋅

−=

−= sers

sw

b

sw RMPashA

QQσ MPa

Due to Items 4.7 and 4.11, ϕl = 1.5; η = 1.3; dw = 8 mm.

00337.0150200

101=

⋅==

s

sw

wb

Aµ ;

24.101005.2

101.24

5

=⋅

⋅==

b

s

E

Eα

Let’s determine the width of inclined cracks opening by formula (266):

( ) ( )219.0

00337.024.10211005.215.0370

8101.2

3.181836.05.1

2115.0

6.0

45

0

=

⋅⋅+⋅⋅+⋅

⋅⋅⋅=

++

=

vb

w

s

wsw

lcrc

Eh

dE

da

αµ

ησϕ

mm which is less than maximum allowable value асrc = 0.3 mm (see Table 1). CALCULATION OF ELEMENTS OF REINFORCED CONCRETE STRUCTURES AS REGARDS THE DEFORMATIONS 4.12(4.22) Deformations (deflections, angle of rotation) of elements of reinforced concrete

structures must be calculated due to formulas of structural theory by means of

Trang 175

determination of curvature values in compliance with instructions of Items 4.13 – 4.21.

4.13(4.23). Curvature is determined:

a) For parts of the element where there are no cracks normal to the longitudinal axis of the element (see Item 4.1) – as for the solid body; b) For the parts of the element where there are cracks normal to the longitudinal axis of the element – as the ratio of the difference of average deformations of the end fibre of the compressed concrete zone and longitudinal stretched reinforcement to the working height of the element section.

During calculation of deformations the force from the concrete settlement Nshr is taken equal to zero. Determination of reinforced concrete elements curvature on the parts without cracks

in the stretched zone

4.14(4.24). On the parts where there are no normal to the longitudinal axis of the element

cracks total value of the curvature must be determined by the following formula

21

111

+

=

rrr (269)

where1

1

r 2

1

r – the curvature caused by short-term loads (determined due to

Item1.12) and by dead loads and long-duration temporary loads determined by the following formulas:

redbb

sh

IE

M

r 11

1

ϕ=

;

(270)

redbb

bl

IE

M

r 1

2

1

1

ϕ

ϕ=

;

ϕb1 – coefficient considering the influence of short-duration creep of concrete and taken equal to: for heavy-weight, light-weight and fine concrete with dense fine aggregate................................................... 0,85 for light-weight and porous concrete with fine aggregate..........................................................0,70

ϕb2 – coefficient considering influence of long-duration creep of concrete on the deformations of the element without cracks taken due to Table 31.

Trang 176

T a b l e 31 (34, 35)

Coefficients ϕb2 and vl by air humidity of the surrounding environment, percents

Kind of concrete 40 – 75 (normal)

lower than 40 (lowered)

more than 75 (increased)

ϕb2 vl ϕb2 vl ϕb2 vl

Heavy-weight, light-weight 2,0 0,15 3,0 0,10 1,6 0,19 Porous 2,0 0,07 3,0 0,04 1,6 0,09 Fine groups:

А 2,6 0,10 3,9 0,07 2,1 0,125 Б 3,0 0,08 4,5 0,05 2,4 0,10 В 2,0 0,15 3,0 0,10 1,6 0,19

N o t e s : 1. Air humidity of the surrounding area is taken due to instructions of Item 1.8. 2. Groups of fine concrete see in Item 2.1. 3. By changing water saturation and drying of concrete values ϕb2, must be multiplied, and values vl – divided into the coefficient 1.2.

Determination of reinforced concrete elements curvature on the parts with cracks in

the stretched zone

4.15 (4.27). On the parts where there are normal to the longitudinal axis cracks in the

stretched zone the curvature of bending, eccentric compressed as well as eccentric stretched by eо ≥ 0,8ho elements of rectangular, T- and I-section must be determined by the following formula

( )ss

s

bb

b

ss

ss

AEh

N

vEbhAEzh

M

r 000

1 ψ

ξϕ

ψψm

++= (271)

For bending elements the last summand of the right part of formula (271) is taken equal to zero. Sign «minus» in this formula is taken by eccentric compression, sign «plus» – by eccentric stretching. In formula (271): Мs – moment relating to the axis which is normal to the plane of action of the moment

and going through the center of gravity of the reinforcement section area S, of all external forces located on one side of the considered section equal to:

- For bending elements Мs = М; - For eccentric compressed and eccentric stretched elements Мs = Nes;

z – distance from the center of gravity of reinforcement section area S to the point of application of the resultant of forces in the compressed zone of the section above the crack (lever arm), determined due to instructions of Item 4.16;

ψs – coefficient considering the work of stretched concrete on the part with cracks and determined due to instructions of Item 4.17;

ψb – coefficient considering unevenness of deformations spread of end compressed fibre of concrete on the length of the part with cracks and taken equal to:

for heavy-weight, fine and light-weight concrete of class В10 and more ...........................................................................0.9

Trang 177

For light-weight and porous concrete of class В7.5 and less... 0.7

ϕf – coefficient determined by formula (277); ξ = x/ho – is determined due to instructions of Item 4.16;

v – coefficient which characterizes the elastic-plastic state of concrete of the compressed zone and taken equal to:

by short-duration action of the load – to coefficient vsh = 0.45; by long-duration action of the load – to the coefficient vl, determined by Table 31.

For bending and eccentric compressed elements made of heavy-weight concrete Mr < Mo curvature can be determined considering the work of stretched concrete above the crack using the following formula

totcrc

crcr

crccrc M

M

MM

MM

rrrr

−

−

−

+

=

00

1111 (272)

where

21

1b

redbb

crc

crc IE

NrM

rϕ

ϕ

+=

; (273)

0

1

r– curvature determined by formula (271) by the moment Мs, which is equal:

for bending moments Мs = Мo; for eccentric compressed moments Ms = Мo + Nуsr;

уsr = yo – а + r – distance from the center of gravity of stretched reinforcement to the axis going through the most distant heart point (see Item 4.2);

Мr – moment determined due to Item 4.2 caused by total load including dead load, long-term and short-term load;

Мo – moment by which stretched concrete above the cracks is excluded out of the work and which is determined by formula (256), where ψ is halved by consideration of long duration of dead loads and long-term loads;

Mcrc,r – see Item 4.2; M, Mtot – moment of external forces relating to the axis going through the center of gravity

of the section caused by the load under review and by total load; ϕb1, ϕb2 – see Item 4.14; by short duration action of the load ϕb2 = 1.0.

4.16.(4.28). Value ξ is determined by formula

( )

55.11

5.1

10

511

0

mh

es

fϕ

µα

λδβ

ξ+

±++

+

= (274)

but taken no more than 1.0, at the same time es/ho is taken no less than 0.5. For bending elements the last summand of the right part of formula (274) is taken equal to zero. In formula (274) top signs are taken by compression force and lower signs – by compression force N. In formula (274): β – coefficient taken equal to:

Trang 178

for heavy-weight and light-weight concrete......... 1,8 for fine concrete.................................................... 1,6 for porous concrete............................................... 1,4

serb

s

Rbh

M

,20

=δ ; (275)

−=

0

'

21

h

h f

fϕλ ; (276)

( )

0

'''

2bh

v

Abbh s

ff

f

α

ϕ+−

= ; (277)

b

ss

E

E

bh

A

0

=µα . (278)

Value z is determined by formula

( )

+

+

−=ξϕ

ξϕ

f

f

f

h

h

hz2

1

2

0

'

0 (279)

For element of rectangular section and T-section with the flange in the stretched zone it is necessary to put values 2a' or h′f = 0 instead of h′f in formulas (276) and (279) if there is or there is no reinforcement S′.

If0

'

h

a<ξ , so for bending moments by ϕf ≥ ϕf1, where

( )

0

'0

1

21

2.0'/2

h

h

ah

f

f

−

−−−=

δβµαϕ ,

values z and r

1 are determined by

0

'

h

a=ξ and ϕf = ϕf1, and by ϕf < ϕf1, values ϕf ,ξ, z

and r

1 are determined without considering S′. For eccentric loaded elements

by0

'

h

a<ξ it is always possible to determine values ϕf , ξ, z and

r

1 without

considering S′.

Calculation of sections which have a flange in the compressed zone by 0h

h f<ξ is

made as for rectangular sections with the width b′f.

Design value of a flange b′f is determined due to Item 3.23. For eccentric compressed elements value z must be taken no more than 0,97es.

4.17.(4.29). Coefficient ψs is determined by formula

Trang 179

( )0

2

8.15.3

125.1

h

es

m

m

mlss

ϕ

ϕϕϕψ

−

−−−=

(280) but no more than 1.0, at the same time it is necessary to take es/ho ≥ 1,2/ϕls. For bending elements the last part of the right part of formula (280) is taken equal to zero. In formula (280): ϕls – coefficient considering influence of duration of the load action and taken in compliance with Table 32;

r

plserbt

mM

WR ,=ϕ (281)

T a b l e 32(36) Duration of the load action Coefficient ϕls by concrete class

В10 and more В7.5 and less 1. Short duration by reinforcement classes:

А-II, А-III 1,1 0,8 А-I, Вр-I 1,0 0,7 2. Long duration 0,8 0,6 But no more than 1.0; Here Wpl – see Item 4.3;

Mr – see Item 4.2.

4.18. Curvature of eccentric stretched elements with longitudinal force N, applied between centers of gravity of areas of reinforcement S and S′, on the parts with normal cracks in the stretched zone is determined by the following formula:

+−=

ss

s

ss

s

s

s

sss

s

AEAEz

Ne

AEz

N

r

'

2

1 ψψψ (282)

where zs = ho – a′ – distance between centers of gravity of reinforcement areas S and S′;

ψs, ψ′s – coefficients which consider the work of stretched concrete for reinforcement S and S′ and determined by formulas:

N

N crc

lss ϕψ −= 1 (283)

N

N crc

lss

'' 1 ϕψ −= (284)

here ϕls – coefficient taken equal to: by short duration of the load action........................................... 0,70 by long duration of the load action............................................ 0,35

Ncrc, N′crc – forces applied in the same point that force N and corresponding to crack formation in the most and the least stretched zone of concrete values Ncrc and N′crc are determined by the following formulas:

Trang 180

re

WRN

plserbt

crc+

=0

, .

(285)

0

,'

'

'

er

WRN

plserbt

crc−

=

and taken no more than N; besides by r' < ео value N′crc is taken equal to N.

In formulas (285): Wpl, W′pl – values Wpl, determined due to Item 4.3 correspondingly for the most and

the least stretched side of the section; r, r′ – distances from the center of gravity of the section to the heart points which

are most distant from the most and the least stretched side of the section; values r and r′ are determined by formula (242).

4.19. Curvature of eccentric stretched elements with longitudinal force N, applied beyond

the centers of gravity of reinforcement areas S and S′, and by eо < 0,8hо is determined

by means of linear interpolation between curvatureIr

1 determined by formula (282)

by еs = 0 (that means by eо = yso, where yso – is the distance from the center of gravity

of reinforcement area S to the center of gravity of the section), and curvatureIIr

1

determined by formula (271) еs =0,8ho - yso (that means by ео = 0,8hо). Than curvature value is:

008.0

1111

s

s

IIII yh

e

rrrr −

−

+

=

(286) 4.20. For elements of rectangular section with symmetric reinforcement under skew

eccentric pressure curvatures are determined by the following formula

βϕ/11

0

=

rr (287)

where 0

1

r– curvature calculated as for the flat eccentric compression due to Item

4.15 – 4.17 on the assumption of the force action N with the eccentricity еo in he plane of he symmetry axis х, at the same time it’s accepted that force plane is located between axis х and the section diagonal;

ϕβ – coefficient considering the influence of the angle of slope of the force plane on the deformation value of skew compressed elements and determined by the following formula

( )

−−−

+

+−−= β

πββ

αµ

αµ

πϕ β 2

cos12.41

2.411

21

3

x

y

x

y

h

h; (288)

A

Asx

x =µ ; A

Asy

y =µ

Trang 181

here sysx AA , – areas of reinforcement located at the surface of the section normal to

axes х and у, at the same time angle rod is considered as by calculation of µх, and µу;

hх, hy – dimensions of the section in the direction of axes х and у;

β – angle of slope of the force plane (plane of the eccentricity of force N) to the plane of axis х, radian.

Deformation plane forms and angle γ with the plane of axis х; angle γ is determined due to the following equation:

y

x

I

Itgtg βγ = (289)

where Ix, Iy – inertia moments of the section relating to axes y and х. Curvatures in planes х and у by skew eccentric compression are:

γcos11

rr x

=

, (290)

γsin11

rr y

=

(291)

wherer

1 is determined by formula (287).

4.21 (4.30). Total curvature r

1 for the part with cracks in the stretched zone must be

determined by the following formula

321

1111

+

−

=

rrrr (292)

where1

1

r curvature caused by all loads of short duration, due to which the

calculation of deformations is made according to the instructions of Item 1.17;

2

1

r curvature caused by dead loads and long-tem loads of short duration;

3

1

rcurvature caused by dead loads and long-term loads of long-duration.

Curvatures 1

1

r ,

2

1

rand

3

1

rare determined by formulas (271), (272), (282),

(286) and (287), at the same time 1

1

r and

2

1

r are calculated by ψs and v

corresponding to the action of the load of short duration, and 3

1

r by ψs and v

corresponding to long duration of the load action. If values 2

1

r and

3

1

r are

negative so they are taken equal to zero.

Trang 182

Determination of deflections

4.22 (4.31). Deflection fm caused by the deflection deformation is determined by the following formula

dxr

Mfx

l

xm

= ∫

1

0

(293)

where xM bending moment in the section х caused by the unit-force applied in the

direction of required displacement of the element in the section along the span length for which the deflection is determined;

xr

1 total value of he element curvature in the section х caused by the load by

which the deflection is calculated.

By determination of the deflection in the middle of the span formula (293) can get the following view:

( ) ( ) ( ) ( )

( )

−+

+

+

+

= ∑

−

= miril

n

irl

mr

nrr

irrn

lf

123

116

11

12

12/

1002

2

(294)

where( )0

1

lr

,

( )0

1

rr

curvatures of the element in the left and the right support;

( )ilr

1,

( )irr

1,

mr

1 curvatures of the element in the section i, in the

symmetrical section i' (draft 87) and in the middle of the span;

п – even number of equal parts on which the span of the element is divided; number п is to be taken no less than 6.

Draft 87. Diagram of the curvature in the reinforced concrete element with variable section along the

perimeter

In formulas (293) and (294) curvatures r

1 are determined by formulas (269) and

(292) correspondingly for the parts without cracks and with cracks; sign r

1 is taken in

compliance with the curvatures diagram. By determination of deflections of statically undeterminable structures it is recommended to consider redistribution of moments caused by cracks formation and non-elastic deformations of concrete. For bending elements of constant section which have cracks on each part within which bending moment has no any sign the curvature can be calculated for the most stressed section taking it for other sections of the same part variable proportionally with the values of bending moment (Draft 88).

Trang 183

Draft 88. Diagrams of bending moments and curvature in the reinforced concrete element of constant

section

а — load distribution scheme; б — bending moments diagram; в curvature diagram

4.23 (4.32, 4.33). For bending elements by 10<h

l it is necessary to consider the influence

of shear forces on their deflection. In that case total deflection ftot is equal to the sum of deflections caused by deformations of deflection fm and deformation of displacement fq.

Deflection fq, caused by deformation of displacement is determined by the following formula

∫=l

xxq dxQf0

γ (295)

where xQ – shear force in the section х caused by the directed to the required

displacement unit force applied in the section where deflection is determined;

γx – shear deformation determined by the following formula

crc

bx

xGbh

Qϕ

ϕγ 25.1

= (296)

here Qx – shear force in the section х caused by external load; ϕb2 – coefficient considering influence of long-duration creep of concrete

and taken according to Table 31; by short-duration action of the load ϕb2 = 1.0;

G – concrete displacement module (see Item 2.12); ϕcrc – coefficient considering influence of cracks on displacement

deformations taken equal to: - On the parts along the length of the element where there are no

normal and inclined to the longitudinal axis of the element cracks.................... 1,0;

- On the parts where there are only inclined to the longitudinal axis of the element cracks....................... 4,8;

- On the parts where there are only normal or normal and inclined to the longitudinal axis of the element cracks – due to formula

xx

redb

crcrM

IE

=

13ϕ (297)

here Mx xr

1– correspondingly moment and curvature in the section caused by the

load by which deflection is determined by its short duration.

4.24.(4.34). For solid slabs no less than 250 mm thick reinforced by flat meshes with cracks in the stretched zone values of deflections calculated by formula (293), are multiplied

by the coefficient 3

0

0

7

−h

htaken no more than 1.5 where h – in mm.

Trang 184

Determination of longitudinal deformations

4.25. Relative deformations εo (elongation or shortening) in the direction of the longitudinal axis of elements are determined in the following manner.

1. Relative deformations of eccentric compressed and eccentric stretched elements with one-valued diagram of stresses: а) for eccentric compressed elements or their single parts – by formula

00 22y

IvE

Ne

AvE

N

redbredb

±=ε (298)

b) for eccentric stretched elements or its parts if there are no cracks – by formula

01

20

1

20 y

IE

Ne

AE

N

redbb

b

redbb

b

ϕ

ϕ

ϕ

ϕε ±−= (299)

In formulas (298) and (299) sign «plus» corresponds to deformations of shortening, sign «minus» – to deformations of elongation; c) for eccentric stretched elements or their parts if there are cracks (that means for elements mentioned in Item 4.18) – by formula

( )

s

sssmssm

z

yzy −+−=

εεε

'

0 (300)

where εsm, ε′sm – average values of elongation of reinforcement S and S', determined by formulas:

( )s

sss

ss

smzAE

ezNψε

−=

(301)

''

's

sss

s

smzAE

Neψε =

here zs, ψs, ψ′s – see Item 4.18. 2. Relative deformations of bending, eccentric compressed and eccentric stretched elements with double-valued diagram of stresses in the section: а) for elements or their parts without cracks in the stretched zone – by formula

01

20 2

yIE

M

AvE

N

redbb

b

redb ϕ

ϕε ±±= . (302)

Rule of signs is the same like for formulas (298) and (299); b) for parts of elements mentioned in Item 4.15 which have cracks in the stretched zone – by formula

( )

0

00

h

yhy ssmsbm −−=

εεε (303)

where εsm, εbm – average values of relative elongation of reinforcement and relative shortening of end compressed fibre of concrete on the part between cracks determined by the following formulas:

Trang 185

= N

z

M

AE

s

ss

s

sm mψ

ε (304)

( ) vEbhz

M

bf

sb

sm

0ξϕ

ψε

+= (305)

where Ms, ψs, z, ϕf, ξ – see Items 4.15 – 4.17; rules of signed see in Item 4.15; c) for the parts of eccentric stretched elements mentioned in Item 4.19 – by means of linear interpolation between value εо determine by formula (300) by еs = 0 (that means by еo = yso) and value еo determined by formula (303) by еs = 0,8ho (that means by еo = 0,8 ho + yso) where yso – see Item 4.19. In formulas (298) – (305):

ys – distance from the fibre under review to the center of gravity of reinforcement S; уо – the same to the center of gravity of the section;

ϕb1, ϕb2 – see Item 4.14; by short-duration action of the load ϕb2 = 1.0; v – see Item 4.15.

Deformations εо, determined by formulas (298) – (303) with sign «plus» correspond to shortening, with sign «minus» – to elongation. By dead loads, long-term and short-term loads acting at the same time value εо is to be determined similar to determination of total curvature due to Item 4.21.

4.26. Shortening (elongation) of elements at the level of considered fibre is determined by the following formula

∑=

=∆n

i

ioill1

ε (306)

where εоi – relative longitudinal deformations in the section, located in the middle of the part with the length li;

п – quantity of parts on which the length of the elements is divided. Approximate methods of deformations calculation

4.27. Deflections of reinforced concrete bending elements made of heavy-weight concrete of constant section, used by normal and high humidity (air humidity more than 40 percent) are obvious less than maximum allowable deflections if the following condition is met

lim0

λ≤h

l (307)

where λlim – limit ratio of the span to the working height of the section less than which examination of deflections is not required (Table 33).

By 10<h

l deflections are les than maximum allowable ones if condition (308) which

considers influence of shear deformations on the element deflection is met:

lim00

:18 λ≤+h

l

h

l (308)

Trang 186

Values λlim given in Table 33 correspond to long duration of distributed load on the

free supported beam by limit deflection equal to200

l.

T a b l e 33

Sections

Coefficients ϕf, ϕft

Values λlim for determination of cases, when it is not necessary to check the deflection of elements of heavy-

weight concrete, by values µα, equal to 0,02 0,04 0,07 0,10 0,15 0,20 0,30 0,40 0,50

ϕf = ϕft = 0 25 17

17 12

14 10

12 9

10 8

9 8

10 10

11 11

11 11

ϕf = 0,2; ϕft = 0

31 22

22 16

18 13

16 11

12 9

10 8

10 10

11 11

11 11

ϕf = 0,4; ϕft = 0

42 25

25 17

23 15

18 12

14 10

11 8

10 9

10 9

11 11

ϕf = 0,6; ϕft = 0

45 30

28 30

24 17

19 14

16 12

13 9

11 9

10 10

11 11

ϕf = 0,8; ϕft = 0

48 32

30 21

25 18

20 15

18 13

15 10

12 9

10 9

10 10

ϕf = 1,0; ϕft = 0

50 35

33 22

26 19

23 17

20 14

17 12

14 9

11 9

10 10

ϕft = 0,2; ϕf = 0

28 18

17 12

14 10

12 9

10 8

9 8

10 10

11 11

11 11

ϕft = 0,6; ϕf = 0

32 20

20 13

15 10

13 9

10 8

9 9

10 10

11 11

12 12

ϕft = 1,0; ϕf = 0

36 22

23 14

16 10

13 9

10 8

9 9

10 10

11 11

12 12

Trang 187

ϕf = ϕft = 0,2 34 23

25 17

19 14

16 11

12 9

10 8

10 10

11 11

11 11

ϕf = ϕft = 0,6 48 33

34 25

26 18

21 14

16 12

15 9

11 8

10 10

11 11

ϕf = ϕft = 1,0 55 42

44 36

36 21

26 17

20 14

17 12

14 9

11 9

10 9

As = A′s

N o t e . Values λlim, given above the line are used by the calculation of the elements reinforced by steel of А-II, under the line – of class A-III.

If maximum allowable deflections f (see Item 1.17) are less than200

l, so values λlim

of Table 33 must be decreased by

l

f:

200

1 times (for example if

300

1=

l

f– by 1,5

times, if 400

1=

l

f– by 2 times).

For solid slabs less than 250 mm thick reinforced by flat meshes values λlim are decreased by means of dividing into the coefficient mentioned in Item 4.24. N o t e . Values λlim can be decreased in the following cases: а) if deflection is determined by action of moment Ml which is the part of the total moment Мtot (positions 2 – 4 of Table 2) – by means of multiplying of λlim Table 33 by the ratio Мtot/ Ml; b) if load is different from the distributed load – by means of multiplying λlim Table 33 by the

ratio mp:48

5 where рm is coefficient taken due to table 35 according to the loading scheme;

в) if deflection is determined according to the combined work of short-term loads, long-term loads and dead loads – by means of multiplying λlim of Table 33 by coefficient ϕθ, determined by the following formula

( )11 −+

=

θ

θϕθ

tot

l

M

M

where θ – ration between the deformation caused by long action of the load and deformation caused by short duration of the same load taken equal to: θ = 1,8 – for elements of rectangular section; θ = 1,5 for elements of T-section with a flange in the compressed zone; θ = 2,2 – for elements of T-section with a flange in the stretched zone.

Trang 188

DETERMINATION OF CURVATURE

4.28. For bending elements made of heavy-weight concrete of constant section mentioned in Item 4.15 and used by environment air humidity higher than 40 percent

curvaturer

1 on the parts with cracks is determined by the following formula

201

,2

21

hAE

RbhM

rss

serbt

ϕ

ϕ−= (309)

where ϕ1, ϕ2 – see Table 34. By dead loads, long-term loads and short-term loads acting at the same time curvature

r

1 is determined by the following formula

−+=

l

serbtll

sh

sh

ss

RbhMM

hAEr 1

,2

2

120

11

ϕ

ϕ

ϕ (310)

где ϕ1sh — coefficient ϕ1 by short duration of the load; ϕ1l, ϕ2l — coefficients ϕ1 and ϕ2 by long duration of the load.

DETERMINATION OF DEFLECTIONS

4.29. For bending elements by 10≥h

l deflection f is determined by the following manner:

а) for elements of constant section which work as free supported or console beams – by formula

21l

rf m

m

ρ

= (311)

where mr

1– curvature in the section with maximum bending moment caused by the

load by which the deflection is determined; рm – coefficient taken due to Table 35;

by the loading scheme of a free supported or a console beam not given in Table 35, the deflection is determined by formulas of materials resistance by rigidity equal to the ratio between maximum moment and maximum curvature; b) if deflection determined due to sub-item «а» is more than maximum allowable value for under-reinforced elements (µ ≤ 0.5 percent) this value is to be determined more exact due to considering of increased rigidity on the parts without cracks by variable rigidity on the parts with cracks; for free supported beams loaded by distributed load it corresponds to the following formula

2

,

111l

rrrf crc

elmm

m

m

−

−

= ρρ (312)

Trang 189

T a b l e 34 Coefficients Coefficient ϕ1 by values µα, equal to Coefficient ϕ2 by values µα, equal to

ϕft ϕf 0,02 0,03 0,04 0,05 0,06 0,07 0,08 0,10 0,13 0,15 0,17 0,20 0,25 0,30 0,35 0,40 0,45 0,50 <0,04 0,04-0,08 0,08-0,15 0,15-0,30 0,30-0,50 Long duration of the load

0,0 0,0 0,0 0,0 0,0 0,0

0,0 0,2 0,4 0,6 0,8 1,0

0,43 0,49 0,52 0,54 0,56 0,57

0,39 0,46 0,49 0,51 0,53 0,54

0,36 0,44 0,47 0,49 0,51 0,52

0,34 0,42 0,46 0,48 0,49 0,51

0,32 0,41 0,45 0,47 0,48 0,50

0,30 0,39 0,44 0,46 0,47 0,49

0,28 0,37 0,42 0,44 0,46 0,48

0,26 0,35 0,40 0,43 0,45 0,47

0,23 0,31 0,38 0,42 0,44 0,46

0,22 0,29 0,35 0,39 0,42 0,44

0,21 0,27 0,33 0,37 0,40 0,42

0,19 0,25 0,31 0,35 0,38 0,41

0,16 0,21 0,26 0,31 0,35 0,38

0,14 0,19 0,24 0,28 0,32 0,35

0,13 0,17 0,22 0,25 0,29 0,32

0,12 0,16 0,20 0,23 0,27 0,30

0,11 0,14 0,18 0,22 0,25 0,28

0,10 0,13 0,17 0,20 0,23 0,26

0,10 0,12 0,13 0,13 0,14 0,15

0,07 0,09 0,10 0,11 0,12 0,13

0,04 0,05 0,06 0,08 0,09 0,10

0,00 0,00 0,02 0,02 0,04 0,06

0,00 0,00 0,00 0,00 0,00 0,00

0,2 0,4 0,6 0,8 1,0

0,0 0,0 0,0 0,0 0,0

0,47 - - - -

0,40 0,42 0,43

- -

0,36 0,36 0,37 0,38 0,40

0,33 0,33 0,33 0,33 0;33

0,31 0,31 0,31 0,30 0,30

0,30 0,30 0,30 0,29 0,29

0,28 0,28 0,27 0,27 0,27

0,26 0,26 0,25 0,24 0,24

0,23 0,22 0,22 0,22 0,22

0,22 0,21 0,21 0,21 0,20

0,21 0,20 0,20 0,20 0,19

0,19 0,19 0,18 0,17 0,17

0,16 0,16 0,15 0,15 0,15

0,14 0,14 0,14 0,14 0,14

0,13 0,13 0,12 0,12 0,12

0,11 0,11 0,11 0,11 0,11

0,11 0,10 0,10 0,10 0,10

0,10 0,10 0,10 0,10 0,10

0,15 0,18 0,20 0,23 0,25

0,12 0,16 0,19 0,22 0,24

0,08 0,13 0,17 0,20 0,23

0,03 0,06 0,09 0,12 0,14

0,00 0,02 0,03 0,05 0,06

0,2 0,4 0,6 0,8 1,0

0,2 0,4 0,6 0,8 1,0

0,51 - - - -

0,45 0,53

- - -

0,43 0,49 0,53

- -

0,40 0,47 0,50 0,53 0,61

0,38 0,45 0,48 0,50 0,53

0,37 0,43 0,46 0,48 0,50

0,36 0,42 0,44 0,46 0,48

0,34 0,39 0,41 0,44 0,45

0,30 0,37 0,39 0,41 0,43

0,28 0,35 0,38 0,39 0,40

0,26 0,33 0,36 0,38 0,39

0,24 0,30 0,34 0,37 0,38

0,21 0,26 0,31 0,34 0,36

0,19 0,23 0,28 0,31 0,34

0,17 0,21 0,25 0,29 0,32

0,16 0,20 0,23 0,26 0,29

0,14 0,18 0,21 0,25 0,27

0,13 0,17 0,20 0,23 0,26

0,16 0,20 0,24

- -

0,13 0,19 0,22 0,25 0,26

0,08 0,14 0,20 0,24 0,25

0,04 0,07 0,12 0,19 0,20

0,00 0,03 0,04 0,08 0,12

Short duration of the load 0,0 0,0 0,0 0,0 0,0 0,0

0,0 0,2 0,4 0,6 0,8 1,0

0,64 0,72 0,76 0,79 0,82 0,84

0,59 0,66 0,69 0,71 0,73 0,74

0,56 0,63 0,66 0,69 0,70 0,71

0,53 0,61 0,65 0,67 0,68 0,69

0,51 0,59 0,63 0,65 0,67 0,68

0,50 0,58 0,62 0,64 0,66 0,67

0,49 0,57 0,61 0,63 0,65 0,66

0,46 0,56 0,60 0,63 0,65 0,66

0,43 0,53 0,59 0,62 0,64 0,66

0,41 0,51 0,57 0,61 0,63 0,65

0,40 0,49 0,56 0,60 0,63 0,65

0,37 0,46 0,53 0,58 0,61 0,63

0,34 0,43 0,49 0,55 0,58 0,61

0,32 0,40 0,46 0,52 0,56 0,59

0,30 0,37 0,44 0,49 0,53 0,56

0,28 0,35 0,41 0,46 0,50 0,54

0,26 0,33 0,39 0,44 0,48 0,52

0,25 0,31 0,37 0,42 0,46 0,50

0,17 0,21 0,23 0,25 0,26 0,27

0,14 0,18 0,20 0,21 0,23 0,24

0,09 0,11 0,14 0,16 0,17 0,18,

0,02 0,03 0,04 0,05 0,06 0,07

0,00 0,00 0,00 0,00 0,00 0,00

0,2 0,4 0,6 0,8 1,0

0,0 0,0 0,0 0,0 0,0

0,74 - - - -

0,60 0,63 0,81

- -

0,56 0,57 0,59 0,63 0,84

0,53 0,54 0,54 0,55 0,57

0,51 0,51 0,51 0,51 0,52

0,49 0,49 0,49 0,49 0,49

0,47 0,47 0,47 0,47 0,47

0,44 0,44 0,44 0,44 0,44

0,42 0,42 0,42 0,42 0,42

0,40 0,40 0,40 0,40 0,40

0,39 0,39 0,39 0,39 0,39

0,37 0,37 0,37 0,37 0,37

0,34 0,34 0,34 0,34 0,34

0,32 0,32 0,32 0,32 0,32

0,30 0,30 0,30 0,30 0,30

0,28 0,28 0,28 0,28 0,28

0,26 0,26 0,26 0,26 0,27

0,25 0,25 0,25 0,25 0,25

0,28 0,35 0,36 0,45 0,50

0,23 0,31 0,39 0,40 0,46

0,16 0,25 0,32 0,38 0,44

0,07 0,14 0,20 0,25 0,29

0,00 0,03 0,08 0,12 0,15

0,2 0,4 0,6 0,8 1,0

0,2 0,4 0,6 0,8 1,0

0,79 - - - -

0,67 0,77

- - -

0,63 0,69 0,76

- -

0,61 0,66 0,70 0,76 0,92

0,59 0,64 0,67 0,71 0,76

0,58 0,62 0,65 0,68 0,71

0,56 0,61 0,64 0,66 0,69

0,55 0,58 0,61 0,64 0,66

0,52 0,56 0,58 0,61 0,63

0,50 0,55 0,57 0,59 0,61

0,48 0,54 0,56 0,58 0,60

0,46 0,52 0,55 0,57 0,58

0,42 0,48 0,53 0,56 0,57

0,39 0,45 0,50 0,53 0,56

0,37 0,43 0,47 0,51 0,54

0,35 0,40 0,45 0,49 0,52

0,33 0,38 0,43 0,47 0,50

0,31 0,37 0,41 0,45 0,48

0,27 0,39 0,50

- -

0,24 0,37 0,46 0,60 0,72

0,17 0,30 0,44 0,57 0,70

0,08 0,16 0,28 0,41 0,55

0,00 0,04 0,11 0,21 0,31

Trang 190

where рcrc – coefficient taken due to Table 36 according to the ratio Mcrc/Mtot (Mcrc – see Items 4.2 and 4.3) ;

elmr ,

1

– curvature in the section with maximum moment determined as for a

solid body by formula (270) caused by the load by which the deflection is determined; it is possible to determine value Ired in formula (270) as for a concrete element.

For other loading schemes value f can be determined by formula (314); For bending elements with built-in supports deflection in the middle of the span is determined by the following formula

( ) ( )

2

00 8

1115.0

1l

rrrf m

rl

m

m

−

+

−

= ρρ (313)

where mr

1

( )0

1

lr

( )0

1

rr

– curvature of the element in the middle of the span, on the

right and on the left support; рm – coefficient determined due to Table 35 as for a free supported beam;

d) for elements of variable section, as well as in cases when more exact determination of deflections is required than by formulas (311) and (313), and the elements and the loads are symmetric relating to the middle of the span deflection is determined by the following formula

+

+

+

=

mrrrr

lf

18

112

16

1

216 210

2

(314)

where 0

1

r 1

1

r 2

1

r mr

1– curvatures on the support, at the distance l

6

1 from the

support, at the distance l3

1 from the support and in the middle of the

span; values of curvatures are calculated with their signs due to the curvatures diagram.

In other cases curvature in the middle of the span is to be determined by formula (294). Curvature values included into formulas (311) – (314) are determined by formulas (271), (272), (282), (286), (309) and (310) if there are cracks in the stretched zone and by formulas (269) and (270) – if there are no cracks. For solid slabs less than 250 mm thick it is necessary to take into account recommendations of Item 4.24.

Trang 191

T a b l e 35 Loading scheme of the

console beam Coefficient рm Loading scheme of the free

supported beam Coefficient рm

4

1

48

5

3

1

12

1

−

l

a

l

a3

6

2

2

68

1

l

a−

N o t e . By loading of the element due to several schemes at the same time

value...

...

21

2211

++

++=

MM

MM mm

m

ρρρ (where pm1 and M1, pm2 and M2, etc – are coefficient pm and maximum

bending moment М for each loading scheme). In that case in formulas (311) – (313) value mr

1 is determined

by value М, equal to the sum of maximum bending moments determined for each loading scheme. T a b l e 36

Mcrc/Mtot 1,00 0,99 0,98 0,96 0,94 0,92 0,90 0,85 0,80 Pcrc 0,104 0,088 0,082 0,073 0,067 0,062 0,058 0,049 0,042

Mcrc/Mtot 0,75 0,70 0,60 0,50 0,40 0,30 0,20 0,10 0,00

Pcrc 0,036 0,032 0,024 0,018 0,013 0,008 0,005 0,002 0,000

( )12

31 crccrc

crc

λλρ

+= where

2

/11 totcrc

crc

MM−−=λ

4.30. For short elements (l/h < 10) of constant section which work as free supported beams deflection is determined due to Item 4.29 and is multiplied by coefficient ρq, considering influence of the shear deformation. Coefficient ρq is determined by the following formula

2

1

+=

l

h

m

q

qρ

ϕρ (315)

where ϕq = 0.5 – if there are no as normal as inclined cracks, that means if conditions (233) and (248) are met;

ϕq = 1.5 – if there are normal or inclined cracks; рm – see Table 35.

Examples of calculation

Example 57. Given: reinforced concrete roof slab of a civil building of rectangular section with dimensions h = 120 mm, b = 1000 mm, ho = 105 mm; the span l = 3.1 m, heavy-weight concrete В25 (Eb =2.7⋅104 MPa; Rbt.ser = 1.6 MPa); stretched reinforcement А-II (Еs = 2.1⋅105 MPa), area of its cross section Аs = 393 mm2 (5 ∅10); total distributed load

Trang 192

qtot = 7 kN/m, including its part of dead loads and long-term loads ql = 6 kN/m; deflection is restricted by esthetic requirements. It is required to calculate the slab as regards deformations. C a l c u l a t i o n . Let’s find out if it’s necessary to calculate the slab as regards deformations due to Item 4.27:

00375.01051000

393

0

=⋅

==bh

Asµ

0292.0107.2

101.200375.0

4

5

=⋅

⋅==

b

s

E

Eµµα

4.88

1.37

8

22

=⋅

==lq

M tot

tot kN·m;

2.78

1.36

8

22

=⋅

==lq

M l

l kN·m

From Table 33 due to µα = 0.029 and ϕf = ϕft = 0 we find λlim = 21. As h < 250 mm, so λlim

is corrected by means of dividing into the coefficient 5.123.17105

105

7

33

0

0 <=

−=

−h

h.

Then 1.1723.1

21lim ==λ

Considering the note to Item 4.27 (case «a»), we have

202.7

4.81.171.17lim ===

l

tot

M

Mλ

As 206.29105

3100lim

0 =>== λh

l so the calculation as regards deformations is required.

Let’s determine the curvature in the middle of the span caused by the moment Мl (as deflection is restricted by esthetic requirements). Let’s take without calculation that the element has cracks in the stretched zone, so curvature is to be determined by formula (309). From Table 34 due to µα = 0.028 and ϕf = ϕft = 0 we find values ϕ1 = 0.393 and ϕ2 = 0.10, corresponding to long duration of the load.

525

26

201

,2

2 1035.1105393101.2393.0

6.112010001.0102.7 −⋅=⋅⋅⋅⋅

⋅⋅⋅−⋅=

−=

hAE

RbhM

r

l

ss

serbtl

m ϕ

ϕ1/mm

Deflection is to be determined due to Item 4.29аtaking due to Table 35: 48

5=mρ

5.13310048

51035.1

1 252 =⋅=

= −

lr

f m

m

ρ mm

As h < 250 mm so total deflection is f = 13.5 ⋅ 1.23 = 16.6 mm, which is more than maximum allowable deflection (see Table 2). As µ = 0.00375 < 0.005 so due to Item 4.296 we define value f more exact by formula

(312). For that we determine values elmr ,

1

and Mcrc.

As we calculate under-reinforced element (µ < 0.01), so Ired and Мcrc are determined as for a concrete section (see Items 4.2 and 4.3):

Trang 193

633

1014412

1201000

12⋅=

⋅==

bhI red mm4;

62,

2 1073.66.11201000292.0292.0 ⋅=⋅⋅⋅== serbtcrc RbhM N·mm

ϕb1 = 0.85 (as for heavy-weight concrete);

564

6

1

2

,

1044.010144107.285.0

2102.71 −⋅=⋅⋅⋅⋅

⋅⋅==

redbb

bl

elm IE

M

r ϕ

ϕ1/mm

Coefficient рсrс is to be determined due to Table 36 by 801.040.8

73.6==

tot

crc

M

M; 042.0=crcρ

( ) 84.9310010042.044.035.148

535.1

111 252

,

=⋅

−−=

−

−

= −

lrrr

f crc

elmm

m

m

ρρ

mm Considering correction for a little height of the section (h < 250 mm) f = 9.84⋅1.23 = 12.1 mm, which is less than maximum allowable deflection f = 15.5 mm. As l/h > 10, so we don’t take influence of the shear deformation into account. Example 58: Given: roof collar beam of a public building of rectangular section with dimensions b = 200 mm, h = 600 mm; a = 80 mm; collar-beam span l = 4.8 m; heavy-weight concrete В25 (Еb = 2.7 × 104 MPa; Rbt,ser = 1.6 MPa); Forking reinforcement А-III (Es = 2 ⋅ 105 MPa), area of its cross section Аs = 2463 mm (4 ∅28); total distributed load qtot = 85.5 kN/m, including its load from dead loads and long-term load ql = 64 kN/m; deflection is restricted by esthetic requirements; air humidity in the room is higher than 40 percent. It is required to calculate the collar beam as regards deformations. C a l c u l a t i o n . Let’s find out if it’s necessary to make the calculation as regards deformations due to Item 4.27.

ho = 600 – 80 = 520 мм ;

0238.0520200

2463

0

=⋅

==bh

Asµ ;

176.0107.2

1020238.0

4

5

=⋅

⋅==

b

s

E

Eµµα ;

2468

8.45.85

8

22

=⋅

==lq

M tot

tot kN·m

1848

8.464

8

22

=⋅

==lq

M l

l kN·m

As l/h = 4.8/0.6 = 8 < 10, so it is necessary to consider the influence of the shear deformations on the deflection of the element: l/ho = 4.8/0.52 = 9.3. Due to Table 33 by µα = 0.176 and ϕf = ϕft = 0 we find λlim = 8.

7.10184

246824.11

3.9

183.9:18 lim

0

==>=+=+l

tot

M

M

h

l

h

lλ

that means the deformations calculation is required. As µ = 0,238 > 0,005, due to Item 4.1 curvature is determined considering the cracks in the stretched zone. As deflection is restricted by esthetic requirements so the calculation is

Trang 194

made as regards the moment Мl. Curvature in the middle of the spanmr

1 is determined by

formula (309). Due to Table 34 by µα = 0,176 and ϕf = ϕft = 0 we find ϕ1 = 0,206 and ϕ2 = 0.

525

6

201

,2

2 10671.05202463102206.0

0101841 −⋅=⋅⋅⋅⋅

−⋅=

−=

hAE

RbhM

rss

serbtl

m ϕ

ϕ1/mm

Total deflection is to be determined due to Items 4.29а and 4.30 considering influence of

the shear deformation. Due to Table 35, 48

5=mρ

225.18.4

6.0

48

55.111

22

=

⋅+=

+=

l

h

m

q

qρ

ϕρ ;

24200

4800

2007.19225.14800

48

510671.0

1 252 ==<=⋅⋅=

= − l

mmlr

f qm

m

ρρ mm

that means deflection of the collar beam is les than the maximum allowable value (see Table 2). Example 59. Given: reinforced concrete slab of the roof with the span 5.7 m; dimensions of the cross section (for a half of the slab section) – due to draft 89; light-weight concrete В25 (Rb,ser = 18.5 MPa; Rbt,ser = 1.6 MPa), average density D1600 (Eb =16.5⋅103 MPa); main reinforcement А-II (Es = 2.1⋅105 MPa), its section area As = 380 mm2 (1∅22); distributed dead load and long-term load on the slab is ql = 8.75 kN/m; deflection of the slab is restricted by esthetic requirements; the room covered with the slab has normal air humidity (40 – 75 percent). It is necessary to calculate the slab as regards the deformations. Draft 89. To the calculation example 59 C a l c u l a t i o n . Moment in the middle of the slab caused by dead loads and long-term loads for a half of the section is:

8.1782

7.5750.8

8

22

=⋅

⋅==

lqM l kN·m

From Draft 89 we have: ho = 300 – 31 = 269 mm; b = (95 + 65)/2 = 80 mm; b'f = 730 mm; hf = 30 mm. As approximate methods of calculation as regards deformations do not belong to structures of light-weight concrete so curvature is to be calculated by formulas of Items 4.15 – 4.17 as

for elements with cracks in the stretched zone ( 005.00176.026980

380

0

>=⋅

==bh

Asµ that

means due to Item 4.1 in the stretched zone there are cracks). Let’s determine curvature of the slab in the middle of the span by formula (271). Moment Ms by action of dead loads and long-term loads is Ms = М = 17.8 kN⋅m. Relative height of the stretched zone of concrete ξ is to be calculated by formula (274). For that we determine the following values:

166.05.1826980

108.172

6

,20

=⋅⋅

⋅==

serb

s

Rbh

Mδ ;

( )905.0

26980

3080730)(

0

''

=⋅

−=

−=

bh

hbb ff

fϕ

Trang 195

855.02692

301905.0

21

0

'

=

⋅−=

−=

h

h f

fϕλ

224.0105.16

101.20176.0

3

5

0

=⋅

⋅==

b

ss

E

E

bh

Aµα

β = 1.8 (as for light-weight concrete), which we insert into the formula (274) and get

( ) ( )221.0

224.010

855.0166.0518.1

1

10

511

=

⋅

+++

=++

+

=

µα

λδβ

ξ

As 112.0269

30221.0

0

'

==>=h

h fξ so calculation is to be continued as for a T-section. Lever

arm z is to be determined by formula (279):

( ) ( )1.251

221.0905.02

221.095.0269

30

12692

1

22

0

'

0 =

+

+−=

+

+

−=ξϕ

ξϕ

f

f

f

h

h

hz mm

Let’s determine coefficient ψs due to Item 4.17. For that by formula (246) we determine value Wpl:

( )63.1

300

269905.022

20

'''1 =⋅==

−=

h

h

bh

hbbf

ffϕγ

202.09.0224.001 =⋅===

h

h

E

E

bh

A

b

ss µααµ

( ) ( ) 51600003008063.1075.0202.05.0292.075.05.0292.0 22'11 =⋅⋅+⋅+=++= bhWpl γαµ

mm3 Coefficient ϕm we determine by formula (281) considering that for bending element Mr = М:

464.0108.17

51600006.16

,=

⋅

⋅==

M

WR plserbt

mϕ

From Table 32 we have ϕls = 0.8; ψs = 1.25 – ϕlsϕm = 1.25 – 0.8 ⋅ 0.464 = 0.879 < 1.0. Curvature of the lab in the middle of the span is to be determined by formula (271), taking coefficients v= vl = 0.15 (see Table 31) and ψ′b = 0.9:

( ) ×⋅

⋅=

++=

1.251269

108.171 6

00 vEbhAEzh

M

r bf

b

ss

ss

m ξϕ

ψψ

( )5

3510685.0

15.0105.1626980221.0905.0

9.0

380101.2

879.0 −⋅=

⋅⋅⋅⋅++

⋅⋅× 1/mm

As l/h = 5700/300 = 19 > 10 due to Item 4.29, deflection f = fm, which is determined by

formula (311). Due to Table 35 48

5=sϕ .

5.282.32570048

510685.0

1 252 <=⋅=

= −

mmlr

f s

m

ϕ mm

That means deflection of the slab is less than the maximum allowable one (see Table 2)

Trang 196

5. CONSTRUCTIVE REQUIREMENTS

GENERAL POSITIONS 5.1.(5.1). During design of concrete and reinforced concrete structures and details to

provide their economic and quality manufacturing, required durability and cooperation work of concrete and reinforcement it is necessary to follow constructive requirements of the present section.

5.2. Structures must be of simple shape. It is recommended to use reinforcement, embedded

elements and lifting loops which are manufactured in form of finished products according to norms and state standards. Reinforcement must be designed in form of enlarged blocks and spatial frameworks to spend less time to put them into the framework.

It is necessary to take minimum possible stripping strength and transporting strength of concrete for effective use of industrial areas. It is necessary to aim at unification of reinforcement and embedded elements in some structures and their series, at little number of different marks and diameters of steel, types of reinforcement elements – mashes and frameworks, spacing of longitudinal and cross rods.

MINIMUM DIMENSIONS OF ELEMENTS SECTIONS 5.3.(5.2). Minimum dimensions of the section of concrete and reinforced concrete elements

determined due to the calculation as regards corresponding forces and corresponding groups of limit states must be specified considering economical requirements, necessity to unify framework forms and reinforcement as well as conditions of accepted technology of structures manufacturing.

Besides dimensions of elements sections must be taken so that to follow the requirements of reinforcement location in the section (thickness of concrete protection layers, distance between rods, etc) and anchorage of reinforcement.

5.4.(5.3). Thickness (Note: Hereinafter values of sections dimensions, thickness of protection layer of concrete and other terms mentioned in the present Guidelines belong to nominal values specified during design and shown on projects. These Nominal Values can vary in the reality but variations must not exceed values given in the corresponding state standards technical specifications and other documents) of solid slabs must be taken in millimeters no less than:

for roofs........................................................................................ 40 for floors of civil and public buildings......................................... 50 for floors of industrial buildings……………………................... 60 for slabs of light-weight concrete В7,and less in all cases ......... 70 Minimum thickness of prefabricated slabs must be determined due to condition of providing of required thickness of concrete protection and conditions of reinforcement location in the slab (see Items 5.32 – 5.41).

Trang 197

Dimensions of sections of eccentric compressed elements must be taken so that their elasticity lo/i in any direction as a rule was not more than: For concrete elements............................................... 200 (for rectangular sections by lo/h ≤ 60) For columns which are elements of buildings…...... 120 (by lo/h ≤ 35) For concrete elements............................................... 90 (for rectangular sections by lo/h ≤ 26)

DIMENSIONS AND SHAPES OF ELMENTS OF STRUCTURES 5.5. Dimensions of prefabricated concrete and reinforced concrete elements must be

specified considering carrying capacity and dimensioning restrictions of technological, transport and installation equipment in the factories and on the building sites. In case of need it is necessary to consider the possibility to lift a reinforced concrete element with the form.

5.6. To avoid damages caused by local concentrations of stresses by abrupt changing of

direction of the element surfaces (for example in internal angles) it is necessary to provide flattening of contours in form of slopes, sloped edges or curving of little dimensions (up to 50 mm) in order not to make local reinforcement (Draft 90, а, б, в). In external sharp corners it is necessary to make sloped edges or curving to avoid chipping of concrete (Draft 90, г).

Draft 90. Sloped edges and curving

а – curving in the ribbed slab; б – sloped edge between a flange and a wall in a T-beam; в – combination of the sloped edge and a curving in the joint of the truss; г – flattening of the sharp corner in the collar-beam; д – curving in the hole for communications, strapping, etc.

5.7. Holes in reinforced concrete elements for communications, stripping, etc must be not

big and located within the cells of reinforcing meshes and frameworks so not to cut the reinforcement and not to make local reinforcement. Corners of the hole must be smooth (Draft 90, д).

5.8. During design of concrete and reinforced concrete structures their shapes must be specified considering the form and conditions of use of the formwork.

By using of forms with a drop side the shape of a detail must not prevent the turning of the side (Draft 91, а) when removing the formwork. By using solid molds to remove the detail out of the formwork it is necessary to design technological slopes no less than 1:10 (Draft 91, б, в). In case of use of solid forms with pressing out the slope must be no less than 1:15 (Draft 91, г). By immediate removal of formworks with fixed vertical moving of the forming element (to avoid crashing of concrete) (Draft 91, д, е) the slope must be no less than 1:50. By using of forms with one fixed and one drop side to provide vertical lifting of the structure by removing of the framework it is used changing from the larger to the less width of the detail [for example from the lower flange to the wall (Draft 91, ж)]; it must

Trang 198

be smooth at the angle no less than 45 degrees. This requirement can be not considered if the framework has a pressing-out system (Draft 91, з). Using of pressing out and immediate remain of the form work must be approved by the manufacturer.

Draft 91. Technological slopes

а – in a form with drop sides; б, в – in the solid form; г – the same by using pressing out; д, е – by immediate removal of the formwork; ж – in a form with a solid side; з – the same by using pressing out; 1 – detail; 2 – form; 3 – drop side; 4 – pressing out; 5 – bearing brass; 6 – forming frame

REINFORCEMENT, MESHES AND FRAMEWORKS Separate reinforcing rods

5.9. Assortment of reinforcing rods for reinforced concrete structures is given in Annex 4. 5.10. During design of reinforced concrete structures especially with great volume of

reinforcement it is necessary to consider the following characteristics of reinforcing rods:

- Dimensions of cross sections of profiled section rods considering allowable deviations;

- Bending radius of rods and dimensions of bent elements; - Allowable deviations from the project dimensions by arrangement of rods

welding meshes and frameworks, embedded elements etc.

5.11. during design of bent rods diameters and angles of bending must meet the requirements of Table 37. Length of bent rods is determined due to the axis of the rod.

T a b l e 37

Reinforcement class

Minimum clear diameter of by the rod diameter d, mm

Maximum angle of bending,

18 and less 20 and more grad А-I, Ас-II, mark 10ГТ 2.5d 2.5d Not limited

А-II 4d 6d 180 A-III 6d 8d 90* Вр-I 4d Not limited

* It is possible to bend the rods at 180 degrees if design resistance against tension is decreased by 10 percent.

Dimensions of hooks for anchorage of plain rods of reinforcement must be taken according to Draft 92.

Draft 92. Dimensions of hooks at the ends of rods of plain main reinforcement

Welded connections of reinforcement

5.12.(5.32). reinforcement of hot-rolled steel of plain and profiled sections, heat-treated

steel-class Ат-IIIС and simple reinforcing wire, are made as a rule by means of

Trang 199

contact welding – point welding and butt-seam welding. It is possible to use semi-automatic arch welding, as well as manual welding due to Item 5.18.

5.13 (5.33). Types of welding connections and welding methods must be chosen

considering use conditions and welding capacity of steel, technical-economical indexes and technological capabilities of the manufacturing factory in compliance with the instructions of state standards and technical specifications for welded reinforcement (Table 38).

Connections not mentioned in the current standard documents can be produced in accordance with the projects approved in compliance wit the established procedure. End joints can be made without using welding by means of swedge casings in compliance with the approval of the manufacturing factory.

5.14.(5.34). In the plant conditions during producing of welded reinforcing meshes,

frameworks and connections it is necessary to use contact welding along the length of some rods – point welding and butt-seam welding (see positions 1, 2 and 5 of Table 38).

5.15.(5.35).During installation of reinforcement details and prefabricated reinforced

concrete structures for connection for butt jointing of rods with diameter 20 mm and more it is necessary to use “tub” welding in special copper or graphite removable forms (see positions 7-9 Table 38), as well as “tub” welding, bath-seam welding and multi-layer welding on steel cover-brackets1 [a cover-bracket – additional constructive detail, taking a part of the axis load, whose section area is no more than 50 percent of the connected rod section area] (see positions 10-13 of Table 38). At the same time first of all it is necessary to use power-activated methods of welding (see positions 7, 8, 10, 12 of Table 38), providing quality control of connections. In special cases it is possible to use welding of vertical rods by means of multi-layer welding without additional elements (see position 14 of Table 38).

5.16. Design of butt welding joints using prefabricated forms and other forming elements is made considering the following requirements:

а) distance between connected rods, as well as distance from the connected rods to the nearest surface of the reinforced concrete element must be taken considering the possibility of installation of forming elements and removal of prefabricated forms. Dimensions and methods of installation of prefabricated forms of steel cover-brackets must be taken due to standard documents for welding. Total free length of reinforcement rods must conform to the distance between surfaces of connected reinforced concrete elements and must be no less than 350 mm. Distance from the butt-ends of free length of reinforcement rods to the surfaces of elements (considering protection of concrete against overheating) is taken no less than 100 mm (Draft 93, а); б) location of connected rods must provide the possibility to insert an electrode at angle no more than 30 degrees to the vertical line (Draft 93, б, в);

Trang 200

в) gaps between connected rods by arch “tub” welding must be made in conformity with the standard state documents and welding norms. By the gaps which are more than maximum allowable ones connection of rods can be made by using of an intermediate element – fill piece of a reinforcing rod of the same diameter and class like connected rods.

Draft 93. Arch “tub” welding of reinforcement outputs.

а – end connection of rods; б – horizontal connection; в – vertical connection 5.17. For connection of rod reinforcement with diameter 10 – 18 mm to each other during

installation, as well as for connection of rod reinforcement with embedded and anchors must be used arch welding with extended joints (see positions 15 and 16 of Table 38 and position 1 of Table 53). By decreased demands for the connection strength (no more than 50 percent of the strength of the connected rod) it is possible to weld rods with diameter 8 mm. Welding of rods with extended joints by diameters 20 an more is possible in special cases.

5.18. If there is no equipment for contact welding it is possible to use arch welding in the

following cases:

а) for connection along the length of reinforcement with diameter 10 mm and more (see position 15 and 16 of Table 38); b) for cross connections of reinforcing meshes with un-normalized strength (see Item 5.19 and position 3 of Table 38). c) for welded connections with normalized strength in meshes and frameworks with required additional constructive elements at points of rods connections (gusset plates, fill pieces, hooks and other) or with forced forming of the joint (see position 4 of Table 38).

5.19. Cross welded connections with non-standardized strength are possible in the

following cases:

а) in meshes with main reinforcement of periodic profile for reinforcement of slabs, walls, etc; б) in connections of longitudinal and cross reinforcing rods of flat meshes with rods which form a spatial framework if the element doesn’t work for torsion and longitudinal rods are not considered in the calculation as compressed ones; в) in points of connection of longitudinal rods of spatial frameworks with cross reinforcement in form of continuous spirals.

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T a b l e 38

Symbol of the welding due to GOST 14098-85

Method of welding

Number of the position of Table 1 of СН 393-78

Position of the rod by the welding

Class (mark) of reinforcement

steel

Diameter of rods, mm

Scheme of the connection structure

Additional instructions

1. Contact point welding of two rods

К1 1А

Horizontal (can be vertical in conductors)

А-I А-II А-III Ат-IIIС Вр-I В-I

640 1040 640

1028 35 35

d'/d = 0.25 – 1.00

2. the same of three rods К2 1Б

The same А-I А-II А-III Ат-IIIС Вр-I В-I

640 1040 640

1028 35 35

d′/d = 0.50 – 2.00

3. Manual arch welding with point tack weld

К3 2

Horizontal and vertical

А-I А-II

(Ст5сп2) А-II

(Ст5пс2) Ас-II А-III

(25 Г2C) Ат-IIIC

1040 1028

1018 1032 1028

1028

In conditions of negative temperatures it is possible to use welded connections only of reinforcement rods of class А-I and Аc-II. Welding is used for connections of non-standardizes strength (see Item 5.19)

4. the same with forced forming of the joint

- 3

Vertical А-I; А-II; А-III Ат-IIIC

1440

1418

Welding is used only for connection of rods basically for cast-in-situ reinforced concrete Welding is performed in prefabricated forms

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Butt joints 5. contact butt seam

welding C1; C2

4A Horizontal А-I

А-П А-III Ат-IIIC

1040 1040 1040 1028

00.185.0

'−=

d

d

30.0'

≥d

d is possible by using of a

special device which provides preliminary worming-up of a rod with big diameter

6. contact butt seam welding with the following mechanical treatment

С3; С4 4Б

Horizontal А-II А-III Ат-IIIC

1040 1040 1022

Recommended for structures working with repeated load

In prefabricated forms

7. bath power-activated welding flux welding

С5; С8-С11 5А; 6А

00.15.0'

−=d

d

For hard-to-reach from the above connections of horizontal rods demanding inclined insertion of an electrode (see Item 5.16б), 00.1

'=

d

d

8. arch power-activated welding with flux cored wire

С6; С9; С12 5Б; 6Б

9. bath mono-electrode welding

С7; С10; С13 5В; 6В

Horizontal and vertical

А-I; А-II; А-III 2040

Welding of horizontal doubled rods of reinforcement A-III with diameter 32–40 mm is allowable by d’/d = 1.0

On the steel cover-bracket

10. arch power-activated welding with flux cored wire

С14; С17 -

Horizontal and vertical

А-I А-П А-III Ат-IIIC

2040 2040 2040 2028

d′/d = 0.5 – 1.0. By welding of reinforcement of class Ат-IIIC d′/d = 0.8 – 1.0

11. bath seam welding С15 9Г

horizontal

12. arch power-activated welding by open-flame arch with alloy wire

С16; С18 9В; 10В

Horizontal and vertical

13. manual arch multi-layer welding

С19 10Б

Vertical

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14. the same without additional technological elements

С20 11

Vertical А-I А-П А-III

2040

00.15.0'

−=d

d

15. manual arch welding with long joints and covers of rods

- 12

Horizontal and vertical

А-I А-П А-III Ат-IIIC

1040 1040 1040 1022

Length of covers l for reinforcement classes: А-I – 6d; А-II, А-III and Ат-IIIC

– 8d. For reinforcement classes А-I, А-II and А-III it is possible to use double-sided joints with the covers length l = 4d

16. manual arch welding with long joints of two rods

- 13

Horizontal and vertical

А-I А-П А-III Ат-IIIC

1040 1025 1025 1018

Length of overlapping l for reinforcement classes: А-I – 6d; А-II, А-III and Ат-IIIC

– 8d. For reinforcement classes А-I, А-II of mark 10ГТ it is possible to use double-sided joints with the overlapping length l = 4d

N o t e . Instructions for welding of high-strength rod reinforcement of classes A-IV, Ат-IVC, A-V and A-VI are given in SNiP 2.03.01-84 (see required Annex 3) and GOST 14098-85.

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In all other cases including by welding of anchor rods of meshes (see Items 3.44, 5.45 and 5.46), welded connections must have standardized strength. Cross connections with non-standardized strength can be made by means of arch welding with point tack weld (see position 3 of Table 38), as well as by means of point manual welding (see positions 1 and 2 of Table38) by decreased demands to the connection strength regulated by GOST 10922 – 75, by decreased demands to minimum relative settlement h/d' (see draft of position 1 Table 38) regulated by GOST 14098 – 85, but no less than mentioned in Table 8 of СН 393-78.

5.20. It is possible to connect the rods by means of welding in any case along the length of the detail. Connections (joints) made by arch welding must be located so that they didn’t prevent concreting, that means it is necessary to make them in points with less reinforcement, to prevent several joints in one section.

Flat welding meshes

1

[1Hereinafter term «flat welding meshes» includes any flat welded reinforcing details (meshes, frameworks)].

5.21. During design of flat meshes it is necessary to consider the requirements of

unification of dimensions, spacing and diameters of cross and longitudinal reinforcement. Meshes must be transportable, convenient for storage and for installation into a form. First of all it is recommended to use ready-mixed meshes or meshes of centralized manufacturing with dimensions corresponding to state standards and norms.

Reinforcing meshes which do not meet the mentioned above requirements must be designed considering their manufacturing by means of point welding on multi-electrode machines.

5.22. Parameters of wide welded meshes are given in Table 39, of narrow meshes – in Table 40.

In order to reduce the number of realignment of multi-electrode machines it is recommended to unify the reinforcement spacing during design, basically longitudinal reinforcement for reinforced concrete details of the present series or catalogue.

T a b l e 39 Data for meshes Parameters of wide welded

meshes produced on multi-electrode meshes

light heavy Additional requirements

Diameters of rods, mm: Longitudinal D

Cross rods d

From 3 to 12

From 3 to 10

From 14 to 32

From 6 to 14

It is recommended one diameter in a mesh. Difference no more than two times is possible. Each pare of rods counting from the edge must be of the same diameter . It is necessary to use rods of one diameter

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Spacing of rods, mm:

Longitudinal v

Cross rods s:

Constant spacing Two different spacing for a strip-mesh: a) bigger b) smaller

100 200 300

any from 100 to 300

any from 140 to 300

any from 60 to 220

200

100; 200; 300; 600

–

–

For light meshes interchange of spacing is possible. Use of spacing which is more than the mentioned ones is possible but they must be divisible by 100 mm. By the mesh width not divisible by 100 mm the rest must be located on one side (see the draft for the table, type II). Type III can by used in compliance with the approval of the producing factory. Strip-mesh1 is produced by diameters of longitudinal rods D ≤ 8 mm. Minimum difference between the value of the bigger and the smaller spacing in one mesh is 80 mm; the smaller spacing (less than 100 mm) is taken as an additional one as well as in points of cutting of the strip-mesh

Minimum length of the rods ends (distance from the butt-end of the rod to the axis of end crossed rod), mm:

Cross rods k

Longitudinal c

20

25

25 but no less than D

25

For meshes produced with longitudinal cutting of the strip k ≥ 50 mm For a strip-mesh c from 30 to 150 mm

Maximum mesh length, L, m 12 9 but no more than the length of not-connected rods

All cross rods must be taken with the same length within one mesh

Width of a mesh, mm: A

B (on axes of end longitudinal rods)

From 800 to 3800 From 1160 to 3750

From 1050 to 3050 From 1000 to 3000

All cross rods must be taken with the same length within one mesh

Maximum number of longitudinal rods

36 16 Number of rods must be taken even

1 Is produced in form of continuous part with the following cutting

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T a b l e 40

Data for meshes

heavy Parameters of wide welded meshes produced on multi-

electrode meshes light

I II

Additional requirements

Diameters of rods, mm: Longitudinal D Cross rods d

From 3 to 8

From 3 to 8

From 10 to 25

From 4 to 12

From 12 to 40

From 6 to 14

In one mesh it is possible to use longitudinal rods of different diameters. It is recommended to use no more than two rods which are different no more than by 2 times. In the mesh it is necessary to use cross rods of the same diameter

Spacing of rods, mm: Longitudinal v

Cross rods s:

From 50 to

390

From 100 to 500

From 75 to

725

From 100 to 400

From 100 to

1400

From 600 (divisible by

50)

For heavy meshes of type I it is possible to use one spacing at the edge of the mesh no less than 50 mm. For heavy meshes of type I: By d ≤ 8 mm s ≥ 100; By d =10 mm s ≥ 150; by d ≥ 12 mm s ≥ 200;

s - s' ≥ 50 Maximum number of different spacing between cross rods

3

2

2 –

Minimum length of the rods ends (distance from the butt-end of the rod to the axis of end crossed rod), mm:

Cross rods k

Longitudinal c

15

25

20

25

25 but no less than D

25

–

For light strip-meshes distance from the butt-end of a longitudinal rod to the axis of the cross rod is recommended to take equal to a half of the spacing of cross rods

Maximum mesh length, L, m

7.2 12 18 –

Width of a mesh, mm: A

B (on axes of end longitudinal rods)

From 80 to

420 From 50 to

390

From 90 to

775 From 50 to

725

From 140 to

1450 From 100 to

1400

–

Number of longitudinal rods

From 2 to 4 From 2 to 6 From 2 to 8 –

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It is possible to take spacing of rods which are different of the values mentioned in Tables 39 and 40 during design of details for a certain producing factory in conformity with the parameters of the equipment under conditions of unification of these spacing on the factory. Welded meshes produced on multi-electrode machines must have rectangular contours with rectangular cells. At the ends of rods there must not be any bendings, hooks or loops. They can be used only in compliance with the approval of the producing plant.

5.23. Welded meshes whose constructive parameters do not allow the produce them on multi-electrode machines must be designed in conformity with manufacturing capabilities of single-point welding machines (Table 41).

5.24. Ready-mixed welded meshes as well as meshes and frameworks produced on multi-

electrode and single-point machines can be used as finished reinforcing details or as a semi finished good which must be followed up (cutting of a mesh, making of holes, bending of a mesh to make a spatial framework and as and exception – welding of additional rods).

Additional rods can be welded by means of contact welding (Draft 94, а, б) considering instructions of Table 41, as well as electric arch welding by means of longitudinal joints (Draft 94, в) considering requirements of Item 5.18. Bending of a mesh is made according to Item 5.27. T a b l e 41 Parameters of welded meshes, produced on single-point welding machines

Parameters values

Maximum diameter of a less of welded rods, mm

For rods of both directions Вр-I, А-II, А-III, Ат-IIIС–25; by rods of classes В-I, А-II–40. See also table 38 (pos. 1 and 2)

Maximum width of welded meshes: recommended allowable

500 1000 + v1 (see draft for the table)

Maximum distances, mm, between axes of rods of one direction by rods diameters, mm:

Up to 10 40 from 12 to 18 50 from 20 to 25 60 28 and 32 70 36 and 40 80 Minimum length of rods ends k to the axis of end cross rod, mm

20, but no less than diameter of the end of the rod

Minimum angle between crossing welded rods, degree

30

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Draft 94. Welding of an additional longitudinal rod to a welding mesh

а – initial mesh; б – welding of an additional rod near the main cross rod by means of point welding; в – the same, close to the main rod by electric arch welding by longitudinal joints; 1 – main rod; 2 – additional rod; 3 – electric arch welding (by length l ≤ 6 m rod 2 can be used only to the ends of the detail)

5.25. By reinforcement of sides of beams of variable height it is recommended:

а) to use frameworks with groups of rods of one height by the slope no more than 1:10 (Draft 95, а); b) to use separate rectangular frameworks (Draft 95, б) or rectangular meshes with the following cutting along the inclined length (Draft 95, в) with a border rod in case of need by the slope no more than 1:10. By reinforcement of rectangular slabs it is recommended to use welding meshes produced of rectangular ones as result of their cutting (Draft 95, г).

Spatial reinforcing frameworks

5.26. Reinforcement of rectangular elements (especially linear ones) must be designed as a rule in form of spatial frameworks.

Spatial frameworks can be made as a whole detail or in form of spatial blocks used in combination with flat or bent meshes, as separate rods, etc. Spatial frameworks must be designed as hard ones for their storage, transportation and to follow their design location by laying them into a form. Their rigidity must be provided by means of installation of required bracings in form of rods, straps, etc.

Draft 95. Reinforcement of details of variable dimensions

а – sides of a beam by means of a mesh with groups of cross rods of the same length; б – the same, by separate rectangular meshes; в – the same, by a rectangular mesh with cutting it along the inclined line and adding of border rods; г – by welded meshes for slabs of variable width, produced by cutting of a rectangular mesh

Embedded elements and tie-down details (hooks, pipes, etc) can be fixed to the spatial framework if required accuracy of the location is provided. If deviations from the design location can cause decrease of bearing capacity of embedded elements so it is necessary to fix these details to the form.

5.27. By forming of spatial frameworks using bent flat meshes it’s recommended to use

bent meshes with contours according to types shown on Draft 96, а and produced by serial bending equipment. At the same time it is necessary to meet the following requirements:

- length of meshes must be no more than 6 m (by approval of a producing factory

length up to 9 m is possible); - length of bent part (see Draft 97, е) – no less than 60 mm and no less than 8d, - diameter of bent rods – no more than 12 mm (approved by producing factory – up to

32 mm). Draft 96. Examples of shapes of bent welding meshes

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а – recommended (meshes are produced by serial equipment); б – allowable (require special equipment or instruments); в – by packing of bent elements of spatial frameworks for storage and transportation (location of straight longitudinal rods is shown relative)

By mass-producing in compliance with approval of a producing factory it can be used bent meshes of other shapes for example according to the type shown on 96, б, whose manufacturing requires special equipment and instruments. Spatial frameworks which must be stored or transported should be designed of elements which can be packed densely (Draft 96, в). Diameter of rods of bent welded meshes, radii and angles of bending, location of longitudinal meshes must be specified considering class of used steel in compliance with Draft 97.

Draft 97. Parameters of bent welded meshes

а, б – point of bending is distant from the longitudinal rods (parameters of bending are taken due to Table 37); в – place of bending coincides with the longitudinal rod located on the external side of the mesh (diameter D is taken due to Table 37 with the increase by 2d), г – place of bending of a mesh coincides with the longitudinal rod which is located outside; д – place of bending of a mesh coincides with the longitudinal rod of a big diameter, located outside of a mesh; е – end parts of a bent rod of a mesh; d – diameter of a bent rod; d1 – diameter of a longitudinal rod; D – diameter of a relative circle of a rod bending

5.28. Formation of reinforcement details into the spatial framework must be made by means of point welding of cruciform crossings of rods by means of welding tongs. Minimum clear distance between rods comfortable for electrodes of welding tongs, for frameworks of reinforced concrete elements are given on Draft 98. Maximum allowable diameters of welded rods are determined due to Table 42.

T a b l e 42

Less diameter of welded rods, mm

Allowable minimum diameters, mm, of rods of classes А-I, А-II, А-III, Ат-IIIС,

welded by tongs with rods of less diameter of classes

А-I А-II, А-III, Ат-IIIС 6 22 18(22) 8 22(32) 16(22)

10 20(36) 10(20) 12 18(36) -(18) 14 14(32) 16 -(32) 18 -(28) 20 -(20)

N o t e . In the brackets there are shown maximum diameters allowable in accordance with the approval of the producing factory.

Draft 98. Examples of spatial frameworks of reinforced concrete elements produced by using of

welding tongs

а – rods of external angles of frameworks of linear structures; б – intermediate rods of frameworks of linear structures; в – rods of a narrow mash with rods of two wide meshes for flat structures; 1 – welding tongs; 2 – wide mesh; 3 – narrow mesh.

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By d1 + d2 ≤ 25 mm a = 60 mm, b = 100 mm; by 28 mm ≤ d1 + d2 ≤ 40 mm а = 75 mm, b = 120 mm 5.29. Spatial frameworks for reinforcement of linear elements (columns, piles, beams, etc.)

should be produced by contact point welding in the following manners: а) connection of plain meshes by separate rods welded to longitudinal rods of meshes by means of welding tongs (Draft 99a) due to Item 5.28; b) connected of bent meshes with rods (Draft 99, б), welded as it’s mentioned above; c) wrapping of spiral cross reinforcement on longitudinal reinforcement (Draft 99, в) with welding of all crosses by welding tongs. Such frameworks are recommended for reinforcement of pipes, piles, columns without consoles and other details of mass-producing; d) stringing on longitudinal rods of stirrups which are bent in advance by means of contact point welding in points of crosses of legs with the following welding by welding tongs of all crosses (Draft 99, г). Cross points of stirrups legs are located staggered along the length of a framework. Such frameworks can be used for reinforcement of columns with intermediate consoles. If there is no welding tongs so it is possible to bind the connections of longitudinal rods and stirrups (in that case it’s recommended to provide spatial rigidity of frameworks by means of welding of additional rods, plates, etc.); e) bending of a flat mesh up to a closed contour and welding of cross rods with a longitudinal rod of an opposite row of an initial mesh by means of welding tongs (Draft 99, д). This way is recommended if there are special instruments or equipment; f) welding of four flat welding meshes by welding tongs (Draft 99, е). This way can be used for producing of column framework when distance between angle and middle rods is less than 75 mm, and number of longitudinal rods is no less than eight; g) connection of two meshes by wiring bars which are perpendicular to the bending plane and welded to the cross reinforcement of meshes (Draft 99ж). This way can be used in beams which don’t work as regards torsion and in columns by total quality of longitudinal reinforcement no more than 3 percent; h) connection of several bent and flat meshes by rods welded by means of welding tongs (Draft 99, и); i) of two diagonally located flat meshes made by welding of longitudinal rods of both meshes with their cross rods by means of single-point machines (Draft 99, к), at the same time it is necessary to provide installation rigidity of a framework by means of welding of rods, plates, etc. This way is possible to use for poles, piles and others by quantity of reinforcement up to 1 percent.

Draft 99. Examples of structures of spatial frameworks of linear elements produced by using of contact

point welding

а – of two meshes and connecting rods welded to the longitudinal reinforcement of meshes; б – of bent meshes and connection rods; в – stringing of spiral cross reinforcement on longitudinal reinforcement; г – of preliminary bent and welded stirrups stringed on longitudinal rods; д –of a mesh bent up to a closed contour; е – of four flat meshes; ж – of two meshes and assembly rods perpendicular to the bending plane and welded to the cross reinforcement of meshes (in beans not working for torsion and in columns by total quantity of reinforcement no more than 3 percent); и – spatial framework of several bent and flat meshes and connection rods welded by means of welding tongs; к – spatial frameworks by total quantity of longitudinal reinforcement up to 1 percent in form of two diagonally located flat meshes; 1 – flat mesh; 2 – connection rod; 3 – bent mesh; 4 – point welding

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5.30. Spatial frameworks of linear elements can be produced without using of welding point welding by means of the following methods: а) connection of meshes by brackets and by arch welding with stirrups (Draft 100, а). In columns, beams working for torsion as well as in compressed zone of beams with the considered in the calculation working reinforcement the length of one-sided welding joints l must be no less than 6d (where d – diameter of a stirrup), and installation connections – 3d; b) connection of flat meshes by means of studs with bending of all crosses (Draft 100, б), at the same time it is necessary to provide installation rigidity of the framework by welding of rods, plates and others; c) connection of flat meshes to each other by means of arch welding of longitudinal rods (Draft 100, в) near all points of stirrups welding. Length of joints l must be no less than 5d (where d – diameter of stirrups). Such connections can be used by quantity of compressed reinforcement in the section no more than 3 percent; d) of longitudinal rods and bent stirrups with binding of crosses (Draft 100, г) and connection of rigidity elements (bound frameworks); e) of one or several bent or flat meshes and connection rods with diameters no more than 6 mm by bending of longitudinal rods of meshes by ends of connection rods making a closed loop by means of a bending instrument (Draft 100, д). This method is recommended if there are special conductors providing safe fixing of frameworks. If there are compressed longitudinal rods so requirements to distances between connection rods are the same as for distances for welded stirrups (see Item 5.59). Due to great labor expenditures frameworks mentioned in the present Item can be used only as exceptions if there are no welding tongs.

Draft 100. Examples of spatial frameworks of linear elements produced without using contact point

welding а – of two flat meshes and brackets welded to the cross reinforcement of meshes; б – of two flat meshes connected by means of studs by binding of all crosses; в – of four flat meshes; г – of longitudinal rods and bent stirrups with binding of crosses; д – framework formed of one or several bent or flat meshes and connecting rods by means of a bending instrument; 1 – flat mesh; 2 – a bracket or a stud; 3 – cross rods of flat meshes; 4 – longitudinal meshes of flat rods; 5 – arch welding; 6 – a bending instrument; 7 – a bent mesh

5.31. Spatial frameworks for reinforcement of flat elements (slabs, wall panels, etc.) should

be produced in the following manner: а) connection of flat meshes of type «steps» by rods welded by welding tongs (Draft 101, а); b) connection of meshes of type «steps» of one direction by means of the same flat meshes of different direction and less height (Draft 101, б). Connection of crosses is made by tongs; binding is possible in case if there are no tongs. To provide stability during installation in both mentioned cases flat meshes of type «steps» can be replaced by bent V-shaped meshes (Draft 101, в). Spatial framework of ribbed slabs or panels is made of bent U-shaped meshes with welding or binding to them of flat meshes of type «steps».

Draft 101. Examples of structures of spatial frameworks for reinforcement of flat elements

а – of flat meshes of type «steps» and connection rods; б – of flat meshes of type «steps» of one direction and the same meshes of other direction and less height; в – of V-shaped meshes of the same direction and the same meshes of other direction and less height

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LOCATION OF REINFORCEMENT, ANCORAGE, BUTT-JOINTS

Protection layer of concrete

5.32. (5.4). Protection layer of concrete for main reinforcement must provide cooperation

work of reinforcement and concrete during all stages of the construction use as well as protection against outside atmospheric, temperature and other effects.

5.33.(5.5, 5.6). Thickness of the protection layer must be as a rule no less than diameter of

a rod and no less than values mentioned in Table 43.

For prefabricated elements of heavy-weight concrete В20 and more thickness of a protection layer can be taken 5 mm less than diameter of a rod but no less than values given in Table 43. For reinforced concrete slabs of heavy-weight concrete of class В20 and more produced on factories in steel forms and protected from the above by topping concrete thickness of the protection layer for top reinforcement must be taken equal to 5 mm. In one-layer structures of light and porous concrete of class В7.5 and less thickness of the protection layer must be, mm:

- for longitudinal main reinforcement – no less than 20; - for outside wall panels (without stamped surface) – no less than 25; - for cross, constructive and distribution reinforcement – no less than 15.

T a b l e 43

Kind of structure

Functions of

reinforcement

Height (thickness) of the section, mm,

Thickness of the protection layer, mm, no less than

1. Slabs, walls, flanges of ribbed slabs

Longitudinal main

Up to 100 More than 100

10 15

2. Beams, ribs of slabs

The same Less than 250 250 and more

15 20

3. Columns, poles ′′ Any 20 3. Foundation

beams and prefabricated foundations

′′ ′′ 30

5. Solid foundations: - if there is concrete base Bottom

main ′′ 35

- if there is no concrete base

The same ′′ 70

б. Any structure cross, distribution, constructive

Less than 250 250 and more

10 15

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N o t e . For structures mentioned in positions 1-3 and for structures which are in contact with the ground values of thickness of the protection layer are increased by 5 mm.

For prefabricated elements of heavy-weight concrete of class В20 and more thickness of the protection layer for longitudinal reinforcement can be taken 5 mm less than diameter of the rod but no less than values mentioned in Table 43. For reinforced concrete slabs of heavy-weight concrete of class В20 and more which are produced on factories in metal forms and protected from the above by means of topping concrete thickness of the protection layer for top reinforcement can be taken equal to 5 mm. In one-layer structures made of light-weight and porous concrete of class В7.5 and less thickness of the protection layer must be, mm: - For longitudinal main reinforcement – no less than 20; - For external wall panels (without textured layer) – no less than 25; - For cross, constructive and distribution reinforcement – no less than 15.

5.34(5.10). For hollow-core elements of ring or box-shaped section distance from rods of

longitudinal reinforcement to internal surface of concrete must meet requirements of Item 5.33.

5.35. In bending, stretched and eccentric compressed by Ml/Nl > 0.3h, except foundations,

thickness of the protection layer for stretched main reinforcement as a rule must not be more than 50 mm. In the protection layer more than 50 mm thick it is necessary to install constructive reinforcement in form of meshes. At the same time section area of longitudinal reinforcement of meshes must be no less than 0.05As, pacing of cross reinforcement must be no more than the height of the element section and conform to instructions of Item 5.54.

5.36. For structures which work in aggressive conditions thickness of the concrete

protection layer must be taken considering requirements of SNiP 2.03.11-85.

By specification of the concrete protection layer it is also necessary to follow instructions of SNiP 2.01.02-85.

5.37 (5.9). To install single-piece reinforcing rods, meshes, frameworks going along the whole length or width of the element the ends of these rods must be distant from the surface of the element if the length of the element is: up to 9 m – by 10 mm, up to 12 m – by 15 mm, more than 12 m – by 20 mm.

Minimum distance between reinforcement rods

5.38 (5.11). Distances between reinforcement rods along the height and the width of the element must provide cooperation work of reinforcement and concrete and must be specified considering the convenience of installing it into a form and concreting.

By choosing of distances between rods in welded meshes as well as in flat and spatial frameworks it is also necessary to consider technological requirements for design of welded reinforcement details mentioned in Items 5.22, 5.23 and 5.28.

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5.39(5.12). Clear distances between separate rods of longitudinal reinforcement and

between longitudinal rods of neighbor flat welded frameworks must be taken no less than maximum diameter of rods, and:

а) if rods hold a horizontal or inclined position during concreting: for bottom

reinforcement – no less than 25 mm; for top reinforcement – no less than 30 mm; by location of bottom reinforcement less than in two rows along the height the distance between rods in the horizontal direction (besides the rods of two bottom rows) must be no less than 50 mm; б) if rods during concreting hold vertical position – no less than 50 mm; by

systematic monitoring of screening of concrete aggregate this distance can be decreased up to 35 mm, but at the same time it must be no less than 1.5-fold of maximum dimension of coarse aggregate. In elements or joints with maximum quantity of reinforcement or embedded elements produced without using vibrating plate compactors or vibrators fixed on the formwork it is necessary to provide in some points clear distance no less than 60 mm for a previbrator head. Distances between such points must be no more than 500 mm.

5.40 (5.12). By space-limited conditions it is possible to locate the rods in pairs without any gap between them or with the distance between the rods of a pair less than the distance required for separate rods. Such pare of rods by specification of the distance between rods (due to Item 5.39) and by determination of anchorage length (due to Items 5.42 – 5.46) must be considered as a relative rod (where d1 and d2 – diameters of rods brought together, с1 – clear distance between these rods, taken no more than diameter of the least rod; Draft 102).

Draft 102. Perimeters of location of one of rods rows of bottom reinforcement (located in one or two

rows along the height) by producing of the detail on the vibrating plate compactor by d = 32 mm and d

= 16 mm

а – even distribution of rods; б – doubled location of rods by decreased distances between the rods of each pair; в – the same if there is no a gap between the rods of each pair

5.41 (5.12). Clear distance between the rods of periodic profile mentioned in Items 5.39

and 5.40 are determined due to the nominal diameter without considering any jutting out and ribs. Taking location of reinforcement in the section with space-limited conditions considering the adjoining of other reinforcing elements and embedded elements it is necessary to take into account diameters of rods considering jutting out and robs, as well as possible deviations from nominal dimensions of reinforcement rods, welded meshes and frameworks, embedded elements, shapes and location of reinforcement and embedded elements in the section.

Anchorage of reinforcement

5.42. To make reinforcement take required forces in the considered section reinforcement

must have enough anchorage. 5.43 (5.13). Rods of periodic profile as well as plain rods used in welded frameworks and

meshes are produced without hooks. Stretched plain rods of bound frameworks and meshes must end with hooks (see Draft 92) or loops.

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5.44 (5.14). Longitudinal rods of stretched and compressed reinforcement must be brought

behind the normal to the longitudinal axis of the element section, where they are considered with total design resistance on the length no less than lan, determined by the following formula

( ) dR

Rll an

b

s

anlan

∆+= λω (316)

but no less than lan = λand, where values ωan ∆λan and λan, as well as possible minimum values lan are determined due to Table 44. At the same time plain reinforcement rods must end with hooks made due to Item 5.11, or have welded cross reinforcement along the length of embedment. To value Rb it is possible to insert coefficients of concrete work conditions except γb2. In elements of fine concrete of group Б (see Item 2.1) values lan, determined by formula (316), must be increased by 10d for stretched concrete and by 5d – for compressed concrete. Values of relative anchorage length λan = lan/d, determined by formula (316) for different classes of concrete and reinforcement are given in Table 45.

T a b l e 44 (37)

Parameters for determination of anchorage length lan of reinforcement without anchors

Periodic profile plain Work conditions of reinforcement

ωan ∆λan λan lan,

мм ωan ∆λan λan lan,

мм

No less No less 1. reinforcement embedment:

а) stretched reinforcement in stretched concrete

0,70 11 20 250 1,20 11 20 250

б) compressed or stretched reinforcement in compressed concrete

0,50 8 12 200 0,80 8 15 200

2. overlapping of reinforcement:

а) in stretched concrete

0,90 11 20 250 1,55 11 20 250

б) in compressed reinforcement

0,65 8 15 200 1,00 8 15 200

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T a b l e 45 Relative length of reinforcement anchorage λan = lan/d by concrete of classes Location of

reinforcement in concrete

Reinforcement class В7.5 В10 B12.5 B15 B20 B25 B30 B35 B40 B45 B50 B55 B60

1. stretched, A-I 72 58 48 42 34 30 27 25 23 22 21 21 20 A-II 56 45 38 34 28 25 23 21 20 20 20 20 20 lan ≥ 250 мм A-III 69 55 46 40 33 29 26 24 22 21 21 20 20 2. compressed, A-I 49 39 33 29 24 20 19 17 16 15 15 14 14 A-II 40 32 27 24 20 18 16 15 14 14 13 13 13 lan ≥ 200 мм A-III 50 40 33 29 24 21 19 17 16 15 15 15 14

In case when anchored rods are installed with the reserve on the section area against the required by the calculation as regards the strength with total design resistance, the anchorage length lan, calculated by formula (316) can be decreased by multiplying it by the ratio between the required due to the calculation and actual section area of reinforcement. If due to the calculation there are cracks caused by the concrete stretching along the anchored rods , so rods must be embedded into the compressed zone of concrete on the length lan calculated by formula (316). At the same time height of compressed zone can be determined due to Item 4.16.

5.45. If it is impossible to meet requirements of Item 5.44 so it is necessary to take special

measures for anchorage of longitudinal rods:

а) special anchors in form of plates, washers, screws, angles, etc at the ends (Draft 103). In that case area of contact of anchor with concrete must correspond the condition of concrete strength to compression (se Item 3.109а), and thickness of anchorage plate must be no less than 1/5 of the whole width (diameter) and correspond to welding conditions (see Table 52); length of the rod embedment must be determined by the calculation as regards chipping (see Item 3.106а) and must be taken no less than 10d;

Draft 103. Anchorage of reinforcement by means of special anchorage at the ends of special anchors

а – welded plate; б – constricted plate; в – button-head; г – button-head with a washer; д – rod welded to an angle; е – a screw with a washer outside; ж – screws inside

b) bending of anchored rod at 90 degrees along the arch of the circle with the clear radius no less than 10d (1 – ll/lan) [where ll – length of the straight part at the beginning of embedment (Draft 104)], corresponding to recommendations of Table 37; on the bent part there are installed additional stirrup against the unbending of rods; c) welding of anchoring cross rods on the length of embedment; in that case anchorage length lan, determined due to Item 5.44, is decreased by the length

ss

w

anAR

Nll =∆ , [where Nw – see formula (82)]; if ∆l ≥ 150 mm, so plain rods can be

made without hooks, at the same time value lan, is not decreased.

Draft 104. Reinforcement anchorage by means of bending

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5.46 (5.15). If longitudinal reinforcing rods going to the end free supports of bending elements have no special anchors and are not welded to support embedded elements, so it I necessary to meet the following requirements:

a) If conditions of Item 3.40 are met, the length of overlapping of stretched rods behind the internal surface of the free support la must be no less than 5d; in welded frameworks and welded meshes with longitudinal main reinforcement of plain rods to each stretched longitudinal reinforcement on the length la it is necessary to weld at least one cross (anchor) rod with diameter dа ≥ 0,5d located at the distance с ≤ 15 mm by d ≤ 10 mm from the end of the framework (mesh) and с ≤ 1,5d by d > 10 mm (Draft 105, а); б) If conditions of Item 3.40 are not met so the length of overlapping of stretched rods behind the internal surface of the free support lа must be no les than 10d; in case if there area used plain rods on the length lа to each longitudinal rod it is necessary to weld no less than two cross (anchorage) rods with diameter dа ≥ 0.5d, at the same time the distance from the end anchor rod to the end of the framework (mesh) must be no more than the mentioned in Item «а» values с (Draft 105, б). If value lаn, determined considering instructions of Item 3.44, less than 10d, the length of overlapping of rods behind the internal surface of the support is decreased up to lаn and taken no less than 5d.

Draft 105. Additional anchorage of reinforcement by means of welding of cross anchor rods

а – in slabs; б – in beams Reinforcement rebated joints (without welding)

5.47 (5.37). Rebated joints of main reinforcement are used for connection of welded and

bound frameworks and meshes, at the same time diameter of main reinforcement must be no more than 36 mm.

It is not recommended to locate rebated joints of rods of main reinforcement in stretched zone of bending and eccentric stretched elements in places of total use of reinforcement. It is not possible to use rebated joints in linear elements whose section is fully stretched.

5.48 (5.38). Joints of stretched or compressed main reinforcement as well as of welded meshes and frameworks in main direction must have the overlapping length l no less than value ll, determined by formula (316).

Values of relative overlapping length d

ll

l =λ calculated by formula (316) for different

classes of reinforcement and concrete are given in Table 46. If connected rods are installed with the reserve on the section area against the required by the strength calculation as regards the maximum forces in the connection zone, the overlapping length ll, calculated by formula (316) can be decreased by multiplying it by the ratio between required by the calculation and actual section area of reinforcement.

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Table 46

Relative length of the lap by concrete of classes Location of reinforcement

in concrete

Reinforcement class

В12,5 B15 B20 B25 B30 B35 B40 B45 B50 B55 B60

stretched, Bp-I 56 49 40 34 30 27 26 24 23 23 22 A-I 59 51 41 35 32 28 27 25 24 23 23

ll ≥ 250 mm A-II 46 40 33 28 26 24 22 21 21 20 20 A-III 56 49 40 34 30 27 26 24 23 23 22

compressed, Bp-I 41 35 29 24 22 20 19 18 17 17 16 A-I 39 26 28 24 21 19 18 17 16 16 16

ll ≥ 200 mm A-II 33 29 24 21 19 17 16 15 15 15 15 A-III 41 35 29 24 22 20 19 18 17 17 16

5.49 (5.39). Joints of welded meshes and frameworks as well as of stretched rods of bound frameworks and meshes without welding must locate as a rule in stagger. At the same time section area of main reinforcement connected in one place or at the distance less than overlapping length l, must be no more than 50 percent of total section area of stretched reinforcement by rods of periodic profile and no more than 25 percent – by plain rods.

Displacement of rods located in different places must be no less than 1.5ll (Draft 106, б). Connection of separate rods, welded meshes and frameworks without stagger is possible by constructive reinforcement (without calculation), as well as on those parts where reinforcement is used no more than 50 percent. In element cross section reinforcement joints must be located symmetrically.

5.50. By rebate joints connected rods must be located as close to each other as it’s possible;

clear distance between connected rods must be no more than 4d, that means 0 ≤ e ≤ 4d (Draft 106, a).

Draft 106. Location of rods connected with overlapping and joints

а – location of rod in the joint; б – location of joints

Neighbor rebate joints must not be located too close to each other. Clear distance between them must be no less than 2d (d – diameter of connected rods) and no less than 30 mm (see Draft 106, б).

5.51 (5.40) Joints of welded meshes in the direction of main reinforcement of plain hot-

rolled steel of class А-I must be made so that in each of welded in the stretched zone meshes there were located no less than two cross rods welded to all longitudinal rods of meshes on the length of overlapping (Draft 107).

Draft 107. Rebate joints of welded meshes in the direction of main reinforcement made of plain rods

а – by location of distributed rods in one plane; б, в – the same in different planes

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Length of overlapping of welded meshes with plain main reinforcement if there are two welded anchor rods on the length of overlapping must be no less than value ll, calculated due to Item 5.48. The same types of joints are used for rebated joints of welded frameworks with one-side location of main rods of all types of steel. Joints of welded meshes in the direction of main reinforcement of class А-II, А-III and Ат-IIIС are made without cross rods within the joint in one or both connected meshes (Draft 108, а, б); at the same time the overlapping length l is taken in compliance with item 5.48. By welding of cross anchor rods to main rods of periodic profile of welded meshes and frameworks (Draft 108, в, г) length of overlapping determined in compliance with Item 5.48, can be decreased by 5d for one cross anchor rod, by 8d – for two cross anchor rods. In all cases length of overlapping must be no less than 15d in stretched concrete and 10d – in compressed concrete. If diameter of main rods connected with overlapping in the stretched zone is more

than 10 mm and distance between rods is less than value bt

s

R

Rd

300 (where d –

minimum diameter of connected rods, mm), in places of joints it is necessary to install additional cross reinforcement in form of stirrups or suspension in form of U-shaped welded meshes brought into the compressed zone. At the same time section area of additional cross reinforcement installed within the joint must be no less than

sw

s

sR

RA4.0 (where А, – section area of all connected rods).

By rebate joints of welded frameworks in beams on the length of the joint independently on the diameter of main rods it is necessary to install additional reinforcement in form of stirrups or U-shaped welded meshes. At the same time spacing of additional cross rods within the joint must be no more than 5d (d – the least diameter of longitudinal main rods).

Draft 108. Rebate joints of welded meshes in the direction of main reinforcement of periodic profile

а – without anchor cross rods on two meshes; б – the same on one of meshes; в – by one anchor rod; г – by two anchor rods

5.52 (5.41). Joints of welded meshes in not main direction are made with overlapping

(counting between axes of end working rods of the mesh): by diameter of distributed reinforcement up to 4 mm – by 50 mm (Draft 109, a, б) the same more than 4 mm – by 100 mm, (Draft 109, в, г) By diameter of main reinforcement 16 mm and more welded meshes in not main direction can be installed close to each other, covering the joints by means of special covering meshes installed with overlapping to each side no less than 15d of distributed reinforcement and no less than 100 mm (Draft 109, д).

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Welded meshes in not main direction can be installed close to each other without overlapping and without additional meshes in the following cases:

- By installation of welded strip meshes in perpendicular directions; - By additional constructive reinforcement in the direction of distributed

reinforcement.

Draft 109. Joints of welded meshes in not main direction

а, б – with overlapping with diameter of distribution reinforcement up to 4 mm; в, г – the same more than 4 mm; д – close by diameter of main reinforcement 16 mm and more

REINFORCEMENT OF REINFORCED CONCRETE ELEMENTS

General positions

5.53 (5.16). Section area of longitudinal reinforcement in reinforced concrete elements

must be taken no less than the one mentioned in Table 47.

Minimum percentage of reinforcement S and S' in eccentric compressed elements whose bearing capacity by design eccentricity is used less than by 50 percent independently on the elasticity of elements is taken equal to 0.05. Requirements of Table 47 are not applied on the reinforcement determined by the calculation of elements for transportation and installation phases; in that case section area of reinforcement is determined only by the strength calculation considering instructions of Item 1.14. Elements not corresponding to requirements of minimum reinforcement, belong to concrete elements. Requirements of the present Item are not considered by specification of the section area of reinforcement installed along the slabs or panels contours according to the bending calculation of the slab (panel).

5.54 (5.22). All surfaces of reinforced concrete elements near which it is installed

longitudinal reinforcement must have also cross reinforcement taking end longitudinal rods to prevent formation of longitudinal cracks. At the same time distance between cross rods at each surface of the element must be no more than 600 mm and no more than double width of the element surface. It is possible not to install cross reinforcement at surfaces of thin ribs 150 mm and less which have only one longitudinal reinforcing rod.

5.55. Correspondence of reinforcement location to its project location must be provided by

means of special measures for reinforcement fixing due to Items 5.118 – 5.120.

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Table 47(38) Work condition of reinforcement

Minimum section area of longitudinal reinforcement in reinforced concrete elements, percent of section area of concrete

1. Reinforcement S in bending as well as in eccentric stretched elements by location of longitudinal force beyond the working height of the section

0,05

2. Reinforcement S, S' in eccentric stretched elements by location of longitudinal force between reinforcement S and S'

0,05

3. Reinforcement S, S' in eccentric compressed elements by:

a) lo/i < 17 (for rectangular sections – by lo/h < 5)

0,05

b) 17 ≤ lo/i ≤ 35 (5 ≤ lo/h ≤ 10) 0,10 c) 35 < lo/i ≤ 83 (10 < lo/h ≤ 24) 0,20 d) lo/i > 83 (lo/h > 24) 0,25

Note. Minimum section area of reinforcement given in Table 47 belongs to the section area of concrete equal to the product of the width of rectangular section or width of the rib of T-section (I-section) b by the main height of the section ho. In elements with longitudinal reinforcement spread evenly along the section contour, as well as in central stretched elements minimum area of all longitudinal reinforcement belongs to the total section area of concrete and must be taken two times more than values mentioned in Table 47.

Reinforcement of compressed elements

LONGITUDINAL REINFORCEMENT

5.56 (5.17). Diameter, mm, of longitudinal rods of compressed elements must be no more

than: for heavy-weight and fine concrete of class less than В25 ………………….. 40 for light-weight and porous concrete of classes: В12,5 and less …………………………………………………………………16 В15 – В25 ……………………………………………………………………..25 В30 and more ………………………………………………………………….40 For high-capacity columns of heavy-weight concrete of class В20 and more and by corresponding equipment for cutting, welding and other it is possible to use rods with diameter more than 40 mm. Diameter of longitudinal rods of eccentric compressed elements of solid structures must be no less than 12 mm. In columns with the dimension of the less side of the section 250 and more diameters of longitudinal rods should be no less than 16 mm.

5.57(5.18). In linear eccentric compressed elements distances between axes of rods of longitudinal reinforcement must be taken in the direction perpendicular to the

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bending plate no more than 400 mm, and in the direction of the bending plane – no more than 500 mm. If distances between axes of main rods in the direction of the bending plane are more than 500 mm so it is necessary to install construction reinforcement with diameter no less than 12 mm, so that distance between longitudinal rods was no more than 400 mm.

5.58(5.19). In eccentric compressed elements whose bearing capacity is used less than by 50 percent as well in elements with elasticity lo/i < 17 (for example in column pockets) where compressed reinforcement is not required by the calculation, and quantity of stretched reinforcement is no more than 0.3 percent it is possible not to install longitudinal and cross reinforcement which is required due to Items 5.54, 5.57, 5.59 and 5.60 along the surfaces parallel to the bending plane. At the same time reinforcement along the surfaces perpendicular to the bending plane is made by welded frameworks and meshes with the concrete protection layer no less than 50 mm and no less than two diameters of longitudinal reinforcement.

CROSS REINFORCEMENT

5.59 (5.22). In eccentric compressed linear elements by considered in the calculation

compressed longitudinal reinforcement to prevent its uplift the stirrups must be installed at distances no more than 500 mm and no more than: by bound frameworks – 15d, by welded frameworks – 20d (d – minimum diameter of compressed longitudinal rods).

Distances between stirrups of eccentric compressed elements in points of rebate joints of main reinforcement without welding must be no more than 10d. If congestion of the element by required by the calculation compressed longitudinal reinforcement S' is more than 1.5 percent so stirrups must be installed at distances no more than 10d and no more than 300 mm. Cross reinforcement structure mutt provide fixing of compressed rods to prevent their lateral uplift in any direction. During control how requirements of the present item are met longitudinal compressed rods which are not considered by the calculation must not be taken into account, if diameter of these rods is no more than 12 mm and no more than a half of cracks of concrete protection layer. Note. By high-strength compressed reinforcement of class A-IV and more distances between stirrups must be no more than 400 mm and for bound frameworks – no more than 12d, for welded frameworks – no more than 15d.

5.60 (5.23). By reinforcement of eccentric compressed elements by flat welded frameworks

two end frameworks (located at opposite surfaces) must be connected to each other to form a spatial framework. For that at the surfaces of elements normal to the plane of frameworks it is necessary to install cross rods welded by contact point welding to angle longitudinal rods of frameworks, or studs connecting these rods at the same distances like cross rods of plane frameworks.

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If end plane frameworks have intermediate longitudinal rods so these rods must be connected with longitudinal rods located at opposite surfaces by means of studs, no less than next nearest and no less than each 400 mm along the width of the element surface. It is possible not to install studs if the width of the present surface of the element is no more than 500 mm and number of longitudinal rods at this surface is any more than four. It is also possible not to install studs at intermediate rods which are distant from the angle ones no farther than 15dw, independently on the width of the element and number of rods. By larger dimensions of the element section it is recommended to install intermediate flat welded meshes (Draft 110, a).

Draft 110. Structure of spatial frameworks in compressed elements

а – welded; б – bound; 1 – connection rods; 2 – plane welded frameworks; 3 – stirrup; 4 – intermediate plane framework; 5 – a stirrup

Structure of bound stirrups in eccentric compressed elements must be so that longitudinal rods (at leas every second one) were located at the bending point of stirrups, and this bendings – at the distance no more than 400 mm along the width of the element surface. By width of the surface no more than 400 mm and number of longitudinal rods at this surface no more than four it is possible to bind all longitudinal rods by one stirrup (Draft 110, б). Independently on the width of the surface and number of rods it is possible not to install intermediate rods in stirrups bendings if these rods are distant from angle rods no more than 15dw. At the ends of bound stirrups it is necessary to install hooks.

5.61 (5.25). Diameter of stirrups of eccentric compressed elements must be taken no less

than 0.25d (d – maximum diameter of longitudinal rods), and in bound frameworks no less than 5 mm.

Reinforcement of bending elements

LONGITUDINAL REINFORCEMENT

5.62 (5.20). Beams and ribs 150 and less mm wide (Draft 111), working for bending can be

reinforced by one flat vertical framework, and more than 150 wide ones and by great loads they must be reinforced by several vertical meshes (frameworks). In beams more than 150 mm wide number of longitudinal main rods brought behind the surface of the support must be no less than two. In ribs of prefabricated panels, decks, multiribbed plates and other 150 and less mm wide it is possible to bring one longitudinal rod to the support.

Draft 111. Reinforcement of beams by plain welded meshes

1 – connecting rods; 2 – flat welded frameworks

5.63 (5.21). In bending elements by the height of the section more than 700 mm at lateral surfaces it is necessary to install constructive longitudinal rods with the distance between them along the height no more than 400 мм and with the section area no less than 0.1 percent of the concrete section area which has the dimensions equal to the distance between these rods along the height, along the width – to a half of the width of the element rib, but no more than 200 mm (Draft 112).

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Draft 112. Installation of constructive longitudinal reinforcement along the height of the beam section

5.64 (5.20). Distance between axes of main rods in the middle part of the slab span and

above the support (on the top) must be no more than 200 mm by the plate width up to 150 mm and no more than 1.5h by the height of the slab more than 150 mm (h – height of the slab).

In slabs more than 350 mm wide distance between axes of main rods can be decreased up to 600 mm. In solid plates distance between rods brought to the surface of the support must be no more than 400 mm, at the same time section area of these rods per 1 m of the slab width must be no less than 1/3 of the section area of rods in the span, determined by the calculation due to maximum bending moment. In hollow-core decks distance between axes of main rods can be decreased due to location of hollows in the section, but no more than up to 2h. By reinforcement of continuous slabs by means of welded roll meshes it is possible near intermediate rods to move all bottom rods to the top zone. Continuous slabs no more than 80 mm thick can be reinforced by means of single flat meshes without bending.

5.65. If main reinforcement of the slab goes parallel to the rib so perpendicular to it it’s necessary to install additional reinforcement with the section no less than 1/3 of maximum section of main reinforcement of the slab in the span and bring it in the slab to each side of the rib at the length no less than 1/4 of design span of the slab.

If main reinforcement of the slab above the support goes perpendicular to the rib, it is necessary to break it no closer than at the distance 1/4 of design span of the slab from the surface of the rib (черт. 113).

Draft 113. Reinforcement of support parts of slabs monolithically connected with beams 1 – main span reinforcement of the slab; 2 – main reinforcement of the slab above the span; l – design span of the slab

In beam slabs section area of distributed reinforcement per width unit of the slab must be no less than 2 percent of the section area of main reinforcement per width unit of the slab in the point of maximum bending moment. Distance between rods of distributed reinforcement of beam slabs must be no more than 600 mm.

CROSS AND DIAGONAL REINFORCEMENT

5.66 (5.26). In beam structures more than 150 mm high, as well as in hollow-core slabs (or

similar multi-ribbed structures) more than 300 mm high it is necessary to install cross reinforcement according to instructions of Item 5.69.

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In solid plates independently on the height, in hollow-core plates (or similar multi-ribbed structures) 300 and less mm high and in beam structures 150 and less mm high it is possible not to install cross reinforcement, at the same time the requirements of the calculation due to Items 3.40 and 3.41 must be met.

5.67. To prevent lateral uplift of rods if compressed zone it is necessary to install cross reinforcement in compliance with requirements of Item 5.59.

5.68 (5.25). Diameter of stirrups in bound frameworks of bending elements must be taken

no less than, mm: by the section height, equal or less than 800 mm 6 the same more than 800 mm 8 The ratio between diameters of cross and longitudinal rods in welded meshes determined due to the welding conditions is taken in conformity with positions 1 and 2 of Table 38.

5.69 (5.27). Cross reinforcement in beams and slabs structures mentioned in Item 5.66 is installed in support parts by distributed loads equal to 1/4 of the span, and by point loads – to the distance from the support to the nearest load but no less than 1/4 of the span with spacing: by section height of the element h, equal or less than 450 mm no more than h/2

and no more than 150 mm by section height of the element h more than 450 mm no more than h/3

and no more than 500 mm On the other part of the span by section height of the element h more than 300 mm it is installed cross reinforcement with spacing no more than 3/4h and no more than 500 mm. For ribbed plates in the middle part of the span if requirements of Item 3.40 are met so mentioned above instructions are not to be taken into account.

5.70. To provide anchorage of cross reinforcement of bending elements the connections of longitudinal and cross rods in welded frameworks must be made in compliance with requirements of positions 1 and 2 of Table 38. In bound frameworks the stirrups must be designed so that at the places of their bending there were longitudinal rods (Draft 114). At the same time in welded as well as in bound frameworks diameter of longitudinal rods must be no less than 0.8 of diameter of cross rods.

By bound reinforcement in intermediate (middle) beams of T-section monolithically connected with a slab, it’s recommended to install open stirrups.

Draft 114. Structure of stirrups of bound frameworks of beams

5.71. Bent rods of reinforcement must be installed in bending elements by their

reinforcement by bound frameworks. The rods must be bent along the arch with radius no less than 10d (Draft 115). In bending elements at the ends of bent rods it is necessary to make straight parts no less than 0,8lan long taken according to instructions of Item 5.44, but no less than 20d in stretched zone and 10d – in compressed zone.

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Draft 115. Structure of reinforcement bend

Straight parts of bent pain rods must end with hooks. Beginning of bend in stretched zone must be distant from the normal, in which bent rod is used due to the calculation, no less than by 0,5ho, and the end of bend must be located no closer than that normal section where bend is not required by the calculation (Draft 116). Distance from the surface of free support to the top end of the first bend (counting from the support) must be no more than 50 mm.

Draft 116. Location of bends, determined due to the moments diagram in the beam

1 – beginning of bend in the stretched zone А; 2 – the same in zone Б; 3 – section where rod а is not required due to the calculation of zone А; 4 – section where rod б is not required due to the calculation of zone Б; 5 – curve of bending moments; 6 – diagram of materials

5.72. Angle of slope of bendings to the longitudinal axis of the element must be taken as a

rule 45 degrees. In beams more than 800 mm high and in beams-walls it is possible to increase angle of slope of bendings up to 60 degrees and in low beams and slabs – to decrease up to 30 degrees.

Rods with bends should be located at the distance no less than 2d from lateral surfaces of the element (d – diameter of bent rod). It is not recommended to bend the rods located directly at lateral surfaces. Rods bends should be located symmetrically relating to longitudinal axis of the beam. Use of bends in form of „floating" rods (Draft 117) is not allowable.

Draft 117. „Floating" rod

Reinforcement of elements working for bending for torsion

5.73 (5.31). In elements working for bending with torsion bound stirrups must be closed

with reliable anchorage at the ends, and for welded frameworks all lateral rods of both directions must be welded to angle longitudinal rods and form closed circle.

Spatial frameworks should be designed considering requirements of Items 5.28, 5.29а-е and 5.30. Distances between cross rods located at the surfaces parallel to the bending plane must correspond to requirements of Item 5.69. Distances between cross rods located at the surfaces normal to the bending plane must be no more than the width of the element section b. At surfaces compressed by bending by Т ≤ М/5 it is possible to increase the distances between cross rods taking them due to Items 5.54 and 5.59. Requirements of the present Item are applied to end beams to which secondary beam or slabs join only from one side, as well as middle beams to which design loads transferred from adjoining to them spans are different and differ more than 2 times.

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Special cases of reinforcement

REINFORCEMENT OF DETAILS WITH HOLES

5.74 (5.50). Big holes in reinforced concrete slabs, panels and others must have additional

reinforcement with the section no less than section of main reinforcement (of the same direction), with is required by the calculation as for a solid one. Replacing reinforcement must be brought behind the edges of a hole at the length no less than overlapping length ll, determined due to Item 5.48.

Holes in sides of elements must have round form and edges of holes must be reinforced.

REINFORCEMENT OF SLABS IN THE PUNCHING ZONE

5.75 (5.29). Cross reinforcement in slabs in a punching zone is installed with the spacing

no more than 1/3h and no more than 200 mm, at the same time the width of the zone of cross reinforcement installation must be no less than 1,5h (h – thickness of a slab).

Anchorage of the mentioned above reinforcement must correspond to requirements of Item 5.70.

DESIGN OF SHORT CONSOLES

5.76. Short consoles can be of constant and variable height with enlargement to the

embedment place.

Consoles of variable height should be designed by great loads.

5.77 (5.30). Cross reinforcement of cross consoles should be made: - by h ≤ 2.5с – by stirrups inclined at the angle 45 degrees (Draft 118, а); - by h > 2.5с – by horizontal stirrups (Draft 118, б).

Draft 118. Short consoles of columns with stirrups

а – inclined; б – horizontal

In all cases spacing of stirrups must be no more than h/4 and no more than 150 mm (h – height of a console). By Limited height of the console it is possible to use hard reinforcement (Draft 119).

Draft 119. Short console with hard reinforcement

CONFINEMENT REINFORCEMENT

5.78 (5.24). Confinement reinforcement prevents cross expansion of concrete, due to that

concrete strength by longitudinal compression is increased.

Confinement reinforcement is used in form of cross welded meshes (Draft 120) or spirals (rings) (Draft 121).

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Draft 120. Confinement reinforcement in form of a pack of cross welded meshes

Draft 121. Spiral confinement reinforcement of reinforced concrete elements

For confinement reinforcement it is necessary to use reinforcement steel of classes А-I, А-II, А-III and Aт-IIIC with diameter no more than 14 mm and steel of class Вр-I. Confinement reinforcement can be used along the whole length of compressed elements (columns, piles) or as local reinforcement of column joints, at points of impact on a pile and others. Besides confinement reinforcement in form of meshes is used by local compression. Meshes and spirals (rings) in columns and piles must surround all main longitudinal reinforcement.

5.79 (5.24). Meshes of confinement reinforcement can be welded of intersectional rods (see Draft 120) or in form of combs. In both cases it is necessary to provide cooperation work of mesh rods and concrete.

By using of confinement reinforcement of welded meshes it is necessary to meet the following conditions: а) section areas of mesh rods per a length unit in one and another direction must not

differ more than by 1.5 times; b) meshes spacing (distance between axes of rods in one direction) must be taken

no less than 60 mm, but no more than 1/3 of the least side of the element section and no more than 150 mm;

c) clear dimensions of meshes cells must be taken no less than 45 mm, but no more than 1/4 of the least side of the element section and no more than 100 mm.

The first welded mesh is located at the distance 15 – 20 mm from the loaded surface of the element. Combs used for confinement reinforcement must be inverted with the overlapping length corresponding to the length given in Item 5.48, and must be made of reinforcement of periodic profile.

5.80 (5.24). By using of confinement reinforcement in form of spirals or rings it is necessary to meet the following conditions (see Draft 121): а) spirals and rings must be round in plan; б) spacing of spiral winding or spacing of rings must be no less than 40 mm, but no more than 1/5 of element diameter and no more than 100 mm; в) diameter of spiral winding or of rings must be no less than 200 mm.

PREFABRICATED FEATURES STRUCTURES

General positions

5.81. Prefabricated reinforced concrete elements must meet the requirements of

produceability: they must have simple shapes (considering slopes in case of necessity), simple reinforcement and little manufacturing content; allow

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mechanization and automatization of production, fast production, as well as transportable and convenient for installation.

Reinforced concrete elements must be designed as a rule for production in forms without following concreting or assembly of separate items before installation.

5.82. Dimensions and shapes of prefabricated elements must conform to requirements of Items 5.5 – 5.8.

5.83. Prefabricated reinforced concrete elements should be designed so that it was possible

to produce them in solid forms. If production of an element in a solid form is impossible so it is recommended to design as big part as it’s possible as a solid one.

5.84. Ribs in sides of beams should be designed only by great point loads or if it is necessary to provide the stability of the side.

5.85. Requirements for accuracy of manufacturing of reinforced concrete elements must be

specified according to the analysis of their adjoining type with other elements. So, for example, by concreting of joints some deviations from nominal dimensions are allowable as they are to be balanced during concreting.

Joints of elements of prefabricated structures

5.86 (5.42). By joins of reinforced concrete elements of prefabricated structures the forces from one element to another are transferred through the abutting main reinforcement, steel embedded elements, filled by concrete or mortar joints, concrete keys or (for compressed elements) directly through concrete surfaces of adjoining elements.

5.87 (5.43). Fixed joints of prefabricated structures must be monolithed by filling of joints between elements by concrete. If during manufacturing of elements the surfaces are adjusted to each other (for example by using of the butt end of one of the elements is used as a framework for the butt end of the other element), so by transferring of only compression force through the joint it is possible to make dry joints.

5.88. It is necessary to make such constructive decisions which provide simplicity of

manufacturing of joint details (embedded elements, meshes and other), their assembly, fixing in the form, forming of the detail as well as installation and connecting of reinforced concrete prefabricated elements.

5.89 (5.44). Joints of elements which take stretching forces must be made:

а) by welding of steel embedded elements; b) welding of reinforcement connecting bars; c) agglutination of elements by constructive polymer mortars using connection derails of rod reinforcement; d) monolithing of connecting rods with overlapping.

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During design of joints of prefabricated structures elements it is necessary to design such connections of embedded elements which prevent unbending of their parts as well as concrete chipping.

5.90. Fixed joints of prefabricated columns should be made by means of “tub” welding of connecting rods located in special cuttings with the following monolithing of these cuttings.

In such joints between butt ends of joined columns it is necessary to design a centralize filler in form of a steel plate anchored in concrete or welded to the spread plate of the embedded element (see Draft 81; Draft 122). Dimensions of a centralize filler are taken no more than 1/3 of the corresponding dimension of the element section. Form and dimensions of cuttings are determined by means of number of joined rods and their diameters (see Draft 122). Total height of cuttings is taken no less than 30 cm and no less than 8d (d – diameter of connecting rods); depth of a cutting allows to install prefabricated rods and to exercise a non-destructive ultrasonic control.

Draft 122. Fixed joint of prefabricated columns by “tub” welding of connecting rods

а – by four angle connecting rods; б – by connecting rods located along the section perimeter; 1 – connecting rods; 2 – concrete for monolithing of cuttings; 3 – centralize filler (meshes of confinement reinforcement in the sections are not shown)

5.91. Joints of prefabricated column with eccentricities of longitudinal forces during the

use stage less than 0,17h can be made with adjoining of butt-ends of columns through the cement layer of polymer mortal layer with break of longitudinal enforcement (contact joints). Different types of contact joints are given on Draft 123.

In the joint of the 1st type a centering pin is jutting out from the butt-end of the column; the centering pin in form of a reinforcing rod with diameter 32 – 36 mm is put into a chase with liquid mortar located in the center of a butt-end of the bottom column. To form a joint filled with mortar at the bottom butt-end it is installed a centering filler with a hole for a pin.

Черт. 123. Types of contact joints of prefabricated columns

1 – centering pin; 2 – centering filler; 3 – mortar; 4 – welding; 5 – tack weld; 6 – rods connected with a plate by welding; 7 – intermediate rods with tack weld; 8 – end plates with stamped holes (cross reinforcement is not shown)

In the joint of the 2nd type the top butt-end has a concrete lug in the center, and the bottom one – a chase corresponding to a lug of round or rectangular form in plan. Rods in connections of the 1st and the 2nd type must be distant from the concrete surface no more than by 10 mm. The 3rd type – joint with steel plates at the butt-ends of columns connected by welding with longitudinal reinforcement rods in stamped or in counter-bores or by means of buttoning plates. Number of rods connected in such manner is determined by the calculation of assembly rods and is taken no less than for (angle) rods. The rest of rods (intermediate) are installed by flat butt-ends fixedly to the plates and are

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welded to them by arch welding. After installation of columns end plates are connected by long welded joints along the perimeter or angles of the section. Thickness of end plates is taken no less than: by connection of reinforcement with the plate in stamped holes or fixedly – 0.25da and 6 mm; in counter bores – 0.35da and 12 mm (da – diameter of longitudinal rods required by the calculation). If in the column shaft there is installed considered in the calculation confinement reinforcement so thickness of the plates should be increased by 2 mm.

5.92 (5.24). At the end parts of connected plates, if special reinforcement is not provided (iron ring, embedded elements), it is necessary to install confinement reinforcement meshes in accordance with the instructions of Items 5.78 and 5.79 – no less than four meshes on the length (counting from the butt-end of the element) no less than 20d, if longitudinal reinforcement is made of plain rods, and no less than 10d – of periodic profile rods (d – maximum diameter of longitudinal reinforcement).

Saturation factor of confinement reinforcement µxy (see Item 3.57) is taken no less than 0.0125. For joints mentioned in Item 5.90, in case of necessity monolithing concrete in zone of cuttings can be reinforced by welded meshes. In zone of cuttings there are installed one or two closed stirrups which bend connection rods. By contact joints of the 3rd type (see Item 5.91) in end parts of joined columns it is possible not to install meshes of confinement reinforcement, if they are not designed in the columns. But at the length 10da of end part it is necessary to install cross reinforcement (stirrups, meshes) of the same structure like in the column shaft taking its spacing no more than: 0.25 of maximum section dimension; 0.6 of spacing of cross reinforcement in the column shaft; 80 mm.

5.93. Dimensions of welded joints made by produce of steel embedded elements and by their connection during installation in the joints of prefabricated elements must be calculated in compliance with the requirements of SNiP II-23-81. Alternative of welding method of connecting rods and constructive elements of these connections should be provided due to Items 5.15 – 5.17, as well as due to state standards and normative documents for welding technologies.

During design of welded joints and embedded elements it is necessary to use welding methods which don’t cause warping of steel details joints.

5.94 (5.51). During design of prefabricated floors elements it is necessary to design the joints between them which are filled with concrete. Width of joints is taken due to condition of their qualitative filling and must be taken no less than 20 mm for elements with the section height 250 mm and no less than 30 mm – for elements with great height. At the same time it is necessary to provide the possibility to arrange to joined reinforcement or embedded elements and to provide their qualitative welding.

Class of concrete for filling of joints which transfer design forces is taken in compliance with Item 2.4.

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For concreting of difficult-to-access or difficult-to-control points of the joint it is recommended their filling by concrete or mortar under pressure as well as use of expanding cement.

Slinging devices

5.95. During design of prefabricated reinforced concrete elements it is necessary to provide comfortable methods of their holding by means of load-handling devices during removal of frameworks, as well as during loading and unloading operations and assembly works.

Methods and points of handling must be chosen considering manufacturing and installation technologies as well as constructive features. The detail must be checked by the calculations.

5.96. In concrete and reinforced concrete details it is necessary to design devices for their strapping: strapping holes (including the ones for prefabricated holes), slots etc or permanent steel strapping loops which must be produced of hot-rolled steel due to Item 2.18.

It is recommended to design handling of details without using any devices which are made of steel by means of making holes, slots, deepenings and others (Draft 124).

5.97. During design of details with strapping loops it is necessary to use common loops. If there are no common loops with required characteristics so it is recommended to design the loops of types shown on Draft 125.

Minimum parameters for loops with straight and bent lags of types П1.1 and П2.1 (see Draft 125) are given in Table 48.

Draft 124. Examples of stripping devices without loops

а – by stripping of a block; б – stripping holes in a column; в – combination of two different stripping devices in one detail; 1 – cargo slings; 2 – hole cutout; 3 – holes; 4 – loops for handling during taking out of the form

Table 48

Loops Symbols of parameters

Размеры, мм

d

R

r

6-12 30 20

14; 16 30 30

18-22 40 40

25 60 60

a1

a2

3d 6d

5.98. Diameter of the loop rod d should be taken due to Item 49 according to the weight of

a detail per a loop. Weight of the detail is determined due to instructions of Item 2.13. By lifting of flat details by four loops weight of the detail is considered to be distributed for three loops.

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Draft 125. Types of strapping loops

а – free arranged in the detail loops of steel of classes А-I and Ас-II; б – arranged in narrow conditions loops of steel of class А-I; в – the same of steel of class Ас-II

Table 49

Diameter of the rod of a loop, mm

Weight of the detail m, kg, per one loop of steel of classes

А-I Ac-II 6 150 8 300

10 700 900 12 1100 1500 14 1500 1900 16 2000 2500 18 2500 3100 20 3100 3900 22 3800 4700 25 4900 6100 28 6100 7600 32 8000 9900

Notes: 1. Values m correspond to the angle between straps and a horizontal line equal to 45 degrees and more; less angle of slope is not allowable. If detail is slinged by means of vertical slings so it is possible to decrease weight of the detail per a loop by 1.4 times. 2. By diameter of the loop rod from 8 to 22 mm it is possible to decrease the mentioned values by 25 percent by special justification.

By lifting by three and more loops located on one butt-end of the detail (for example on the wall panel) weight of the detail is taken as distributed only on two loops, that’s why in that case it is not recommended to install more than two loops. By using of devices (balanced lifting frame) providing self-balancing of forces among slings it is possible to distribute the weight of the detail among loops in accordance with the structure of the device.

5.99. The height of the lug eye he (see Draft 125) corresponding to the dimensions of pulling hooks of loading slings must be taken equal to, mm: 60 by diameter of the loop rod from 6 to 16 mm. 80 by diameter of the loop rod 18 and 22 mm 150 by diameter of the loop rod from 25 to 32 mm Length ls and depth of the embedment hb of ends of the loop lugs into the concrete of the detail (see Draft 125) should be taken due to Table 50. By location of slinging loops in standard deepenings (Draft 126, а) value hb can be counted from the top surface of a concrete element. In all cases value ls should be taken no less than 200 mm. For loops made of reinforcement steel ∅25А-I and ∅28А-III and more values ls and hb should be increased by 20 percent.

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Table 50 Standard cube-strength of

concrete at the moment of the first lifting of the detail, MPa

Length of embedment

into the concrete ls

Depth of embedment

into the concrete hb

from 3 to 5 45d (50d) 35d (40d) More than 5 up to 8 35d (40d) 25d (30d)

More than 8 up to 15 30d (35d) 20d (25d) More than 15 up to 25 25d (30d) 15d (20d)

More than 25 20d (25d) 15d (20d) Note. Values given in brackets belong to the lifting cases in vertical direction of one-layer thin-walled elements (wall panels of heavy-weight concrete) no more than 220 mm thick.

Branches of a loop of steel of class А-I, as well as straight (without bendings) branches of loops of steel of class Ас-II should end with hooks. In case of necessity it is possible to locate branches at the angle one to another no more than 45 degrees. For details of light-weight concrete sling loops should be reinforced by a cross rod located at the level of hooks of loop branches. Distance between a lateral surface of the tail of the loop hook and the surface of the detail measured in the plain of the hook should be taken no less than 4d (Draft 125a). In case if it’s impossible to make the embedment of loop ends at the required length, anchorage of the loop must be made by different methods, for example by welding to embedded elements etc. safety of accepted anchorage should be confirmed by calculations or tests.

Draft 126. Dimensions of holes for embedded location of sling loops eyes

a – closed hole: б – open hole (at the edge of the detail) by diameter of the loop rod 6 – 16 mm: R1=125 mm, а = 30 mm, b1 = 50 mm, l1=25 mm, l2=30 mm; by diameter of the loop rod 18 – 22 mm: R1=150 mm, а=40 mm, b1=65 mm, l1=30 mm, l2=30 mm;

5.100. It is possible to locate sling loops in the holes so that their eyes were located below

the surface of a concrete or reinforced concrete detail. This location is recommended by mechanized finishing of the concrete surface, when the loops prevent such finishing. Holes for loops can be closed (see Draft 126, a) or open (Draft 126, б). In the last case they don’t collect water which can freeze, as well as conditions of fixing of loops are better.

EMBEDDED ELEMENTS

General positions

5.101. During design of reinforced concrete structures it is recommended to use mainly

unified welded, stamped and stamped-welded embedded elements. 5.102. Embedded elements must be anchored in concrete.

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Welded embedded elements usually consist of plates (section of strip steel, angle or shaped steel) with welded to them normal or inclined anchors (Draft 127). Stamped embedded elements consist of parts which perform a function of plates and strip anchors which have stamped jutting for reinforcement of anchorage (Draft 128, а, б). If during manufacturing of embedded elements there are used pressing and welding, so such details are called stamped-welded ones (Draft 128, в, г). Constructive requirements for stamped embedded elements and stamped-welded details made with welding of anchor rods to tamped embedded elements are given in „Recommendations for design of steel embedded elements of reinforced concrete structures " (Мoscow., Stroyizdat, 1984). Embedded elements can also have supports for work against shear (see Draft 127, в), devices for fixing to forms, bolts for connection of prefabricated elements and other.

5.103. For mechanized finishing of the reinforced concrete element surface it is recommended to embed the plates into concrete at the depth no less than for 5 mm.

In the plates of embedded elements located on the top surface of the element (by concreting) with the least dimension more than 250 mm and in the plates with cover the whole or the larger part of the concreted element it is necessary to make the wholes for air out during laying and compaction of concrete and for concrete work quality control.

5.104. To provide project location of the embedded element in the detail it is necessary to fix it by temporary fixing devices to elements of the form before concreting. Examples of such fixing are given in Recommendations mentioned in Item 5.102. By location of the detail on the open during concrete works surface of the element when its fixing to the form is not expedient the detail can be welded to the reinforcement.

Draft 127. Examples of structures of welded embedded elements

а – with inclined and normal anchors „open table"; б – „closed table"; в – „open table" with supports; г – with angle steel; 1 – normal anchors (T-welded); 2 – inclined anchors (welded with overlapping); 3 – support working in two directions; 4 – the same in one direction; 5 – holes for fixing

Draft 128. Examples of embedded elements structures

а, б – stamped; в, г – stamped-welded 5.105. To provide the working life of embedded elements it is necessary to design their

anticorrosive protection. Way of protection is chosen according to the corrosive power in compliance with requirements of SNiP 2.03.11-85 and Guidelines to it, as well as Recommendations mentioned in Item 5.102.

5.106. In main rods steel consumption for embedded elements must be calculated

separately for reinforcement and connection details. At the same time into the embedded elements weight it is included also weight of anchors and other welded rods considering technological allowance required for welding (washing and settlement into the melt, anchorages embedment and other).

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Plates

5.107. Choosing of steel grade for plates of embedded elements is made due to Table 13. 5.108. Thickness of plates or other external details of embedded elements is determined

due to Item 3.105, as well as due to technological requirements for welding mentioned in Items 5.116 and 5.117 but no less than 4 mm. Besides by welding of steel elements to the plate which has contact with concrete its thickness must be taken according to the height of assembly angle joint no less than values mentioned in Table 51.

5.109. By dimensions of the plate similar to dimensions of the reinforced concrete element

it is necessary to consider their positive deviation limit and to provide the possibility of free installation of embedded element by negative deviations of form dimensions.

Table 51 Thickness of a plate of

the embedded element, t, mm

Maximum allowable height kf, mm, of the angle joint

Single-pass Double-pass 4 5 6 5 6 8 6 8 10 8 12 16

≥10 1.8t Anchors

5.110. Anchors of embedded elements must be designed mainly of classes of reinforcement

А-II and А-III with diameter 8 – 25 mm. It is possible to use reinforcement steel Aт-IIIC for anchors welded with overlapping.

Steel grade for anchor rods must be taken in compliance with Table 12. Anchors of plain reinforcement of reinforcement of class А-I can be used only if they are reinforced by plates or button-heads at the ends. For constructive details it is possible to use anchors of the same reinforcement with hooks at the ends.

5.111. Number of normal anchor rods in the embedded element is taken no less than four, but if three are no tearing forces and bending moments it can be decreased up to two. By action of tearing forces and bending moments if these forces are applied in the plane of anchors location it is possible to use embedded elements with two normal anchor rods.

Number of inclined anchors must be taken no less than two. At the same time it is necessary to design minimum one normal anchor. If there are no less than four normal anchors in the embedded element so it is possible to take one inclined anchor. For anchor rods welded with overlapping their angle of inclination towards the shear force must be taken equal to 15 to 30 degrees.

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Distances between anchor axes required by the calculation must be no less than values shown on Draft 129.

Draft 129. Minimum distances between anchors and from anchors to concrete edge

By anchors made of steel of classes А-I and А-II: a = 4dd, b = 6dd, с = 3dd, е = 3dd; the same of class А-III: a = 5dd, b = 7dd, с = 3.5dd, е = 4dd (dd – anchor diameter required by the calculation)

5.112 (5.14). Length of anchor rods of embedded elements by stretching forces applied to

them must be no less than value lап, determined due to instructions of Item 5.44. At the same time length of stretched anchor rods embedded into the stretched concrete or in compressed concrete by σbc/Rb > 0.75 or σbc/Rb < 0.25 must be determined by formula (316), using values wап, ∆λап and λап due to position 1а of Table 44. In other cases the mentioned values should be taken due to position 16 of Table 44 (here σbc – compression force in concrete acting perpendicular to the anchor rod and determined as for elastic material by the safety factor γf = 1.0

If part of the anchor with the length а is located in the zone with the stress in concrete corresponding to the condition 0.75 ≥ σbc/Rb ≥ 0.25, so ωап is determined due to the following formula

( )

a

a

anl

dal 5.07.0 +−=ω , (317)

where la – actual anchor length. Other parameters of table 44 are determined in the same manner.

By acting of stretching and shear forces on normal anchor rods the right part of formula (316) is multiplied by the coefficient δ3, equal to:

7.0/1

3.0

113 +

+=

anan NQδ , (318)

where Qan1, Nan1 – stretching and shearing forces in the anchor rod determined due to Item 3.101.

At the same time value lan must be no less than minimum value lan сdue to Item 5.44. For normal anchors the length is calculated from the internal surface of plates, for inclined ones – from the beginning of bending or butt-end edge of the plate.

5.113 (5.45). Length of anchors of embedded elements by action of stretching forces on them (see Item 3.101) can be decreased if anchor plates are welded to the ends of rods or if there are hot-made button-heads with diameter no less than 2d – for rods of reinforcement class А-I and А-II and no less than 3d – for rods of reinforcement class А-III. In such cases length of the anchor rod is determined due to the calculation of chipping and compression of concrete (see Items 3.106, 3.107 and 3.109) and are taken no less than 10d (d – anchor diameter).

Anchor plates must meet the requirements of Item 5.45 а. In case if crack formation in concrete along anchors is possible (σbt > Rbt) within their design length ends of anchors must be reinforced by welded plates or button-heads. At the same time ends of anchors must be located in compressed zone of elements. In

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eccentric stretched elements if longitudinal force is located between reinforcement S

and S' ends of anchors must be located at the opposite surface of the element and brought behind the longitudinal reinforcement.

5.114. By acting of pressing force on the embedded element a part of the shear force can be transferred to concrete through the supports of strip steel or of reinforcement lugs (see Draft 127, в). Height of supports should be taken no less than 10 and no more than 40 mm by the ration between the thickness of the support and its height no less than 0.5. Distance between supports in the direction of the shear force is taken no less than six heights of the support.

5.115. Embedded elements in light-weight concrete of classes В5 – В10 should be designed

so that tearing forces were taken by normal anchors and shearing forces – by inclined ones. Anchors of embedded elements in such cases should be taken of reinforcement steel of periodic profile of class А-II or plain reinforcement steel of class А-I with diameter no more than 16 mm. At the ends of anchors it is necessary to provide reinforcement in form of button-heads and welded plates. Length of anchor rods and dimensions of strengthening are determined due to the calculation of chipping and compression of concrete (see Items 3.106, 3.107 and 3.109), at the same time length of concrete is taken no less than 15d, and diameter of a button-head – no less than 3d.

Welded connections of embedded elements

5.116. Welded connections of anchors with plates should be designed due to Table 52. During manufacturing of T-connections of anchors with flat elements there are used: flux arch welding (position 1 – 3); contact welding (positions 4, 5); mechanized welding in carbonic acid CO2 (positions 6, 7); “tub” one-electrode welding in prefabricated forms (position 9); manual arch welding with bead joints (position 8). All mentioned processes can be used by welding of embedded elements of type „open table" (see Draft 127, а, в), and welding methods due to positions 6-9 – also for manufacturing of embedded elements of type „closed table" (see Draft 127, б).

5.117. Welded overlapped connections of anchors and reinforcing rods with plates should

be designed according to the instructions of Table 53. It is recommended to use mainly contact projection welding (positions 2 and 3 of Table 53).

Welded joints by welding of flat elements (plate, angles and other) should be made due to SNiP II-23-81.

FIXING OF REINFORCEMENT

5.118 (5.49). Location of reinforcement must correspond to its project location and this

provides by means of fixing means. Fixing of reinforcement is performed by means of: а) one-use devices which stay in concrete; б) prefabricated devices which are taken out of concrete before and after its hardening: в) special details fixed to main surface of a form or formwork which not prevent removing of reinforced concrete element out of the form.

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5.119. It is recommended to use the following one-use devices: а) to provide required thickness of concrete protection layer – due to Draft 130; б) to provide required distance between separate reinforcement details or rods – due to Draft 131; в) to meet requirements mentioned in sub-items „а" and „б", - due to Draft 132. Kind of a fixing devise to provide thickness of concrete protection layer at external surfaces of elements should be chosen due to requirements of Table 54. It’s not allowed to use off-cuts of reinforcing rods as fixing devices. In stretched zone of concrete elements used in conditions of corrosive environment it is not possible to install plastic fillers under rods of main reinforcement or close to them – under rods of distributed reinforcement. In such elements there should be used mainly fillers of dense cement-sand mortar, concrete or asbestos cement.

5.120. In case of use of one-use fixing devices it in compliance with requirements of Table 54 is necessary to show which of these devices are allowable in the given element.

Thickness of protection layer of concrete in the point of installation of the filler-holder should be taken divisible into 5 mm.

Table 52

Welding methods of T-connections anchors and

reinforcing rods with the plates

Symbol of the connection due to GOST 14098-85

Position number of Table 1 of СН

393-78

Reinforcement class

Rod diameter d, mm

Thickness of the

element t, mm

Minimum ratio t/d

Distance between rods axes

z, мм

Distance from the rod axis to the

plate edge

Rod length,

mm

Additional instructions

Т1 17

А-I 8-40

А-II 10-25 28-40

1. arch mechanized flux welding

А-III 8-25 28-40

≥6

0,50 0,55 0,70 0,65 0,75 0,65

by d≤22 mm z=25 + d;

by d≥25 mm

z = 2d

≥ l,5d

≥80

Maximum length of the rod is 400 mm

2. flux arch welding with small-scale mechanization

Т2 17

AT-IIIC 10-18

≥10

0,75

The same

≥l,5d

≥80

Connections of type Т2 of reinforcement of class AТ-IIIC are not allowed

3. mechanized flux welding on the rigidity element (relief)

Т3 *

А-I А-II А-III AТ-IIIC

8–25 10–25 8–25

10–18

≥4

0,40 0,40 0,50 0,50

For d = 8–16 z≥d+25;

For d = 18–25 z≥2d+10

≥l,5d ≥80

4. contact projection resistance welding

Т6 **

А-I А-II А-III

6–20 10–20 6–20

≥4 ≥4 ≥6

0,40 0,40 0,50

≥50 ≥ 2d ≥80

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5. contact straight flash welding

Т7 ***

А-I А-II А-III AТ-IIIC

10-20 10-20 22-40 10-22

≥4 ≥6

≥12 ≥6

0,40 0,50 0,50 0,50

≥80 ≥1,5d

6. arch mechanized welding in СО2, in a stamped hole

Т8; Т9 **

А-I А-II А-III AТ-IIIC

10-36 10-36 10-36 10-22

≥4 0,30 0,30 0,30 0,30

≥50 ≥2d ≥80 Manual welding is possible

7. arch mechanized welding in СО2 in a galvanized hole

T10; T11 20

А-I А-II А-III AТ-IIIC

12-25 12-25 12-25 12-18

≥8 50 ≥2,5d ≥l,5d ≥7d

8. arch manual welding in a counterbored hole

Т12 21

А-I А-II А-III AТ-IIIC

8-40 10-40 8-40

10-18

≥6 ≥8 ≥6 ≥8

0,50 0,65 0,75 0,75

≥3d ≥2d

9. “tub” single-electrode welding in a prefabricated form

T13 18

А-I А-II А-III

16-40 16-40 16-40

≥8 0,50 ≥80 ≥2d ≥150

* Welding technology is given in „Recommendations for flux welding technologies of inclined connections of embedded elements and T-connections as regards the rigidity element " (ПЭМ ВНИИС Госстроя СССР, 1982). ** Welding technology is given in „Instructions for technologies of light stamped-welded embedded elements of reinforced concrete structures *** Welding technology is given in „ Instructions for technologies of contact welding of embedded elements of type „open table" (ВСН 65), Киев, 1985.

Table 53

Welding methods of overlapped

connections of anchors and

reinforcing rods with plates

Symbol of the connection due to GOST 14098-85 Position number of 1 of СН 393-78

Reinforcement class

Rod diameter d, mm

Thickness of the element t, mm

Minimum ratio t/d

Distance between rods axes z, мм

Distance from the rod axis to the plate edge

Overlapping length in diameters d

Additional instructions

1. manual arch side-lap weld

Н2 19

A-I A-II; A-III Aт-IIIC AT-IV;AT-IVK AT-VCK;A-VI; Àт-IVC;AT-V A-V

10-40 10-40 10-28 10-22 10-28 10-28 10-32

≥4 ≥4 ≥4 ≥5 ≥5 ≥5 ≥5

0,3 0,3 0,3 0,4 0,4 0,4 0,4

≥3d ≥ d 3d

4d 4d 5d 5d

5d

5d

Distance from the butt-end of the rod to the edge of the plate must be no less than d

2. contact welding on one projection

НЗ 15

A-I A-II A-III

6-16 10-16 6-16

4–5 0,3 ≥4d ≥2d 4d Distance from the center of the projection to the butt-end of the rod must be no less than 2d

3. contact welding on two projections

Н4 16

A-I A-II A-III Aт-IIIC

12-16 12-16 12-16 12-16

4–6 0,3 ≥7d ≥2d 7d Connection must be used when influence of occasional moments is possible

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Черт. 130. One-use holders which provide required thickness s of concrete protection layer

а - в – with big contact surface with a form, produced of cement-sand mortar; г – with small contact surface with a form, produced of cement-sand mortar; д – the same of asbestos cement; е - з – the same of plastic (perforated); и – the same of aluminum perforated strip; к - м – the same of reinforced steel; 1 – main surface of the form; 2 –

holder; 3 – fixed reinforcement; 4 – twisting of binding wire; 5 – binding wire embedded into the holder; 6 – possible elastic ring; 7 – supports welded to reinforcement

Table 54 Holder kind

Mortar holder, concrete holder, asbestos-cement

holder

plastic (polyethylene)

Steel holder Element

application conditions

Kind of s face side of the element

РМ РБ ПМ ПБ СЗ СН In the open air

Clean concrete surface for painting; tiled during the concrete works and other

+ – + – + –

Mechanically treated + – – – – – In rooms with normal humidity conditions

Clean concrete surface + – + – + –

Concrete surface for painting with water compositions

+ х + х + х

Concrete surface for painting with oil, enamel and synthetic paints; concrete surface for tiling

+ + + + + +

Concrete surface for wall-paper hanging

+ + + + + –

Notes: 1. Symbols: Р – Mortar, concrete and asbestos-cement holders; П – plastic and polyethylene holders; С – steel holders; М – small contact surface of a holder and form (formwork); Б – big contact surface of a holder and form (formwork); З – holders protected against corrosion; Н – holders not protected against corrosion. 2. Sign „+" means possible; sign „–" means not possible; sign „х" – possible but not recommended.

Draft 131. One-use holders which provide required distance

а-в – between single reinforcement details; г – between rods; 1 – a separator of reinforcement steel which is installed between rows of meshes; 2 – holder-filler to provide a concrete protection layer; 3 – elongated cross rods of the framework bent around rods of the mesh; 4 – a holder for connection of crossing rods (spatial spiral of spring wire) ; 5 – binding point

Draft 132. One-use holders providing required thickness of concrete protection layer and distance

between separate reinforcement elements at the same time

а – in flat slabs; б, в – in rectangular section beams; г – in ring section elements; 1 – a holder of U-shaped framework; 2 – reinforcement meshes; 3 – main surface of the form; 4 – a comb-framework holder; 5 – flat reinforcement framework; 6 – rods-holders welded to frameworks in addition; 7 – holder in form of a cramp of reinforcing wire; 8 – concentric frameworks; 9 – point of bending

For one-use holders made of reinforcement steel it is necessary to design drawings. On working projects of reinforcement details and in case if it is necessary to show

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general views of reinforcing of reinforced concrete elements it is necessary to show location of these holders or support rods, and in the specification it is necessary to show steel consumption for their production. It is possible not to show location and quantity of non-metallic holders-fillers on the drawings.

INDIVIDUAL CONSTRUCTIVE REQUIREMENTS

5.121 (5.47). Settlement joints are designed as a rule in cases if the building is constructed

up on non-homogeneous grounds (collapsible soils and other), in points of abrupt changing of loads etc.

If in mentioned cases settlement joints are not provided so the foundations must have enough strength and rigidity preventing damage of located above structures or they must have a special structure for it. Settlement joints as well as contraction joints in solid concrete and reinforced concrete structures must be performed as open-ended one, cutting the structure up to the foundation base. Distances between contraction joints in concrete foundations and walls can be taken I compliance with the distances between joints of the located above structures.

5.122 (5.48). In concrete structures it is necessary to provide constructive reinforcement: а) in points of abrupt changing of dimensions of elements sections; b) in points of walls height variation (on the part no less than 1 m); c) in concrete walls under and above the openings of each storey; d) in structures under effect of dynamic loads; e) at a less stressed surface of eccentric compressed elements if maximum stress of the section determined as for elastic body is more than 0.8Rb, and minimum stress is less than 1 MPa or if it becomes a stretching one, at the same time reinforcement coefficient µ is taken no less than 0.025 percent.

Requirements of the present Item are not applied on prefabricated concrete elements which are tested at stages of transportation and installation.

REQUIREMENTS GIVEN ON PRODUCTION DRAWINGS OF REINFORCED CONCRETE STRUCTURES General requirements

5.123. On working projects of reinforced concrete structures or in explanation notes to

them the following information must be given: а) reinforcement class as regards the strength against compression, and in cases mentioned in Item 2.5 it is necessary to show concrete grade as regards resistance to frost and water proofing capacity, and for light-weight concrete – grade as regards average density; b) reinforcement type (rod or wire), its profile, class, and in case of necessity (for example for structures working by low temperatures) steel grade; number of the state

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standard, or number of technical specifications for the reinforcement type; numbers of state standards (or technical specifications) ready-mixed reinforcement details (meshes or frameworks) if they are used; work conditions of welded connections (low temperatures or variable loads); normative documents for welding; in complicated cases – methods of manufacture of a spatial reinforcement framework and its installation order; materials consumption; c) measures for anti-corrosion protection and for protection against high temperatures if necessary; d) concrete protection layer for main reinforcement, as well as necessity of holders installation which provide design location of reinforcement, their types; e) design schemes, loads; design forces in main sections, included forces caused by dead loads and long-term loads.

Additional requirements mentioned on production drawings of elements of

prefabricated structures

5.124. On production drawings of elements of prefabricated structures or in explanation

notes to them except the data mentioned in Item 5.123, it is necessary to give the following information: а) minimum dimensions of support parts; b) degree (quality) of surface finishing (if necessary); c) points for elements holding by removing out of the form, lifting and installation, points of their support during transportation and storage; d) requirements for undercutting by the producer factory to provide qualitative site installation (if necessary), and for elements with a dim top or butt-ends (for example rectangular section with single or asymmetric double reinforcement) – requirements for marking (note) application by the producer factory, providing correct location of such elements during their lifting, transportation and installation; e) for elements whose examples due to requirements of GOST 8829–85 or their standard documents are to be examined by loading it is necessary to show test schemes, values of loads, deflections and other parameters under control; f) value of specified delivery strength of concrete for specified installation and loading conditions; g) weight of a prevaricated element determined due to Item 2.13.

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ANNEX 1

KIND OF LIGHT AND POROUS CONCRETE AND THIR APPLICATION AREA

Kind of aggregate Kind of concrete coarse fine

Field of application

Light-weight concrete with artificial aggregates

1. Expanded clay concrete Expanded clay All concrete and reinforced concrete structures with non-stressed reinforcement, with the exception of special structures

2. Shungite concrete Shungite Wall panels, slabs of floors and roofs 3. Agloporite concrete Agloporite The same like in position 1 4. Foam slag concrete Foam slag The same like in position 2, except the

slabs of roofs and floors under the load more than 10000 Pa

5. Perlite concrete Perlite (1000 kilogram-force/m2) 6. Slag-concrete Foam cinder aggregate The same like in position 2 7. Expanded clay perlite concrete

Expanded clay Perlite The same like in position 5

8. Agloporite concrete Agloporite 9. Foam slag concrete with granulated slag

Foam slag Light granulated slag

10. Expanded clay concrete Expanded clay Light granulated slag

То же

Light-weight concrete with natural aggregates

11. Pumeconcrete Анийская or lithoid pumice The same like in position 1 12. Slag concrete Slaggy lava The same like in position 5 13. Tuffcrete Slaggy lava The same like in position 1 14. Concrete with shelly limestones

Shelly limestone The same like in position 5

Porous concrete (foam)

15. Expanded clay concrete Expanded clay Without sand with porous and glass

Wall blocks and panels

16. Agloporite concrete Agloporite sand

Notes: 1. Light-weight concretes can have mixed fine aggregate – porous of different types (including ash ТЭС and ash-slag mixtures) and glass sand. 2. Light-weight concretes due to positions 1 – 5 can have fine aggregate only of glass sand. 3. Light-weight concretes of class В7.5 and less with glass sand can be used as an exception by correspondent justification by porization more than 6 percent. 4. By test data approved in compliance with the established procedure, for design of concrete and reinforced concrete structures except mentioned above light-weight concrete there can be used also the following kinds of concrete: thermolite concrete with thermolite crushed stone or gravel and thermolite or glass sand; concrete with agloporite gravel and agloporite or glass sand; concrete with ash gravel and glass sand. 5. Concretes foamed by gas (expanded clay gas concrete and other), can be used in walling structures by corresponding justification approved in accordance with the established procedure. 6. For heat insulation structures of multi-layer structures can be used light-weight concretes with intergranular porosity (coarse-pored and finepored – without fine graded sand).

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ANNEX 2

VALUES ζ AND Ao FOR CALCULATION OF BENDING ELEMENTS STRENGTH

µ, % Reinforcement class A-II A-III Concrete class В12,5 B15 B20 B15 B25 B30 ζ Ao ζ Ao ζ Ao ζ Ao ζ Ao ζ Ao

0,10 0,979 0,274 0,982 0,275 0,987 0,276 0,976 0,356 0,986 0,360 0,988 0,361 0,15 0,969 0,407 0,973 0,409 0,980 0,412 0,964 0,528 0,979 0,536 0,982 0,538 0,20 0,958 0,536 0,964 0,540 0,973 0,545 0,953 0,696 0,972 0,710 0,976 0,713 0,25 0,948 0,664 0,955 0,669 0,967 0,677 0,941 0,859 0,965 0,880 0,971 0,886 0,30 0,937 0,787 0,945 0,794 0,960 0,806 0,929 1,017 0,958 1,049 0,965 1,056 0,35 0,927 0,908 0,936 0,918 0,953 0,934 0,917 1,171 0,951 1,215 0,959 1,225 0,40 0,916 1,026 0,927 1,038 0,947 1,060 0,905 1,321 0,944 1,378 0,953 1,391 0,45 0,906 1,142 0,918 1,157 0,940 1,184 0,893 1,467 0,937 1,539 0,947 1,555 0,50 0,896 1,254 0,909 1,273 0,933 1,307 0,882 1,610 0,930 1,697 0,941 1,718 0,55 0,885 1,363 0,900 1,386 0,927 1,427 0,870 1,747 0,923 1,852 0,935 1,877 0,60 0,875 1,470 0,891 1,497 0,920 1,546 0,858 1,879 0,916 2,006 0,929 2,035 0,65 0,864 1,572 0,882 1,605 0,913 1,662 0,846 2,007 0,909 2,156 0,923 2,191 0,70 0,854 1,674 0,873 1,711 0,907 1,777 0,834 2,131 0,902 2,304 0,918 2,344 0,75 0,843 1,770 0,864 1,814 0,900 1,890 0,822 2,250 0,895 2,449 0,912 2,496 0,80 0,833 1,866 0,855 1,917 0,893 2,001 0,810 2,365 0,888 2,592 0,906 2,645 0,85 0,822 1,956 0,845 2,011 0,887 2,110 0,799 2,479 0,881 2,732 0,900 2,792 0,90 0,812 2,046 0,836 2,107 0,880 2,218 0,787 2,585 0,874 2,870 0,894 2,937 0,95 0,801 2,131 0,827 2,206 0,873 2,323 0,775 2,687 0,867 3,005 0,888 3,080 1,0 0,791 2,215 0,818 2,290 0,867 2,427 0,763 2,785 0,860 3,138 0,882 3,220 1,1 0,770 2,372 0,800 2,464 0,853 2,628 0,739 2,967 0,846 3,395 0,870 3,495 1,2 0,749 2,517 0,782 2,628 0,840 2,822 0,716 3,136 0,832 3,642 0,859 3,761 1,3 0,728 2,650 0,764 2,781 0,827 3,009 0,692 3,384 0,818 3,879 3,847 4,019 1,4 0,707 2,771 0,745 2,920 0,813 3,188 — — 0,803 4,106 0,835 4,268 1,5 0,687 2,885 0,727 3,053 0,800 3,360 — — 0,789 4,322 0,823 4,508 1,6 0,666 2,984 0,709 3,176 0,787 3,524 — — 0,775 4,528 0,812 4,740 1,7 — — 0,691 3,289 0,773 3,681 — — 0,761 4,724 0,800 4,963 1,8 — — 0,663 3,392 0,760 3,830 — — 0,747 4,910 0,788 5,178 1,9 — — 0,655 3,485 0,747 3,972 — — 0,733 5,085 0,776 5,384 2,0 — — — — 0,733 4,107 — — 0,719 5,250 0,765 5,581 2,1 — — — — 0,720 4,234 — — 0,705 5,405 0,753 5,770 2,2 — — — — 0,707 4,353 — — — — 0,741 5,950 2,3 — — — — 0,693 4,465 — — — 0,729 6,122 2,4 — — — — 0,680 4,50 — — — — 0,717 6,285

1000bh

As

µ= ;

0hR

MA

s

sζ

= ; 20

0bh

MA = , МПа

Notes: 1. Values ζ and Ao determined by formulas:

1005.01

µζ

b

s

R

R−= ; ζ

µsRA

1000 =

2. By choosing of reinforcement of class А-III with diameter 6 and 8 mm values µ are multiplied by 1.03.

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ANNEX 3

DIAGRAMS OF BEARING CAPACITY OF ECCENTRIC COMPRESSED ELEMENTS OF RECTANGULAR SECTION WITH SYMMETRIC REINFORCEMENT OF

HEAVY-WEIGHT AND LIGHT-WEIGHT CONCRETE

Symbols:

by М1l/М1 = 1.0 (see Item 3.54); - - - - - - - - - by М1l/М1 = 0.5.

Draft 1. Diagrams for elements of heavy-weight concrete

h

l0=λ ; 0bhR

N

b

nα ;20bhR

M

b

mα ; 0bhR

AR

b

ss

sα

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Symbols:

by М1l/М1 = 1.0 (see Item 3.54); - - - - - - - - - by М1l/М1 = 0.5.

Draft 1. Diagrams for elements of heavy-weight concrete (end)

h

l0=λ ; 0bhR

N

b

nα ;20bhR

M

b

mα ; 0bhR

AR

b

ss

sα

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Symbols:

by М1l/М1 = 1.0 (see Item 3.54); - - - - - - - - - by М1l/М1 = 0.5.

Draft 2. Diagrams for elements of light-weight concrete by average density grade no less than D 1800

h

l0=λ ; 0bhR

N

b

nα ;20bhR

M

b

mα ; 0bhR

AR

b

ss

sα

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Symbols:

by М1l/М1 = 1.0 (see Item 3.54); - - - - - - - - - by М1l/М1 = 0.5.

Draft 2. Diagrams for elements of light-weight concrete by average density grade no less than D 1800

(end)

h

l0=λ ; 0bhR

N

b

nα ;20bhR

M

b

mα ; 0bhR

AR

b

ss

sα

Notes (for Drafts 1 and 2): 1. Diagrams of Draft 1 can be used by concrete class from В15 to В50 by а = а' from 0.05ho to 0,15ho. 2. Diagrams of Draft 2 can be used by concrete class from В10 to В40 by а = а' from 0,05ho to 0,15ho. 3. By М1l/М1 < 0.5 values as are determined by means of linear interpolation. 4. Values М are determined due to the calculation as regards non-deformed scheme without considering coefficient η.

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ANNEX 4

СОРТАМЕНТ АРМАТУРЫ

Design area of the cross rod, mm2, if quantity of rods is Diameter of reinforcement of classes

Nominal diameter of a rod, mm 1 2 3 4 5 6 7 8 9

Theoretic weight of

1 m of reinforce

ment length,

kg

A-I A-II А-III Ат-IIIC

Bp-I

3 7,1 14,1 21,2 28,3 35,3 42,4 49,5 56,5 63,6 0,052 — — — — + 4 12,6 25,1 37,7 50,2 62,8 75,4 87,9 100,5 113 0,092 — — — — + 5 19,6 39,3 58,9 78,5 98,2 117,8 137,5 157,1 176,7 0,144 — — — — + 6 28,3 57 85 113 141 170 198 226 254 0,222 + — + — — 8 50,3 101 151 201 251 302 352 402 453 0,395 + — + — —

10 78,5 157 236 314 393 471 550 628 707 0,617 + + + + — 12 113,1 226 339 452 565 679 792 905 1018 0,888 + + + + — 14 153,9 308 462 616 769 923 1077 1231 1385 1,208 + + + + — 16 201,1 402 603 804 1005 1206 1407 1608 1810 1,578 + + + + — 18 254,5 509 763 1018 1272 1527 1781 2036 2290 1,998 + + + + — 20 314,2 628 942 1256 1571 1885 2199 2513 2828 2,466 + + + + — 22 380,1 760 1140 1520 1900 2281 2661 3041 3421 2,984 + + + + — 25 490,9 982 1473 1963 2454 2945 3436 3927 4418 3,84 + + + — — 28 615,8 1232 1847 2463 3079 3685 4310 4926 5542 4,83 + + + — — 32 804,3 1609 2413 3217 4021 4826 5630 6434 7238 6,31 + + + — — 36 1017,9 2036 3054 4072 5089 6107 7125 8143 9161 7,99 + + + — — 40 1256,6 2513 3770 5027 6283 7540 8796 10053 11310 9,865 + + + — — 45 1590,4 3181 4771 6362 7952 9542 11133 12723 14313 12,49 — + — — — 50 1963,5 3927 5891 7854 9818 11781 13745 15708 17672 15,41 — + — — — 55 2376 4752 7128 9504 11880 14256 16632 19008 21384 18,65 — + — — — 60 2827 5654 8481 11308 14135 16962 19789 22 616 25443 22,19 — + — — — 70 3848 7696 11544 15392 19240 23088 26936 30784 34632 30,21 — + — — — 80 5027 10055 15081 20108 25135 30162 35190 40216 45243 39,46 — + — — —

Notes: 1. Nominal diameter of rods for reinforcing steels of periodic profiles corresponds to nominal diameter of plain rods with equal cross section. Actual dimensions of rods of periodic profile are determined in GOST 5781-82. 2. Sign "+" means presence of diameter for reinforcement of the present class. 3. theoretic weight of 1 m of the length of reinforcement В-I is taken equal to: by d = 3 mm – 0.055 kg; by d = 4 mm – 0.099 kg; by d = 5 mm – 0.154 kg.

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ANNEX 5

MAIN LETTER SYMBOLS

FORCES CAUSED BY EXTERNAL LOADS AND EFFECTS IN THE CROSS

SECTION OF THE ELEMENT

М – bending moment or moment of external forces relating to the center of gravity of a section; N – longitudinal force; Q – shear force; Т – torsion moment; Мsh, Ml, Mtot – moment relating to the center of gravity of the section caused by short-term loads, dead loads and long-term loads, all loads.

MATERIALS FEATURES

Rb, Rb,ser – design resistances of concrete against axial compression for limit states of the first and the second groups; Rbt, Rbt,ser – design resistances of concrete against axial tension for limit states of the first and the second groups; Rs, Rs,ser – design resistances of concrete against tension for limit states of the first and the second groups; Rsw – design resistance of cross reinforcement against tension, determined due to Item 2.21; Rsc – design resistance of reinforcement against compression for limit states of the first group; Eb – initial elasticity module of concrete by compression and tension; Еs – reinforcement elasticity module; а – ration between corresponding elasticity moduli of reinforcement Еs and concrete Eb.

CHARACTERISTIC OF LOCATION OF LONGITUDINAL REINFORCEMENT IN

THE CROSS SECTION OF THE ELEMENT

S – symbol of longitudinal reinforcement: а) if there is a stretched and compressed by external loads section zones – located in the stretched zone; b) by section completely compressed by external load– located at a less compressed surface of the element; c) by section completely stretched by external load:

- for eccentric stretched elements – located at a more stretched surface of the section; - for centrally stretched elements – all reinforcement located in the cross section;

S' – symbol of longitudinal reinforcement: а) if there are stretched and compressed by external loads zones of the section – located in compressed zone; b) by section completely compressed by external load – located at a more compressed surface of the section; c) by section of eccentric stretched elements completely stretched by external load – located at a less stretched surface of the section.

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GEOMETRIC FEATURES

b – width of a rectangular section; width of a rib of I- and T-sections; bf, b'f – width of a flange of I- and T-sections in stretched and compressed zones; h – height of rectangular, I- and T-sections; hf, h′f – height of a flange of I- and T-sections in stretched and compressed zones; а, а′ – distances from the resultant of forces in reinforcement S and S' to the nearest surface of the section; ho – main height of the section equal to h – a; х – height of compressed zone of concrete;

ξ – relative height of compressed zone of concrete, equal to 0h

x;

s – distance between stirrups measured along the element length; eo – eccentricity of longitudinal force N relating to the center of gravity of the section, determined due to Item 3.3; е, е′ – distances from the point of application of longitudinal force N to the resultant of forces in reinforcement S and S′; es – distance from the point of application of longitudinal force N to the center of gravity of the section area of reinforcement S; l – span of the element; lo – design length of the element under compression longitudinal force; value is taken due to Table17 and Item 3.55; i – radius of inertia of the cross section of the element relating to the center of gravity of the section; d – nominal diameter of rods of reinforcement steel; As, A′s – areas of sections of reinforcement S и S′; Аsw – section areas of stirrups located in one plane which is normal to the longitudinal axis of the element and which crosses the inclined section; Asw1 – section area of the one rod of a stirrup; Аs1 – section area of the one rod of longitudinal reinforcement; µ – reinforcing coefficient determined as a ratio between the section area of reinforcement S and area of the cross section of the element bho without considering compressed and stretched flanges; А – concrete area in the cross section; Ab – section area of the compressed zone of concrete; Ared — section area of the element including area of concrete as well as area of all longitudinal reinforcement multiplied by the ratio of elasticity module of reinforcement and concrete; Ired – inertia moment of the section of the element relating to its center of gravity; Wred – resistance of the section of the element for end stretched fibre determined as for elastic material; D – diameter of ring or round section.

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INDEXES OF LETTER SYMBOLS AND EXPLANATION WORDS

One-letter indexes

а — (anchor),

а — (accidental); а (axial); b — (beton) – concrete; с — (compression); d — (depth); d — (designed),

е — (eccentricity); е (ear); f (flange); f — (force); h — (horizontal); k — (key); l — (long); l (level); l — (left); l — (lap); т — (middle); т — (moment); п — (normal) longitudinal force; п — (normative); р — (partition); q cross force Q; R — design resistance R; r — (right); s — (in situ); s (steel) reinforcement; t — (tension); t — (transverse); t — (torsion); t — (temperature); и (ultimate); v — (vertical) rib or side of a beam; w — (web); w — (welding); x — in the direction of axis х or in the section х; у — in the direction of axis у; у — (yield point).

Double- and three-letter indexes

an — (anchoring); col — (column); cir (circular) round, ring; cr (critical); crc — (cracking) crack formation; el — (elastic); ef (effective);

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fact — actual; inc — bent, inclined (inclined); inf bottom (inferior); int — internal (interior); lim — (limit); loc — (local); max — (maximal); min — (minimal); ov — (overhang); out (output); pl (non-elastic) (plastic); red (redacted); ser (service); sh — (short); shr (shrinkage) settlement; sup — (support); sup (super) top; tot (total); web — rib or side of a beam.

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