Concept of Reinforced Cement Concrete Design · PDF fileConcept of Reinforced Cement Concrete Design (RCC) Based on Working Stress Method (WSM) Page 1 of 22 Concrete: Concrete is a

Concept of Reinforced Cement Concrete Design

Based on Working Stress Method (WSM)

Suraj Kr Ray B. Tech (civil)

Concept of Reinforced Cement Concrete Design (RCC) Based on Working Stress Method (WSM)

Page 1 of 22

Concrete:

Concrete is a substance which obtained by mixing Cement, Sand, Aggregates & water in

a suitable proportion. And it can be cast in any desired shape & size for bear external loads.

Reinforced Cement Concrete (RCC):

Concrete is strong in Compression but weak in tension. Tensile Strength of concrete can

be increase by providing steel inside concrete according to requirement. So, it can be define as:

“Reinforced Cement Concrete is a combined substance in which concrete forms a layer

surrounding Steel and also, concrete and steel both are hardly bonded to each other is termed

as Reinforced Cement Concrete (RCC).”

Advantage and Disadvantage of Reinforced Cement Concrete (RCC)

Advantage of RCC:

It has high compressive strength as compare of simple Concrete.

It has better fire resistance capacity than steel.

It needs less maintenance as compare of steel.

It can be cast in any required shape at site too.

By using steel in concrete cross-sectional dimension of structure can be reduced.

In some structures like pile, dams, etc it is more economical.

Disadvantage of RCC:

It needs mixing, casting and curing, all of which affect the final strength of concrete.

The cost of the forms used to cast concrete is relatively high.

It has low compressive strength as compared to steel.

In columns/beams of multi-storey buildings Cracks develop in concrete due to

shrinkage and the application of live loads.

Singly Reinforced Beam Based on Working Stress Method

Page 2 of 22

Structure of Concrete containing Reinforcements (RCC Structures):

Slab:

A concrete slab is common element of structures/ buildings. It presents as horizontally

in form of roof or floors. They may carry gravity load as well as lateral loads. Depth of slabs

is very small as compare of its length and width.

Beams:

A member of structure in which length is greater than other two Dimensions (breadth,

depth) is termed as Beam.

Column:

A member of structure in which Depth is greater than other two Dimensions (breadth,

length) is termed as Column.

Frames:

A frame is structural member consists of Slab, Beams and columns.

Walls:

Walls are vertical plate elements resisting gravity as well as lateral loads e.g. retaining

walls, basement walls etc.


Page 3 of 22

Loads working on structures:

Members of structure must be design for support specific loads. The loads which used

to occur on structural member are divided into three categories,

1. Dead loads

2. Live loads

3. Environmental loads

Dead loads:

A load which is constant in magnitude and fixed in location throughout the lifetime of

the structure is termed as Dead loads. E.g. load of beams, columns, walls, slab, etc

Live loads:

A load which is get changing in magnitude and varies in location with certain time is

termed as Live loads.

Environmental loads:

Consists mainly of snow loads, wind pressure and suction, earthquake loads (i.e inertial

forces) caused by earthquake motions. Soil pressure on subsurface portion of structures,

loads from possible ponding of rainwater on flat surfaces and forces caused by temperature

differences. Like live loads, environmental loads at any given time are uncertain both in

magnitude and distribution.

Reinforced Cement Concrete Design Methods

Design of is an art to provide a safe, economical, serviceable & functional structure.

According to art of design and uses of materials there are following methods:

1. Ultimate/Strength design Method

2. Working stress Design Method

3. Limit State Design Method


Page 4 of 22

Such type of system of design in which we consider the elastic limit in all calculation is termed as Working Stress Method of design. In short it is called WSM design method. The base of WSM is the concept of modular ratio, which is the ratio of young’s modulus of steel by young’s

modulus of concrete and it is adopted as m =

.

ASSUMPTIONS FOR THE WORKING STRESS METHOD OF DESIGN:-

Plane section remains plane before and after bending.

Concrete in tension zone is ignored during the analysis of beam.

Obey Hook’s law so the stress-strain relationship of steel and concrete will be straight

line under working load.

Steel and concrete form a composite structure.

SOME SYMBOLES USED IN RCC :-

cbc = Allowable stress in concrete in bending in compression in direct stress.

st = Allowable stress in steel in bending in tension in direct stress.

sc = Allowable stress in steel in bending in compression in direct stress.

y = Characteristic strength of steel.

ck = Characteristic strength of concrete.

Ast = Area of steel in tension zone.

Asc = Area of steel in compression zone.

m = Modular ratio (m = 280/3 cbc)

ECC = effective concrete cover

CCC = clear concrete cover

WORKING STRESS METHOD OF DESIGN


Page 5 of 22

Methods of R.C.C. design

Working Stress Method Limit Stress Method

1. The stress in an element is obtained from 1. The stress is obtained from design

load

The working load and compare with and compare with design strength.

Permissible stresses.

(Load = working load considered.) (Load = Working load x Load factor)

2. This method follows linear stress-strain 2. In this method, it follows linear

Behaviour of both materials. Relationship (one of the major diff.).

3. Modular ratio can be used to determine 3. The ultimate stresses of materials

allowable stresses. itself are used as allowable stress.

4. Material capabilities are under estimated 4. The material capabilities are not under

to large extent. Factor of safety are used estimated as much as they are in WSM.

in working stress method. Partially factor of safety are used.

Allowable stress in conc., cbc = ck/FOS cbc ck/partial FOS

Allowable stress in steel, st = y/FOS st = y/partial FOS

5. Its result gives an uneconomical section. 5. This method of design gives

economical section.


Page 6 of 22

Types of RCC sections

Unbalanced section

Over Reinforced Under Reinforced

Balanced Section


Page 7 of 22

Balanced sections Under reinforced sections Over reinforced sections

Such type of section In which concrete and Steel attain its Permissible strength is termed as balanced section or critical section or economical section.

Such types of section in which steel attain its permissible stress but concrete attain stress lower than its permissible stress.

Such type of section in which concrete attain its permissible strength but steel remain below to its permissible strength.

This type sections occurs when amount of provided steel is neither less nor more than the steel required for a critical section.

This type of sections occurs when area of provided steel is less than the area of steel required for balanced section.

This type sections occurs when area of provided steel is more than the area of steel required for balanced section.

In this type of sections critical Neutral axis and Actual neutral axis are same line. Actual neutral axis

Actual Neutral axis remains above than the critical neutral axis.

Actual NA Critical NA

Actual neutral axis remains below than critical neutral axis.

Critical NA

Actual NA

Stress diagram

cbc st/m

Stress diagram

’cbc st/m ’cbc< cbc

Stress diagram

cbc ’st/m ’st< st

Moment of resistence,

MR=bnc cbc –

= st Ast (d – n/3) = Qbd2


MR=bn ’cbc –

= st Ast (d – n/3)


MR=bn cbc –

= ‘st Ast (d – n/3)


Page 8 of 22

SOME FORMULAS AND THEIR DERIVATIONS:-

Neutral axis:-

Neutral axis is an imaginary axis about which the moment of area of tension zone

and compression zone is equal.

In fig;

b= Breadth of cross-section n

d= Effective depth of cross-section d

n= Depth of neutral axis NA

Ast= steel area provided to that section

Now ,

Total area of compression zone= bn b

b

Total area of tension zone= Ast (in tension zone concrete also present but it is not

considered)

Taking moment of area of compression zone about neutral axis:

=Area of compression zone x distance of C.G. (compression zone) from NA

= (b.n).n/2

Again taking moment of tension zone about NA

= Area of tension zone x distance of C.G. (tension zone) from NA

= mAst x (d-n)

.’. From assumption of R.C.C.

bn(n/2) = mAst (d-n)

Compression zone

Tension zone


Page 9 of 22

MOMENT OF RESISTANCE:

When a beam bends under loads, its top fibres become shorter than its actual size and bottom fibres become larger than its actual size. Due to this bending maxm compressive stress occur on the outer most top fibre and maxm tensile stress occur on most outer bottom layer to prevent beam from this bending. There should be provided some additional strength to the beam.

Such type of moment which works against the Bending Moment (B.M.) is termed as “MOMENT OF RESISTANCE (M.R.)”.

M.R. = Resultant compressive force x leaver arm

= Resultant tensile force x leaver arm

(Leaver arm is the distance between two points one of them is resultant tensile force and another one is resultant compressive force)

Mathematically it can be represent as

MR =

= σst Ast

This formula is for balanced section in case of unbalanced section

In over reinforced section st will replace with ’st in under reinforced section cbc will replace with ’cbc

for balanced section

M.R. = Qbd2

here Q =

J = (1-

) (J is known as co-efficient of lever arm)

K =

or

(here r =

and m =

.)

% Steel = p =


Page 10 of 22

Balanced or Critical or Economical section:-

A section of RCC in which most distant concrete fibre in compression and steel in

tension attain its permissible stresses simultaneously, is termed as Balanced section.

Since in this type of section, used concrete and steel both get fully utilized that’s

why it is also known as Economical or Critical section.

Bending stress:-

Bending stress is the longitudinal stress that is introduced at a point in a body

subjected to loads (perpendicular to the length) that cause it to bend.

In RCC beam compression zone get compressed and tension zone get elongate as

a result of bending stress.

Beam Before Bending Beam after Bending

Neutral Axis (nc or n):-

Neutral axis is the axis through a beam where the stress is zero; that is there is

neither compression nor tension.

Compression zone

Tension zone


Page 11 of 22

Value of Neutral axis (nc) for Balanced section:-

=

⇒

=

⇒

= 1+

⇒ =

Let

=r (‘.’

is constant for any grade of steel & concrete)

⇒ =

=

=

) d

Let

= k (k =co-efficient of Neutral axis)

.’. (For a critical section)


Page 12 of 22

Moment of Resistance (M.R.) = Qbd2 for balanced/Critical section:-

MR = b nc cbc –

= b k d cbc –

( ‘.’ nc = kd)

= b k d cbc –

= b d2 [

j k cbc ] (let

–

= j = co-efficient of lever arm for a critical section)

MR = Q b d2 ( Q =

j k cbc )

Percentage of steel for a Critical section (p):-

Taking moments of area of concrete and steel about Neutral axis:

Area of concrete x distance of CG from NA = Area of steel x distance of CG from NA

⇒ b nc x

= m Ast x ( d - nc )

⇒ b

= m Ast x ( d - nc )

⇒ b

= m Ast x ( d - kd )

⇒ b

= m Ast x d( 1 - k)


Page 13 of 22

⇒ b

= m Ast x d( 1 - k)

⇒ b

= m Ast x ( 1 - k)

⇒

=

⇒ p =

x 100 =

TYPES OF PROBLEMS IN SINGLY REINFORCED BEAM:

1. To find Moment of resistance of singly reinforced beam. 2. To find stress in steel and concrete if M.R. is given. 3. To design a beam for given conditions.

For solving problems…

Step 1:

Write down all given values under a title “Given Data”.

Step 2:

Determine actual depth of Neutral axis (i.e. n) for given section by taking Moment of Compression zone & Tension zone about NA.

b

= m Ast x ( d - n )

Step 3:

Calculate Critical NA (i.e. nc).

For this we can use this relationship

nc = k d or,

=

(this relationship came from similar triangle of stress diagram)

Step 4:

Compare values of n and nc

If n < nc ; section is under reinforced use this formula

MR=bn ’cbc –

= st Ast (d – n/3)

If n > nc ; section is over reinforced use this formula

MR=bn cbc –

= ‘st Ast (d – n/3)


Page 14 of 22

Example 1:

The cross-section of a singly reinforced concrete beam is 300 mm wide and 500 mm depth to the centre of the tension reinforcement which consist of four bars of 16 mm diameter. If the stress in concrete and steel are not exceed 7 N/mm2 and 140 N/mm2 respectively, determine the moment of resistance of the beam. Take m = 13.33

Sol:

Given Data:

b= 300mm = wide of beam

d= 500mm = depth of beam

N= 4 = no. of bars

Ø= 16mm = dia of bars

Ast= no. of bars x

d2 = 4 x

162 = 804.248 ≈804 mm2

= 7 N/mm2

= 140 N/mm2

m= 13.33

To find Actual Natural axis (n):-

b

= m Ast x ( d - n )

⇒ 300 x

= 13.33 x 804 x (500 – n)

⇒ 150 n2 =5358660 – 10717.32 n

⇒ 150 n2–+10717.32 n – 5358660 =0

⇒ n = 156.63 mm ( another value of n (-228.079mm) is not possible so ignored)


Page 15 of 22

To find critical Natural axis (nc):-

nc = kd =0.4 x 500 = 200mm ( k =

=

=

= 0.3999 ≈ 0.4 )

Comparison between n and nc:-

n = 156.63 mm < nc = 200 mm

So, it is the case of under-reinforced

To find MR:-

MR = st Ast

= 140 x 804 x

= 50418772.07 N-mm

.‘. MR = 50.4 KN-m

Example 2 :-

The moment of resistance of a rectangular singly reinforced beam of width b mm and

effective depth d mm is 0.9bd2 N-mm. If the stress in the outside fibbers of concrete and steel do

not exceed 5 N/mm2 and 140 N/mm2 respectively and the modular ratio equals 18, determine

the ratio of depth of natural axis from the extreme compression fibres to the effective depth of

the beam and the ratio of the area of the tensile steel to the effective area of the beam.

soln:-


Page 16 of 22

Given Data:-

b = b mm = width of beam

d = d mm = effective depth of beam

MR= 0.9bd2 N-mm

≯ 5 N/mm2

≯ 140 N/mm2

m= 18

To find:-

1.

= ?

2.

= ?

For Balanced section:

MR = Qbd2

= 0.87 bd2

m =

r =

K =

j = 1

Q =

j k

5 N/mm2 140 N/mm2 18.67 28 0.4 0.87 0.87

But given MR = 0.90 bd2

Given MR ( = 0.90 bd2 ) > MR for Balanced section (0.87 bd2)

.’. Given section is Over reinforced

So Concrete will attain its permissible stress earlier than steel

Now,

MR =

⇒ 0.9 bd2 =

⇒ 0.90 d2 =

⇒

⇒ 1.08 d2 = 3dn – n2

⇒ 1.08 =

⇒ 1.08 = 3n’ – n’2 (Let

)

⇒


Page 17 of 22

.’.

.’.

=

=

=

Now from stress diagram:

∆BOC ≈ ∆ AOD

.’.

Since section is Over reinforced

So, concrete will reach at its permissible stress

Which is 2

.’.

[‘.’ (

]

.’.

⇒

⇒ 2

For equilibrium Resultant Compression Force must be equal to Resultant Tension Force

i.e.

⇒

(

⇒

=

= 0.00834


Page 18 of 22

Example 3 :-

Find the percentage of tensile reinforcement necessary for a singly reinforced balanced rectangular section if the permissible stress in concrete and steel are c and t Newton/mm2 respectively and modular ratio is m.

soln:-

Given Data:

2

2

m = m

for equilibrium;

Total compression force = Total tension force

i.e.

(1)

for balanced section;

nc = k d

=

=

=

=

Putting this value of nc in eqn (1)

⇒

⇒

⇒

⇒

⇒

⇒

.’. Required percentage of Reinforcement =


Page 19 of 22

Example 4 :-

A reinforcement concrete beam 300mm wide by 600mm total depth has a span of 8 metres. Find the necessary tension reinforcement at the centre of the span to enable the beam to carry a load of 6000N/m in addition to its own weight;

Consider cover below the steel centre= 40mm Weight of beam = 25000N/m3

Permissible stress in concrete = 7N/mm2 Permissible stress in steel = 230N/mm2 Modular ratio = 13.33 soln:

Self weight of beam:-

Weight of beam = 25000N/m3

.’. Weight of unit length of beam = 25000 x 1 x b x d

= 25000 x 1 x

x

N/m

= 4500 N/m

Total load resist by beam = U. D.L. + self weight of beam

⇒ W = 6000 + 4500 = 10500 N/m

Max bending moment (B.M.) for a UDL =

BMmax =

= 84000 N-m = 84 x 106 N-mm

For M20 & Fe 415 ; j = 0.9

Moment or resistance ( MR ) = Stress Force x Lever arm

⇒ MR = st Ast

⇒ MR = st Ast

( ‘.’ nc = kd)

⇒ MR = st Ast x jd (

)


Page 20 of 22

⇒ MR = st Ast x jd

.’. Ast =

2

Example 5 :-

Design a rectangular beam section for a simply supported beam of 7m span and having a UDL of 35KN/m run by working stress method of design. Use M20 & Fe415 steel rods.

Soln:

Given Data: 7m (length (or span) of the beam) UDL = 35 KN/m

For M20 & Fe415

Maxm Bending Moment for given UDL =

=

= 214.375 KNm = 214.375 x 106 Nm

Since this is a Design Problem & during design always consider a balanced section. So; for a Balanced section: MR = Q b d2 = 0.9 b d2

(Since Breadth and depth of Beam is not defined so it depends upon the designer that what value he want to take A/C to experience.) For a balanced section,

we assumed, b =

.’. MR = 0.9 x

⇒ 214.375 x 106 = 0.9 x

⇒ 781.0 mm

.’. b =

=

So we take; b = 390 mm d = 780 mm

Area of steel =

=

2

m =

r =

K =

j = 1

Q =

j k

7 N/mm2 230 N/mm2 13.33 32.857 0.288 0.9 0.9


Page 21 of 22

Suppose provide 20 mm bars

Area of one bar =

=314 mm2

No of Bars required =

So providing 5 bars of 20 mm diameter Area of steel provided = 5 x 314 = 1570 mm2 2 Hence ok Check for the designed beam: From assumption: Moment of area of compression zone about NA = Moment of area of tension zone about NA

.’.

195.25n2 = (13.33 x 1570 x 780) – (13.33 x 1570 x n) For balanced section: nc = kd = 0.22 x 780 = 224.64mm n nc

It is clear that concrete will get its permissible strength before steel so;

MR =

=

= 229.82 KNm > 214.375 KNm Hence ok.

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Concept of Reinforced Cement Concrete Design · PDF fileConcept of Reinforced Cement Concrete Design (RCC) Based on Working Stress Method (WSM) Page 1 of 22 Concrete: Concrete is a