STRAP Eurocode

8/20/2019 STRAP Eurocode

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© ATIR Engineering Software Ltd.

Version 2010

January 2010


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All rights reserved. No parts of this work may be reproduced in any form or by any means - graphic, electronic, or

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Products that are referred to in this document may be either trademarks and/or registered trademarks of the

respective owners. The publisher and the author make no claim to these trademarks.

While every precaution has been taken in the preparation of this document, the publisher and the author ass ume no

responsibil ity for errors or omis sions, or for damages resulting from the use of information contained in this

document or from the use of programs that may accompany it. In no event shall the publis her and the author be

liable for any loss of profit or any other commercial damage caused or alleged to have been caus ed directly or

indirectly by this document.

Codes - EU


Special thanks to:

All the people who contributed to this document, the programmers, secretaries, STRAP dealers and users.

Last not least, we want to thank EC Software who wrote the help tool called HELP & MANUAL that was used to

create this document.

Disclaimer The STRAP program has been written by a team of highly qualified engineers

and programmers and have been extensively tested. Nevertheless, the authorsof the software do not assume responsibility for the validity of the results

obtained from the programs or for the accuracy of this documentation

The user must verify his own results

The authors remind the user that the programs are to be used as a tool for structural design and analysis, and that the engineering judgement of the user is the final arbiter in the development of a suitable model and the interpretation

of the results.


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3Contents


Table of Contents

Foreword 0

Part I Codes - Eurocode 5................................................................................................................................... 61 Eurocode 2

.......................................................................................................................................................... 6Beams1.1

.......................................................................................................................................................... 9Columns1.2

.......................................................................................................................................................... 13Walls1.3

.......................................................................................................................................................... 15Seismic - Beams1.4

.......................................................................................................................................................... 16Seismic - Columns1.5

.......................................................................................................................................................... 17Punching1.6

.......................................................................................................................................................... 18Slab deflections1.7

.......................................................................................................................................................... 19Slab re inforceme nt1.8

................................................................................................................................... 222 Eurocode 2 - Post-tensioned beams

.......................................................................................................................................................... 22Losses2.1

.......................................................................................................................................................... 25Shear2.2

.......................................................................................................................................................... 26Ultimate mom ent2.3

.......................................................................................................................................................... 27Deflections2.4

................................................................................................................................... 293 Eurocode 3

.......................................................................................................................................................... 29Classification of sections3.1

.......................................................................................................................................................... 30Strength of steel3.2

.......................................................................................................................................................... 31Shear3.3

.......................................................................................................................................................... 31Bending3.4

.......................................................................................................................................................... 32Lateral-tors ional buckling3.5

.......................................................................................................................................................... 33Axial force - compre ssion3.6

.......................................................................................................................................................... 34Axial force - tension3.7

.......................................................................................................................................................... 35Combined bending & axial force3.8

.......................................................................................................................................................... 36Deflections3.9

.......................................................................................................................................................... 36Castellated/cellular beams3.10

................................................................................................................................... 404 Eurocode 3 - cold-formed

.......................................................................................................................................................... 40Axial force - compre ssion4.1

.......................................................................................................................................................... 41Axial force - tension4.2

.......................................................................................................................................................... 42Bending - without LTB4.3

.......................................................................................................................................................... 42Combined axial force & bending4.4

.......................................................................................................................................................... 43Deflections4.5

.......................................................................................................................................................... 43Effective section properties4.6

.......................................................................................................................................................... 44Lateral buckling strength4.7

.......................................................................................................................................................... 44Shear4.8

.......................................................................................................................................................... 45Strength of steel4.9

................................................................................................................................... 465 Eurocode 4

.......................................................................................................................................................... 46Composite beams5.1

.......................................................................................................................................................... 50Composite columns5.2

Index 51


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Part

ICodes - Eurocode


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Codes - Eurocode 5


1 Codes - Eurocode

Select one of the following Codes:

Eurocode 2

Eurocode 3Eurocode 3 - cold-formed

Eurocode 4

6

29

40

46


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Codes - EU 6


1.1 Eurocode 2

The longitudinal reinforcement calculation is based on the design assumptions outlined in § 4.2.1.3.3;

the following rectangular stress block is used.

The modulus of elasticity:

Values in Code Table 3.2 are used.

Es = 200 kN/mm² (§ 3.2.4.3)

The stress in reinforcement is calculated as Es · strain, but not greater than f yd.

Refer to:

Beams

Columns

Beams - seismic

Columns - seismic

1.1.1 Beams

The beam design procedure includes:

calculation of moment and shear envelopescalculation of redistributed moments and reduced shear (option)

calculation of reinforcement steel areas

calculation of links with variable spacing or bent-up bars with constant links

check of allowable deflections

Note that axial forces are ignored by the program.

Moment redistr ibut ion (optional):

Moments in continuous beams are redistributed as permitted in § 5.5, according to the following

guidelines:

The support moments in the envelope are reduced up to the maximum percentage specified by the

user, but not less than the minimum percentage specified.

The maximum span moments in the envelope remain constant or are decreased (unless the minimumredistribution requirement forces an increase in the span moment, which will generally occur in

exterior spans with stiff columns).

The shear forces in the spans are adjusted so as to maintain equilibrium of forces and moments.

For beams with columns, the moment transferred by the beam into the column before and after

redistribution is constant. This prevents redistribution in the columns and maintains equilibrium in

loading cases with horizontal loads.

Note that the program checks that the redistribution percentage does not exceed the allowable (§ 5.5)

after the reinforcement is calculated, and displays warnings if required.

6

9

15

16


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Codes - Eurocode 7


Longi tudinal re inforcem ent:

For the calculation of reinforcement for beams with moment and axial force , refer to slab reinforcement

.

Minimum reinforcement:min = As,min / Ac = 0.26 (f ctm /f yk) but not less than 0.0013 (Eq. 9.1N)

where:

Ac = b`d for rectangular sections and T-sections with the flange in compression

Ac = bw d + (bf - bw) tf for T-sections with the flange in tension

Maximum reinforcement is limited to 4%. (§ 9.2.1.1 - 3)

Rectangular beams:

Referring to the following Figure, the reinforcement area is calculated as follows:

As = M / f yd z

K = M / b d² f ck

The maximum depth of the compression block is defined by:

for f ck

50

(x/d) max = ( - k1) / k2 (Eq. 5.10a)

for f ck

> 50

(x/d) max = ( - k3) / k4 (Eq. 5.10b)

where:

= the ratio of the moment after redistribution to the moment before redistribution.

k1,k2,k3,k4 are defined by the user in the Files - Setup option

Rearranging, the compression block is at its maximum when:for f

ck 50

K = K' =cc

[k2( - k1) - 0.5 ( - k1)²] / k22

c

for f ck

> 50

K = K' =cc

[k4( - k3) - 0.5 ( - k3)²] / k42

c

With no redistribution:

for f ck

50

19


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Codes - EU 8


K = K' = 0.196 (for the default values of cc

, , etc in the Code)

for f ck

> 50

K = K' is also dependent on the value of f cu

If K > K' compression reinforcement is required and is calculated from:

A's = (K - K') f ck bd² / f yd (d`-d')

As = (`K' f ck bd² / f yd z) + A's

Tee beams:

compression block entirely in flange: designed as rectangular beams.

compression block in web:

As = Asf + Asw

where:

Asf = Cf / (f yd)

Cf = f cd tf (b - bw)

Mf = Cf ( d - tf / 2 )

Mw = M - Mf

and

Asw is calculated from Mw as outlined for rectangular sections.

The maximum resistance moment of the concrete without compression reinforcement is:

Mc = K' f cu bw d² + f cd (b-bw ) (d - hf /2) tf

where K' was derived for rectangular sections.

Shear Reinforcement:

The shear stress at the face of the support is reduced to the value at a distance 'd' from the face, as

specified in § 6.2.1(8). (optional)

Links only (variable spacing) - Truss Method: (§ 6.2.3)

VRd,s = (Asw /s)z f ywd cot (6.8)VRd,max = bw z 1 f cd / (cot + tan ) (6.9)

The program selects the maximum value of cot so that VRd,s does not exceed VRd,max

1 = 0.6(1.0 - f ck/250) (6.6N)

w,min = (0.08 f ck)/f yk (9.5N)

The program ignores axial forces when designing for shear.


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Codes - Eurocode 9


Torsion:

Torsion reinforcement is calculated according to Section 6.3 of the Code.

The torsion force is always assumed by the program to be carried by the rectangular web of the beam:

The following terms are used:

The program calculates the torsion reinforcement as follows:

Transverse reinforcement (stirrups)

(6.28)

Longitudinal reinforcement:

Combined shear and torsion:

torsion is ignored if :

(TEd /TRd,c) + (VEd /VRd,c) 1.0 (6.31)

TRd,c = f ctd tef 2Ak (6.26)

VRd,c = [Crd,c(k)(100 1f ck)1/3

] bwd (6.2a)k = 1 + (200/d) < 2.0 (d in mm)

Crd,c = 0.18/ c (0.18 may be modified in the Setup option)

VRd,c,min = (Vmin)bwd (6.2b)

Vmin = 0.035 k3/2 f ck

1/2 (6.3N) (0.035 may be modified in the Setup option)

The program checks the following interaction diagram:

(TEd /TRd,max) + (VEd /VRd,max) 1.0 (6.29)

TRd,max = f cd tef 2Ak 0.5 (6.30)

= 0.6(1-fck/250) (6.6N)

VRd,max= bwz 1 f cd /2 (6.9)

1

=

refer to "Shear" for explanation of other terms

Longitudinal spacing of links is limited to uk/8 (9.2.3 - 3)

1.1.2 Columns

minimum: the program limits the longitudinal reinforcement to no less than:

0.10 NEd / f yd but not less than 0.002 Ac (Eq. 9.12N)


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Codes - EU 10


Reinforcement > 8%: the program displays a warning that the reinforcement exceeds

the allowable (§ 9.5.2-3)Reinforcement > 10%: the program stops the design and displays a "No solution"

warning.

Slenderness:

A column is considered as short when:

x < lim and y < lim (5.13N)

where:

= lo / i (5.14)

lo = k·l (5.8.3.2)

k = defined by the user

l = clear column length = member length between ends of restraint (5.8.3.2-3)

i = radius of inertia of the gross cross-section measured in the plane under consideration.

= 20 A B C/

A = 0.7

B = 1.1

C = 0.7

n = NEd,max /(Ac f cd)

Note:

is calculated separately for both directions. If the column is slender in one direction and short in

the other, the column is considered as slender.

The program designs for a minimum eccentricity eo = h/30 but not less than 20 mm (6.1-4)

Shor t columns:

The program designs for M = MEd + NEd e i NEd eowhere:

ei =

= 1/(100 . .

1/300 i 1/200

l = actual length of member

Slender colum ns:

The initial end moments are MoEd = MEd + NEd e i NEd eo

The total design moment MEd = MoEd + M2 (Eq. 5.31)

where M2 = is the additional 2nd-order moment

The user may define the additional moment (as a magnifier factor) in each of the two major directions or

specify that the program calculate the value automatically.

The program calculates the additional moment as follows:

M2 = NEd e2 (Eq. 5.33)

e2 = lo² / 10 (1/r) (5.8.8.2-3)

where:

1/r = Kr K (1/r o) (Eq. 5.34)

K = 1 + ef 1.00 (Eq. 5.37)

= 0.35 +f ck /200 - /150 (5.8.8.3-4)


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Codes - Eurocode 11


= is defined by the user

1/r o = (5.8.8.3-1)

= f yd / Es

d = h/2 + is (Eq. 5.35)

Kr = ( Nu - N ) / ( Nu - Nbal) < 1 (Eq. 5.36)N = design ultimate axial loadNbal = design axial load capacity of a balanced section;

calculated exactly by the program.Nu = design ultimate capacity of a section subject to

axial load only = 0.85 f cd Ac + f yd As

For both short and slender columns the imperfection moment (ei) and additional moments (eo) are

superimposed on the design moments as follows:

Braced columns:

The initial moment at the point of maximum additional moment is calculated as:

for columns not subject to transverse loading:

Moe = 0.6 Mo2 + 0.4 Mo1 0.4 Mo2 (5.32)

where:

Mo2 = larger factored end moment

Mo1 = smaller factored end moment

for columns subject to transverse loading:

Moe = maximum moment.

The additional moment has an approximately parabolic shape; refer to the figure below.

Unbraced columns:

The additional moment is assumed to be distributed approximately linearly over the height of the

column, with Madd applied at both ends in a manner that increases the moment at these section.

The following figure shows an example for a slender column consisting of one STRAP member, braced inthe M2 direction and unbraced in the M3 direction. The design moments at top/middle/bottom are

superimposed on the moment diagrams.

Load condi t ion s:

The program calculates design moments at three locations in each column - top, middle and bottom. In

slender columns, the additional moment is calculated and the initial moments are modified as explained

above.

If the column is subject to transverse loads, the middle moment is taken as the maximum moment along


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Codes - EU 12


the span, but not less than 0.6·M2 + 0.4·M1

The three design moments are calculated separately for the M2 and M3 moments (M2/M3 refer to the

STRAP results). The program then checks the capacity for all three locations, i.e.

top: Ptop, M2,top, M3,top

middle: max(Ptop,Pbot), M2,mid, M3,mid

bottom: Pbot, M2,bot, M3,bot

Columns are designed either for uniaxial bending or biaxial bending

space frame columns with moments about both axes are always designed for biaxial bending

plane frame columns or space frame columns with moments about one axis are designed for uniaxial

bending if:

x / y 2 and y / x 2 (5.38a) and -

(ey /heq)/(ex /beq) 0.2 or (ex /beq)/(ey /heq) 0.2 (5.38b)

Uniaxia l bendin g :

The program adds additional moments to the M3 moments, and create separate load conditions with

the minimum M2 moments, where both positive and negative values are generated (important for

unsymmetric sections).

For example:

M2i M3i M2 M3

0.0 15.7 0.0 15.7

-7.3 0.0

7.3 0.0

Biaxial bending :

The program checks for design moments about both axes; if the moments equal zero about one axis,

the program designs the column as for plane frames.

The program adds additional moments simultaneously about both axes.

In many cases, two or more STRAP members may be combined to form a column. In addition, the

support locations in the M2 and M3 directions may not be identical.

The program searches for supports at the nodes in either direction and defines "design spans". Each

"design span" is calculated separately. For example:

In each design span, the program creates the combinations of M2 and M3 moments at top/middle/

bottom of the span. The object is to ensure that the program creates a load situation that includes the


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Codes - Eurocode 13


maximum moments.

The program calculates the additional moment for the combined span. It then checks whether one of the

ends is in the middle third of the combined span. If yes, the program uses the additional moment at

that end and at the middle. If both ends are outside the middle third, the program uses the actual

moments.

Referring to the example in the Figure above, it calculates the load situations as follows:

Span A:

Design Design moment

Location M2 M3

Top Mtop or

Mtop-Madd/2.

Mi+Madd

Middle Mi+Madd Mi+Madd

Bottom Mbottom+Madd/2. Mbottom+Madd/2.

Tension:

Columns with a tension axial force are always designed as short columns.

The capacity of a column in pure tension is equal to As·f yd.

The capacity of columns with combined tension and bending is calculated identically to that of column

with combined compression and bending.

Shear:

The program searches for the load case with (V - VRd,c)max

VRd,c is calculated according to Equation (6.2a):

VRd,c = [Crd,c(k)(100 1f ck)1/3 + k1 cp ] bwd

where:

bwd = gross area for round sections (from ACI).

As = 0.5 As for1

k1 : defined by the user in the Files - Setup option.

k = 1 + (200/d) < 2.0 (d in mm)

Crd,c = 0.18/ c (0.18 may be modified in the Setup option)

VRd,c,min = (Vmin)bwd (6.2b)

Vmin = 0.035 k3/2 f ck

1/2 (6.3N) (0.035 may be modified in the Setup option)

Links are detailed with uniform spacing according to the user specified parameters.minimum diameter : maximum of - 0.25*(longitudinal bar diameter), 6 mm (9.5.3-1)

maximum spacing : minimum of - minimum dimension, 20*( min. long. bar diameter), 400 mm.

reduced by 0.6 at column ends (9.5.3-3/4)

1.1.3 Walls

The program calculates the minimum eccentricity and slender wall additional moments for weak axis

bending according to the methods used for columns.


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Codes - EU 14


Minimum eccentricity = min(H/400, h/30, 20) mm.

where:

H = storey/wall height

h = wall thickness

The effective length factor for a wall segment, k, is calculated as follows (if not defined by the user):

wall consisting of a single segment: k = 1.0

another segment attached at one end:

segments attached at both ends:

where:

lw = net segment length

H = storey height

Reinforcem ent details:

No seismic loads

minimum reinforcement area = 0.002 * Aw all

maximum bar spacing = min(3*b, 400 mm).

where b = wall thickness

Seismic loads

Concentrated reinforcement at the ends of walls:

minimum area = 0.005 *Aw all - As,dist

in non-critical regions, Lconc = b.

in critical regions: Lconc = maximum of -a. 1.5*bb. 0.15*wall lengthc. L'conc = xu * (1 - cu2/ cu2,c)

xu = (vd + v) Lw

ecu2 = 0.0035

ecu2,c = 0.0035 + 0.1 w d

w d = is calculated according to Eq. (5.20) The program assumes when calculating

according to Eq.(5.4) that qo=3x1

Ductility.v : the program assumes that v = 0.4%

The program checks the the concentrated reinforcement 0.005 Aconc

Note:

Concentrated reinforcement is not designed if the wall length< 4 * thickness

Distributed reinforcement:

minimum area = 0.20% * Adist


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Codes - Eurocode 15


but not less than v, where:

for < 2.0 - v (V - P)/0.8 Ac Fsd

for 2.0 - v 0.002

where:

= M/Vlw 3.0

maximum spacing:High ductility: min(20 , 200) mm

Medium ductility: min(25 , 250) mm

Low ductility: min(3b, 400) mm

Transverse reinforcement:

minimum h:

for 2.0 - h Vd/0.8 Ac Fsd 0.2%

for 2.0 - h Vd-Vcd)/0.75 Ac Fsd 0.2%

where:

= M/Vlw 3.0

Vcd:- critical region: compression: as in EC2, ignoring the axial force.

tension Vcd = 0.- non-critical region: as in EC2, considering the axial force (tension and compression)

1.1.4 Seismic - Beams

Main reinforcem ent: (403.3.1.5)

tension reinforcement at sections where required:

min max

min 0.5·f ctm/f yk (Eq. 2.12)

max:

The program always provide minimum reinforcement but only displays a warning if the required

reinforcement exceeds max

at all sections:

at supports:

Mcap,pos > 0.5·Mcap,neg (2.7.1.2.2-2b)

at spans:

Mcap > 0.25·(Mcap,neg),max (from other Codes)

Transverse reinfo rcem ent:

The program does not calculate the value of as specified in Sections 2.7.2.2.(3). The program does not

calculate diagonal reinforcement but only stirrups according to EC2, i.e. the program always assumes

that > -0.5 and that |Vs|max > 3(2+ )· Rd·bw ·d.

Minimum requirements:


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Codes - EU 16


1.1.5 Seismic - Columns

Column seismic moments and shears are calculated as explained in Seismic - general, except for the

following:

Bending moments:

Referring to EC8 - 2.1.1.1, the program assumes that1 = 1.00. Therefore, the sum of the beamcapacity moments at a joint are distributed to the columns above and below according to the ratio of

the design moments in the columns (instead of the relative stiffness)

Shear:

Referring to EC8 - 2.8.1.1.2 - Equation (2.24), the program assumes that MCRd = MDRd = the

maximum seismic design moment in the column (calculated from the beam capacities as explained

above), rather than the actual resisting moment of the column.

Longi tudinal re inforcem ent:

Flexural strength at joint: (2.8.1.1)

High:

(2.8.2.1-1)

Medium:

(2.8.3.1-1)

Low: No special requirement

Reinforcement ratio:0.01 < < 0.04 (2.8.1.3-8)

Transverse reinfo rcem ent:

Hoops (spiral or circular) required over length lcr:

High: max (1.5·dc, span/5, 600 mm) (Eq. 2.28)

Medium: max (1.5·dc, span/6, 450 mm) (Eq. 2.31)

Low: max (1.0·dc, span/6, 450 mm) (Eq. 2.34)

Maximum spacing within lcr:

High: min (bo/4, 5 L, 100 mm) (Eq. 2-30)

Medium: min (bo/3, 7 L, 150 mm) (Eq. 2-33)

Low: min (bo/2, 9 L, 200 mm) (Eq. 2-35)

Maximum spacing outside lo:min (b, 12 L, 300

mm)

(EC2 - non-

seismic)

Minimum diameter:

High: min (bo/4, 5 L, 100 mm) (Eq. 2-30)

Medium: min (bo/3, 7 L, 150 mm) (Eq. 2-33)

Low: min (bo/2, 9 L, 200 mm) (Eq. 2-35)

Additional requirements for High moment frames:


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Codes - Eurocode 17


1.1.6 Punching

The program calculates the punching stress and reinforcement according to EC2 (2002) Sect ion 6.4 -

"Punching"

The program calculates the area of single- or multiple-leg stirrup type slab shear reinforcement:

The program assumes that the distance between the radial lines of shear reinforcement at the control

perimeter is always less than 2d.

vRd,c, the punching shear resistance provided by the concrete in slabs reinforcement, is calculated as

follows:

The shear is checked at the basic control perimeter at 2d from the face of the column, as follows:

Centre column : u1, according to Section 6.4.2.

Edge column : u1*, according to Figure 6.20a

Corner column : u1*, according to Figure 6.20b

When vEd > vRd,c , shear reinforcement is calculated from the equation:

where:

s is always assumed = d/2

Av = the cross sectional area of all legs on one peripheral line that is geometrically s imilar to the

perimeter of the column section, i.e. the sum of the vertical legs of stirrups 1,2,3 and 4 in the

figure above.

represents the portion of the unbalanced moment at the column-slab connection assumed to be


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Codes - EU 18


transferred by shear:

Centre column:

Edge column, moment only about axis parallel to edge (towards interior):

=1.0 (6.4.3-4)

Edge column, moment about both axes (towards interior):

Corner column (moments towards interior):

= u1 /u1* (6.48)

Edge/corner column, moment towards exterior:

uout, the 'control perimeter' at which shear reinforcement is no longer required, is calculated from:

The program also checks that adjacent to the column:

1.1.7 Slab deflections

The deflections are calculated using on the effective moment of inertia, Ie, according to section 7.4.3 and

equations(7.18) and (7.19) in the Code.

For each selected deflection combination the program calculates Ie for each element in both directions,

based on the elastic moments, thereby creating new orthotropic elements with the effective properties.

The program then solves the model again to determine the slab deflection.

where:

Ig = gross moment-of-inertia, including reinforcement

Icr = cracked moment-of-inertia

M = service moment

Mcr = cracking moment

The total deflection has three components:

dt = immediate deflection due to long-term loads (dead + sustained live)

The program uses Ie calculated from long-term loads

dt = long-term deflection due to long-term loads

The program uses Ie calculated from long-term loads

= creep factor.

di = immediate deflection due to other live loads (not sustained)The program uses Ie calculated from al l immediate loads

The total deflection d = dt + dt + di

= dt(1 + ) + di


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Codes - Eurocode 19


1.1.8 Slab reinforcement

This section describes the calculation method for reinforced concrete elements, beams and walls with

combined moment and axial force.

Rectangular sections : The stress block for the case of bending only is:

where:

Md = design moment. The value may be the STRAP moment or the Wood & Armer moment

(elements)

Mcd = the moment corresponding to the Code compression block height in pure bending.

=

For the general case:

Nd = design axial force. The value may be the STRAP force or the Wood & Armer force

(elements).

Transposing the axial force to the tension steel level:

Reinforcement:

There are three design cases:

Section in tension and compression:

Msd > Mcd

M = Msd - Mcd

A's = M/(d - d') 0.87f y

As = A's + Mcd /(zmin·0.87f y) - Nd /0.87f y

C12/15 to C35/45: zmin = 0.82d

C40/50 and greater: zmin = 0.86d


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Codes - EU 20


Msd < Mcd

}

As = Msd / (1 - 0.5 )·d·0.87f y - Nd /0.87f y

Entire section in tension:

where: e = Md /Nd

Entire section in compression

the program initially checks where the concrete capacity is sufficient for equilibrium. If so,

reinforcement is not required and only minimum reinforcement is provided.

the program then tries to provide reinforcement only at the face with the greater compression. If

equilibrium of forces and moments cannot be satisfied -

the program provides reinforcement at both faces.

Note:

all sections are designed for minimum eccentricity

for small loads the program calculates the reinforcement based on an elastic stress distribution.

Mi nim um eccentr ic i ty:

Walls:

max (h/30, 20 mm) Section 6.1-4 (columns)

Beams, Slabs:

max (h/30, 20 mm) Section 6.1-4 (columns)

Mi nim um reinforcem ent:

Slabs,beams: (Section 9.2.1.1) As,min = 0.26(f ctm/f yk) bd

Walls: (Section 9.6.2)

As,min = 0.002 bh

Tee section s

Tee sections are designed similarly to rectangular sections:


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Codes - Eurocode 21


Note:

the program assumes that the entire flange width is effective.


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Codes - EU 22


1.2 Eurocode 2 - Post-tensioned beams

The program designs the beams according to Eurocode 2, April 2002.

Losses

Ultimate moment

Shear

Deflections

1.2.1 Losses

Frict ion L oss:

The losses due to friction, Pu(x), are calculated according to:

(5.45)

The loss consists of two components:

where:Pmax =prestressing force in the tendon at the jacking end

=the coefficients of friction, defined by the user

=cumulative angle in radians through which the tangent to the cable profile has turned, between

the jack and the current point.

x =distance from the jack

If the user specified that the program use the estimate, then

where:

a different estimate may be specified for wobble and curvature loss.

x = distance along the strand from the point where P = Pmax

L = Strand length

Note:

losses are calculated for a specific time "t" . The number of strands jacked up to "t" and the

percentage of jacking force are defined by the user in the "jacking sequence" option.

Elast ic shortenin g loss:

Elastic losses result from the instantaneous deformation of the concrete when the prestressing force is

applied. In posttensioned beams, a cable is shortened by the subsequent jacking of other cables in the

beam, i.e. there is no elastic shortening loss if all cables are jacked simultaneously.

In the following example, cable 'i' is jacked after cable 'j'.

The stress in the concrete at level 'j' due to the jacking of cable 'i' is:

22

26

25

27


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Codes - Eurocode 23


The loss of force in cable 'j" is:

where:

Pi = the initial force in cable 'i', after friction losses

Ac,Ic = Area and moment-of-inertia of the concrete section

Es,Ec = modulus-of-elasticity of cable 'j' and the concrete

A j = the area of cable 'j'

Note:

If the user specified that the program use the estimate for elastic shortening, then

Prior to solving the model, the program calculates Pi = (jacking force - friction loss estimate). The

elastic shortening loss in cables in other beams in the model is not calculated.

When the model is solved, a load case is created for each cable; the axial force result for this case is

equal to the axial force solution from the relevant STRAP load case less the friction loss and this value

is used to calculate the elastic shortening loss. The elastic shortening loss for cables in other beamsis also calculated because the axial loads in the entire model are calculated for each cable load case.

'draw-in' losses are considered as a post-transfer loss and are not subtracted from the jacking force

when calculating Pi.

Ec = Ec(28) for all times = 't',

Draw-in lo ss:

Draw-in loss is calculated taking into account the friction losses along the strand:

The program creates a mirror image of the cable force diagram at the jacking end. The loss is

calculated from

where:

P = the shaded area in the diagram above

= the draw-in length specified by the user.

Note:

If the user specified that the program use the estimate for draw-in loss, then

if a cable is jacked several times (jacking sequence), the draw in losses of the previous jacking are

eliminated by the subsequent jacking.

Tim e depend ent lo sses - General:

Time dependent losses are calculated by the 'time-step' method whereby the losses are calculated at

the end of each time step and the prestressing force is reduced accordingly for the next time step.


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Codes - EU 24


Note that the losses at the end of any time step are calculated relative to time at jacking to; therefore

the loss between steps tm and tn is P(tn,to) - P(tm ,to)

The time-dependent losses - creep, shrinkage and relaxation - are not independent of each other; the

total time dependent loss is calculated according to Code equation (5-46):

where:

= relaxation and shrinkage at location x, at time t.the absolute value of the variation of stress in

the cables due to creep, and all other symbols are defined in the Code.

cs, Ct and f re are calculated as follows:

Shrin k age loss:

There are three options available in the program for calculating the shrinkage loss - the 'Code method'

and two general code independent methods:

Code method:

The shrinkage loss is calculated as:

Pshr (t) = sh Ep Ap (f,user )

where:

= the shrinkage strain calculated according to Eq.(3.8):

=

= the drying shrinkage strain, calculated according to Eq. (B-11).

= the autogenous shrinkage strain, calculated according to Eq. (3-11).

The strain may be modified by a user defined factor.

Alternate method 1:

The user defines the total shrinkage strain eshr,tot and the time (in days) when one-half of the total

strain occurs = t0.5. The shrinkage loss is calculated as:

Alternate method 2:

The user defines the total shrinkage strain eshr,tot and a constant value Cst. The shrinkage loss is

calculated as:

Creep lo ss:

There are three options available in the program for calculating the creep coefficient - the 'Code method'

and two general code independent methods:

Code method:

The general creep coefficient is calculated as:

(B.1)where:

all values are calculated according to equations(B.2) to (B.10), i.e. including the influence of cement

type, humidity, etc.

t = time when creep is calculated (days) from beam casting

to = time when load was applied (days), from beam casting

The coefficient may be modified by a user defined factor.

Alternate method 1:

The user defines the total creep factor Cu (total/elastic strain) and the time (in days) when one-half of


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Codes - Eurocode 25


the total creep occurs = t0.5. The creep coefficient is calculated as:

Alternate method 2:

The user defines the total creep factor Cu (total/elastic strain) and the value of the expression in thefollowing equation. The creep coefficient is calculated as:

Relaxatio n lo ss:

pr at time 't' is calculated according to Equations (3.30) and (3.31) for ordinary and low-relaxation

prestressing steel, respectively.

The value of 1000, (relaxation loss at 1000 hours), may be specified according to one of the following

methods:User-defined value (e.g. when certificate is available)

Code recommended values, from Section 3.3.2(6).

The user specifies the total relaxation loss at 500,000 hours; the program then calculates 1000 from

equations (3.30) and (3.31).

1.2.2 Shear

The program calculates the shear capacity at the time requested for every point along the length of the

beam:

the effective cable forces are calculated taking into account the losses and jacking sequence at the

time requested.

at each point along the length of the beam the capacity is checked at the centre-of-gravity of thesection and at any point on the section height where there is a jump in section width; the maximum

value only is displayed for each point.

The program first calculates the cracking moment, Mcr :

Mcr = Wfull f ten + Mo

where:

Wfull = modified elastic modulus, taking into account the shift of the center-of-gravity of the section

due to the cables and regular reinforcement.

f ten = f ctk0.5 / c

Mo = pty Wfull

pty = stress from the cables at the tension face of the section

The shear capacity is calculated as follows:

uncracked sections (M < Mcr)

where:

cp = N/Ac < 0.2f cd

cracked sections (M Mcr)

where:


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Codes - EU 26


CRd,c = 0.18/ c

k = 1 + (200/d) 2

= As /bd 0.02

(VRd,c)min = (vmin + 0.15 cp)bwd (6.2.b)

VRd,max = cw bw z f cd (cot + tan ) (6.9)

where:

cw : from Eq. (6.11)

= 0.6 (1 - f ck /250) (6.6)

The program calculates shear reinforcement when V > Vc

Inclined strut method:

The reinforcement is calculated from the equations:

VRd,s = (Asw /s) z f ywd cot (6.8)

VRd,max = cw bw z f cd (cot + tan ) (6.9)

standard method:

(Asw /s) = (V - Vc) / (0.9d f ywd)

Note:

Shear capacity is compared to the design shear only for those combinations defined as "Factored" or

"Service & factored".

The program adds/subtracts the vertical component of the cable force to the shear capacity.

1.2.3 Ultimate moment

The program calculates the ultimate moment capacity at the time requested for every point along the

length of the beam, based on the stress-strain parameters defined by the Code.

where:

= 0.8 f ck 50 MPa (3.19)

= 0.8 - (f ck - 50)/400 f ck > 50 MPa (3.20)

= 1.0 f ck 50 MPa (3.21)

= 1.0 - (f ck - 50)/200 f ck > 50 MPa (3.22)

ecu3 from Table 3.1cc = 1 (3.1.6)

c = 1.5 (Table 2.1N)

s = 1.15 (Table 2.1N)

s,max = 0.02

The neutral axis location at equilibrium is calculated by an iterative procedure:

the program calculates the contribution of the concrete and tension/compression cables and regular

reinforcement, each cable and bar according to its eccentricity.


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Codes - Eurocode 27


losses and section properties are calculated for the specified point in time, considering only that

cables that have already been jacked.

maximum concrete strain is assumed at the compression face and the strains in the cables and

reinforcement are calculated assuming a linear distribution.

For bonded cables, the stress is calculated from the stress-strain diagram defined in "Parameters".

For unbonded cables, the program calculates effective stress, plus stress increase for the ultimatecondition, defined by the user in "Parameters".

Note:

Ultimate moment capacity is compared to the design moment only for those combinations defined as

"Factored" or "Service & factored".

1.2.4 Deflections

The displacements are calculated based on effective moment-of-inertia values at each point.

For long-term deflections, the deflections at each 'step' are calculated according to the cable forces at

that step.

Deflection due to cable forces are calculated based on the losses at the time the deflections are

calculated.Changes in supports at certain stages change the boundary conditions of those stages for integration

of the curvature along the beam.

Imm ediate deflect ion:

the program calculates the moment-of-inertia for the full section, using modified properties due to

cables and regular reinforcement.

the program calculates the cracked moment-of-inertia using an iterative method to determine the

center-of gravity location and equilibrium condition ( C = T) at each point. A linear stress distribution

over the height of the section is assumed because the service limit state is checked.

The program first calculates the cracking moment, Mcr :

Mcr = Wfull f ten + Mo

where:

Wfull = modified elastic modulus, taking into account the shift of the center-of-gravity of the sectiondue to the cables and regular reinforcement.

f ten = f ctm (Table 3.1)

Mo = 0.90 pty Wfull bonded (5.10.9)

0.95 pty Wfull unbonded

pty = stress from the cables at the tension face of the section

The program then calculates an effective curvature, Ceff , at every point along the beam, where Ceff =

M/EIeff

for M < Mcr : Ceff = Cfull

for M > Mcr : Ceff = Cfull Ccr

where:

M = Mloads + Mcable

Cfull = curvature for the uncracked sectionCr = curvature for the cracked section

The curvature is integrated twice, using the supports at each stage to compute the deflection.

Lon g-term deflect ion:

The program sums up the deflections for different stages. For each stage it searches for the

combination that gives the maximum creep. Only "service" or "Factored and service" combinations are

considered. The program takes into account that supports may be added/removed at different stages.


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Codes - EU 28


The deflections at each stage are calculated as described in "Immediate deflection".

Note:

the program adds the shrinkage curvature to Ceff (refer to EC2 - losses for calculation of the

shrinkage strain)

22


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Codes - Eurocode 29


1.3 Eurocode 3

The design is based on EN1993-1-1 : Eurocode 3 - Part 1-1 - General Rules for Buildings - May 2005.

The design is modified by the "UK National Annex to Eurocode 3" if the Annex is selected by the user inthe STRAP main menu Setup - Miscellaneous - Code factors option.

1.3.1 Classification of sections

Referring to Tables 5.2, the program determines the classification of every member for each combination

as follows:

in the case of biaxial bending, the program classifies the section according to "minor axis", which

always governs.

when calculating , the program assumes that the section is fully stressed and that the part of fy not

resulting from the actual axial force is caused by bending moments.


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Codes - EU 30


in a "Combined beam", the program calculates the classification for each segment and uses the worst

case.

in a tapered beam, the program calculates the classification at each end and uses the worst case.

Effective Cross-sectio n Prop erties of Class 4 Sectio ns: - § 6.2.2.5

The effective cross section properties of Class 4 sections are calculated using the effective widths of the

compression elements. Referring to EN 1993-1-5:

and k is calculated according to section type as follows:

The program calculates Aeff , Weff and the eccentricities enx, eny based on the reduced cross sectionproperties.

1.3.2 Strength of steel

The program allows design with all steel grades (Fe360, 430, 510, Fe E 275, Fe E 355) - a different

grade may be assigned to each member.

The program calculates the design strength, f y, for various thicknesses according to Table 3.1.


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Codes - Eurocode 31


1.3.3 Shear

The shear limit state is calculated as follows:

where Av is calculated as:

For tapered sections, the program calculates Fv and Av at 20 intervals along the member length

Note:

the program does not check shear buckling resistance of sections with thin webs (d/t > 69 )

according to EN 1993-1-5.

1.3.4 Bending

The program calculates all result values at: - 1/10 of span intervals, and at points of intermediate

supports. For tapered sections or combined beams with different properties, the program uses the

actual section at each point.

Note:the program does no t check built-up sections with thin webs (d/t > 69 ) according to EN 1993-1-5.

Moment Capacity with Low Shear Load (VEd 0.5·Vpl,Rd) - § 6.2.5

The moment capacity is calculated according to section classification as required by the Code.

Class 1 or Class 2 sections: Mc = Wpl f y / MO

Class 3 sections: Mc = Wel f y / MO

Class 4 sections: Mc = Weff f y / M1

where:

Wpl = plastic section modulus

Wel = elastic section modulus

Weff = effective section modulus, calculated as per § 5.3.5.

For T-sections (major axis bending) and [-sections (minor axis bending), the program uses the value of

Wel calculated using the larger distance from the centre-of-gravity of the sect ion.

Moment Capacity with High Shear Load (VEd > 0.5·Vpl,Rd) - § 6.2.8


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Codes - EU 32


The moment capacity is calculated according to section classification as required by the Code.

Mc is calculated according to the section classification (refer to 'low shear load')

Av as calculated in Shear .

1.3.5 Lateral-torsional buckling

Lateral-torsional buckling is calculated individually for each segment between intermediate supports, and

separately along the top and bottom flanges.

The designer can specify the exact location of intermediate supports for each member.

The effective length of individual segments for LTB is calculated according to the 'k' value specified by the

user in the "End Conditions" option of the Postprocessor.

The program checks that:

** UK Natio nal An nex ** (NA .2.17)

rolled sections:, LT,0 - as specified in STRAP - Setup

welded sections

= 1.00 , LT,0= 0.2

** UK Natio nal An nex ** (NA .2.18)

kc = 1/ C1

C1 = (Mcr for the actual moment diagram)/(Mcr for a uniform moment diagram)

where Mcr is the elastic critical moment for lateral-torsional buckling and is calculated according to the

31


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Codes - Eurocode 33


equations in Annex F of the 1992 version of EC3:

where:

C1, C2 and C3 are calculated from Tables F.1.1, F.1.2 (see below) based on the bending momentdiagram shape factor

warping factor kw = 1 (§ F.1.2 (4))

for I, RHS and Pipe sections: Z j = 0

for T sections: Z j is calculated according to F.1.4 (2)

If the flange direction of T,[ sections is not defined, or if a destabilizing load is defined, the program

specifies the sign of Zg such that the value of Mcr is decreased.

Referring to Table F.1.2, the program automatically determines whether a member is loaded in the

segment between adjacent intermediate supports. Members with a maximum Note that if you COMBINE

a series of unloaded members which have individual bending moment diagrams of varying slopes, the

program will consider the COMBINED beam as "loaded".

Unloaded segments: C2 = 0

Loaded segments:

Only the first two diagrams from Table F.1.2 are used to calculate C1, C2 and C3. For more general

bending moment diagrams, the program estimates which of the two cases is closer to the actual

shape, based on the end moment values.

1.3.6 Axial force - compression

The compression capacity of a member is calculated as per § 6.3.1, but not greater than the value

calculated in § 6.3.4.

NSd X A A f y / M1

but:

A f y / M0 (Class 1, 2 and 3 sections)

Aeff f y / M1 (Class 4 sections)

where:

A = 1 for Class 1, 2 and 3 sections

A = Aeff /A for Class 4 sections

Table 5.5.3 is used to determine which buckling curve in Table 6.1 is applicable, according to the

buckling axis, thickness and section type.

Slenderness = kl/r :

The program may be instructed to calculate kx and ky, the effective length factors, according to

Appendix E or the values may be input directly by the designer (a default value of 1.00 is assumed by

the program for all members).

Tapered members or combined beams with different properties:

The program calculates the exact Euler buckling load for the member and then finds an equivalent

length l1 for a member with the minimum area which gives the same Euler buckling load. The

minimum area and the length l1 are used in all the equations.

Angles, Channels and T-section Struts - § 4.7.10

Back-to-back struts - § 4.7.13


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Codes - EU 34


The calculation is carried out according to § 4.7.10 and § 4.7.13 of BS 5950, which is a more detailed

version of the method in Eurocode 3:

Single Angle Struts - § 4.7.10.2

Slenderness is the maximum of

kx · Lx/rvv

ky · Ly /r vv

0.7 · Lx /r xx + 30 Chec

ked only if the member is defined as "pinned" at both ends

0.7 · Ly /r yy + 30 " "

0.7 · Lx /r vv + 15 " "

0.7 · Ly /r vv + 15 " "

where kx and ky are defined by the user - normally 0.85 or 1.00, depending on the connection

condition in § 4.7.10.2 (a) and (b).

Double Angle Struts - § 4.7.10.3


kx · Lx/rxx

0.7 Lx /r xx + 30

1.4 c (§ 4.7.9c)

All four conditions are checked even if the member if not defined as "pinned" at both ends.

c is calculated for a single angle (using r vv). When calculating c, the program assumes Lc = L/3 to

comply with the requirement that the member should be divided into at least three segments (§

4.7.13.1e) and ( c) max = 50.

Single Channel Struts - § 4.7.10.4


kx · Lx/rxx

ky · Ly /r yy

where kx,ky are defined by the user.

T-section Struts -§ 4.7.10.5

Effective Length is the maximum of

kx · Lx/rxx

ky · Ly /r yy

0.7 · L/r xx + 30

where kx,ky are defined by the user.

Note that the third condition is applied only if the member was defined as "pinned" at both ends.

1.3.7 Axial force - tension

Tension capacity is calculated as:

NEd A f y / M0 (6.6)

0.9 Anet f y / M2 (6.7)

The designer may define an area reduction factor for tension members (a default value may be specified)

in order to specify the "Net Area of Simple Tension Members" as outlined in §6.2.2.2.


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Codes - Eurocode 35


1.3.8 Combined bending & axial force

Relationships are calculated according to section classification.

Local

Class 1 and Class 2 sections (6.4.1)

where:

and n = NEd / Npl,Rd

MNx,Rd, MNy,Rd are calculated from the reduced plastic section moduli in the CONSTRADO tables,

except as follows:

T-sections, angles, double angles and double channels:

The equation with Mnx, Mny cannot be used for these sections because the relevant data is not

available; the program calculates the capacity using the equation for Class 3 sections.

RHS (from CONSTRADO tables)

Pipe (from CONSTRADO tables)

Class 3 sections (6.2.9.2)

Class 1 and 3 sections - T, L, 2L

Class 4 sections (6.2.9.3)

where:

Npl, Mpl,x and Mpl,y are calculated according to the section classification

eN is the shift of the centroidal axis in sections (Figure 5.3.1 and 5.3.2)


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Codes - EU 36


Overall - bending and axial compression (6.3.3)

where:

NEd, My,Ed, Mz,Ed

= design axial compression force and moments

= moments due to the shift in the centroidal axis for Class 4 section according to

Table 6.7

= flexural buckling reduction factors according to 6.3.1

= lateral-torsional buckling reduction factor according to 6.3.2

kyy,kyz,kzy,kzz = interaction factors, calculated according to Annex B - Method 2.

** UK Natio nal Annex ** (NA .3.2)

Sections other than I, H or hollow: Class 1 and Class 2 sections are designed as Class 3 sections.

1.3.9 Deflections

The deflection check is a Serviceability calculation so the program uses the user-defined load

combinations without multiplying the elastic deflections by the load factors.

When checking the maximum deflection along the span of the member, the program ignores the

deflection of the end nodes, except in cantilevers where the maximum deflection is calculated at the free

end (unsupported node).

Maximum allowable deflections per member must be entered by the designer (a default value may be

specified).

Note:

The deflections calculated in the results are based on the section input in STRAP geometry. When

checking a different section, the Postprocessor modifies the deflection value by Inew /Iold, where:

Inew = moment-of-inertia of section being checked.

Iold = moment-of-inertia of STRAP geometry property.

When checking deflections of a "combined" beam, the program uses the length of the entire

combined beam and ignores any possible deflection support at the combined nodes. If such supports

exists, the allowable deflection parameter should be modified accordingly.

the program ignores intermediate supports (buckling and lateral-torsional) when checking deflections.

1.3.10 Castellated/cellular beams

The design is based on the following publication:

Large Web Openings for Service Integration in Composite Floors

Eurocode Design Method for Composite and Non-Composite Beams with Web Openings

The Research Fund for Coal and Steel, 2006.

The program designs the following section types:


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Codes - Eurocode 37


Composite sections may be defined.

The program assumes that all of the openings are vertically centered in the beams.

Section classification

Flange:

Class

1 b/tf < 9

2 b/tf < 10

3 b/tf < 14

Web:

Class

2 lt or

3 lt or

The program carries out the same regular beam design checks at both the full section and the cut

section locations.

In addition, the program does the following four design checks:

Shear resistance of the top and bo ttom tees

Ved /Vc,rd 1.0

Ved

=

maximum shear force acting at the opening

Vc,rd

=

shear capacity

=

Av Fy / ( mo 3)

rolled sections: Av = Anet - 2bf tf + (2r + tw) tf fabricated

sections:

Av = (dt+db) tw

Vierendeel ben ding


38/52

Codes - EU 38


Class1,2 : Med /Mc 1.0 Ned /Nc,rd 1.0

Class 3: Med /Mc + Ned /Nc,rd 1.0

Med = V (e/4)

V = maximum shear at the openinground opening: e = 0.45ho = effective opening length

hexagon opening: e = s

Class 1,2 : Mc = Wpl,Rd Fy / mo

Class 3,4 : Mc = Wel,Rd Fy / mo

Mc = reduced capacity of the T-section. The reduction is calculated according to the axial force and the

shear force acting at the section.

Ned = M/h1 + P/2

P = design axial force

h1 = distance the centers of the two T-sections

The reduced plastic/elastic modulus is calculated by reducing the section about the plastic/elastic axis

and calculating the reduced section properties

reduced area due to axial force: Aa = Ned /Fyreduced area due to shear force: As = (2 - 1)

2 Aw(only when the shear stress is greater than 50% of the allowable, > 0.5)

Nc,rd = A Fy / mo

Web post bu ckling

V / Nb,rd 1.0

widely spaced opening (s > ho):

le = 0.7hoV = Ved = maximum shear force acting at the opening

closely spaced opening (s ho):

le = (s2 + ho

2)

V = Vh = horizontal shear force

= 12 le /tw

according to buckling curve (c)

Nb,rd = buckling resistance - EC3 - Section 6.3

Horizon tal shear acting on the w eb pos t

Vh /Vn 1.0


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Codes - Eurocode 39


Vh =maximum shear force acting on the post

=V do /h'

Vn =

shear capacity

= s tw Fy / (

Com posi te beams

A modified shear value V' is used in all calculations, where -

V' = V = VcncVcnc = vRd,c bw hsbw = min (bf , bt+2hs)

bf = user-defined concrete flange width

bt = steel flange width

Vierendeel bending :

The tensile force in the bottom tee due to bending, NEd,b , is calculated as follows:

NEd,b = [MEd - Mcnc] / [h - zt - zb] + P/2

but not less than MEd / [h + hs - 0.5hc - zb]

where:

MEd is calculated using he modified shear value

Mcnc = Nc [zt + hs - 0.5hc]

hc = depth of compression block in the slab

Nc = compression force in the concrete flange

"No. of studs" specified by the user:

Nc= nsc Prd = 0.85 f cd beff nsc = the number of shear studs between the point in question and the point of zero moment

Prd = the design capacity of a shear stud.

full capacity:

Nc= 0.85 f cd beff

Note:

the program calculates more accurate results if the Design at each 1/10 of span option is

selected.


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Codes - EU 40


1.4 Eurocode 3 - cold-formed

Code clauses in the Help refer to EC3 - Part 1.3- 1996 - “Supplementary Rules for Cold-formed Thin

Gauge Members & Sheeting" , February 1996.

The program designs the following section shapes:

and additional User-defined shapes.

Sections are added to the steel section library using the Files / Utilities / Create/edit a steel sections

table option in the STRAP main menu.

Effective section properties

Strength of steel

Shear

Bending - without LTB

Axial force - tension

Axial force - compression

Deflections

Lateral buckling strength

1.4.1 Axial force - compression

The program checks that the applied compression force does not exceed the design buckling resistance

of the cross-section, calculated according to 6.2:

where:

Nsd = factored axial compression force

Nb,Rd = design buckling resistance for axial compression, calculated according to:

Nb,Rd = Aeff f y / M1 (6.1)

where:

Aeff = effective sect ion area, obtained from Section 4 by assuming a uniform compressive stress

com,Ed = f yb / M1

= reduction factor for buckling resistance, determined from

= imperfection factor, obtained from Table 6.1

= relevant slenderness for relevant buckling mode.

(6.2.b)

The lowest value of for flexural buckling of the member about any relevant axis, or for torsional or

43

45

44

42

41

40

43

44


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Codes - Eurocode 41


torsional-flexural buckling, is used.

Flexural buck l ing:

The appropriate buckling curve is determined from Table 6.2 according to the section type and axis of

buckling. The relative slenderness about the x and y axis is determined as follows:

l = buckling length for flexural buckling about the relevant axes (lx or ly)

i = radius of inertia about the corresponding axes (ix or iy), based on the properties of the gross

section

The relevant slenderness is calculated as follows:

} (6.4a)

where:

=

= elastic critical stress for torsional buckling, determined from 6.5a

= elastic critical stress for torsional-flexural buckling, calculated for all open section types

(symmetric and non-symmetric) by solving the following equation from the AISI Cold-Formed

Steel Manual (Part III, Supplementary Information - Section 4):

xo,yo = the distances from the centroid to the shear centre along the x,y axes of the

section

r o = polar radius of inertia of the section about the shear centre

The program first searches for the longest segment between axial supports and calculates ex

for this segment. It then searches for the longest segment defined for LTB (i.e. between +z

and/or -z supports) that overlaps (even partially) the critical axial force segment and calculates

t.

1.4.2 Axial force - tension

The program checks that the applied tension force does not exceed the design tension resistance of the

cross-section, calculated according to 5.2:

where:

Nsd = factored axial tension forceNt,Rd = design tension resistance, equal to the lesser of:

Nt,Rd = f ya Ag / Mo (5.1)

Fn,Rd = Anet f u / M2 (Table 8.1)Ag = gross section areaFn,Rd = net section resistance (Section 8.4)


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Anet = net section areaf ya = average yield strengthf u = ultimate tensile strength

= 1.25

1.4.3 Bending - without LTB

The program checks that the applied moment about the relevant axis is less than the moment capacity

of the cross-section calculated according to Section 5.4 of the Code, i.e.

where:

Msd = moment (factored loads)

Mc,Rd = moment capacity of the cross-section, calculated from:

Mc,Rd = f y Weff / - when Weff < Wel (5.3a)

Mc,Rd = f ya Wel/ - when Weff = Wel (5.3b)

where:

Weff = effective section modulus, based on the the effective cross-section subjected only to bendingmoment

Wel = gross elastic section modulus

Note:


supports.

When computing the moment-of-inertia and elastic section modulus of flexural members, the program

uses the effective section properties calculated according to Section 5.6.2.

1.4.4 Combined axial force & bending

Compression and bending

The program checks the combined bending and compression stress according to Section 6.5:

where:

Nb,min = min f yb Aeff / M1

Mx = Mx,sd + Nsd eNx

My = My,sd + Nsd eNy

Mcx = f yb Weff,x,com / M1

Mcy = f yb Weff,y,com / M1

Mbx = LT f yb Weff,x,com / M1

All other values are calculated as explained in Section 6.5.

Tension and bending

The program checks the combined bending and tension stress according to Section 5.5:

where:

Weff,x,ten, Weff,y,ten = effective section modulus for maximum tensile stress if subject only to


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moment about the x-x or y-y axis respectively. = 1.1

Note that when selecting a sect ion the program also checks bending without the axial force. The

equation for combined tension and bending is less conservative than bending alone; if a section is

inadequate for bending, then combined tension and bending will not be checked.

The program checks combined bending and shear according to Section 5.10:

where:

Msd factored bending moment

Vsd factored shear force

Mc,Rd moment resistance of the section

Vw,Rd shear resistance of the section

1.4.5 DeflectionsThe deflection is calculated based on the effective section properties determined according to Section 4.

The effective width of a compression element is calculated according to the compressive stress com,Ed,

se r in the element subjected to the service limit state loading.

1.4.6 Effective section properties

The effective section properties are calculated on the basis of the effective width beff of the compression

elements. The effective width is calculated according to the compression stress com,Ed in the relevant

element, where:

com,Ed = f yb / M1

f yb = basic yield strengthM1 = 1.1 = partial factor for buckling failure (section 2.2)

Plane com pression elements withou t st if feners

The effective width is calculated according to Section 4.2 and is obtained from Table 4.1 (Doubly

supported compression elements) and Table 4.2 (Outstand compression elements).

Plane com pression elements with edge stif feners

The first and last elements of a section are assumed to be stiffeners (lips) if the length of the segment

is less than 28 mm; otherwise they are assumed to be unstiffened elements.

The effective width is calculated according to Section 4.3.2:

only single-edge folds may be calculated (Figure 4.2.a)

The effective section properties are calculated according to Section 4.3.2.2 - General Procedure

Plane com pression elements with intermedi ate stif feners

The program assumes that the following elements have intermediate stiffeners if the projection is

smaller than 28 mm:

The effective width is calculated according to Section 4.3.3. The effective section properties are

calculated according to Section 4.3.3.2 - General Procedure.


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1.4.7 Lateral buckling strength


supports.

The designer can specify the exact location of intermediate supports for each member.

Lateral-torsional buckling is calculated individually for each segment between intermediate supports, and

separately along the top and bottom flanges; the calculation is done separately for positive moments

(supports on bottom flange only are considered) and for negative moments (supports on top flange only

are considered).

For laterally unbraced segments subject to lateral buckling, the factored load moment shall not exceed

the design buckling resistance moment:

where:

Msd = design moment

Mb,Rd = design buckling resistance moment, calculated according to section 6.3.

1.4.8 Shear

The program checks that the actual shear force is less than the allowable shear resistance of the web,

i.e.

Vsd Vw,Rd

according to Section 5.8.

The shear resistance of the web Vw,Rd is the lesser of the shear buckling resistance Vb,Rd and the

plastic buckling resistance Vpl,Rd:

shear buck lin g resistance - Vb,Rd:

Vb,Rd = (hw /sin ) t f bv / M1 (5.13)

where:

hw = web height between midlines of flanges

= slope of the web relative to flanges

t = web thickness

M1 = 1.1

f bv = shear buckling strength, is calculated according to Table 5.2, based on the value of } ,

where

sw = slant height of web

f yb = basic yield strength

E = 21,000 N/mm²

pl astic shear resistance - Vpl ,Rd:

Vpl,Rd = (hw /sin ) t (f bv / 3)/ Mo (5.14)

where:

M0 = 1.1

and all other symbols are explained above.

Vpl,Rd is checked only if sw /t 72 (f yb /f y) ( M0 / M1)


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where:

=

f y = f ya if the conditions of Section 3.1.2 are satisfied; otherwise f y = f yb

Note:

For sections with a number of segments parallel to the direction of shear, the program calculates Vr for each segment and uses the sum (lips are not considered as segments).

For combined shear and moments, refer to Combined forces .

1.4.9 Strength of steel

The program allows for design in all grades of steel - a different grade may be assigned to each member.

The increase in strength due to cold-forming is calculated according to Section 3.1.2 if the option is

selected by the user. The increased average yield strength f ya is substituted for f y when calculating

axial capacity, flexural capacity and combined stresses

f ya = f yb + (f u - f yb) knt²/Ag

(f u + f yb)/2 (3.1)

where:

Ag = gross cross-sectional area

k = 5

n = number of 90° bends in the cross-section with internal radius r

t = nominal thickness

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1.5 Eurocode 4

Composite beams

Composite columns

1.5.1 Composite beams

This section details the method used by the program to select steel sections for composite beams when

using the Eurocode 4 composite design code.

The user specifies the topping dimensions, properties and reinforcement, parameters that specify the

type of shear connection, details on short term vs. long term loading, etc. The program then selects the

lightest steel section that provides the required composite section capacity (the topping dimensions are

not modified by the program during the steel beam selection process).

The program differentiates between areas of positive (sagging) and negative (hogging) moment:

designed as composite sections

designed as non-composite steel sections according to Eurocode 4 (with the

addition of reinforcement in the slab, if specified).

Materials

Structural steel:

p y , the design strength of the structural steel, is determined according to Eurocode 4.

Concrete:

The nominal concrete grade (C25 to C50) is specified by the user. The program assumes that the

topping is stressed to a uniform compression of 0.85 f cu/1.5.

C25 concrete is assumed to be "lightweight"

Modulus of elasticity :

Modular ratio = Es /Ec

Section types

The program can design the following section types:

Classification

Negative moments:

Classification according to Eurocode 4.

Positive moments:

Classification according to Eurocode 4, except that the section is assumed to be in compression from

the topping to plastic neutral axis of the composite section and in tension below the plastic neutral

axis.

Shear - ver tical (s ectio n 4.4.2)

The steel beam is designed to resist the entire vertical shear force in accordance with Eurocode 3.

Refer to Eurocode 3- Shear .

46

50

31


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Sh ear - lo ngitu din al (s ectio n 6.2)

The user specifies the following input in the Default/Parameters option:

capacity of a single shear connector

actual number of connectors per beam in the positive moment region, or

program to calculate number of connectors required in the positive moment region for full shear

connection.

If the user specifies the number of connectors in the positive moment region and the number is less than

the number required to develop the positive moment capacity of the sect ion, the program designates the

beam as having a "partial shear connection" and reduces the bending capacity accordingly (see below).

The number of connectors in negative moment regions cannot be specified. The program assumes full

shear connection in the negative moment region if topping reinforcement is specified; hence the user

should not specify topping reinforcement if the required number of connectors cannot be provided.

The program calculates the number of studs as follows:

positive moment region : np = 2 F/Fstud

where:

Fstud = capacity of a single shear connector * 0.85

F = min(F1, F2)

F1 = maximum force in the steel section = A f y /1.1

F2 = maximum force in the concrete section = 0.85 f c Ac /1.5

The force must be taken on both sides of the point of maximum positive moment; hence the factor '2'.

negative moment region (only if the user specifies topping reinforcement):

nn = k FA /Fstud

where:Fstud = capacity of a single shear connector * 0.85

F = As f y /1.15

k = 2 if there a negative moment at both ends; otherwise k = 1

Output:

topping As =0 :

np = total number of shear connectors required in the posit ive moment region

topping As specified :

np+nn = total number of studs required in the positive and negative moment

regions, and

nn = total number of studs required in the negative moment regions (on a

separate line)

Note:If multiple load cases are defined, detailed results for all load cases must be displayed to determine

the required distribution of shear connectors.

it is the user's responsibility to detail the spacing of the studs along the length of the beam.

Bending - Posi t ive moment

The moment capacity of the section for positive moment is calculated according to section 4.4:

Class 1 plastic web: plastic moment capacityClass 2 compact web: plastic moment capacity


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Class 3 semi-compact

web:

reduced plastic moment capacity

Class 4 slender web: reduced elastic moment capacity

The calculation of S, the plastic section modulus, assumes that:

the topping width is reduced by the factor 0.85 f cu/1.5/pytopping area in tension (below the 'plastic neutral axis') is ignored

for partial shear connections (see above), the program calculates the theoretical topping height that

provides a full shear connection and uses this reduced width to calculate the plastic section modulus.

for Class 1 or Class 2 webs with high shear load (Fv > 0.5 Pv), the program calculates the reduced

moment capacity Mcv according to the Code.

The calculation of Z, the elastic section modulus, assumes that:

the topping width is reduced by the factor:

short-term loading:n = Es /Eclong-term loading: n =

topping area in tension is ignored

For partial shear connection, the flange width bf is reduced by the factor (Vcap/Vreq) 1, the degree of

partial connection.

The program carries out the following design checks:

where:

M1 = moment due to load applied to steel (non-composite) beamM2 = moment due to short-term loadM3 = moment due to long-term loadMcs = moment capacity of steel beam only, calculated according to Eurocode 3 .Mc1, Mc2, Mc3 = moment capacity corresponding to M1, M2 and M3, respectively, calculated as

explained above, but subject to the following limitation: S < kZ:

Bending - Negative momen t

The Default/Parameters option allow the user to define topping reinforcement that will increase the

negative moment capacity.

The moment capacity of the section for negative moment is calculated assuming:

no topping reinforcement: steel section capacity according to Eurocode 3

topping reinforcement: the area of the reinforcement is added to the section when calculating S:

Reinforcement should not be specified if the shear connectors required for full shear connection

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cannot be provided in the region of negative moments.

The program carries out the following design checks:

where:

M = M1 + M2 + M3Mc = capacity of section with reinforcementM1, M2, M3, Mcs = as defined above

Lateral-torsio nal buckl ing

The program assumes that the top flange is continuously supported by the concrete topping. The LTB

capacity of the bottom flange is calculated according to Eurocode 3 .

Deflections

The program calculates the total span deflection as follows:

= (M1) + (M2) + (M3)

where:

M1, M2, M3 = as defined above (bending)

: calculated using the moment of inertia of the steel beam only

= the moment-of-inertia is calculated assuming that the topping width is reduced according toa modular ratio n = Es /Ec

= the moment-of-inertia is calculated assuming that the topping width is reduced according toa modular ratio

The increased deflection arising from partial shear connection is determined from EC4, assumingunpropped construction:

= c + 0.3 (1 - Na /Np) ( s - c

where:

Na = actual number of shear connectorsNp = number of shear connectors required for full shear connectionc = deflection for full shear connection as calculated aboves = deflection for the steel beam acting alone

Axial force

The Default/Parameters option allow the user to specify one of the following design options:

ignore axial force

axial force taken by steel beam only (capacity according to Eurocode 3 )

Com bined bending and axial force

Design check according to equations in Eurocode 3

Compo site castel lated/cellular b eams

Refer to Castellated beams - composite

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Codes - EU 50


1.5.2 Composite columns

The program calculates the axial capacity of composite columns according to section 4.8.3 - "Simplified

Method of Design".

Note that the concrete is ignored when calculating the flexural capacity of the section.

The plastic resistance to compression is calculated by adding the plastic resistances of the column

components:

Npl,Rd = Aa f y /1.1 + Ac (0.85f ck)/1.5 + As f sk /1.15

for filled circular sections:

Npl,Rd = Aa f y /1.1 + Ac (f ck)/1.5 + As f sk /1.15

where:

Aa, f y - refers to the structural steel section

Ac, f ck - refers to the concrete area

As, f sk - refers to the additional reinforcement

The effective elastic flexural stiffness of cross sections is calculated as:

(EI)e = EaIa + 0.8 (Ec /1.35) Ic + Es Is

where:

Ea, Ia - refers to the structural steel section

Ec, Ic - refers to the concrete area

Es, Is - refers to the additional reinforcement


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Index 51


Index

- C -Codes

Eurocodes 5

Cold-formed 40

- E -Eurocode 2 6, 22

Eurocode 3 29

Cold-formed 40

Eurocode 4 46


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