Composite Construction Design (ULS Only)

Date or reference

Composite DesignComposite Design

Beams and SlabsBeams and Slabs

2

British Standards to Eurocode

Composite Design hasn’t really changed significantly – just a few new formulae to consider.

(and a little bit more work as usual!)

3

Some Useful References

1. R.P.Johnson. Composite structures of steel and concrete

3rd Ed 2004, Blackwell.

2. W.H.Mosley, J.H.Bungey, R.Hulse, Reinforced concrete design to Eurocode 2, 2007, Palgrave Macmillan.

4

Composite Systems

Typical composite sections

Shear Studs

5

Composite Systems

6

Composite Systems

Composite Profiled Slimdec System

7

Composite Systems

Composite Slab with Steel Decking

8

Design Procedure

Apart from EN1990 (Actions), the following codes of practice

are required for the design of Composite Beams.

(a)EC2, (EN1992-1-1) for the design of concrete structures

(b)EC3, (EN 1993-111) for the design of steel structures

(c)EC4, (EN 1994-1-1) for the design of composite steel and

concrete structures

Also the National Annexes may be required.

9

Effective Width of Concrete Flange

EC4 CL 5.4.1.2

Effective breadth beff =>

b0 is the distance between centres of adjacent shear studs

bei is the effective width of the concrete flange on each side of the steel web.

bei = Le/8 but < half the distance to the centre of the adjacent beam

Le is the approximate distance between points of zero bending moment (L/2 for midspan of a continuous beams or L for a single simply supported beam.)

oeieff bbb + Σ=

10

Effective Width of Concrete Flange:

Basic Example

Continuous beam, with spans equal to 12m, with adjacent beams at 4m centres.

The effective breadth of the concrete flange is:

beff = 2 x Le/8 = 2 x 0.5 x 12/8 = 1.5m

This is less than half the distance between adjacent beams (2m), so this is o.k.

If the beam was simply supported, the effective

breadth would be 3m

11

Design Procedure

Principal Stages in the Design are:

1. Preliminary sizing – the depth of the beam (UB) initially approximated as:

Simply supported => Span/20Continuous => Span/24

The yield strength and classification should then be determined to EC3.

12

Design Procedure

2. During construction (Unpropped only)

Loads considered =>

Beam self weight

Shuttering/steel decking weight

Weight of the wet concrete

Imposed construction load => usually 0.5~0.75 kN/m2

Check Bending, Shear at ULS, check deflection at SLS

13

Design Procedure

3. Bending and shear of composite section at ULS

Compare the moment of resistance of the composite section with the ultimate design moment.

Check the shear strength of the steel beam (alone)

4. Shear connectors and Transverse steel at ULS

Design shear connectors at the concrete/steel beam

interface, for either full or partial interaction

Provide transverse reinforcement to resist the longitudinal shear in the concrete flange.

14

Design Procedure

5. Deflection

Check the deflection of the beam to protect against cracking of architectural finishes.

Note that for slabs, deflection due to pondingshould also be considered.

15

Design Procedure

Design of the steel beam during construction =>

At ULS bending,

The plastic section modulus Wpl,y for the steel beam may be calculated

from:

Where

MEd is the ultimate design momentfy is the design strength of the steel (EC3 table 3.1)

This assumes the compression flange for the steel beam is adequately restrained against buckling by the steel decking and the steel section used can be classified as a plastic or compact section as defined in EC3 sections 5.5 and 5.6

y

Ed

yplf

MW =,

16

Design Procedure


At ULS Shear

The shear (as usual) is considered to be carried by the steel beam alone at the construction stage and also for the final composite stage.

The ultimate shear strength of a rolled I-beam is based on the following shear area Aw

Aw = Aa – 2btf + (tw + 2r)tf but not less than ηhwtw

where Aa is the cross sectional area of the steel beam and hw is the overall height of the web. η can be taken as 1.0

The other dimensions are shown in the next figure.

17

Design Procedure


At ULS Shear

18

Design Procedure


At ULS Shear

However, you can still take the shear area conservatively as

the web area

Av = d x tw

where d is the depth of the straight portion of the web.

The design plastic shear resistance is:3

,

MO

yv

Rdpl

fAV

γ=

MOγ = 1.0

19

Design Procedure


At SLS Deflection

Deflections at midspan are calculated, i.e.

EI

wL

384

5 4

=δ

These deflections at the construction stage due to the permanentloads are locked into the beam as the concrete hardens.

20

Design Procedure

Example => Design of steel beam for construction loads

21

Design Procedure


Steel strength and Classification (EC3 tables 3.1 and 5.2)

Web thickness tw = 9mm

Flange thickness tf = 14.4mm

Both < 40mm

From EC3 section 3.2 table 3.1, yield strength fy = 355 N/mm2

From EC3, section 5.6, table 5.2

3.58723.450.9

6.40781.0

235=×<===== εε

wy t

d

t

c

f

Therefore the steel section is class 1

22

Design Procedure


Loading at Construction=>

Average depth of concrete slab and ribs = 90 + 50/2 = 115 mm

Weight of concrete =0.115 x 25 x 3 =8.62 kN/mSteel deck =0.15 x 3 =0.45 kN/mSteel beam =74 x 9.81 x 10-3 =0.73 kN/m

Total dead load =9.8 kN/m

Imposed construction load = 0.75 x 3 =2.25 kN/m

Ultimate load = 1.35Gk + 1.5Qk = (1.35 x 9.8 + 1.5 x 2.25) = 16.6 kN/m

23

Design Procedure


Bending.

Maximum bending moment = wL2/8 = (16.6 x 92)/8 = 168 kNm

Moment of resistance of steel section = Wpl,yfy = 1653 x 355 x 10-3

= 587kNm > 168 kNm OK

24

Design Procedure


Shear

Maximum shear force V = wL/2 = 16.6 x 9/2 = 74.7 kN

Shear resistance of section =

Using conservative approach, with web depth d = 407.6mm, web thickness t=9mm

Av = dtw = 407.6 x 9 = 3.67 x 103 mm2

Shear resistance of section =

(Capacity should be considerably larger here - OK)

3,

MO

yv

Rdpl

fAV

γ=

kNkNV Rdpl 7.747521030.1

3551067.3 33

, >=××

××= −

25

Design Procedure

Composite Section at ULS

Check moment capacity and shear strength. Define tensile and compressive strength of the elements as follows (no major difference to BS approach):

Resistance of the concrete flange Rcf = 0.567fckbeff(h-hp)Resistance of the steel section Rs = fyAa

Resistance of the steel flange Rsf = fybtfResistance of overall web depth Rw = Rs-2Rsf

Resistance of clear web depth Rv = fydtwResistance of the concrete above the neutral axis Rcx = 0.567fckbeffxResistance of the steel flange above the neutral axis Rsx = fybx1

Resistance of the web over distance x2 Rwx = fytwx2

26

Design Procedure


General Section dimensions =>

(depth between fillets)

27


Neutral axis in the concrete flange = >

28


Neutral axis in the steel flange = >

29


Neutral axis in the steel web = >

30


Equations for Moment Capacity =>

( ) ( )sf

fcfspcfasc

R

tRRhhRhRM

422

2−

−+

+=

( )v

cfpacf

scR

dRhhhRMM

42

2

−++

+=

Neutral axis in concrete flange =>

Neutral axis in steel flange =>

Neutral axis in steel web =>

( )

−

−+=22

p

cf

sa

sc

hh

R

Rh

hRM

31


Example – Moment of Resistance of Composite Section

32



(1)From first principles:

Resistance of concrete flange:

Rcf = 0.567fckbeff(h-hp) = 0.567 x 25 x 3000 x (140 – 50) x 10-3 = 3827kN

Resistance of steel beam: Rs = fy Aa = 355 x 9460 x 10-3 = 3358kN

As Rs < Rcf, the neutral axis is within the concrete flange:

Neutral axis depth: 0.567fckbeffx = Rs = 3358kN

Therefore: x = (3358 x 103)/(0.567 x 25 x 3000) = 79mm

33



Moment of resistance =>

Lever arm z to the centre of the steel section is:

z = (ha/2 + h - x/2) = 457/2 + 140 – 79/2 = 329 mm

Therefore

Mc = Rsz =3358 x 329 x 10-3 = 1105 kNm

34



(2) Alternatively, using the design equations =>

( ) ( )kNm

hh

R

Rh

hRM

p

cf

sa

sc 1105102

50140

3827

3358140

2

4573358

22

3 =

−

−+=

−

−+= −

Nothing has really changed here, the basic equilibriumequations are the same

35


Shear strength VRd of the composite section

The resistance to vertical shear is assumed to be taken by the steel beam alone (as for the construction stage).

Shear resistance =>

Shear area Av is given by:

(For most cases this is conservatively taken as the web area, d x tw)

3,

MO

yv

Rdpl

fAV

γ=

Av = Aa – 2btf + (tw + 2r)tf

36


Up to this stage everything is (almost) as

before in terms of the design procedures, with

a very few formula changes to EC

37


Design of Shear Connectors

Design shear resistance PRd of automatically welded shear studs is the lesserof the following two equations (cl. 6.6.1.3):

38


Design of Shear Connectors

39


Design of Shear Connectors – reduction factors (orientation of steel sheeting)

Further reduction factors kl and kt to be applied to PRd (CL 6.6.4)

Depends on whether the ribs of the profiled sheet are parallel or transverse

to the supporting beam.

For ribs parallel to the supporting beam:

For ribs transverse to the supporting beam:

40



There is an upper limit kt,max given in table 6.2 of EC4

41



There is an upper limit kt,max given in table 6.2 of EC4:

42


Degree of Shear Connection – Full Connection

The change in horizontal shear (between zero and maximum moment)

will be the lesser of Rs or Rc.

To develop the full moment of resistance of the composite section,

the number of shear connectors nf required over the half span is the lesser of:

where

43


Degree of Shear Connection – Partial Connection

Sometimes the full shear connection is not required to provide adequate

moment capacity. Hence the number of shear connectors can be reduced

providing partial shear connection (normally for simpler stud layout/detailing)

Degree of shear connection =>

n is the number of shear connectors for full shear connection over a length of beam

nf is the number of shear connectors provided in that length

44


Degree of Shear Connection – Partial Connection

Cl 6.6.1.2 provides limits to the degree of shear connection according to the distance

Le, the distance in sagging between the points of zero moment.

1. The nominal diameter d of the shank of the headed stud is within the range

16mm < d < 25mm, and the overall length of the stud after welding is > 4d

2. The nominal diameter d of the shank of the headed stud is d=19mm

and the overall length of the stud after welding is > 76mm

For Le <25

For Le <25

( ) 4.003.075.0355

1 ≥−

−≥ ηη e

y

Lf

( ) 4.004.000.1355

1 ≥−

−≥ ηη e

y

Lf

45


Ultimate Moment of Resistance – Partial Shear Connection

Other conditions are presented in Cl 6.6.1.2, but the moment capacity for partial

shear connection is derived from stress blocks.

In the analysis the depth of the concrete stress blocks sq is:

effck

q

qbf

Rs

567.0=

Where Rq is the shear resistance of the studs provided.

46



1. Neutral axis in the steel flange h < x < h+tf : Rs > Rq > Rw

47



2. Neutral axis in the steel web x > h+tf : Rc < Rw

48



The moment capacities are:

Neutral axis in steel flange: (a)

Neutral axis in steel web: (b)

( ) ( )sf

fqs

cf

pq

q

as

cR

tRR

R

hhRhR

hRM

422

2−

−

−−+=

( )v

q

cf

pqa

qscR

dR

R

hhRh

hRMM

422

2

−

−−++=

49



Good News – can still use the linear interaction method (method b)

50


Concentrated Loads – Shear Connection

Shear connectors need to be spaced closer together between the

concentrated load and adjacent supports.

Number of shear connectors between

Load and adjacent support

Where:

Nt is the total number of shear connectors required between the support and

the point of maximum moment

Mi is the bending moment at the concentrated load

Ms is the capacity of the steel member

Mc is the moment capacity of the composite section

( )( )sc

sit

iMM

MMNN

−

−=

51


Concentrated Loads – Shear Connection

52


Example – Partial Shear Connection

Using previous example for the composite section ULS =>

Span = 9 m

80 shear studs in pairs at 225 mm (19mm dia, 100mm height)

Wpl,y = 1653 N/mm2

fy = 355 N/mm2

fck = 25 N/mm2

Calculate the degree of shear resistance and moment of resistance of thecomposite section based upon the shear connectors provided.

53



Design shear resistance PRd is the lesser of the equations in Cl.6.6.1.3, withfu = 450 N/mm2

kNdf

Pv

u

Rd 7.811025.1

4/194508.04/8.0 322

=××××

== −π

γ

π

Calculate the reduction factor kt (transverse sheeting) with an upper limit of 0.8 from table 6.2 EC4 for the profiled sheeting

79.0150

100

50

80

2

7.01

7.0 0 =

−×=

−=

p

sc

pr

th

h

h

b

nk

54



Hence the design shear resistance PRd of a stud = 0.79 x 81.7 = 64.5 kN

For full shear connection, the number of studs required over the half span is:

525.64

3358===

Rd

s

fP

Rn

Hence for full shear connection, the total number of studs required overthe whole span = 104

55



The degree of shear connection, η, is

77.0104

80==η

( ) ( ) OKLf

e

y

77.052.0903.075.0355

355103.075.0

3551 <=×−

−=−

−≥η

The lower limit for η is calculated from:

56



Use the linear interaction method to calculate the moment of resistance Mp

For the steel beam

Hence

( ) sscp MMMM +−= η

kNmfWM yypls 587103551653 3

, =××== −

( ) kNmM p 986587587110577.0 =+−×=

57


Longitudinal Shear Resistance for Steel Sheeting

Use either:

1) m-k method

2) Partial interaction method

Both deal with parameters concerned with empirical behaviour

(i.e. lots of tests in the lab!)

58



For the m-k method Cl. 9.7.3 applies (mechanical and frictional interlock)

The maximum design vertical shear in a slab of width b should not exceedthe design shear resistance given by:

59



The partial interaction method should only be used when the longitudinalshear failure is ductile (EC4 contains caveats).

60


Vertical Shear Resistance

For partially anchored sheeting, vertical shear resistance is provided by

Equation 6.2b) in EC2 (for resistance of members not requiring designshear reinforcement =>

Normally σcp can be taken as zero for simply supported members

21

ck2

3

min 035.0 fkv =

61


Transverse Reinforcement

Transverse reinforcement is supplied to resist the longitudinal shear in a flanged beam, and follows the variable strut inclination method as outlined in EC2

Uses the rules in Cl.6.2.4 EC2

Longitudinal shear stress =

62



63



∆x is half the distance between point of maximum moment and zero moment

Transverse reinforcement per unit length:

For compression flanges, 26.50 < θ <450 ,

so for minimum reinforcement take θ = 26.50

64



To prevent crushing of the concrete flange, the following must be satisfied:

(6.22)

65

Deflections

1) Calculate deflections at the construction stage

Deflection due to permanent load plus variable loading is checked.Permanent load deflection is locked into the beam at this stage

2) Calculate deflections for the composite stage based on a transformedsection (not shown, refer to Ref 1 for example)

Deflection due to composite stage plus construction stage gives totalDeflection.

Check < span/250 (for longer spans a pre camber can be incorporated)

66

Example:

Simply Supported Composite Beam

6m Long Composite Beam, UDL at 3m centres:

67

Example:


Profiled Sheeting => Comflor 51 (Corus 2002)

68

Example:


Design Data:

Beam Spacing => L = 6mBeam Spacing => s = 3mTotal Slab Depth => h = 130 mmDepth of Concrete above profile => hc = 79 mmDeck Profile Height => hp = 51 mmWidth of the bottom trough => bbot = 122.5 mmWidth of the top trough => btop = 112.5 mm

Average width of rib = 30mm (equates to 6.55 ribs per metre)

69

Example:


Shear Connectors:

Diameter => d = 19 mm

Overall height before welding =>hsc = 100mm

Height after welding => 95 mm

70

Example:


Material Properties of Steel (EN1993-1-1, Table 3.1)

Structural Steel:

Assume S275, maximum thickness < 40mm

Yield Strength, fy = 275 N/mm2

Ultimate Strength, fu= 430 N/mm2

Steel Reinforcement:

Yield Strength fy = 500N/mm2

71

Example:


Material Properties of Concrete (EN1992-1-1, Table 3.1)

Normal Weight concrete, strength class C25/30

Wet Density => 26 kN/m2

Dry Density => 25 kN/m2

Cylinder strength => fck = 25 N/mm2

Secant modulus of elasticity => Ecm = 31 kN/mm2

72

Example:


Actions – Permanent Loading (gk)

Self weight of concrete slab =>

Wet UDL => [(130x1000)-(7x30x51)]x26x10-6 = 3.10 kN/m2

Dry UDL => [(130x1000)-(7x30x51)]x25x10-6 = 2.98 kN/m2

Construction stage Composite stage

Concrete slab => 3.10 kN/m2 2.98 kN/m2

Steel Deck => 0.15 kN/m2 0.15kN/m2

Steel Beam=> 0.2 kN/m2 0.2 kN/m2

-----------------------------------------------------------------------------------------Totals => 3.45 kN/m2 3.48 kN/m2

73

Example:


Actions – Variable Loading (qk)

Construction stage:

Construction loading => 0.5kN/m2

Composite stage:

Floor Loading => 3,8kN/m2 (from structural arrangement)

74

Example:


Ultimate Limit State – Construction Stage

Combination of actions (EN1990, Eqn 6.10b)

Distributed load on beam (0.85 x 1.35 x 3.45) + (1.5 x 0.5) = 4.71 kN/m2

Total Load Fd = 4.71 x 6.0 x 3.0 = 84.78 kN

Maximum design moment at midspan

My,Ed = FdL/8 = 84.78 x 6 / 8 = 63.59 kNm

Maximum shear => beam assumed to take shear from composite loading, therefore not calculated here.

75

Example:


Ultimate Limit State – Composite Stage

Combination of actions (EN1990, Eqn 6.10b)

Distributed load on beam (0.85 x 1.35 x 3.48) + (1.5 x 3.8) = 9.69 kN/m2

Total Load Fd = 9.69 x 6.0 x 3.0 = 174.42 kN

Maximum design moment at midspan

My,Ed = FdL/8 = 174.42 x 6 / 8 = 130.82 kNm

Maximum design shear

VEd = Fd/2 = 174.42 / 2 =87.21 kN

76

Example:


Partial Factors for Resistance

Structural Steel (EN1993-1-16.1(1)) γM0=1.0

Concrete (EN 1992-1-1, Table 2.1N) γC=1.5

Reinforcement γS=1.15

Shear connectors (EN 1994-2.4.1.2) γv=1.25

Longitudinal shear γvs=1.25

77

Example:


Steel beam – trial sectionRequired Plastic Modulus (construction stage loading) =>

Wpl,y = My,EdγM0/fy = 63.6x103x1.0/275 = 231cm3

Try => 254x102x22 UKB, S275 Wpl,y = 259cm3

From Section Tables =>

ha = 254.0 mm tf = 6.8mm

b = 101.6 mm r = 7.6mm

d = 225.2 mm Aa = 28cm3

tw = 5.7mm Wpl,y = 258cm3

Elastic Modulus E = 210kN/mm2

Section is Class 1 under bending (EN 1993-1-1,3,2,6(1))

78

Example:


Composite stage member resistance checks

Compressive resistance of concrete slab:

Effective width at midspan of compression flange (EN1994 5.4.1.2) =>

beff = b0 + Σ bei ; bei = Le/8 = L/8 = 6/8 = 0.75m (for SS beam)

Assume single shear studs, hence b0 = 0m

beff = 0 + (2 x 0.75) = 1.50m<3m (beam spacing)

Compression resistance (EN1994 6.2.1.2) =>

Nc,slab = 0.85fckbeffhc/γc = 0.85x25x1500x79x10-3/1.5 = 1679 kN

79

Example:



Tensile resistance of steel section:

Npl,a = fdAa =fyAa/γM0 = 275 x 28 x102/1.0 = 770 kN

Since Npl,a < Nc,slab, the plastic neutral axis lies in the concrete flange

Design bending resistance - full shear connection (EN1994 6.2.1)

kNmh

N

Nh

hNM c

slabc

apla

aplRdpl 184102

79

1679

770130

2

254770

22

3

,

,

,, =×

×−+=

×−+= −

80

Example:



Bending moment at mid span My,Ed = 131 kNm

Hence, design bending resistance of the composite beam is adequate

0.171.0184

131

,

,<==

Rdpl

Edy

M

M

81

Example:


Shear connector resistance (EN1994 6.6.3.1)

Design shear resistance of a single shear connector is the smaller of:

andv

cmck

Rd

EfdP

γ

α 229.0=

( )v

u

Rd

dfP

γ

π 4/8.0 2

=

26.519

100==

d

hsc0.4>

d

hsc 0.1=αAs Then

82

Example:


Shear connector resistance

So

or

As 73.7 kN < 81.7kN, Hence PRd = 73.7 kN

kNPRd 7.731025.1

103125190.129.0 332

=×××××

= −

( )kNPRd 7.8110

25.1

4/194508.0 32

=××××

= −π

83

Example:


Influence of deck shape (EN1994 6.6.4.2)

Ribs are transverse to beam,

Also assume 1 stud per trough, nr = 1.0

Reduction factor

Hence, Kt =1.0, no reduction in shear resistance i.e. PRd = 73.7 kN

0.117.0 0 ≤

−

=

p

sc

pr

th

h

h

b

nk

48.1151

100

51

5.112

1

7.0=

−

=tk > 1

84

Example:


Number of shear studs in half span

Using 1 shear connector per trough =>

Stud spacing along the beam = 152.5 mm

Assuming either a primary beam width or column width of 254mm say, then:

( )18

5.152

2/2543000=

−=n Shear stud connectors

per half span

85

Example:


Degree of shear connection

Hence, full shear connection is provided, and no reduction in bending resistance is required.

kNPR Rdq 13277.731818 =×==

0.17.1770

1327

,

>==apl

q

N

R

86

Example:


Shear buckling resistance of the uncased web

For unstiffened webs if the shear buckling

resistance of the web should be checked.

Where:

εη

72>

t

hw

92.0275

235235===

yfε

With η = 1.0 ( ) mmthh 4.2408.622542 faw =×−=−=

2.6692.00.1

7272=×

=ε

η2.42

7.5

4.240

w

ww ===t

h

t

h

As 42.2 < 66.2 the shear buckling resistance of the web does not need to be checked.

87

Example:


Resistance to vertical shear

Shear resistance of the composite beam is:( )

3

M0

yvRda,pl,Rdpl,

γ

fAVV ==

( )rttbtAA 22 wffv ++−= wwthη

[ ])6.72(7.58.6)8.66.1012(2800v ×+×+××−=A 1560v =A

For rolled I and H sections loaded parallel to the web:

but not less than

mm2

0.1=η 13707.54.2400.1ww =××=thη(Conservatively)

1560 mm2 > 1370 mm2 Therefore, Av = 1560 mm2 247kN 100.13

2751560 3-

Rdpl, =××

×=V

88

Example:


Resistance to vertical shear

35.0247

87

Rdpl,

Ed ==V

V

< 1.0, Therefore the design resistance to vertical shear is adequate.

As there is no shear force at the point of maximum bending moment (mid span) no reduction (due to shear)in bending resistance is required

89

Example:


Design of Transverse Reinforcement

f

ydsf

s

fA

θcotfEdhv

f

sf

s

Aθcotyd

fEd

f

hv

The area of reinforcement (Asf) can be determined using the following equation:

therefore,

Neglect the contribution of the decking and check the resistance of the concrete flange to splitting:

> >

4351.15

500

s

yyd ===

γ

ff

where: hf is the depth of concrete above the metal decking, therefore, hf = hc = 79 mm

For compression flanges 26.5° ≤ θ ≤ 45°

N/mm2

90

Example:



The longitudinal shear stress is the stress transferred from the steel beam to the

concrete. This is determined from the minimum resistance of the steel, concrete and

shear connectors. In this case, the plastic neutral axis lies in the concrete flange,

and there is full shear connection, so the plastic resistance of the steel section to

axial force needs to be transferred over each half-span. As there are two shear

planes (one on either side of the beam, running parallel to it), the longitudinal shear

stress is:

83.13000702

1000770

2 f

apl,

EdL, =××

×==

xh

Nv

∆N/mm²

91

Example:



For minimum area of transverse reinforcement assume θ = 26.5°

f

sf

s

A14710

6.52cot435

7983.1

cot

3

yd

fEdL,=×

×

×=

θf

hv≥

Therefore, provide A193 mesh reinforcement (193mm2/m) in the slab.

mm2/m

92

Example:


Crushing of the concrete flange

Where

ffcdEdL, cossin θθνfv ≤

25016.0 ckf

- 54.0250

25 -16.0 =

59.35.26cos5.26sin5.1

2554.0cossincd =×××=fff θθν

83.1EdL, =v

Verify that: fcd = fck/γc

ν =

N/mm²

N/mm² < 3.59 N/mm²

Therefore the crushing resistance of the concreteis adequate.

where

=

93

Example:


Serviceability limit state

Performance at the serviceability limit state should be verified. However, no verification is included here. The National Annex for the country where the building is to be constructed should be consulted for guidance.Considerations would be:

Short-term, long-term and dynamic modular ratiosServiceability combinations of actionsComposite bending stiffness of the beamTotal deflection and deflection due to imposed loadsStresses in steel and concrete (to validate deflection assumptions)Natural frequency.

Composite Construction Design (ULS Only)

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Transcript of Composite Construction Design (ULS Only)