Chapter 4 Steel and concrete composite beamsSimply supported composite beams钢-混凝土组合梁——简支组合梁

By Professor Chen ShimingLecture Notes for Presentation

2015

4.1 Introduction• In buildings/bridges it is common to find concrete

slabs supported on steel beams. If slip at the interface is eliminated, or at least reduced, the slab and steel member will act together .

• In practice, such action is achieved by studs(栓钉）

or other connectors which are welded to the structural steel and embedded in the concrete slab.

• A composite beam has greater stiffness and generally a higher load capacity than its non-composite counterpart. Consequently, a smaller steel section is usually required.

Composite beams（组合梁截面形式）

Steel encased composite sections

Prestressed Composite Beams（预应力组合梁）

Prestressing of composite steel girders with tendon体外预应力组合梁

Some shear connection devices

4.2 Beam response and failure modes• Simply supported beams are subjected to positive

(sagging) moment and shear

Typical failure modes for composite beam: critical sections.

Potential shear failure planes.

4.3 Construction methods（施工方式）

• Construction methods / affect structural performance

• ‘Propped construction（有支撑） ’ to support the steel beam until the concrete hardened, (propped at quarter-span points and mid-span); Props are usually left in place until the concrete slab has developed three-quarters of its design strength.

• ‘unpropped construction （无支撑） ’ casting of concrete without props; Initially, the steel beam alone resists its own weight, and that of the formwork, wet concrete, and placement loads.

Bending moment relationship for unshored (curve A) and shored (curve B) composite beams and steel beam (curve C).

• The construction procedures can affect the structural behavior of the composite beam施工方法会影响组合梁

受力变形.

• In the case of the unshored construction, the steel section has to be checked for all the possible loading condition arising during construction.

• In the case of shored constructions, the overall load, including self weight, is resisted by the composite member.

4.4 Effective width of concrete slab混凝土板有效翼缘宽度

• A typical compositeconstruction consists ofa slab connected to aseries of parallel steelmembers, and theconcrete flange widthmay not be fullyeffective in resistingcompression due to‘shear lag’（剪力滞）

Effect of shear lag（剪力滞）

• Effective width is adopted as a simplification of the shear lag problem. The effective width be is defined as ‘that width of slab which acted on by the actual maximum stress would cause the same static effect as the variable stress which exists in fact’.

∫−

=2/

2/

max

b

b

effx bdx σσ

dimensions of the system, but also on the type of loading, the support conditions and the cross section considered.

the ratio be/B depends not only the relative

• Effective width of concrete slab

• In most Codes of practice, very simple formulae are given for the calculation of effective width.

• EC4 （欧洲组合结构规范）proposed that the effective width on each side of the steel web should be L/8, but not greater than half the distance to the next adjacent web, not greater than the projection of the cantilever slab for edge beams.

• In Chinese design practice（中国规范） of composite beams, similar effective width is adopted given as the following:

• where b1, b2 are effective width on each side of the steel web, be L/6,

021 bbbbe ++=

0 1 2eb b b b= + +Effective width of concrete slab

b0 determination

4.5 Elastic analysis（弹性分析）

• For the service limit state（正常使用极限状态）

• Checks must be made to ensure that at service loads, its deflection is not excessive and the stresses are within the elastic range.

• When elastic theory is used, it may be necessary to include the effects of creep and shrinkage of concrete, and of temperature, if accurate results are required.

Properties of concrete（混凝土特性）

• Time dependence （时变特性）of concrete properties, i.e., shrinkage and creep, should be considered when determining the response of composite structures under sustained loads（持续作用荷载), with particular reference to member stiffness.

• A value of the modular is defined to account for the creep effect in the analysis:

Full-interaction elastic analysis of cross-sections（transformed section换算截面）

• Assume that plane section remain plane, and tensile strength of concrete is neglected.

• The composite section to be replaced in the analysis by an equivalent all-steel cross-section. The breadth of the equivalent steel slab depends upon the modular ratio, αe, defined by:

where Ec is elastic modulus of concrete and Es is elastic modulus of steel.

cse / EE=α

Equivalent Steel slab

• the cross-sections remain plane • interface slip is neglect• concrete tensile strength is neglected• Theory of the transformed sections

Elastic stress distribution with neutral axis in slab.

Elastic stress distribution with neutral axis in steel beam.

“elastic moment of resistance” as the moment at which the strength of either structural steel or concrete is achieved

• “elastic moment of resistance” as the moment at which the strength of either structural steel or concrete is achieved

4.6 Ultimate strength , Plastic analysis（塑性极限分析） (sagging moment region)Possible failure modes:

• Failure of the composite member in flexure by the development of a plastic hinge;

• Reduction or complete loss of composite action, leading to collapse of the member in flexure; due to excessive slip or complete failure of the shear connection;

• Local shear failure in the slab in regions of high stress around shear connectors

• Failure in vertical shear. When designing composite beams, it is assumed that vertical shear is resisted by the structural steel member acting alone.

• Failure in longitudinal shear.

Strength criteria for simply supported beam: I-positive bending moment,

II-vertical shear

III-longitudinal shear

Plastic Analysis（塑性分析，基本假定：）

• The shear connectors are able to transfer the forces occurring between the steel and the concrete at failure (full shear connection).

• No slip occurs between the steel and the concrete (complete interaction).

• Tension in concrete is neglected.• The strains caused by bending are directly proportional

to the distance from the neutral axis; or plane cross-sections remain plane after bending, even at failure

• The relationship between the stress σa, and the strain εaof steel; The relation between the stress σc and the strain εc of concrete.

relationship between the stress and the strain

• When the plastic neutral axis lies in the slab :

• When the plastic neutral axis lies in the steel :

Vertical shear resistance• The vertical shear resisted by a composite beam

should satisfy the following requirement:

where t w — thickness of steel webh w — clear depth of steel webf v — design shear strength of

The influence of vertical shear on the momentresistance is neglected here.

vww fhtV ≤

Vertical Shear in Combination with Bending Moment

Figure : Resistance in bending and vertical shear

Load-slip behaviour of the connector

In the case of simply supported beams, it is easy to determine the distribution of bending moments from the equilibrium conditions. Plastic behaviour is assumed in the stress distribution over the cross-section.

Longitudinal shear resistance and transverse reinforcement

• To achieve composite action, force must be transferred from the shear connection to the full effective width of the slab, so that shear failure of concrete surrounding the connectors must be prevented.

• Check resistance of the slab to longitudinal shear

• A slab must have adequate transverse reinforcement to transmit the stress from the connectors and ensure that there is no risk of premature failure of the concrete due to longitudinal shear.

• .The design shear per unit length, VSc1

must not be greater than the shear resistance VRd of the failure surface .

DESIGN CALCULATION • Ultimate Limit State • To check the resistance of the critical cross-sections

Sd ≤Rdwhere Sd is the design value of an internal force or moment; Rd is the design value of the resistance.

• Serviceability Limit State• under service conditions, the deflections and vibrations

do not exceed allowable values and that cracking of the concrete is limited.

Ed ≤Cd

Creep, shrinkage and temperature effect

• Under sustained compressive stress, concrete undergoes a time-dependent increase in strain. n as ‘Creep’ is a function of the age at which loading is applied, its duration and intensity, the relative humidity of the environment.

• In composite beams, creep of the concrete causes an increase in deflection, and accompanied by a relaxation of stress in the concrete and an increase of stress in the steel member.

• Concrete also undergoes long-term deformation due to shrinkage, causing additional deflection in the composite beam. Shrinkage occurs irrespective of whether the concrete is subjected to stress.

• Shrinkage depends mainly on the relative humidity of the environment. In the fairly dry environment of a building, an unrestrained concrete slab could be expected to shrink by 0.03% of its length, or more. In a composite beam, restrained by the steel member which exerts a tensile force on it through the shear connection, so shrinkage in the concrete slab is less than the ‘free’ shrinkage.

• In a simply supported beam, the shrinkage causes sagging curvature, with compressive stress in the upper part of the steel section.

Calculations for deflections• Deflection of a composite beam is checked by elastic

theory, and the effects of creep should be included in theelastic analysis, by using a reduced modulus of elasticityfor concrete, Ec, under sustained loading. Saying:

• By transformed cross-section method, the secondmoment of area of the equivalent all-steel section iscalculated. The equivalent breadths of concrete slab arebe /αe, and be /2αe for the short term load and long termload respectively, where αe is modular ratio.

φ+=

11 cc

EE

The use of Limiting span/depth ratios

• An alternative to calculation, deflections can be controlled by limiting span-depth ratios.

• For simply-supported composite beams, the maximum deflection due to distributed load should not exceed span/250. These are given in terms of the density of concrete, the ratio of depth of slab to depth of steel member, the ratio of area of concrete to area of steel, and the grade of structural steel.

• There are also modification factors to allow for various degrees of end fixity, to ensure lower deflection ratios in long spans and for beams where the resistance to bending exceeds that required at the ultimate limit state.

4.7 Partial-interaction design（部分抗剪连接设计）

• When the number of shear connectors provided, N, is such that the degree of interaction between the slab/steel beam is just sufficient to provide the required flexural strength, and is less than the number N f required for a full-interaction design（完

全抗剪组合梁）.• If the number of shear connectors is reduced below

that required, the effect of slip on stresses and deflection can no longer be neglected.

• influence on bending moment strength and on deflection

At about 50% connection the ductility of the connectors becomes important and, in practice, the fully plastic moment can only be achieved by the provision of 1/2 as many again shear connectors. This is shown in the line C to D and E to F to G when strain hardening of the steel beam is taken into account. The dotted line in the figure provides a connection ratio of about 1.25 that virtually guarantees the full plastic moment capacity of the beam. Hence the 0.8 value in the determination of stud strength.

The resisting moment depends upon the number of connectors between any point and the support. In order to overcome this the 0.8 reduction factor artificially ensures that the actual resisting moment is 25% higher than the design resistance moment as shown in the figure.

• The method is applicable only to simply supported beams not subjected to heavy point loads.

• For the ultimate state: For the deflection

)( ff

papau MMNNMM −+= )]/(1)[(5.0 ffsf NN−δ−δ+δ=δ

4.8 Worked examples• Example No.1 Simply-supported beam

Design Data: Span L = 12.0 m; spacing of steel beam3.6 m; concrete depth hc= 125 mm

• Design strength for structural steel fs = 330 N/mm2

• Concrete grade = C30Loading: Design values for ultimate limit state:• Dead load = 21.0 kN/m; Imposed load = 29.0 kN/mBending moment: Ultimate design moment M = wL2/8 =

(21+29)×122/8 = 900 kNmEffective breadth of concrete flange be

be is the lesser of : b = 3.6 m or L/3 = 4.0 m

• (L/6 on each side of the steel web, but not greater thanhalf the distance to the next adjacent web)

then be = 3.6 mSelect 457×191×82UB (rolled) (with As = 104.5 cm2)• Assuming that the neutral axis is in the slab (assuming

that fc = 0.4 fcu) , then:

As xp ≤ 125 mm, the neutral axis lies in slab.• Ultimate moment of resistance Mu:• Mu = Asfy (D/2 + hc – xp/2 ) = 104.5×102×330×(460/2

+125 –79.8/2) = 1087 kNm• ∴ adopt 457×191×82UB in Grade 50 steel.

cue

ysp fb

fAx

4.0= mm8.79

3036004.0330105.104 2

=××××

=

• Example No.2 Design of composite beams:further topics on beam of Example No.1Elastic properties of composite section:

• Modulus of elasticity for steel: Es = 205 kN/mm2

• Short term modulus of elasticity for concrete grade C30:Ec = 28 kN/mm2

∴ short term modulus ratio: αE = Es / Ec = 205/28 =7.3• Long term modulus of elasticity for concrete grade C30:Ec’ = Ec /2 = 28/2 =14 kN/mm2

∴ long term modulus ratio: αE = Es / Ec’ = 205/14 = 14.6Second moment of area for short-term loading

Assume neutral axis lies in slab∴ bexe

2 + 2 αEAsxe –2 αE Asd =0

i.e 3600×xe2+2×7.3×10450×xe- 2×7.3×10450×

(460.3/2 + 125 ) = 0xe = 103.3 mm; As xe < 125, xe does lies in slab

Second moment of area for long-term loadingAssume neutral axis lies in steel section, then:

unit)steel(incm121479

)3.1032

2.460(1045010371033.73

3.1033600

)(3

4

243

23

=

−+×+××

=

−++α

= essE

ee xdAIxbI

mm7.1361253600104506.14

2/1263600)1252/460(104506.14

2/

2

/

2/

=×+×

×++×=

+α+α

=cesE

cesEe hbA

hbdAx

• As xe > 125, neutral axis does lie in steel

• Loading: Values for serviceability limit stateSelf weight of slab and steel section = 11.5 kN/mFinish = 3.6 kN/m

=15.1 kN/mImposed load (live) =18.0 kN/m

unit)steelin(cm107931

)7.1361252

2.460(10450

1037103)5.627.136(6.141253600

6.14121253600

)()2/(12

4

2

423

22//

3

=

−++

×+−×

+××

=

−++−α

+α

= essceE

ce

E

ce xdAIhxhbhbI

Calculation of deflection• a) Assuming propped construction

Dead load deflection =

Imposed load deflection =Total deflection = 18.4 + 19.5 = 38 mm = span /316b) Assuming unpropped construction

construction load of 11.5 kN on steel section

deflection =

mm4.1810107931205384

12000)121.15(54

3

=×××

××

mm5.1910121479205384

12000)1218(54

3

=×××

××

mm8.401037103205384

12000)125.11(53

3

=×××

××

Dead load deflection on composite section due to finishes

=Imposed load deflection as (a) = 19.5 mmTotal deflection = 40.8 + 4.4 + 19.5 = 65 mm = span /185Calculation of stress at serviceability limit state

Bending moment due to :Self-weight of slab and steel section =Finishes =

Imposed =

mm4.410107931205384

12000)126.3(54

3

=×××

××

kNm2078

125.11 2

=×

kNm658

126.3 2

=×

kNm32481218 2

=×

a) Assuming propped constructionMaximum tensile stress in steel=

= 113 +129 = 242 N /mm2

Maximum compressive stress in concrete=

b) Assuming unpropped constructionMaximum tensile stress in steel

)3.1031252.460(10121479

10324)7.1361252.460(10107931

10)65207(4

6

4

6

−+×

×+−+

××+

24

6

4

6

mm/N2.68.34.2101214793.7

3.10310324101079316.14

7.13610)65207(=+=

××××

+××××+

)3.1031252.460(10121479

10324

)7.1361252.460(10107931

106510161210207

4

6

4

6

3

6

−+×

×+

−+×

×+

××

=

= 128 + 27 +129 =284 N/mm2

Maximum compressive stress in concrete= 0.6 + 3.8 = 4.4 N/mm2

Example No.3 Design of shear connectionContinuation of Example No.1Proposed shear connectorsMinimum overall height ≮ 100 mmTry 19 mm diameter headed studs 100 mm longNominal static strength in grade 30 concrete: Pu = 100kNDesign static strength at ultimate limit statePd = 0.8 Pu / γ m = 0.8×100/1.1 = 72.7 kN

=××

××+

××××

4

6

4

6

101214793.73.10310324

101079316.147.1361065

Number of connectorsNumber per half-span = = 48

∴ propose 24 pairs for each half-span spacedlongitudinally at 250 mm centers with end distance of125 mm. ≯ maximum permitted ( 600 mm, or 3×125 =375 mm, or 4×100 = 400 mm)∴ Adopt 19 mm φ headed studs 100 mm long in pairs,spaced longitudinally at 250 mm centers with enddistance of 125 mm.

3107.7233010450

÷×

=d

ys

PfA