Chapter 4.2

BS5950:Part 3: Section 3.1

Simply Supported Composite Beams

1

Summary: The procedures for determining the section classification are similar to those for bare steel

sections, although some modifications may be made. The moment resistance of class 1 and 2 sections is calculated using plastic analysis, the

details depending on the neutral axis position. The moment resistance of class 3 sections is calculated using elastic analysis, with due

account for creep and special consideration of buildings used mainly for storage. The vertical shear strength is based on that of the bare steel section. The details of the longitudinal shear connection (number and type of connector, and slab

reinforcement) are determined on the basis of the longitudinal force transmitted between the steel section and concrete slab.

Where insufficient connectors are provided, the beam may be designed on the basis of partial interaction, the moment resistance calculated on the basis of the longitudinal force transmitted between the steel section and concrete slab.

Deflection limits are as stated in EC3 for bare steel sections. Concrete cracking can be controlled by ensuring a minimum amount of slab reinforcement

and limiting bar size and spacing.

Notes: This material comprises one 60 minute lecture.

Objectives: To outline the design checks which are necessary for both ultimate and serviceability limit

states.

To describe the procedures for determining the section classification, and the modifications which may be made.

To explain the procedures for calculating the plastic moment resistance of class 1 and 2 sections in relation to neutral axis position.

To explain the procedures for calculating the elastic moment resistance of class 3 sections in relation to the method of construction, with special consideration of buildings used mainly for storage.

To describe the simplified procedures for checking the vertical shear strength of composite beams.

To explain how the longitudinal shear connection is designed. To introduce the concept of partial interaction, and describe how it affects the calculated

moment of resistance.

To outline the requirements for controlling deflections and concrete cracking at the serviceability limit state.

2

CONTENTS 1 Introduction 2. Advantages and Disadvantages of Composite Construction 3 Construction Methods 4 Effective Width 5 Shear Connection: Full and Partial Interaction 6 Moment Capacity 7 Moment Capacity at High Shear 8 Design of Shear Connections 8.1 Shear Stud Connectors in Solid Slab 8.2 Shear Stud Connectors in Composite Slab

8.3 Reduction Factor for Deck Shape 8.4 Full Shear Connection

8.5 Partial Shear Connection 8.6 Minimum degree of shear connection

9 Moment Capacity Mc with Partial Shear Connection Based on Plastic Theory 10 Check Capacity At Other Locations 11 Transverse Reinforcement 12 Composite Beams - Serviceability

12.1 Modular Ratio e 12.2 Second Moment of Area Ig 12.3 Elastic Section Modulus 12.4 Serviceability Stresses 12.5 Deflections 12.6 Deflection for Partial Composite Section

13 Examples

3

Composite Beams 1 Introduction Composite construction in buildings is covered by BS5950:Part 3: Section 3.1 which deals with the design of steel beams, usually I section, to act "compositely" with concrete or composite slab by use of shear connectors. If slip at the interface between concrete and steel is free to occur, each component will act independently, as shown in Figure 1. If slip at the interface is eliminated, or reduced, the slab and steel member will act together as a composite unit. The resulting increase in strength and stiffness will depend on the extent to which slip is prevented.

Figure 1 Non composite and composite beams

Various forms of composite section are shown in Figure 2.

Figure Variious forms of composite beams

4

Figure 3 (a) composite beam with composite slab (b) Composite beam supporting precast slab

One way of achieving a bond between beam and slab is to weld stud shear connectors on to the upper flange of the steel beam (Figure 3a). These form an anchorage for the concrete slab, preventing any movement in position between the underside of the slab and the beam flange. The concrete slab, which is necessary anyway to support area loads, acquires an additional function; it forms the compression chord of the composite cross-section. The tensile bending stresses are borne by the steel beam. A higher degree of stiffness thus ensures minimal deflection. 2. Advantages and Disadvantages of Composite Construction

Advantages Disadvantages Saving in Steel Weight up to

30 to 50% Require connectors

Greater stiffness leading to shallower steel beams for the same span

Complexity in Design

Increase floor stiffness Increased span length for a given member

Rapid construction 3 Construction Methods Propped Construction Propping is generally required to support the steel beam until the concrete has hardened. Propping at the quarter-span points and mid-span is generally adequate. The props are usually left in place until the concrete slab has developed three-quarter of

its design strength. Unpropped Construction In order to reduce the time of construction, "unpropped construction" is usually

preferred. Initially the steel beam alone resists its own weight, and that of the formwork, wet concrete and placement loads. Other loads are added later and so are carried by the composite member when the concrete is hardened.

5

Beam Span To Overall Depth It is normally found that strength and serviceability design limit states are just satisfied when the ratio of beam span to overall depth (including the concrete or composite slab) is between 18 to 22. This represents the optimum design of simple composite beams. 4 Effective Width A composite beam behaves structurally as one of the series of parallel T beams with thin wide flanges as shown in Figure 4. The effective width Be of the slab is defined which acts in conjunction with the steel beam. The concrete flange is in compression and the steel section is largely in tension when the beam is subject to sagging moment, as in simply-supported construction. The effective width Be is calculated as followed: 4.6 Secondary beams (The slab is perpendicular to the beam span)::

Be = 25% the beam length (L/4) but not greater than the beam spacing (bo) Primary beams (The slab & the beam span in the same direction): Be = 25% the beam length (L/4) but not greater than 80% the beam spacing (0.8bo) Edge beams: Half of the above values (L/8), as appropriate, plus any projection of the

slab beyond the centre-line of the beam. where bo = Average spacing of adjacent beams. L = Beam effective span length

e L/8 or bo /2

Be Be = smaller of L/4 or bo

bo

smaller of

FIGURE 4

6

Primary beamSecondary beams

L

bobo

bo

bo shown in the figure is the beam spacing for the secondary beam. For primary beam the beam spacing is L (i.e, bo = L) The effective widths are used in both strength and serviceability calculations. For the primary beam (i.e., slab spanning parallel to the steel beam) the factor 0.8 accounts for the co-existing effects of bending in the slab and beam where both span in the same direction and utilise a proportion of the compressive strength of the concrete. The effective span length is taken as the original span length for beams simple supported at both ends. In continuous beams L is taken as 0.8 x span length for end spans, or 0.7 x span length for internal spans.

5 Shear Connection: Full and Partial Interaction Full Strength shear connection: Sufficient shear studs are provided to develop the plastic strength of the beam. However, this may result in a large number of shear connections, particularly if the dimensions of the cross-section or the material properties are governed by factors other than the strength of the composite beam. Partial Strength Shear Connection: A economical design may be achieved in which the number of shear connections is such that the degree of interaction is just sufficient to provide the required flexural strength.

7

6 Moment Capacity Compressive and Tensile Capacities of Concrete and Steel The tensile capacity of the steel section is obtained from Clause 4.4.2 as: R As y= where A = Cross-sectional area of the steel section y = Design strength of the steel (BS5950:Part 1, Table 5) and it is dependent on the thickness and grade of steel. For deck laid perpendicular to the beam, the compressive capacity of the concrete slab over its effective width is obtained from Clause 4.4.2 as: R f B D Dc cu e s p= 0 45. ( ) where Be = the effective width of the slab fcu = Cube strength of concrete Ds = Overall slab depth Dp = Depth of the deck profile.

0.45fc

Ds - Dp

D Rs

Rc

STUDS

8

DpDs

Parellel DeckingPerpendicular Decking

DpDs

Concrete SlabSteel deck

Be

Concrete SlabSteel deck

Be

This equation also conservatively applies for cases where the deck is laid parallel to the beam. Plastic Moment Capacity of a Full Composite Beam B2.2 Only positive moment, full shear connection and compact steel beam, are considered in the following derivation. The plastic moment capacity of a symmetrical composite section, Mc, is derived based on the relative magnitudes of Rs and Rc which determines the position of the plastic neutral axis (PNA). Design equations, which are given in BS5950:Part 3 Appendix B, are presented in the following order using the case numbers as given in the code. Case 2b: R Rs c (PNA in concrete flange)

xD - Ds p

R s

D

D s D p

Be

R s

Tension =

Compression =

Neutral axis depth, x Equilibrium of forces Tension = Compression A f B xy cu e = 0 45. x

Af B

y

cu e=

0 45.

x RR

D Dsc

s p= ( ) because R f B D Dc cu e s p= 0 45. ( ) Taking moment about the top of the slab, and substituting for x

M R D D R xc s s s= + ( )2 2 = + R D D R RR D Ds s s sc s p2 2 ( )

9

where D = Depth of the steel beam, as shown in the figure Case 2a: R R Rs c w> (PNA lies in steel flange) Rw is the axial capacity of the web of the steel beam and is equal to R Rs f 2 . Rf is the axial capacity of one steel flange of thickness T; Rf = BTy

x

Ds - Dp

D

Ds Dp

Be

Rc

y (Ds-Dp)/2+DpPNA

T Neutral axis depth, x From equilibrium: 2y c sBx R R+ = 2y s cBx R R= x R R

BR R

R Ts c

y

s c

f= =

2 2 /

where Rf = resistance of steel flange = BTy B2.1 Moment about top flange of beam

M R D R DD D

R R xc s c ps p

s c= + + 2 2 2( )

or

M R D D RR D Dc s ss

cs p= +

2 2 ( )

M R D RD D R R

RT

c s cs p s c

f= + +

2 2 4

2( )

10

Case 1a: R Rc w< (PNA lies in web of steel beam)

x

Ds - Dp

D

Ds Dp

Be

PNA

T

Rc

y

y Neutral axis from the centroidal axis of the beam, x 2y ctx R= x R

tR

R dd R

Rc

y

c

v

c

v= = =

2 2 2 /

where R dtv y= = resistance of the clear web depth Moment about the centroid of the beam

M M R D DD D

R xc s c ps p

c= + + + 2 2 2

( )

M M RD D D D

R d RRc s cp s p

cc

v= + + +

22 2

12

)

where Ms = the moment capacity of the steel section

Rv = the axial capacity of the web of depth d. This differs only slightly from Rw (due to the root radius) and may be taken as Rw = Dty as a good approximation.

The above equations assume that the web is compact (i.e., the depth of web in compression does not exceed 38t, where t is the web thickness and is 275 / y ), and the beam is fully composite. An equivalent expression is also given for a partially composite beam, and it will be discussed in Section 4.8.

M M RD D D d R

Rc s cs p c

v= + + +

)2 4

2

11

7 Moment Capacity at High Shear 5.3.4 For beams subject to point load, high shear and moment may co-exist at the point load positions. Where the applied shear force Fv exceeds 0.5Pv, the moment capacity of the composite section should be reduced to account for the effects of shear as given in Clause 5.3.4 as: M M M M FPcv c c f

v

v=

( )

2 12

where Mc = Moment capacity of the composite section

Mf = Plastic moment capacity of the section having deducted the web area; Pv = shear stress of 0.6y times the shear area of the beam when d t/ 63

(Pv is the lesser of the shear capacity and web buckling capacity determined from BS5950:Part 1 Clause 4.2.3.

Pv = 0.6yDt, as shown in the figure.

Example 1: Design a simply supported composite beam using BS5950:Part3.1 Design data: Grade 43 Universal Beams Span = 12.0m Spacing of steel beams = 3.0m Concrete slab depth = 125mm Concrete Grade = 30 Loading Dead Load = 15.0kN/m Imposed Load = 16.0KN/m Bending Moment Design Moment = M= WL kNm

2 2

814 15 16 16 12

8839= + =( . . )

D t

Beam3.0m

3.0m

12m

Ds

Be = 3m

12

Effective width of the concrete flange Be is the lesser of bo = 3.0m or L/4 = 3.0m Assuming that the natural axis is in the concrete flange i.e., Rc > Rs where R As y= M R D D RR D Dc s s

s

cs p= +

2 2 ( )

Substituting for D D mmp s= =0 125, and R kNc = =045 30 3000 125 10 50633. ( ) we have M R D Rc s s= +

0125 2 2 5063 0125. ( . )

Trial and error design Procedure: (1) select a steel section, and find R As y= from the section table (2) check Rc = 5063kN > Rs (3) evaluate Mc from Eq. (2) and check that Mc is greater than 839kNm.

Section D (m) Rs(kN) Mc(kNm) UB406 x 178 x 67 0.4094 2350 741 UB406 x 178 x 74 0.4128 2620 783.4 *UB457 x 191 x 74 0.4572 2620 841.8

*Work Example: Select UB 457 x 191 x 74 Rs = 2620kN < Rc = 5063kN, i.e., neutral axis lies in the slab

M Rc s= +

0125

045722

26152 5063 0125.

. ( . ) =841.8kNm>839kNm OK

Check for High Shear Note that high shear does not co-exist with maximum moment, therefore moment reduction is not required. The design assumes that serviceability criteria do not control and the beam is fully composite. 8 Design of Shear Connections 5.4.6 8.1 Shear Stud Connectors in Solid Slab The characteristic resistance Qk of a headed shear stud with the dimensions and properties given in Clause 5.4.6 (Table 5) as shown below.

13

The design capacities of shear connectors to resist longitudinal shear are taken as 80% of their characteristic resistance in sagging (positive) moment regions and 60% in the hogging (negative) moment region. Design Capacity = Q = 0.8Qk 5.4.3 For lightweight aggregate concrete, the characteristic resistance should be reduced and taken as 90% the value given in Table 5. For lightweight concrete: Q = 0.9(0.8Qk) 8.2 Shear Stud Connectors in Composite Slab There is a potential reduction in the resistance of the shear connectors when used in a composite slab consisting of deck profile. In principle, the smooth flow of force into concrete depends on the projected angle from the base of the connector to the top of the adjacent profile, as shown in the figure below.

14

8.3 Reduction Factor for Deck Shape The following reduction factor, k, are to be applied to the characteristic resistance in Table 5 of the code. (i) Decking perpendicular to beam 5.4.7.2 For one shear connector per trough:

k bDh

Dr

p p=

085 1 10. .

For two shear connectors per trough:

k bDh

Dr

p p=

06 1 08. .

For three shear connectors per trough:

k bDh

Dr

p p=

05 1 06. .

br

h Dp

h Dp

br

15

where

br is the average trough width (for trapezoidal deck profile) or minimum trough width (for re-entrance deck profile).

h is the as-welded height of shear connector In all cases h D and h Dp p + 35 2 . The lower limit is to ensure an adequate projection of the head of the stud above the trough. The upper limit is to avoid the above formulae becoming unconservative. Restrictions on br Restrictions are placed on the dimension br of that may be used in the above equations when shear connectors are welded non-centrally in the troughs. For shear connectors placed in the "unfavourable" location such that the zone of the concrete in compression in front of the stud is maximised, then b er = 2 where e = minimum distance from the centre of the shear connector to the mid-height of the adjacent web of the profile, as shown in the following figures. For shear connectors placed in pairs, but in off-set pattern, alternatively on the unfavourable and favourable sides on the trough, br may be determined as for centrally loaded shear connectors.

16

(ii) Decking parallel to beam 5.4.7.3 For b D kr p/ . ; . =1 5 1 0 For b Dr p/ .< 1 5 k bD

hD

r

p p=

06 1 10. .

Design capacity of shear connector with decking Normal weight concrete R = (0.8Qk ) k Light weight concrete R = 0.9 (0.8Qk ) k k = reduction factor for decking effect. 8.4 Full Shear Connection In order to develop full composite action, the longitudinal force Rq to be transferred by the shear connectors should exceed the smaller of Rc or Rs. This is known as "full shear connection." R Smaller of R or Rq s c> where Rq = n kQs k( )

ns = number of shear connectors between the points of zero and maximum moment k = reduction factor for decking Qk = characteristic resistance from Table 5.

If Rq is less than the smaller of Rc or Rs then the beam should be designed as "partial shear connection". The moment capacity of the section needs to be evaluated to account for the effects of partial connection. Example 2: From Example 1, determine the number of shear studs required for full shear connection. Design Data: UB457x191x74 Span = 12.0m Concrete slab depth = 125mm Concrete Grade = 30 Shear Studs: 19mm, 95mm long Loading Dead Load = 15.0kN/m Imposed Load = 16.0KN/m

M = 839kNm

280kN

280kN

17

Design moment = 839kNm Design Shear force = 280kN For UB 457 x 191 x 74 R A kNs y= = 2615 Smaller of Rc and Rs is 2615kN. Capacity of shear connector (19mm diameter and 95mm long) Q kNk = 100 Table 5 Design capacity Q = 0.8Q kNk = 80 No. of connectors per half span = 2615/(80) = 32.7 Use 34 connectors with two connectors per trough in pairs with spacing as shown 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 8.5 Partial Shear Connection If the choice of cross-section is governed by considerations other than ultimate flexural resistance, it may be unnecessary to provide as many shear connectors as required by full shear connection. Indeed such partial connection may be unavoidable where a slab is constructed with profiled sheeting. If this spans onto the steel section, the number of connectors may be limited by only being able to place them in the troughs of the profile. The reduced interaction between the slab and the steel beam will result in a reduction in the load carrying capacity of the composite member. Where ductile shear connectors are used, the resistance moment of the critical cross-section of the beam Mcr may still be calculated by means of plastic theory but now taking account of the reduced value of the compressive force that will be developed in the concrete flange.

18

8.6 Minimum degree of shear connection To avoid brittle failure of shear connector due to excessive slip, it is necessary to impose a minimum degree of shear connection, dependent of the span L of the beam. Let the degree of shear connection, Na/Np, be defined as the actual of shear connections provided (Na) divided by the number required for full shear connection (Np). Limits are placed on the degree of shear connections 5.5.2. For spans up to 10m, Na/Np 0.4. For spans between 10 to 16m, Na/Np > (L-6)/10, 0.4 where L is the beam span in metres. For spans greater than 16m, full shear connection should be used. The moment capacity need o be calculated based on either the plastic method or by the simplified method. Moment Capacity

The reduced interaction between the slab and the steel beam will result in a reduction in the load carrying capacity of the composite member. Where ductile shear connectors are used, the

Na/Np

Mcr/Mc

Ms

Mr

Mc

Ms/Mc

Rq

Min. of Rc or Rs

19

resistance moment of the critical cross-section of the beam MRd may still be calculated by means of plastic theory givenin the subsequent sections. but now taking account of the reduced value of the compressive force that will be developed in the concrete flange. The relation between Mrc and degree of shear connection is qualitatively given by the convex curve ABC in the above Figure where Ms and Mc are the design plastic resistances to sagging bending of the structural steel section alone, and of the composite section with full shear connection, respectively.

Alternatively, a conservative value of Mrc may be determined by the straight line AC in the above Figure:

( ) arc s c sp

NM M M MN

= +

9 Moment Capacity Mc with Partial Shear Connection Based on Plastic Theory Appendix B

Partial shear connection applies when Rq is less than R and Rc s. For a compact section with equal steel flanges, the moment capacity of a composite section, Mc, is determined by the following: Case 3a: R Rq w< (PNA in web) B.2.3

Rq

y

y

y

x

Ms

2y=

yBe

DPNA

Rq

Rq

DpDs

xt R and R tdy q v y2 = = x

Rt

d RR

q

y

q

v= =

2 2

Also, R R yD Dq c s p

=

= y RR

D Dqc

s p( )

Take moment about the centroid of steel section Mc = Ms +Rq[D/2+Ds-y/2]-Rqx/2 Substituting for x & y and rearranging terms:

20

Rw =Dt y is the axial capacity of the web between the extreme fibre of the flanges. Rv is the axial capacity of the web of depth d. Rv may be taken as equal to Rw. Case 4: R Rq w (PNA in steel flange)

Ds - Dpy

Rq

y

y

2yx

D

PNA

Rs

RqDs

Be

Rs -Rq

Dp

xR R

BR R

R Ts q

y

s q

f= =

2 2 / in which Rf = 2BTy

yRR

D Dqc

s p= ( ) Moment about top of steel flange

2x)RR(

2yDR

2DRM qssqsc

+=

The last term in this equation is small, and can be neglected. Note that the expression in case 3a assume that the web is compact (see Table 2 of BS5950:Part 3 for web classification) If d

t R Rq v>

761

/

, the web is non-compact.

A modified expression for Mc is given in BS5950: Part 3 Clause B2.3 Case 3 (b) when the web is non-compact. 10 Check Capacity At Other Locations

M M R D DRR

D D RR

dc s q s

q

c

s p q

v= + +

2 2 4

2

4T

R)RR(

2)DD(

RR

DR2DRM

f

2qsps

c

qsqsc

+=

21

For beams subject to uniform loads, no check is required on moment capacity at other locations except at the maximum moment location. For beams subject to concentrated loads, the number of shear connectors calculated shodl be distributed to ensure that all intermediate locations have adequate moment capacity. The following formula may be used (5.4.5.5):

Ni = Np (Mi-Ms) / (Mc-Ms) Where Ni = number of shear connectors between the intermediate load point and the adjacent support. Np = number of shear connectors provided. Mi = moment at the intermediate point i. Ms = moment capacity of the steel section. Mc = moment at the composite section

N2

Ms

N1 N2 N1

N1 N3 =0 N2 N1 N1

M1 M2

N2 N1

22

Example: Moment capacity of partial composite beam Moment capacity of steel section Ms = 278 kNm Moment capacity of full composite section Mc = 612 kNm No. of connectors from zero to max. moment, Np = 38

11 Transverse Reinforcement Composite design requires a shear transfer from the shear connectors to the effective width of the slab. It is therefore needed to prevent shear failure of concrete surrounding the connectors. The following figure shows the critical planes for shear failure, i.e., planes 1-1, 2-3-3-2.

Figure: Potential shear planes through slab. The shear resistance per unit length of shear plane in a solid slab of normal concrete is given by

cucvysvr fA03.0fA7.0 += + Vp subject to an upper limit of

15 35-15 = 23 23 15

M1 = 410 M2 = 565N1

N2 Ni = Np (Mi-Ms) / (Mc-Ms) = 38(410-278)/(612-278) = 15 N2 = 38

1

1 1

12 2

33

Transverse Reinforcement

23

r cv cuA f= 08. + Vp fcu = characteristic cube strength of concrete in N/mm2 ( 40 N/mm2 ) fy = yield strength of the reinforcement Asv = steel reinforcement area per unit length of the beam, crossing the shear plane Acv = cross-sectional area per unit length of beam of the concrete shear surface under consideration. = 1.0 for normal weight concrete = 0.8 for lightweight concrete. Vp = the contribution of the profile steel decking if applicable.

= thickness of steel deck x design strength of steel deck Shear to be Resisted The total longitudinal shear per unit length V to be resisted should be determined from

V = NQ/S N = no. of shear connectors in a group S = longitudinal spacing C/C of shear connectors Q = 0.8Qk Qk = characteristic resistance of shear connector from Table 5 of BS5950Part 3.1 The designer must ensure that V Vr.

24

Dp = 50mm

Ds = 130mm

Metal Decking

1

1

1

1

Calculation of Shear Plane Example: Light-weight concrete Grade 30 to be used K = 0.8 = reduction facor due to metal decking Design shear force V = NQ/S N = 2 studs per rib Q = 0.8 (0.9 x 0.8Qk) = 58 kN (0.8 is the rduction factor for decking perpendicular to the beam; 0.9 is rduction fator for light-weight concrte; Qk = 100 kN for 19mm stud)

25

Spacing of stud = 375 mm For intermediate beam, there are two shear planes For each shear plane 1-1 as shown in the figure V = 309/2 = 155 N/mm Sehar Resistance

r sv y cv cu0.7A f 0.03 A f = + + Vp Acv = (80 + 50/2) x 1mm = 105 mm2 / mm = 0.8 fcu = 30 N/mm2 For sheeting continuous across the beam: Vp = tp x py = 1 x 280 N/mm2 = 280 N/mm Assume A142 mesh: Asv = 142 mm2/m or 0.142 mm2/mm, Fy = 460 N/mm2 0.7Asvfy = 0.7 x 0.142 x 460 = 46 N/mm 0.03Acvfcu = 0.03 x 0.8 x 105 x 30 = 76 N/mm

cucvysvr fA03.0fA7.0 += + Vp = 46 + 76 + 280 = 402 N/mm

r cv cu p0.8 A f V = + = 0.8 x 0.8 x 105 (30)0.5 + 280 = 648 N/mm Therefore Vr = 402 N/mm > V = 155 N/mm OK However, for edge beam, there is only one shear plane. In this case V = 155 x 2 = 310 N/mm, hence same reinforcement can be used for the edge beam. 12 Composite Beams - Serviceability 12.1 Modular Ratio e 4.1 The modular ratio e is defined as the ratio of the elastic moduli of steel to concrete . e s s= + ( ) where is the modular ratio for long term loading; s is the modular ratio for short term loading; is the proportion of the total loading which is long term. and s can be obtained from Clause 4.1, Table 1 of BS5950 Part 3.1. Imposed loads on floors should be assumed to be two-thirds short term and one-third long

term in building of normal usage. Storage loads and loads which are permanent in nature should be taken as long term

26

Appropriate values of e for office-type buildings are 10 and 15 for normal and lightweight concrete respectively. e = Es / Ec 12.2 Second Moment of Area Ig The second moment of area of the section (often termed inertia) is used in Establishing the moments and forces in the structures in elastic global analysis of

continuous beams. Calculation of deflections. The second moment of area is normally derived assuming the concrete is uncracked, although allowance can be made for the effects of concrete cracking in as given in the code Clause 5.2.3. The composite section is transformed into an equivalent steel section by dividing the cross-sectional area of concrete by the appropriate modular ratio e as shown in the figure. Take moments about the upper surface of the concrete gives a neutral axis depth, defining the part of the composite section in compression of

yA D D B D D

A B D Dge s e s p

e e s p= + + +

( ) ( )[ ( )]

22

2

Elastic Stress Distribution Transformed Section The second moment of area of the uncracked composite section Ig as in Appendix B.3 is given as:

Be/e Stress <

Dp yg

D

Ds

Equivalent Steel Area

Stress less than y

27

I IB D D AB D D D D D

A B D Dg xe s p

e

e s p s p

e e s p= + + + ++

( ) ( )( )[ ( )]

3 2

12 4 where Ix is the second moment of area of the steel section of cross-section area, A. e is the ratio of the elastic moduli of steel to concrete (modular ratio). This may be obtained from Clause 4.1, Table 1. This expression ignores the contribution of the concrete within the ribs of the decking which is conservative for the case where the ribs run parallel to the beam. 12.3 Elastic Section Modulus The elastic section modulus is used to determine the stresses in the section at the serviceability limit state. If the concrete is uncracked, the elastic section modulus is (i) At the lower fibre of the bottom Steel flange Z

ID D ys

g

s g= +

(ii) At the upper surface of the concrete slab Z

Iycg e

g=

where yg is the depth of the elastic neutral axis below the upper surface of the concrete slab such that:

yA D D B D D

A B D Dge s e s p

e e s p= + + +

( ) ( )[ ( )]

22

2

Checking of cracked and uncracked section Appendix B.4 of the code gives alternative expressions for the cases where the concrete is cracked. In most cases, the concrete will be uncracked at the serviceability limit state. If A

D D BD D

s p e

p e< +

( )( )

2

2 the section is cracked If A

D D BD D

s p e

p e +

( )( )

2

2 it is uncracked. B4.1

28

12.4 Serviceability Stresses The stresses in the steel and concrete are determined using the elastic section moduli calculated from the above step for the case where the beam is subject to unfactored imposed loads. For unpropped beams add the steel stresses to those calculated for the steel section subject to the self-weight of the floor. Compare the total steel stresses with the stress limit of y (Clause 2.4.3). For propped beams, determine the stresses in the concrete due to self weight of the floor applied to the composite section. Compare the total concrete stress due to self weight and imposed loads with the stress limit of 0.5fcu (Clause 2.4.3). Compare the steel stresses with y . At the extreme steel fibre, the bending stress is

f MZbs s y= <

At the extreme concrete fibre, the bending stress is

f MZ fbc c cu= < 05. 12.5 Deflections Deflection limits for beams are specified in BS5950:Part 1. Deflections are calculated for the case of unfactored imposed loads applied to the composite section. It is also necessary to consider the effect of deflection of the steel beam due to self weight of the floor in the unpropped construction. For unpropped construction: Total deflection = Deflection of composite section due to unfactored imposed load + Deflection of steel beam due to self-weight of the floor. Deflection due to self weight may be offset by precambering of the steel beam. For propped construction. Total deflection = Deflection of composite section due to unfactored imposed load For the case where the composite beam is subject to unfactored imposed loading, the beam deflection is calculated using the second moment of area given in the previous section Compare this deflection with a limit of Span/360 typical of buildings of normal usage. It may be necessary to reduce this deflection limit for beams supporting cladding or heavy partition. 12.6 Deflection for Partial Composite Section For partial shear connection, the deflection of the beams should be determined from Clause 6.1.4 which is given below: For propped construction: = + c a p s cN N0 5 1. ( / )( ) For unpropped construction: = + c a p s cN N0 3 1. ( / )( )

29

where s is the deflection of the steel beam acting alone c is the deflection of a composite beam with full shear connection for the same

loading. NN

a

p

= degree of shear connection Clause 5.4.4

Np = number of shear connectors required for full composite Na = actual number of shear connectors provided For full composite beams NN

a

p

= 1

= c

30

13 Examples 13.1 Example A From Example 1, determine the serviceability limit state of the simply-supported beam during (a) construction stage (b) composite stage. Design Data: UB457x191x74 Span = 12.0m Beam spacing = 3.6m Concrete slab depth = 125mm Concrete Grade = 30 Desity of concrete = 24 kN/m3 Shear Studs: 19mm, 95mm long Loading (1) During construction (Unpropped): Dead Load Floor (Concrete slab) 0.125m x 24 kN/m3 x 3.6m =10.8kN/m Steel Beam 74kg/m x 9.81 x 10-3 = .73kN/m Total Dead Load =11.53kN/m Imposed construction loads = 0.50kN/m2 x 3.6m = 1.8kN/m (2) Composite Stage (After construction) Total Dead Load (including floor, beam, ceiling etc.) = 15.0kN/m ......................................in which 13kN/m is long term loading. Total Imposed Load (including partitions etc.) = 16.0KN/m .....................in which (1/3x16kN/m = 5.33kN/m) is long term loading. Serviceability Deflection Deflection of beam at the construction stage:

= 5384

4WLEI

mm5.451033400205000384

1200053.1154

4

== > L/360 = 33.33mm

Precambering for dead load is required. Deflection of beam at the composite stage: s = =6 18 e s s= + ( ) = Long term loadingTotal Loading long term loading = dead load + 1/3 Imposed load

656.0161533.515 =+

+= 9.13= e

beams

12m

3.6m

3.6m

31

I IB D D AB D D D D D

A B D Dg xe s p

e

e s p s p

e e s p= + + + ++

( ) ( )( )[ ( )]

3 2

12 4 A = 95cm2 Ix = 33400cm4 Be = 3600mm Ds = 125mm Dp = 0 D = 457.2mm I cmg = + + =33400 4473 49522 87395 4 = 5

384

4WLEIg

W = Imposed load + additional dead Load after construction = 16kN/m + (15-6.13) = 24.9

= =5 24 9 12000

384 205000 87395 10 37 44

4.

. mm

Total deflection for fully composite beam: Construction stage (assumed pre-cambering for 45mm).... 0 mm Composite stage .............................................................. .. 37.4mm

Total Deflection = 37.4 mm = 321L

This is slightly higher than the allowable limit of L/360. Check for serviceability stresses To check whether the section is cracked or uncracked. A = 9500mm2 Be = 3600mm Ds = 125mm Dp = 0 D = 457.2mm

222

73769.13)02.457(

3600125)2(

)(mm

DDBDD

ep

eps =+=+

Since A > ( )( )D D BD D

s p e

p e

+

2

2 elastic neutral axis is in the steel member and the section is

uncracked. Bending stress in steel section at construction stage: M = WL kNm

2 2

8613 12

8 110 3= =. .

Bending stress fMZ N mmbf x

= = =110 3 101460 10 7556

6

32. . /

Stresses in steel and concrete at composite stage Depth of neutral axis below top of the concrete flange: y

A D D B D DA B D Dg

e s e s p

e e s p= + + +

( ) ( )[ ( )]

22

2

)]5.12(3609.1395[2)5.12(360)5.12272.45(1.1395 2

+++=gy =13.83cm

(i) At the lower fibre of the Steel flange

ZI

D D ysg

s g= +

= 3217383.135.1272.45

87395 cm=+

(ii) At the upper surface of the concrete slab

ZIycg e

g= = 39693883.13

1.1387395 cm=

32

W = Imposed load + additional dead Load after construction W = 24.9kNm Bending moment M = WL kNm

2 2

824 9 12

8 448 2= =. .

Bending stress in concrete 236

/64.41096938102.448 mmN

ZMf

cbc =

== < 0.5f N mmcu = 15 2/ Bending stress in steel 23

6

/206102173102.448 mmN

ZMf

sbs =

== Total stress in steel: Construction Stage...............................................75.56 N/mm2 Composite Stage...................................................206 N/mm2 282 N/mm2 Which is greater than y N mm= 275 2/ NG! Use Grade 50 steel or larger steel section!!

33

13.2 Example B The floor layout consisting of simply supported composite beams of 12m span length is shown below. Design the beam for strength and serviceability.

beams

12m

3m

3m

Floor Plan

Unpropped Construction

Floor Span

Materials: Structural steel Grade 50 y N mm= 355 2/ E = 205000N/mm2 Concrete: Lightweight Grade 30 f N mmcu = 30 2/ Density 1800 3kg m/

Deck Slab depth D mms = 130 Profile height D mmp = 50 Trough spacing = 300mm Average trough width = 150mm Design strength of profile steel sheets: yp N mm= 280 2/ Thickness t mmp = 1 Shear connectors 19mm diameter studs 95mm as welded length

DsDp

300mm

150mm

1mm thicksteel deck

95mm

19mm diameter stud

34

Floor Loading: (i) Concrete slab and steel decking Assuming slab thickness of 80 + 25 =105mm Weight of concrete = 1800 x 9.81 x 105 x 10-6 = 1.85 kN/m2 Weight of steel deck = 0.15 kN/m2 Total weight = 2.00 kN/m2 (ii) Construction Stage Floor = 2.00 kN/m2 Steel beam = 0.22 kN/m2 Total dead load = 2.22 kN/m2 Imposed construction load 0.50 kN/m2 (iii) Composite Stage Floor = 2.00 kN/m2 Steel beam = 0.22 kN/m2 Ceiling = 0.50 kN/m2 Total dead load = 2.72 kN/m2 Total Imposed load (including partitions) 6.0kN/m2 Effective Width L/4 = 12000/4 = 3000 mm , bo = 3000mm Be = 3000mm Select initial size of steel beam At composite stage Design load W = (1.6 x 6 + 1.4 x 2.72 ) x 3 = 40.2 kNm Design shear force F kNv = =40 2 12 2 241. / Design moment M WL kNm= = =

2 2

840 2 12

8724.

Assuming that the plastic neutral axis lies in the concrete slab i.e., R Rc s> Full shear connection: M R D D RR D Dc s s

s

cs p= +

2 2 ( )

Resistance of concrete flange R f B D Dc cu e s p= 0 45. ( ) B2.1 D D mmp s= =50 130, R kNc = =0 45 30 3000 130 50 10 32403. ( ) M R D Rc s s= +

0130 2 2 3240 0130 005. ( . . )

M R D Rc s s= +

0130 2 2 3240 008. ( . ) (1)

Normally the depth of the beam may be taken approximately as Span length / 20 =12000 20 600/ mm By trial and error, check R Rc s> , and Mc from Eq. (1) is greater than 724kNm.

Section (Grade 50 Steel)

D (m) Rs(kN) Mc(kNm)

UB356 x 171 x 67 0.364 3032 834.4 *UB457 x 191 x 67 0.4537 3032 968

Note design strength for Grade 50 steel is 355N/mm2

35

*Select UB 457 x 191 x 67 Grade 50 Since (Rc = 3240kN) > (Rs = 3032kN), neutral axis lies in the slab

+= )08.0(324023032

24537.013.03032cM = 968kNm > 724kNm OK

Check for Shear

F WL kNv = = + =21 4 2 72 3 1 6 6 3 12

2241( . . . )

05 05 0 6 05 821 411. . ( . ) .P Dt kNv yw= = = > 241kN Note that high Shear does not coincide with the maximum moment. Shear Connection R kNc = 3240 R A kNs y= = 3032 Smaller of Rc and Rs is 3032kN. Capacity of shear connector (19mm diameter and 95mm long) in lightweight concrete 5.4.6 Q kN kNk = =0 9 100 90. Design capacity Q = 0.8Q kNk = 72 Reduction factor for deck profile

k bDh

Dr

p p=

060 1 08. .

for two studs per rib 5.4.7.2

br= Average trough width = 150 mm h = overall height of the stud = 95 mm

k = = >060

15050

9550 1 162 08. . .

k = 0.8 Resistance of a shear connector = 0.8 x 72 = 57.6kN For full composite, no. of connectors per half span required = 3032/(57.6) = 52 Since the trough spacing is 300 mm, no. of connectors that can be accommodated in half span, assuming two connectors per trough = 2 x 6000/300 = 40 as shown below.

CL

2 studs @300mm CLhalf span length = 6000mm

36

Degree of shear connection = NN

a

p

= No of actual connectors providedNo of connectors in full composite

..

.= =4052

0 77

NN

Lap

= ( ) . .610

0 6 0 4 OK

Therefore, the beam needs to be redesigned for partial shear connection. Redesign for Beam Moment Capacity with Partial Shear Connection Resistance of overall web depth R R Rw s f= B2.1 R kNw = =3032 2 189 9 12 7 3551000 1320

. .

As R kNq = =40 57 6 2304. Since R Rq w> , plastic neutral axis is in flange B2.3 M R D R D

RR

D D R RR

Tc c q s

q

c

s p s q

f= +

2 2 4

2( ) ( ) B2.3

R kNf = =189 9 12 7 355 10 8563. . Mc = +

3032 45362 2304 13023043240

802

3032 2304856

12 74

2. ( ) .

= 920kNm > M = 724kNm OK Serviceability Deflection Deflection of beam at the construction stage due to dead load:

= 5384

4WLEI

= =5 2 22 3 12000

384 205000 29410 1029 8

4

4

( . ). mm < L/360 = 33.33mm

Deflection of beam at the composite stage: e s s= + ( ) s = =10 25 for lightweight concrete Table 1 Long term loading: Dead load 2.72 kN/m2 1/3 Imposed Load 2.00 kN/m2 4.72 kN/m2 Total Loading 8.72 kN/m2

= =4 728 72 0 541..

.

= + =e 10 0 541 25 10 18 1. ( ) .

37

I IB D D AB D D D D D

A B D Dg xe s p

e

e s p s p

e e s p= + + + ++

( ) ( )( )[ ( )]

3 2

12 4 A = 85.4cm2 Ix = 29410cm4 Be = 3000mm Ds = 130mm Dp = 50 D = 453.6mm

423

4 80900)80300012.188540(4

)501306.453(803000854012.18128030001029410 cmIg =+

++++=

= 5384

4WLEIg

Total imposed load W = (6+0.5) x 3m = 19.3kN/m Dead Load due to ceiling is small (0.5kN/m2) and it is neglected in the service check.

mmc 7.31108090020500038412000)3.19(5

4

4

==

Deflection when steel beam is acting alone

mms 2.87294107.3180900 ==

= 31.7 + 0.3(1 - 0.77) (87.2 31.7) = 35.5 mm Total deflection for partial composite beam: Construction stage (no pre-cambering, and unpropped).... 29.8mm Composite stage .............................................................. .. 35.5mm

Total Deflection = 65.3mm = 184

L

This is probably unacceptable. However, we will proceed with the calculation. Check for serviceability stresses To check whether the section is cracked or uncracked. B.4 A = 8540mm2

222

191412.18)5026.453(

300080)2(

)(mm

DDBDD

ep

eps =+=+

Since A > ( )( )D D BD D

s p e

p e

+

2

2 elastic neutral axis is in the steel member and the concrete is

uncracked. Bending stress in steel section at construction stage:

M = WL kNm2 2

82 22 3 12

8120= =.

For 457 x 191 x 67 UB Grade 50 Bending stress f M

ZN mmbf

x= = =

120 101300 10

926

32/

Stresses in steel and concrete at composite stage since A > ( )

( )D D BD D

s p e

p e

+

2

2

Depth of neutral axis below top of the concrete flange:

38

yA D D B D D

A B D Dge s e s p

e e s p= + + +

( ) ( )[ ( )]

22

2

)]80(300012.188540[2)80(3000)13026.453(3.198540 2

+++=gy =169 mm

(i) At the lower fibre of the Steel flange Z

ID D ys

g

s g= +

= 80900 10453 6 130 169

1950 3+ =. cm (ii) At the upper surface of the concrete slab Z

Iycg e

g= = 80900 19 3

169 10924001

3 =

. cm

Additional service load after composite = 6 + 0.5 = 6.5 kN/m2

Bending moment M = kNm3518

12)35.6(8

WL 22 ==

Bending stress in concrete f MZ

N mmbcc

= = =324 10

92400 103 51

6

32. /

< 0.5f N mmcu = 15 2/ Bending stress in steel f M

ZN mmbs

s= = =

324 101950 10

166 26

32. /

Total stress in steel: Construction Stage................................................92.0 N/mm2 Composite Stage...................................................166.2 N/mm2 258.2 N/mm2 As 258.2 N/mm2 is less than y N mm= 355 2/ OK

13.3 Example C

The floor plan in Fig. 1 shows the layout of the secondary composite beams of 8m span length, simply supported at their ends. The beams are spaced at 3.0m centres. The steel decking acts compositely with the concrete slab and the deck ribs run perpendicular to the secondary beams. The materials used for the composite beams are Grade 43 steel and Grade 25 normal weight concrete. Shear studs of 19mm diameter and 95mm welded length are to be used to provide the composite action. Based on the following loading data: Slab weight of concrete and decking = 2.1 kN/m2 Weight of Steel beam (approximated) = 0.22 kN/m2 Imposed Construction Load = 0.50 kN/m2 Ceiling, services, and raised floor after construction = 0.7 kN/m2 (additional dead load) Total imposed load after construction = 5.0 kN/m2 (a) Design a composite beam section and determine the number of shear

connectors required to carry the factored loading. (b) Determine whether propped construction method is required if the maximum

service deflection is limited to L/360 (consider dead and imposed load)..

39

Figure 1 Solution:

Floor Loading: (i) Construction Stage Dead Load = 2.1 + 0.22 = 2.32 kN/m2 Imposed Load = 0.50 kN/m2 (ii) Composite Stage Dead Load = 2.1 + 0.22 + 0.7 = 3.02 kN/m2 Imposed Load = 5.0 kN/m2 Effective Width L/4 = 8000/4 = 2000 mm , bo = 3000 mm Be = 2000 mm Select initial size of steel beam At composite stage Design load W = (1.6 x 5 + 1.4 x 3.02 ) x 3 = 36.7 kNm Design shear force F kNv = =36 7 8 2 146 8. / . Design moment M WL kNm= = =

2 2

836 7 8

8293 6. .

Resistance of concrete flange R f B D Dc cu e s p= 0 45. ( ) B2.1 D D mmp s= =50 120, R kNc = =0 45 25 2000 120 50 10 15753. ( ) Assuming that the plastic neutral axis lies in the concrete slab R Rc s> Full shear connection: M R D D RR D Dc s s

s

cs p= +

2 2 ( )

M R D Rc s s= +

0120 2 2 1575 0120 005. ( . . )

M R D Rc s s= + 0120 2 2 222 10 5. . (1)

Select a steel section and check the neutral axis depth lies with the concrete slab, i.e., R Rc s> , and Mc from Eq. (1) is greater than 293.6kNm.

Secondary beams

L = 8m

Floor Plan

3m 3m 3m 3m 3m

A

A

Deck span

160

300

70

50

120

Di mensi ons are i n mm

Sect i on A-A

40

Section (Grade 43) D (m) Rs(kN)

< Rc Mc (kNm)

UB254 x 146 x 37 0.2595 1304 287.9 UB305 x 165 x 40 0.3038 1419 341.8 UB356 x 127x 39 0.3529 1359 361.8

Select UB356 x 127x 39 R kNc = 1575 R A kNs y= = 1359 Smaller of Rc and Rs is 1359 kN. Capacity of shear connector (19mm diameter and 95mm long) in normal weight concrete is Q kNk = 95 Design capacity Q = 0.8Q kNk = 76 Reduction factor for deck profile

kbD

hD

r

p p=

0 6 1 08. .

for two stud per rib

br = Average trough width = 170 mm Dp = 50 mm h = overall height of the stud = 95 mm

k = = >060

17050

9550 1 1836 08. . .

k = 0.8 Resistance of a shear connector = 0.8 x 76 =60.8 kN For full composite, No. of connectors per half span required = 1359/(60.8) = 22.4 = 23 Since the trough spacing is 300mm, max. no. of connectors can be accommodated

(assuming two connectors per trough) = 2x 4000/300 = 27.

i.e., full composite is possible!! Total number of studs per span for full composite action = 2 x 23 = 46 studs OK Serviceability Deflection Deflection of beam at the construction stage:

= 5384

4WLEI

W = 2.32 x 3 kN/m I = 10100 cm4

= =5 2 32 3 8000

384 205000 10100 1017 93

4

4

( . ) . mm

Deflection of beam at the composite stage: e s s= + ( ) s = =6 18 for normal weight concrete Long term loading: Dead load 3.02 kN/m2 1/3 Imposed Load 5/3 = 1.67 kN/m2 4.69 kN/m2 Total Loading 5 + 3.02 = 8.02 kN/m2 = =4 698 02 0 585

.

..

= + =e 6 0 585 18 6 13 0. ( ) .

41

To check whether the section is cracked or uncracked. A = 4940 mm2 ( )( ) ( . )D D BD D m

s p e

p e

+ =

+ =

2 22

270 2000

352 9 2 50 13 1664

Since A > ( )( )D D BD D

s p e

p e

+

2

2 elastic neutral axis is in the steel member and the section is uncracked.

I IB D D AB D D D D D

A B D Dg xe s p

e

e s p s p

e e s p

= + + + + ( ) ( )( )

[ ( )]

3 2

12 4 A =49.4 cm2 Ix = 10100 cm4 e = 13 Be = 2000mm Ds = 120mm Dp = 50 D = 352.9 mm

I g = + + + +

+ (

( . )( )

)10100 102000 70

12 134940 2000 70 352 9 120 50

4 4940 13 2000 70104

3 24

=33689cm4

= 5384

4WLEIg

Total imposed load W = 5 x 3m = 15kN/m additional dead load = 0.7kN/m2 x 3 = 2.1kN/m W = 15 + 2.1 = 17.1kN/m

c mm= =5 171 8000

384 205000 33689 10 1324

4( . )

.

Total deflection for fully composite beam for unpropped construction: Construction stage (no pre-cambering, and unpropped).... 17.93 mm Composite stage .............................................................. .. 13.2 mm Total Deflection = 31.1 mm

which is higher than allowable deflection of L/360 = 22.2mm. For propped construction: total deflection = 13.2 mm < L/360 = 22.2mm Hence propped construction is recommended.