Composite Construction and Design - colincaprani.com

3rd Architecture 

Introduction 

Composite Construction and Design 

Composite construction refers to any members composed of more than 1 material. 

The parts of these composite members are rigidly connected such that no relative 

movement can occur. Examples are: 

Timber and steel ‘flitch’ beams Timber-reinforced concrete 

Typical steel and concrete composite construction 

Composite construction aims to make each material perform the function it is best at, 

or to strengthen a given cross section of a weaker material. 

Name and explain another form of composite construction. 

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Behaviour of Composite Beams 

In the following, we consider only the case of structural steel sections and reinforced 

concrete slabs. A comparison of behaviours is: 

The non-composite beam deflects further, hence it is less stiff. Note that the E-value 

hasn’t changed so it is the I-value that changes. In addition to the increase in stiffness 

there is also a large increase in moment capacity leading to reduced section sizes. The 

metal decking can also be used as permanent formwork, saving construction time. 

Non-composite behaviour 

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The concrete slab is not connected to the steel section and therefore behaves 

independently. As it is weak in longitudinal bending, it deforms to the curvature of 

the steel section and has its own neutral axis. The bottom surface of the concrete slab 

is free to slide over the top flange of the steel section and slip occurs. The bending 

resistance of the slab is often so small that it is ignored. 

Composite Behaviour 

In this case, the concrete slab is connected to the steel section and both act together in 

carrying the load. Slip between the slab and steel section is now prevented and the 

connection resists a longitudinal shear force similar in distribution to the vertical 

shear force shown. 

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Composite Construction Layout 

Composite deck floors using shallow profiles are usually designed to span 2.5 to 4.5 

m between supports. When the deck is propped during construction the spans are 

around 4 to 5 m. 

Long span floors (12 to 18 m) are achieved by primary beams at 6 to 9 m centres. 

Shorter secondary beams support the slab (Diagram A). The type of grid shown in 

Diagram B offers services integration within the depth of the floor. Alternatively the 

secondary beams can be designed to span the longer distance so that the depths of the 

primary and secondary beams can be optimized. 

The Asymmetric Beam (ASB) system from Corus allows a squarer panel (Diagram 

C) and is designed to compete with RC flat-slab construction. 

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Propped Construction 

5 

Note that the beam layouts all 

describe simply-supported spans 

and this is usual. Continuous 

spans of composite beams can 

cause problems, though can be 

very useful nonetheless. 

Over the support the concrete 

cracks (and these can be large); 

the steel must take the majority 

of the bending alone, and so a 

portion of the section is in 

compression. Slender sections 

are prone to local buckling in 

and any intervening column may 

need to be strengthened to 

absorb the compression across 

its web. Lateral-torsional 

buckling of the beam may also 

be a problem. 

The steel beam is supported at mid- or quarter-span until the concrete slab has 

hardened sufficiently to allow composite action. Propping affects speed of 

construction but allows smaller steel sections. 

Unpropped Construction 

The steel beams must carry the weight of the wet concrete on its own. By the time 

construction loads can be applied to the slab, some composite behaviour can be used. 

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Elements of Composite Construction 

The elements that make up composite construction are: 

There are two main forms of deck: shallow and deep. The figure above illustrates a 

typical shallow deck (50–100 mm) and below is a deep deck (225 mm) supported on 

an ASB. The deep deck systems are proprietary; we will only consider the design of 

shallow deck systems, though the principles are the same. 

The beams are ordinary structural steel sections (except for the ASB). 

The shear studs are normally 19 mm diameter 100 mm high studs, though there are 

different sizes. 

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Design of Composite Beams 

The design involves the following aspects: 

1. Moment capacity: 

Design the section such that the moment capacity is greater than that required. 

2. Shear capacity 

To ensure adequate capacity; this based on the steel section alone – as per usual 

structural steel design. 

3. Shear connector capacity 

To enable full composite action to be achieved; these must be designed to be 

adequate. 

4. Longitudinal shear capacity 

Check to prevent possible splitting of the concrete along the length of the beam. 

5. Serviceability checks: 

a. Deflection; 

b. Elastic behaviour, and; 

c. Vibration. 

These checks are to ensure the safe and comfortable use of the beam in service. We 

check to ensure it does not cause cracking of ceilings and is not dynamically ‘lively’. 

Also, we verify that it is always elastic when subjected to service loads to avoid 

problems with plastic strain (i.e. permanent deflection) of the beam. We will not 

consider checks on vibration and will only outline the calculations for the elastic 

check. 

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Design of Composite Beams: Moment Capacity 

Just as in ordinary steel and RC design, the composite moment capacity is derived 

from plastic theory. There are three cases to consider, based on the possible locations 

of the plastic neutral axis (PNA), shown below. 

When calculating the PNA location, we assume a stress of py in the steel and 0.45fcu 

in the concrete. The tensile capacity of the beam of area A is: 

Fs= pyA The compression capacity of the slab depends on the orientation of the decking (Dp), 

and is: 

( ) 

F = 0.45 f D − 

D B 

c cu s p e 

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where Be is the effective breadth of the slab. We also define the axial capacities of the 

flange and web as: 

Ff= BTpy 

F = F − 2 F or F = Dtp 

w s f w y 

Using the notation given, where the depth of the PNA is yp, we have three capacities: 

• Case (a): PNA is in the slab; occurs when Fc > Fs: 

⎡ D F ⎛ D s s − Dp 

⎞⎤ 

Mc = Fs ⎢ + Ds − ⎜ ⎟⎥ 

⎣ 2 Fc 

⎝ 2 ⎠⎦ 

• Case (b): PNA is in the steel flange; occurs when Fs> Fc 

9 

( ) 2 

D ⎛ Ds − Dp ⎞ ⎧⎪ Mc = Fs + Fc⎜ 

⎟−⎨ 

2 ⎝ 2 ⎠ ⎩⎪ Fs − Fc 

Ff 

T⎫⎪ 

⋅ ⎬ 

4 

⎭⎪ 

(the term in the braces is small and may be safely ignored). 

• Case (c): PNA is in the steel web; occurs when Fw > Fc 

2 

⎛ Ds + Dp + D⎞ Fc D 

Mc = Ms + Fc⎜ 

⎟− 

⋅ 

⎝ 2 ⎠ Fw 

4 

where M s = pS y x is the moment capacity of the steel section alone. 

The effective breadth Be is taken as: 

B ≤ B = 0.25L≤ 

S 

e 

where B is the width of the steel section and S 

is the centre-to-centre spacing of the 

composite beams (2.5 to 4.5 m) and L is the 

(simply-supported) span of the beam. 

Don’t Panic! 

Case (a) is frequent; (b) less so, but (c) is very 

rare. Therefore, for usual design, only Fc and 

Fs are required (ignoring the term in the 

braces). Note that if Fs> Fc, 

check that Fw >/ Fc 

to ensure that you are using Case (b). 

C. Caprani


Design of Composite Beams: Shear Capacity 

The shear capacity is based on the capacity of the steel section only. 

The capacity is: Pv = 0.6 pyAv where Av= tD. 

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Design of Composite Beams: Shear Connector Capacity 

The shear connectors used in ordinary composite construction are dowel-type studs. 

Other forms used to be used, but headed-studs are now standard. They allow easy 

construction as they can be shot fixed or welded through the deck onto the beam, 

after the deck has been laid. In addition to the shear strength, the headed studs 

prevent the vertical separation, or uplift, of the concrete from the steel. 

Note that although some slip does occur (which reduces the capacity slightly) we 

usually design for full shear connection, though partial interaction is also possible. 

Shear plane 

The shear force to be transmitted is the smaller of 

Fc and Fs as calculated earlier. We only need to 

transfer shear in the zones between zero and 

maximum moment. Therefore the number of shear 

connectors required in each half of the span (see 

diagram above) is: 

11 

N 

p 

( F F ) 

min , 

= 

Q 

c s 

p 

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Where Qp is the force in each shear connector, and 

Q / 4 D ; > / 600 mm; longitudinally and as shown in the figure: 

s 

p 

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Design of Composite Beams: Longitudinal Shear Capacity 

The force transmitted by the shear studs can potentially split the concrete along the 

weakest failure plane. Some such planes are shown: 

Perpendicular Deck Parallel Deck 

Failure planes a-a, b-b and c-c are usually critical; d-d has no strength contribution 

from the decking itself (which is possible, though we will always ignore this safely). 

Any reinforcement in the slab that crosses these planes is taken to contribute. The 

force per unit length to be resisted is: 

NQ T p 

v = 

s 

where s is the shear-stud spacing and NT is the number of studs across the width of 

the beam (1 or 2). This must be less than the capacity which is: 

( ) 

v = 0.03 f L + 0.7A 

f 

r cu s sv y 

< 

( 0.8 ) 

f L 

cu s 

where Asv is the area of reinforcement, per unit length, crossing the failure plane and 

Ls is the length of the failure plane: 

• Plane a-a and c-c: L = 2D 

and A = 2A 

s p 

sv s 

• Plane b-b: L = 2h+ 

d + s where st is the transverse spacing of the 2 studs and s = 0 

s t 

for only 1 stud. Also, A = 2( 

A + 

A ) 

sv s sc 

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t


Design of Composite Beams: Serviceability checks 

For these checks we define the following: 

- the depth to the elastic neutral axis: 

x 

e 

Ds − Dp ⎛D⎞ + αer⎜ 

+ Ds⎟ 

2 ⎝ 2 

= 

⎠ 

( 1+ 

α r) 

A 

r = 

B D D 

14 

e 

( − ) 

e s p 

- the second moment of area of the un-cracked composite section: 

( ) 

( + r) 

( ) 

2 3 

A D+ D + D B D −D 

Ig = Ix 

+ + 

41 α 12α 

s p e s p 

e e 

where α e is the effective modular ratio which can be taken as 10 for most 

purposes; Ix is the second moment of area of the steel section alone; and the other 

symbols have their previous meanings. 

- the section modulus for the steel and concrete: 

Z 

s 

I g 

= 

D+ D −x 

s e 

Z 

c 

α I 

= 

x 

The composite stiffness can be 3–5 times, and the section modulus 1.5–2.5 times that 

of the steel section alone. 

e g 

e 

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a. Deflection 

Deflection is checked similarly to ordinary steel design, the allowable deflection is: 

L 

δ allow = 

360 

Assuming a uniformly distributed load, the deflection is: 

4 

5wL 

q 

δ = 

384EI 

where wq is the imposed UDL only and E = 205 kN/mm 2 . 

b. Elastic behaviour 

We check that the stresses in the steel or concrete remain elastic under the service 

w = w + w : 

loads, that is, under ser g q 

where 

c. Vibration 

2 

wserL M ser = . 

8 

We will not check this. 

M 

M ser 

σ = < p 

σ , = < 0.45 f 

Z 

s, ser 

ser 

Zs 

y 

15 

g 

c ser cu 

c 

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Design Example 

Check that the proposed scheme shown is adequate. 

Design Data 

Beam: 

- 457×152×52 UB Grade 43A (py = 275 N/mm 2 ) 

- Span: 7 m simply supported; beams at 6 m centre to centre. 

- Section properties: 

Slab: 

A = 66.5 cm 2 ; D = 449.8 mm; tw = 7.6 mm; 

Ix = 21345×10 4 mm 4 ; tf = 10.9 mm; b = 152.4 mm 

- Ds = 250 mm 

T12-150 

Qk = 6.5 kN/m 2 

- Grade 30N concrete (fcu = 30 N/mm 2 ) 

- Reinforcement T12-150: Asv = 754 mm 2 /m = 0.754 mm 2 /mm 

16 

457×152×52 UB 

250 

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Solution 

Loading: 

The dead load of slab is: 

G = 0.25× 24 = 6 kN/m 

k 

Hence, the UDL to beam, including self weight: 

w 

w 

g 

sw 

= 6× 6 = 36 kN/m 

52× 9.81 

= = 0.5 kN/m 

3 

10 

So the serviceability and ultimate loads are: 

2 

17 

w = 6× 6.5 = 39 kN/m 

w = 36 + 0.5 + 39 = 75.5 kN/m 

w = 1.4( 36 + 0.5) + 1.6( 39) = 113.5 kN/m 

ser 

Design moments and shear: 

M 

V 

ult 

ser 

Moment Capacity: 

2 2 

wser L 75.5× 7 

= = = 462.4 kNm 

8 8 

wultL 113.5× 7 

= = = 397.3 kN 

2 2 

Effective width; B = 0.25L= 0.25× 7000 = 1750 mm 

e 

( ) 

F = 0.45 f D −D 

B 

c cu s p e 

( )( )( ) 

0.45 30 250 1750 

= 

3 

10 

= 5906.25 kN 

q 

M 

ult 

ult 

2 2 

wultL 113.5× 7 

= = = 695.2 kNm 

8 8 

Fs = pyA 275 66.5 10 

= 

3 

10 

= 1828.75 kN 

2 ( × ) 

Thus we have Case (a): PNA is in the slab because Fc > Fs: 

Thus: 

⎡DF⎛ Ds − Dp 

⎞⎤ 

s 

Mc = Fs ⎢ + Ds − ⎜ ⎟⎥ 

⎣ 2 Fc 

⎝ 2 ⎠⎦ 

⎡449.8 1828.75 ⎛250 − 0 ⎞⎤ 

= ⎢ + − ⎜ ⎟ × 

2 5906.25 2 

⎥ 

⎣ ⎝ ⎠⎦ 

= 797.7 kNm 

−3 

( 1828.75) 250 10 

M c > Mult ∴OK 

C. Caprani


Shear Capacity: 

Thus: 

P = 0.6 p A 

v y v 

−3 

( )( )( ) 

= 0.6 275 

= 564 kN 

449.8 7.6 × 10 

Pv > Vult ∴ OK 

Shear Connector Capacity: 

Assuming a 19 mm × 100 mm high connector: 

Q = 100 kN from the table of characteristic stud strengths 

k 

( ) 

Q < / 0.8Q = 0.8 100 = 80 kN 

p k 

Hence the number required in each half of the span is: 

min ( Fc, Fs) 

N p = 

Qp 

min ( 5906.25,1828.75) 

= 

80 

1828.75 

= 

80 

= 22.8 

We will use 24 studs as we are putting NT = 2 studs at each position along the beam. 

L 

s = 

2Np−1 7000 

= 

212 ( ) −1 

= 304.3 mm 

Hence use 300 mm c/c evenly spaced along the length of the beam. 

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Longitudinal Shear Capacity: 

Consider these failure planes: 

The 110 mm transverse spacing is v ∴ OK 

110 

19 

a-a 

b-b 

100 

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Serviceability: Deflection 

Calculate the area ratio: 

2 

A 66.5× 10 

r = = = 0.0152 

B D D 

( − ) 1750( 250 − 0) 

e s p 

Hence the second moment of area of the section is: 

( ) 

( ) 

2 3 

s p e s p 

A D+ D + D B D −D 

Ig = Ix 

+ + 

41 12 

( + αer) αe 

4 ( 2 

)( 

( + × ) 

) 

6 4 

20 

( ) 

( ) 

2 3 

66.5× 10 449.8 + 250 + 0 1750 250 − 0 

= 21345× 10 + + 

4 1 10 0.0152 12 10 

= 1148× 10 mm 

Therefore the deflection is: 

4 

5wL 

q 

δ = 

384EI 

g 

( )( ) 

5 39 7000 

= 

384 205 10 1148 10 

= 5.2 mm 

And the allowable is: 

Thus 

3 6 

( × )( × ) 

L 

δ allow = 

360 

7000 

= 

360 

= 19.4 mm 

δ < δ ∴ 

OK 

allow 

4 

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Serviceability: Elastic behaviour 

In addition to our previous calculations, we need: 

2 

wserL M ser = 

8 

2 

75.5× 7 

= 

8 

= 462.4 kNm 

And the section properties, first the depth to the elastic neutral axis: 

x 

e 

Ds − Dp ⎛D⎞ + αer⎜ 

+ Ds⎟ 

2 ⎝ 2 

= 

⎠ 

= 171.2 mm 

( 1+ 

α r) 

e 

250 − 0 ⎛449.8 ⎞ 

+ 10( 0.0152) ⎜ + 250⎟ 

2 ⎝ 2 

= 

⎠ 

( 1+ 10( 0.0152) 

) 

And the elastic section modulii; 

Z 

s 

I g 

= 

D+ D −x 

And the stresses; 

s e 

6 

1148× 10 

= 

449.8 + 250 −171.2 

6 3 

= 2.17× 10 mm 

M 

σ = < p 

sser , 

ser 

Zs 

y 

6 

462.4× 10 

= < 275 

6 

2.17× 10 

= ∴ 

2 

213.1 N/mm OK 

21 

Z 

c 

α I 

= 

x 

e g 

e 

6 ( × ) 

10 1148 10 

= 

171.2 

= 67× 10 mm 

6 3 

M 

σ = < 0.45 f 

c, ser 

ser 

Zc 

cu 

6 

462.4× 10 

= 6 

67× 10 

< 0.45 30 

= < ∴ 

( ) 

2 

6.9 13.5 N/mm OK 

Hence both the steel and concrete stresses remain elastic under the service loads, and 

so not permanent plastic deformations will occur. 

This design has passed all requirements and is therefore acceptable. 

C. Caprani


Problem 1 



Beam: 

- UB is Grade 43A 

- Span: 7.5 m simply supported; beams at 6 m centre to centre; 

- Use 1 shear stud at each location. 

Slab: 

T10-100 

Qk = 3.0 kN/m 2 


22 

406×140×46 UB 

250 

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Problem 2 

An allowance of 2.7 kN/m 2 extra dead load is required for ceilings/services and floor 

tiles. 



Beam: 

- UB is Grade 43A 

- Span 8 m simply supported; beams at 5 m centre to centre; 

- Use 2 shear studs at each location. 

Slab: 

T10-180 

T10-180 

Qk = 3.5 kN/m 2 


23 

457×152×52 UB 

180 

C. Caprani

Composite Construction and Design - colincaprani.com

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