Composite beams-and-slabs1

1. COMPOSITE BEAMS AND SLABS
Introduction
The term composite can be used of any structural medium in which two or more materials
interact to provide the required strength and stiffness. In steelwork construction, the term
refers to cross-section which combine steel section with concrete in such a way that the two
act together as one unit. Examples are shown below.
In situ concrete in situ concrete precastunit
Headed stud connector
The performance of composite beams is similar to that of reinforced concrete (rc) beams, but
there are two main differences.
Firstly, the steel section has a significant depth and its second moment of area may not be
ignored, unlike that of steel bar reinforcement.
Secondly, the concrete to reinforcement bond which essentially for rc action is absent, in
composite beams generally must be provided by shear connection.
Design method for composite beams therefore follows those methods for rc with modification
as indicated. Due to the presence of the concrete slab, problems of steel compression flange
instability and local bucking of the steel members are not usually relevant in simply
supported member except doing erection.
Recommendations for design in composite constructions are not included in part 1 of BS
5950 but are included in part 3 and 4 of BS 5950.
Advantages of Composite Beam (CB) Construction
The advantages of CB over normal steelwork beams are
1. The increased moment capacity and
2. Stiffness, or alternatively the reduced steel sizes for the same moment capacity.
Disadvantages of CB
The disadvantage of composite construction is the need to provide shear connectors to ensure
interaction of the parts.

2. The following are considered when dealing with composite structures
a) Shear and moment capacity
Essentially composite beams are T- beams with wide concrete flanges. Effective breadth
(bs) may be taken as one- fifth of the span for simply supported. While continuous and
cantilever beams they are treated separately (see BS 5950 part 3).
Shear capacity is based on the resistance of the web of the steel section alone.
Pv = 0.6Py Av
Moment capacity (Mc) is based on the assumed ultimate steel conditions as shown in Figs 1
and 2
i. When NA (𝑋 𝑝) is within the concrete slab depth (𝑑 𝑐)
Mc=AbPy (𝑑 𝑐 +
𝐷
2
-
𝑋𝑝
2
)
ii. When NA (𝑋 𝑝) is within the steel beam
Mc = AbPy (
𝐷
2
+
𝑑𝑐
2
) - 2𝐴 𝑏𝑐 𝑃𝑦 (𝑑 𝑠𝑐 –
𝑑𝑐
𝑐
)
𝑏𝑠
𝑑 𝑐 𝑋 𝑝 𝐹𝑐 = 0.4𝑓𝑐𝑢 𝑏𝑠 𝑋 𝑝
𝑑 𝑐 +
𝐷
2
D 𝐴 𝑏 𝑃𝑦
Where, 𝑋 𝑝= AbPy
0.4𝑓𝑐𝑢 𝑏𝑠
Ab = steel area
Fig 1; NA is within the concrete slab

3. 𝑏𝑠
𝑑 𝑐 𝑑 𝑠𝑐 0.4𝑓𝑐𝑢 𝑏𝑠 𝑋 𝑝
𝐷
2
𝑋𝑝 𝐴 𝑏𝑐 𝑃𝑦
(𝐴 𝑏− 𝐴 𝑏𝑐 )𝑃𝑦
Where, 𝐴 𝑏𝑐 =
𝐴𝑏
2
- 0.2 𝑓𝑐𝑢 𝑏 𝑠 𝑑 𝑐
𝑃𝑦
Ab = steel area
Fig. 2; NA within the steel beam
Shear Connectors
There are various forms of it but the preferred type is the headed stud.
Shear connectors must perform the primary function of:
a) Transferring of shear at the steal /concrete interface (equivalent to bond), hence
controlling slip between the two parts.
b) Secondly, carries the tensile force between the parts, hence controlling separation.
The performance of all shear connectors is affected by
a) Lateral restraint of the surrounding concrete
b) Presence of tension in the concrete, and
c) Type of concrete used (normal or light weight)
Shear force (𝑃𝑐) =
𝐹𝑐
𝑁𝑠𝑐
Where, 𝐹𝑐= 0.4𝐹𝑐𝑢 𝑏𝑠 𝑋 𝑝 (where NA is in concrete)
𝐹𝑐= 0.4𝐹𝑐𝑢 𝑏𝑠 𝑑 𝑐 (where NA is in steel)
And Nsc = No of studs required
The connector force (𝑃𝑐) ≯ 0.75𝑃 𝑘

4. Shear strength (𝑃 𝑘)of headed studs
Diameter
(mm)
Height
(mm)
Shear strength 𝑃 𝑘 in (KN) for 𝑓𝑐𝑢in (N/mm2)
20 30 40 50
22 100 112 126 139 153
19 100 90 100 109 119
16 75 6 74 82 90
Local shear in Concrete
The total shear connection depends on
a) The shear connector itself and
b) The ability of the surrounding concrete to transmit the shear stresses.
Therefore, longitudinal shear failure is possible as such transverse reinforcement must be
provided with strength greater than the applied shear per unit length (q):
q≯ 0.15 𝐿 𝑠 𝑓𝑐𝑢
And q≯ 0.9𝐿 𝑠 + 0.7𝐴 𝑒 𝑃𝑟𝑦
Where, 𝐴 𝑒 is either (𝐴 𝑟𝑡+𝐴 𝑟𝑏) or 2 𝐴 𝑟𝑏 depending on the shear path
𝑃𝑟𝑦 = the design strength of the reinforcement
𝑓𝑐𝑢 = the concrete cube strength
𝐿 𝑠 is either (2 (connector width + stud height))
Or 2(slab depth)
Deflection
As in steel beam design, defection ought to be checked for at the serviceability limit state
(un- factored loads)
The values of Neutral Axis (NA) depth (𝑋 𝑒) and the Equivalent Second Moment area
(𝐼𝑏𝑐) allows defection to be calculated using normal elastic formulae with a value of
𝐸𝑠 = 205 KN/mm2.
Modular ratio (m)
𝑓𝑐𝑢(N/mm2) Short term Sustained
20 8.2 16.4
30 7.3 14.6
40 6.6 13.2
50 6.0 12.0

5. 𝑏𝑠
𝑑 𝑐 𝑋 𝑒
𝐷
2
Steel 𝐼𝑏
Strain diagram
Area 𝐴 𝑏 and r =
𝐴 𝑏
𝑏 𝑠 𝑑 𝑐
Fig 3 Transformed section
𝑋 𝑒 = (
𝑑 𝑐
2
+ mr (
𝐷
2
+ 𝑑 𝑐 )) / (1 +mr)
𝐼𝑏𝑐 = 𝐴 𝑏 (D + 𝑑 𝑐) 2 / 4 (1 + mr) +
𝑏 𝑠 𝑑 𝑐
3
12𝑚
+ 𝐼𝑏
Actual deflection (ℓ) =
𝑤𝑙3
60𝐸 𝐼𝑏𝑐
Example 1
Yodebees Consult Limited based in Jos was contracted to design a steel structure. The plan,
section and other details are shown in Fig. Q1
Dimension:
Span of beam = 10 m
Beam centers = 6 m
Concrete slab thickness (dc) = 200 mm spanning in two ways
Screed thickness (ts) = 40 mm
Loading:
Concrete slab unit weight ( 𝛾𝑐)) = 23.8 KN/m3
Screed unit weight ( 𝛾𝑠) = 22 KN/m3
Imposed load = 5.0 KN/m2
Characteristics cube strength (𝑓𝑐𝑢 ) = 30 N/mm2
Self weight of beam = 6 KN (Assumed)
Young modulus of steel = 205 KN/m2
Characteristic strength of steel (𝑃𝑟𝑦) = 460 N/mm2
Others:
Area of reinforcement (Ae) = 0.800 mm2/m (ϴ10mm@200mm𝑐
𝑐⁄
Modular ratio (m) = 13.2 sustained
Length of shear path (Ls) = 380 mm
Use 22mm diameter by 100 mm high headed stud, (shear strength, Pk = 119 Kn)

6. Question
a) Design the most economical composite I- section beam to BS 5950 Part 3, given that
Zx (calculated) be reduced by 59 %
b) Check the suitability of the connectors and
c) Check for deflection.
10m
6m Screed
Beam Slab
6m
6m
Fig Q1
Solution
3m
3m
3m 4m 3m
Load computation
Dead load due to slab = 𝛾𝑐 (dc) = 4.76 KN/m2
Dead load due to screed= 𝛾𝑠(𝑡𝑠) = 0.88 KN/m 2
Total (gk) = 5.64KN/m2
Area Calculation
= 2 (bh) = 24m2
= (½bh) 4 = 18.m2
Dead load (𝑤 𝑑)
On = gk(A) =135.4KN
On = gk(A) =101.5KN

7. Imposed load (𝑤𝑖)
On = qk(A) = 120KN
On =qk(A) = 90KN
Ultimate load (w)
Uniform dead load = 1.4x6 = 8.4KN
On = 1.4wd +1.6wi = 382KN
On =1.4wd +1.6wi = 286KN 191 191
143 143
8.4
4.2 4.2 334 334
10m 3m 2m 2m 3m
∴ 𝑀 𝑥 = 1061KNm = 𝑀 𝑚𝑎𝑥
𝐹 𝑚𝑎𝑥 = 338KN
A) Design Aspect
Assume 𝑝 𝑦 = 275 N/mm2 (Table 6)
∴ 𝑍 𝑥 =
𝑀 𝑥
𝑃 𝑦
= 3858 cm3
But 𝑍 𝑥 should be reduced by 59%
∴ 𝑍 𝑥 = 1582cm3
Use 457 x 191 x 82 Kg/m UB (𝑍𝑡𝑎𝑏𝑙𝑒 = 1612 cm3)
Other parameters are:
𝐴 𝑏= 104.5cm2; D= 460.2mm; t = 9.9mm and T= 16mm
Check the following
a) Shear capacity (𝑃𝑣 ) = 0.6𝑃𝑦 𝐴 𝑣= 752 KN
But
𝐹𝑥
𝑃𝑣
= 0.45 < 1.0 (Section adequate)
b) Moment capacity (𝑀𝑐 )
Assume that 𝑋 𝑝 is within the concrete slab as shown
𝑏𝑠
𝑑 𝑐 𝑋 𝑝
457 x 191 x 82 (UB)

8. Calculate, 𝑋 𝑝= AbPy = 119mm< dc
0.4𝑓𝑐𝑢 𝑏𝑠
∴ NA is within the slab.
Moment capacity (Mc) =AbPy (𝑑 𝑐 +
𝐷
2
-
𝑋𝑝
2
) = 1063KNm > Mx
𝑀 𝑥
𝑀 𝑐
= 0.99 < 1.0 (Section adequate)
B) Shear connectors
Force in the concrete @ Mid –span (𝐹𝑐 ) = 0.4𝑓𝑐𝑢 𝑏𝑠 𝑋 𝑝 = 2880KN
But Shear strength (𝑃 𝑘)= 126 KN (given).
No of studs required (𝑁𝑠𝑐) =
𝐹𝑐
𝑃𝑐
= 30 studs
But the connector force (𝑃𝑐) ≯ 0.75𝑃 𝑘 = 94.5KN
The studs are to be evenly distributed in each half span
Spacing =
𝐿
2⁄
𝑁𝑠𝑐
= 167mm
Shear per unit length (q) =
𝐹𝑐
𝐿
2⁄
= 576N/mm
But q≯ 0.15 𝐿 𝑠 𝑓𝑐𝑢
And q≯ 0.9𝐿 𝑠 + 0.7𝐴 𝑒 𝑃𝑟𝑦
But (Ls) = 380mm (given)
0.15 𝐿 𝑠 𝑓𝑐𝑢 = 1710N/mm
And
0.9𝐿 𝑠 + 0.7𝐴 𝑒 𝑃𝑟𝑦 = 600N/mm
∴ Shear connector is adequate
C) Defection
Use the un-factored imposed load (W) =210KN
But r =
𝐴 𝑏
𝑏 𝑠 𝑑 𝑐
= 0.026
M = 13.2 (given)
𝐶𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒 𝑁𝐴 (𝑋 𝑒 ) = (
𝑑 𝑐
2
+ mr (
𝐷
2
+ 𝑑 𝑐 )) / (1 +mr) = 184mm< 𝑑 𝑐
∴ Use the transformed formula to obtain moment of inertia(𝐼𝑏𝑐)
𝐼𝑏𝑐 = 𝐴 𝑏 (D + 𝑑 𝑐) 2 / 4 (1 + mr) +
𝑏 𝑠 𝑑 𝑐
3
12𝑚
+ 𝐼𝑏 = 514185 cm4
∴ The actual deflection is given by the formula (ℓ) =
𝑤𝑙3
60𝐸 𝐼𝑏𝑐
= 3.3mm

9. But max defection = 𝐿
360⁄ = 27.7mm
∴ Deflection is adequate
The section chosen is adequate to sustain the loads.