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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 precast unit

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 and2. 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.

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 (dc ¿

Mc=AbPy (dc + D2

-Xp2

)

ii. When NA (X p) is within the steel beam

Mc = AbPy ( D2

+ dc2

) - 2Abc P y(d sc – dcc

)

bs

dc X p F c = 0.4f cu bs X p

dc + D2

D Ab Py

Where, X p= AbPy 0.4f cu bs

Ab = steel area

Fig 1; NA is within the concrete slab

bs

dc d sc 0.4f cu bs X p

D2

X p Abc P y

¿¿

Where, Abc= Ab2

- 0.2f cu bs dc

P y

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, andc) Type of concrete used (normal or light weight)

Shear force (Pc) = Fc

N sc

Where, F c= 0.4 Fcu bs X p(where NA is in concrete)

F c= 0.4 Fcu bs dc (where NA is in steel)

And Nsc = No of studs required

The connector force (Pc ¿≯� 0.75 Pk

Shear strength (P¿¿k )¿ of headed studs

Diameter(mm)

Height(mm)

Shear strength Pk in (KN) for f cuin (N/mm2)20 30 40 50

22 100 112 126 139 15319 100 90 100 109 11916 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 Ls f cu

And q≯ 0.9Ls + 0.7Ae P ry

Where, Ae is either (Art+Arb) or 2 Arb depending on the shear path

Pry = the design strength of the reinforcement

f cu = the concrete cube strength

Ls 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 ¿¿allows defection to be calculated using normal elastic formulae with a value of

E s = 205 KN/mm2.

Modular ratio (m)

f cu(N/mm2) Short term Sustained20 8.2 16.430 7.3 14.640 6.6 13.250 6.0 12.0

bs

dc X e

D2

Steel I b

Strain diagram

AreaAb and r = Ab

bs dc

Fig 3 Transformed section

X e = (dc

2 + mr (

D2

+dc )) / (1 +mr)

I bc = Ab (D +dc) 2 / 4 (1 + mr) + bs dc

3

12m +I b

Actual deflection (ℓ) = w l3

60 E I bc

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. Q1Dimension:

Span of beam = 10 m Beam centers = 6 m Concrete slab thickness (dc) = 200 mm spanning in two ways Screed thickness (ts) = 40 mmLoading:

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 (f cu ) = 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@200mmcc

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)

Questiona) 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 (wd ¿ On = gk(A) =135.4KN

On = gk(A) =101.5KN

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 1438.4

4.2 4.2 334 334 10m 3m 2m 2m 3m

∴M x = 1061KNm ¿ M max

Fmax = 338KN

A) Design Aspect Assume py = 275 N/mm2 (Table 6)

∴ Zx = M x

P y = 3858 cm3

But Zx should be reduced by 59% ∴ Zx=¿ 1582cm3

Use 457 x 191 x 82 Kg/m UB (Z table= 1612 cm3)Other parameters are:

Ab= 104.5cm2; D= 460.2mm; t = 9.9mm and T= 16mm

Check the following

a) Shear capacity (Pv ) = 0.6P y Av= 752 KN

But F x

Pv = 0.45 ¿1.0 (Section adequate)

b) Moment capacity (M c) Assume that X p is within the concrete slab as shown

bs

dc X p

457 x 191 x 82 (UB)

Calculate, X p= AbPy = 119mm¿ dc 0.4f cu bs

∴ NA is within the slab.

Moment capacity (Mc) =AbPy (dc + D2

-Xp2

) = 1063KNm¿ Mx

M x

M c

= 0.99¿ 1.0 (Section adequate)

B) Shear connectors Force in the concrete @ Mid –span¿¿ ) = 0.4f cu bs X p = 2880KN But Shear strength (P¿¿k )¿ = 126 KN (given).

No of studs required (N sc) = Fc

Pc = 30 studs

But the connector force (Pc ¿≯� 0.75 Pk = 94.5KN

The studs are to be evenly distributed in each half span

Spacing = L2

N sc

= 167mm

Shear per unit length (q) = Fc

L2

= 576N/mm

But q≯ 0.15 Ls f cu

And q≯ 0.9Ls + 0.7Ae P ry

But (Ls) = 380mm (given)

0.15 Ls f cu = 1710N/mm And 0.9Ls + 0.7Ae P ry = 600N/mm ∴ Shear connector is adequate

C) Defection Use the un-factored imposed load (W) =210KN

But r = Ab

bs dc = 0.026

M = 13.2 (given)

Calculate NA (X ¿¿e)¿ = (dc

2 + mr (

D2

+dc )) / (1 +mr) = 184mm¿dc

∴ Use the transformed formula to obtain moment of inertia¿)

I bc = Ab (D +dc) 2 / 4 (1 + mr) + bs dc

3

12m +I b = 514185 cm4

∴ The actual deflection is given by the formula (ℓ) = w l3

60 E I bc = 3.3mm

But max defection = L

360 = 27.7mm

∴ Deflection is adequate The section chosen is adequate to sustain the loads.