Composite slab with profiled metal decking provides economical solutions for floors of steel framed building systems. This is because they are easier to install, lighter in weight, and faster to execute when compared with precast, prestressed, and solid slab system for steel-framed buildings. The composite action of this floor system is achieved by welding steel studs to the top flange of the steel beams and embedding the studs in the concrete during concrete pouring.

Composite construction reduces frame loadings and results in a cheaper foundation system. Cold-formed thin-walled profiled steel decking sheets with embossments on top flanges and webs are widely used as the profiles. The use of this profiled metal decking eliminates the need for mat reinforcement in the slab and acts as the permanent shuttering for the concrete. Props are therefore not usually required during the process of concreting. This support scheme is usually suitable for spans that are less than 4m (spacing of the supporting beams). The supporting beams themselves can, however, span up to 12m.

Structural engineers usually rely on load/span tables produced by metal deck manufacturers in order to determine the thickness of slab and mesh reinforcement required for a given floor arrangement, fire rating, method of construction, etc. The table below shows an example of a typical load/span table available from one supplier of metal decking.

To download the full SMD technical data sheet for different types of profiled metal decking, click below.

Once the composite slab has been designed, the design of the primary and secondary composite beams (i.e. steel beams plus slab) can begin. This is normally carried out in accordance with the recommendations in Part 3: Section 3.1 of BS 5950. In Europe, composite sections are designed according to the requirements of Eurocode 4 (EN 1994 â€“ 1- 1).

The steps in the design of profile metal decking for composite floors are;

- Determine the effective breadth of the concrete slab.
- Calculate the moment capacity of the section.
- Evaluate the shear capacity of the section.
- Design the shear connectors.
- Assess the longitudinal shear capacity of the section.
- Check deflection.

**Design Example**

The figure below shows a part plan of a composite floor. The slab is to be constructed using profiled metal decking and normal weight, grade 30 concrete. The longitudinal beams are of grade S275 steel with a span of 7.5 m and spaced 3 m apart. Design the composite slab and verify the suitability of **406 x 178 x 67Â UKB** as the internal beams. The required fire-resistance is 1 hour.

Imposed load = 4 kN/m^{2}

Partition load = 1 kN/m^{2}

Weight of finishes = 1.2 kN/m^{2}

Weight of ceiling and services = 1 kN/m^{2}**Total applied load = 7.2 kN/m ^{2}** (to be used for slab design i.e. selection from manufacturerâ€™s span-load table)

**SLAB DESIGN**

From the span-load table above, the configuration below will be satisfactory for the unpropped slab.

Beam span = 7.5 m

Beam spacing = 3.0 m

Total depth of slab h_{s} = 130 mm

Depth of profile h_{p} = 60 mm

Overall height of profile h_{d }= 72 mm

Depth of concrete above profile = 58 mm

Profile: SMD TR60^{+} (1.2 mm gauge)

Gauge = 1.2 mm

Mesh: A142

From the manufacturerâ€™s technical data sheet;

Volume of concrete = 0.096 m^{3}/m^{2}

Weight of concrete (wet) = 2.26 kN/m^{2}

Weight of concrete (dry) = 2.21 kN/m^{2}

Weight of profile = 0.131 kN/m^{2}

Height to neutral axis = 33 mm

**Shear connector**

Connector diameter *d* = 19 mm

Overall welded height of *h _{sc}* = 95 mm

Ultimate tensile strength

*f*= 450 N/mm

_{u }^{2}

**Concrete**

Normal weight C25/30 concrete

Cylinder trength f_{ck }= 25 N/mm^{2}

Cube strength f_{ck,cube} = 30 N/mm^{2}

Secant modulus of elasticity E_{cm} = 31 kN/mm^{2}

**Actions at Construction Stage**

**Permanent Actions**

Self weight of sheeting = 0.131 kN/m^{2} x 3 m = 0.393 kN/m

Allowance for beam self-weight = 1.0 kN/m

Allowance for mesh = 0.05 kN/m^{2} x 3m = 0.15 kN/m

Total g_{k} = 0.393 + 1 + 0.15 = 1.543 kN/m

**Variable Actions**

Self-weight of fresh concrete = 2.26 kN/m^{2 }x 3m = 6.78 kN/m (note that fresh concrete is treated as variable action in the construction stage)

Construction load = 0.75 kN/m^{2} x 3m = 2.25 kN/m

Total q_{k} = 6.78 + 2.25 = 9.03 kN/m

At ultimate limit state = 1.35g_{k} + 1.5q_{k} = 1.35(1.543) + 1.5(9.03) = 15.63 kN/m

Design moment M_{Ed} = ql^{2}/8 = (15.63 x 7.5^{2})/8 = 109.898 kNm

Design shear force V_{Ed} = ql/2 = (15.63 x 7.5)/2 = 58.61 kN

**Actions at the composite stage**

**Permanent Actions**

Self weight of sheeting = 0.131 kN/m^{2} x 3 m = 0.393 kN/m

Allowance for beam self-weight = 1.0 kN/m

Allowance for mesh = 0.05 kN/m^{2} x 3m = 0.15 kN/m

Self-weight of dry concrete = 2.21 kN/m^{2 }x 3m = 6.63 kN/m

Weight of finishes = 1.2 kN/m^{2}

Weight of ceiling and services = 1 kN/m^{2}

Total g_{k} = 0.393 + 1 + 0.15 + 6.63 + 1.2 + 1 = 10.373 kN/m

**Variable Actions**

Imposed load on floor = 4 kN/m^{2 }x 3m = 12 kN/m

Movable partition allowance = 1 kN/m^{2} x 3m = 3 kN/m

Total q_{k} = 12 + 3 = 15 kN/m

At ultimate limit state = 1.35g_{k} + 1.5q_{k }= 1.35(10.373) + 1.5(15) = 36.5 kN/m

Design moment M_{Ed} = ql^{2}/8 = (36.5 x 7.5^{2})/8 = 256.6 kNm

Design shear force V_{Ed} = ql/2 = (36.5 x 7.5)/2 = 136.875 kN

An advanced UK beam S275 is to be used for this design.

F_{y} = 275 N/mm^{2}

Î³_{m0Â }= 1.0 (Clause 6.1(1) NA 2.15 BS EN 1993-1- 1:2005)

From steel tables, the properties of **406 x 178 x 67Â UKB** are;

Depth h = 409.4 mm

Width b = 178.8 mm

Web thickness t_{w} = 8.8mm

Flange thickness t_{f} = 14.3 mm

Root radius r = 10.2 mm

Depth between fillets d = 360.4 mm

Second moment of area y axis I_{y} = 24300 cm^{4}

Elastic modulus W_{el,y} = 1190 cm^{3}

Plastic modulus W_{pl,y} = 1350 cm^{3}

Area of section A = 85.5 cm^{2}

Height of web h_{w} = h â€“ 2t_{f} = 380.8 mm

E_{s} (Modulus of elasticity) = 210000 N/mm^{2Â }(Clause 3.2.6(1))

*Check out also â€¦.*

Design of Steel Beams to BS 5950 â€“ 1: 2000

Structural Analysis of Compound Arch-Frame Structure

**Classification of section**

Îµ = âˆš(235/F_{y}) = âˆš(235/275) = 0.92 (Table 5.2 BS EN 1993-1- 1:2005)

**Outstand flange **

Flange under uniform compression c = (b â€“ t_{w} â€“ 2r)/2 = [178.8 â€“ 8.8 â€“ 2(10.2)]/2 = 74.8 mm

c/t_{f} = 74.8/14.3 = 5.23

The limiting value for class 1 is c/t_{f}Â Â â‰¤ 9Îµ = 9 Ã— 0.92

5.23 < 8.28

Therefore, outstand flange in compression is class 1

**Internal Compression Part (Web under pure bending)**

c = d = 360.4 mm

c/t_{w}Â = 360.4/8.8 = 40.954

The limiting value for class 1 is c/t_{w}Â â‰¤ 72Îµ = 72 Ã— 0.92 = 66.24

40.954 < 66.24

Therefore, the web is plastic. Therefore, the entire section is class 1 plastic.

**Member Resistance Verification** â€“ Construction Stage

__Moment Resistance__

For the structure under consideration, the maximum bending moment occurs where the shear force is zero. Therefore, the bending moment does not need to be reduced for the presence of shear force (clause 6.2.8(2))

M_{Ed}/M_{c,Rd} â‰¤Â 1.0 (clause 6.2.5(1))

M_{c,Rd}Â = M_{pl,RdÂ }= (M_{pl,yÂ }Ã— F_{y})/Î³_{m0}

M_{c,Rd}Â = M_{pl,RdÂ }= [(1350 Ã— 275)/1.0] Ã— 10^{-3} = 371.25 kNm

At the construction stage;

M_{Ed}/M_{c,Rd}Â = 109.898/371.25 = 0.296 < 1.0 Ok

__Shear Resistance (clause 6.6.2)__

The basic design requirement is;

V_{Ed}/V_{c,Rd}Â â‰¤Â 1.0

V_{c,Rd}Â = V_{pl,RdÂ }= A_{v}(F_{yÂ }/ âˆš3)/Î³_{m0Â }(for class 1 sections)

For rolled I-section with shear parallel to the web, the shear area is;

A_{v}Â = A â€“ 2bt_{f} + (t_{w} + 2r)t_{f} (for class 1 sections) but not less than Î·h_{w}t_{w}

A_{v}Â = (85.5 Ã— 10^{2} â€“ (2 Ã— 178.8 Ã— 14.3) + [8.8 + 2(10.2)] Ã— 14.3 = 3854 mm^{2}

Î· = 1.0 (conservative)

Î·h_{w}t_{wÂ }= (1.0 Ã— 380.8 Ã— 8.8) = 3351.04 mm^{2}

3854 > 3351.04

Therefore, A_{v}Â = 3854 mm^{2}

The shear resistance is therefore;

V_{c,Rd}Â = V_{pl,RdÂ }= [3854 Ã— (275/ âˆš3)/1.0]Â Ã— 10^{-3}Â = 612 kN

At the construction stage;

V_{Ed}/V_{c,Rd}Â = 58.61/612 = 0.095 < 1.0 Ok

**Shear Buckling**

Shear buckling of the unstiffnened web will not need to be considered if;

h_{w}/t_{wÂ }â‰¤Â 72Îµ/Î·

h_{w}/t_{wÂ }= 380.8/8.8 = 43.27

72Îµ/Î·Â = (72 Ã—Â 0.92)/1.0Â = 66

43.27 < 66 Therefore shear buckling need not be considered.

**Design Resistance of Shear Connectors**

**Shear connector in a solid slab**

The design resistance of a single headed shear connector in a solid concrete slab automatically welded in accordance with BS EN 14555 should be determined as the smaller of;

P_{Rd} = (0.8 x *f _{u}* x Ï€ x 0.25d

^{2})/

*Î³*

_{v}(Clause 6.6.3.1(1) Equ(6.18) or

P

_{Rd}= [0.29 x Î± x d

^{2}x âˆš(f

_{ck}x E

_{cm})]/

*Î³*

_{v}

Where;

Î± = 1.0 as h

_{sc}/d = 95/19 > 4 (Equation 6.21)

P

_{Rd}= (0.8 x 450 x Ï€ x 0.25 x 19

^{2})/1.25 = 81.7 kN

P

_{Rd}= [0.29 x 1.0 x 19

^{2}x âˆš(25 x 31 x 10

^{3})]/1.25 = 73.7 kN

**Shear connectors in profiled steel sheeting**

For profiled sheeting with ribs running transverse to the supporting beams, P_{Rd,solid} should be multiplied by the following reduction factor;

k_{t }= (0.7/âˆšn_{r}) x (b_{0}/h_{p}) x (h_{sc}/h_{p} â€“ 1)

b_{0} = width of a trapezoidal rib at mid height of the profile = (133 + 175)/2 = 154 mm

h_{sc} = 95 mm

h_{p} = 60 mm

n_{r} = 1.0 (for one shear connector per rib)

k_{t }= (0.7/âˆš1.0) x (154/60) x (95/60 â€“ 1) = 1.0

Therefore P_{Rd} = k_{t}P_{Rd,solid} = 1.0 x 73.7 = 73.7 kN

The design resistance per rib = n_{r}P_{Rd} = 1 x 73.7 = 73.7 kN

**Degree of shear connection**

For composite beams in buildings, the headed shear connectors may be considered as ductile when the minimum degree of shear connection given in clause 6.6.1.2 is achieved.

For headed shear connectors with;

h_{sc} â‰¥ 4d and 16mm â‰¤ d â‰¤ 25 mm

The degree of shear connection may be determined from;

Î· = N_{c}/N_{c,f}

Where;

N_{c} is the reduced value of the compressive force in the concrete flange (i.e. force transferred by the shear connectors)

N_{c,f} is the compressive force in the concrete flange at full shear connection (i.e. the minimum of the axial resistance of the concrete and the axial resistance of the steel)

For steel sections with equal flanges and *L _{e}* < 25 m;

Î· â‰¥ 1 â€“ (355/f_{y}) x (0.75 â€“ 0.03*L _{e}*) where â‰¥ 0.4

*L*= distance between points of zero moment = 7.5 m

_{e}Î· â‰¥ 1 â€“ (355/275) x (0.75 â€“ 0.03 x 7.5) = 0.322, therefore Î· = 0.4

**Degree of shear connection present**

To determine the degree of shear connection present in the beam, the axial resistances of the steel and concrete are required (N_{pl,a} and N_{c,f }respectively)

Determine the effective width of the concrete flange

At the mid-span, the effective width of the concrete falnge is;

b_{eff }= b_{0 }+ âˆ‘b_{ei}

For n_{r} = 1.0, b_{0} = 0 mm

b_{ei} = L_{e}/8 but ot greater than b_{i}

L_{e} = 7.5 m (point of zero moment)

b_{i }= distance from the outside shear connector to a point between adjacent webs. Therefore;

b_{1} = b_{2} = 1.5 m

b_{e1 }= b_{e2} = L_{e}/8 = 7.5/8 = 0.9375 m

The effective flange width is therefore

b_{eff} = b_{0} + b_{e1} + b_{e2} = 0 + 0.9375 + 0.9375 = 1.875 m = 1875 mm

**Compressive resistance of the concrete flange**

The design strength of the concrete f_{cd} = 25/1.5 = 16.7 N/mm^{2}

The TR60^{+} profile has a 12 mm deep re-entrant above the stiffener making the overall profile depth h_{d} = 12 mm + 60 mm = 72 mm

The compressive resistance of the concrete flange is therefore;

N_{c,f} = 0.85f_{cd}b_{eff}h_{c} = 0.85 x 16.7 x 1875 x 58 x 10^{-3} = 1543.7 kN

Tensile Resistance of the steel member

N_{pl,a} = f_{y}.A = 275 x 85.5 x 10^{2} x 10^{-3} = 2351.25 kN

The compressive force in the concrete at full shear connection is the lesser. Therefore N_{c }= 1543.7 kN

**Resistance of the shear connectors**

*n* is the number of shear connectors present to the point of maximum bending moment. In this example, there are 7.5(2 x 0.333) = 12 ribs available for positioning shear connectors per half span.

N_{c} = *n* x P_{Rd} = 12 x 73.7 = 884.4 kN

The degree of shear connection present therefore is;

Î· = N_{c}/N_{c,f} = 884.4/1543.7 = 0.572 > 0.40 (Okay)

**Design Resistance of the Cross-section at the composite stage**

**Bending Resistance**

According to clause 6.2.1.2, the plastic rigid theory may be used for one connector per trough. With partial shear connection, the axial force in the concrete flange N_{c }is less than N_{pl,a} (884.4 kN < 2351.25 kN). Therefore, the plastic neutral axis lies within the steel section. Assuming that the plastic neutral axis lies a distance *x _{pl}* below the top of the flange of the section, where;

*x _{pl}* = (N

_{pl,a}â€“ N

_{c})/2f

_{y}b = (2351.25 â€“ 884.4)/(2 x 275 x 178.8) = 0.0149 m = 14.916 mm > t

_{f}(14.3 mm)

Therefore the plastic neutral axis lies below the top flange.

y_{c}Â = N_{c}/[0.85f_{ck}b_{eff}Â /*Î³*_{c}]Â â‰¤ h_{c}

y_{c}Â = (884.4 x 1000)/[0.85 x 25 x 1875/*1.5*]Â = 33.28 mm

M_{Rd }Â = N_{c}(h_{c} + d_{a} â€“ y_{c}/2) + 2bt_{f}f_{y}(d_{a} â€“ t_{f}/2) + t_{w}(y_{a} â€“ t_{f})(f_{y})(2d_{a} â€“ y_{a} â€“ t_{f})

M_{Rd }Â = (884.4 x 10^{3}) x (130 + 204.7 â€“ 33.28/2) + 2 x 178.8 x 14.3 x 275 x (204.7 â€“ 14.3/2) + 8.8(14.92 â€“ 14.3) x 275 x (2 x 204.7 â€“ 14.92 â€“ 14.3) = 559669744.2 Nmm = 559.669 kNm

M_{Ed} = 256.6 kNm

M_{Ed}/M_{Rd} = 256.6/559.669 = 0.458 < 1.0 (Okay)

**Shear Resistance at the Composite Stage**

The shear resistance is therefore;

V_{c,Rd}Â = V_{pl,RdÂ }= [3854 Ã— (275/ âˆš3)/1.0]Â Ã— 10^{-3}Â = 612 kN

V_{Ed} = 136.875 kN

V_{Ed}/V_{c,Rd}Â = 136.875/612 = 0.223 < 1.0 Ok

**Longitudinal shear resistance of the slab**

Neglecting the contribution of the steel, we ned to verify that;

A_{sf}f_{sd}/S_{f } > v_{Ed}h_{f}/cotÎ¸

Where;

v_{Ed} is the design longitudinal shear stress in the concrete slab

f_{sd} is the design yield strength of the reinforcing mesh = 0.87f_{yk} = 0.87 x 500 = 434.8 N/mm^{2}

h_{f} = depth of the concrete above the profiled sheeting = 70 mm

Î¸ angle of failure (try 26.5^{o})

A_{sf}/S_{f} = A_{t} (for the plane of failure shown as section a-a)

A_{t} is the cross-sectional area of transverse reinforcement mm^{2}/m)

The verification equation therefore becomes;

A_{t}f_{yd} > v_{Ed}h_{f}/cotÎ¸

The required area of tensile reinforcement At must satisfy the following;

At > v_{Ed}h_{f}/f_{yd}cotÎ¸

The longitudinal shear stress is given by;

v_{Ed} = âˆ†F_{d}/h_{f}âˆ†x

Where;

âˆ†x is the critical length under consideration, which is usually taken as the distance between the maximum bending moment and the support = L/2 = 7.5/2 = 3.75m

âˆ†F_{d} = N_{c}/2 = 884.4/2 = 442.2 kN

v_{Ed} = âˆ†F_{d}/h_{f}âˆ†x = (442.2 x 10^{3})/(70 x 3750) = 1.68 N/mm^{2}

v_{Ed}h_{f}/f_{yd}cotÎ¸ = (1.68 x 70)/(434.8 x cot 26.5^{o}) = 0.134 mm^{2}/mm

For the arrangement, the area of tensile reinforcement required is 134 mm^{2}/m

Therefore A142 mesh provided is adequate (A_{sprov }= 142 mm^{2}/m)

**Crushing of the concrete flange**

It is important to verify that

v_{Ed} < vf_{cd}sinÎ¸_{f}cosÎ¸_{f}

v = 0.6(1 â€“ f_{ck}/250) = 0.6 x (1 â€“ 25/250) = 0.54

vf_{cd}sinÎ¸_{f}cosÎ¸_{f} = 0.54 x 16.67 x sin(26.5) x cos(26.5) = 3.59 N/mm^{2}

(v_{Ed}) 1.68 N/mm^{2} < 3.59 N/mm^{2} (Okay)

**Serviceability limit state**

**Modular ratios**

For short term loading, the secant modulus of elasticity should be used. E_{cm} = 31 kN/mm^{2}. This corresponds to a modular ratio of;

n_{0} = E_{s}/E_{cm} = 210/31 = 6.77 (clause 5.4.2.2)

For long term loading;

n_{L} = n_{0}(1 + Ïˆ_{L}Ï•_{t})

Where Ïˆ_{L} is the creep multiplier taken as 1.1 for permanent loads and Ï•_{t} is the creep coefficient taken as 3.0.

n_{L} = 6.77 x (1 + 1.1 x 3) = 29.11

When calculating deflection due to variable action, the modular ratio is taken as;

n = 0.333n_{L} + 0.667n_{0} = 0.333(29.11) + 0.667(6.77) = 14.22

Since the beam is simply supported, use the gross value of second moment of area, I_{c}, of the uncracked section to calculate deflection.

I_{c }= I_{y} + [b_{eff}(h_{s} â€“ h_{p})^{3}/12âˆ™n_{i}] + [Aâˆ™b_{eff}(h_{s} â€“ h_{p})(h + h_{s} + h_{p})^{2}]/4[Aâˆ™n_{i} + b_{eff}(h_{s} â€“ h_{p})]

For n_{0} = 6.77

I_{c }= 24300 x 10^{4} + [1875(130 â€“ 60)^{3}/(12 x 6.77)] + [85.5 x 10^{2} x 1875(130 â€“ 60) x (409.4 + 130 + 60)^{2}]/4[85.5 x 10^{2} x 6.77 + 1875(130 â€“ 60)] = 78385 x 10^{4} mm^{4}

For n_{L} = 29.11

I_{c }= 24300 x 10^{4} + [1875(130 â€“ 60)^{3}/(12 x 29.11)] + [85.5 x 10^{2} x 1875(130 â€“ 60) x (409.4 + 130 + 60)^{2}]/4[85.5 x 10^{2} x 29.11 + 1875(130 â€“ 60)] = 50999 x 10^{4} mm^{4}

For n = 14.22

I_{c }= 24300 x 10^{4} + [1875(130 â€“ 60)^{3}/(12 x 14.22)] + [85.5 x 10^{2} x 1875(130 â€“ 60) x (409.4 + 130 + 60)^{2}]/4[85.5 x 10^{2} x 14.22 + 1875(130 â€“ 60)] = 64543 x 10^{4} mm^{4}

**Deflection due to actions on the steel at the construction stage;**

Actions = self weight of fresh concrete + mesh + sheeting + steel section

w_{1} = 5gl^{4}/384EI = (5 Ã— 8.323 Ã— 7500^{4})/(384 Ã— 210000 Ã— 24300 Ã— 10^{4}) = 6.719 mm

**Deflection due to permanent action on the steel at the composite stage;**

Actions = weight of finishes + ceiling and services

w_{2} = 5gl^{4}/384EI = (5 Ã— 6.6 Ã— 7500^{4})/(384 Ã— 210000 Ã— 50999 x 10^{4}) = 2.54 mm

**Deflection due to variable action on the steel at the composite stage;**

Actions = Imposed load + partition allowance

w_{3} = 5ql^{4}/384EI = (5 Ã— 15 Ã— 7500^{4})/(384 Ã— 210000 Ã— 64543 x 10^{4}) = 4.559 mm

Total deflection = w_{1} + w_{2} + w_{3} = 6.719 + 2.54 + 4.559 = 13.818 mm

Allowable deflection = L/360 = 7500/360 = 20.833 mm.

13.818 < 20.833 Therefore deflection is okay.

Good morning Engineer Ubani. How can I purchase a copy of the publication on profile metal deck

Hi Engr. Mike. Kindly send an e-mail to info@structville.com

Thanks.

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