Waffle slabs can be described as the equivalent of 2-way solid slabs especially when the spans are large, and ribbed system is to be adopted. The design of waffle slabs is the same as that of ribbed slabs, with the difference being that waffle slabs have ribs spanning in both directions, and the coefficients used for analysing the slab is similar to those used for two-way restrained slab. Waffle slabs are supported on beams or columns, where the support zones are made to be uniformly thick.

**Design Example**

A church building has a square grid of 7.5m x 7.5m of waffle slab, and is to support an imposed load of 5 kN/m

^{2}, design an interior panel of the waffle slab, and a supporting beam. f

_{ck}= 30 Mpa, f

_{yk}= 500 Mpa.

For details of waffle mould, see details of technical data sheet attached.

**Solution**

From basic span/effective depth ratio of 26, the trial depth of slab to be adopted = 7500/26 = 288 mm

From the table, let us select the following mould details;

Mould height = 225 mm

Top layer thickness = 75 mm

Total height of slab = 300 mm

Average rib width = 176 mm

**Load Analysis**

Self weight of concrete = 5.2 kN/m

^{2}(see technical data sheet)

Finishes = 1.2 kN/m

^{2}

Partition allowance = 1.5 kN/m

^{2}

Total permanent action = 7.9 kN/m

^{2}

Imposed load = 5 kN/m

^{2}

At ultimate limit state = 1.35gk + 1.5qk = 1.35(7.9) + 1.5(5) = 18.165 kN/m

^{2}

Load per rib = 0.9 × 18.165 = 16.35 kN/m

^{2}

**Analysis of the slab**

Treating as all sides fixed:

k = Ly/Lx = 1.0

Span coefficient = +0.024

Support coefficients = -0.032

Design of the span

**Designing the span as a T-beam;**

M

_{Ed}= 0.024 × 16.35 × 7.5

^{2 }= 22.07 kN.m

Effective depth (d) = h – C

_{nom}– ϕ/2 - ϕ

_{links}

Assuming ϕ12 mm bars will be employed for the main bars, and ϕ10 mm bars for the stirrups (links)

d = 300 - 25 - (12/2) -10 = 259 mm

k = M

_{Ed}/(f

_{ck}bd

^{2}) = (22.07 × 10

^{6})/(30 × 500 × 259

^{2}) = 0.0219

Since k < 0.167, no compression reinforcement required

z = d[0.5+ √(0.25 - 0.882k)]

k’ = 0.0219

z = d[0.5+ √((0.25 - 0.882(0.0219))] = 0.95d = 246.05 mm

Depth to neutral axis

*x*= 2.5(d - z) = 2.5(259 - 246.05) = 32.375 mm < 1.25h

_{f}(93.75 mm)

Therefore, we design rib as a rectangular section

Area of tension reinforcement A

_{s1}= M

_{Ed}/ (0.87f

_{yk}z)

A

_{s1}= M

_{Ed}/ (0.87f

_{yk}z) = (22.07 × 10

^{6}) / (0.87 × 500 × 0.95 × 259) = 206 mm

^{2}

Provide 2H12 Bot (A

_{Sprov}= 226 mm

^{2})

**Check for deflection**

ρ = A

_{s,req}/bd = 206 / (900 × 259) = 0.0008837

ρ

_{0}= reference reinforcement ratio = 10

^{-3}√(f

_{ck}) = 10

^{-3}√(30) = 0.00547

Since ρ ≤ ρ

_{0};

L/d = k [11 + 1.5√(f

_{ck}) ρ

_{0}/ρ + 3.2√(f

_{ck}) (ρ

_{0}/ρ - 1)

^{(3⁄2)}]

k = 1.5 (continuous system)

L/d = 1.5 [11 + 1.5√(30) × (0.00547/0.0008837) + 3.2√(30) × [(0.00547 / 0.0008837) - 1]

^{(3⁄2)}]

L/d = 1.5[11 + 50.855 + 207.227] = 269.082

β

_{s}= (500 As

_{prov})/(f

_{yk}As

_{req}) = (500 × 226) / (500 × 206) = 1.09

b

_{eff}/b

_{w}= 900/176 = 5.11

Therefore multiply basic length/effective depth ratio by 0.8

Therefore limiting L/d = 1.09 × 0.8 × 269.082 = 234.639

Actual L/d = 7500/259 = 28.96

Since Actual L/d (28.96) < Limiting L/d (234.639), deflection is satisfactory.

**Design of Supports;**

M

_{Ed}= 0.032 × 16.35 × 7.5

^{2 }= 29.43 kN.m

k = M

_{Ed}/(f

_{ck}bd

^{2}) = (29.43 × 10

^{6})/(30 × 176 × 259

^{2}) = 0.083

Since k < 0.167, no compression reinforcement required

z = d[0.5+ √(0.25 - 0.882k)]

k = 0.083

z = d[0.5+ √((0.25 - 0.882(0.083))] = 0.92d

Area of tension reinforcement A

_{s1}= M

_{Ed}/ (0.87f

_{yk}z)

A

_{s1}= M

_{Ed}/ (0.87f

_{yk}z) = (29.43 × 10

^{6}) / (0.87 × 500 × 0.92 × 259) = 283 mm

^{2}

Provide 3H12 TOP (A

_{Sprov}= 339 mm

^{2})

**Shear Design**

**Maximum shear force in the rib**V

_{Ed}= 0.33 × 16.35 × 7.5 = 40.466 kN/m

V

_{Rd,c}= [C

_{Rd,c}.k.(100ρ

_{1}f

_{ck})

^{(1/3)}+ k

_{1}.σ

_{cp}]b

_{w}.d ≥ (V

_{min}+ k

_{1}.σ

_{cp}) b

_{w}.d

C

_{Rd,c}= 0.18/γ

_{c}= 0.18/1.5 = 0.12

k = 1 + √(200/d) = 1 + √(200/259) = 1.88 < 2.0, therefore, k = 1.88

V

_{min}= 0.035k

^{(3/2)}f

_{ck}

^{0.5}

V

_{min}= 0.035 × (1.88)

^{1.5}× 30

^{0.5}= 0.494 N/mm

^{2}

ρ

_{1}= As/bd = 339/(176 × 259) = 0.00743 < 0.02; Therefore take 0.00743

V

_{Rd,c}= [0.12 × 1.88 (100 × 0.00743 × 30 )

^{(1/3)}] × 176 × 259 = 28941.534 N = 228.94 kN

Since V

_{Rd,c}(28.94 kN) < V

_{Ed}(40.466 kN), shear reinforcement is required.

The compression capacity of the compression strut (V

_{Rd,max}) assuming θ = 21.8° (cot θ = 2.5)

V

_{Rd,max}= (b

_{w}.z.v

_{1}.f

_{cd}) / (cotθ + tanθ)

V

_{1}= 0.6(1 - f

_{ck}/250) = 0.6(1 - 30/250) = 0.528

f

_{cd}= (α

_{cc}f

_{ck}) / γ

_{c}= (0.85 × 30) / 1.5 = 17 N/mm

^{2}

Let z = 0.9d

V

_{Rd,max }= [(176 × 0.9 × 259 × 0.528 × 17) / (2.5 + 0.4)] × 10

^{-3}= 126.981 kN

Since < V

_{Ed}< V

_{Rd,c}< V

_{Rd,max}

_{}Hence A

_{sw}/ S = V

_{Ed}/ (0.87 F

_{yk}z cot θ) = 40466 / (0.87 × 500 × 0.9 × 259 × 2.5 ) = 0.159

**Minimum shear reinforcement;**

A

_{sw}/ S = ρ

_{w,min}× b

_{w}× sinα (α = 90° for vertical links)

ρ

_{w,min}= (0.08 × √(f

_{ck})) / f

_{yk}= (0.08 × √30) / 500 = 0.000876

A

_{sw}/S

_{min}= 0.000876 × 176 × 1 = 0.154

Maximum spacing of shear links = 0.75d = 0.75 × 259 = 194.25 mm

Provide H8mm @ 175 mm c/c as shear links (A

_{sw}/ S = 0.574)

**Slab Topping**

A142 BRC Mesh can be provided or H8 @ 250mm c/c

Watch out for design of supporting beams...

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