# Structural Design of Cantilever Slabs

Cantilever slabs are common features in residential and commercial buildings due to the need to have bigger spaces on upper floors. To achieve this, architects normally extend the slab beyond the ground floor building line, thereby forming a cantilever. Cantilever slabs in reinforced concrete buildings are usually characterised by a reinforced concrete slab projecting from the face of the wall and having a back span that extends into the interior panels of the building.

The design of reinforced concrete cantilever slab is usually governed by serviceability limit state requirements. Typically, the design involves the provision of adequate concrete thickness and reinforcement to prevent excessive deflection and vibration.

During the detailing of reinforced concrete cantilever slabs, the main reinforcements are provided at the top and they extend into the back span by at least 1.5 times the length of the cantilever or 0.3 times the length of the back span, whichever is greater. Furthermore, it is important to provide at least 50% of the reinforcement provided at the top, and at the bottom.

In this article, we are going to show how we can analyse and design cantilever slabs subjected to floor load and block wall load according to the requirements of EN 1992-1-1:2004 (Eurocode 2).

## Worked Example – Design of Cantilever Slabs

A cantilever slab 200 mm thick is 1.715m long, and it is supporting a blockwork load at 1.0m from the fixed end. Design the slab using the data given below;

Purpose of building – Residential
fck = 25 Mpa
fyk = 460 Mpa
Concrete cover = 25 mm
Height of block wall = 2.75 m
Unit weight of concrete = 25 kN/m3
Unit weight of block with renderings = 3.75 kN/m2

Self weight of slab = (25 × 0.2) = 5 kN/m2
Finishes (assume) = 1.2 kN/m2
Partition allowance = 1.0 kN/m2
Total characteristic permanent action (pressure load) gk  = 7.2 kN/m2

Permanent action from wall Gk = 3.75  × 2.75 = 10.3125 kN/m

Variable action on slab qk = 1.5 kN/m2

At ultimate limit state;
n = 1.35gk + 1.5qk
n = 1.35(7.2) + 1.5(1.5) = 11.97 kN/m2

Ultimate load from wall = 1.35 x 10.3125 =  13.92 kN/m

### Design Forces

MEd = (13.92 × 1) + (11.97 × 1.7152)/2 = 31.523 kNm
VEd = (13.92) + (11.97 × 1.715) = 34.45 kN

### Flexural design

MEd  = 31.523 kNm
Effective depth (d) = h – Cnom – ϕ/2 – ϕlinks
Assuming ϕ12 mm bars will be employed for the main bars
d = 200 – 25 – 6 = 169 mm

k = MEd/(fckbd2) = (31.523 × 106)/(25 × 1000 × 1692) = 0.044

Since k < 0.167, no compression reinforcement required
z = d[0.5+ √(0.25 – 0.882K) ]
z = d[0.5+ √(0.25 – 0.882(0.044))] = 0.95d

As1 = MEd/(0.87fyk z) = (31.523 × 106)/(0.87 × 460 × 0.95 × 169) = 490 mm2/m
Provide X12@200 c/c  TOP (ASprov = 565 mm2/m)

### Check for deflection

ρ = As,prov /bd = 565 / (1000 × 169) = 0.0033
ρ0 = reference reinforcement ratio = 10-3√(fck) = 10-3√(25) = 0.005
Since if ρ ≤ ρ0;

L/d = K [11 + 1.5√(fck) ρ0/ρ + 3.2√(fck) (ρ0 / ρ – 1)(3⁄2)

k = 0.4 (Cantilevers)

L/d = 0.4 [11 + 1.5√(25) × (0.005/0.0033) + 3.2√(25) × [(0.005 / 0.0033) – 1](3⁄2)
L/d = 0.4[11 + 11.363 + 5.9159] = 11.311

βs = (500 Asprov)/(fyk Asreq) = (500 × 565) / (460 × 490) = 1.253

Therefore limiting L/d = 1.253  × 11.311 = 14.172
Actual L/d = 1715/169 = 10.147

Since Actual L/d (10.147) < Limiting L/d (14.172), deflection is satisfactory.

Exercise for Students
(1) Provide distribution bars
(2) Verify the section for shear
(3) Check for cracking
(4) Do the detailing sketches

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