Cantilever Slab Design Step-by-Step as per IS 456:2000 By Mohan Dangi (Gold medalist)

Comprehensive Cantilever Slab Design Example (IS 456:2000) – Complete Step-by-Step Guide

📋 Table of Contents

• Project Overview
• Step 1 – Depth & Effective Span Calculation
• Step 2 – Load Analysis & Internal Forces
• Step 3 – Reinforcement Design
• Step 4 – Design Verification Checks
• Design Summary Table
• Key Design Considerations

Project Overview

This comprehensive example demonstrates the complete design procedure for a one-way cantilever slab commonly used in balconies, roof overhangs, and architectural projections.

Parameter	Value
Slab width	4.5 m
Clear span	1.5 m
Supporting beam size	250 mm × 450 mm
Concrete grade	M30 (fck = 30 MPa)
Steel grade	Fe 500 (fy = 500 MPa)
Live load	4.0 kN/m²
Floor finish	1.2 kN/m²
Exposure condition	Moderate
Clear cover	25 mm

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📏 Step 1 – Depth & Effective Span Calculation

1.1 Trial Depth Calculation

According to IS 456:2000, Clause 23.2.1(c), the depth calculation for deflection control uses:

Formula: $d = \frac{L_x}{7 \times M_f}$

Where:
• $L_x$ = Clear span = 1500 mm
• $M_f$ = Modification factor = 1.5 (for cantilever slab)

Calculation:

$$d = \frac{1500}{7 \times 1.5} = \frac{1500}{10.5} = 142.9 \text{ mm}$$

1.2 Overall Depth Determination

The overall depth includes effective depth, cover, and bar diameter:

$D = d + \text{cover} + \frac{\phi}{2}$

Assuming 10 mm diameter bars:

$$D = 142.9 + 25 + \frac{10}{2} = 172.9 \text{ mm}$$

Provide D = 180 mm

Available effective depth: $d = 180 - 25 - 5 = 150 \text{ mm}$

1.3 Effective Span

As per IS 456:2000, Clause 22.2(c):

$L_e = L_{\text{clear}} + \frac{d}{2}$

$$L_e = 1500 + \frac{150}{2} = 1575 \text{ mm} = 1.575 \text{ m}$$ ⬆️ Back to top

Step 2 – Load Analysis & Internal Forces

2.1 Load Computation

Load Type	Calculation	Value (kN/m²)
Self-weight	25 × 0.18	4.50
Floor finish	Given	1.20
Live load	Given	4.00
Total service load (w)	Sum	9.70
Factored load (wu = 1.5w)	1.5 × 9.70	14.55

2.2 Maximum Bending Moment

For cantilever slab with uniformly distributed load:

$M_u = \frac{w_u \times L_e^2}{2}$

$$M_u = \frac{14.55 \times 1.575^2}{2} = \frac{14.55 \times 2.48}{2} = 18.07 \text{ kN·m}$$

2.3 Maximum Shear Force

$V_u = w_u \times L_e$

$$V_u = 14.55 \times 1.575 = 22.92 \text{ kN}$$ ⬆️ Back to top

Step 3 – Reinforcement Design

3.1 Depth Check for Flexure

Using limit state method for Fe 500 steel (SP-16 Chart 46):

$M_u = 0.133 \times f_{ck} \times b \times d^2$

Required depth:

$$d_{\text{required}} = \sqrt{\frac{M_u}{0.133 \times f_{ck} \times b}} = \sqrt{\frac{18.07 \times 10^6}{0.133 \times 30 \times 1000}} = 67.3 \text{ mm}$$

Since d_provided (150 mm) > d_required (67.3 mm) → SAFE

3.2 Main Reinforcement Design

Steel area calculation using balanced section formula:

$A_{st} = 0.5 \times \frac{f_{ck}}{f_y} \times \left[1 - \sqrt{1 - \frac{4.6 \times M_u}{f_{ck} \times b \times d^2}}\right] \times b \times d$

Substituting values:

$$A_{st} = 0.5 \times \frac{30}{500} \times \left[1 - \sqrt{1 - \frac{4.6 \times 18.07 \times 10^6}{30 \times 1000 \times 150^2}}\right] \times 1000 \times 150$$ $$A_{st} = 256.2 \text{ mm}^2\text{/m}$$

3.3 Minimum Reinforcement Check

As per IS 456:2000, Clause 26.5.1.1 for cantilever slabs:

$A_{st,min} = \frac{0.85 \times b \times d}{f_y}$

$$A_{st,min} = \frac{0.85 \times 1000 \times 150}{500} = 255 \text{ mm}^2\text{/m}$$

Since calculated area (256.2 mm²) > minimum (255 mm²), use calculated value.

3.4 Bar Spacing Calculation

Using T10 mm bars (Area = 78.54 mm² per bar):

$\text{Spacing} = \frac{\text{Area of bar} \times 1000}{\text{Required area per meter}}$

$$\text{Spacing} = \frac{78.54 \times 1000}{256.2} = 306.5 \text{ mm}$$

Provide T10 @ 300 mm c/c

Actual area provided: $\frac{78.54 \times 1000}{300} = 261.8 \text{ mm}^2\text{/m}$

3.5 Distribution Reinforcement

Minimum distribution steel:

$A_{s,min} = \frac{0.12}{100} \times b \times D$

$$A_{s,min} = \frac{0.12}{100} \times 1000 \times 180 = 216 \text{ mm}^2\text{/m}$$

Provide T10 @ 350 mm c/c (224.4 mm²/m)

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Step 4 – Design Verification Checks

4.1 Shear Check

Nominal shear stress:

$\tau_v = \frac{V_u}{b \times d}$

$$\tau_v = \frac{22.92 \times 10^3}{1000 \times 150} = 0.153 \text{ MPa}$$

Steel percentage: $p_t = \frac{261.8}{1000 \times 150} \times 100 = 0.175\%$

From IS 456:2000 Table 19, for pt = 0.175%: τc = 0.31 MPa

Since τv (0.153 MPa) < τc (0.31 MPa) → SAFE in shear

4.2 Deflection Check

Steel stress calculation:

$f_s = 0.58 \times f_y \times \frac{A_{st,required}}{A_{st,provided}}$

$$f_s = 0.58 \times 500 \times \frac{256.2}{261.8} = 284.1 \text{ MPa}$$

Modification factor from IS 456 Figure 4: Kt ≈ 1.68

Permissible L/d ratio: $(\frac{L}{d})_{permissible} = 7 \times K_t = 7 \times 1.68 = 11.76$

Actual L/d ratio: $(\frac{L}{d})_{actual} = \frac{1500}{150} = 10.0$

Since actual < permissible → SAFE in deflection

4.3 Development Length Check

For M30 concrete and T10 bars with Fe 500:

$L_d = \frac{\phi \times f_y}{4 \times \tau_{bd}} \approx 470 \text{ mm}$

Available anchorage length at support = 500 mm > 470 mm

Development length requirement satisfied

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Design Summary Table

Design Parameter	Provided Value	Code Reference
Overall Depth (D)	180 mm	IS 456 Cl. 23.2.1
Effective Depth (d)	150 mm	Calculated
Main Reinforcement	T10 @ 300 mm c/c	IS 456 Cl. 26.5.1.1
Distribution Steel	T10 @ 350 mm c/c	IS 456 Cl. 26.5.2
Clear Cover	25 mm	IS 456 Table 16
Steel Percentage	0.175%	Above minimum
Factored Moment	18.07 kN·m	Calculated
Factored Shear	22.92 kN	Calculated

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Key Design Considerations

• Reinforcement Detailing: Ensure proper anchorage of main bars at the fixed support to resist negative moments effectively.
• Serviceability Control: Deflection often governs cantilever slab design; maintain adequate depth-to-span ratios.
• Construction Quality: Proper concrete consolidation and curing are critical, especially at the tension face.
• Load Path: Understand that maximum moment and shear occur at the support, requiring careful detailing.
• Code Compliance: All calculations follow IS 456:2000 provisions for safety and serviceability.
• Minimum Steel: Cantilever slabs require enhanced minimum reinforcement due to their structural behavior.

Note: This design example provides a comprehensive methodology that can be adapted for similar cantilever slab projects with different dimensions and loading conditions while maintaining code compliance. Written By Mohan Dangi (Gold medalist)

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