Design Example of One-Way Slab (IS 456:2000)

A slab is one of the most important structural elements in reinforced concrete buildings. It transfers loads from floors to beams and columns.

When the length of the slab is more than twice the shorter span, the slab mainly bends in one direction. Such slabs are called one-way slabs.

In this article, we will learn the step-by-step design of a one-way slab according to IS 456:2000 using a practical example.

This example will help civil engineering students and site engineers understand the complete slab design process including depth calculation, load calculation, bending moment, reinforcement design, shear check, and deflection check.

Step-by-Step One Way Slab Design

Step 1: Given Data

Clear span = 3.5 m
Width of supports = 200 mm

Concrete grade = M20
Steel grade = Fe415

Live load = 4 kN/m²
Floor finish = 1 kN/m²

Step 2: Depth of Slab

Effective depth is assumed using span/depth ratio.$$d = \frac{span}{25}$$​ $$d = \frac{3500}{25} = 140 \, mm$$Assuming clear cover = 20 mm

Overall depth$$D = 165 \, mm$$

Step 3: Effective Span

Effective span is the smaller of the following values:

1. Clear span + effective depth
2. Centre to centre distance of supports

$$L = 3.64 \, m$$Step 4: Load Calculation

Self weight of slab$$= 0.165 \times 25$$ $$= 4.125 \, kN/m^2$$Floor finish$$= 1 \, kN/m^2$$Live load$$= 4 \, kN/m^2$$Total service load$$w = 9.125 \, kN/m^2$$Ultimate load$$wu = 1.5 \times w$$$$wu = 13.69 \, kN/m^2$$

Step 5: Bending Moment

For simply supported slab$$Mu = \frac{wuL^2}{8}$$$$Mu = 22.67 \, kN-m$$

Step 6: Check Limiting Moment

$$Mu,lim = 0.138 fck bd^2$$Mu,lim=0.138fckbd2 $$Mu,lim = 54 \, kN-m$$Mu,lim=54kN−m

Since$$Mu < Mu,lim$$Mu<Mu,lim

The section is under-reinforced.

Step 7: Main Reinforcement

Required steel area$$Ast = 524 \, mm^2$$Ast=524mm2

Provide

10 mm diameter bars @ 150 mm c/c

Step 8: Distribution Reinforcement

Minimum steel$$Ast = 0.12\% \text{ of gross area}$$Provide

8 mm diameter bars @ 250 mm c/c

Step 9: Shear Check

Shear stress$$\tau_v = 0.178 \, N/mm^2$$Permissible shear stress$$\tau_c = 0.30 \, N/mm^2$$Since$$\tau_v < \tau_c$$The slab is safe in shear.

Step 10: Deflection Check

$$(L/d)_{max} = 28$$Actual ratio$$(L/d) = 26$$(L/d)=26

Since$$26 < 28$$The slab satisfies deflection criteria.

Final Reinforcement

Main reinforcement
10 mm bars @ 150 mm c/c

Distribution reinforcement
8 mm bars @ 250 mm c/c

Overall slab thickness
165 mm

IS Code References

The following clauses of IS 456:2000 are used in slab design:

• Clause 23.2 – Deflection control
• Clause 26.5 – Minimum reinforcement
• Table 19 – Shear strength of concrete
• Clause 22 – Bending moment calculation
• SP-16 design tables for reinforcement

The design of a one-way slab involves several important steps such as calculating loads, determining slab thickness, checking bending moment, designing reinforcement, and verifying shear and deflection criteria.

By following the procedures given in IS 456:2000, engineers can design safe and efficient slab structures.

Understanding this design process is essential for civil engineering students, structural engineers, and site engineers working on reinforced concrete structures.

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