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One Way Slab

This document provides a detailed example of one-way slab design, including given data such as span, loads, and material properties. It outlines calculations for effective depth, ultimate moments, shear forces, and reinforcement requirements, confirming that the design meets safety and deflection limits. The slab is determined to be under-reinforced and shear stress is within safe limits.

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Md Rihan Maaz
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33 views4 pages

One Way Slab

This document provides a detailed example of one-way slab design, including given data such as span, loads, and material properties. It outlines calculations for effective depth, ultimate moments, shear forces, and reinforcement requirements, confirming that the design meets safety and deflection limits. The slab is determined to be under-reinforced and shear stress is within safe limits.

Uploaded by

Md Rihan Maaz
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as PDF, TXT or read online on Scribd
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One-Way Slab Design Example

1 Given Data
• Clear span, L = 3.5 m
• Width of supports = 200 mm
• Live load = 4 kN/m2

• Floor finish = 1 kN/m2


• Grade of Concrete: M20, fck = 20 N/mm2
• Steel Reinforcement: Fe415, fy = 415 N/mm2

2 Depth of Slab
Assuming depth:
span 3500
d= = = 140 mm
25 25
Using 10 mm diameter bars and a clear cover of 20 mm:

Effective depth, d = 140 mm

Overall depth, D = 165 mm

3 Effective Span
The least value of:

• Clear span + effective depth: 3.5 + 0.14 = 3.64 m


• Centre-to-centre of supports: 3.5 + 0.20 = 3.70 m
⇒ Lef f = 3.64 m

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4 Loads
• Self-weight of slab: 0.165 × 25 = 4.125 kN/m2
• Floor finish = 1.000 kN/m2
• Live load = 4.000 kN/m2
Total service load, w = 9.125 kN/m
Ultimate load, wu = 1.5 × 9.125 = 13.69 kN/m

5 Ultimate Moments and Shear Forces


Mu = 0.125wu L2 = 0.125 × 13.69 × 3.642 = 22.67 kN.m
Vu = 0.5wu L = 0.5 × 13.69 × 3.64 = 24.92 kN

6 Limiting Moment of Resistance


Mu,lim = 0.138fck bd2
= (0.138 × 20 × 1000 × 1402 ) × 10−6
= 54 kN.m
Since Mu < Mu,lim , the section is under-reinforced.

7 Main Reinforcements
 
Ast fy
Mu = (0.87fy Ast d) 1 −
bdfck
 
6 415Ast
22.67 × 10 = (0.87 × 415Ast × 140) 1 −
1000 × 140 × 20
Solving for Ast :
Ast = 524 mm2
Using 10 mm diameter bars:
1000 × Ast 1000 × 78.5
sv = = = 150 mm
Ast 524
Adopt spacing of **150 mm**, with alternate bars bent up at supports.

8 Distribution Reinforcement
Ast = 0.12% of gross cross-sectional area
(0.0012 × 1000 × 165) = 198 mm2
Provide **8 mm bars at 250 mm c/c** (Ast = 201 mm2 ).

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9 Check for Shear Stress
Vu 24.92 × 103 2
τv = = = 0.178 N/mm
bd 1000 × 140
100Ast 100 × 0.5 × 524
pt = = = 0.187
bd 1000 × 140
Permissible shear stress in slab (Table 19, IS:456):
2
τc = (1.27 × 0.28) = 0.30 N/mm > τv

Shear stress is **within safe limits**.

10 Check for Deflection Control


(L/d)max = (L/d)basic × Kt × Kc × Kf
Kt = (100 × 524)/(1000 × 140) = 0.37
From IS456 tables:

• Fig. 8.1, K1 = 1.40


• Fig. 8.2, Kc = 1.00
• Fig. 8.3, Kf = 1.00

(L/d)max = 20 × 1.40 × 1.00 × 1.00 = 28


(L/d)actual = 3600/140 = 26 < 28
**Deflection limit is satisfied.**

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Figure 1: Detialing of One Way Slab

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