Answer Posted / aanand babasaheb chougule

Design steps of one way slab
1. Depth of slab
d= lx/20 * M.F. (M.F. = Modification factor)

Assume the dia. of bar (8, 10, 12) and
Final overall depth
D= d+15+0/2
d= d-15-0/2
2. Effective Span
This is taken as the lesser of the followings
a) Center to center distance between followings
L= lx + t
L= lx + d
3. Load
Consider 1m width of the slab and find out the total udl
a) W = (Self weight + Live Load + Floor finish)
b) Find factor load
Wu = 1.5 * w
4. Factor bending moment
Mu = Wu * l²/8
5. Equating Mu limit to Mu find the depth req. if it is
less than‘d’ than O.K otherwise
Increase D & repeat 2, 3, and 4
Mulimit = Mu
6. Area of main steel
Pt = 50fck/fy (1-√1 – 4.6 Mu / fck bd²)
Ast = Pt/100 *100* d
7. Spacing of main steel
S = Area one bar * 1000/ Ast
Check: - a) 3d
b) 450 mm
8. Distribution Steel
Astd = 0.15 % of Ag -Fe 250 (mild)
= 0.12% of Ag – Fe 415/5000 (torque)
Ag = b.D
9. Spacing of distribution steel
S= Area one bar * 1000/ Astd
Check: - a) 5d
b) 450 mm

10. Check for depth
a) Žv = V/bd
b) Pt —› Zc
c) Zv < Zc …….. No shear reinforcement req.

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