Question

Design one way simply supported slab .Given the following

data. Span=4m; superimposed load=2KN/m2

floor

finish=1KN/m2

; concrete M15 and steel mild steel

.Support=345mm wide wall.​

Answer 1

Answer:

Design a one way slab with a clear span of 5m, simply supported on 230mm thick masonry walls and subjected to a live load of 4kN/m2 and a surface

finish of 1kN/mm2.Assume Fe 415 steel. Assume that the slab is subjected to moderate exposure conditions.

Step 1: Type of Slab.

ly/lx = 5/1 = 5>2.it has to be designed as one way slab.

Step 2:Effective depth calculation.

d = span/(basic value x modification factor) = 5000/(20x0.95) = 270mm D = 270 + 20 + 10/2 = 295mm

Step 3: Effective Span.

Le = clear span + effective depth = 5000 + 270 = 5.27m (or) Le =c/c distance b/w supports = 5000 + 2(230/2) = 5.23m Adopt effective span = 5.23m least value.