Design of RCC Slab -Limit Stress Method

Design of RCC Slab -Limit Stress Method

Design an interior panel of RC slab 3mX6m size, supported by wall of 300mm thick. Live load on the slab is 2.5kN/m2 .the slab carries 100mm thick lime concrete (density 19kN/m3 ).Use M15 concrete and Fe 415 steel.

Step 1: Type of Slab. ly/lx = 6/3 = 2 = 2.

it has to be designed as two way slab.

 Step 2:Effective depth calculation. 

For Economic consideration adopt shorter span to design the slab.

 d = span/(basic value x modification factor) 

= 3000/(20x0.95) = 270mm

 D = 270 + 20 + 10/2 = 295mm 

Step 3: Effective Span. 

For shorter span: Le = clear span + effective depth

 = 3000 + 270 = 3.27m 

(or) 

 Le =c/c distance b/w supports = 3000 + 2(230/2) =3.23m

Adopt effective span = 3.23m least value.

 For longer span: 

Le = clear span + effective depth = 6000 + 270 = 6.27m

 (or)

 Le =c/c distance b/w supports 

= 6000 + 2(230/2) = 6.23m 

 Adopt effective span = 6.23m least value. 

Step 4: load calculation 

Live load = 2.5kN/m2 

Dead load = 1x1x0.27x25 = 6.75kN/m2 

Dead load = 1x1x0.1x19 = 1.9kN/m2 

Floor Finish = 1kN/m2 

Total load = 12.15kN/m2 

Factored load = 12.15 x 1.5 = 18.225kN/m2 

Step 5: Moment calculation.

 Mx αx . w . lx 0.103x18.225x3.23 = 9.49kNm 

 My αy . w. lx 0.048 x18.225x3.23 = 4.425kNm 

Step 6: Check for effective depth. 

M = Qbd2

 d2 = M/Qb = 9.49/2.76x1 = 149.39mm say 150mm.

 For design consideration adopt d = 150mm. 

Step 7: Area of Steel. 

For longer span: 

Mu = 0.87 fy Ast d (1- (fy ast)/(fck b d)) 

4.425x106 = 087x415xAstx150(1-(415 Ast)/(20x1000x150))

 Ast = 180mm2 

Use 10mm dia bars 

Spacing ,S = ast/Astx1000 = (78.53/300)1000 = 261mm say 260mmc/c 

 Provide 10mm dia @260mm c/c.

 For shorter span: 

Mu = 0.87 fy Ast d (1- (fy ast)/(fck b d))

 9.49x106 = 087x415xAstx150(1-(415 Ast)/(20x1000x150)) 

Ast = 200mm2 Use 10mm dia bars Spacing ,

S = ast/Astx1000 = (78.53/300)1000 = 281mm say 300mmc/c 

 Provide 10mm dia @300mm c/c