Design Calculations For Two Way Slab Page
Slab
Sample Calculations (By Rankine - Grashoff Method) Using IS 456 : 2000:
Slab Designation : S2 Two Way Slab
Live Load on Slab (wL): 2 kN/m2
Floor Finish Load on Slab (wF): 1 kN/m2
Clear Span of Slab in Long Direction N-S (Ly) : 5.9 m
Clear Span of Slab in Short Direction E-W (Lx) : 4.9 m
For continuous two way slab for serviciability condition
Modification Factor (Assumed): 1.4
Span to depth retio shall be (26xMF) 135 mm
Use Depth of Slab (D) : 135 mm
Effective Cover : 20 mm
Effective depth os slab ( d ) : 115 mm
Concrete Strength : 20 N/mm2
Yield Strngth of Steel : 500 N/mm2
Dead load of the slab : 0.135 x 25 kN/m2 take,density of Concrete 25kN/m 3
= 3.375 kN/m2
Total Working Load on Slab (wT) = 2+1+3.375 = 6.375 kN/m2
Total Ultimate Load on Slab (wT) = 6.375 x 1.5 = 9.563 kN/m2
Support Conditions: ( Ignoring Continuity ) N
North Edge Continuous (Y/N) : N
South Edge Continuous (Y/N) : N
East Edge Continuous (Y/N) : N W E
West Edge Continuous (Y/N) : N
Case no for the given support from I.S. 456 : 9 S
Effective span in long direction (Ly) is lesser of
I ) Clear span + d = 5.02 m
II ) Centre to centre of supports = 5.13 m Hence Lx = 5.02 m
Effective span in long direction (Ly) is lesser of
I ) Clear span + d = 6.015 m
II ) Centre to centre of supports = 6.13 m Hence Ly = 6.015 m
Retio of long span to short span ( Ly/Lx ) : 6.015 / 5.02
= 1.2
The co-efficient of possitive and negative span are interpolated from the chart
given in the Code.
Short span + ve 0.072
Short span - ve 0.000
Long span + ve 0.056
Long span - ve 0.000
Using these co-efficients the moments are calculated as followes.
M = a Wu.Lx² For midspan moment possitive co-efficients are used.
For support moments negative co-efficients are used.
0.5 x fck 4.6 x M
Ast = ( ----------------
0.5 fck/Fy ){1-SQRT[1-
[ 1 - SQRT{1- --------------- --------- }] b d
fy fck b d²
Short span + ve moment = 0.072x6.375 x 1.5 = 9.563x5.02x5.02
= 17.35 kN-m
0.5 x20 4.6 x 17.35e6
Area of steel required( Ast ) = -----------------{1 - SQRT[1- ---------------------------- ]}1000x115
500 20x 1000x115x115
= 378 mm²
Also minimum reinforcement is
Min Steel = 0.12 %
Min Steel = 0.12 / 100 x 1000 x 115
= 138 mm²
As steel required is more than min. steel hence actual calculated steel is to be provided.
Using bar dia = 10 mm A= 78.54 mm²
Spacing of bars = 78.54 x 1000/ 378
= 207.778 mm
Say = 205 mm
So use 10 mm bars at 205 mm centre centre
All other calculations for moments and steel requirements are tabulated below.
Directions Moment Coeff Moment R/F Reqd Dia Spac. R/F Provided
kN.m mm2 mm mm mm2 SAY
Short (+ve) : 0.072 17.35 378.11 10 207.7 10 @ 200 125
(-ve) : 0.000 0.00 138.00 10 569.1 10 @ 300 125
Long (+ve) : 0.056 13.50 287.93 10 272.8 10 @ 270 150
(-ve) : 0.000 0.00 138.00 10 569.1 10 @ 300 150
Check for effective depth ( d )
Mu = Ru x b x d² Ru = 2.98
d = SQRT[Mu/(Ru x b )]
= SQRT[17.35 / (2.98 x 1000 )]
= 0.076 m 76.31 mm The depth provided is more than required hence safe
Check for Under reinforced section :
Ultimate moment of resistance ( Mu ) = Ru x b x d²
= 2.98 x 1000 x 115 x 115
= 3.9E+07 N-mm
= 39.411 kN-m
As all the moments are less than the Ultimate Moment of Resistance SECTION IS UNDER REINFORCED.
Check for deflection :
Ast prov = 628 mm2
Ast reqd = 378.11 mm2
Pt used = 0.55 %
fs = 0.58*fy*(Ast reqd/ Ast prov) . . IS456,Pg38
fs = 174.604 N/mm2
MF (actual) = 1 / ( 0.225+(0.003 x fs)+0.625(log10(pt)) ) . . SP24,para 22.2.1
MF (actual) = 1.72 <= 2
MF (actual Used) = 1.72
d reqd. = 109.57 mm
d prov = 115 mm
Total Depth Provided = 135 mm
The depth provided is more than required hence safe
Check for Shear :
Ult. Shear force (Vu) = 23.43 kN
Ast prov = 628 mm2
Ast reqd = 378.11 mm2
Pt used = 0.55 %
fck = 20 N/mm2
k= 1.2 . .( IS456,cl.40.2.1.1 )
Design shear strength of concrete (tc) = 0.48 N/mm2 . .( IS456,tb19 )
Per. Shear Capacity = k.tc.b.d = 66.24 kN > Vu
Tc,max= 2.8 As, tc < tc,max , Hence ok.
The depth provided is more than required hence safe in shear
