Hi everyone in this post i am providing you a detailed solution for a numerical of “Design of Two Way Slab” step by step.
Numerical:
Design a simply supported two way slab over a room 4.8 m x 4.0 m
effective, subjected to UDL 5 kN/m2
(inclusive of self wt.) Use M20 and
Fe 415. Draw reinforcement detail check for shear may not be given.
Take αx = 0.084 and αy = 0.059
Solution:
Given:
Effective dimensions of the two-way slab: 4.8 m×4 m4.8 , text{m} times 4 , text{m}
Load w=5 kN/m2w = 5 , text{kN/m}^2 (inclusive of self-weight)
Characteristic compressive strength of concrete fck=20 N/mm2f_{ck} = 20 , text{N/mm}^2
Yield strength of steel fy=415 N/mm2f_y = 415 , text{N/mm}^2
Coefficient αx=0.084alpha_x = 0.084
Coefficient αy=0.059alpha_y = 0.059
Design Constants:
For Fe 415, Mu lim=0.138 fck b d2
Estimation of Slab Thickness:
Assume MF = 1.4
Therefore,
d=span20×1.4d = frac{text{span}}{20 times 1.4}
d=400020×1.4d = frac{4000}{20 times 1.4}
d=142.86 mm say 150 mmd = 142.86 , text{mm} , text{say} , 150 , text{mm}
Assuming 10 mm Ø main bars, & c = 20 mm,
D=d+c+(Ø2)D = d + c + left(frac{text{Ø}}{2}right)
D=150+20+(102)D = 150 + 20 + left(frac{10}{2}right)
D=175 mmD = 175 , text{mm}
Effective Span:
Lxe=4000 mmL_{xe} = 4000 , text{mm}
Lye=4800 mmL_{ye} = 4800 , text{mm}
Consider 1 m wide strip
Load: wgiven=5 KN/m2w_{text{given}} = 5 , text{KN/m}^2
w=1×1×5=5 KN/m2w = 1 times 1 times 5 = 5 , text{KN/m}^2
Factored load:
wd=1.5×5=7.5 KN/m
w_d = 1.5 times 5 = 7.5 , text{KN/m}
Factored Bending Moment:
αx=0.084alpha_x = 0.084
αy=0.059alpha_y = 0.059
Mxd=αx⋅wd⋅Lxe2M_{xd} = alpha_x cdot w_d cdot L_{xe}^2
Mxd=0.084×7.5×42M_{xd} = 0.084 times 7.5 times 4^2
Mxd=10.08 KNmM_{xd} = 10.08 , text{KNm}
Myd=αy⋅wd⋅Lxe2M_{yd} = alpha_y cdot w_d cdot L_{xe}^2
Myd=0.059×7.5×42M_{yd} = 0.059 times 7.5 times 4^2
Myd=7.08 KNm
M_{yd} = 7.08 , text{KNm}
Effective Depth of Slab:
0.138 fck b d2=Mxd0.138 , f_{ck} , b , d^2 = M_{xd}
0.138×20×1000 d2=10.08×1060.138 times 20 times 1000 , d^2 = 10.08 times 10^6
dreqd=60.43 mm<(davailable=150 mm) Hence OKd_{text{reqd}} = 60.43 , text{mm} < (d_{text{available}} = 150 , text{mm}) , text{Hence OK}
Area and Spacing of Steel:
Astx=0.5fckfy[1−1−(4.6 Mxdfck⋅b⋅d2)]⋅b⋅dA_{stx} = 0.5 frac{f_{ck}}{f_y} left[ 1 – sqrt{1 – left( frac{4.6 , M_{xd}}{f_{ck} cdot b cdot d^2} right)} right] cdot b cdot d
Astx=0.5×20415[1−1−(4.6×10.08×10620×1000×150×150)]×1000×150A_{stx} = frac{0.5 times 20}{415} left[ 1 – sqrt{1 – left( frac{4.6 times 10.08 times 10^6}{20 times 1000 times 150 times 150} right)} right] times 1000 times 150
Astx=191.278 mm2
Area and Spacing of Distribution Steel:
Select 8 mm Ø bars, Spacing = min of a, b, c
a) Calculation of spacing SxS_x:
Sx=(π/4×82×1000191.27)S_x = left(frac{pi/4 times 8^2 times 1000}{191.27}right)
Sx=262.76 mm (say 260 mm)S_x = 262.76 , text{mm} , text{(say 260 mm)}
Sx=260 mm c/cS_x = 260 , text{mm c/c}
b) Maximum spacing based on depth:
3d=3×150=450 mm3d = 3 times 150 = 450 , text{mm}
c) Given maximum spacing:
300 mm300 , text{mm}
Therefore, spacing Sx=260 mm c/cS_x = 260 , text{mm c/c}.
d′=d−∅=150−8=142 mmd’ = d – varnothing = 150 – 8 = 142 , text{mm}
Asty=0.5×fckfy[1−1−(4.6×Mydfck×b×d2)]×b×d′A_{st_y} = frac{0.5 times f_{ck}}{f_y} left[ 1 – sqrt{1 – left( frac{4.6 times M_{yd}}{f_{ck} times b times d^2} right)} right] times b times d’
Asty=0.5×20415[1−1−(4.6×7.08×10520×1000×142×142)]×1000×142A_{st_y} = frac{0.5 times 20}{415} left[ 1 – sqrt{1 – left( frac{4.6 times 7.08 times 10^5}{20 times 1000 times 142 times 142} right)} right] times 1000 times 142
=141.07 mm2= 141.07 , text{mm}^2
Astmin=(0.12100)×b×DA_{st_{text{min}}} = left(frac{0.12}{100}right) times b times D
=(0.12100)×1000×175= left(frac{0.12}{100}right) times 1000 times 175
=210 mm2= 210 , text{mm}^2
Since AstyA_{st_y} calculated is very low, take AstxA_{st_x} and AstyA_{st_y} as 210 mm2^2.
Therefore, spacing of 8 mm diameter bars:
Sx=Sy=(π4×82×1000Astmin)S_x = S_y = left( frac{pi}{4} times frac{8^2 times 1000}{A_{st_{text{min}}}} right)
Sx=Sy=(50.26×1000210)S_x = S_y = left( frac{50.26 times 1000}{210} right)
=239.33 mm (say 230 mm c/c, which is less than 3d or 300 mm)= 239.33 , text{mm (say 230 mm c/c, which is less than 3d or 300 mm)}
Provide 8 mm Ø bars at 230 mm c/c in both directions.
Reinforcement Details:
