Hi everyone in this post i am providing the detailed RCC numerical solution in step by step. Please follow the steps to get the proper solutions.
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| Simply Supported Beam Calculate depth and area of steel |
Numerical:
Calculate depth and area of steel at mid span of a simply
supported beam over as clear span 6 m. The beam is carrying all
inclusive load 20 KN/m. Assume 300 mm bearings. Use M20 and
Fe500
Solution:
Given:
- Length of beam (l): 6 m
- Effective length (Le): 6 + 0.3/2 + 0.3/2 = 6.3 m
- Load (w): 20 kN/m
- Characteristic compressive strength of concrete (fck): 20 N/mm²
- Yield strength of steel (fy): 500 N/mm²
Solution:
- Moment Calculation:
M=w⋅Le28M = frac{w cdot L_e^2}{8}
=20⋅6.328= frac{20 cdot 6.3^2}{8}
=99.225 KNm= 99.225 text{ KNm}
- Factored Moment:
Md=γf⋅M
M_d = gamma_f cdot M
=1.5⋅99.225
= 1.5 cdot 99.225
=148.8375 KNm= 148.8375 text{ KNm}
- Maximum Moment Capacity:
Mu lim=0.133⋅fck⋅b⋅d2M_u text{ lim} = 0.133 cdot f_{ck} cdot b cdot d^2
- Assume beam width (b):
b=d/2
Equating Mu lim to Md:
0.133⋅fck⋅b⋅d2=148.83×106
0.133 cdot f_{ck} cdot b cdot d^2 = 148.83 times 10^6
0.133⋅20⋅(d/2)⋅d2=148.83×106
0.133 cdot 20 cdot (d/2) cdot d^2 = 148.83 times 10^6
d3⋅1.33=148.83×106
d^3 cdot 1.33 = 148.83 times 10^6
d3=111908270.7
d^3 = 111908270.7
d=481.89 mmd = 481.89 text{ mm}
Say, d=490 mmd = 490 text{ mm}
Assume effective cover (d′d’) = 40 mm:
D=d+d′=490+40=530 mmD = d + d’ = 490 + 40 = 530 text{ mm}
Therefore:
b=490/2=245 mmb = 490 / 2 = 245 text{ mm}
- Percentage of tensile reinforcement:
Pt lim=Astb⋅d⋅100
P_{t text{ lim}} = frac{text{Ast}}{b cdot d} cdot 100
=0.038⋅fck
= 0.038 cdot f_{ck}
=0.038⋅20
= 0.038 cdot 20
=0.76%
= 0.76 %
Ast=0.76⋅b⋅d100
text{Ast} = frac{0.76 cdot b cdot d}{100}
=0.76⋅245⋅490100
= frac{0.76 cdot 245 cdot 490}{100}
=912.38 mm2= 912.38 text{ mm}^2
Therefore:
Ast=912.38 mm2<Astmin=0.85⋅b⋅dfy=0.85⋅245⋅490500=204.085 mm2→Oktext{Ast} = 912.38 text{ mm}^2 < text{Ast}_{text{min}} = frac{0.85 cdot b cdot d}{f_y} = frac{0.85 cdot 245 cdot 490}{500} = 204.085 text{ mm}^2 rightarrow text{Ok}
Assuming 20 mm Ø bars to be used:
No. of bars:
=912.38(π4⋅202)=2.90→Say 3 Nos= frac{912.38}{left(frac{pi}{4} cdot 20^2right)} = 2.90 rightarrow text{Say 3 Nos}
Provide 3 bars of 20 mm Ø with area = 942.48 mm²
