 |
| RCC Numerical - Moment of Resistance |
Numerical:
Find the moment of resistance Mu for a beam 300 x 600 mm, effective provided with 3 bars of 16 mm diameter and 2 bars of 12 mm diameter on tension side. M20& Fe500 are used.
Solution:
Given:
b = 300 mm
d = 600 mm
fck = 20 N/mm²
Ast = 3 x Ï€/4 x 16² + 2 x Ï€/4 x 12²
= 829.38 mm²
Solution:
Xu = (0.87fy Ast) / (0.36 fck.b)
= (0.87 x 500 x 829.38) / (0.36 x 20 x 300)
= 167.03 mm
Xu max = 0.46d
= 0.46 x 600 = 276 mm
As Xu < Xu max, the section is under-reinforced.
Mu = 0.87 fy Ast (d – 0.42 Xu)
= 0.87 x 500 x 829.38 [600 – 0.42(167.03)]
= 191.158 x 10⁶ Nmm = 191.158 KNm
b = 300 mm
d = 600 mm
fck = 20 N/mm²
Ast = 3 x Ï€/4 x 16² + 2 x Ï€/4 x 12²
= 829.38 mm²
Solution:
Xu = (0.87fy Ast) / (0.36 fck.b)
= (0.87 x 500 x 829.38) / (0.36 x 20 x 300)
= 167.03 mm
Xu max = 0.46d
= 0.46 x 600 = 276 mm
As Xu < Xu max, the section is under-reinforced.
Mu = 0.87 fy Ast (d – 0.42 Xu)
= 0.87 x 500 x 829.38 [600 – 0.42(167.03)]
= 191.158 x 10⁶ Nmm = 191.158 KNm
0 Comments