Civil Engineering - RCC Structures Design - Discussion
Discussion Forum : RCC Structures Design - Section 2 (Q.No. 44)
44.
A simply supported beam , 6 m long and of effective depth 50 cm, carries a uniformly distributed load 2400 kg/m including its self weight. If the lever arm factor is 0.85 and permissible tensile stress of steel is 1400 kg/cm2, the area of steel required, is
Discussion:
14 comments Page 1 of 2.
Walde said:
9 years ago
Ast = M/j * Qst * d.
= 18 cm^2.
= 18 cm^2.
(1)
Babar said:
8 years ago
At = (0.5 * 6 * 1.05 * 7200) ÷ 1400 = 16.2 = ~ 16.
(1)
Marwat said:
5 years ago
Ast = M/0.85(tensile stress * lever arm * depth).
M = wl^2/8,
Ast = 15.725.
M = wl^2/8,
Ast = 15.725.
(1)
Manoj said:
1 decade ago
What the formula for this?
AMRO said:
9 years ago
I think it must be 18.
A = M/J * FY * D.
A = M/J * FY * D.
Ankita said:
9 years ago
Somebody please explain it.
Krishna said:
9 years ago
Ast = M/(tensile stress * lever arm * depth).
M= wl^2/8,
Ast = 18.15.
M= wl^2/8,
Ast = 18.15.
Pramit said:
8 years ago
I think the answer is 18cm2.
Kushvaha said:
8 years ago
Ast =M /sigma st x j x d.
= 10800000 /1400 x.85 x500.
= 18.15 cmsq.
= 10800000 /1400 x.85 x500.
= 18.15 cmsq.
Xain said:
8 years ago
Here M stands for?
Can anyone explain in detail?
Can anyone explain in detail?
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