Civil Engineering - Theory of Structures - Discussion
Discussion Forum : Theory of Structures - Section 1 (Q.No. 14)
14.
Keeping the depth d constant, the width of a cantilever of length l of uniform strength loaded with a uniformly distributed load w varies from zero at the free end and
at the fixed end
at the fixed end
at the fixed end
at the fixed endDiscussion:
23 comments Page 1 of 3.
Jeganathan said:
4 years ago
Sigma is nothing but f (bending stress).
And Z(section modulus) = bd2/6.
Then,
B.M. at fixed support is wl2/2.
M= f.(I/Y)
M=f.Z
wl2/2 = f * Z.
wl2/2 = f * bd2/6.
b= (3wl2)/fd2.
Answer C.
And Z(section modulus) = bd2/6.
Then,
B.M. at fixed support is wl2/2.
M= f.(I/Y)
M=f.Z
wl2/2 = f * Z.
wl2/2 = f * bd2/6.
b= (3wl2)/fd2.
Answer C.
(6)
ALEX said:
1 year ago
M/I = F/Y.
WL2/2 = F * Z = F * BD2/6.
B = 3WL2/(F*D2).
WL2/2 = F * Z = F * BD2/6.
B = 3WL2/(F*D2).
(1)
Ravi said:
9 years ago
This is not uniform load so b.m is not (w*l^2)/2 varying load so b.m is (w*i^2)/6 answer.
(1)
Abiy said:
3 years ago
The right Answer is A.
(1)
Patil said:
8 years ago
Yes, @Ravi
You are correct, bm not wl^2/2. It is wl^2 /6 n final and is wl^2/(d^2*s).
You are correct, bm not wl^2/2. It is wl^2 /6 n final and is wl^2/(d^2*s).
(1)
Abiy said:
3 years ago
@Gupta.
Thanks for your explanation but for load w uniformly distributed from free end zero.
W=wl/2.
Which it's Mmax=wl/3 At the fixed end.
So, the answer is C.
Thanks for your explanation but for load w uniformly distributed from free end zero.
W=wl/2.
Which it's Mmax=wl/3 At the fixed end.
So, the answer is C.
(1)
Bishwo said:
4 years ago
Bending moment at fixed =1/3 (wl^2) /2.
1/3 is because it's a triangular slab of thickness d and with b. 1 part of whole rectangle will give BM at the fixed end.
1/3 is because it's a triangular slab of thickness d and with b. 1 part of whole rectangle will give BM at the fixed end.
(1)
Anshul Gupta said:
1 decade ago
Sigma denoted by S.
And Z(section modulus) = bd2/6.
Then,
B.M. at fixed support is wl2/2.
wl2/2 = S*Z.
wl2/2 = S*bd2/6.
b= (3wl2)/Sd2.
Answer C.
And Z(section modulus) = bd2/6.
Then,
B.M. at fixed support is wl2/2.
wl2/2 = S*Z.
wl2/2 = S*bd2/6.
b= (3wl2)/Sd2.
Answer C.
(1)
Narayan said:
7 years ago
s= σ(direct stress).
M= wl^2-1/2*(l*w)*2/3(l)
= wl^2/6.
But,
M=S * Z.
wl^2/6 = S * b * d^2/6.
b = wl^2/s * d^2.
M= wl^2-1/2*(l*w)*2/3(l)
= wl^2/6.
But,
M=S * Z.
wl^2/6 = S * b * d^2/6.
b = wl^2/s * d^2.
Vishal Singh said:
2 years ago
If it is zero at the free end then it should be a uniformly varying load with zero at the free end.
How can uniformly distributed load is zero at free?
How can uniformly distributed load is zero at free?
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