Civil Engineering - Theory of Structures - Discussion

Discussion Forum : Theory of Structures - Section 1 (Q.No. 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 end
Answer: Option
No answer description is available. Let's discuss.
22 comments Page 1 of 3.

Vishal Singh said:   10 months 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?

Abiy said:   2 years ago
The right Answer is A.

Abiy said:   2 years ago

Thanks for your explanation but for load w uniformly distributed from free end zero.
Which it's Mmax=wl/3 At the fixed end.
So, the answer is C.

Jeganathan said:   3 years ago
Sigma is nothing but f (bending stress).
And Z(section modulus) = bd2/6.

B.M. at fixed support is wl2/2.
M= f.(I/Y)
wl2/2 = f * Z.
wl2/2 = f * bd2/6.
b= (3wl2)/fd2.

Answer C.

Bishwo said:   3 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.

Michael said:   3 years ago
Thanks @Anshul Gupta.

SHUBHAM SIDDHARTH NAYAK said:   3 years ago
Cantilever carrying UDL- BM at fixed end = wl^2 /2.
We know, M = fz.
Wl^2/2= f *(bd^2/6),
b =3wl^2/fd^2.

Syed Mazhar said:   4 years ago
From bending eq
M/I= f/y=E/R

Dhananjay said:   5 years ago

How can you equate BM with S*Z? Here, units are different. Please explain.

Nilraj. said:   5 years ago

This is uniformly Distributed load.

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