Civil Engineering - UPSC Civil Service Exam Questions - Discussion
Discussion Forum : UPSC Civil Service Exam Questions - Section 20 (Q.No. 9)
9.
A fixed beam and a simply supported beam having same span and develop same maximum bending moment due to uniformly distributed load on entire span. What is the ratio of uniformly distributed load on fixed beam to that on simply supported beam ?
Discussion:
8 comments Page 1 of 1.
Santu naskar said:
8 years ago
According to me, it is (12M/L^2)/(8M/L^2) = 12/8 = 1.5.
(3)
Chhaya said:
8 years ago
@Santu.
12M/L^2 at support on fix beam on center fix beam 24M/L^2.
So answer is right.
12M/L^2 at support on fix beam on center fix beam 24M/L^2.
So answer is right.
(2)
Rocky Patel said:
8 years ago
Maximum bending moment in the fixed beam is at support which is 12 M/L^2. So, 1.5 is the right answer.
(1)
Roy said:
8 years ago
1/3 should be the answer. Because it's the ratio of fixed to s.s.b (wl2/24÷wl2/8)=1/3.
(1)
Phani said:
7 years ago
In question, the ratio of UDL should be find.
Not to find FEM if it is FEM then answer ratio is 1.
FEM for UDL acting on both the beams is M = wl^2/12.
Not to find FEM if it is FEM then answer ratio is 1.
FEM for UDL acting on both the beams is M = wl^2/12.
Manisha thakur said:
7 years ago
Option 3 is correct.
Max bending moment at fixed beam =wl^2/24.
Max bending moment at simply-supported beam =wl^2/8.
And the ratio of these two will be = 3.
Max bending moment at fixed beam =wl^2/24.
Max bending moment at simply-supported beam =wl^2/8.
And the ratio of these two will be = 3.
(1)
Nandu said:
6 years ago
I agree with @Santu Naskar and @Rocky Patel. It is 1.5.
(1)
A Ali said:
5 years ago
The correct answer is 1.5.
(1)
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