# Civil Engineering - Strength of Materials - Discussion

Discussion Forum : Strength of Materials - Section 4 (Q.No. 33)
33.
A beam of length L supported on two intermediate rollers carries a uniformly distributed load on its entire length. If sagging B.M. and hogging B.M. of the beam are equal, the length of each overhang, is
0.107 L
0.207 L
0.307 L
0.407 L
0.5 L.
Explanation:
No answer description is available. Let's discuss.
Discussion:
11 comments Page 1 of 2.

Yousuf said:   8 years ago

Praveen said:   8 years ago

Shakeel said:   7 years ago
How to solve it?

Kajal Tomar said:   6 years ago
Max. Sagging B.M = w/8( L^2 -4a^2).
Max hogging B.M = wa^2/2.
Sagging B.M = Hogging B.M ( acc to ques);
a = 0.207L.

Naitik Trivedi said:   6 years ago
@Kajal.

After solving this formula;
Answer is l= 2a.
a=0.5l (E) is correct answer.

Nilraj said:   6 years ago
Kajal is right @Naitik.

l=2a is a condition for zero bending moment at centre.

Jinz said:   4 years ago
@Kajal how does it come out .207 L.

Pravin said:   4 years ago
@All.

Max. Sagging B.M = w/8( L^2 -4a^2).
Max hogging B.M = wa^2/2.
Sagging B.M = Hogging B.M,
w/8( L^2 -4a^2)=wa^2/2,
After solving;
a=0.353l.

Hamza cheema said:   4 years ago
You are right, Thanks @Pravin.

Prakash basyal said:   3 years ago
Sagging B.M at the centre = Hogging B.M at support.
-W*L/2*L/4+WL/2(L/2-X) = WX^2/2.
L^2-4LX = 4X^2.
L^2-4LX+4X^2 = 8X^2.
(L-2X)^2 = 8X^2.
X = 0.207L. .
So, B is correct.