# 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, isDiscussion:

11 comments Page 1 of 2.
Sadulla said:
2 years ago

@Pravin.

Yes, you are correct.

Yes, you are correct.

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.

-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.

Hamza cheema said:
4 years ago

You are right, Thanks @Pravin.

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.

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.

Jinz said:
4 years ago

@Kajal how does it come out .207 L.

Nilraj said:
5 years ago

Kajal is right @Naitik.

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

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

Naitik Trivedi said:
6 years ago

@Kajal.

After solving this formula;

Answer is l= 2a.

a=0.5l (E) is correct answer.

After solving this formula;

Answer is l= 2a.

a=0.5l (E) is correct answer.

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.

Max hogging B.M = wa^2/2.

Sagging B.M = Hogging B.M ( acc to ques);

a = 0.207L.

Shakeel said:
7 years ago

How to solve it?

Praveen said:
8 years ago

Please Explain the answer.

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