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
Discussion:
11 comments Page 1 of 2.
Yousuf said:
8 years ago
Please explain clearly.
Praveen said:
8 years ago
Please Explain the answer.
Shakeel said:
8 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.
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.
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.
l=2a is a condition for zero bending moment at centre.
Jinz said:
5 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.
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.
-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.
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers