Question:-
Match List-I (End conditions of columns) with List-II (Lowest critical load) and select the correct answer using the codes given below the list:
List- I
A. Column with both ends hinged
B. Column with both ends fixed
C. Column with one ends fixed and the other end hinged
D. Column with one ends fixed and the other end free
List –II
1. (p2EI)/L2
2. (2p2EI)/L2
3. (4p2EI)/L2
4. (p2EI)/4L2
(E is the young’s modulus of elasticity of column material, L is the length and I is the second moment of area of cross-section of the column)
Option (A)
A B C D
1 2 3 4

Option (B)
A B C D
3 2 1 4

Option(C)
A B C D
1 3 2 4

Option(D)
A B C D
2 4 3 1

Correct Option:
(C)
Question Solution:
Critical load PE = (p2El/L2eq)
Column with both ends pinned leq = l
\ PE = (p2El/l2)
Column with both ends fixed; leq = 1/2
\ PE = (4p2El/l2)
Column with one end fixed and other end hinged
Leq = (1/Ö2)
\ PE = (2p2El/l2)
Column with one end fixed and other end free Leq = 2l
\ PE = (p2El/4l2)
question-answer-faq-2924