Mechanical Engineering - Strength of Materials - Discussion
Discussion Forum : Strength of Materials - Section 6 (Q.No. 28)
28.
A column of length (l) with both ends fixed may be considered as equivalent to a column of length __________ with one end fixed and the other end free.
Discussion:
10 comments Page 1 of 1.
Deepak said:
9 years ago
It must be option c) l/2.
(2)
DURGESH KUMAR DUBEY said:
9 years ago
Option B is correct because there compare from both end fixed instead of both end hinged.
Neeraj said:
9 years ago
You are right @Durgesh.
Vikram said:
8 years ago
According to me, The correct option is C.
(2)
Sweta said:
8 years ago
Yes right @Durgesh.
Shilpa said:
8 years ago
option B is right.
The equivalent length for both ends fixed beam(Le)=l/2.
The equivalent length for one end fixed and other ends free(Le)=2L.
crippling load of column=(pie)^2.E.i/(Le)^2.
now equating both columns.
(π)^2.E.i/(l/2)^2=(π)^2.E.i/(2L)^2
4/l^2=1/4L^2.
By cross multiplication
16L^2=l^2.
sqrt(16L^2) =l.
4L=l.
L=l/4 hence proved.
A column of lenghth 'l' having both ends fixed is equal to a column of length l/4 having on end fixed and other end free.
The equivalent length for both ends fixed beam(Le)=l/2.
The equivalent length for one end fixed and other ends free(Le)=2L.
crippling load of column=(pie)^2.E.i/(Le)^2.
now equating both columns.
(π)^2.E.i/(l/2)^2=(π)^2.E.i/(2L)^2
4/l^2=1/4L^2.
By cross multiplication
16L^2=l^2.
sqrt(16L^2) =l.
4L=l.
L=l/4 hence proved.
A column of lenghth 'l' having both ends fixed is equal to a column of length l/4 having on end fixed and other end free.
(2)
Harender said:
7 years ago
B is right and well explained @Shilpa.
(1)
Gohilbhumirajsinhji said:
7 years ago
Well explained, thanks @Shilpa.
(2)
Srk said:
6 years ago
You're right @Shilpa.
Vinay BEL said:
5 years ago
L = l/4 hence proved.
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