Civil Engineering - Strength of Materials - Discussion
Discussion Forum : Strength of Materials - Section 5 (Q.No. 9)
9.
If all the dimensions of a bar are increased in the proportion n : 1, the proportion with which the maximum stress produced in the prismatic bar by its own weight, will increase in the ratio
1 : n
n : 1
1 : 
: 11 : n.
Discussion:
13 comments Page 1 of 2.
CK kumar said:
3 years ago
Stress due to self weight= Density * length of bar.
So stress is directly proportional to length only.
So, now if you increase the length by n and compare it with old stress then you will get the answer.
So stress is directly proportional to length only.
So, now if you increase the length by n and compare it with old stress then you will get the answer.
Dheeraj said:
4 years ago
Stress under own weight = wl^2/2E.
Where w is unit weight,
So, it might be 4 times.
Where w is unit weight,
So, it might be 4 times.
Rot said:
5 years ago
Thanks @Suraj.
Rot said:
5 years ago
Thank you for explaining @ M K.
Amit kushwaha said:
5 years ago
Thanks @Suraj.
M K said:
5 years ago
Everyone is calculating Deflection but in the question they have asked the maximum stress not deflection. So, answer B is correct.
Miller Jonson said:
5 years ago
It should be B.
Krish said:
6 years ago
Elongation due to self-weight, √L=WL/2AE.
√L/L =W/2AE.
Stress=(√L/L)*E= W/2A.
Hence, in my opinion, the answer would be 1/n^2:1.
√L/L =W/2AE.
Stress=(√L/L)*E= W/2A.
Hence, in my opinion, the answer would be 1/n^2:1.
Suraj said:
6 years ago
The answer is absolutely right,
if you consider a section at " x " distance from the bottom , weight of that portion W= density x volume = density X area X length = d X A X x, for x= L, W= d X A X L, now stress = w/A, stress = d * L, so only L is variable here.
if you consider a section at " x " distance from the bottom , weight of that portion W= density x volume = density X area X length = d X A X x, for x= L, W= d X A X L, now stress = w/A, stress = d * L, so only L is variable here.
(1)
Bharat said:
7 years ago
As stress = WL/2AE.
If all dimensions increased in ration n:1 then length becomes n times original length and area becomes n^2 times original area and hence stress becomes 1/n times. So the answer is 1/n:1.
If all dimensions increased in ration n:1 then length becomes n times original length and area becomes n^2 times original area and hence stress becomes 1/n times. So the answer is 1/n:1.
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