Civil Engineering - Strength of Materials - Discussion
Discussion Forum : Strength of Materials - Section 3 (Q.No. 22)
22.
An open-ended cylinder of radius r and thickness t is subjected to internal pressure p. The Young's modulus for the material is E and Poisson's ratio is μ. The longitudinal strain is
zero
none of these.
Discussion:
18 comments Page 1 of 2.
Ram Prasad Poudel said:
8 years ago
The answer given above is right i.e zero because it is the case of open-ended cylinder where no longitudinal stress is developed hence no longitudinal strain.
@Abhi.
This case is for a cylinder with closed ends.
@Abhi.
This case is for a cylinder with closed ends.
Tejasva Mudgal said:
6 years ago
The Answer will be D
Direct longitudinal strain = 0
Poision effect due to lateral/hoop strain = Pd/(2t) * (poisson ratio /E) (compressive).
So, "None of these " will be the answer.
Direct longitudinal strain = 0
Poision effect due to lateral/hoop strain = Pd/(2t) * (poisson ratio /E) (compressive).
So, "None of these " will be the answer.
MJM Gcek said:
4 years ago
@Taral.
Let me tell you that strain=stress/E and the stress is actually Pd/4t and strain = pd/4tE.
But here as it is an open cylinder so the strain is considered 0.
Let me tell you that strain=stress/E and the stress is actually Pd/4t and strain = pd/4tE.
But here as it is an open cylinder so the strain is considered 0.
Taral Bhaumik said:
6 years ago
Those who are saying pd/4tE, let me tell you that pd/4tE longitudinal stress and question asked to find strain. Strain and stress.
Strain= 0.
Stress=pd/4tE.
Strain= 0.
Stress=pd/4tE.
Taral Bhaumik said:
6 years ago
Those who are saying pd/4tE, let me tell you that pd/4tE longitudinal stress and question asked to find strain. Strain and stress.
Strain= 0.
Stress=pd/4tE.
Strain= 0.
Stress=pd/4tE.
(1)
Azzu said:
7 years ago
The answer is correct because longitudinal stress developed only in the closed-ended cylinder.
In question, clearly mentioned that open-ended.
In question, clearly mentioned that open-ended.
Debanjan Ghosh said:
7 years ago
Right @Akib,
It's open end cylinder so it's pressure is zero, because here no longitudinal stress & strain, so answer will be zero.
It's open end cylinder so it's pressure is zero, because here no longitudinal stress & strain, so answer will be zero.
Bibhuprasad said:
7 years ago
The stress=PD/4t.
D=2r,strain=stress/E where E= young's modulus.
longitudinal strain=2Pr/4tE = Pr/2tE.
D=2r,strain=stress/E where E= young's modulus.
longitudinal strain=2Pr/4tE = Pr/2tE.
(1)
Abhi said:
8 years ago
Longitudinal stress= PD/4t.
E=stress/strain.
D=2r.
Substitute. You'll get the answer as option C.
E=stress/strain.
D=2r.
Substitute. You'll get the answer as option C.
Yogesh said:
9 years ago
Due to Internal pressure, only radial strain is caused, So longitudinal strain should be ZERO.
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