Mechanical Engineering - Strength of Materials - Discussion
Discussion Forum : Strength of Materials - Section 4 (Q.No. 41)
41.
A thin cylindrical shell of diameter (d), length (l) and thickness (t) is subjected to an internal pressure (p). The ratio of longitudinal strain to hoop strain is
Discussion:
7 comments Page 1 of 1.
Tanveer shahzad said:
5 years ago
Circumferential strain= pd/2tE(1-1/2m).
Longitudinal strain= pd/2tE(1/2-1/m).
Simply required= Longitudinal strain/ Circumferential strain.
= {pd/2tE(1/2-1/m)}/{pd/2tE(1-1/2m)}.
= {m-2/2m}/ { 2m-1/2m}.
= m-2/2m-1.
Option A is correct.
Longitudinal strain= pd/2tE(1/2-1/m).
Simply required= Longitudinal strain/ Circumferential strain.
= {pd/2tE(1/2-1/m)}/{pd/2tE(1-1/2m)}.
= {m-2/2m}/ { 2m-1/2m}.
= m-2/2m-1.
Option A is correct.
Dip said:
5 years ago
Option A is the correct Answer.
Biri singh said:
6 years ago
Option is B correct.
Harika said:
6 years ago
The correct option is B.
Nikhil goyal said:
6 years ago
The answer is correct as u = 1/m.
GURU said:
6 years ago
The correct answer is option B.
Krunal degamadiya said:
8 years ago
Sigma hoop=Pd/2t,
Sigma longi=Pd/4t,
Sigma long/Sigma hoop=1/2.
Now,
Strain longi = ( Sigma longi + M * Sigma hoop)/E.
Strain hoop=(Sigma hoop+M*Sigma longitudinal)/E.
Take ratio.
Sigma longi=Pd/4t,
Sigma long/Sigma hoop=1/2.
Now,
Strain longi = ( Sigma longi + M * Sigma hoop)/E.
Strain hoop=(Sigma hoop+M*Sigma longitudinal)/E.
Take ratio.
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