Civil Engineering - Strength of Materials - Discussion
Discussion Forum : Strength of Materials - Section 5 (Q.No. 22)
22.
A solid circular shaft of diameter d is subjected to a torque T. The maximum normal stress induced in the shaft, is
zero
none of these.
Discussion:
21 comments Page 2 of 3.
Siddhant said:
7 years ago
The max normal stress for a combination of b.m and torsion is given by;
16(M+(M(sq)+T(sq))^.5/(π)D^3.
put M = 0 in this equation.
16(M+(M(sq)+T(sq))^.5/(π)D^3.
put M = 0 in this equation.
Anita said:
10 years ago
But answer B is for the max shear stress.
Ram said:
8 years ago
Please, explain the right answer.
Azzu said:
8 years ago
There is only torsional shear stress developed no normal will be developed so the answer is A.
Manu said:
8 years ago
C is the right answer. B is for shear stress.
Vipul said:
8 years ago
Option D is correct.
Pratul Sharma said:
8 years ago
There will be only shear stress developed due to torsion. So, normal stress is zero only.
Havila said:
9 years ago
@ Arbor Michigan.
Option C is correct.
Option C is correct.
Arbor Michigan said:
9 years ago
Anyone say the correct answer?
Attri said:
9 years ago
Option A is the right answer.
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