Mechanical Engineering - Strength of Materials - Discussion
Discussion Forum : Strength of Materials - Section 1 (Q.No. 46)
46.
The strain energy stored in a solid circular shaft subjected to shear stress (τ) is (where C = Modulus of rigidity for the shaft material)
x Volume of shaft
x Volume of shaft
x Volume of shaft
x Volume of shaftDiscussion:
38 comments Page 2 of 4.
Vanraj Gajarotar said:
9 years ago
Right answer is B, Option D is for shaft subjected to pure torque.
Prakash said:
9 years ago
Right answer is B.
Shubham said:
9 years ago
I also think that option B is correct.
PRAKASH SINGH said:
9 years ago
Opt is B, because shear stress is given not a torsional stress so, the right answer is B.
Sridevi said:
9 years ago
Answer is b because it is given as only shear is applied on shaft. Not torque.
Er. Aman said:
9 years ago
Shear stress is due to torsion. So, d is correct. Refer the text book.
Somesh said:
9 years ago
According to Rs Khurmi. Option B is right.
Bunnnysunny said:
9 years ago
Guys, I think option B is correct. Option D is for torsional resilience.
Chahat sharma said:
8 years ago
Siimply, solve strain energy =0.5*T*θ.
Put T=(τ*J)/R.
And theta = (τ*L)/(C*R).
Note: separate volume as (π*d2*L).
You will get the same answer as D option.
Put T=(τ*J)/R.
And theta = (τ*L)/(C*R).
Note: separate volume as (π*d2*L).
You will get the same answer as D option.
Mohan said:
8 years ago
If it is purely shear stress than the answer is B.
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