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.
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
21 comments Page 2 of 3.

Shilpa said:   7 years ago
A is the right answer.

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.

Utsav said:   7 years ago
A is correct.
B is for max shear stress.
C is for max bending stress.

Shubham Rawat said:   7 years ago
Answer B.

It is based on pure shear case σ = τ.

Nirz said:   6 years ago
[T/{j=(Ix+Iy)}]=(Ï„/R),
> (T*R)÷j = τ.
>(T*2D)÷(π*D^4/32) = Ï„.
>(T*16)/π*D^3 = Ï„.

Bipin Dhungana said:   6 years ago
As per torsional formula :

T = Stress*polar modulus.
Stress = T/Z.
Z = pi*d3/16.
Stress = 16T/pi*d3.
(1)

Kvk said:   6 years ago
In the case of pure torsion, the beam element is under pure shear.
In pure shear case, the principal stresses are equal to shear stress acting on the element.

So, here maximum normal shear stress is equal to maximum shear stress.
(1)

Naveen said:   6 years ago
Torque equation T/J =F/R.
F=TR/J.
We know R=d/2.
Polar modulus J=pi d^4/32.
Then substitute we can get the given answer.

Dipunku said:   5 years ago
It should be the shear stress.
(1)

Pankaj Pokhrel said:   2 years ago
For pure torsion in a shaft, normal stress is zero. So, A should be the right answer.


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