Mechanical Engineering - Strength of Materials - Discussion

Discussion Forum : Strength of Materials - Section 1 (Q.No. 23)
23.
The torque transmitted by a solid shaft of diameter (D) is (where τ = Maximum allowable shear stress)
x τ x D3
x τ x D3
x τ x D3
x τ x D3
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
9 comments Page 1 of 1.

Praddep gk said:   7 years ago
T/j=q/r tortion equation.
J=(π*d^4)/32.
r=d/2.
Substitute in the equation we get the answer ie B.
(1)

Fri said:   7 years ago
T=16(S)/πD^3.

Karthik s said:   7 years ago
Torque = F*(D/2).
F = stress*area.

Stess = T/2, area =(pi * d * d)/4,
Therefore torque = (T * pi * d * d * d)/16.

SAURABH GANGWAR said:   8 years ago
T/J = q/R Torsion equation.

T = q*J/R.

Polar modulus of solid shaft J = pi D^4/32.

Where D is diameter of solid shaft.

Than T = (pi D^4/32)*q/(D/2).

= 2*pi*D^4*q/32 D.

= pi*q*D^3/16 answer.

Answer is D.

MD HUSSAIN said:   8 years ago
T/J = τ/R Torsion equation.

T = j τ/R where j = (pi D^4)/32, D = 2R substituting.

T = ((pi D^4)τ)/16D.

T = ((pi D^3)τ)/16.

Senjal maurya said:   9 years ago
T = (J/R)*Q, where Q = T.

Meghashyam.raju said:   9 years ago
J = pi*d^4/32.

And keep R = d/2 as he told max shear stress.

Harish said:   9 years ago
How to solve this sum?

Schngrdhr said:   9 years ago
T = (pi)/32*D^4*T/R.

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