Civil Engineering - UPSC Civil Service Exam Questions - Discussion

Discussion Forum : UPSC Civil Service Exam Questions - Section 23 (Q.No. 4)
4.
A circular shaft subjected to torsion undergoes a twist of 1° in a length of 120 cm. If the maximum shear stress induced is limited to 1000 kg/cm2 and if modulus of rigidity G = 0.8 x 106 kg/cm2, then the radius of the shaft should be
π/18
18/π
π/27
27/π
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
3 comments Page 1 of 1.

POOJA N V said:   4 years ago
r/R = G * angle &teeta;/L.
120/R = (0.8*10^5*1*π)/(1*10^3*180).
21.6 * 10^6 = 80*10^3*R*π.
R = 270/π.

PARASHURAM KAGI said:   8 years ago
from The Torsion Equation we have,
T/J=GO/L=Fs/R here consider only GO/L= Fs/R,
From the data we have theta = 1, l = 120cm, toue = 1000kg/cm^2 and G = 0.8 X 10^6,
0.8 x 10^6 x 1/120 = 1000/R then Radius of shaft R = 27/3.142(Phi).

Shyama charan shukla said:   8 years ago
Use formula t / r = T / J = G0/L.

0 angle in radian.

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