If a solid shaft (diameter 20 cm, length 400 cm, N = 0.8 x 10 5 N
| If a solid shaft (diameter 20 cm, length 400 cm, N = 0.8 x 105 N/mm2) when subjected to a twisting moment, produces maximum shear stress of 50 N/mm2, the angle of twist in radians is
A. 0.001
B. 0.002
C. 0.025
D. 0.004
Please scroll down to see the correct answer and solution guide.
Right Answer is: C
SOLUTION
\(\frac{{\rm{T}}}{{\rm{J}}} = \frac{{\rm{\tau }}}{{\rm{r}}} = \frac{{{\rm{G\theta }}}}{{\rm{L}}} \to {\rm{Torsion\;Formula}}\)
Where,
G = Modulus of rigidity, L = length, θ = twisting angle, T = Twisting moment, J = Polar moment of inertia, τ = shear stress, and r = radius
Calculation:
d = 20 cm, r = 10 cm = 100 mm
L = 400 cm, N/G = 0.8 × 105 N/mm2
τ = 50 N/mm2
\(\frac{{50}}{{10}} = \frac{{0.8 \times {{10}^5} \times {\rm{\theta }}}}{{400}}\)
\( \Rightarrow {\rm{\theta }} = \frac{{400 \times 50}}{{10 \times 0.8 \times {{10}^5}}} = 0.025\)