Determine the diameter of the shaft, so that the angle of twist does not exceed 0.5 degree and the shear stress of shaft material does not exceed 40MPa when F=25N. Take modulus of rigidity G=27 GPa. Refer the fig.

Data: Diameter of shaft (d) = ?

Angle of twist ($\theta)$ = 0.5°

Shear stress ($\zeta)$ = 40 Mpa

Force (F) = 25N

Modulus of Rigidity (G) = $ 37 Gpa = 37 \times 10^3 N/mm^2$

$\rightarrow$ $Torque = Force \times distance = f \times d^1 $

$= 25 \times 150$

(T) = 3750 N mm

L is the length of shaft.

1] For Strength:

$\frac{T}{J_p} = \frac{\zeta}{r}$

$\therefore$ $\frac{3750}{\frac{\pi}{32} \times d^4} = \frac{40}{(\frac{d}{2})}$

$\therefore$ d = 7.81 mm

2] For stiffness:

$\frac{T}{J_p} = \frac{G \ \theta}{L}$

$\therefore$ $\frac{3750}{\frac{\pi}{32} \times d^4} = \frac{37 \times 10^3}{150} \times (0.5 \times \frac{\pi}{180})$

$\therefore$ d = 11.54 mm.

Value of d is greater of above two values i.e. d = 11.54 mm.