Determine the diameter of the shaft AB, if the allowable shear stress in shaft AB is 74 MPa and vibration limit the angle of twist of shaft AB to 0.05 radians. The shaft is 0.6 m long and modulus of rigidity for the shaft material =75GPa.

Data: $Power \ (P) = 450 \ KW = 450 \times 10^3 \ W$

N = 1200 Revolution/Minute (rpm)

Shear Stress = 74 mpa = 74 $N/mm^2$

Angle of twist = 0.05 radians.

Length of shaft = 0.6 m = 600 mm.

Modulus of Rigidity = $75 \ Gpa = 75 \times 10^3 \ N/mm^2$

Diameter of shaft = ?

$\rightarrow$ 1] Power transmitted is,

$P = \frac{2 \pi \ N \ T}{60} \ (W)$

$450 \times 10^3 = \frac{2 \pi \times 1200 \times T}{60}$

$\therefore$ $T = 3580.98 \ Nm = 3580.98 \times 10^3 \ Nmm$

2] For strength:

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

$\frac{ 3580.98 \times 10^3}{ \frac{\pi}{32} \times d^4} = \frac{74}{(\frac{d}{2})}$

$\therefore$ d = 62.69 mm.

3] For stiffness:

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

$\therefore$ $\frac{3580.98 \times 10^3}{ \frac{\pi}{32} \times d^4} = \frac{75 \times 0.05 \times 10^3}{600}$

$\therefore$ d = 49.15 mm.

Value of diameter of shaft such that it must satisfies both condition strength and stiffness, for that we are taking greater values of two i.e. d = 62.69 mm.