Question

solid shaft of 10cm diameter transmits 74 kW at 150 rev/min. Calculate (a) the torque on the shaft, (b) the maximum shear stress developed, (c) the angle of twist in a length of 1.50m, and (d) the shear stress at a radius of 3cm. Take G = 8 MN/cm2.

Answer 1

a)The torque on the shaft

$P=Tw$

Where P is the power transmitted, T is the torque, and ω is the angular velocity (units are in terms of rad/s).

We can get the angular velocity from the rotation speed of the shaft (150 rev/min) knowing that $1 rev=2\pi rad$

Solving for the torque,

$74 kW= 74 (\frac {kJ}{S}) =T *( 150\frac{rev}{min}) (\frac{2\pi rad}{1 rev}) (\frac{1 min}{60 sec})$

$T= 4.711 kJ$

b)The maximum shear stress developed

$T_{max} =\frac {Tc}{J}$

Where $T_{max}$ is the maximum shear stress, T is the torque applied, c is the radius of the circular shaft (half of the shaft diameter, 5 cm = 0.05 m), and J is the polar moment of inertia of the shaft.

J is calculated for a circular solid shaft via the following:

$J= \frac{\pi c4}{2}$

$T_{max} =\frac{Tc}{J} =\frac{2T}{\pi(c^{3})} =\frac{(2)(4.711 kJ)}{\pi(0.05 m)^{3}}$

$T_{max} =239992.9 kPa$

c) The angle of twist in a length of 1.50 m

 $\theta= \frac{TL}{GJ}$

Where $\theta$ is the angle of twist, L is the length of the shaft, and G is the Modulus of Rigidity (provided as$G= 8mN/cm^{2} =8*10^{10} Nm^{2}=80 GPa)$

With $T=4.711kJ =4711 Nm$, solving for the angle o twist:

$J= \frac{\pi c^{4}}{2} =\frac{\pi(0.05 m)^{4}}{2} =9.817*10^{-6} m^{4}$

$\theta=\frac{TL}{GJ}=\frac{(4711Nm)(1.50 m)}{(8*10^{10} \frac{N}{m^{2})} (9.817*10^{-6}m^{4})}$

$\theta=8.997*10^{-3} rad$

d)The shear stress at a radius of 3 cm

$T=\frac{Tp}{J}$

Where p is the radial distance considered for the analysis. Solving,

$T=\frac{TP}{J}=\frac{(4.711kJ)(0.03 m)}{9.817*10^{-6}m{^4}}$

$T=14395.7 kPa$