A hollow circular shaft of inner radius 10 mm, outer radius 20 mm and length 1 m is to be used as a torsional spring. If the shear modulus of the material of the shaft is 150 GPa, the torsional stiffness of the shaft (in kN-m/rad) is ________ (correct to two decimal places).

The correct answer is 35.34.

To find the torsional stiffness of a hollow circular shaft, we can follow these step-by-step calculations:

1. Identify the given parameters:
Inner radius of the shaft, $r_i=10\text{mm}=0.01\text{m}$
Outer radius of the shaft, $r_o=20\text{mm}=0.02\text{m}$
Inner diameter of the shaft, $d_i=2\times r_i=20\text{mm}=0.02\text{m}$
Outer diameter of the shaft, $d_o=2\times r_o=40\text{mm}=0.04\text{m}$
Length of the shaft, $L=1\text{m}$
Shear modulus of the material, $G=150\text{GPa}=150\times {10}^9{\text{N/m}}^2$

2. Formula for torsional stiffness:
The torsional stiffness ($k_t$) of a shaft is defined as the torque required to produce a unit radian of twist. It is given by:
$k_t=\frac{G{I}_p}{L}$
where $I_p$ is the polar moment of inertia of the hollow shaft's cross-section.

3. Calculate the polar moment of inertia ($I_p$):
For a hollow circular shaft, the polar moment of inertia is:
$I_p=\frac{\pi }{32}\left(d_o^4-d_i^4\right)$
Substituting the diameter values in meters:
$I_p=\frac{\pi }{32}\left({0.04}^4-{0.02}^4\right)$
$I_p=\frac{\pi }{32}\left(2.56\times {10}^{-6}-1.6\times {10}^{-7}\right)$
$I_p=\frac{\pi }{32}\left(2.4\times {10}^{-6}\right)\approx 2.3562\times {10}^{-7}{\text{m}}^4$

4. Calculate the torsional stiffness ($k_t$):
Now substitute $G$, $I_p$, and $L$ into the stiffness formula:
$k_t=\frac{150\times {10}^9\times 2.3562\times {10}^{-7}}{1}$
$k_t\approx 35343\text{N-m/rad}$
Convert this value into kN-m/rad by dividing by 1,000:
$k_t=35.343\text{kN-m/rad}$

Rounding to two decimal places, the torsional stiffness of the shaft is 35.34 kN-m/rad.