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A hollow circular shaft of inner radius $10 \: mm$ outer radius $20 \: mm$ 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).
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The torsional stiffness of a straight section is

$k = \frac{GJ}{L}$, where G is Shear/Rotational Modulus, J is Torsional Constant, and L is the length of the section.

Now, $J$ for a circular tube is defined as

$J = \frac{2}{3} \pi r t^3$, where $r$ is the mean radius and $t$ is the wall thickness

So, the torsional stiffness can be calculated from

 $k = \frac{2G \pi r t^3}{3L}$

$k = \frac{2 \times 150GPa \times 3.141/rad \times 15mm \times 10mm^3}{3 \times 1mm}$

or, $k = \frac{2 \times (150\times 10^3 MPa) \times 3.141/rad \times 15mm \times 10mm^3}{3 \times 1mm}$

or, $k = \frac{2 \times (150\times 10^3 N/mm^2) \times 3.141/rad \times 15mm \times 10mm^3}{3 \times 1mm}$

or, $k = \frac{2 \times (150~kN/mm^2) \times 3.141/rad \times 15mm \times 10mm^3}{3 \times 1mm}$

or, $k = \frac{2\times3.141\times22500}{3} kn-mm/rad$

or, $k = 47123.89 kn-mm/rad$

Finally, changing units to $m$ from $mm$, $k = 47.12 kn-m/rad$
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