A rod with a length of 1 m and a radius of 20 mm is made of high-strength steel.The rod is subjected to a torque T, which produces a shear stress below the proportional limit. If the cross section at one end is rotated 45 degrees in relation to the other end, and the shear modulus G of the material is 90 GPa, what is the amount of applied torque?

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Step 1Given data:\begin{array}{l}l=1 \mathrm{~m}=1000 \mathrm{~m} \\ r=20 \mathrm{~mm} \\ \theta=45^{\circ}=0.7854 \mathrm{Rad} \\ G=90 \mathrm{GPa}=90 \times 10^{3} \mathrm{~N} / \mathrm{mm}^{2}\end{array}Step 2Calculation of polar moment of inertia:J=\frac{\pi d^{4}}{32}Substituting the value in the formula:\begin{aligned} J & =\frac{\pi \times(40)^{4}}{32} \\ & =251327.41 \mathrm{~mm}^{4}\end{aligned}Step 3Calculation of torque:T=\frac{G \theta}{l} \times JSubstituting the value in the formula:\begin{aligned} T & =\frac{\left(90 \times 10^{3} \mathrm{~N} / \mathrm{mm}^{2}\right) \times(0.7854 \mathrm{Rad})}{1000 \mathrm{~mm}} \times(251327.41 \mathrm{~mm}) \\ T & =17765329.3 \mathrm{~N}-\mathrm{mm} \\ & =17.7 \mathrm{KN}-\mathrm{m}\end{aligned}Step 4The amount of torque applied is 17.7 KN-m. ...