Question 3.4: A solid circular shaft, of diameter 50 mm and length 300 mm,...
A solid circular shaft, of diameter 50 mm and length 300 mm, is subjected to a gradually increasing torque T . The yield stress in shear for the shaft material is 120 MN/m^{2} and, up to the yield point, the modulus of rigidity is 80 GN/m^{2}.
(a) Determine the value of T and the associated angle of twist when the shaft material first yields.
(b) If, after yielding, the stress is assumed to remain constant for any further increase in strain, determine the value of T when the angle of twist is increased to twice that at yield.
(a) For this part of the question the shaft is elastic and the simple torsion theory applies,
T=\frac{\tau J}{R} =\frac{120\times 10^{6} }{25\times 10^{-3} }\times \frac{\pi \left(25\times 10^{-3} \right)^{4} }{2} =2950
=2.95 kNm
\theta =\frac{\tau L}{GR} =\frac{120\times 10^{6}\times 300\times 10^{-3} }{80\times 10^{9}\times 25\times 10^{-3} } =0.018 radian
=1.03°
If the torque is now increased to double the angle of twist the shaft will yield to some radius R_{1} . Applying the torsion theory to the elastic core only,
\theta =\frac{\tau L}{GR}
2\times 0.018=\frac{120\times 10^{6}\times 300\times 10^{-3} }{80\times 10^{9}\times R_{1} }
Therefore partially plastic torque, from eqn. (3.12),
T_{pp}=\frac{\pi \tau _{y} }{6} \left[4R^{3}-R_{1} ^{3} \right] (3.12)
=\frac{\pi \tau _{y} }{6} \left[4R^{3}-R_{1} ^{3} \right]
=\frac{\pi \times 120\times 10^{6} }{6} \left[4\times 25^{3}-12.5^{3} \right] 10^{-9}