Question 3.6: A hollow circular bar of 100 mm external diameter and 80 mm ...
A hollow circular bar of 100 mm external diameter and 80 mm internal diameter (Fig. 3.42) is subjected to a gradually increasing torque T. Determine the value of T:
(a) when the material of the bar first yields;
(b) when plastic penetration has occurred to a depth of 5 mm;
(c) when the section is fully plastic.
The yield stress in shear of the shaft material is 120 MN/m^{2}.
Determine the distribution of the residual stresses present in the shaft when unloaded from conditions (b) and (c).
(a) Maximum elastic torque from eqn. (3.1 1)
T_{E} =\frac{\pi R^{3} }{2} \tau _{y} (3.11)
=\frac{\pi \tau _{y} }{2R} \left[R^{4}-R^{4}_{1} \right] =\frac{\pi \times 120\times 10^{6} }{2\times 50\times 10^{-3} } \left(625-256\right) 10^{-8}
=13900 Nm=13.9 kNm
(b) Partially plastic torque, from eqns. (3.1 1) and (3.13),
T_{E} =\frac{\pi R^{3} }{2} \tau _{y} (3.11)
T_{FP} =\frac{2\pi }{3} R^{3}\tau _{y} (3.13)
=\frac{\pi \tau _{y} }{2R_{2}} \left[R^{4}_{2}-R^{4}_{1} \right]+\frac{2\pi \tau _{y} }{3} \left[R^{3}-R^{3}_{2} \right]
=\frac{\pi \times 120\times 10^{6} }{2\times 45\times 10^{-3} }\left(4.5^{4}-256 \right) 10^{-8} + \frac{2\pi \times 120\times 10^{6} }{3}\left(125-91\right) 10^{-6}
=6450+8550=15000 Nm=15kNm
(c) Fully plastic torque from eqn. (3.16) or eqn. (3.13)
T_{FP} =\frac{\pi \tau _{y} }{6R_{1} }\left[ 4R^{3}R_{1}-4R^{4}_{1} \right] =\frac{2\pi \tau _{y} }{3} \left[R^{3}-R^{3}_{1} \right] (3.16)
T_{FP} =\frac{2\pi }{3} R^{3}\tau _{y} (3.13)
=\frac{2\pi \tau _{y} }{3} \left[R^{3}-R^{3}_{1} \right]
= \frac{2\pi \times 120\times 10^{6} }{3}\left[125-64\right] 10^{-6} =15330=15.33 kNm
In order to determine the residual stresses after unloading, the unloading process is assumed completely elastic.
Thus, unloading from condition (b) is equivalent to applying a moment of 15 kNm of opposite sense to the loading moment on a complete elastic bar. The effective stress introduced at the outer surface by this process is thus given by the simple torsion theory
\frac{T}{J} =\frac{\tau }{R}
\tau =\frac{T R}{J} =\frac{15\times 10^{3} \times 50\times 10^{-3}\times 2 }{\pi \times \left(50^{4}-40^{4} \right)10^{-12} }
=\frac{15\times 10^{3} \times 50\times 10^{-3}\times 2}{\pi \times \left(5^{4}-4^{4} \right)10^{-8}}
=129 MN/m^{2}
The unloading stress distribution is then linear, from zero at the centre of the bar to 129 MN/m^{2} at the outside. This can be subtracted from the partially plastic loading stress distribution as shown in Fig. 3.43 to produce the residual stress distribution shown. Similarly, unloading from the fully plastic state is equivalent to applying an elastic torque of 15.33 kNm of opposite sense. By proportion, from the above calculation,
equivalent stress at outside of shaft on unloading =\frac{15.33}{15} \times 129=132 MN/m^{2}
Subtracting the resulting unloading distribution from the fully plastic loading one gives the residual stresses shown in Fig. 3.44.
