Question 3.5: A 50 mm diameter steel shaft is case-hardened to a depth of ...
A 50 mm diameter steel shaft is case-hardened to a depth of 2 mm. Assuming that the inner core remains elastic up to a yield stress in shear of 180 MN/m^{2} and that the case can also be assumed to remain elastic up to failure at the shear stress of 320 MN/m^{2}, calculate:
(a) the torque required to initiate yielding at the outside surface of the case;
(b) the angle of twist per metre length at this stage.
Take G = 85 GN/m^{2} for both case and core whilst they remain elastic.
Since the modulus of rigidity G is assumed to be constant throughout the shaft whilst elastic, the angle of twist θ will be constant.
The stress distribution throughout the shaft cross-section at the instant of yielding of the outside surface of the case is then as shown in Fig. 3.41, and it is evident that whilst the failure stress of 320 MN/m^{2} has only just been reached at the outside of the case, the yield stress of the core of 180MN/m^{2} has been exceeded beyond a radius r producing a fully plastic annulus and an elastic core.
By proportions, since G_{ case}= G_{core} then
\left(\frac{\tau }{r} \right) _{case} =\left(\frac{\tau }{r} \right) _{core}
\frac{180}{r} =\frac{320}{25}
r=\frac{180}{320} \times 25=14.1 mm
The shaft can now be considered in three parts:
(i) A solid elastic core of 14.1 mm external radius;
(ii) A fully plastic cylindrical region between r = 14.1 mm and r = 23mm;
(iii) An elastic outer cylinder of external diameter 50 mm and thickness 2 mm.
Torque on elastic core =\frac{\tau _{y}J }{R} =\frac{180\times 10^{6} }{14.1\times 10^{-3} } \times \frac{\pi \left(14.1\times 10^{-3} \right) ^{4} }{2}
=793 Nm = 0.793 kNm
Torque on plastic section =2\pi \tau _{y} \int_{r_{1} }^{r_{2} }{r^{2} dr }
=\frac{2\pi \times 180\times 10^{6} }{3}\left[23^{3} – 14.1^{3} \right]10^{-9}
=\frac{2\pi \times 180\times 10^{6}\times 9364\times 10^{-9} }{3}
=3.53 kNm
Torque on elastic outer case =\frac{\tau _{y}J }{r} =\frac{320\times 10^{6} }{25\times 10^{-3} }\pi \left[\frac{25^{4}-23^{4} }{2} \right] 10^{-12}
=2.23 kNm
Therefore total torque required =\left(0.793+3.53+2.23\right)10^{3}
=6.55 kNm
Since the angle of twist is assumed constant across the whole shaft its value may be determined by application of the simple torsion theory to either the case or the elastic core.
For the case: \frac{\theta }{L} =\frac{\tau }{GR} =\frac{320\times 10^{6} }{85\times 10^{9}\times 25\times 10^{-3} }
=0.15 rad =8.6°
