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/m² and that the case can also be assumed to remain elastic up to failure at the shear stress of 320 MN/m², calculate: (a) the torque required to

Question

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 [latex]180 MN/m^{2}[/latex] and that the case can also be assumed to remain elastic up to failure at the shear stress of [latex]320 MN/m^{2}[/latex], 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 [latex]G = 85 GN/m^{2}[/latex] for both case and core whilst they remain elastic.

Accepted Answer

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 [latex]320 MN/m^{2}[/latex] has only just been reached at the outside of the case, the yield stress of the core of [latex]180MN/m^{2}[/latex] has been exceeded beyond a radius r producing a fully plastic annulus and an elastic core.

By proportions, since [latex]G_{ case}= G_{core}[/latex] then

[latex]\left(\frac{ au }{r} ight) _{case} =\left(\frac{ au }{r} ight) _{core}[/latex]
[latex]\frac{180}{r} =\frac{320}{25}[/latex]
[latex]r=\frac{180}{320} imes 25=14.1 mm[/latex]

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 [latex]=\frac{ au _{y}J }{R} =\frac{180 imes 10^{6} }{14.1 imes 10^{-3} } imes \frac{\pi \left(14.1 imes 10^{-3} ight) ^{4} }{2}[/latex]
[latex]=793 Nm = 0.793 kNm[/latex]

Torque on plastic section [latex]=2\pi au _{y} \int_{r_{1} }^{r_{2} }{r^{2} dr }[/latex]
[latex]=\frac{2\pi imes 180 imes 10^{6} }{3}\left[23^{3} - 14.1^{3} ight]10^{-9}[/latex]

[latex]=\frac{2\pi imes 180 imes 10^{6} imes 9364 imes 10^{-9} }{3}[/latex]
[latex]=3.53 kNm[/latex]

Torque on elastic outer case [latex] =\frac{ au _{y}J }{r} =\frac{320 imes 10^{6} }{25 imes 10^{-3} }\pi \left[\frac{25^{4}-23^{4} }{2} ight] 10^{-12}[/latex]
[latex]=2.23 kNm[/latex]

Therefore total torque required [latex]=\left(0.793+3.53+2.23 ight)10^{3}[/latex]
[latex]=6.55 kNm[/latex]

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: [latex]\frac{ heta }{L} =\frac{ au }{GR} =\frac{320 imes 10^{6} }{85 imes 10^{9} imes 25 imes 10^{-3} }[/latex]
[latex]=0.15 rad =8.6°[/latex]

Question 3.5: A 50 mm diameter steel shaft is case-hardened to a depth of ...

Related Answered Questions

Verified Answer:

Verified Answer:

Verified Answer:

Verified Answer:

(a) A rectangular-section steel beam, 50 mm wide by 20 mm deep, is used as a simply supported beam over a span of 2 m with the 20 mm dimension vertical. Determine the value of the central concentrated load which will produce initiation of yield at the outer fibres of the beam. (b) If the central ...

Verified Answer:

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/m² and, up to the yield point, the modulus of rigidity is 80 GN/m² (a) Determine the value of T and the associated angle of ...

Verified Answer: