A 25-mm-diameter shaft is statically torqued to 230 N · m. It is made of cast 195-T6
aluminum, with a yield strength in tension of 160 MPa and a yield strength in compression of 170 MPa. It is machined to final diameter. Estimate the factor of safety of
the shaft.

Question

Accepted Answer

The maximum shear stress is given by

[latex] au =\frac{16T}{\pi d^3} =\frac{16\left(230 ight) }{\pi \left[25\left(10^{-3} ight) ight]^3 } =75\left(10^6 ight) \ N/m^2=75 \ MPa [/latex]

The two nonzero principal stresses are 75 and −75 MPa, making the ordered principal stresses [latex] \sigma _1= 75, \sigma _2= 0, and \ \sigma _3= −75 [/latex]MPa.
From Eq. (5–26),

[latex] \frac{\sigma _1}{S_t} -\frac{\sigma _3}{S_c} =\frac{1}{n} [/latex]

for yield,

[latex]n=\frac{1}{{\sigma _1}/{S_{yt} }-{\sigma _3}/{S_{yc} }} =\frac{1}{{75}/{160}-{\left(-75 ight)}/{170} } =1.10 [/latex]

Alternatively, from Eq. (5–27),

[latex]S_{sy} =\frac{S_{yt}S_{yc}}{S_{yt}+S_{yc}} =\frac{160\left(170 ight) }{160+170} =82.4[/latex] MPa

and [latex] au _{max} = 75 [/latex] MPa. Thus,

[latex] n = \frac{S_{yc}}{ au _{max}} = \frac{82.4}{75}=1.10 [/latex].

Question 2.5.2: A 25-mm-diameter shaft is statically torqued to 230 N · m. I...

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