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{16 T}{\pi d^{3}}=\frac{16(230)}{\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, ext { and } \sigma_{3}=-75 MPa[/latex]. From Eq. (5–26), for yield,

[latex]\frac{\sigma_{1}}{S_{t}}-\frac{\sigma_{3}}{S_{c}}=\frac{1}{n}[/latex] (5–26)

[latex]n=\frac{1}{\sigma_{1} / S_{y t}-\sigma_{3} / S_{y c}}=\frac{1}{75 / 160-(-75) / 170}=1.10[/latex]

Alternatively, from Eq. (5–27),

[latex]S_{s y}=\frac{S_{y t} S_{y c}}{S_{y t}+S_{y c}}[/latex] (5–27)

[latex]S_{s y}=\frac{S_{y t} S_{y c}}{S_{y t}+S_{y c}}=\frac{160(170)}{160+170}=82.4 MPa[/latex]

and [latex] au_{\max }=75 MPa[/latex]. Thus,

[latex]n=\frac{S_{s y}}{ au_{\max }}=\frac{82.4}{75}=1.10[/latex]

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