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

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.

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The maximum shear stress is given by

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

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

 

\frac{\sigma _1}{S_t} -\frac{\sigma _3}{S_c} =\frac{1}{n}

 

for yield,

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

 

Alternatively, from Eq. (5–27),

 

S_{sy} =\frac{S_{yt}S_{yc}}{S_{yt}+S_{yc}} =\frac{160\left(170\right) }{160+170} =82.4 MPa

 

and \tau _{max} = 75 MPa. Thus,

 

n = \frac{S_{yc}}{\tau _{max}} = \frac{82.4}{75}=1.10 .

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