Question 5.2: A 25-mm-diameter shaft is statically torqued to 230 N · m. I...

The maximum shear stress is given by

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

The two nonzero principal stresses are 75 and −75 MPa, making the ordered principal stresses σ_{1} = 75, σ_{2} = 0,  and  σ_{3} = −75 MPa. From Eq. (5–26), for yield,

\frac {σ_{1}}{S_{t}} −\frac {σ_{3}}{S_{c}} =\frac {1}{n}           (5–26)

n =\frac {1}{σ_{1}/S_{yt} − σ_{3}/S_{yc}} =\frac  {1}{75/160 − (−75)/170} = 1.10

Alternatively, from Eq. (5–27),

S_{sy} =\frac {S_{yt} S_{yc}}{S_{yt} + S_{yc}}           (5–27)

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

and τ_{max} = 75  MPa. Thus,

n =\frac {S_{sy}}{τ_{max}} =\frac {82.4}{75} = 1.10