Question 11.2: A low carbon steel shaft of 30 mm diameter is to be designed...
A low carbon steel shaft of 30 mm diameter is to be designed to carry an axial load P = 30 kN, a bending moment M = 150 Nm and a torque T = 250 Nm. If the yield point for the material is 280 MPa, calculate the factor of safety in design of the shaft based on the maximum shear stress criterion of failure against yielding.
The shaft material is subjected to the normal stresses due to P and M and shear stress due to T.
Therefore, resultant normal stress, \sigma_{x x} is
\sigma_{x x}=\frac{4 P}{\pi d^2}+\frac{32 M}{\pi d^3}
or \sigma_{x x}=\frac{4}{\pi}\left\lgroup \frac{P}{d^2}+\frac{8 M}{d^3} \right\rgroup =\frac{4}{\pi}\left\lgroup \frac{30 \times 10^3}{30^2}+\frac{8 \times 150 \times 10^2}{30^3} \right\rgroup
Therefore,
\sigma_{x x}=99.03 MPa
and shear stress, \tau_{x y} is
\tau_{x y}=\frac{16 T}{\pi d^3}=\frac{16(250) \times 10^3}{\pi(30)^3} MPa
Therefore,
\tau_{x y}=47.16 MPa
Clearly, according to maximum shear stress criterion
\tau_{\max }=\frac{1}{\text { factor of safety }}\left(\tau_{ yp }\right)=\left\lgroup \frac{\sigma_{ yp }}{2} \right\rgroup\left\lgroup \frac{1}{\text { factor of safety }} \right\rgroup
\sqrt{\left\lgroup \frac{\sigma_{x x}}{2} \right\rgroup^2+\tau_{x y}^2}=\left\lgroup \frac{\sigma_{ yp }}{2} \right\rgroup\left\lgroup \frac{1}{\text { factor of safety }} \right\rgroup
Hence,
\text { Factor of safety }=\frac{\left(\sigma_{y p } / 2\right)}{\sqrt{\left(\sigma_{x x} / 2\right)^2+\tau_{x y}^2}}
or \text { Factor of safety }=\frac{\sigma_{ yp }}{\sqrt{\sigma_{x x}^2+4 \tau_{x y}^2}}=\frac{280}{\sqrt{99.03^2+4(47.16)^2}}=2.047
Hence, in the design required factor of safety against yielding as 2.047.