We first calculate resultant normal stress on the shaft element as

\sigma_n=\frac{4 P}{\pi d^2}+\frac{32 M}{\pi d^3}=\frac{4 P}{\pi d^2}\left[P+\frac{8 M}{d}\right]

=\frac{4}{\pi(20)^2}\left[25\left(10^3\right)+\frac{8}{20}(50)\left(10^3\right)\right]

=143.24 MPa

and the shearing stress developed on the shaft is

\tau=\frac{16 T}{\pi d^3}=\frac{(16)(120)\left(10^3\right)}{\pi(20)^3}=76.39  MPa

(a) Hence, by octahedral shear stress theory

\sqrt{\sigma_n^2+3 \tau^2}=\frac{\sigma_{y p }}{\text { factor of safety }}

or        \text { Factor of safety }=\frac{\sigma_{ yp }}{\sqrt{\sigma_n^2+3 \tau^2}}=\frac{400}{\sqrt{143.24^2+3(76.39)^2}}=2.05

So, the required factor of safety is 2.05.

(b) According to maximum shear stress criterion, we get

\tau_{\max }=\sqrt{\left(\frac{\sigma_n}{2}\right)^2+\tau^2}=\frac{\tau_{ yp }}{\text { factor of safety }}=\frac{\sigma_{ yp }}{2(\text { factor of safety })}

Therefore,

\text { Factor of safety }=\frac{\sigma_{ yp }}{2 \sqrt{\left(\sigma_n / 2\right)^2+\tau^2}}

=\frac{\sigma_{ yp }}{\sqrt{\sigma_n^2+4 \tau^2}}=\frac{400}{\sqrt{143.24^2+4(76.39)^2}}=1.91

So the required factor of safety is 1.91.