Given P = 1 kN           S_{y t}=380 N / mm ^{2} \quad(f s)=2 .

Step I Calculation of permissible shear stress
According to maximum shear stress theory,

S_{s y}=0.5 S_{y t}=0.5(380)=190 N / mm ^{2} .

The permissible shear stress is given by,

\tau_{\max }=\frac{S_{s y}}{(f s)}=\frac{190}{2}=95 N / mm ^{2}           (i).

Step II Calculation of bending and torsional shear stresses
The stresses are critical at the point A, which is subjected to combined bending and torsional moments.
At the point A,

M_{b}= P \times(250)=(1000)(250)=250 \times 10^{3} N – mm .

M_{t}= P \times(500)=(1000)(500)=500 \times 10^{3} N – mm .

\sigma_{b}=\frac{M_{b} y}{I}=\frac{\left(250 \times 10^{3}\right)(d / 2)}{\left(\pi d^{4} / 64\right)} .

=\left(\frac{2546.48 \times 10^{3}}{d^{3}}\right) N / mm ^{2} .

\tau=\frac{M_{t} r}{J}=\frac{\left(500 \times 10^{3}\right)(d / 2)}{\left(\pi d^{4} / 32\right)} .

=\left(\frac{2546.48 \times 10^{3}}{d^{3}}\right) N / mm ^{2} .

Step III Calculation of maximum shear stress
The stresses at point A and corresponding Mohr’s circle are shown in Fig. 4.40(b) and (c) respectively.
In these figures,

\sigma_{x}=\sigma_{b}=\left(\frac{2546.48 \times 10^{3}}{d^{3}}\right) N / mm ^{2} \quad \sigma_{z}=0 .

\tau=\tau_{x z}=\tau_{z x}=\left(\frac{2546.48 \times 10^{3}}{d^{3}}\right) N / mm ^{2} .

From Mohr’s circle,

\tau_{\max .}=\sqrt{\left(\frac{\sigma_{x}}{2}\right)^{2}+\left(\tau_{x z}\right)^{2}} .

=\left[\sqrt{\left(\frac{2546.48}{2 d^{3}}\right)^{2}+\left(\frac{2546.48}{d^{3}}\right)^{2}}\right] \times 10^{3} .

=\frac{2847.05 \times 10^{3}}{d^{3}}           (ii).

Step IV Calculation of shaft diameter
Equating (i) and (ii),

\frac{2847.05 \times 10^{3}}{d^{3}}=95 \quad \therefore d=31.06 mm .