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## Q. 4.13

The shaft of an overhang crank subjected to a force P of 1 kN is shown in Fig. 4.40(a). The shaft is made of plain carbon steel 45C8 and the tensile yield strength is 380 N/mm². The factor of safety is 2. Determine the diameter of the shaft using the maximum shear stress theory.

## Verified Solution

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),

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