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## Q. 4.14

The dimensions of an overhang crank are given in Fig. 4.41. The force P acting at the crankpin is 1 kN. The crank is made of steel 30C8 $\left(S_{v t}=400 N / mm ^{2}\right)$ and the factor of safety is 2. Using maximum shear stress theory of failure, determine the diameter d at the section – XX. ## Verified Solution

$\text { Given } P=1 kN \quad S_{y t}=400 N / mm ^{2} \quad(f s)=2$.

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

$S_{s y}=0.5 S_{y t}=0.5(400)=200 N / mm ^{2}$.

The permissible shear stress is given by,

$\tau_{\max .}=\frac{S_{s y}}{(f s)}=\frac{200}{2}=100 N / mm ^{2}$      (i).

Step II Calculation of bending and torsional shear
stresses
The section of the crankpin at XX is subjected to combined bending and torsional moments. At the section XX,

$M_{b}=1000 \times(50+25+100)=175 \times 10^{3} N – mm$.

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

$\sigma_{x}=\sigma_{b}=\frac{M_{b} y}{I}=\frac{\left(175 \times 10^{3}\right)(d / 2)}{\left(\pi d^{4} / 64\right)}$.

$=\left(\frac{1782.54 \times 10^{3}}{d^{3}}\right) N / mm ^{2}$.

$\sigma_{y}=0$.

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

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

Step III Calculation of maximum shear stress
The problem is similar to the previous one and the maximum shear stress is given by,

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

$=\sqrt{\left(\frac{1782.54 \times 10^{3}}{2 d^{3}}\right)^{2}+\left(\frac{2546.48 \times 10^{3}}{d^{3}}\right)^{2}}$.

$=\frac{2697.95 \times 10^{3}}{d^{3}} N / mm ^{2}$           (ii).

Step IV Calculation of diameter at section-XX
Equating (i) and (ii),

$\frac{2697.95 \times 10^{3}}{d^{3}}=100 \therefore d=29.99 \text { or } 30 mm$.