Question 4.20: A curved link of the mechanism made from a round steel bar i...

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

(fs) = 3.5.

Step I Calculation of permissible tensile stress

\sigma_{\max }=\frac{S_{y t}}{(f s)}=\frac{400}{3.5}=114.29 N / mm ^{2} .

Step II Calculation of eccentricity (e)
At the section XX,

R = 4D

R_{i}=4 D-0.5 D=3.5 D .

R_{o}=4 D+0.5 D=4.5 D .

From Eq. (4.60),

R_{N}=\frac{\left(\sqrt{R_{o}}+\sqrt{R_{i}}\right)^{2}}{4}             (4.60).

R_{N}=\frac{\left(\sqrt{R_{o}}+\sqrt{R_{i}}\right)^{2}}{4} .

=\frac{(\sqrt{4.5 D}+\sqrt{3.5 D})^{2}}{4}=3.9843 D .

e=R-R_{N}=4 D-3.9843 D=0.0157 D .

Step III Calculation of bending stress

h_{i}=R_{N}-R_{i}=3.9843 D-3.5 D=0.4843 D .

A=\frac{\pi}{4} D^{2}=\left(0.7854 D^{2}\right) mm ^{2} .

M_{b}=1000 \times 4 D=(4000 D) N – mm .

From Eq. (4.56), the bending stress at the inner fibre is given by,

\sigma_{b i}=\frac{M_{b} h_{i}}{A e R_{i}}             (4.56).

\sigma_{b i}=\frac{M_{b} h_{i}}{A e R_{i}}=\frac{(4000 D)(0.4843 D)}{\left(0.7854 D^{2}\right)(0.0157 D)(3.5 D)}.

=\left(\frac{44886.51}{D^{2}}\right) N / mm ^{2}             (i).

Step IV Calculation of direct tensile stress

\sigma_{t}=\frac{P}{A}=\frac{1000}{\left(0.7854 D^{2}\right)}=\left(\frac{1273.24}{D^{2}}\right) N / mm ^{2}            (ii).

Step V Calculation of dimensions of link Superimposing the bending and direct tensile stresses and equating the resultant stress to permissible stress, we have

\sigma_{b i}+\sigma_{t}=\sigma_{\max } .

\left(\frac{44886.51}{D^{2}}\right)+\left(\frac{1273.24}{D^{2}}\right)=114.29 .

D = 20.10 mm.