Question 16.4: Design a four-bar mechanism when the motions of the input an...

The angular displacement of input link is governed by x whereas that of the output link by y, θ varies from 40° to 120° (i.e. through 80°) and Φ from 60° to 132° (i.e., through 72°). x=2, 3, 4.

\text { The corresponding values of } y \text { are: } 2 \times 2^{2}=8,2 \times 3^{2}=18,2 \times 4^{2}=32 .

r_{x}=\frac{\theta_{f}-\theta_{i}}{x_{f}-x_{i}}=\frac{120^{\circ}-40^{\circ}}{4-2}=\frac{80}{2}=40 .

\frac{\theta_{2}-\theta_{i}}{x_{2}-x_{i}}=r_{x}, \frac{\theta_{2}-40^{\circ}}{3-2}=40, \theta_{2}=80^{\circ} .

r_{y}=\frac{\phi_{f}-\phi_{i}}{y_{f}-y_{i}}=\frac{132-60}{32-8}=\frac{72}{24}=3 .

\frac{\phi_{2}-\phi_{i}}{y_{f}-y_{i}}=r_{y}=\frac{\phi_{2}-60}{18-8}=3, \phi_{2}=90^{\circ} .

\cos 40^{\circ}=0.7660, \cos 60^{\circ}=0.5000, \cos \left(40^{\circ}-60^{\circ}\right)=0.9397 .

\cos 80^{\circ}=0.1736, \cos 90^{\circ}=00, \cos \left(80^{\circ}-90^{\circ}\right)=0.9848 .

\cos 120^{\circ}=-0.5, \cos 132^{\circ}=0.6691, \cos \left(120^{\circ}-132^{\circ}\right)=0.9781 .

A=\left|\begin{array}{rrr} 0.5000 & 0.7660 & 1 \\ 0.0000 & 0.1736 & 1 \\ -0.6691 & -0.5000 & 1 \end{array}\right|=-0.0596, A_{1}=\left|\begin{array}{ccc} 0.9397 & 0.7660 & 1 \\ 0.9848 & 0.1736 & 1 \\ 0.9791 & -0.5000 & 1 \end{array}\right|=-0.03455 .

A_{2}=\left|\begin{array}{rrr} 0.5000 & 0.9397 & 1 \\ 0.0000 & 0.9848 & 1 \\ -0.6691 & 0.9781 & 1 \end{array}\right|=0.0335, A_{3}=\left|\begin{array}{rrr} 0.5000 & 0.7660 & 0.9397 \\ 0.0000 & 0.1736 & 0.9848 \\ -0.6691 & -0.5000 & 0.9781 \end{array}\right|=-0.0645 .

k_{1}=\frac{A_{1}}{A}=0.5763=\frac{d}{a}, a=\frac{1}{0.5763}=1.73 .

  k_{2}=\frac{A_{2}}{A}=-0.562=-\frac{d}{c}, c=1.78 .

  k_{3}=\frac{A_{3}}{A}=\frac{-0.0645}{-0.0596}=1.0802 .

=\frac{a^{2}-b^{2}+c^{2}+d^{2}}{2 a c}=\frac{(1.73)^{2}-b^{2}+(1.78)^{2}+1}{2 \times 1.73 \times 1.78}, \quad b=0.7 .

The mechanism is shown in Fig.16.19.