A 4-bar mechanism is required such that the input and out angles are coordinated as given below:

80° 50° 30° Input Crank angle

50° 30° 0° Output follower angle

Synthesize the four-bar mechanism.

Question

A 4-bar mechanism is required such that the input and out angles are coordinated as given below:

80°	50°	30°	Input Crank angle
50°	30°	0°	Output follower angle

Synthesize the four-bar mechanism.

Accepted Answer

The Freudenstein’s equation for displacement of a four-bar mechanism is,

[latex] k_{i} \cos \phi_{i}+k_{2} \cos heta_{i}+k_{3}=\cos \left( heta_{i}-\phi_{i} ight) [/latex].

[latex]\cos \left( heta_{i}-\phi_{i} ight)[/latex]	[latex] heta_{i}-\phi_{i}, deg[/latex]	[latex]\cos \phi_{i}[/latex]	[latex]\phi_{i}, deg[/latex]	[latex]\cos heta_{i}[/latex]	[latex] heta_{i} deg[/latex]	Position
0.866	30	1.000	0	0.866	30	1
0.940	20	0.866	30	0.643	50	2
0.940	20	0.500	60	0.174	80	3

[latex] \left[\begin{array}{lll} 1.000 & 0.866 & 1 \ 0.866 & 0.643 & 1 \ 0.500 & 0.714 & 1 \end{array} ight]\left\{\begin{array}{l} k_{1} \ k_{2} \ k_{3} \end{array} ight\}=\left\{\begin{array}{l} 0.866 \ 0.940 \ 0.940 \end{array} ight\} [/latex].

Solving by Cramer’s rule,

[latex] \Delta=\left[\begin{array}{lll} 1.000 & 0.866 & 1 \ 0.866 & 0.643 & 1 \ 0.500 & 0.174 & 1 \end{array} ight]=-0.01877 [/latex].

[latex] \Delta_{1}=\left[\begin{array}{lll} 0.866 & 0.866 & 1 \ 0.940 & 0.643 & 1 \ 0.940 & 0.714 & 1 \end{array} ight]=-0.0347 [/latex].

[latex] \Delta_{2}=\left[\begin{array}{lll} 1.000 & 0.866 & 1 \ 0.866 & 0.940 & 1 \ 0.500 & 0.940 & 1 \end{array} ight]=-0.02708 [/latex].

[latex] \Delta_{3}=\left[\begin{array}{lll} 1.000 & 0.866 & 0.866 \ 0.866 & 0.643 & 0.940 \ 0.500 & 0.174 & 0.940 \end{array} ight]=-0.005 [/latex].

[latex] k_{1}=\frac{\Delta_{1}}{\Delta}=\frac{-0.0347}{-0.01877}=1.8487 [/latex].

[latex] k_{2}=\frac{\Delta_{2}}{\Delta}=\frac{-0.02708}{-0.01877}=-1.4427 [/latex].

[latex] k_{3}=\frac{\Delta_{3}}{\Delta}=\frac{-0.005}{-0.01877}=0.2664 [/latex].

[latex] k_{1}=\frac{d}{a}, a=\frac{1}{1.8487}=0.541 ext { units for } d=1 ext { unit } [/latex].

[latex] k_{2}=-\frac{d}{c}, c=\frac{-1}{-1.4427}=0.693 ext { units } [/latex].

[latex] k_{3}=\frac{a^{2}-b^{2}+c^{2}+d^{2}}{2 a c} [/latex].

[latex] 0.2664=\frac{(0.541)^{2}-b^{2}+(0.693)^{2}+1}{2 imes 0.541 imes 0.693} [/latex].

[latex] b^{2}=1.5732 [/latex].

b = 1.254 units.

The lengths of various links are:
a=0.541 units, b=1.254 units, c=0.693 units, d=1 unit.
The mechanism is shown in Fig.16.27.

Question 16.13: A 4-bar mechanism is required such that the input and out an...

Related Answered Questions

$\cos \left(\theta_{i}-\phi_{i}\right)$	$\theta_{i}-\phi_{i}, deg$	$\cos \phi_{i}$	$\phi_{i}, deg$	$\cos \theta_{i}$	$\theta_{i} deg$	Position
0.866	30	1.000	0	0.866	30	1
0.940	20	0.866	30	0.643	50	2
0.940	20	0.500	60	0.174	80	3