Question 16.11: For the four-bar linkage, the following data are given:θ2=60...

The Freudenstein’s equation for the four-bar mechanism shown in Fig.16.25 is:

k_{1} \cos \theta_{4}+k_{2} \cos \theta_{2}+k_{3}=\cos \left(\theta_{2}-\theta_{4}\right)                     (1).

where            k_{1}=\frac{d}{a}, k_{2}=\frac{d}{c}, k_{3}=\frac{a^{2}-b^{2}+c^{2}+d^{2}}{2 a c}

Taking the first time derivative of Eq. (1), we have

k_{1} \omega_{4} \sin \theta_{4}+k_{2} \omega_{2} \sin \theta_{2}=\left(\omega_{2}-\omega_{4}\right) \sin \left(\theta_{2}-\theta_{4}\right) .

Taking the second time derivative of Eq. (2)

k_{1}\left(\omega_{4}^{2} \cos \theta_{4}+\alpha_{4} \sin \theta_{4}\right)+k_{2}\left(\omega_{2}^{2} \cos \theta_{2}+\alpha_{2} \sin \theta_{2}\right) .

=\left(\alpha_{2}-\omega_{4}\right) \sin \left(\theta_{2}-\theta_{4}\right)+\left(\omega_{2}-\omega_{4}\right)^{2} \cos \left(\theta_{2}-\theta_{4}\right)              (3).

Substituting the values of various terms in Eqs. (1) and (3), we get

k_{1} \cos 90^{\circ}+k_{2} \cos 60^{\circ}+k_{3}=\cos \left(60^{\circ}-90^{\circ}\right) .

or            0.5 k_{2}+k_{3}=\frac{\sqrt{3}}{2} .

or              k_{2}+2 k_{3}=\sqrt{3}                (4).

k_{1} \times 2 \times \sin 90^{\circ}+k_{2} \times 3 \sin 60^{\circ}=(3-2) \sin \left(60^{\circ}-90^{\circ}\right) .

or          2 k_{1}+\frac{3 \sqrt{3}}{2} k_{2}=-\frac{1}{2}.

or          4 k_{1}+3 \sqrt{3} k_{2}=-1               (5).

=(-1-0) \times \sin \left(60^{\circ}-90^{\circ}\right)+(3-2)^{2} \times \cos \left(60^{\circ}-90^{\circ}\right) .

or      k_{2}\left[9 \times \frac{1}{2}-\frac{\sqrt{3}}{2}\right]=(-1) \times\left(-\frac{1}{2}\right)+1 \times \frac{\sqrt{3}}{2} .

k_{2}(4.5-0.866)=\frac{1}{2}+\frac{\sqrt{3}}{2} .

k_{2}=0.376                 (6).

Solving Eqns. (4) to (6), we get

k_{1}=-0.738, k_{3}=0.678 .

k_{2}=\frac{d}{a}, a=\frac{-1}{0.738}=-1.355.

k_{2}=-\frac{d}{c}, c=\frac{-1}{0.376}=-2.659 .

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

0.678 \times 2 \times(-1.355) \times(-2.659)=(-1.355)^{2}-b^{2}+(-2.659)^{2}+1 .

b^{2}=5.0207, b=2.24 .

Thus          d=1.0, a=-1.355, b=2.24, c=-2.659 .