Question 16.2: Derive Freudenstein’s equation for a four bar linkage.

Consider a four-bar mechanism as shown in Fig.16.16 in equilibrium. The magnitudes of the links AB, BC, CD and DA are a, b, c and d respectively. θ, β and Φ are the angles of AB, BC and DC respectively with the x-axis. AD is the fixed link. AB is the input link and DC the output link.
The displacement along x-axis is,

a \cos \theta+b \cos \beta=d+c \cos \phi .

or        b \cos \beta=c \cos \phi-a \cos \theta+d .

or        (b \cos \beta)^{2}=(c \cos \phi-a \cos \theta+d)^{2} .

=c^{2} \cos ^{2} \phi+a^{2} \cos ^{2} \theta+d^{2}-2 a c \cos \theta \cos \phi-2 a d \cos \theta+2 c d \cos \phi             (1).

The displacement along y-axis is,

a \sin \theta+b \sin \beta=c \sin \phi .

or          b \sin \beta=c \sin \phi-a \sin \theta .

or          (b \sin \beta)^{2}=(c \sin \phi-a \sin \theta)^{2} .

=c^{2} \sin ^{2} \phi+a^{2} \sin ^{2} \theta-2 a c \sin \theta \sin \phi        (2).

Adding Eqs. (1) and (2), we get

b^{2}=c^{2}+a^{2}+d^{2}-2 a c(\sin \theta \sin \phi+\cos \theta \cos \phi)-2 a d \cos \theta+2 c d \cos \phi .

or    2 c d \cos \phi-2 a d \cos \theta+a^{2}-b^{2}+c^{2}+d^{2}=2 a c(\sin \theta \sin \phi+\cos \theta \cos \phi) .

Dividing throughout by 2ac, we get

\frac{d}{a} \cos \phi-\frac{d}{c} \cos \theta+\frac{a^{2}-b^{2}+c^{2}+d^{2}}{2 a c}=\cos (\phi-\theta)=\cos (\theta-\phi) .

or    k_{1} \cos \phi+k_{2} \cos \theta+k_{3}=\cos (\theta-\phi)         (16.9).

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

Eq. (16.9) is known as Freudenstein equation.