Force Analysis of Fourbar Linkage

For a fourbar linkage, the problem set-up first requires that we know:

• Angular and linear accelerations (determined through kinematic analysis)
• Masses of links
• Position information (link lengths, link positions, location of center of gravity for each link)
• Mass moment of inertia with respect to CG, $I_G$, for each link

We then write out the forces and moments for each link.

Link 2:

$$\begin{array}{rl}{F}_{12x}& +F_{32x}=m_2{a}_{G_{2x}}\\ F_{12y}& +F_{32y}=m_2{a}_{G_{2y}}\\ T_{12}& +(R_{12x}{F}_{12y}-R_{12y}{F}_{12x})+(R_{32x}{F}_{12y}-R_{32y}{F}_{32x})=I_{G_2}{\alpha }_2\end{array}$$

Link 3:

$$\begin{array}{rl}{F}_{43x}& -F_{32x}+F_{Px}=m_3{a}_{G_{3x}}\\ F_{12y}& -F_{32y}+F_{Py}=m_3{a}_{G_{3y}}\\ (R_{43x}& F_{43y}-R_{43y}{F}_{43x})+(R_{23x}{F}_{32y}-R_{23y}{F}_{32x})+(R_{Px}{F}_{Py}-R_{Py}{F}_{Px})=I_{G_3}{\alpha }_3\end{array}$$

Link 4:

$$\begin{array}{rl}{F}_{14x}& -F_{43x}=m_4{a}_{G_{4x}}\\ F_{14y}& -F_{43y}=m_4{a}_{G_{4y}}\\ (R_{14x}& F_{14y}-R_{14y}{F}_{14x})-(R_{34x}{F}_{43y}-R_{34y}{F}_{43x})+T_4=I_{G_4}{\alpha }_4\end{array}$$

This can be written in matrix form as a simultaneous system of equation with 9 unknowns:

$$\left[\begin{array}{ccccccccc}1& 0& 1& 0& 0& 0& 0& 0& 0\\ 0& 1& 0& 1& 0& 0& 0& 0& 0\\ -R_{12y}& R_{12x}& -R_{32y}& R_{32x}& 0& 0& 0& 0& 1\\ 0& 0& -1& 0& 1& 0& 0& 0& 0\\ 0& 0& 0& -1& 0& 1& 0& 0& 0\\ 0& 0& R_{23y}& -R_{23x}& -R_{43y}& R_{43x}& 0& 0& 0\\ 0& 0& 0& 0& -1& 0& 1& 0& 0\\ 0& 0& 0& 0& 0& -1& 0& 1& 0\\ 0& 0& 0& 0& R_{34y}& -R_{34x}& -R_{14y}& R_{14x}& 0\end{array}\right]\times \left[\begin{array}{c}{F}_{12x}\\ F_{12y}\\ F_{32x}\\ F_{32y}\\ F_{43x}\\ F_{43y}\\ F_{14y}\\ F_{14x}\\ T_{12}\end{array}\right]=\left[\begin{array}{c}{m}_2{a}_{G_{2x}}\\ m_2{a}_{G_{2y}}\\ I_{G_2}{\alpha }_2\\ m_3{a}_{G_{3x}}-F_{Px}\\ m_3{a}_{G_{3y}}-F_{py}\\ I_{G_3}{\alpha }_3-R_{px}{F}_{Py}+R_{Py}{F}_{Px}\\ m_4{a}_{G_{4x}}\\ m_4{a}_{G_{4y}}\\ I_{G_4}{\alpha }_4-T_4\end{array}\right]$$ $$\left[\begin{array}{c}\mathbf{A}\end{array}\right]\left[\begin{array}{c}\mathbf{B}\end{array}\right]=\left[\begin{array}{c}\mathbf{C}\end{array}\right]$$

We can then take the inverse of the matrix to solve:

$$\left[\begin{array}{c}\mathbf{B}\end{array}\right]={\left[\begin{array}{c}\mathbf{A}\end{array}\right]}^{-1}\left[\begin{array}{c}\mathbf{C}\end{array}\right]$$