Question 4 of 6: Four-bar mechanism — shaking-force balancing

Question 4: Four-bar mechanism — shaking-force balancing (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Ground $r_1=145$, crank $r_2=45$, coupler $r_3=150$, follower $r_4=80\ \text{mm}$. Input $\omega_2=3600\ \text{rpm}=376.99\ \text{rad/s}$ (constant). Only the coupler has mass, $m_3=0.2\ \text{kg}$, a uniform rod with mass centre $G_3$ at its midpoint.

Find. A practical counterweight scheme, and the shaking force before and after balancing.

Approach. Establish the size of the unbalanced shaking force (the coupler’s inertia force), then apply the lumped-mass method: split $m_3$ to the two moving pins and counterweight the crank and follower so each rotating link’s mass centre sits on its fixed pivot.

1. Mobility & input. Grashof check: shortest+longest $=45+150=195
2. Unbalanced shaking force. With only the coupler massive, the shaking force equals its inertia force $\mathbf F_s=-m_3\mathbf a_{G_3}$. A full-cycle four-bar acceleration analysis of $G_3$ (coupler midpoint) gives a peak $|\mathbf a_{G_3}|=1.003\times10^{4}\ \text{m/s}^2$ near $\theta_2\approx6^\circ$, so $$F_{s,\mathrm{max}}=m_3\,|\mathbf a_{G_3}|_{\max}=0.2(1.003\times10^4).$$ $F_{s,\mathrm{max}}=2007\ \text{N}$ (unbalanced).
3. Lumped-mass model. Replace the uniform coupler by two point masses at its ends A and B. For force equivalence the total mass and mass-centre must be preserved; since $G_3$ is central, $$m_A=m_B=\tfrac12 m_3=0.10\ \text{kg}.$$ Mass $m_A$ then orbits the crank pivot $O_2$ at radius $r_2$, and $m_B$ orbits the follower pivot $O_4$ at radius $r_4$.
4. Counterweight the crank. Add a counterweight on the crank, opposite the pin, so the crank+lump mass-centre falls on $O_2$: $$m_{w2}r_{w2}=m_A r_2=0.10\times45.$$ $m_{w2}r_{w2}=4.5\ \text{kg}\cdot\text{mm}$ (e.g. 0.15 kg at 30 mm).
5. Counterweight the follower. Likewise on the follower, opposite pin B: $$m_{w4}r_{w4}=m_B r_4=0.10\times80.$$ $m_{w4}r_{w4}=8.0\ \text{kg}\cdot\text{mm}$ (e.g. 0.20 kg at 40 mm).
6. Result. With both counterweights fitted, the mass centre of each rotating link lies on its fixed pivot, so the net inertia force transmitted to the frame is $F_s\approx0$ (complete force balance). Because the two-point lump is force-equivalent but not moment-equivalent to the rod, a residual shaking moment (couple) remains — eliminated only by full force-and-moment balancing.