Question 4 of 6: Four-bar shaking force and balancing

Question 4: Four-bar shaking force and 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. Grashof crank-rocker: ground $r_1=0.145$ m, crank $r_2=0.035$ m, coupler $r_3=0.150$ m (uniform rod, $m_3=0.15$ kg, centre $G_3$ at its midpoint), rocker $r_4=0.080$ m. $\omega_2=872.5\ \text{rpm}=91.37\ \text{rad/s}$, constant; crank and rocker massless.

Find. Maximum shaking force $F_s$; a counterweight scheme that cuts the peak by ≈40%.

Approach. With only the coupler having mass, the shaking force is $\mathbf F_s=m_3\,\mathbf a_{G_3}$. Sweep the crank through a full turn (constant $\omega_2$), solve the four-bar position at each angle, and take $\mathbf a_{G_3}=\omega_2^2\,\mathrm{d}^2\mathbf r_{G_3}/\mathrm{d}\theta_2^2$; the peak magnitude gives $F_{s,\mathrm{max}}$. For balancing, split the coupler into equal point masses at pins $A$ and $B$ (valid since $G_3$ is central) and counter-rotate them with crank/rocker counterweights.

1. Kinematics and peak shaking force. Solving the loop $\mathbf r_2+\mathbf r_3=\mathbf r_1+\mathbf r_4$ over $\theta_2\in[0,2\pi)$ and differentiating $\mathbf r_{G_3}=\mathbf r_A+\tfrac12\mathbf r_3$ twice: $$a_{G_3,\mathrm{max}}=394\ \text{m/s}^2\ (\text{near }\theta_2\approx9^\circ),\qquad \boxed{F_{s,\mathrm{max}}=m_3\,a_{G_3,\mathrm{max}}=0.15(394)\approx59.1\ \text{N}}.$$
2. Lump the coupler to its pins. Because $G_3$ is at the coupler midpoint, two equal point masses reproduce the coupler’s mass and centre of mass exactly: $$m_A=m_B=\tfrac12 m_3=0.075\ \text{kg}.$$ Mass $m_A$ then executes pure rotation about $O_2$ (radius $r_2$) and $m_B$ about $O_4$ (radius $r_4$); the shaking force is exactly $m_A\mathbf a_A+m_B\mathbf a_B$.
3. Full-balance counterweight products. A counterweight opposite each rotating point mass cancels its inertia force when $m_{cw}r_{cw}=m\,r$: $$(mr)_{\text{crank}}=m_A r_2=0.075(0.035)=2.63\times10^{-3}\ \text{kg}\!\cdot\!\text{m},$$ $$(mr)_{\text{rocker}}=m_B r_4=0.075(0.080)=6.00\times10^{-3}\ \text{kg}\!\cdot\!\text{m}.$$ Applying these fully drives the shaking force to zero.
4. Size for a 40% reduction. Because the two point-mass forces add linearly, counterweighting each by a common fraction $f$ leaves a residual $(1-f)$ of the peak. For a 40% cut, $f=0.40$: $$\boxed{(m_{cw}r_{cw})_{\text{crank}}=1.05\times10^{-3},\quad (m_{cw}r_{cw})_{\text{rocker}}=2.40\times10^{-3}\ \text{kg}\!\cdot\!\text{m}}$$ (each opposite its link), giving $F_{s,\mathrm{max}}:59.1\to35.5\ \text{N}$ (−40%).