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.

Find. the maximum shaking force $F_s$ over a cycle, and a counterweight scheme giving $\ge35\%$ reduction at the min- and max-transmission-angle configurations.

Shaking force

With massless crank and follower, the only inertia force is the coupler’s, so the shaking force transmitted to the frame is

$$\vec F_s=-m_3\,\vec a_{G_3}.$$

$G_3$ is the mid-point of the coupler, $\vec a_{G_3}=\vec a_A+\vec\alpha_3\times\vec r_{G_3/A}-\omega_3^2\vec r_{G_3/A}$, where $\vec a_A=-\omega_2^2 r_2\hat r_2$ (crank at constant speed) and $\omega_3,\alpha_3$ come from the four-bar velocity and acceleration loop equations $r_2e^{i\theta_2}+r_3e^{i\theta_3}=r_1+r_4e^{i\theta_4}$. Sweeping the crank through a full turn, the coupler-CG acceleration peaks near the crank’s alignment with the frame:

$$|a_{G_3}|_{\max}=5.39\times10^{3}\ \text{m/s}^2\quad(\text{at }\theta_2\approx9^\circ),$$

$$\boxed{F_{s,\max}=m_3\,|a_{G_3}|_{\max}=0.25(5.39\times10^{3})\approx1.35\ \text{kN}.}$$

The very high speed (3450 rpm) is what turns a mere 0.25 kg rod into a kilonewton-level shaking force ($F\propto\omega^2$).

Transmission-angle configurations

The transmission angle $\mu$ (between coupler and follower) is extreme when the crank lies along the frame line ($\theta_2=0^\circ,180^\circ$). Evaluating the loop closure:

$$\mu_{\min}=41.9^\circ\ (\theta_2=0^\circ),\qquad \mu_{\max}=86.0^\circ\ (\theta_2=180^\circ).$$

Both lie in the acceptable band ($>40^\circ$), and these are the two positions at which the problem asks for the shaking-force reduction.

Balancing scheme (complete force balance)

Model the coupler as two point masses at its pins. Because $G_3$ is central, static equivalence (matching mass and centre of mass — all that the shaking force depends on) gives

$$m_A=m_B=\tfrac12 m_3=0.125\ \text{kg},\quad\text{at pins } A\ (\text{on the crank}) \text{ and } B\ (\text{on the follower}).$$

$m_A$ then orbits $O_2$ at radius $r_2$ and $m_B$ orbits $O_4$ at radius $r_4$ — each a pure rotating unbalance, removable by a counterweight on that link (Berkof–Lowen full-force balancing):

$$ (m_{cw}r_{cw})_{\text{crank}}=m_A r_2=0.125(0.030)=3.75\times10^{-3}\ \text{kg}\cdot\text{m},$$

$$ (m_{cw}r_{cw})_{\text{follower}}=m_B r_4=0.125(0.080)=1.00\times10^{-2}\ \text{kg}\cdot\text{m},$$

placed diametrically opposite $A$ on the crank and opposite $B$ on the follower. This makes the total mass centre of the moving links stationary, so the net shaking force falls to (ideally) zero at every position — in particular a ≈100% reduction at both the $\mu_{\min}$ and $\mu_{\max}$ configurations, comfortably exceeding the required 35%. (If only a partial counterweight is desired, a fraction $f$ of each product reduces the shaking force by the same fraction $f$, so $f=0.5$ already meets the 35% target with lighter counterweights.)