22-Mec-A2 Kinematics and Dynamics of Machines · Undated paper
Question 5 of 7: Shaking force of a four-bar crank-rocker and its balancing
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper: National Exams — 16-Mec-A2 Kinematics and Dynamics of Machines · 3 hours, open book · Rubric: answer Q1 (compulsory) plus any four of Q2–Q7. Marks: 20 each. All seven questions are solved here.
Reference texts (subject). R. L. Norton, Design of Machinery, 6th ed. (mobility Ch. 2, synthesis Ch. 3, velocity/acceleration Ch. 6–7, cams Ch. 8, epicyclic trains §9.6–9.9, balancing Ch. 13); Uicker, Pennock & Shigley, Theory of Machines and Mechanisms, 5th ed.; C. E. Wilson & J. P. Sadler, Kinematics and Dynamics of Machinery, 3rd ed.; S. S. Rao, Mechanical Vibrations, 6th ed. (Part B multi-DOF torsional Ch. 6, impact Ch. 1).
Note — figures printed without dimensions. Every figure on this paper is printed without dimensions, and several carry inconsistent numbers (e.g. Q4 shows two conflicting tooth-count sets, and a ring $N_3=20$ cannot mesh a sun $N_1=32$; Q5 prints $AG_3=BG_3=82.5$ mm with $AB=179$ mm, a geometrically impossible triangle). The methods below are exact and complete; where a numerical answer depends on figure geometry (Q2, Q5) or on internally inconsistent data (Q4), the adopted reading is stated explicitly and the value is flagged as representative. Scale-free results (mobility, Grashof class, angular-velocity ratios, transmission-angle location, mode shapes, non-dimensional natural frequencies) are exact regardless of the figure dimensions.
Question 5: Shaking force of a four-bar crank-rocker and its balancing (20 marks)
Find. (i) the shaking-force magnitude $|F_s|$ at the smallest-transmission-angle pose; (ii) a counterweight balancing scheme and its effect.
Check. The paper prints $AG_3=BG_3=82.5$ mm, which cannot coexist with $AB=179$ mm ($82.5+82.5=165<179$ — an impossible triangle). The physically-consistent reading for a uniform coupler is $G_3$ at the coupler midpoint ($=89.5$ mm from each pin), adopted here. Since the crank and rocker are massless (per the statement), the shaking force is $F_s=m_3\,a_{G3}$ alone.
Approach. Do a full position–velocity–acceleration analysis of the four-bar; locate the crank angle giving the minimum transmission angle; evaluate $a_{G3}$ there; then $|F_s|=m_3|a_{G3}|$ and design counterweights.
Grashof check. $s+l=57+228=285crank-rocker (crank fully rotates), consistent with the statement.
Smallest transmission angle. The transmission angle $\mu$ (between coupler and rocker at $B$) reaches its extremes when the crank is collinear with the ground line. Sweeping $\theta_2$ over $360^\circ$, the minimum occurs at $\theta_2=180^\circ$ (crank pointing away from $B_0$), where the coupler–rocker diagonal is longest:
$$\cos\mu=\frac{L_3^2+L_4^2-|AB_0|^2}{2L_3L_4},\quad |AB_0|=L_1+L_2=285\ \text{mm}\ \Rightarrow\ \boxed{\mu_{\min}=45.2^\circ}.$$
Acceleration of the coupler CG at that pose. With $\omega_2=30$ rad/s, $\alpha_2=0$, position/velocity/acceleration closure gives, at $\theta_2=180^\circ$: $\omega_3=6.00$ rad/s, $\alpha_3=289.3$ rad/s$^2$. The crank pin acceleration is purely centripetal, $a_A=\omega_2^2L_2=(30)^2(0.057)=51.3$ m/s$^2$; propagating through the coupler to its midpoint $G_3$,
$$\mathbf a_{G3}=\mathbf a_A+\boldsymbol\alpha_3\times\mathbf r_{AG_3}-\omega_3^2\,\mathbf r_{AG_3}\ \Rightarrow\ |a_{G3}|=\boxed{46.3\ \text{m/s}^2}.$$
(i) Shaking force. Crank and rocker massless, so the only inertial force transmitted to the frame is the coupler’s:
$$|F_s|=m_3\,|a_{G3}|=0.5(46.3)=\boxed{23.2\ \text{N}}.$$
The coupler also produces an inertia (shaking) couple $I_{G3}\,\alpha_3=0.05(289.3)=14.5\ \text{N}\cdot\text{m}$ about $G_3$; this is a moment, not a force, and is reacted at the bearings.
(ii) Balancing strategy — lump-and-counterweight. Model the coupler by two point masses at its pins (valid for force balance when $G_3$ is central): $m_A=m_B=m_3/2=0.25$ kg. Attach a counterweight to the crank opposite $A$ and to the rocker opposite $B$ so each rotating link’s mass-moment is cancelled:
$$m_{cw,2}\,r_{cw,2}=m_A\,L_2=0.25(0.057)=14.3\ \text{kg}\cdot\text{mm},$$
$$m_{cw,4}\,r_{cw,4}=m_B\,L_4=0.25(0.129)=32.3\ \text{kg}\cdot\text{mm}.$$
Complete two-point counterbalancing drives the net first-order shaking force toward zero (a small residual remains because the two-point model is exact for the force but leaves an inertia couple). Applying only a fraction $f$ of each counterweight scales the shaking force down by that same fraction (the contributions superpose linearly), so partial balancing is available if the added mass/couple penalty of full balance is unacceptable. A complete-force-balance design (Berkof–Lowen) fixes both counterweights simultaneously to null the total centre-of-mass acceleration.
Fig. 5. Four-bar at the smallest-transmission-angle pose ($\theta_2=180^\circ$, $\mu_{\min}=45.2^\circ$). The coupler CG $G_3$ (midpoint) has $|a_{G3}|=46.3$ m/s$^2$, giving a shaking force $m_3a_{G3}=23.2$ N.
