22-Mec-A2 Kinematics and Dynamics of Machines · December 2018
Question 5 of 7: Shaking force and balancing of an inverted crank-slider
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Reference texts (subject). R. L. Norton, Design of Machinery, 6th ed. (number synthesis & mobility Ch. 2, position/velocity Ch. 4–6, cam design Ch. 8, epicyclic trains §9.6–9.9, balancing Ch. 13); J. J. Uicker, G. R. Pennock & J. E. 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: single-DOF transient Ch. 2–4, multi-DOF Ch. 5–6). Open-book exam; everyone answers Q1, then any three of Part A (Q2–Q5) and one of Part B (Q6–Q7). Every question is solved here.
Question 5: Shaking force and balancing of an inverted crank-slider (20 marks)
Given. Crank $r_2=0.50$ m, ground $r_1=0.90$ m, $\omega_2=1450\ \text{rpm}=151.8$ rad/s (constant, $\alpha_2=0$). Only the slider-coupler carries mass: $m_3=0.5$ kg with centre $G_3$ at the crank pin $A$, $I_{G_3}=0.175$ kg·m$^2$.
Find. the general kinematic expressions, the shaking force at $\theta_2=75^\circ$ and $135^\circ$, and a 50% balancing scheme.
Inverted crank-slider (not to scale). Only slider-coupler 3 has mass ($m_3=0.5$ kg, centre $G_3$ at pin $A$). Because $G_3$ sits on the crank pin, its acceleration is purely centripetal, so the shaking force $F_s=m_3\omega_2^2 r_2$ has constant magnitude and rotates with the crank.
Approach. Place $G_2$ at the origin and $G_4$ at $(r_1,0)$; write the loop closure for the inverted slider-crank to get $\theta_4$, the slip distance $b$, and their derivatives; then note the coupler mass centre sits on the crank pin, so its acceleration is purely centripetal and the shaking force follows immediately.
(i) Position. Crank pin $A=(r_2\cos\theta_2,\,r_2\sin\theta_2)$. Link 4 passes through $G_4$ and $A$, so
$$\theta_4=\operatorname{atan2}\!\big(r_2\sin\theta_2,\ r_2\cos\theta_2-r_1\big),\qquad b=|G_4A|=\sqrt{r_1^2+r_2^2-2r_1r_2\cos\theta_2}.$$
Velocity / acceleration. Differentiating the loop,
$$\omega_4=\frac{r_2(r_2-r_1\cos\theta_2)}{b^2}\,\omega_2,\qquad \dot b=-r_2\omega_2\sin(\theta_2-\theta_4),$$
and since $\alpha_2=0$ the mass-centre acceleration is purely centripetal:
$$\mathbf a_{G_3}=\mathbf a_A=-\omega_2^2 r_2(\cos\theta_2,\ \sin\theta_2),\qquad |\mathbf a_{G_3}|=\omega_2^2 r_2.$$
(ii) Shaking force. With crank and follower massless, the only inertia force is that of the coupler, $\mathbf F_s=-m_3\mathbf a_{G_3}$, of magnitude
$$|\mathbf F_s|=m_3\,\omega_2^2\,r_2=0.5\,(151.8)^2(0.50)=\boxed{5.76\ \text{kN}}.$$
Because $\mathbf a_{G_3}$ is centripetal, this magnitude is independent of $\theta_2$: at both $\theta_2=75^\circ$ and $\theta_2=135^\circ$ the shaking force is $5.76$ kN, directed radially along the crank (at $75^\circ$ and $135^\circ$ respectively). The force is a constant-magnitude vector that simply rotates with the crank — a pure rotating unbalance.
(iii) Balancing scheme. A rotating-unbalance force is cancelled by a counter-rotating mass. Mount a counterweight on the crank, $180^\circ$ opposite $A$, with $m_{cw}r_{cw}=m_3 r_2$ for full cancellation. For the requested 50% reduction,
$$m_{cw}\,r_{cw}=0.5\,m_3\,r_2=0.5(0.5)(0.50)=\boxed{0.125\ \text{kg}\cdot\text{m}}$$
(e.g. $m_{cw}=0.5$ kg at $r_{cw}=0.25$ m). Reason: the shaking force rotates rigidly with the crank at constant magnitude, so a single crank-mounted counterweight produces an equal-and-opposite rotating force at every instant — the ideal, fully-effective case for counterweight balancing. A residual shaking moment $I_{G_3}\alpha_3$ from the coupler’s angular acceleration remains and would need a separate measure (it is not a force).
