Shaking force and balancing of a 3‑cylinder single‑plane engine: 22-Mec-A2 Kinematics and Dynamics of Machines, December 2019, Q4

Question 4 — Shaking force and balancing of a 3‑cylinder single‑plane engine (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.

Crank radius $r$	0.06 m	Connecting rod $l$	0.6 m ⇒ $n=l/r=10$
Crank speed $\omega$	100 rad/s (ccw)	Reciprocating mass/cyl $m$	2.5 kg
Cylinder axes	$0^\circ,120^\circ,240^\circ$	Rotating masses	already balanced

Find. Unbalanced shaking force at the three crank–centreline positions; a balancing scheme; residual force; effectiveness.

Three cylinders $120^\circ$ apart on a single crankpin. When the crank aligns with a cylinder axis, that cylinder is at TDC and its inertia force is maximal along its own centreline.

Approach. Each cylinder’s reciprocating inertia force acts along its own centreline with magnitude $m\omega^2 r(\cos\phi + \tfrac1n\cos 2\phi)$, $\phi$ measured from that cylinder’s TDC. Sum the three force vectors; the primary ($\cos\phi$) terms and secondary ($\cos2\phi$) terms each collapse to a constant‑magnitude rotating resultant.

Force scale. $m\omega^2 r = (2.5)(100)^2(0.06)=\boxed{1500\ \text{N}}$. Each cylinder’s axial inertia force is $F_i = 1500\left(\cos\phi_i + \tfrac{1}{10}\cos2\phi_i\right)$ N, along axis $\hat u_i$ at $\alpha_i\in\{0^\circ,120^\circ,240^\circ\}$, with $\phi_i=\theta-\alpha_i$.
Primary resultant. Summing $\sum_i 1500\cos(\theta-\alpha_i)\,\hat u_i$ over three axes $120^\circ$ apart gives a vector of constant magnitude that rotates with the crank: $$F_{p}=\tfrac{N}{2}\,m\omega^2 r=\tfrac{3}{2}(1500)=\boxed{2250\ \text{N}},$$ independent of $\theta$ — a “direct” rotating primary force (verified at $\theta=0^\circ,30^\circ,60^\circ,90^\circ$: all 2250 N).
Secondary resultant. The $\tfrac1n\cos2\phi$ terms sum, for three equally‑spaced axes, to another constant‑magnitude vector rotating at $2\omega$ (in the opposite sense): $$F_{s}=\tfrac{N}{2}\,\frac{m\omega^2 r}{n}=\tfrac{3}{2}\frac{1500}{10}=\boxed{225\ \text{N}}.$$
(i) Unbalanced shaking force at a centreline. When the crank coincides with a cylinder axis ($\theta=0^\circ$, $120^\circ$ or $240^\circ$), the primary and secondary resultants both point along that same axis and add directly: $$F_{\text{shake}}=F_p+F_s=2250+225=\boxed{2475\ \text{N}}$$ — the same magnitude at all three centreline positions (by symmetry).
(ii) Balancing scheme. Because the primary is a pure rotating vector (unlike an in‑line engine, where the primary oscillates along one line), it is cancelled completely by a single rotating counterweight on the crankshaft, placed opposite the crank with $$m_c r_c = \tfrac{3}{2} m r = 1.5(2.5)(0.06)=\boxed{0.225\ \text{kg}\cdot\text{m}}$$ (e.g. 3.75 kg at 60 mm, or a lighter mass at a larger radius). After this counterweight the primary is zero at every crank angle, leaving only the secondary: $$F_{\text{shake,balanced}}=F_s=\boxed{225\ \text{N}}\ \text{(constant, rotating at }2\omega).$$ To remove the secondary as well would require two counter‑rotating balance shafts spinning at $2\omega$.
(iii) Effectiveness. The single counterweight removes $2250/2475 = \boxed{90.9\%}$ of the peak shaking force. The $9.1\%$ residual is the second‑order (secondary) force, small because $1/n=0.1$; it is smooth and rotating, so for most service it is acceptable, and if vibration limits demand it, a pair of $2\omega$ balance shafts eliminates it entirely.

Quantity	Value
Force scale $m\omega^2 r$	1500 N
Primary resultant (rotating with crank)	2250 N
Secondary resultant (rotating at $2\omega$)	225 N
(i) Unbalanced shaking force at each centreline	2475 N
(ii) Counterweight $m_c r_c$	0.225 kg·m (opposite crank)
(ii) Balanced residual (secondary only)	225 N
(iii) Peak force removed	90.9 %

Topic: Reciprocating‑mass balancing of a multi‑cylinder single‑crank (radial) engine; primary and secondary shaking forces.

Key relations: axial inertia force $F=m\omega^2r(\cos\phi+\tfrac1n\cos2\phi)$; radial‑engine primary $F_p=\tfrac{N}{2}m\omega^2r$ (rotating, balanceable by one counterweight); secondary $F_s=\tfrac{N}{2}m\omega^2r/n$ (rotates at $2\omega$).

Why this works: For $N\ge3$ equally‑spaced cylinders on one crankpin, the vector sum of the per‑cylinder primary forces is a rotating vector of fixed length $\tfrac{N}{2}m\omega^2r$ — identical to a rotating unbalance — so a single rotating counterweight cancels it exactly. The secondaries similarly sum to a fixed‑length vector but at twice speed, which a plain counterweight cannot follow.

Common pitfall: Treating the primary like an in‑line engine’s (a line‑oscillating force that a rotating counterweight only trades between axes). In a radial single‑crank engine the primary genuinely rotates, so it is fully balanceable by one weight — the elegant result of this problem.

Source: Norton, Design of Machinery 6e, Ch. 13 (balancing of reciprocating machinery, multi‑cylinder engines).

Question 4 of 8: Shaking force and balancing of a 3‑cylinder single‑plane engine

Question 4 — Shaking force and balancing of a 3‑cylinder single‑plane engine (20 marks)