90° V-Twin — Primary Shaking Force: 22-Mec-A2 Kinematics and Dynamics of Machines, May 2013, Q4

Question 4: 90° V-Twin — Primary Shaking Force (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. Two pistons on a common crankpin, bank (included) angle $\psi=90^\circ$ so each cylinder axis is $\beta=45^\circ$ from the bisector; reciprocating mass $m=2.5\ \text{kg}$ per piston; crank radius $r=0.090\ \text{m}$, rod length $l=0.500\ \text{m}$; crank speed $N=4000\ \text{rpm}$, crank angle $\phi=35^\circ$ measured from the cylinder-1 axis.

Find. The resultant primary shaking force at $\phi=35^\circ$.

Fig. 4 — 90° V-twin with both connecting rods on one crankpin. The two cylinder axes are ±45° from the vertical bisector; the crank is drawn at 10° from the bisector (φ=35° from the cyl-1 axis).

Approach. Each cylinder contributes a primary reciprocating inertia force $m r\omega^2\cos\theta_i$ directed along its own axis, where $\theta_i$ is the crank angle from that cylinder’s TDC. Resolve both along common $x$–$y$ axes (bisector vertical) and sum.

Crank speed and inertia scale. $$\omega=\frac{2\pi N}{60}=418.9\ \text{rad/s},\qquad m r\omega^2=2.5(0.090)(418.9)^2=3.95\times10^{4}\ \text{N}.$$
Crank angle at each cylinder. Cylinder 1: $\theta_1=\phi=35^\circ$. Cylinder 2 lies $\psi=90^\circ$ away, so $\theta_2=\phi-90^\circ=-55^\circ$, and $\cos\theta_2=\cos(\phi-90^\circ)=\sin\phi$.
Primary force of each cylinder along its axis. With unit axes $\hat u_1=(\sin45^\circ,\cos45^\circ)$ and $\hat u_2=(-\sin45^\circ,\cos45^\circ)$: $$\mathbf F=m r\omega^2\big[\cos\phi\,\hat u_1+\sin\phi\,\hat u_2\big].$$
Components (bisector = y). $$F_x=m r\omega^2\sin45^\circ(\cos\phi-\sin\phi)=6.86\ \text{kN},$$ $$F_y=m r\omega^2\cos45^\circ(\cos\phi+\sin\phi)=38.9\ \text{kN}.$$
Resultant. $$|\mathbf F|=\sqrt{F_x^2+F_y^2}=\boxed{39.5\ \text{kN}},\qquad \text{at }10^\circ\text{ from the bisector.}$$
Elegant check. For a 90° V-twin on one crankpin the primary force reduces exactly to $m r\omega^2\sqrt{\tfrac12}\sqrt{2}=m r\omega^2$ — constant magnitude, rotating with the crank. Indeed $39.5\ \text{kN}=m r\omega^2$, and its line lies along the crank (here $45^\circ-35^\circ=10^\circ$ from the bisector). This is why a 90° V-twin’s primary shake can be fully balanced by a single rotating counterweight.

Quantity	Value
$m r\omega^2$	39.5 kN
Primary force components ($F_x,F_y$)	6.86 kN, 38.9 kN
Resultant primary shaking force	39.5 kN
Direction	along the crank, 10° from bisector

Topic. Reciprocating-mass balancing of a multi-cylinder (V) engine — primary shaking force.

Key relations. Reciprocating acceleration $a\approx r\omega^2(\cos\theta+\tfrac{r}{l}\cos2\theta)$; primary term $m r\omega^2\cos\theta$ per cylinder; vector sum along the two bank axes.

Why this works. The shaking force is the net inertia reaction of the reciprocating masses on the frame. Keeping only the $\cos\theta$ (once-per-rev) term isolates the primary force. For a 90° bank the two cosine terms combine (via $\cos^2+\sin^2=1$) into a constant-magnitude vector that simply rotates with the crank — a special, fully balanceable case.

Common pitfall. Using the wrong crank reference for the second cylinder (it trails by the bank angle, not by 180°), and confusing the primary force with the secondary ($2\omega$) force, which the question does not ask for.

Source. Norton, Design of Machinery, §13 (engine dynamics: shaking forces, V-engine balancing); Wilson & Sadler, ch. on balancing.

Question 4 of 6: 90° V-Twin — Primary Shaking Force

Question 4: 90° V-Twin — Primary Shaking Force (20 marks)