22-Mec-A2 Kinematics and Dynamics of Machines · December 2014

Question 4 of 6: 90° V-twin engine — primary shaking force

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

National Exams, December 2014 — 07-Mec-A2, Kinematics and Dynamics of Machines. 3 hours, OPEN BOOK. Six questions of 20 marks each; candidates answer any FIVE. All SIX are solved in full below (Part A — mechanisms and machine dynamics, Q1–Q4; Part B — mechanical vibration, Q5–Q6).

Reference texts: R. L. Norton, Design of Machinery, 5th ed. (instant centres and velocity analysis, four-bar position/velocity/acceleration, planetary gear kinematics, engine balancing); S. S. Rao, Mechanical Vibrations, 6th ed. (single- and multi-degree-of-freedom free and forced response, modal analysis); J. E. Shigley & C. R. Mischke, Mechanical Engineering Design (epicyclic train-value cross-check).

Question 4: 90° V-twin engine — 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. V-twin, included angle $\psi=90^{\circ}$, both connecting rods on one crankpin. Reciprocating mass per cylinder $m=0.5\ \text{kg}$; crank radius $r=0.120\ \text{m}$; rod length $L=0.300\ \text{m}$ (so $n=L/r=2.5$); constant crank speed $\omega=60\ \text{rad/s}$; crank angle $\phi=30^{\circ}$ measured from one bank axis.

Find. The primary resultant shaking force $\mathbf F_s$ at $\phi=30^{\circ}$.

Figure 4.1 — symmetric 90° V-twin on a single crankpin. Each piston’s primary inertia force acts along its own bank; their sum (green) has constant magnitude $mr\omega^2$ and rotates with the crank.

Approach. Write each cylinder’s primary reciprocating inertia force along its own bank axis, resolve both onto a common frame, and add. For the 90° angle the sum collapses to a single rotating vector.

Primary force of one slider-crank. The reciprocating inertia force along a cylinder axis is $F=mr\omega^2\cos\theta$ (primary) $+\,mr\omega^2\tfrac{\cos2\theta}{n}$ (secondary), with $\theta$ the crank angle from that cylinder’s dead-centre. Only the primary part is required here.
Two banks 90° apart. Take bank 1 along a direction $\hat{\mathbf a}_1$ and bank 2 along $\hat{\mathbf a}_2$ (perpendicular). With the crank at $\phi$ from bank 1, its angle from bank 2 is $\phi-90^{\circ}$, so $$\mathbf F_s=mr\omega^2\bigl[\cos\phi\,\hat{\mathbf a}_1+\cos(\phi-90^{\circ})\,\hat{\mathbf a}_2\bigr] =mr\omega^2\bigl[\cos\phi\,\hat{\mathbf a}_1+\sin\phi\,\hat{\mathbf a}_2\bigr].$$
Resultant magnitude. Since $\hat{\mathbf a}_1\perp\hat{\mathbf a}_2$, $$|\mathbf F_s|=mr\omega^2\sqrt{\cos^2\phi+\sin^2\phi}=mr\omega^2,$$ a value independent of $\phi$. Numerically $$|\mathbf F_s|=mr\omega^2=(0.5)(0.120)(60)^2.$$ $|\mathbf F_s|=216\ \text{N}$ (constant).
Direction at $\phi=30^{\circ}$. The components are $mr\omega^2\cos30^{\circ}=187\ \text{N}$ and $mr\omega^2\sin30^{\circ}=108\ \text{N}$; the resultant therefore points along the crank direction and rotates with the crankshaft. Resolved on a horizontal reference (banks at $\pm45^{\circ}$) it makes $75^{\circ}$ with the horizontal at this instant. Because the force is a constant-magnitude rotating vector, it can be completely balanced by a single crankshaft counterweight of $mr$ opposite the crankpin.

Quantity	Value
$n=L/r$	2.5
Primary force magnitude $mr\omega^2$	216 N (constant, all $\phi$)
Components at $\phi=30^{\circ}$	187 N & 108 N (along the two banks)
Character	rotating force — fully balanceable by a counterweight

Check. The clean constant-magnitude result is exact only for a 90° V-angle; it is the reason 90° V-twins (and V8s) can be primary-balanced with counterweights alone. The secondary forces ($\propto\cos2\theta/n$) do not cancel and would add a smaller $2\omega$ disturbance — but the question asks only for the primary force.

Topic. Primary shaking-force balance of a multi-cylinder (V) reciprocating engine.

Key relations. Single-cylinder primary force $F=mr\omega^2\cos\theta$ along the bore; for a $2\beta$ V-engine the bank forces sum vectorially. For $2\beta=90^{\circ}$ the sum is $mr\omega^2$ at the crank angle — a pure rotating force.

Why this works. Resolving two perpendicular harmonic forces of amplitudes $\cos\phi$ and $\sin\phi$ produces a vector of fixed length that simply rotates — the classic phasor identity $\cos^2+\sin^2=1$. A rotating unbalance is cancellable by an equal, opposite rotating mass (the counterweight), unlike a fixed-line (single-cylinder) shaking force.

Common pitfall. Measuring both crank angles from the same reference (they differ by the V-angle), or assuming counterweighting removes the shaking force of an inline single/twin — there the primary force is along one line and a rotating counterweight only trades vertical for horizontal shake. (Norton, Design of Machinery, ch. 13.)

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