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

Question 1 of 6: Six-bar mechanism — instant centres & output angular velocity

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 1: Six-bar mechanism — instant centres & output angular velocity (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. A Stephenson-III six-bar: input crank 2 (ground pivot O₂, moving pin A) → coupler 3 (A–P) → grounded ternary link 4 — the straight bar carrying pins O₄, P and B, which the drawing shows collinear → coupler 5 (B–U, the large solid link) → output rocker 6 (U–O₆). The input rotates at $\omega_2=1800\ \text{rpm}=188.5\ \text{rad/s}$. Joint centres scaled from the 1:5 drawing (drawing units, origin arbitrary, $x$ right, $y$ up):

Point	x	y	Role
O₂	945	−2094	frame pivot of input crank 2
A	815	−2336	pin 2–3
P	1106	−1958	pin 3–4
O₄	882	−1668	frame pivot of ternary link 4
U	1157	−1819	pin 5–6
B	1311	−2221	pin 4–5 (third pin of the ternary bar)
O₆	1577	−2374	frame pivot of output link 6

Find. (i) the 15 instant centres, and (ii) the angular velocity of output link 6, obtained through the input–output instant centre I_2,6.

Figure 1.1 — the mechanism at the given position. Links 2–6 with the seven pin/pivot centres; the common instant centre I₂₆ (red) lies on the line joining O₂ and O₆, and the input–output velocity ratio follows from the two dashed segments.

Approach. Count the centres with $N=n(n-1)/2$; the primary centres follow by inspection and the secondaries by Kennedy’s three-centre theorem. Two relative-velocity loops then give $\omega_4$ and $\omega_6$, and the same $\omega_6$ is recovered from the I_2,6 distance ratio.

Number of instant centres. With $n=6$ links, $$N=\tfrac{n(n-1)}{2}=\tfrac{6\cdot5}{2}=15.$$ Seven are found by inspection (the seven revolute pins): I₁₂=O₂, I₂₃=A, I₃₄=P, I₁₄=O₄, I₄₅=B, I₅₆=U, I₁₆=O₆. The remaining eight (I₁₃, I₂₄, I₁₅, I₂₅, I₃₅, I₄₆, I₂₆, I₃₆) follow from Kennedy’s theorem.
Locating I_2,6 (Kennedy). The three mutual centres of any three links are collinear, so I₂₆ lies simultaneously on line I₁₂I₁₆ (i.e. O₂O₆) and on line I₂₄I₄₆. Their intersection is I₂₆ = (1019, −2127) in drawing units — a point between the two ground pivots, $80.5$ units from O₂ and $610.7$ units from O₆.
Velocity loop 1 (links 3–4). The crank gives $\mathbf v_A=\boldsymbol\omega_2\times\mathbf r_{A/O_2}$. Pin P is shared by coupler 3 (about A) and ternary link 4 (about O₄): $$\mathbf v_A+\boldsymbol\omega_3\times\mathbf r_{P/A}=\boldsymbol\omega_4\times\mathbf r_{P/O_4}.$$ The two scalar components give $\omega_3=+102.5$ and $\omega_4=+23.73\ \text{rad/s}$ (CCW).
Velocity loop 2 (links 5–6). Link 4 is rigid, so pin B has $\mathbf v_B=\boldsymbol\omega_4\times\mathbf r_{B/O_4}$. Pin U is shared by coupler 5 (about B) and rocker 6 (about O₆): $$\mathbf v_B+\boldsymbol\omega_5\times\mathbf r_{U/B}=\boldsymbol\omega_6\times\mathbf r_{U/O_6}.$$ Solving, $\omega_5=-1.66$ and $\omega_6=-24.85\ \text{rad/s}$ — magnitude $24.85\ \text{rad/s}\ (\approx 237\ \text{rpm})$, sense opposite to the input.
Cross-check by the I_2,6 ratio. The point I₂₆ has one velocity whether viewed on link 2 (about O₂) or link 6 (about O₆), so $$\frac{\omega_6}{\omega_2}=\frac{\overline{O_2\,I_{26}}}{\overline{O_6\,I_{26}}}=\frac{80.5}{610.7}=0.1318 \;\Rightarrow\; \omega_6=0.1318\times188.5=24.85\ \text{rad/s},$$ confirming the loop result.

Quantity	Value
Number of instant centres	15 (7 by inspection, 8 by Kennedy)
Input–output centre I_2,6	(1019, −2127), on line O₂O₆
$\omega_3,\ \omega_4$	+102.5, +23.73 rad/s
$\omega_5$	−1.66 rad/s
Output angular velocity $\omega_6$	24.85 rad/s (≈237 rpm), opposite to input

Check. This is a scaled graphical problem (Scale 1:5) with no printed dimensions; the seven joint centres were read from the printed figure of the drawing, so $\omega_6$ carries a graphical tolerance of a few percent. The ratio $\omega_6/\omega_2$ (and hence $\omega_6$) is dimensionless and independent of the 1:5 scale — only the absolute I_2,6 distances scale. The sense of $\omega_6$ (here CW for a CCW input) matches the rotation arrow drawn on link 6.

Topic. Instant centres of velocity, Kennedy’s theorem, and the instant-centre angular-velocity ratio.

Key relations. $N=n(n-1)/2$; the three mutual centres of any three links are collinear (Kennedy); for two moving links $j,k$ and the frame, $\omega_k/\omega_j=\overline{I_{1j}I_{jk}}/\overline{I_{1k}I_{jk}}$.

Why this works. I_2,6 is the one point that, at this instant, belongs to both link 2 and link 6 with a single velocity. Since that velocity equals $\omega_2\!\cdot\!\overline{O_2I_{26}}$ on link 2 and $\omega_6\!\cdot\!\overline{O_6I_{26}}$ on link 6, equating them gives the output speed directly — no intermediate links needed.

Common pitfall. Because the input link here rotates about a pin, I₁₂ is at O₂ (a finite point) — not at infinity as it would be for a sliding input; putting it at infinity mislocates I_2,6. Also, ternary link 4 carries three centres (O₄, P, B), which must all be used when tracing the Kennedy lines. (Norton, Design of Machinery, ch. 6.)

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