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

Question 3 of 6: Compound planetary gear train — output speed

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 3: Compound planetary gear train — output speed (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. Input sun gear 1 at $\omega_1=700\ \text{rpm}$. Stage A: sun 1 → compound planet 2·3 (on carrier C) → internal ring 4. Stage B: sun 5 → planet 6 (on carrier D) → fixed internal ring 7. The compound wheel {4,5} spins freely on the output axis; carriers C and D are both the output shaft ($\omega_C=\omega_D=\omega_o$); $\omega_7=0$. Tooth numbers $N_1{=}22, N_2{=}17, N_3{=}20, N_4{=}96, N_5{=}36, N_6{=}18$.

Find. (a) $\omega_o$ (magnitude and sense); (b) $\omega_{2}-\omega_C$.

Figure 3.1 — kinematic skeleton. Input sun N₁ drives the compound planet N₂·N₃ (carrier C) against internal ring N₄ (stage A); the free compound wheel {N₄,N₅} then drives sun N₅ → planet N₆ (carrier D) → fixed ring N₇ (stage B). Both carriers are the output shaft.

Approach. Fix the missing ring N₇ from the sun-planet-ring coaxial condition, then write the train-value equation $\dfrac{\omega_L-\omega_{arm}}{\omega_F-\omega_{arm}}=e$ for each stage and solve the two coupled equations for $\omega_o$.

Missing ring N₇. In stage B, planet 6 meshes sun 5 (external) and ring 7 (internal) on the same axis, so the coaxial condition gives $$N_7=N_5+2N_6=36+2(18)=72.$$ (N₇ is not printed on the paper; it is fixed by geometry.)
Stage A train value (carrier C held). Gear 1–2 is an external mesh ($-$), gear 3–4 is an internal (ring) mesh ($+$): $$e_A=\Bigl(-\frac{N_1}{N_2}\Bigr)\Bigl(+\frac{N_3}{N_4}\Bigr)=-\frac{22\cdot20}{17\cdot96}=-0.2696.$$
Stage B train value (carrier D held). Sun 5 to ring 7 through planet 6 (external then internal): $$e_B=\Bigl(-\frac{N_5}{N_6}\Bigr)\Bigl(+\frac{N_6}{N_7}\Bigr)=-\frac{N_5}{N_7}=-\frac{36}{72}=-0.5.$$ With $\omega_7=0$: $\dfrac{0-\omega_o}{\omega_{45}-\omega_o}=-0.5\Rightarrow$ $\omega_{45}=3\,\omega_o$.
Couple the stages. The compound wheel is the “last” member of stage A, so $\dfrac{\omega_{45}-\omega_o}{\omega_1-\omega_o}=e_A$. Substituting $\omega_{45}=3\omega_o$: $$\frac{3\omega_o-\omega_o}{700-\omega_o}=-0.2696\;\Rightarrow\;2\omega_o=-0.2696(700-\omega_o).$$ Solving, $\omega_o=-109.1\ \text{rpm}$ — i.e. $109\ \text{rpm}$ opposite to the input (an overall reduction of $700/109.1=6.42:1$, reversing). Then $\omega_{45}=3\omega_o=-327\ \text{rpm}$.
(b) Gear 2 relative to carrier C. In the frame of carrier C the sun–planet mesh gives $\dfrac{\omega_2-\omega_C}{\omega_1-\omega_C}=-\dfrac{N_1}{N_2}$, so $$\omega_2-\omega_C=-\frac{22}{17}\bigl(700-(-109.1)\bigr).$$ $\omega_2-\omega_C=-1047\ \text{rpm}$ — the planet spins at $1047\ \text{rpm}$ relative to its carrier.

Quantity	Value
Fixed ring N₇ (from coaxial N₅+2N₆)	72
Stage train values $e_A,\ e_B$	−0.2696, −0.500
Compound-wheel speed $\omega_{45}$	−327 rpm
Output shaft $\omega_o$	109.1 rpm, opposite to input (reverse)
Gear 2 relative to carrier C	1047 rpm

Check. N₇ is not given, so gear 7 must be an internal ring (the only reading consistent with a fixed member coaxial to sun 5 and planet 6), fixing $N_7=72$. Gear 4 is likewise read as an internal ring on the compound wheel (large, $N_4=96$, meshing the small planet 3 internally); an all-external reading of stage A would instead give $\omega_o\approx+83\ \text{rpm}$ (same sense). The ring interpretation matches the figure’s large top gear 4 and its web across to sun 5.

Topic. Kinematics of a two-stage (compound) epicyclic gear train with a shared carrier.

Key relations. Train-value (superposition) equation $\dfrac{\omega_L-\omega_{arm}}{\omega_F-\omega_{arm}}=e=\prod(\pm N_{\text{driver}}/N_{\text{driven}})$ (external mesh $-$, internal/ring mesh $+$); coaxial ring condition $N_{ring}=N_{sun}+2N_{planet}$.

Why this works. Observed from a frame rotating with the carrier, an epicyclic train becomes an ordinary train whose ratio is the tabulated $e$. Because both stages share one carrier (the output shaft), the ring/sun of one stage and the sun/ring of the next are linked through the free compound wheel, giving two equations in the single unknown $\omega_o$.

Common pitfall. Sign errors at ring meshes (a planet inside a ring does not reverse sense → positive ratio), and dropping a fixed member from the equation — a fixed ring still appears with $\omega=0$, which is exactly what pins the arm speed. (Shigley & Mischke, Mechanical Engineering Design, planetary gears; Norton ch. 9.)

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