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23-Mechatronics-A5 Mechanical Design · December 2019

Question 8 of 10

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

Notes on this paper

16-Mex-A5 — Kinematics and Dynamics of Machines · National Exams, December 2019 · 3 hours, open book · Part A: Q1 compulsory, choose 3 of Q2–Q6; Part B: choose 1 of Q7/Q8. Every question is solved below for study purposes.

Reference texts: Norton, Design of Machinery (mechanism DOF, Grashof, velocity/acceleration analysis, cam design, balancing, gear trains); Wilson & Sadler, Kinematics and Dynamics of Machinery; Rao, Mechanical Vibrations (Q7–Q8).

Check: Several figures on this paper are hand-drawn, to-scale sketches with no printed dimensions (Q1a/b, Q2, Q3, Q6). Pin locations and link lengths were digitised directly from the printed figure and are reported. Absolute lengths for Q3 assume the stated "Scale 1:10" maps drawn millimetres to real millimetres (a labelling convention, not printed on the page); ratios and angular results (ω, α, Grashof class) do not depend on this assumption. The 8-gear train topology in Q6 (which planet meshes which sun/ring) is likewise read from the cross-section drawing; the specific combination adopted is the only one of the readings tried that gives a physically consistent (non locked) gear train.

Question 6 (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 two-stage compound planetary gear train. Stage 1: ring gear 1 fixed to the housing, sun gear 2 on the input shaft, planet 3 meshing both, carried by carrier $C_1$. Stage 2: sun gear 4 rigidly driven by $C_1$; a compound planet (gears 5–6, rigid on one pin) with gear 5 meshing sun 4 and gear 6 meshing ring gear 7; carrier $C_2$ (holding the 5–6 compound planet) is the output shaft. $N_1{=}100,N_2{=}31,N_3{=}15,N_4{=}60,N_5{=}58,N_6{=}50,N_7{=}90,N_8{=}29$.

Find. The speed ratio $\omega_{\text{out}}/\omega_{\text{in}}$.

1 (ring, fixed) 2 3 4 5/6 7 (ring)+8 (sun), fixed housing carrier2 Input Output
Fig. 6 — the two-stage compound PGT read from the cross-section: gear 1's fixed ring and gear 7/gear 8's coupled ring+sun form the two grounded members; carrier $C_2$ is the output.

Approach. Solve each simple stage with the standard epicyclic (Willis) relation, taking $\omega_{\text{in}}=1$ (normalised); Stage 1's carrier output feeds Stage 2's sun input.

  1. Stage 1 (ring 1 fixed, sun 2 = input, carrier $C_1$ unknown). Relative to the arm: $(\omega_2-\omega_{C_1})N_2=-\omega_{p1}N_3$ (sun–planet, external) and $\omega_{p1}N_3=(0-\omega_{C_1})N_1$ (planet–ring, internal). Eliminating $\omega_{p1}$: $$(\omega_2-\omega_{C_1})N_2 = -\omega_{C_1}N_1\ \frac{N_3}{N_3}\ \Rightarrow\ \frac{\omega_{C_1}}{\omega_2} = \frac{N_2}{N_1+N_2} = \frac{31}{131}$$ $$\boxed{\omega_{C_1} = \tfrac{31}{131}\,\omega_{\text{in}} = 0.23664\,\omega_{\text{in}}}$$ This becomes the input to stage 2: $\omega_4=\omega_{C_1}$.
  2. Stage 2 (sun 4 = input from $C_1$; ring 7 — coupled with sun 8 — taken as the second grounded member of this stage, sharing the housing with ring 1; carrier $C_2$ = output). Relative-to-arm relations for the compound planet: $$(\omega_4-\omega_{C_2})N_4 = -\omega_{p2}N_5\quad(\text{sun4–planet5, ext.}),\qquad \omega_{p2}N_6 = (0-\omega_{C_2})N_7\quad(\text{planet6–ring7, int.})$$ Eliminating $\omega_{p2}$: $$\frac{\omega_{C_2}}{\omega_4} = \frac{N_4N_6}{N_4N_6+N_5N_7} = \frac{60\times50}{60\times50+58\times90}=\frac{3000}{8220}=\frac{50}{137}=0.36496$$
  3. Combine both stages. $$\frac{\omega_{\text{out}}}{\omega_{\text{in}}} = \frac{\omega_{C_2}}{\omega_4}\cdot\frac{\omega_{C_1}}{\omega_{\text{in}}} = \frac{50}{137}\times\frac{31}{131} = \frac{1550}{17{,}947}$$ $$\boxed{\frac{\omega_{\text{out}}}{\omega_{\text{in}}} = 0.08637 \quad\Longleftrightarrow\quad \frac{\omega_{\text{in}}}{\omega_{\text{out}}} = 11.58\ (\text{reduction, same sense of rotation})}$$
QuantityValue
Stage 1: $\omega_{C_1}/\omega_{\text{in}}$31/131 = 0.23664
Stage 2: $\omega_{C_2}/\omega_4$50/137 = 0.36496
Overall speed ratio $\omega_{\text{out}}/\omega_{\text{in}}$1550/17947 = 0.08637
Overall reduction $\omega_{\text{in}}/\omega_{\text{out}}$11.58 : 1
Check: the cross-section shows gear 5 apparently close enough to both sun 4 and sun 8 to suggest it meshes both; that literal reading was checked algebraically and found to force the whole train into a rigid lock ($\omega_{C_2}\equiv\omega_4$ for any tooth numbers) — not a valid reduction gearbox. The topology adopted above (ring 7 and sun 8, both housing-mounted and rotating together, form the fixed member of stage 2, matching how nested epicyclic stages are commonly grounded to one shared housing at two different diameters) is the only reading tried that gives a physically consistent, non-degenerate train, and is used here with that caveat disclosed.