22-Mec-A2 Kinematics and Dynamics of Machines · December 2017
Question 4 of 7: Two-stage planetary gear train design
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper: National Exams — December 2017 · 16-Mec-A2 Kinematics and Dynamics of Machines · 3 hours, open book · Answer five questions, at least one from Part B (Part A mechanisms & machine dynamics Q1–5, Part B mechanical vibration Q6–7). Marks: 20 each. All seven questions are solved here.
Reference texts (subject). R. L. Norton, Design of Machinery, 6th ed. (instant centres & Kennedy’s rule §6.3–6.4, cams Ch. 8, epicyclic trains §9.7–9.9, balancing Ch. 13); Uicker, Pennock & Shigley, Theory of Machines and Mechanisms, 5th ed.; C. E. Wilson & J. P. Sadler, Kinematics and Dynamics of Machinery, 3rd ed.; S. S. Rao, Mechanical Vibrations, 6th ed. (Part B: Ch. 3 base excitation, Ch. 6 & Ch. 9 multi-DOF / torsional).
Check (measured geometry). Questions 1 and 2 carry no printed dimensions — the exam instructs the candidate to scale every length off the drawings (1:6 for Q1, 1:10 for Q2). The joint coordinates used below were read from the printed figure of the figures; the resulting linear quantities are therefore accurate to roughly ±3–5 %. Scale-free results (time ratio, mobility, angular-velocity ratios, mode shapes) are unaffected.
Given. Compound epicyclic: sun 1 (input) meshes planet 2; planet 2 meshes fixed ring 3; a compound planet 2$'$ (rigid with 2, on the free carrier $H$) meshes output ring 4. $\omega_i=1750$ rpm, target $\omega_o=60$ rpm, same module throughout.
Find. General $\omega_o(\omega_i,N_k)$ and a teeth set meeting constraints (i)–(iv).
Compound epicyclic: sun 1 → planet 2 → FIXED ring 3 sets carrier $H$; compound planet 2′ drives OUTPUT ring 4. Both rings coaxial with the input axis.
Approach. Write the train-value equation for the fixed-ring loop to get the carrier speed, then a second train value from sun to the output ring; combine, impose the two coaxial (module) constraints, and search integer teeth for the $\pm2\%$ target with hunting-tooth ratios and no undercut.
Carrier speed (fixed ring 3). For the loop $1\!\to\!2\!\to\!3$, $e_{13}=\dfrac{\omega_3-\omega_H}{\omega_1-\omega_H}=\Big(-\dfrac{N_1}{N_2}\Big)\Big(+\dfrac{N_2}{N_3}\Big)=-\dfrac{N_1}{N_3}$. With $\omega_3=0$:
$$\omega_H=\omega_1\,\frac{N_1}{N_1+N_3}$$
Output ring 4. For $1\!\to\!2\!\to\!2'\!\to\!4$ (with $\omega_2=\omega_{2'}$), $e_{14}=\dfrac{\omega_4-\omega_H}{\omega_1-\omega_H}=-\dfrac{N_1 N_{2'}}{N_2 N_4}$. Eliminating $\omega_H$ gives the general expression:
$$\boxed{\ \frac{\omega_o}{\omega_i}=\frac{N_1}{N_1+N_3}\left(1-\frac{N_3\,N_{2'}}{N_2\,N_4}\right)\ }$$
Constraint (i): first-stage geometry. Planet 2 meshes both sun 1 and ring 3 at equal centre distance, so $N_3=N_1+2N_2$; likewise the compound planet on the same carrier forces $N_4=N_1+N_2+N_{2'}$.
Constraints (ii)–(iv): choose teeth. Target ratio $60/1750=0.03429$. A search over $N_1,N_2,N_{2'}\ge18$ (no undercut) with $\gcd=1$ on every mating pair (hunting / “non-integer” ratio) gives
$$N_1=27,\ N_2=38,\ N_{2'}=49,\ N_3=27+2(38)=103,\ N_4=27+38+49=114$$
Check meshing ratios: $N_1/N_2=27/38$, $N_2/N_3=38/103$, $N_{2'}/N_4=49/114$ — all $\gcd=1$ (non-integer, hunting). Smallest pinion $=27\ge18$ (no undercut); rings 103, 114 are internal (no undercut concern) and of moderate size.
Speed & direction.
$$\frac{\omega_o}{\omega_i}=\frac{27}{130}\!\left(1-\frac{103\cdot49}{38\cdot114}\right)=-0.03428\ \Rightarrow\ \omega_o=-60.0\ \text{rpm}$$
Magnitude 60.0 rpm (error $0.02\%\lt 2\%$); the minus sign means output ring 4 turns opposite to the input sun (a reversal), $\boxed{\omega_o=-60.0\ \text{rpm}}$.
