NivaarExam PrepOfficial exam papers ↗

22-Mec-A2 Kinematics and Dynamics of Machines · December 2013

Question 3 of 6: Compound planetary gear reduction

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

Notes on this paper

Paper: National Examinations — 07-Mec-A2 Kinematics and Dynamics of Machines, December 2013. Open-book, 3-hour paper; answer FIVE of six questions, all of equal value (20 marks). Part A (mechanisms & machine dynamics, Q1–Q4) and Part B (mechanical vibration, Q5–Q6). All six questions are solved here.

Reference texts: R. L. Norton, Design of Machinery (6th ed., McGraw-Hill) — velocity/acceleration analysis, cams, gear trains, engine balancing; C. E. Wilson & J. P. Sadler, Kinematics and Dynamics of Machinery (3rd ed.); S. S. Rao, Mechanical Vibrations (6th ed., Pearson) — Part B; J. E. Shigley & C. R. Mischke, Mechanical Engineering Design.

Note on the figure-based questions. Q1 (six-bar) is a scaled graphical problem (“Scale 1:5”) with no printed dimensions; its link geometry is read from the drawing, so angular results carry a graphical tolerance of a few percent and the linear accelerations carry the 1:5 length calibration (±5 %). Q3 (planetary box) is solved from the kinematic topology inferred from the sectioned schematic. Both interpretations are stated explicitly with each answer.

Question 3: Compound planetary gear reduction (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. Sun gear 1 is the input (1800 rpm ccw). Gear 1 meshes planet 2; planet 3 is compound with 2 (same planet shaft) and meshes the fixed internal ring gear 6 and also planet 4; planet 4 meshes output sun 5, which is rigidly connected to output gear 7. Planets 2, 3, 4 are carried by the Arm (carrier); suns 1, 5 and ring 6 turn on fixed axes.

Find. $\omega_7$ (magnitude and sense).

1input→236 (ring, fixed)455=7 (out)Arm (carrier)Sun 1 in; compound planet 2–3; 3 meshes fixed ring 6 & planet 4; 4 meshes output sun 5(=7). Schematic topology, not to scale.
Topology of the compound planetary train: input sun 1, compound planet 2–3, fixed ring 6, idler-planet 4, output sun 5 = 7, all planets on the Arm.

Approach. Use the tabular (train-value) relation $\dfrac{\omega_L-\omega_a}{\omega_F-\omega_a}=e$ around the fixed-ring loop to get the Arm speed, then cascade the same relation through planet 4 to the output sun.

  1. Train value, sun 1 → fixed ring 6. External mesh 1–2 contributes $-z_1/z_2$; the compound carries $\omega_2=\omega_3$; the external-to-internal mesh 3–6 contributes $+z_3/z_6$: $$e_{16}=\left(-\frac{z_1}{z_2}\right)\!\left(+\frac{z_3}{z_6}\right)=-\frac{26\cdot18}{50\cdot35}=-0.2674.$$
  2. Arm speed (ring fixed, $\omega_6=0$). $$\frac{\omega_6-\omega_a}{\omega_1-\omega_a}=e_{16}\;\Rightarrow\;\omega_a=\frac{e_{16}}{e_{16}-1}\,\omega_1=\frac{-0.2674}{-1.2674}(1800)=\;$$$\omega_a=379.8$ rpm (ccw).
  3. Planet 3 spin. From $\dfrac{\omega_3-\omega_a}{\omega_1-\omega_a}=-\dfrac{z_1}{z_2}=-0.52$: $\;\omega_3=\omega_a-0.52(\omega_1-\omega_a)=-358.7$ rpm (cw).
  4. Cascade 3 → 4 → output sun 5. Two external meshes (3–4 and 4–5) make planet 4 a pure idler: $$\frac{\omega_5-\omega_a}{\omega_3-\omega_a}=\left(-\frac{z_3}{z_4}\right)\!\left(-\frac{z_4}{z_5}\right)=\frac{z_3}{z_5}=\frac{18}{18}=1,$$ so $\omega_5=\omega_3=-358.7$ rpm. Since gear 7 is rigid with gear 5, $\omega_7=\omega_5=358.7$ rpm, CW — opposite to the input.
QuantityValue
Arm (carrier) speed379.8 rpm, ccw
Planet 3 (=2) speed358.7 rpm, cw
Output gear 7358.7 rpm, CW (ratio ≈ 5.02:1 reduction, reversed)
Check: The section skeleton is read as “output taken from sun 5 (= gear 7)”, consistent with the printed note that gear 5 connects to gear 7. If instead the Arm were the output member, the box would deliver the carrier speed, 379.8 rpm ccw. The tooth data (in particular $z_3=z_5$, making gear 4 an idler) support the sun-5 output reading.