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

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

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

Notes on this paper

National Exams, May 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 mechanical advantage, cam design, gear-train kinematics, linkage balancing); S. S. Rao, Mechanical Vibrations, 6th ed. (impulsive loading, single- and two-degree-of-freedom free and forced response); J. E. Shigley & C. R. Mischke, Mechanical Engineering Design (planetary-train cross-check).

Question 3: Compound planetary gear reduction — 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 $\omega_1=1200\ \text{rpm}$ (ccw, taken positive). Read from the printed sectional schematic: the outer drum is a single body — its web (labelled “Arm”) is journalled on the input shaft and carries the compound planet 2·3, while its rim carries the internal teeth labelled 7, so the carrier and gear 7 are the same output member. Internal ring 4 is joined through its web and the central hub to sun gear 5, and that compound wheel {4, 5} floats in a frame bearing. Gear 6 runs in a frame bearing of its own, so it is an idler on a fixed axis meshing sun 5 externally and ring 7 internally. Tooth numbers as listed.

Find. $\omega_7$ and its sense of rotation.

Figure 3.1 — kinematic layout read from the printed section: sun 1 drives compound planet 2·3 against ring 4; the free wheel {4, 5} drives fixed-axis idler 6, which drives internal ring 7 — and ring 7 is itself the arm that carries planet 2·3, closing the loop.

Approach. Settle from the drawing which members are actually held, confirm the coaxial (ring) conditions, then write the epicyclic train-value equation of the planet-2·3 stage (whose carrier is gear 7 itself) together with the ordinary fixed-axis ratio of the 5–6–7 stage, and solve the two simultaneously — the train is a closed loop, not a cascade.

Reading the section. All three ground symbols on the drawing are rolling elements drawn between hatched frame lines, i.e. bearings: they carry the input shaft, the {4, 5} hub and gear 6’s shaft, so no gear is clamped to the frame. The only rotating carrier drawn is the drum, whose web is labelled “Arm” and whose rim sits exactly at radius $r_5+2r_6$ and is labelled 7; hence $\omega_{arm}=\omega_7$. Two checks confirm this is the only consistent reading: (i) were gear 6 a planet on that same drum, its stage equation would read $(\omega_7-\omega_7)/(\omega_5-\omega_7)=-z_5/z_7$, forcing $\omega_5=\omega_7$ and locking the train solid; (ii) were {4, 5} grounded, $\omega_5=0\Rightarrow\omega_6=0\Rightarrow\omega_7=0$, contradicting $\omega_7=\omega_{arm}\ne 0$.
Coaxial check. $z_1+z_2=z_4-z_3$ ($26+50=76=94-18$ &checkmark) and $z_7=z_5+2z_6$ ($88=18+70$ &checkmark), confirming gears 4 and 7 are internal rings.
Epicyclic stage (sun 1 → ring 4, carrier = gear 7). Mesh 1–2 is external ($-$) and mesh 3–4 planet-to-ring internal ($+$): $$e_A=\left(-\frac{z_1}{z_2}\right)\!\left(+\frac{z_3}{z_4}\right)=\left(-\frac{26}{50}\right)\!\left(\frac{18}{94}\right)=-0.09957,\qquad \frac{\omega_4-\omega_7}{\omega_1-\omega_7}=e_A.$$
Fixed-axis stage (sun 5 → idler 6 → ring 7). Both centres are fixed in the frame, so $\omega_6=-\omega_5 z_5/z_6$ and $\omega_7=+\omega_6 z_6/z_7$, i.e. $$\omega_7=-\frac{z_5}{z_7}\,\omega_5=-0.2045\,\omega_5\quad\Longleftrightarrow\quad \omega_4=\omega_5=-\frac{z_7}{z_5}\,\omega_7=-4.889\,\omega_7.$$
Solve the loop. Substituting $\omega_4=-4.889\,\omega_7$ into step 3, $$-4.889\omega_7-\omega_7=-0.09957\,(1200-\omega_7)\ \Rightarrow\ -5.9885\,\omega_7=-119.49.$$ $\omega_7=19.95\ \text{rpm}$, positive ⇒ ccw (the same sense as the input), an overall reduction of $60.1{:}1$. The floating wheel {4, 5} then turns at $\omega_{4,5}=-97.5$ rpm (cw) and the compound planet at $\omega_{2,3}=-594$ rpm.

Quantity	Value
Epicyclic-stage train value $e_A$	−0.09957
Fixed-axis stage ratio $\omega_7/\omega_5$	−0.2045
Free compound wheel {4, 5}	−97.5 rpm (cw)
Compound planet 2·3	−594 rpm
Output gear 7 (= the arm)	19.95 rpm, CCW
Overall reduction	60.1 : 1

Check. The topology was taken from the printed section, not inferred from the tooth numbers. The drum carrying both the “Arm” label and the gear-7 teeth is one continuous outline (web plus both rims, the lower rim ending in a tooth tick labelled 7 at exactly $r_5+2r_6$ from the axis); the ring-4 web runs into the central hub that carries sun 5, and that hub and gear 6’s shaft each have a frame bearing of their own. Reading the drawing instead as two independently grounded members (ring 4 and sun 5) driving a separate carrier turns the train into a simple cascade and gives $\omega_7=130.9$ rpm, but that reading needs a seventh body the drawing does not contain and leaves gear 6 attached to no carrier. Both coaxial tooth-sum identities hold under either reading, so they do not discriminate — the drawing does.

Topic. Kinematics of a closed-loop (feedback) compound planetary gear train.

Key relations. Train-value (superposition) equation $\dfrac{\omega_L-\omega_{arm}}{\omega_F-\omega_{arm}}=e$ with $e=\prod(\pm z_{\text{driver}}/z_{\text{driven}})$ (external mesh $-$, internal/ring mesh $+$); for a pair on fixed centres that same product is the speed ratio directly; coaxial ring condition $z_{ring}=z_{sun}+2z_{planet}$.

Why this works. Viewed from the rotating carrier the planet stage becomes an ordinary train with train value $e_A$. Because that carrier is also the output ring, the two stages are not independent: the fixed-axis stage feeds its output straight back to the carrier of the first, so the two equations must be solved together. That feedback is what turns a nine-to-one cascade into a sixty-to-one reduction.

Common pitfall. Assuming any two-stage planetary must be a cascade of two grounded-ring stages — read the section for which members ride bearings (free) and which are clamped (held), and check whether the carrier and a ring are the same part. Also the usual sign trap: a pinion meshing the inside of a ring does not reverse rotation, so that ratio is positive. (Shigley & Mischke, Mechanical Engineering Design, planetary gears; Norton, Design of Machinery, ch. 9.)

← Question 2 Next: Question 4 →