23-Mechatronics-A5 Mechanical Design · December 2019
Question 3 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.
Given. A planar 8-bar linkage with 10 full joints and DOF = 1 must satisfy Gruebler's equation for the stated link census $B{=}6,T{=}0,Q{=}2,P{=}0,H{=}0$ (link count check: $6+0+2+0+0=8$✓).
Find. Two topologically distinct, valid mechanisms in this family, mixing revolute (R) and prismatic (P) joints, each confirmed to give DOF = 1.
Approach. Confirm the family satisfies Gruebler's equation for any mix of full joints (revolute or prismatic each remove 2 DOF identically), then propose two link arrangements — a six-bar-plus-slider "Stephenson-type" chain and a Watt-type chain with two sliders — each built from 6 binary and 2 quaternary links with 10 simple full joints.
Confirm mobility for the stated census. $n=8$, and with 10 full joints (all $j_1$, no half joints since revolute/prismatic both remove 2 DOF):
$$\text{DOF}=3(n-1)-2j_1 = 3(7)-2(10)=21-20=\boxed{1}$$
This holds independent of the R/P mix, since a prismatic joint is still a full (lower) pair.
Design 1 (all-revolute-plus-one-slider). Ground (part of the two Q links' pin count) – crank(B) – Q-link (4 pins: ground, crank, two couplers) – two couplers(B) – second Q-link (4 pins: two couplers, ground, output slider) – a rocker(B) to a second ground pivot – an output slider(B) riding in a fixed prismatic guide. Every joint R except the final link–slider connection, which is P.
Design 2 (two sliders, symmetric). Two input sliders (B each), each pinned to one arm of a quaternary rocker-plate; the two quaternary plates are cross-braced by two coupler binaries(B); a final output rocker(B) is grounded and pinned to one quaternary plate. This gives 2 prismatic joints (the input sliders) and 8 revolute joints — still 10 full joints, DOF=1.
Verify each design's joint count directly (sum of pins over all links, divided by 2, must equal 10): Design 1: $B$ links contribute 2 pins each ($6\times2=12$), $Q$ links contribute 4 each ($2\times4=8$); total pin-instances $=20$, joints $=20/2=10$.✓ Design 2: identical pin-count bookkeeping (link types are unchanged from the stated census), so it also gives exactly 10 joints.✓
Fig. 1(c) — two distinct topologies in family $B{=}6,T{=}0,Q{=}2,P{=}0,H{=}0$: Design 1 uses one prismatic (output slider) and 9 revolute joints; Design 2 uses two prismatic (input sliders) and 8 revolute joints. Both reduce to DOF = 1.
