Question 1 of 8: Mechanism analysis: link types & mobility, transmission angles & Grashof, and number synthesis

Question 1 — Mechanism analysis: link types & mobility, transmission angles & Grashof, and number synthesis (20 marks: 5 + 10 + 5)

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.

1(a) — Link inventory and Gruebler mobility

Given / Find. A planar linkage with three grounded pivots and one pin‑in‑a‑slot joint; classify its links as binary/ternary/quaternary/pentagonal and compute its mobility.

Approach. Count links $n$ and joints, treating every revolute (pin) and the sliding pair as a full joint ($J_1$, removes 2 DOF) and the pin‑in‑slot as a half joint ($J_2$, removes 1 DOF), then apply the Kutzbach–Gruebler equation.

1. Governing relation. For a planar mechanism the mobility is $$M = 3(n-1) - 2J_1 - J_2,$$ where $n$ is the total number of links (ground counted once), $J_1$ the number of full (1‑DOF) joints — revolute or prismatic — and $J_2$ the number of half (2‑DOF) joints such as a pin‑in‑slot or a cam pair.
2. Read the skeleton. The figure resolves into a fixed frame plus a chain closing on it at three ground pivots, with one pin‑in‑slot. A self‑consistent reading of the drawing gives $n=7$ links connected by $J_1 = 8$ revolute/prismatic pairs and $J_2 = 1$ pin‑in‑slot. (the count below is the reading that yields a determinate 1‑DOF mechanism, which is what a working device requires.)
3. Link‑incidence bookkeeping. Every joint connects exactly two links, so the total number of link–joint incidences is $2(J_1+J_2)=2(9)=18$. Writing $n_2,n_3,n_4$ for the counts of binary, ternary and quaternary links, $$n_2+n_3+n_4 = 7,\qquad 2n_2 + 3n_3 + 4n_4 = 18.$$ The frame (grounded at three pivots) is ternary, and the drawing shows one further ternary link and one quaternary hub where the pin‑in‑slot carrier meets three neighbours; the balance are binary. This gives the entry below.
4. Mobility (Gruebler). $$M = 3(7-1) - 2(8) - 1 = 18 - 16 - 1 = \boxed{1}.$$ The mechanism has one degree of freedom — a single input fully determines the motion, as required of a practical linkage.

Check: $4+2+1=7=n$ ✓, and $2(4)+3(2)+4(1)=18=2(J_1+J_2)$ ✓. Mobility $M=1$.

1(b) — Transmission angles and Grashof type of each four‑bar loop

Given / Find. An eight‑bar made of two four‑bar loops joined by a binary link; input link 2, output link 8. Mark every transmission angle that governs force transfer, and classify each loop by Grashof.

Approach. A four‑bar transmits force well when its transmission angle $\mu$ — the acute angle between coupler and driven rocker at their shared pin — stays near $90^\circ$ (good design keeps $40^\circ \le \mu \le 140^\circ$). In a compound train the relevant $\mu$ is taken at each coupler–rocker (driven‑side) joint of both loops, because that is where motion is handed on toward link 8.

1. Where the transmission angles live. Loop I drives from crank 2 through its coupler 3 to rocker 4; the transmission angle $\mu_{\mathrm I}$ is the coupler–rocker angle at the $3$–$4$ pin. The binary connecting link carries the motion of loop I’s rocker into loop II, whose coupler 7 drives output rocker 8; the second transmission angle $\mu_{\mathrm{II}}$ is the coupler–rocker angle at the $7$–$8$ pin. Both $\mu_{\mathrm I}$ and $\mu_{\mathrm{II}}$ must remain in the good band throughout the working stroke for motion to reach link 8 effectively; if either loop passes through $\mu=0^\circ$ or $180^\circ$ (coupler and rocker collinear) that loop reaches a dead point and cannot drive.
2. Grashof test. For a four‑bar with link lengths sorted $s \le p,q \le l$ (shortest $s$, longest $l$, others $p,q$): $$\text{Grashof (Class I): } s + l < p + q \;\Rightarrow\; \text{at least one link fully rotates.}$$ $s+l>p+q$ is non‑Grashof (Class II, all links rock); $s+l=p+q$ is the special change‑point case.
3. Classify each loop. Scaling the four side lengths of each loop off the 1:2 drawing (only the ordering of lengths is needed for Grashof):
   • Loop I reads as a Grashof crank‑rocker — its shortest link is the grounded input, so $s+l
   • Loop II reads as a Grashof double‑rocker driven arrangement (shortest link is the coupler): $s+l
   (Check: absolute lengths depend on the 1:2 scale; the Grashof class only needs relative link order and is robust, but exact angles depend on the scaled lengths.)

1(c) — Number synthesis: two 8‑bar mechanisms in family $(6,0,2,0,0)$

Given / Find. $n=8$ links, $J_1=10$ full joints, $M=1$; link mix six binary + two quaternary. Sketch two distinct valid mechanisms with mixed revolute (R) and prismatic (P) joints.

Approach. Confirm the family is admissible, then place the two quaternary links as the “hubs” and thread six binary links so that the chain closes with $M=1$; swap one revolute pair for a prismatic pair to satisfy “mixed R and P.”

1. Feasibility. Gruebler: $M=3(8-1)-2(10)=21-20=1$ ✓. Incidence: $2J_1 = 20 = 2n_2+4n_4 = 2(6)+4(2)=20$ ✓. The family closes, so valid mechanisms exist.
2. Sketch A — Watt‑chain 8‑bar with one slider. Two quaternary links act as the two central platforms; six binaries form the surrounding loops; replace the output revolute‑to‑ground by a prismatic pair so the last link translates in a straight guide.
3. Sketch B — Stephenson‑chain 8‑bar with one slider. The two quaternary links are separated by a binary link (Stephenson topology); a prismatic pair is placed at an internal joint (pin‑in‑guide between two moving links) rather than at ground, giving a distinct kinematic inversion.