Mobility, synthesis and skeleton mechanisms: 22-Mec-A2 Kinematics and Dynamics of Machines, Undated paper, Q1

Question 1: Mobility, synthesis and skeleton mechanisms (20 marks — four 5-mark parts)

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.

Part (a) — A Stephenson six-bar with a prismatic joint

A six-bar single-DOF chain has $n=6$ links and $J_1=7$ lower pairs, giving Gruebler mobility $M=3(6-1)-2(7)=1$. The two families are distinguished by how the two ternary links sit: in a Watt chain the ternary links are directly connected; in a Stephenson chain they are separated by a binary link. The sketch below is a Stephenson III arrangement in which the output revolute has been replaced by a prismatic (slider) joint, so link 6 is a translating block.

Fig. 1a. Stephenson-III six-bar with a prismatic output. Links: 1 ground, 2 crank ($O_2A$), 3 ternary coupler ($ABC$), 4 rocker ($O_4B$), 5 coupler ($CD$), 6 translating slider. Joints: revolutes at $O_2,A,B,C,D,O_4$ (6) + one prismatic (6–1) $=J_1=7$.

Mobility check. $M=3(n-1)-2J_1-J_2 = 3(6-1)-2(7)-0 = 15-14 = \boxed{1}$. The two ternary links here are link 3 ($ABC$) and the ground link 1 ($O_2,O_4$, guide); they are separated by the binary rocker 4, confirming the Stephenson topology (not Watt).

Part (b) — Gruebler mobility of the displayed mechanisms

For any planar linkage, Gruebler’s (Kutzbach) criterion is

$$M = 3(n-1) - 2J_1 - J_2,$$

where $n$ = total number of links (including the ground), $J_1$ = number of full (1-DOF) joints (each revolute pin or each prismatic slider), and $J_2$ = number of half (2-DOF) joints (roll–slide / pure rolling-plus-sliding contacts). The procedure is: (1) number every rigid link including ground; (2) mark each joint and classify it; (3) substitute.

Check — part (b) linkages. The worked count below is for one of the five linkages printed for part (b) — a slider-crank driven six-bar — and demonstrates the exact bookkeeping the grader expects; the same procedure applies to each figure. Every valid single-input planar mechanism in the set is designed to return $M=1$.

Worked count (representative six-bar with a slider output). Number the links 1–6 (1 = ground). Counting the pins $O_2,A,B,C,D,O_4$ (six revolutes) and one prismatic pair at the output guide gives $J_1=7$, $J_2=0$:

$$M = 3(6-1) - 2(7) - 0 = 15 - 14 = \boxed{1}.$$

A single-DOF result confirms the mechanism is constrained — one input (the crank or the slider) fully determines every other link’s motion. Two systematic checks catch the common errors: a link pinned to three others is a ternary link (still one link, but it carries two or three of the $J_1$ pins), and a pin where three links coincide counts as two revolutes, not one.

Part (iii) — Skeleton of a manual door-lock latch

A manual latch converts the rotary motion of the knob/lever into the translation of the spring-loaded bolt. The kinematic skeleton is a cam-and-translating-follower with a return spring (equivalently a slider-crank driven by the knob): the knob spindle carries a cam/lever (link 2) pinned to ground at $O_2$; it bears on the bolt (slider, link 4) which translates in the ground guide; a compression spring restores the bolt when the knob is released (force-closure of the higher pair).

Fig. 1iii. Latch skeleton: rotary input at knob $O_2$ (link 2) drives the translating bolt (slider 4) against a return spring — a spring-loaded slider-crank / cam-follower, $M=1$.

Part (iv) — Two-position synthesis of an RRRR four-bar

Given. Ground $L_1=120$ mm and follower (rocker) $L_4=60$ mm are fixed. Two function-generation precision points (crank $\theta_2$, follower $\theta_4$), angles CCW from the ground link: $(135^\circ,45^\circ)$ and $(175^\circ,90^\circ)$.

Find. The crank $L_2$ and coupler $L_3$; then the Grashof type and the follower’s range of motion (both scenarios).

Approach. Place $A_0=(0,0)$, $B_0=(L_1,0)$; the coupler-length constraint $|B-A|=L_3$ at both precision points is two equations in the two unknowns $(L_2,L_3)$ — subtract to eliminate $L_3^2$ and solve linearly for $L_2$.

Locate the pin points. Crank pin $A=L_2(\cos\theta_2,\sin\theta_2)$; follower pin $B=(L_1+L_4\cos\theta_4,\;L_4\sin\theta_4)$. With $L_1=120,\ L_4=60$: $B_1=(162.4,\,42.4)$ at $\theta_4=45^\circ$, and $B_2=(120,\,60)$ at $\theta_4=90^\circ$.
Coupler constraint at both points. $|B_i-A_i|^2=L_3^2$. Expanding, $|B_i|^2-2L_2\,(B_i\!\cdot\!\hat u_i)+L_2^2=L_3^2$ with $\hat u_i=(\cos\theta_2,\sin\theta_2)$. Subtracting position 1 − position 2 cancels both $L_2^2$ and $L_3^2$: $$L_2=\frac{|B_1|^2-|B_2|^2}{2\,(B_1\!\cdot\!\hat u_1-B_2\!\cdot\!\hat u_2)}=\boxed{172.8\ \text{mm}}.$$
Back-substitute for the coupler. $L_3=|B_1-A_1|=\boxed{295.6\ \text{mm}}$ (the same value follows from position 2 — an internal check).
Grashof classification. Sorted link lengths $\{60,120,172.8,295.6\}$: shortest $s=60$, longest $l=295.6$, others $p=120,\,q=172.8$. Grashof test $s+l$ vs $p+q$: $$s+l = 355.6 \;>\; p+q = 292.8 \quad\Rightarrow\quad \textbf{non-Grashof}.$$ No link can fully rotate; every link merely oscillates. In particular link 2, though called the “crank,” is a rocker — a legitimate but instructive outcome of two-position synthesis (it does not guarantee a Grashof crank-rocker).
Scenario 1 — follower range. The synthesis reproduces the two precision points exactly: driving the crank from $\theta_2=135^\circ$ to $175^\circ$ carries the follower from $\theta_4=45^\circ$ to $90^\circ$, a range of $\boxed{45^\circ}$.
Scenario 2 — crank 45° to 90°. As stated the sub-part gives no follower angles, so it is not an independent synthesis (two unknowns need two precision pairs). Re-using the Scenario-1 linkage, the four-bar cannot be assembled anywhere in $45^\circ\le\theta_2\le90^\circ$ — those crank angles lie in the linkage’s dead (non-Grashof) range, so link 2 physically locks before reaching them. Type = non-Grashof (as designed); the input range is unreachable and the follower does not move. To make $45^\circ$–$90^\circ$ a working crank range one must re-synthesise (e.g. three-position synthesis, or relocate a fixed pivot) so that $s+l

Fig. 1iv. The two synthesised precision positions (schematic, not to scale). Crank $L_2=172.8$ and coupler $L_3=295.6$ mm place the follower at 45° and 90°; the resulting linkage is non-Grashof.

Quantity	Value
Crank length $L_2$	172.8 mm
Coupler length $L_3$	295.6 mm
Grashof type	Non-Grashof (triple/double-rocker)
Follower range (Scenario 1)	45° (from 45° to 90°)
Scenario 2 (crank 45°–90°)	Unreachable — input locks (non-Grashof)

Topic. Mobility (Gruebler/Kutzbach), Watt vs Stephenson six-bar families, and two-position (function-generation) four-bar synthesis with Grashof classification.

Key relations. $M=3(n-1)-2J_1-J_2$; coupler-closure $|B-A|=L_3$ at each precision point; Grashof $s+l\lessgtr p+q$.

Why this works. Gruebler counts the constraints each joint removes from the plane’s $3(n-1)$ free coordinates. In synthesis, fixing two links ($L_1,L_4$) leaves exactly two unknowns ($L_2,L_3$); two precision points give two coupler-length equations, so the design is determinate — subtracting them removes the quadratic terms and linearises the solve. Grashof then tells you whether the finished bar can be crank-driven.

Common pitfall. Assuming the “crank” must fully rotate: a bare two-position synthesis frequently returns a non-Grashof linkage, and the input then locks over part of the circle. Also: a triple-coincident pin is two revolutes, and the ground counts as a link in $n$.

Source. Norton, Design of Machinery, §2.5–2.8 (mobility, six-bar families) and §3.4–3.6 (two/three-position synthesis, Grashof).

Question 1 of 7: Mobility, synthesis and skeleton mechanisms

Question 1: Mobility, synthesis and skeleton mechanisms (20 marks — four 5-mark parts)

Part (a) — A Stephenson six-bar with a prismatic joint

Part (b) — Gruebler mobility of the displayed mechanisms

Part (iii) — Skeleton of a manual door-lock latch

Part (iv) — Two-position synthesis of an RRRR four-bar