Question 1 of 7: Mobility, skeleton notation and Grashof analysis

Question 1: Mobility, skeleton notation and Grashof analysis (40 marks — four 10-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.

1(i) — Synthesis of an 8-bar mechanism with $M=1$, $J_1=9$, $J_2=2$

Given / Find. Design an eight-link ($n=8$) planar chain whose Gruebler count is exactly $M=1$ with nine full (1-DOF) pairs, two half (2-DOF) pairs, and at least one prismatic (slider) pair.

Approach. Fix the number of links at $n=8$ and back-solve the required joint mix from Gruebler’s equation, then realise it with a physical chain that includes a slider (prismatic full pair) and two higher pairs (a cam contact and a gear mesh give the two $J_2$).

1. Impose Gruebler with the target freedoms. For a planar mechanism $M=3(n-1)-2J_1-J_2$. With $n=8$ and $M=1$: $$1=3(8-1)-2J_1-J_2=21-2J_1-J_2\;\Longrightarrow\;2J_1+J_2=20.$$ The specified $J_1=9,\;J_2=2$ satisfy this: $2(9)+2=20.\;\checkmark$
2. Choose a realisable joint set. Use eight revolute-type full pairs and one prismatic full pair ($J_1=9$, one of them a slider), plus two higher pairs ($J_2=2$): one cam–follower roll/slide contact and one gear mesh. This keeps one slider as required and supplies the two 2-DOF pairs.
3. Assemble the chain (figure). Ground 1 carries a crank 2 (revolute at $O_2$); coupler 3 drives a block 4 that translates in a guide on ground (the prismatic pair); a ternary link 5 pinned near the block carries a roller that rides a cam face and also meshes a small gear on link 2 (the two higher pairs); links 6 and rocker 7 close the loop back to a second ground pivot $O_7$.
4. Confirm the freedom. $n=8$, $J_1=9$ (8 R + 1 P), $J_2=2$: $\;M=3(7)-2(9)-2=21-18-2=\boxed{1}.$ A single input (the crank) drives the whole chain.

1(ii) — Mobility of the backhoe/loader linkage by Gruebler

Given. The excavator-bucket linkage: a ground body, a main boom, an upper rocker, the bucket, and two hydraulic cylinders (each a body + a rod joined by a slider). Find. the number and type of links and joints, and $M$.

1. Count the links ($n=8$). 1 ground, 2 main boom, 3 upper rocker, 4 bucket, 5 upper-cylinder body, 6 upper-cylinder rod, 7 lower-cylinder body, 8 lower-cylinder rod.
2. Count and classify the joints. Eight revolute pairs — ground–boom, boom–rocker, rocker–bucket, boom–bucket, boom–cyl5, rod6–rocker, boom–cyl7, rod8–bucket — plus two prismatic pairs (each cylinder body–rod slider). So $J_1=10$ (8 R + 2 P) and $J_2=0$.
3. Apply Gruebler. $$M=3(n-1)-2J_1-J_2=3(8-1)-2(10)-0=21-20=\boxed{1}.$$

The linkage has one degree of freedom: with the boom held, either hydraulic cylinder acting alone (its slider is the input) fully determines the bucket pose, which is exactly how such an implement is controlled.

1(iii) — Skeleton diagram of the bed-box mechanism

Stripped to standard notation, the lid mechanism is a planar four-bar: the box frame is the fixed link 1 (two ground pivots $O_2,O_4$), a support arm 2 and a support arm 4 are the cranks/rockers, and the lid itself is the coupler 3. All four joints are revolute (pin) pairs. (The physical unit uses two such linkages, one on each side, working as mirror images; kinematically they are a single four-bar.)

As a check, $n=4$, $J_1=4$ R, $J_2=0\Rightarrow M=3(3)-2(4)=1$ — a proper single-DOF lid mechanism.

1(iv) — Grashof analysis of the four-bar

Given. Fixed pivots $A$ (input) and $D$ (output), with $A$ located $0.75$ in horizontally and $1.0$ in vertically from $D$. Input crank $AB=1.25$ in, coupler $BC=2.5$ in, output link $DC=2.5$ in.

Find. (a) mechanism type by Grashof; (b) output-link range of motion; (c) extreme transmission angles.

1. Establish the four link lengths. The ground link joins the two fixed pivots: $$L_1=AD=sqrt{0.75^2+1.0^2}=\sqrt{1.5625}=1.25\ \text{in}.$$ So the set is $L_1=1.25$ (ground), $L_2=1.25$ (input), $L_3=2.5$ (coupler), $L_4=2.5$ (output).
2. Apply the Grashof criterion. With shortest $s=1.25$, longest $l=2.5$, and the remaining two $p=1.25,\,q=2.5$: $$s+l=1.25+2.5=3.75,\qquad p+q=1.25+2.5=3.75.$$ $$\boxed{s+l=p+q}\;\Rightarrow\;\textbf{special-case (change-point) Grashof mechanism.}$$ Because the two equal short links ($1.25$: ground and input) are adjacent and the two equal long links ($2.5$: coupler and output) are adjacent, it is specifically a deltoid (kite) change-point linkage. The input crank (a shortest link, joined to ground) can fully rotate, so it is a crank; the mechanism reaches change-points where all four links line up and the assembly can switch branches.
3. Range of motion of the output link (b). Closing the loop for every input angle and tracking the branch that contains the drawn configuration, the output link $DC$ swings between the two change-point limits: $$\theta_4\in[\,127^\circ,\;307^\circ\,]\quad\Rightarrow\quad \boxed{\Delta\theta_4\approx 180^\circ}$$ i.e. the output rocker sweeps roughly half a turn (through the drawn straight-down position $\theta_4=270^\circ$) before reaching a dead/change position at each end of travel.
4. Extreme transmission angles (c). The transmission angle $\mu$ (between coupler 3 and output 4 at $C$) is extreme when the input crank is collinear with the ground link, giving a triangle $B\text{-}C\text{-}D$ with $BD=L_1\pm L_2$: $$\cos\mu=\frac{L_3^2+L_4^2-BD^2}{2L_3L_4}.$$ Extended crank, $BD=L_1+L_2=2.5$: $\;\cos\mu=\dfrac{2.5^2+2.5^2-2.5^2}{2(2.5)(2.5)}=0.5\Rightarrow\mu=60^\circ.$
   Folded crank, $BD=L_1-L_2=0$: $\;\cos\mu=\dfrac{6.25+6.25-0}{12.5}=1\Rightarrow\mu=0^\circ.$ $$\boxed{\mu_{\min}=0^\circ,\qquad \mu_{\max}=60^\circ.}$$

The transmission angle collapsing to $0^\circ$ (at the folded change-point, where $B$ coincides with $D$) confirms the special-case nature: the mechanism passes through a singular position where it transmits no useful force and the branch is indeterminate — the classic drawback of a change-point linkage.