22-Mec-A2 Kinematics and Dynamics of Machines · December 2017
Question 1 of 7: Six-bar punch mechanism — time ratio, instant centres, redesign
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper: National Exams — December 2017 · 16-Mec-A2 Kinematics and Dynamics of Machines · 3 hours, open book · Answer five questions, at least one from Part B (Part A mechanisms & machine dynamics Q1–5, Part B mechanical vibration Q6–7). Marks: 20 each. All seven questions are solved here.
Reference texts (subject). R. L. Norton, Design of Machinery, 6th ed. (instant centres & Kennedy’s rule §6.3–6.4, cams Ch. 8, epicyclic trains §9.7–9.9, balancing Ch. 13); Uicker, Pennock & Shigley, Theory of Machines and Mechanisms, 5th ed.; C. E. Wilson & J. P. Sadler, Kinematics and Dynamics of Machinery, 3rd ed.; S. S. Rao, Mechanical Vibrations, 6th ed. (Part B: Ch. 3 base excitation, Ch. 6 & Ch. 9 multi-DOF / torsional).
Check (measured geometry). Questions 1 and 2 carry no printed dimensions — the exam instructs the candidate to scale every length off the drawings (1:6 for Q1, 1:10 for Q2). The joint coordinates used below were read from the printed figure of the figures; the resulting linear quantities are therefore accurate to roughly ±3–5 %. Scale-free results (time ratio, mobility, angular-velocity ratios, mode shapes) are unaffected.
Given. A crank-shaper (quick-return) six-bar driving a horizontal punch slider. Link 2 is the crank, pinned to ground at $O_2$; its pin $A$ carries slider-block 3, which slides on the oscillating bar 4. Bar 4 is grounded by a pin at $A_u$; its upper end $B$ drives connecting rod 5, which drives the punch slider 6 (point $C$) along the horizontal ground guide. Scaling off the 1:6 drawing gives crank radius $r=O_2A\approx 73$ mm and pivot offset $d=O_2A_u\approx 185$ mm ($O_2$ and $A_u$ lie on a common vertical line), so the governing ratio is $r/d\approx 0.396$.
Find. (i) the time ratio TR; (ii) the seven primary instant centres located by inspection; (iii) $I_{24}$ and $I_{46}$ by Kennedy’s theorem; and the cheapest change that raises TR by 15%.
Six-bar crank-shaper: crank 2 ($O_2A$) drives block 3 on oscillating bar 4 ($A_uB$); rod 5 drives punch slider 6 (point C). Dashed red = Kennedy line $I_{12}I_{14}$; red dots = secondary centres $I_{24}$, $I_{46}$.
Approach. Recognise the quick-return geometry to get the crank angles that bound the two strokes (hence TR), tabulate the seven joint (primary) instant centres, then apply Kennedy’s three-centres rule to intersect two known centre-lines for each requested secondary centre.
Mechanism & mobility. Six links, seven lower pairs (five revolutes at $O_2,A,A_u,B,C$ plus two prismatic pairs: block 3 on bar 4, and slider 6 on ground). Gruebler: $M=3(6-1)-2(7)=15-14=1$ — a single-DOF quick-return.
Seven primary instant centres (by inspection). A revolute is the pin itself; a prismatic pair puts the centre at infinity, perpendicular to the sliding direction:
Centre
Joint
Location
$I_{12}$
crank 2 – frame
at $O_2$ (pin)
$I_{23}$
crank 2 – block 3
at $A$ (pin)
$I_{34}$
block 3 – bar 4 (slide)
$\infty$, $\perp$ to bar 4
$I_{14}$
bar 4 – frame
at $A_u$ (pin)
$I_{45}$
bar 4 – rod 5
at $B$ (pin)
$I_{56}$
rod 5 – slider 6
at $C$ (pin)
$I_{16}$
slider 6 – frame (slide)
$\infty$, vertical ($\perp$ horizontal travel)
Time ratio. Block 3 forces bar 4 to always pass through the crank pin $A$, which travels a circle of radius $r$ about $O_2$. Bar 4 reaches an extreme (a dead centre of the punch) when the line $A_uA$ is tangent to that circle, i.e. when $O_2A\perp A_uA$. With $O_2A_u=d$ vertical, the tangent crank positions satisfy $\sin\theta=-r/d$, symmetric about the downward vertical at $\alpha_0=\arcsin(r/d)$:
$$\alpha_0=\arcsin\frac{r}{d}=\arcsin(0.396)=23.4^\circ$$
The crank sweeps $(180^\circ+2\alpha_0)$ over the slow (working) stroke and $(180^\circ-2\alpha_0)$ over the fast return, so
$$\mathrm{TR}=\frac{180^\circ+2\alpha_0}{180^\circ-2\alpha_0}=\frac{226.7^\circ}{133.3^\circ}=\boxed{1.70}$$
$I_{24}$ by Kennedy. Centres $I_{24}$, $I_{12}$, $I_{14}$ are collinear (links 1,2,4), and $I_{24}$, $I_{23}$, $I_{34}$ are collinear (links 2,3,4). Line 1 is $I_{12}I_{14}=O_2A_u$ (the vertical centreline). Line 2 passes through $I_{23}=A$ toward $I_{34}=\infty$, i.e. through $A$ perpendicular to bar 4. Their intersection gives $\boxed{I_{24}}$ on the centreline, about $9.3$ mm above $O_2$ as drawn, i.e. $\approx 9.3\times 6=$ ≈ 56 mm above $O_2$ on the real machine. (Sanity check: $I_{24}$ can never be further from $O_2$ than the crank radius — algebra gives $\overline{O_2I_{24}}=r\,(r+d\sin\theta_2)/(d+r\sin\theta_2)$, whose magnitude is bounded by $r=73$ mm because $r<d$.)
$I_{46}$ by Kennedy. $I_{46}$, $I_{14}$, $I_{16}$ collinear (links 1,4,6) and $I_{46}$, $I_{45}$, $I_{56}$ collinear (links 4,5,6). Since $I_{16}$ is at infinity in the vertical direction, line 1 is the vertical through $I_{14}=A_u$. Line 2 is $I_{45}I_{56}=BC$ (the axis of rod 5). Their intersection gives $\boxed{I_{46}}$ — on the $A_u$ vertical, level just above $B$ (≈ 18 mm above $B$, i.e. ≈ 154 mm above $O_2$, real). Note this is a much higher point than $I_{24}$; the two must not be confused.
Redesign for +15% TR (lowest cost). Target $\mathrm{TR}'=1.15(1.70)=1.96$. Invert the TR relation:
$$2\alpha_0'=180^\circ\frac{\mathrm{TR}'-1}{\mathrm{TR}'+1}=58.2^\circ\ \Rightarrow\ \frac{r}{d}=\sin 29.1^\circ=0.487$$
TR depends only on $r/d$, so the cheapest change is to lengthen the crank from $r/d=0.396$ to $0.487$, a factor $0.487/0.396=1.23$ — re-drill the crank pin $A$ about $23\%$ farther out (crank $\approx 73\to 90$ mm). No link is added and no ground bearing is relocated. Reducing $d$ instead (moving the $A_u$ pivot up $\approx 19\%$) achieves the same ratio but requires re-machining the frame — more costly.
