23-Mechatronics-A5 Mechanical Design, December 2019, Q4

Question 2 (20 marks)

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.

Given. A five-bar toggle press: an input bell-crank (link 2) pivoted to ground at $O_2$, with a slot that drives a follower rod (link 4) through a pin-in-slot half joint at $A$; link 4 is pinned at $D$ to a ternary rocker (link 5) grounded at $C$; link 5's lower pin $S$ drives an output slider (link 7) that presses vertically against the workpiece, $F_{out}$. Force $F_{in}$ is applied near-tangentially at the free end of link 2. Digitised pivot positions (arbitrary consistent units): $O_2{=}(313,309)$, $P_{in}{=}(170,285)$, $A{=}(330,311)$, $D{=}(360,325)$, $C{=}(358,288)$, $S{=}(360,380)$ (source $y$ measured downward; the vertical output guide is confirmed by the drawn wall hatching and by $F_{out}$'s arrow direction).

Find. The mechanical advantage $MA=F_{out}/F_{in}$ at the drawn configuration, via the kinematic (velocity) route $MA=v_{in}/v_{out}$.

Fig. 2 — the toggle press at the drawn instant. Ground pivot $C$ sits (to within the drawing's precision) directly above pin $S$: rotating link 5 about $C$ can only move $S$ horizontally at this instant, which the vertical slider guide forbids — the classic toggle (dead-centre) condition.

Approach. Set $\omega_2=1\,\text{rad/s}$, propagate it through the pin-in-slot joint to link 4, close the loop at $D$ against link 5's rotation about $C$, and enforce that $S$'s velocity is purely vertical (the slider constraint) — then compare the resulting output speed with the input point's speed via $MA=v_{in}/v_{out}$ (virtual-work equivalent, valid since $F_{in}$ is drawn tangential to $O_2P_{in}$ and $F_{out}$ is drawn along $S$'s constrained direction).

Input point speed. With $\vec r_{O_2P_{in}} = P_{in}-O_2$ (in the standard math frame, $y$ flipped) $=(-143,24)$, $|\vec r_{O_2P_{in}}|=145.0$ units, and $\omega_2=1\,\text{rad/s}$: $$v_{in} = \omega_2\,|\vec r_{O_2P_{in}}| = 145.0\ \text{units/s (tangential, since } F_{in}\perp O_2P_{in} \text{ to within } 9.6^\circ)$$
Locate $C$ relative to the slider pin $S$. $$\vec r_{CS} = S-C = (2,\,-92)\quad\text{(std frame)}$$ The $x$-offset (2 units) is negligible next to the $y$-offset (92 units): $C$ is, for practical purposes, directly above $S$.
Velocity of $S$ as a point on link 5 (pure rotation about $C$). For any $\omega_5$, $$\vec V_S = \omega_5 \times \vec r_{CS} = \omega_5\,(-r_{CS,y},\,r_{CS,x}) = \omega_5\,(92,\,2)$$ This vector is almost entirely horizontal for any nonzero $\omega_5$.
Impose the vertical-slider constraint. Link 7 translates only vertically, so $V_{S,x}=0$ is required: $$\omega_5 \times 92 = 0 \;\Rightarrow\; \omega_5 = 0 \;\Rightarrow\; \vec V_S = (0,0)$$ Because link 4–link 5 close at $D$ with link 5 momentarily not rotating, the closure equation at $D$ forces $\vec V_D=(0,0)$ as well — consistent with $\omega_5=0$ (the full 3-equation velocity-loop solve confirms $V_D=(0,0)$ exactly, with the slot-slip speed at $A$ absorbing all of link 2's input motion).
Interpret and compute $MA$. $v_{out}=|\vec V_S| = 0$ at this exact instant while $v_{in}=145\,\text{units/s}\neq0$: the mechanism is at (or immediately adjacent to) its toggle / dead-centre position. $$\boxed{MA = \dfrac{v_{in}}{v_{out}} \to \infty \quad\text{at the position shown}}$$

Quantity	Value
Input point speed $v_{in}$ ($\omega_2=1\,\text{rad/s}$)	145.0 units/s (tangential)
$C$-to-$S$ offset	$(2,-92)$ — nearly vertical
Output slider speed $v_{out}$	$\approx 0$ (toggle condition)
Mechanical advantage $MA$	$\to\infty$ at the drawn position

Check: the toggle condition found here is a direct, robust consequence of the drawing ($C$ sits almost exactly above $S$, independent of the small reading tolerance on the 2-unit horizontal offset) rather than of measurement noise — any press mechanism drawn with its output rocker's ground pivot vertically above the ram pin is, by definition, being shown at its dead-centre stroke position, which is exactly where a toggle press is designed to deliver its peak force. In practice the real press runs a small, finite distance short of exact dead-centre, where $MA$ is very large but finite.

Topic: Mechanical advantage of linkages via kinematic (virtual-work) analysis; toggle mechanisms.

Key relations: $F_{in}\,v_{in}=F_{out}\,v_{out}$ (ideal, lossless linkage) $\Rightarrow MA=F_{out}/F_{in}=v_{in}/v_{out}$; for a point on a rigid body rotating about a fixed pivot, $\vec V=\vec\omega\times\vec r$.

Why this works: a toggle mechanism deliberately drives its output link's ground pivot into near-alignment with the output ram's line of action so that, right at the working stroke, the output velocity collapses toward zero for a finite input velocity — trading stroke speed for an enormous, momentary force gain. This is precisely why toggle linkages are used in presses, clamps, and circuit breakers.

Common pitfall: computing $MA$ from lever-arm ratios alone without checking whether the mechanism is near a singular (toggle) configuration, which can make even a coarse velocity-ratio estimate wildly wrong; or forgetting that $F_{in}$'s useful component is only the part perpendicular to its own lever arm.

Source: Norton, Design of Machinery, Ch. 3 & 11 (mechanical advantage, toggle mechanisms).

Question 4 of 10

Question 2 (20 marks)