Mechanisms: count the freedom, close the loop
Introducido| English | Significado |
|---|---|
| links | The rigid bodies of a mechanism, joined to each other by kinematic pairs. |
| mechanism/ˈmekənɪzəm/ | An assembly of links and joints that transforms one motion into another. |
| kinematic pairs | Joints between two links that allow certain relative motions, such as a pin or a slider. |
| Gruebler's equation | The formula that counts a planar mechanism's degrees of freedom from its links and joints. |
| Grashof condition | The test on link lengths, s + l ≤ p + q, that tells whether a four-bar linkage has a link that can rotate fully. |
| inversions | The different mechanisms obtained from one chain by fixing a different link. |
| transmission angle | The angle between a mechanism's coupler and its output link; near 90 degrees the force drives the output most directly. |
How many motors does it need?
- A dosing pump turns a motor's rotation into a piston stroke of exactly the right length. Choosing the links, and knowing how many inputs the mechanism needs, comes before any force calculation.
- A mechanism is a chain of links joined by kinematic pairs: lower pairs (pin, slider) allow one relative motion; higher pairs (cam, gear contact) allow two.
Gruebler's count
- Gruebler's equation for planar mechanisms: $M = 3(n - 1) - 2j_1 - j_2$, with $n$ links (ground included), $j_1$ one-freedom joints and $j_2$ two-freedom joints.
- The slider-crank has 4 links and 4 joints: $M = 9 - 8 = \mathbf{1}$, one motor. A five-bar with 5 pins has $M = 2$ and needs two inputs.
How many inputs?
Classify each linkage by its degrees of freedom or Grashof type.
Degrees of freedom of a planar chain with 4 links and 4 pin or slider joints?
Dificultad 2/5 · 140 monedas + un pequeño bono en el primer intento de hoy
3(4 - 1) - 2 x 4.
Of a five-bar with 5 pins?
Dificultad 2/5 · 140 monedas + un pequeño bono en el primer intento de hoy
3 x 4 - 2 x 5.
Where is the piston?
- Close the vector loop: $r\sin\theta = l\sin\phi$ and $x = r\cos\theta + \sqrt{l^2 - r^2\sin^2\theta}$.
- With $r = 40$, $l = 160$ mm and $\theta = 60°$: $\phi = \mathbf{12.5°}$ and $x = 20 + 156.2 = \mathbf{176.2}$ mm.
- The stroke is $2r = \mathbf{80}$ mm.
Rod angle phi for r = 40, l = 160 mm at 60 degrees (one decimal, degrees)?
Dificultad 2/5 · 140 monedas + un pequeño bono en el primer intento de hoy
arcsin of 40 sin 60 over 160.
Piston position x (one decimal, mm)?
Dificultad 2/5 · 140 monedas + un pequeño bono en el primer intento de hoy
r cos theta + root(l^2 - r^2 sin^2 theta).
Can a link turn fully?
- The Grashof condition: shortest plus longest $\le$ the sum of the other two. Links 40, 100, 80 and 110 give $150 \le 180$, so it is Grashof, and with the shortest as crank it is a crank-rocker.
- Ground a different link and the same chain gives a different mechanism: these are its inversions.
Links 40, 100, 80 and 110 mm satisfy Grashof because:
Dificultad 1/5 · 100 monedas + un pequeño bono en el primer intento de hoy
s + l = 150 against p + q = 180.
The rod pushes sideways too
- The rod meets the piston at angle $\phi$, so an axial force $F$ in the stroke needs a side force $F\tan\phi$ from the cylinder wall. The transmission angle 传动角, here $90° - \phi$, measures how directly the rod drives.
- $\phi$ peaks at $\theta = 90°$, where $\sin\phi = r/l = 0.25$, so $\phi = 14.5°$: a 1 kN piston load presses the wall with $\mathbf{258}$ N. A longer rod lowers the side load and the wear.
Slider-crank $r/l = 0.25$, 1 kN piston load at $\theta = 90°$: side force on the cylinder wall (N, nearest whole)?
Dificultad 2/5 · 140 monedas + un pequeño bono en el primer intento de hoy
$\phi = \sin^{-1}0.25 = 14.48°$; $1000\tan 14.48° = 258$ N.
A longer connecting rod raises the side force on the piston.
Dificultad 1/5 · 100 monedas + un pequeño bono en el primer intento de hoy
$\sin\phi_{max} = r/l$ falls as $l$ grows, so the side force falls.
Worked example: how far does the rocker swing?
- In the crank-rocker (crank 40, coupler 100, rocker 80, ground 110 mm) the rocker's extremes come when crank and coupler line up: stretched to 140 mm, folded to 60 mm.
- The cosine rule in each triangle with the 110 mm ground and 80 mm rocker gives $93.6°$ and $32.2°$: the rocker swings $\mathbf{61.4°}$ for every turn of the crank.
Crank-rocker 40, 100, 80, 110 mm: rocker swing angle (one decimal, degrees)?
Dificultad 2/5 · 140 monedas + un pequeño bono en el primer intento de hoy
$\cos^{-1}[(110^2 + 80^2 - 140^2)/(2 \cdot 110 \cdot 80)] = 93.6°$; with 60 mm, $32.2°$; difference $61.4°$.
Three counting slips
- Forgetting the ground. It is a link.
- Three links at one pin counted once. That is two joints.
- The wrong root. The other sign gives the crossed assembly.
In Gruebler's equation the ground counts as a link.
Dificultad 1/5 · 100 monedas + un pequeño bono en el primer intento de hoy
n includes the fixed frame.
Recap
- $M = 3(n - 1) - 2j_1 - j_2$: the slider-crank has one degree of freedom.
- Loop closure: $\phi = 12.5°$, $x = 176.2$ mm, stroke 80 mm.
- Grashof $s + l \le p + q$; inversions change the ground link.