Kinematic Analysis of Mechanisms

Instantaneous centre (IC)

For two bodies in relative plane motion, the instantaneous centre is the point (common to both) that has no relative velocity at that instant. Each body appears to rotate about it. It is found at the intersection of the perpendiculars to the velocity directions of two points.

Number of ICs

For a mechanism of $n$ links, each pair of links has one IC:

$$N = \frac{n(n-1)}{2}$$

A four-bar mechanism ($n = 4$) has $N = 6$ ICs.

Types

• Fixed IC: its location is fixed in space (for example, the pivot of a link with the frame).
• Permanent IC: its location moves, but it is a pin joint between two moving links (for example, the pin between crank and coupler).
• Neither fixed nor permanent: both its position and its type change with the configuration, for example the IC between coupler and frame.

Kennedy's theorem (three-centre theorem)

Any three bodies in relative plane motion have three instantaneous centres, and they lie on a straight line.

Proof. Let the three bodies be 1, 2, 3 with $I_{12}$ and $I_{13}$ known. Assume $I_{23}$ is not on the line joining $I_{12}$ and $I_{13}$ and take it as the point $P$ (as a point of body 2 and body 3). The velocity of $P$ as a point of body 2 is perpendicular to $I_{12}P$, and as a point of body 3 is perpendicular to $I_{13}P$. These two directions are different (since the lines $I_{12}P$ and $I_{13}P$ are different), while at the IC $I_{23}$ the velocities of the point of body 2 and body 3 must be identical in magnitude and direction. This is possible only if $P$ lies on the line through $I_{12}$ and $I_{13}$, where the two perpendicular directions coincide. Hence all three ICs are collinear.

Four-bar mechanism ABCD (links: 1 = AD frame, 2 = AB, 3 = BC, 4 = CD)

Obvious ICs (pin joints): $I_{12}$ at A, $I_{23}$ at B, $I_{34}$ at C, $I_{14}$ at D.

Remaining two by Kennedy's theorem:

• $I_{13}$: lies on the line $I_{12}I_{23}$ (AB extended) and on $I_{14}I_{34}$ (DC extended). It is the intersection of AB and DC produced.
• $I_{24}$: lies on $I_{23}I_{34}$ (BC extended) and on $I_{12}I_{14}$ (AD extended). It is the intersection of BC and AD produced.

   I13 *
        \        .
         \   B---C
          \ /     \
        A-----------D  ---- * I24

Use: $\omega_2 \cdot I_{12}I_{24} = \omega_4 \cdot I_{14}I_{24}$ (both equal the speed of $I_{24}$), and $\omega_2 \cdot I_{12}I_{23} = \omega_3 \cdot I_{13}I_{23}$ (both equal the speed of the pin B).

Configuration

Take O as origin and the line of stroke along the $x$-axis.

$$\omega_{OA} = \frac{2\pi \times 300}{60} = 31.416\ \text{rad/s}, \qquad v_A = \omega_{OA}\cdot OA = 31.416 \times 0.150 = 4.712\ \text{m/s}$$

$v_A$ is perpendicular to OA. Obliquity of the rod: $\sin\phi = \dfrac{OA\sin 40^\circ}{AB} = 0.1607 \Rightarrow \phi = 9.25^\circ$.

Coordinates: A $= (114.9,\ 96.4)$ mm, B $= (707.1,\ 0)$ mm.

Locating the instantaneous centre $I_{13}$ (connecting rod)

• The velocity of A is perpendicular to OA, so $I$ lies on the line OA produced.
• The velocity of B is along the line of stroke, so $I$ lies on the perpendicular to the line of stroke at B.

The intersection gives $I$ at $(707.1,\ 593.3)$ mm.

           I  (instantaneous centre)
          /|
        /  |
      A/   |        IA = 773.1 mm
     /\    |        IB = 593.3 mm
   O/__\___|__B ->

$$IA = \sqrt{(707.1 - 114.9)^2 + (593.3 - 96.4)^2} = 773.1\ \text{mm}, \qquad IB = 593.3\ \text{mm}$$

(b) Angular velocity of the rod

The rod rotates about $I$, so $v_A = \omega_{AB}\cdot IA$:

$$\omega_{AB} = \frac{v_A}{IA} = \frac{4.712}{773.1/1000} = 6.096\ \text{rad/s}$$

(clockwise: A lies to the lower left of $I$ and moves to the upper left.)

(a) Velocity of slider

$$v_B = \omega_{AB}\cdot IB = 6.096 \times \frac{593.3}{1000} = 3.617\ \text{m/s}$$

(towards the crank end, towards O.)

Check with the analytic result $v_B = \omega r\left[\sin\theta + \dfrac{r\sin2\theta}{2\sqrt{l^2 - r^2\sin^2\theta}}\right] = 3.617$ m/s. It agrees.

(c) Velocity of midpoint G

Distance from I to G (G is the midpoint of AB): $IG = 620.4$ mm.

$$v_G = \omega_{AB}\cdot IG = 6.096 \times \frac{620.4}{1000} = 3.781\ \text{m/s}$$

perpendicular to IG.

Answer: $v_B = 3.617$ m/s, $\omega_{AB} = 6.096$ rad/s (clockwise), $v_G = 3.781$ m/s.

Principle

For two points A and B on the same rigid link, the velocity of B relative to A is perpendicular to AB, and its magnitude is $\omega \cdot AB$:

$$\vec v_B = \vec v_A + \vec v_{BA}, \qquad v_{BA} = \omega_{AB}\cdot AB \ \ (\perp AB)$$

The vector equation is solved by drawing a velocity polygon: each vector has known direction, and the magnitude of at least one is known.

Four-bar mechanism ABCD (crank AB with $\omega_2$, AD fixed)

1. Draw the configuration diagram to scale (space diagram).
2. Compute the velocity of B: $v_{B} = \omega_2 \cdot AB$, perpendicular to AB. Choose a velocity scale (for example 1 cm = 0.5 m/s).
3. Choose a pole $o$ (the zero velocity point for fixed A and D). Draw $ob$ perpendicular to AB, of length proportional to $v_B$.
4. Vector for C relative to B: from $b$ draw a line perpendicular to BC (direction of $v_{CB}$).
5. Vector for C relative to D: from $o$ (as D is fixed) draw a line perpendicular to CD (direction of $v_C$).
6. The intersection of these two lines is point $c$. Then $oc$ is the velocity of C, and $bc$ is the velocity of C relative to B.
7. Scale the lengths:

$$\omega_3 = \frac{v_{CB}}{BC} = \frac{bc}{BC}, \qquad \omega_4 = \frac{v_C}{CD} = \frac{oc}{CD}$$

Sense: $\omega_3$ is found by placing the vector $bc$ at C on the space diagram (viewed from B); $\omega_4$ by placing $oc$ at C (about D).

 velocity polygon        space diagram
      b                  B-------C
     /|\                 |        \
    / | \                A         D
   o--+--c

Velocity of a point E on BC (velocity image)

The triangle $bce$ in the polygon is similar to the triangle BCE on the link (the polygon of points on a rigid link is an image of the link, rotated by $90^\circ$). So:

1. Draw $be$ so that $\dfrac{be}{bc} = \dfrac{BE}{BC}$ on the line $bc$ (for E on BC).
2. For E off the line BC, make triangle $bce$ similar to BCE with the same letter order.
3. Join $o$ to $e$. Then $oe$ gives the velocity of E.

Points to remember

• Points of fixed pivots are at the pole $o$.
• Same letter order and same sense in the image as on the link.
• The angular velocity of a link is the same for all its points: $\omega = \dfrac{\text{relative velocity vector}}{\text{length of link}}$.

(a) Ladder

Velocity of A is horizontal, velocity of B is vertical (along the wall). Draw perpendiculars to these velocities at A and B (a vertical line through A and a horizontal line through B). They meet at the instantaneous centre $I$, which is the corner of the rectangle formed by the floor, wall and the two perpendiculars.

   wall
   |  B
   |  |\
   I--+-\----  <- horizontal line through B
   |    \ \
   |     \ A --> 2 m/s
   +------------- floor

Distances from $I$: $IA = L\sin 60^\circ = 3.464$ m (equal to the height of B), $IB = L\cos 60^\circ = 2.0$ m (equal to the distance of A from the wall).

$$\omega_{AB} = \frac{v_A}{IA} = \frac{2}{3.464} = 0.577\ \text{rad/s (anticlockwise)}$$ $$v_B = \omega_{AB}\cdot IB = 0.577 \times 2.0 = 1.155\ \text{m/s (downwards)}$$

The midpoint M of the ladder is at the centre of the circle through A, B and I, so $IM = L/2 = 2$ m.

$$v_M = \omega_{AB}\cdot IM = 0.577 \times 2 = 1.155\ \text{m/s}$$

directed perpendicular to $IM$.

(b) Rolling wheel

For pure rolling, the point of contact with the road is the instantaneous centre. Angular velocity:

$$\omega = \frac{v_C}{r} = \frac{6}{0.3} = 20.0\ \text{rad/s}$$

• Topmost point (distance $2r = 0.6$ m from the IC): $v_{top} = 20 \times 0.6 = 12.0$ m/s, horizontal, in the direction of travel.
• Rim point at the level of the centre (distance $r\sqrt{2}$ from IC): $v = 20 \times 0.3\sqrt{2} = 8.485$ m/s, at $45^\circ$ to the horizontal.

Answer: (a) $\omega = 0.577$ rad/s, $v_B = 1.155$ m/s, $v_M = 1.155$ m/s; (b) top $= 12.0$ m/s, side point $= 8.485$ m/s.

Vector equation

Let B$_2$ be the crank pin (block) and B$_3$ the coincident point on the lever. The block slides along the lever, so

$$\vec v_{B_2} = \vec v_{B_3} + \vec v_{B_2B_3}$$

• $v_{B_2}$: magnitude $\omega\cdot O_1B = 10 \times 0.12 = 1.20$ m/s, perpendicular to $O_1B$ (horizontal, towards the left, since the crank is vertical and rotates anticlockwise).
• $v_{B_3}$: perpendicular to $O_2B$ (unknown magnitude).
• $v_{B_2B_3}$: along the slot $O_2B$ (unknown magnitude).

Geometry

$O_2B = \sqrt{300^2 + 120^2} = 323.1$ mm. The slot makes an angle $\alpha = \tan^{-1}(120/300) = 21.80^\circ$ with the horizontal, so $\cos\alpha = 0.9285$, $\sin\alpha = 0.3714$.

Velocity polygon (resolving $v_{B_2}$)

         b2 <---- o      o: pole
          \ ' |
  along    \ ' | perp. to O2B
  slot      \  |
             b3

Draw $o b_2$ = 1.2 m/s horizontal. Through $o$ draw a line perpendicular to $O_2B$ (direction of $v_{B_3}$) and through $b_2$ draw a line parallel to $O_2B$ (direction of $v_{B_2B_3}$). They meet at $b_3$. Resolving $v_{B_2}$ in these two directions:

$$v_{B_2B_3} = v_{B_2}\cos\alpha = 1.2 \times 0.9285 = 1.114\ \text{m/s}$$ $$v_{B_3} = v_{B_2}\sin\alpha = 1.2 \times 0.3714 = 0.446\ \text{m/s}$$

Angular velocity of lever

$$\omega_{lever} = \frac{v_{B_3}}{O_2B} = \frac{0.446}{323.1/1000} = 1.379\ \text{rad/s (anticlockwise)}$$

The sense: the block moves left and the lever is on its right side, so the lever turns anticlockwise about $O_2$.

Velocity of D

$$v_D = \omega_{lever}\cdot O_2D = 1.379 \times 0.5 = 0.690\ \text{m/s}$$

perpendicular to the lever (D is on the opposite side of $O_2$, so it moves in the sense opposite to $B_3$).

Answer: $\omega_{lever} = 1.379$ rad/s anticlockwise, sliding velocity $= 1.114$ m/s, $v_D = 0.690$ m/s.

Comparison

Centrode

The locus of the instantaneous centre of a moving body is called its centrode. The path of the IC in the fixed body is the fixed centrode; the path in the moving body is the moving centrode. The motion of the body is equivalent to rolling of the moving centrode on the fixed centrode without slipping.

Example: a ladder sliding down a wall: the fixed centrode is a circular arc (of radius equal to ladder length), and the moving centrode is a semicircle on the ladder as diameter.

Types of instantaneous centres

1. Fixed IC: remains in the same place for all positions of the mechanism, for example the IC of the crank and frame at the crank pivot.
2. Permanent IC: changes its place as the mechanism moves, but joins two links by a pin joint, for example the crank-pin joint (crank and connecting rod).
3. Neither fixed nor permanent: both position and type change with configuration, for example the IC of the connecting rod and the frame of a slider crank.

Instantaneous centres of the four links (1 = AD, 2 = AB, 3 = BC, 4 = CD)

Number of ICs $= \dfrac{4 \times 3}{2} = 6$. Four are pin joints:

$I_{12} = A\,(0, 0)$, $I_{23} = B\,(50.0, 86.6)$, $I_{34} = C\,(273.7, 198.3)$, $I_{14} = D\,(300, 0)$ mm.

By Kennedy's theorem the remaining two are:

• $I_{13}$ is the intersection of line AB (through $I_{12}$, $I_{23}$) and line DC (through $I_{14}$, $I_{34}$).
• $I_{24}$ is the intersection of line BC (through $I_{23}$, $I_{34}$) and line AD (through $I_{12}$, $I_{14}$).

Solving the line equations:

$$I_{13} = (243.9,\ 422.5)\ \text{mm}, \qquad I_{24} = (-123.5,\ 0)\ \text{mm}$$

   I13 *  (243.9, 422.5)
        \  .
         \    C
          B  /  \
         /      \
 I24 *--A---------D------

(Not to scale; $I_{24}$ lies on AD produced beyond A.)

Angular velocity of rocker CD

$I_{24}$ is a point common to links 2 and 4, so its velocity is the same whether taken on AB or on CD:

$$\omega_2 \cdot I_{12}I_{24} = \omega_4 \cdot I_{14}I_{24}$$ $$I_{12}I_{24} = 123.5\ \text{mm}, \qquad I_{14}I_{24} = 300 + 123.5 = 423.5\ \text{mm}$$ $$\omega_4 = 10 \times \frac{123.5}{423.5} = 2.916\ \text{rad/s}$$

$I_{24}$ lies outside A and D on the same side of both, so $\omega_4$ has the same sense as $\omega_2$: anticlockwise.

Angular velocity of coupler BC

$I_{13}$ is the IC of links 1 and 3. The pin B is common to links 2 and 3:

$$v_B = \omega_2 \cdot AB = 10 \times 0.100 = 1.0\ \text{m/s}$$ $$\omega_3 = \frac{v_B}{I_{13}B} = \frac{1.0}{387.8/1000} = 2.578\ \text{rad/s}$$

B lies between $I_{12}$ and $I_{13}$ on the same line, so the senses of $\omega_2$ and $\omega_3$ are opposite: $\omega_3$ is clockwise.

Velocity of C

$$v_C = \omega_4 \cdot CD = 2.916 \times 0.200 = 0.583\ \text{m/s}$$

perpendicular to CD (check: $v_C = \omega_3 \cdot I_{13}C$ gives the same value).

Answer: $\omega_3 = 2.578$ rad/s (clockwise), $\omega_4 = 2.916$ rad/s (anticlockwise), $v_C = 0.583$ m/s.