1

Velocity & Acceleration Diagrams for

Simple Mechanisms

2

Velocity Diagrams for Simple Mechanisms

Consider the following link, AO, rotating about O with an angular velocity rad/s. Velocity of point

A relative to O = .AO and this may be represented by a vector ao, perpendicular to link, drawn to

scale.

Since ao = .AO it follows that = AO

ao

Consider now point B on the link

Velocity of B relative to O = bo = .BO and therefore = BO

bo

It therefore follows that AO

ao =

BO

bo and hence

BO

AO

bo

ao - the use of this will become apparent later.

Four Bar Chain

We now apply the above to a four-bar chain similar to that shown below.

AB, BC, CD are free to move, AD is fixed.

Velocity of B relative to A = ba = BA . BA

VAO = Velocity of A relative to O = ao

O

B

A

Rad/s

Configuration

a

o

Vector

Representation

A

B

C

D (A)

BA

a

b

3

Velocity of C relative to B = cb (magnitude unknown)

Velocity of C relative to D(A) = cd (magnitude unknown)

The construction of the velocity diagram is as follows.

1. Draw ba perpendicular to BA, proportional to BA . BA.

2. Add point (d).

3. Draw cd perpendicular to CD, magnitude unknown.

4. Draw cb perpendicular to CB.

5. Intersection of cd and cb fixes point c.

From the diagram

CB = CB

cb and CD =

CD

cd

Reciprocating Mechanisms

This is a four-bar chain in which the link CD is infinitely long

A = main bearing; AB = crank; B = crankpin; BC = con rod; C = piston.

odc idc A

B

C (D)

c

b

a(d)

Velocity diagram to scale

4

Velocity of B relative to A = ba = .BA

Velocity of B relative to C = bc (magnitude unknown)

Velocity of C relative to D = cd (magnitude unknown)

From this information, the velocity diagram shown below can be drawn. Hence the velocity of piston,

ac, and angular velocity of connection-rod, BC, may be found.

a(d)

b

c

b

a(d)

Velocity diagram

ac = velocity of piston

BC = bc/BC

5

Tutorial Problems

1. The crank and con rod of a pump are respectively 35mm and 150mm, determine the plunger

velocity when the crank is driven anticlockwise at 3rev/s and has turned through;

a) 60o;

b) 130o from idc.

(0.64m/s; 0.44m/s)

2. For the four bar chain given below, determine the angular velocity of the link CD and the linear

velocity of G relative to A.

AB = 350; BC = 600; CD = 400; AD = 800; BG = 250mm

(19.5rad/s; 10.5m/s)

3. For the configuration shown below determine the linear velocity of pistons C and E

AB = 75; BC = 300; BD = 120; DE = 140mm

(12m/s; 0.68m/s)

C

D A

45O

40 rad/s

G

C

E

120 mm

A

120O

A

3 rev/s

B

C(D)

B

2000 rev/min

D

6

4. Determine the velocity of the piston at C and the centre of gravity G in the slider-crank

mechanism shown in the figure below. (Use a scale of 1cm = 1m/s)

(14.8m/s; 13.8m/s)

5. In the mechanism shown below, OA rotates clockwise about the fixed centre O at 10rad/s. If

the angle AOB is 45o, draw the velocity diagram to scale and hence determine the velocities of

C and D relative to O and the angular velocity of the link CD.

B

A

O

C

D 0.6 m

OA = 0.3; AB = 1.5; AC = 0.6; CD = 0.9 m

C

G

B

A 2700

rev/min

60O

AB = 50 mm

BC = 200 mm

BG = 75 mm

25 mm

7

tang

Acceleration Diagrams

Consider the rotating link shown below

Centripetal acceleration of B relative to A is ABAcent = BA

bar

r

v 22

2

and is represented by a

vector ba drawn parallel to link as shown.

Tangential acceleration of B relative to A = ABAtang = .BA and is represented by another vector

drawn perpendicular to the link.

b

For the link the total effect is the combined effect of the centripetal and tangential accelerations. This

is shown below.

Centripetal

acceleration1

A

rad/s

rad/s2

B

Tangential

acceleration

cent

a

b

tang

a

b

cent

a

b

a

Resultant or total

acceleration of B

relative to A

8

Now consider the four-bar chain shown below

The procedure for drawing the acceleration diagram is as follows:

Draw the velocity diagram in order to establish all velocities.

Centripetal acceleration of B relative to A = ABAcent = BA

ba 2

Tangential acceleration of B relative to A = ABAtang = .BA

Centripetal acceleration of C relative to B = ACBcent = CB

cb2

Centripetal acceleration of C relative to D(A) = ACDcent = CD

cd 2

With the above information, we should be able to draw the acceleration diagram. This is shown

below.

c

cent

d

A

B

C

D(A)

cent

a

b

tang b

a

b cent

c

9

cent

cent

tang

b

c

cent

cent

d tang

cent c

tang

From the diagram CB = CB

cb gtan

and CD = CD

cd gtan

Reciprocating Mechanisms

Diagram shows a reciprocating mechanism consisting of a crank, connecting rod, and piston. The

procedure for drawing the acceleration diagram is shown below the figure.

Draw velocity diagram to determine all velocities.

Centripetal acceleration of B relative to A = ABAcent = BA

ba 2

Tangential acceleration of B relative to A = ABAtang = .BA = 0 ( constant)

Centripetal acceleration of B relative to C = ABCcent = BC

bc 2

b

a

b

c

a

b

a(d)

c

b

A

B

C(D)

(constant)

Acceleration diagram to scale

10

The construction of the acceleration diagram is given below.

b

cent

cent c b

(total)

a c

Acceleration diagram to scale

From the diagram CB = CB

cb gtan

1. For information on centripetal acceleration see:

http://www.freestudy.co.uk/dynamics/centripetal%20force.pdf

11

B

Tutorial Problems - Acceleration Diagrams

1. For the configuration shown below, determine the acceleration of C (and its direction) and the

angular acceleration of DC and BC.

AB = 20; BC = 30; CD = 25; AD = 40mm

(0.445m/s2; 15.6rad/s; 8.66rad/s2)

2. For the reciprocating mechanism shown below, determine the angular acceleration of the con

rod.

AB = 20; BC = 45mm

(5rad/s2)

3. Two links PA and QB turn about fixed parallel axes through P and Q. The ends A and B are

pin-jointed to a connecting link AB. The length of the four links are PA = 40mm, AB = 140mm,

BQ = 80mm and PQ = 160mm. PA turns at a uniform speed of 120 rev/min in an anti-clockwise

direction. For the position in which A and B are on opposite sides of PQ and the angle APQ is

30o, find the angular velocity and angular acceleration of BQ.

(6.5 rad/s; 18 rad/s2)

60O A

2 rad/s2

5 rad/s

C

D(A)

30O

= 5 rad/s

A

B

C(D)

5 mm

P

120 rev/min A

B

Q

30O

12

90O

B

4. In the given mechanism AB = 50mm, BC = 100mm, BD = 75mm and the line of stroke of D is

50mm from A and is perpendicular to the line of stroke of C. When the angle CAB is 45o, D

has an upward velocity of 1m/s and an upward acceleration of 50m/s2. Determine the velocity

and acceleration of slider C for this condition and the corresponding angular velocity and

angular acceleration of crank AB.

(1.8m/s; 103m/s2; 1160rad/s2)

(The accuracy of these answers cannot be guaranteed.)

5. Determine the acceleration of the piston C and the angular acceleration of AB and BC. AB =

50mm, BC = 200mm.

45O

A

D

C

C

B

A

60O

25

mm

2 rad/s2

6 rad/s

13

Page left blank for you to make notes