Relative velocity method notes, velocity & acceleration analysis of mechanism, Graphical method

1. Mr. Keshav R Pagar TOM I Mechanical Engineering
1Unit V: Velocity and Acceleration Analysis: Graphical Methods-I
THEORY OF MACHINE I
VELOCITY AND ACCELERATION ANALYSIS OF SIMPLE
MECHANISMS: GRAPHICAL METHODS-I
Syllabus
 Relative velocity method: Relative velocity of a point on a link, Angular velocity of a
link, Sliding velocity, Velocity polygons for simple mechanisms.
 Introduction
There are 3 Graphical methods to find velocity and acceleration of the mechanism.
1. Relative velocity and Acceleration Method
2. Instantaneous centre of rotation ICR
3. Kleins Construction Method
In this unit we are going to study the 1st
two methods i.e Relative velocity and
Acceleration Method and Instantaneous centre of rotation ICR
A. Relative Velocity and Acceleration Method
1. Relative Velocity Method:
The application of vectors for the relative velocity of two bodies moving along parallel
lines and inclined lines, as shown in Fig. (a) and (b) respectively.
Consider two bodies A and B moving along parallel lines in the same direction with
absolute velocities VA and VB such that VA > VB, as shown in Fig. (a). The relative
velocity of A with respect to B
VAB= Vector difference of VA and VB =

2. Mr. Keshav R Pagar TOM I Mechanical Engineering
2Unit V: Velocity and Acceleration Analysis: Graphical Methods-I
From Fig.(b), the relative velocity of A with respect to B (i.e. VAB) may be written in
the vector form as follows:
VAB= Relative velocity of A with respect to B =
Similarly, the relative velocity of B with respect to A (i.e. VBA) may be written in the
vector form as follows:
VBA= Relative velocity of B with respect to A =
1.1 Motion of a Link
Consider two points A and B on a rigid link AB, as shown in Fig. Let one of the
extreme point (B) of the link move relative to A , in a clockwise direction.
Since the distance from A to B remains the same, therefore there can be no relative
motion between A and B, along the line AB.
The relative motion of B with respect to A must be perpendicular to AB.

3. Mr. Keshav R Pagar TOM I Mechanical Engineering
3Unit V: Velocity and Acceleration Analysis: Graphical Methods-I
Hence velocity of any point on a link with respect to another point on the same
link is always perpendicular to the line joining these points on the configuration
(or space) diagram.
A machine or a mechanism is represented by a skeleton or a line diagram, commonly
known as configuration diagram.
The relative velocity of B with respect to A (i.e. VBA) is represented by the vector
ab and is perpendicular to the line AB as shown in Fig.(b).
Let ω = Angular velocity of the link AB about A .
We know that the velocity of the point B with respect to A ,
...(i)
Similarly, the velocity of any point C on AB with respect to A ,
...(ii)
From equations (i) and (ii),
Thus, we see from above equation, that the point c on the vector ab divides it in the
same ratio as C divides the link AB.

4. Mr. Keshav R Pagar TOM I Mechanical Engineering
4Unit V: Velocity and Acceleration Analysis: Graphical Methods-I
1.2 Rubbing Velocity at a Pin Joint
The links in a mechanism are mostly connected by means of pin joints.
The rubbing velocity is defined as the algebraic sum between the angular velocities
of the two links which are connected by pin joints, multiplied by the radius of the
pin.
Consider two links OA and OB connected by a pin joint at O as shown in Fig.
ω1= Angular velocity of the link OA or
the angular velocity of the point A with respect to O.
ω2= Angular velocity of the link OB or
the angular velocity of the point B with respect to O, and
r = Radius of the pin.
Rubbing velocity at the pin joint O
= (ω1 – ω2) *r, if the links move in the same direction
= (ω1 + ω2) *r, if the links move in the opposite direction
Note : When the pin connects one sliding member and the other turning member,
the angular velocity of the sliding member is zero. In such cases,
Rubbing velocity at the pin joint = ω*r
where ω = Angular velocity of the turning member, and
r = Radius of the pin.
FORMULAE:
1. Velocity of link V= Angular velocity of link * Length of link

5. Mr. Keshav R Pagar TOM I Mechanical Engineering
5Unit V: Velocity and Acceleration Analysis: Graphical Methods-I
V= ω*r
ω=V/r
2. Velocity of any point on a link (When there is a point on a link)
3. Each line in velocity diagram represents the velocity of that particular link
General Procedure to draw the velocity diagram by Relative Velocity Method
Step 1: Draw the Mechanism as per given length of link called configuration diagram,
or space diagram.
Step 2: Find the velocity of input link.
Step 3: Now the velocity diagram, Mark all fixed point on as a single point
Step 4: Draw the velocity of input link (with suitable scale) as per the direction of
angular velocity given in the configuration diagram from the marked point (step 3) of
velocity diagram.
Step 5: From the last point (Step 4) of the input link draw the line perpendicular to the
next link connecting to the last point of the input link
Step 6: Similarly draw the perpendicular lines (Parallel line in case of sliding motion
i.e motion of slider) from previously obtained points (that points may be fixed marked
point or obtained point during drawing)
Step 7: Determine the values that needs to be find.
Note:1. Solve this example on sheet or on drawing book with necessary drawing
equipment.
Numerical on Relative Velocity Method
Ex. In a four-bar chain ABCD, AD is fixed and is 150 mm long. The crank AB is 40 mm
long and rotates at 120 r.p.m. clockwise, while the link CD = 80 mm oscillates about

6. Mr. Keshav R Pagar TOM I Mechanical Engineering
6Unit V: Velocity and Acceleration Analysis: Graphical Methods-I
D. BC and AD are of equal length. Find the angular velocity of link CD when angle
BAD = 60°.
Soln
:- Given – 4 Bar Mechanism ABCD,
AB- Crank= 40mm, ⦟BAD=600
, NBA=120rpm (CW)
ωBA=
2∗𝛱∗𝑁
60
= 12.568 rad/sec
BC- Coupler =150mm,
CD- Follower=80mm,
DA- Fixed Link=150mm
To Find - ωCD= ?
Step 1: Draw the Mechanism as per given length of link,
Step 2: Find the velocity of input link
VBA= ωBA*AB = 12.568*0.04 = 0.503m/sec
Step 3: Now the velocity diagram,
As link AD is fixed, therefore points “a and d” are taken as one point in the
velocity diagram,
Step 4: Draw the velocity of input link with suitable scale
Here consider scale: 2cm=0.25m/sec
So, VBA =VB=0.503m/sec = 4.02 cm (Direction perpendicular to AB),
After drawing this point, you will get point “b”.

7. Mr. Keshav R Pagar TOM I Mechanical Engineering
7Unit V: Velocity and Acceleration Analysis: Graphical Methods-I
Step 5: from “b” point just draw perpendicular to link BC on both side of point
“b” in velocity diagram.
Step 6: draw a line perpendicular to CD from point d (marked fixed point) that
will intersect the previous line mark that point as “c”. (because the line BC & CD are
joined at C in configuration diagram).
Step 7: Determine the required value
The angular velocity of link CD
As we know the relation between ωCD= VCD/CD
For VCD form the velocity diagram measure the distance cd in mm
convert the value in m/sec from assumed scale.
So, VCD = 0.385m/s.
Then, ωCD= VCD/CD = 0.385/0.08 = 4.8 rad/s.
Ex. The crank and connecting rod of a theoretical steam engine are 0.5 m and 2 m long
respectively. The crank makes 180 r.p.m. in the clockwise direction. When it has turned
45° from the inner dead centre position, determine: 1. velocity of piston, 2. angular
velocity of connecting rod.
Soln
:- Solve in similar way as above example, just the velocity of slider is drawn as
same as the motion of slider from fixed point.(Because the slider is sliding on the
fixed surface.)
Sequence to draw velocity diagram:

8. Mr. Keshav R Pagar TOM I Mechanical Engineering
8Unit V: Velocity and Acceleration Analysis: Graphical Methods-I
1. Write the given data and terms to be find.
2. Draw Configuration diagram (space diagram) as given in example.
3. Calculate VA From given data. VBO = VB = ωBO × OB = 18.852 × 0.5 = 9.426 m/s
4. Mark all fixed point as a single point o.
5. Draw VB (i.e ob distance in velocity diagram) Perpendicular to OB from space
diagram by considering the scale.
6. Draw a line perpendicular to BP from “b” in velocity diagram.
7. For velocity of slider (i.e Point P) draw a parallel line to the motion of slider from
fixed point, it will intersect the previously drawn line mark that point “p”.
8. Name each vector which represents the velocity of each link.
9. Measure the length from velocity diagram to find the velocity of different links
of the mechanism by converting with suitable scale

9. Mr. Keshav R Pagar TOM I Mechanical Engineering
9Unit V: Velocity and Acceleration Analysis: Graphical Methods-I
Ans: VP = vector op = 8.15 m/s
Ex: In Fig. the angular velocity of the crank OA is 600 r.p.m. Determine the linear
velocity of the slider D and the angular velocity of the link BD, when the crank is
inclined at an angle of 75° to the vertical. The dimensions of various links are : OA =
28 mm ; AB = 44 mm ; BC 49 mm ; and BD = 46 mm. The centre distance between the
centres of rotation O and C is 65 mm. The path of travel of the slider is 11 mm below
the fixed point C. The slider moves along a horizontal path and OC is vertical.
Soln
:-
Sequence to draw velocity diagram:

10. Mr. Keshav R Pagar TOM I Mechanical Engineering
10Unit V: Velocity and Acceleration Analysis: Graphical Methods-I
1. Write the given data and terms to be find.
2. Draw Configuration diagram (space diagram) as given in example.
3. Calculate VA From given data. VAO = VA = ωAO × OA = 62.84 × 0.028 = 1.76 m/s
4. Mark all fixed point as a single point “o,c”
5. Draw VA (i.e “oa” distance in velocity diagram)Perpendicular to OA from space
diagram by considering the scale.
6. Draw a line perpendicular to AB from “a” in velocity diagram.
7. Draw line perpendicular to BC from “c” (Fixed marked point) in velocity diagram,
it will intersecting the previous line, mark the intersection point as “b”.
8. From the point “b” in velocity diagram draw perpendicular to line BD.
9. For velocity of slider (i.e Point D) draw a parallel line to the motion of slider from
fixed point, it will intersect the previously drawn line mark that point “d”.
10.Name each vector which represents the velocity of each link.
11.Measure the length from velocity diagram to find the velocity of different links of
the mechanism by converting with suitable scale.
Ans: VD = vector od = 1.6 m/s
Ex. In the toggle mechanism, as shown in Fig. the slider D is constrained to move on a
horizontal path. The crank OA is rotating in the counter-clockwise direction at a speed

11. Mr. Keshav R Pagar TOM I Mechanical Engineering
11Unit V: Velocity and Acceleration Analysis: Graphical Methods-I
of 180 r.p.m. The dimensions of various links are as follows : OA = 180 mm ; CB = 240
mm ; AB = 360 mm ; and BD = 540 mm. For the given configuration, find : 1. Velocity
of slider D, 2. Angular velocity of links AB, CB and BD.
Soln
:-
Sequence to draw velocity diagram:
1. Write the given data and terms to be find.
2. Draw Configuration diagram (space diagram) as given in example.

12. Mr. Keshav R Pagar TOM I Mechanical Engineering
12Unit V: Velocity and Acceleration Analysis: Graphical Methods-I
3. Calculate VA From given data. VAO = VA = ωAO × OA = 18.85 × 0.18 = 3.4 m/s
4. Mark all fixed point as a single point “o,c”
5. Draw VA (i.e “oa” distance in velocity diagram) Perpendicular to OA from space
diagram by considering the scale.
6. Draw a line perpendicular to AB from “a” in velocity diagram.
7. Draw line perpendicular to BC from “c” (Fixed marked point) in velocity diagram,
it will intersecting the previously drawn line, mark the intersection point as “b”.
8. From the point “b” in velocity diagram draw perpendicular to line BD.
9. For velocity of slider (i.e Point D) draw a parallel line to the motion of slider from
fixed point, it will intersect the previously drawn line mark that point “d”.
10.Name each vector which represents the velocity of each link.
11.Measure the length from velocity diagram to find the velocity of different links of
the mechanism by converting with suitable scale.

13. Mr. Keshav R Pagar TOM I Mechanical Engineering
13Unit V: Velocity and Acceleration Analysis: Graphical Methods-I
Ans: VD = vector cd = 2.05 m/s
Ex. The dimensions of the various links of a mechanism, as shown in Fig, are as follows
:AB = 30 mm ; BC = 80 mm ; CD = 45 mm ; and CE = 120 mm.
The crank AB rotates uniformly in the clockwise direction at 120 r.p.m. Draw the
velocity diagram for the given configuration of the mechanism and determine the
velocity of the slider E and angular velocities of the links BC, CD and CE.

14. Mr. Keshav R Pagar TOM I Mechanical Engineering
14Unit V: Velocity and Acceleration Analysis: Graphical Methods-I
Ans: VE=120 mm/s ;
Angular Velocities of the links BC, CD and CE are 2.8 rad/s ; 5.8 rad/s ; 2 rad/s
Ex: Fig, shows a toggle mechanism in which link D is constained to move in horizontal
direction. For the given configuration, find out : 1. velocities of points B and D; and 2.
angular velocities of links AB, BC, and BD. The rank OA rotates at 60 r.p.m. in
anticlockwise direction.
Ans. velocities of points B and D 0.9 m/s; 0.5 m/s;
Angular velocities of links AB, BC, and BD 0.0016 rad/s (ccw) 0.0075 rad/s
(ccw),0.0044 rad/s (ccw).
Take examples from any book an try to solve.