Basics of Instantaneous Center rotation with steps to solve numericals

1. Theory of Machines-I
Kinematic Analysis of Mechanisms
Basics of Instantaneous Centre of Rotation (ICR)
Prof. K N Wakchaure
Department of Mechanical Engineering
Sanjivani College of Engineering, Kopargaon

2. Basics of ICR
• The instant center of rotation (ICR), also called instantaneous velocity center,
or also instantaneous center or instant center, is the point fixed to a body
undergoing planar movement that has zero velocity at a particular instant of
time.
“Point on a rigid body whose velocity is zero at a given instant”
ICR method is used to find velocity of different links and points in the Mechanism.
Steering Gear MechanismFan

3. • Pin Joint
ICR is Located at the center of Pin
• For Sliding Pair: Straight/ Flat Surface
ICR lies at Infinity on the common normal at the point of
contact.
• For Sliding Pair: Curved Surface
ICR lies at centre of curvature of curved surface.
• For Rolling Pair
• ICR lies at the point of contact of two bodies.
Locations of ICR

4. Types of ICR
1. Fixed instantaneous centres
• I12 AND I14 are Fixed ICR which remain in the same place for all
configurations of the mechanism.
2. Permanent instantaneous centres
• I23 AND I34 as they move when the mechanism moves, but the joints are of
permanent nature.
3. Neither fixed nor permanent Centres
• I13 and I24 are neither fixed nor permanent instantaneous centres as they
vary with the configuration of the mechanism.
• Fixed and Permanent ICR can be find by viewing the mechanism but
for Neither fixed nor permanent Centres require to use Kennedys
Theorem

5. Kennedy’s Theorem
• The Aronhold Kennedy’s theorem states that if three bodies (Links) move relatively to
each other, they have three instantaneous centres and lie on a straight line.
1 2 3
I12 I23
I13
Suppose link 1,2,3 are having relative motion
According to Kennedys Theorem I12, I23, I13
should be along straight line irrespective of
sequence.
2 3 4
I23 I34
I24
I24, I23, I34
1 3 4
I13 I34
I24
I13, I34, I14

6. Procedure to Solve Numericals
Step-1:-Read the problem statement Carefully.
Step-2:-Draw the given Mechanism with Suitable Scale.
Step-3:-Give the numbers to the link in the mechanism starting with fixed LINK.
Step-4:-Find No. of ICR using Formula: N=
𝒏∗(𝒏−𝟏)
𝟐
N= no of ICR
n=Total no. of links in the Mechanism

7. Procedure to Solve Numericals
Step-5:-Draw the table of Instantaneous centres.
1 2 3 4
I12 I23 I34
I13 I24
I14
3 2 1
Write no. of links
Total 6 ICR
Step-6:-Locate Fixed and Permanent ICR.

8. Procedure to Solve Numericals
Step-7:-Circle/ highlight the known ICR in the table..
1 2 3 4
I12 I23 I34
I13 I24
I14
3 2 1
Write no. of links
Total 6 ICR
I13 AND I24 are neither fixed nor permanent ICR
Kennedy's theorem is required to locate I13 and I24

9. Procedure to Solve Numericals
Step-7:-Draw one circle of Arbitrary diameter, divide it n number of Parts.
1 2 3 4
I12 I23 I34
I13 I24
I14
3 2 1
Step-8:-Draw one circle (Keneddy’s Circle) of Arbitrary diameter, divide it ‘n’
number of Parts.
n= No of Links in the Mechanism here n=4

10. Procedure to Solve Numericals
1 2 3 4
I12 I23 I34
I13 I24
I14
Step-9:- Join the points in Keneddy’s circle for known ICR

11. Procedure to Solve Numericals
1 2 3 4
I12 I23 I34
I13 I24
I14
Step-10:- Use Kennedy’s theoremm to locate I13 and I24
Now concentrate on I13,I13 should be diagonal of quadrilateral or
common side for two triangles
Here I13 is diagonal of Quad. 1234.
Join I13
I13 is making two triangles ∆I12.I23.I13
And ∆I14.I34.I13
I12 I23
I14 I34
I13
Join ICR I12 and I23 and extend line
Join ICR I14 and I34 and extend line
Above two line intersects at common point which is I13

12. Procedure to Solve Numericals
1 2 3 4
I12 I23 I34
I13 I24
I14
Step-10:- Use Kennedy’s theoremm to locate I13 and I24
Now concentrate on I24,I24 should be diagonal of quadrilateral or
common side for two triangles
Here I24 is diagonal of Quad. 1234.
Join I24
I24 is making two triangles ∆I12.I14.I24
And ∆I23.I34.I24
I12 I𝟏𝟒
I𝟐𝟑 I34
I24
Move ICR I14 at I12 and extend line
Join ICR I23 and I34 and extend line
Above two line intersects at common point which is I24
I34

13. Procedure to Solve Numericals
Step-11:- Find Velocity and Angular Velocity
Use Angular Velocity Ratio
Angular Velocity of any link in the mechanism if angular velocity
off any link/ input link is known to us
Find the angular velocity of link y when angular velocity of link x
is known to us
𝝎 𝒚
𝝎 𝒙
=
𝑰𝟏𝒙 − 𝑰𝒙𝒚
𝑰𝟏𝒚 − 𝑰𝒙𝒚
angular velocity of link 3 when angular
velocity of link 2 is known
𝜔3
𝜔2
=
𝐼12 − 𝐼23
𝐼13 − 𝐼23
𝜔2

14. Procedure to Solve Numericals
Step-12:- Find Velocity and Angular Velocity
Velocity of any point the mechanism
𝜔2
Velocity of any point P which is on link Q
Then 𝑉𝑃 = (𝐼1𝑄 − 𝑃)X𝜔 𝑄*S.F
Velocity of any point P which is on link 3
Then 𝑉𝑃 = (𝐼13. 𝑃)X𝜔3*S.F
P
𝑽 𝑷
Velocity of point C is the velocity of Slider
Velocity of any point C which is on link 3
Then 𝑉𝐶 = (𝐼13. 𝐶)X𝜔3*S.F

15. Procedure to Solve Numericals
Step-12:- Find Velocity and Angular Velocity
Velocity of slider in the mechanism
𝜔2
Velocity of Slider (N) in mechanism is the velocity of point(Say
M) at the contact between slider and adjacent link(k).
Velocity of any point M which is on link K
Then 𝑉𝑁 = 𝑉 𝑀 = (𝐼1𝐾. 𝑀)X𝜔 𝐾*S.F
In this case velocity of slider(link 4) is the velocity of point
C(Point at the joint of link 3 and link 4).
By considering point C is on link 3
Then 𝑉4 = 𝑉𝑐 = (𝐼13. 𝐶)X𝜔3*S.F
𝑽 𝑪 = 𝑽 𝟒

16. •ThankYou
Prof. K NWakchaureTheory of Machines-I