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Theory of Machines-I
Velocity and Acceleration Analysis of Mechanisms
Numerical:- Instantaneous Centre of Rotation (ICR)
Prof. K N Wakchaure
Department of Mechanical Engineering
Sanjivani College of Engineering, Kopargaon
Problem Statement
• A 6bar mechanism, as shown in Fig. 6, has the following dimensions: OA = 200
mm; AB = 1.5 m; BC = 600 mm; CD = 500 mm and BE = 400 mm, OE=1.35m.
Locate all the instantaneous centres. If crank OA rotates uniformly at 120 r.p.m.
clockwise, find 1. the velocity of B, C and D, 2. the angular velocity of the links
AB, BC and CD.
Procedure to Solve Numericals
Step-1:-Read the problem statement Carefully.
Step-2:-Draw the given Mechanism with Suitable Scale.
SCALE 1:8Given:𝑁𝟐=120 rpm
𝜔 𝟐 =
2 ∗ 𝜋 ∗ 𝑁𝟐
60
= 𝟏𝟐. 𝟓𝟕 𝑟𝑎𝑑/𝑠
Basics of ICR
Scale 1:8
Step-3:-Give the numbers to the link in the mechanism starting with fixed LINK.
Step-5:-Draw the table of Instantaneous centres.
1 2 3 4 5 6
I12 I23 I34 I45 I56
I13 I24 I35 I46
I14 I25 I36
I15 I26
I16
Step-6:-Locate Fixed and Permanent ICR.
I16@Ꚙ
Step-4:-Find No. of ICR using Formula: N=
𝒏∗(𝒏−𝟏)
𝟐
n=6
N=15 (no. of ICR)
Step-7:-Circle/ highlight the known ICR in the table..
1 2 3 4 5 6
I12 I23 I34 I45 I56
I13 I24 I35 I46
I14 I25 I36
I15 I26
I16
I16@Ꚙ
Step-8:-Draw one circle (Kennedy’s Circle) of Arbitrary diameter, divide it ‘n’
number of Parts.
n= No of Links in the Mechanism here n=6
1 2 3 4 5 6
I12 I23 I34 I45 I56
I13 I24 I35 I46
I14 I25 I36
I15 I26
I16
I16@Ꚙ
Step-9:- Join the points in Kennedy’s circle for known ICR
1 2 3 4 5 6
I12 I23 I34 I45 I56
I13 I24 I35 I46
I14 I25 I36
I15 I26
I16
I12 I𝟏𝟒
I𝟐𝟑 I34
I24
Join ICR I14 and I12 and extend line
Join ICR I23 and I34 and extend line
I16@Ꚙ
Step-10:- Use Kennedy’s theorem to locate Remaining ICR
Consider Quad. 1-2-3-4
At the intersection locate I24
1 2 3 4 5 6
I12 I23 I34 I45 I56
I13 I24 I35 I46
I14 I25 I36
I15 I26
I16
I12 I𝟐𝟑
I𝟏𝟒 I34
I13
Join ICR I12 at I23 and extend line
Join ICR I14 and I34 and extend line
I16@Ꚙ
Step-10:- Use Kennedy’s theorem to locate Remaining ICR
Consider Quad. 1-2-3-4
At the intersection locate I13
1 2 3 4 5 6
I12 I23 I34 I45 I56
I13 I24 I35 I46
I14 I25 I36
I15 I26
I16
I1𝟒 I𝟒𝟓
I𝟏𝟔 I𝟓𝟔
I15
Join ICR I14 and I45 and extend line
Join ICR I16 and I56 and extend line
Step-10:- Use Kennedy’s theorem to locate Remaining ICR
Consider Quad. 1-4-5-6
At the intersection locate I15
1 2 3 4 5 6
I12 I23 I34 I45 I56
I13 I24 I35 I46
I14 I25 I36
I15 I26
I16
I1𝟒 I𝟏𝟔
I𝟒𝟓 I𝟓𝟔
I46
Move ICR I16 at I14 and extend line
Join ICR I45 and I56 and extend line
Step-10:- Use Kennedy’s theorem to locate Remaining ICR
Consider Quad. 1-4-5-6
At the intersection locate I46
1 2 3 4 5 6
I12 I23 I34 I45 I56
I13 I24 I35 I46
I14 I25 I36
I15 I26
I16
I1𝟐 I𝟏𝟔
I𝟐𝟒 I𝟒𝟔
I26
Move ICR I16 at I12 and extend line
Join ICR I24 and I46 and extend line
Step-10:- Use Kennedy’s theorem to locate Remaining ICR
Consider Quad. 1-2-4-6
At the intersection locate I26
1 2 3 4 5 6
I12 I23 I34 I45 I56
I13 I24 I35 I46
I14 I25 I36
I15 I26
I16
I1𝟐 I𝟏𝟓
I𝟐𝟒 I𝟒𝟓
I25
Join ICR I12 at I15 and extend line
Join ICR I24 and I45 and extend line
Step-10:- Use Kennedy’s theorem to locate Remaining ICR
Consider Quad. 1-2-4-5
At the intersection locate I25
1 2 3 4 5 6
I12 I23 I34 I45 I56
I13 I24 I35 I46
I14 I25 I36
I15 I26
I16
I1𝟑 I𝟏𝟔
I𝟑𝟒 I𝟒𝟔
I36
Move ICR I16 at I13 and extend line
Join ICR I34 and I46 and extend line
Step-10:- Use Kennedy’s theorem to locate Remaining ICR
Consider Quad. 1-3-4-6
At the intersection locate I36
1 2 3 4 5 6
I12 I23 I34 I45 I56
I13 I24 I35 I46
I14 I25 I36
I15 I26
I16
I𝟑𝟔 I𝟔𝟓
I𝟑𝟒 I𝟒𝟓
I35
Join ICR I36 at I65 and extend line
Join ICR I34 and I45 and extend line
Step-10:- Use Kennedy’s theorem to locate Remaining ICR
Consider Quad. 1-3-5-6
At the intersection locate I35
Procedure to Solve Numericals
Step-11:- Find Velocity and Angular Velocity
Use Angular Velocity Ratio
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
Procedure to Solve Numericals
Step-11:- Find Velocity and Angular Velocity
Use Angular Velocity Ratio
𝜔3
𝜔2
=
𝐼12 − 𝐼23
𝐼13 − 𝐼23
𝜔 𝟒
𝜔2
=
𝐼12 − 𝐼2𝟒
𝐼1𝟒 − 𝐼2𝟒
𝜔 𝟓
𝜔2
=
𝐼12 − 𝐼2𝟓
𝐼1𝟓 − 𝐼2𝟓
𝜔2
𝝎 𝟐=12.57rad/sec
𝜔 𝟑 = 𝟑. 𝟎𝟕𝒓𝒂𝒅/𝒔𝒆𝒄
𝜔 𝟒 = 𝟖. 𝟑𝟔𝒓𝒂𝒅/𝒔𝒆𝒄
𝜔 𝟓 = 𝟐. 𝟐𝟖𝒓𝒂𝒅/𝒔𝒆𝒄
Procedure to Solve Numericals
Step-12:- Find Velocity and Angular Velocity
Velocity of any point the mechanism
Velocity of any point P which is on link Q
Then 𝑉𝑃 = (𝐼1𝑄 − 𝑃)X𝜔 𝑄*S.F
Velocity of any point B which is on link 3
Then 𝑉𝐵 = (𝐼13. 𝐵)X𝜔3*S.F
Velocity of any point B which is on link 4
Then 𝑉𝐵 = (𝐼14. 𝐵)X𝜔4*S.F
Velocity of any point C which is on link 4
Then 𝑉𝐶 = (𝐼14. 𝐶)X𝜔4*S.F
Velocity of any point D which is on link 5
Then 𝑉𝐷 = (𝐼15. 𝐷)X𝜔5*S.F
S.F=8
Procedure to Solve Numericals
Step-12:- Find Velocity and Angular Velocity
Velocity of slider in the mechanism
𝜔2
In this case velocity of slider(link 6) is the velocity of point
D(Point at the joint of link 5 and link 6).
By considering point D is on link 5
Then 𝑉6 = 𝑉𝐷 = (𝐼15. 𝐷)X𝜔5*S.F
𝑉𝐵 =3.444m/sec
𝑉𝐶 = 1.67m/sec
𝑉6 = 𝑉𝐷= 1.11m/sec
𝑆𝑖𝑚𝑢𝑙𝑎𝑡𝑖𝑜𝑛 𝑢𝑠𝑖𝑛𝑔 𝐺𝑒𝑜𝑔𝑒𝑏𝑟𝑎 𝑆𝑜𝑡𝑤𝑎𝑟𝑒
•ThankYou
Prof. K NWakchaureTheory of Machines-I

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Instantaneous Centre of Rotation (ICR) Analysis and Velocity Calculation of a 6-Bar Mechanism

  • 1. Theory of Machines-I Velocity and Acceleration Analysis of Mechanisms Numerical:- Instantaneous Centre of Rotation (ICR) Prof. K N Wakchaure Department of Mechanical Engineering Sanjivani College of Engineering, Kopargaon
  • 2. Problem Statement • A 6bar mechanism, as shown in Fig. 6, has the following dimensions: OA = 200 mm; AB = 1.5 m; BC = 600 mm; CD = 500 mm and BE = 400 mm, OE=1.35m. Locate all the instantaneous centres. If crank OA rotates uniformly at 120 r.p.m. clockwise, find 1. the velocity of B, C and D, 2. the angular velocity of the links AB, BC and CD.
  • 3. Procedure to Solve Numericals Step-1:-Read the problem statement Carefully. Step-2:-Draw the given Mechanism with Suitable Scale. SCALE 1:8Given:𝑁𝟐=120 rpm 𝜔 𝟐 = 2 ∗ 𝜋 ∗ 𝑁𝟐 60 = 𝟏𝟐. 𝟓𝟕 𝑟𝑎𝑑/𝑠
  • 4. Basics of ICR Scale 1:8 Step-3:-Give the numbers to the link in the mechanism starting with fixed LINK.
  • 5. Step-5:-Draw the table of Instantaneous centres. 1 2 3 4 5 6 I12 I23 I34 I45 I56 I13 I24 I35 I46 I14 I25 I36 I15 I26 I16 Step-6:-Locate Fixed and Permanent ICR. I16@Ꚙ Step-4:-Find No. of ICR using Formula: N= 𝒏∗(𝒏−𝟏) 𝟐 n=6 N=15 (no. of ICR) Step-7:-Circle/ highlight the known ICR in the table..
  • 6. 1 2 3 4 5 6 I12 I23 I34 I45 I56 I13 I24 I35 I46 I14 I25 I36 I15 I26 I16 I16@Ꚙ Step-8:-Draw one circle (Kennedy’s Circle) of Arbitrary diameter, divide it ‘n’ number of Parts. n= No of Links in the Mechanism here n=6
  • 7. 1 2 3 4 5 6 I12 I23 I34 I45 I56 I13 I24 I35 I46 I14 I25 I36 I15 I26 I16 I16@Ꚙ Step-9:- Join the points in Kennedy’s circle for known ICR
  • 8. 1 2 3 4 5 6 I12 I23 I34 I45 I56 I13 I24 I35 I46 I14 I25 I36 I15 I26 I16 I12 I𝟏𝟒 I𝟐𝟑 I34 I24 Join ICR I14 and I12 and extend line Join ICR I23 and I34 and extend line I16@Ꚙ Step-10:- Use Kennedy’s theorem to locate Remaining ICR Consider Quad. 1-2-3-4 At the intersection locate I24
  • 9. 1 2 3 4 5 6 I12 I23 I34 I45 I56 I13 I24 I35 I46 I14 I25 I36 I15 I26 I16 I12 I𝟐𝟑 I𝟏𝟒 I34 I13 Join ICR I12 at I23 and extend line Join ICR I14 and I34 and extend line I16@Ꚙ Step-10:- Use Kennedy’s theorem to locate Remaining ICR Consider Quad. 1-2-3-4 At the intersection locate I13
  • 10. 1 2 3 4 5 6 I12 I23 I34 I45 I56 I13 I24 I35 I46 I14 I25 I36 I15 I26 I16 I1𝟒 I𝟒𝟓 I𝟏𝟔 I𝟓𝟔 I15 Join ICR I14 and I45 and extend line Join ICR I16 and I56 and extend line Step-10:- Use Kennedy’s theorem to locate Remaining ICR Consider Quad. 1-4-5-6 At the intersection locate I15
  • 11. 1 2 3 4 5 6 I12 I23 I34 I45 I56 I13 I24 I35 I46 I14 I25 I36 I15 I26 I16 I1𝟒 I𝟏𝟔 I𝟒𝟓 I𝟓𝟔 I46 Move ICR I16 at I14 and extend line Join ICR I45 and I56 and extend line Step-10:- Use Kennedy’s theorem to locate Remaining ICR Consider Quad. 1-4-5-6 At the intersection locate I46
  • 12. 1 2 3 4 5 6 I12 I23 I34 I45 I56 I13 I24 I35 I46 I14 I25 I36 I15 I26 I16 I1𝟐 I𝟏𝟔 I𝟐𝟒 I𝟒𝟔 I26 Move ICR I16 at I12 and extend line Join ICR I24 and I46 and extend line Step-10:- Use Kennedy’s theorem to locate Remaining ICR Consider Quad. 1-2-4-6 At the intersection locate I26
  • 13. 1 2 3 4 5 6 I12 I23 I34 I45 I56 I13 I24 I35 I46 I14 I25 I36 I15 I26 I16 I1𝟐 I𝟏𝟓 I𝟐𝟒 I𝟒𝟓 I25 Join ICR I12 at I15 and extend line Join ICR I24 and I45 and extend line Step-10:- Use Kennedy’s theorem to locate Remaining ICR Consider Quad. 1-2-4-5 At the intersection locate I25
  • 14. 1 2 3 4 5 6 I12 I23 I34 I45 I56 I13 I24 I35 I46 I14 I25 I36 I15 I26 I16 I1𝟑 I𝟏𝟔 I𝟑𝟒 I𝟒𝟔 I36 Move ICR I16 at I13 and extend line Join ICR I34 and I46 and extend line Step-10:- Use Kennedy’s theorem to locate Remaining ICR Consider Quad. 1-3-4-6 At the intersection locate I36
  • 15. 1 2 3 4 5 6 I12 I23 I34 I45 I56 I13 I24 I35 I46 I14 I25 I36 I15 I26 I16 I𝟑𝟔 I𝟔𝟓 I𝟑𝟒 I𝟒𝟓 I35 Join ICR I36 at I65 and extend line Join ICR I34 and I45 and extend line Step-10:- Use Kennedy’s theorem to locate Remaining ICR Consider Quad. 1-3-5-6 At the intersection locate I35
  • 16. Procedure to Solve Numericals Step-11:- Find Velocity and Angular Velocity Use Angular Velocity Ratio 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
  • 17. Procedure to Solve Numericals Step-11:- Find Velocity and Angular Velocity Use Angular Velocity Ratio 𝜔3 𝜔2 = 𝐼12 − 𝐼23 𝐼13 − 𝐼23 𝜔 𝟒 𝜔2 = 𝐼12 − 𝐼2𝟒 𝐼1𝟒 − 𝐼2𝟒 𝜔 𝟓 𝜔2 = 𝐼12 − 𝐼2𝟓 𝐼1𝟓 − 𝐼2𝟓 𝜔2 𝝎 𝟐=12.57rad/sec 𝜔 𝟑 = 𝟑. 𝟎𝟕𝒓𝒂𝒅/𝒔𝒆𝒄 𝜔 𝟒 = 𝟖. 𝟑𝟔𝒓𝒂𝒅/𝒔𝒆𝒄 𝜔 𝟓 = 𝟐. 𝟐𝟖𝒓𝒂𝒅/𝒔𝒆𝒄
  • 18. Procedure to Solve Numericals Step-12:- Find Velocity and Angular Velocity Velocity of any point the mechanism Velocity of any point P which is on link Q Then 𝑉𝑃 = (𝐼1𝑄 − 𝑃)X𝜔 𝑄*S.F Velocity of any point B which is on link 3 Then 𝑉𝐵 = (𝐼13. 𝐵)X𝜔3*S.F Velocity of any point B which is on link 4 Then 𝑉𝐵 = (𝐼14. 𝐵)X𝜔4*S.F Velocity of any point C which is on link 4 Then 𝑉𝐶 = (𝐼14. 𝐶)X𝜔4*S.F Velocity of any point D which is on link 5 Then 𝑉𝐷 = (𝐼15. 𝐷)X𝜔5*S.F S.F=8
  • 19. Procedure to Solve Numericals Step-12:- Find Velocity and Angular Velocity Velocity of slider in the mechanism 𝜔2 In this case velocity of slider(link 6) is the velocity of point D(Point at the joint of link 5 and link 6). By considering point D is on link 5 Then 𝑉6 = 𝑉𝐷 = (𝐼15. 𝐷)X𝜔5*S.F 𝑉𝐵 =3.444m/sec 𝑉𝐶 = 1.67m/sec 𝑉6 = 𝑉𝐷= 1.11m/sec