Instantaneous Centre of Rotation (ICR) Analysis and Velocity Calculation of a 6-Bar Mechanism
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INSTANTANEOUS CENTER ROTATION numerical. 6 BAR MECHANISM. VELOCITY ANALYSIS
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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 β’ 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:ππ=120 rpm π π = 2 β π β ππ 60 = ππ. ππ πππ/π
- 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@κ Step-4:-Find No. of ICR using Formula: N= πβ(πβπ) π 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@κ Step-8:-Draw one circle (Kennedyβs Circle) of Arbitrary diameter, divide it βnβ 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@κ Step-9:- Join the points in Kennedyβs 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 Iππ Iππ I34 I24 Join ICR I14 and I12 and extend line Join ICR I23 and I34 and extend line I16@κ Step-10:- Use Kennedyβs 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 Iππ Iππ I34 I13 Join ICR I12 at I23 and extend line Join ICR I14 and I34 and extend line I16@κ Step-10:- Use Kennedyβs 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 I1π Iππ Iππ Iππ I15 Join ICR I14 and I45 and extend line Join ICR I16 and I56 and extend line Step-10:- Use Kennedyβs 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 I1π Iππ Iππ Iππ I46 Move ICR I16 at I14 and extend line Join ICR I45 and I56 and extend line Step-10:- Use Kennedyβs 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 I1π Iππ Iππ Iππ I26 Move ICR I16 at I12 and extend line Join ICR I24 and I46 and extend line Step-10:- Use Kennedyβs 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 I1π Iππ Iππ Iππ I25 Join ICR I12 at I15 and extend line Join ICR I24 and I45 and extend line Step-10:- Use Kennedyβs 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 I1π Iππ Iππ Iππ I36 Move ICR I16 at I13 and extend line Join ICR I34 and I46 and extend line Step-10:- Use Kennedyβs 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 Iππ Iππ Iππ Iππ I35 Join ICR I36 at I65 and extend line Join ICR I34 and I45 and extend line Step-10:- Use Kennedyβs 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 π π π π = π°ππ β π°ππ π°ππ β π°ππ angular velocity of link 3 when angular velocity of link 2 is known π3 π2 = πΌ12 β πΌ23 πΌ13 β πΌ23 π2
- 17. Procedure to Solve Numericals Step-11:- Find Velocity and Angular Velocity Use Angular Velocity Ratio π3 π2 = πΌ12 β πΌ23 πΌ13 β πΌ23 π π π2 = πΌ12 β πΌ2π πΌ1π β πΌ2π π π π2 = πΌ12 β πΌ2π πΌ1π β πΌ2π π2 π π=12.57rad/sec π π = π. πππππ /πππ π π = π. πππππ /πππ π π = π. πππππ /πππ
- 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 ππ = (πΌ1π β π)Xπ π*S.F Velocity of any point B which is on link 3 Then ππ΅ = (πΌ13. π΅)Xπ3*S.F Velocity of any point B which is on link 4 Then ππ΅ = (πΌ14. π΅)Xπ4*S.F Velocity of any point C which is on link 4 Then ππΆ = (πΌ14. πΆ)Xπ4*S.F Velocity of any point D which is on link 5 Then ππ· = (πΌ15. π·)Xπ5*S.F S.F=8
- 19. Procedure to Solve Numericals Step-12:- Find Velocity and Angular Velocity Velocity of slider in the mechanism π2 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 π6 = ππ· = (πΌ15. π·)Xπ5*S.F ππ΅ =3.444m/sec ππΆ = 1.67m/sec π6 = ππ·= 1.11m/sec
- 20. ππππ’πππ‘πππ π’π πππ πΊπππππππ πππ‘π€πππ
- 21. β’ThankYou Prof. K NWakchaureTheory of Machines-I