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Mechanics of Machines Dr. Mohammad Kilani

Mechanics of Machines Dr. Mohammad Kilani. Class 4 Acceleration Analysis. Acceleration of a Point on an Ground Pivoted Link. Acceleration of a Point. Let the vector r = r u θ describe the position of a point. The velocity and acceleration of the point are given b. r. sin θ j. θ.

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Mechanics of Machines Dr. Mohammad Kilani

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  1. Mechanics of MachinesDr. Mohammad Kilani Class 4 Acceleration Analysis

  2. Acceleration of a Point on an Ground Pivoted Link

  3. Acceleration of a Point • Let the vector r = r uθ describe the position of a point. The velocity and acceleration of the point are given b r sinθj θ cosθi aslipanormalatangaCoriolis

  4. Acceleration of a Point on an Ground Pivoted Link y • If the point is taken to be on a rigid link, that is pivoted to the ground, and the origin of the coordinate is taken to be the of the pivot point, then d2r/dt2 =dr/dt = 0, and we obtain • Define the vector ω = dθ/dtkand the vector α = d2θ/dt2kthen the acceleration of point P is given by A x

  5. Acceleration of a Point Sliding on a Ground Pivoted Link aCoriolis r, dr/dt, d2r/dt2 ω

  6. ACCELERATION ANALYSIS OF FOUR BAR MECHANISMS

  7. Acceleration Analysis of a Four Bar Mechanism • For a 4-bar kinematic chain with all links of constant lengths, the acceleration loop closure equation is found by starting with the position loop closure equation as follows: • Design Parameters: r1, r2, r3, r4, θ1 • Position Analysis Parameters: θ2, θ3, θ4 • Velocity Analysis Parameters: ω2, ω3, ω4 • Acceleration Analysis Parameters: α2, α 3, α4

  8. Acceleration Analysis of a Four Bar Mechanism Design Parameters: r1, r2, r3, r4, θ1 Position Analysis Parameters: θ2, θ3, θ4 Velocity Analysis Parameters: ω2, ω3, ω4 Acceleration Analysis Parameters: α2, α3, α4 • To eliminate α3 dot product with uθ3

  9. ACCELERATION ANALYSIS OF SLIDER-CRANK MECHANISMS

  10. Acceleration Analysis of a Slider-Crank Mechanism • The acceleration loop closure equation is found by differentiating the position loop closure equation twice as follows equation • Design Parameters: r2, r3, r4, θ1 , θ4 • Position Analysis Parameters: θ2, θ3, r1 • Velocity Analysis Parameters: ω2, ω3, dr1/dt • Acceleration Analysis Parameters: α2, α 3, d2r1/dt2 r3 r4 rp r2 r1

  11. Acceleration Analysis of a Slider-Crank Mechanism Design Parameters: r2, r3, r4, θ1 , θ4 Position Analysis Parameters: θ2, θ3, r1 Velocity Analysis Parameters: ω2, ω3, dr1/dt Acceleration Analysis Parameters: α2, α3, d2r1/dt2 • To find d2r1/dt2, eliminate α3 by dot product with uθ3 r3 r4 rp r2 r1

  12. Acceleration Analysis of a Slider-Crank Mechanism Design Parameters: r2, r3, r4, θ1 , θ4 Position Analysis Parameters: θ2, θ3, r1 Velocity Analysis Parameters: ω2, ω3, dr1/dt Acceleration Analysis Parameters: α2, α3, d2r1/dt2 • To find α3 eliminate d2r1/dt2, by dot product with uθ1+π/2 r3 r4 rp r2 r1

  13. Acceleration Analysis of an Inverted Slider-Crank Mechanism Design Parameters: r1, r2, θ1 Position Analysis Parameters: θ2, θ3, r3 Velocity Analysis Parameters: ω2, ω3, dr3/dt Acceleration Analysis Parameters: α2, α3, d2r3/dt2 • The acceleration loop closure equation is found by differentiating the position loop closure equation twice as follows equation θ3 r2 r3 r1 θ2 θ1

  14. Acceleration Analysis of an Inverted Slider-Crank Mechanism Design Parameters: r1, r2, θ1 Position Analysis Parameters: θ2, θ3, r3 Velocity Analysis Parameters: ω2, ω3, dr3/dt Acceleration Analysis Parameters: α2, α3, d2r3/dt2 • To find d2r3/dt2 eliminate α3 by dot product with uθ3 • To find α3 eliminate d2r3/dt2 by dot product with uθ3+π/2 θ3 r2 r3 r1 θ2 θ1

  15. Example: (Problem 7-48) F • L2 = 0.8 in, L3= 2.97 in, L5 = 2.61 inθ2 = 241°, O2O4 = 1.85 @ 278.5° • The problem is an inverted slider-crank mechanism whose output link is the input link to an ordinary slider-crank mechanism. The problem is solved in two stages: (I) an inverted slider-crank mechanism, and (II) an ordinary slider-crank mechanism. • Stage I. For the inverted slider-crank mechanism 5 E 6 O2 ω2 = 40 rad/min α2 = 1500 rad/min2 2 4 3 Design : r1 = O2O4 = 1.85 in, r2 = L2 = 0.8 in, θ1= 278.5° Position: θ2 = 241° Velocity ω2= 40 rad/min= 2/3 rad/s Acceleration : α2 = 1500 rad/min2 = 0.416 rad/s2 O4 θ3 r2 r3 r1 θ2 θ1

  16. Example: (Problem 7-48) – Stage I F • L2 = 0.8 in, L4 = 2.97 in, L5 = 2.61 inθ2 = 241°, O2O4 = 1.85 @ 278.5° • Design: r1 = O2O4 = 1.85 in, r2 = L2 = 0.8 in, θ1= 278.5° • Position: θ2 = 241° 5 E 6 O2 ω2 = 40 rad/min α2 = 1500 rad/min2 2 4 3 O4 θ3 r2 r3 r1 θ2 θ1

  17. Example: (Problem 7-48) – Stage I F • L2 = 0.8 in, L4 = 2.97 in, L5 = 2.61 inθ2 = 241°, O2O4 = 1.85 @ 278.5° • Design: r1 = O2O4 = 1.85 in, r2 = L2 = 0.8 in, θ1= 278.5° • Position: θ2 = 241°, θ3 = 120.34°, r3 = 1.31 in. • Velocityω2= 40 rad/min= 2/3 rad/s 5 E 6 O2 ω2 = 40 rad/min α2 = 1500 rad/min2 2 4 3 O4 θ3 r2 r3 r1 θ2 θ1

  18. Example: (Problem 7-48) – Stage I F • L2 = 0.8 in, L4 = 2.97 in, L5 = 2.61 inθ2 = 241°, O2O4 = 1.85 @ 278.5° • Design: r1 = O2O4 = 1.85 in, r2 = L2 = 0.8 in, θ1= 278.5° • Position: θ2 = 241°, θ3 = 120.34°, r3 = 1.31 in. • Velocityω2= 2/3 rad/s, ω3= -0.21 rad/s, dr3 /dt = -0.46 in/s • Acceleration: α2 = -1500 rad/min2 = -0.4167 rad/s2 5 E 6 O2 ω2 = 40 rad/min α2 = 1500 rad/min2 2 4 3 O4 θ3 r2 r3 r1 θ2 θ1

  19. Example: (Problem 7-48) – Stage II F G • For the ordinary slider-crank mechanism, the following information are now available • Design: r2 = O4E= 2.97 in, r3 = EF= 2.61 in, r4 = O4G 3.25 in θ1= 0°, θ4= 90° • Position: θ2 = θO4E = 120.34°, θ3 = θEF , r1 = GF. • Velocityω2= ω O4E = 0.21 rad/s, ω3= ω EF, dr1 /dt = dGF/dt • Acceleration: α2 = α O4E = -0.25 rad/s2, α 3= α EF, d2r1 /dt2= d2GF/dt2 5 E 6 O2 ω2 = 40 rad/min α2 = 1500 rad/min2 2 r3 4 r4 3 O4 rp r2 r1

  20. Example: (Problem 7-48) – Stage II F G • For the ordinary slider-crank mechanism, the following information are now available • Design: r2 = O4E= 2.97 in, r3 = EF= 2.61 in, r4 = O4G 3.25 in θ1= 0°, θ4= 90° • Position: θ2 = θO4E = 120.34°, θ3 = θEF , r1 = GF. 5 E 6 O2 ω2 = 40 rad/min α2 = 1500 rad/min2 2 r3 4 r4 3 O4 rp r2 r1

  21. Example: (Problem 7-48) – Stage II F G • For the ordinary slider-crank mechanism, the following information are now available • Design: r2 = O4E= 2.97 in, r3 = EF= 2.61 in, r4 = O4G 3.25 in θ1= 0°, θ4= 90° • Position: θ2 = θO4E = 120.34°, θ3 = θEF =13.34°, r1 = GF = 1.41 in • Velocityω2= ω O4E = 0.21 rad/s, ω3= ω EF, dr1 /dt = dGF/dt 5 E 6 O2 ω2 = 40 rad/min α2 = 1500 rad/min2 2 r3 4 r4 3 O4 rp r2 r1

  22. Example: (Problem 7-48) – Stage II F G • For the ordinary slider-crank mechanism, the following information are now available • Design: r2 = O4E= 2.97 in, r3 = EF= 2.61 in, r4 = O4G 3.25 in θ1= 0°, θ4= 90° • Position: θ2 = θO4E = 120.34°, θ3 = θEF =13.34°, r1 = GF = 1.41 in • Velocityω2= ω O4E = 0.21 rad/s, ω3= ω EF = 0.124 rad/s dr1 /dt = dGF/dt = 2 in/s • Acceleration: α2 = α O4E = -0.25 rad/s2, α 3= α EF, d2r1 /dt2= d2GF/dt2 5 E 6 O2 ω2 = 40 rad/min α2 = 1500 rad/min2 2 r3 4 r4 3 O4 rp r2 r1

  23. Example: (Problem 7-48) – Stage II F G • For the ordinary slider-crank mechanism, the following information are now available • Design: r2 = O4E= 2.97 in, r3 = EF= 2.61 in, r4 = O4G 3.25 in θ1= 0°, θ4= 90° • Position: θ2 = θO4E = 120.34°, θ3 = θEF =13.34°, r1 = GF = 1.41 in • Velocityω2= ω O4E = 0.21 rad/s, ω3= ω EF = 0.124 rad/s dr1 /dt = dGF/dt = 2 in/s • Acceleration: α2 = α O4E = -0.25 rad/s2, α 3= α EF, d2r1 /dt2= d2GF/dt2 5 E 6 O2 ω2 = 40 rad/min α2 = 1500 rad/min2 2 r3 4 r4 3 O4 rp r2 r1

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