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DESIGN OF MACHINERY " SOLUTION MANUAL 5-2-1 PROBLEM 5-2 Statement: Design a fourbar mechanism to give the two positions shown in Figure P3-1 of coupler motion. Given: Coordinates of the points A1, B1, A2, and B2 with respect to A1: A1x := 0.0 B1x := 1.721 A2x := 2.656 B2x := 5.065 A1y := 0.0 B1y := −1.750 A2y := −0.751 B2y := −0.281 Assumptions: Use the points A1 and A2 as the precision points P1 and P2. Define position vectors in the global frame whose origin is at A1. Solution: See solution to Problem 3-4 and Mathcad file P0502. 1. Note that this is a two-position motion generation (MG) problem because the output is specified as a complex motion of the coupler, link 3. See Section 5.3 in which the equations for the two-position motion generation problem are derived. 2. Two solution methods are derived in Section 5.3 and are presented in equations 5.7 and 5.8 for the left dyad and equations 5.11 and 5.12 for the right dyad. 3. Method 1 (equations 5.7 and 5.11) requires the choosing of three angles for each dyad. Method 2 (equations 5.8 and 5.12) requires the choosing of two angles and a length for each dyad. Method 1 is used in this solution. 4. In order to obtain the same solution as was done graphically in Problem 3-4, the necessary assumed values were taken from that solution as shown below. 5. Define the position vectors R1 and R2 and the vector P21 using Figure 5-1 and equation 5.1. p 21 := 2.760 6. 7. p 21 = 2.760 From the trigonometric relationships given in Figure 5-1, determine α2 and δ2. α 2 := 56.519⋅ deg α 2 = 56.519 deg δ 2 := −15.789⋅ deg δ 2 = −15.789 deg From the graphical solution (see figure below), determine the values necessary for input to equations 5.7. θ := 94.394⋅ deg β 2 := −40.366⋅ deg φ := −45.479⋅ deg DESIGN OF MACHINERY SOLUTION MANUAL 5-2-2 O4 93.449° jY 54.330° 2.760 u A1 X 45.479° 0.281 P 21 15.789° B2 0.751 v 1.750 A2 134.521° 2.656 2.409 B1 w 40.366° 94.394° 75.124° O2 8. Solve for the WZ dyad using equations 5.7. ( ( ) ) ( ) ( ( ) ) ( ) ( ( ) ) ( ) ( ( ) ) ( ) A := cos( θ ) ⋅ cos β 2 − 1 − sin( θ ) ⋅ sin β 2 B := cos( φ) ⋅ cos α 2 − 1 − sin( φ) ⋅ sin α 2 ( ) C := p 21⋅ cos δ 2 D := sin( θ ) ⋅ cos β 2 − 1 + cos( θ ) ⋅ sin β 2 ( ) E := sin( φ) ⋅ cos α 2 − 1 + cos( φ) ⋅ sin α 2 F := p 21⋅ sin δ 2 w := 4.000 z := 0.000 w = 4.000 z = 0.000 These are the expected values of w and z based on the design choices made in the graphical solution and the assumptions made in this problem. 9. From the graphical solution (see figure above), determine the values necessary for input to equations 5.11. σ := −93.449⋅ deg γ 2 := 54.330⋅ deg ψ := 134.521 ⋅ deg 10. Solve for the US dyad using equations 5.11. ( ( ) ) ( ) ( ( ) ) ( ) A' := cos( σ ) ⋅ cos γ 2 − 1 − sin( σ ) ⋅ sin γ 2 B' := cos( ψ ) ⋅ cos α 2 − 1 − sin( ψ ) ⋅ sin α 2 ( ) C := p 21⋅ cos δ 2 DESIGN OF MACHINERY SOLUTION MANUAL 5-2-3 ( ( ) ) ( ) ( ( ) ) ( ) D' := sin( σ ) ⋅ cos γ 2 − 1 + cos( σ ) ⋅ sin γ 2 ( ) E' := sin( ψ ) ⋅ cos α 2 − 1 + cos( ψ ) ⋅ sin α 2 F := p 21⋅ sin δ 2 u := 4.000 s := 2.454 u = 4.000 s = 2.454 These are the expected values of u and s based on the design choices made in the graphical solution and the assumptions made in this problem. 11. Solve for the links 3 and 1 using the vector definitions of V and G. Link 3: θ 3 := −45.479⋅ deg θ 3 = −45.479 deg v := 2.454 v = 2.454 θ 1 := 75.120⋅ deg θ 1 = 75.120 deg g := 6.447 g = 6.447 Link 1: 12. Determine the initial and final values of the input crank with respect to the vector G. θ 2i := θ − θ 1 θ 2i = 19.274 deg θ 2f := θ 2i + β 2 θ 2f = −21.092 deg 13. Define the coupler point with respect to point A and the vector V. rp := z δ p := φ − θ 3 rp = 0.000 δ p = 0.000 deg which is correct for the assumption that the precision point is at A. 14. Locate the fixed pivots in the global frame using the vector definitions in Figure 5-2. O2x := 0.306 O2x = 0.306 O2y := −3.988 O2y = −3.988 DESIGN OF MACHINERY SOLUTION MANUAL 5-2-4 O4x := 1.961 O4x = 1.961 O4y := 2.243 O4y = 2.243 15. Determine the rotation angle of the fourbar frame with respect to the global frame (angle from the global X axis to the line O2O4. θ rot := atan2( O4x − O2x) , ( O4y − O2y) θ rot = 75.125 deg 16. Determine the Grashof condition. Condition( S , L , P , Q) := SL ← S + L PQ ← P + Q return "Grashof" if SL < PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise Condition( v , g , w , u ) = "non-Grashof" 17. DESIGN SUMMARY Link 2: w = 4.000 θ = 94.394 deg Link 3: v = 2.454 θ 3 = −45.479 deg Link 4: u = 4.000 σ = −93.449 deg Link 1: g = 6.447 θ 1 = 75.120 deg Coupler: rp = 0.000 δ p = 0.000 deg Crank angles: θ 2i = 19.274 deg θ 2f = −21.092 deg