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CSU MECH 324 - DESIGN OF MACHINERY

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DESIGN OF MACHINERY SOLUTION MANUAL 5-8-1! PROBLEM 5-8Statement:Design a linkage to carry the body in Figure P5-1 through the two positions P1 and P2 at the angles shown in the figure. Use analytical synthesis without regard for the fixed pivots shown. Use the free choices given below.Given:Coordinates of the points P1 and P2 with respect to P1:P1x0.0 P1y0.0 P2x1.236 P2y2.138Angles made by the body in positions 1 and 2:θP1210 deg.θP2147.5 deg.Free choices for the WZ dyad :z 1.075β227.0 deg.φ204.4 deg.Free choices for the US dyad :s 1.240γ240.0 deg.ψ74.0 deg.Two argument inverse tangentatan2 x y,( ) 0.5π.return x 0ifatanyxreturn x 0>ifatanyxπotherwiseSolution:See Figure P5-1 and Mathcad file P0508.1. 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. Because of the data given in the hint, the second method of Section 5.3 will be used here.2. Define the position vectors R1 and R2 and the vector P21 using Figure 5-1 and equation 5.1.R1P1xP1yR2P2xP2yP21xP21yR2 R1 P21x1.236=P21y2.138=p21P21x2P21y2p212.470=3. From the trigonometric relationships given in Figure 5-1, determine α2 and δ2.α2θP2θP1α262.500 deg=δ2atan2 P21xP21y, δ2120.033 deg=4. Solve for the WZ dyad using equations 5.8.Z1xz cosφ().Z1x0.979=Z1yz sinφ().Z1y0.444=2nd Edition, 1999DESIGN OF MACHINERY SOLUTION MANUAL 5-8-2A cosβ21 A 0.109=D sinα2D 0.887=B sinβ2B 0.454=Ep21cosδ2.E 1.236=C cosα21 C 0.538=Fp21sinδ2.F 2.138=W1xACZ1x.DZ1y.E.BCZ1y.DZ1x.F.2 A.W1x1.462=W1yACZ1y.DZ1x.F.BCZ1x.DZ1y.E.2 A.W1y3.367=wW1x2W1y2w 3.670=θatan2 W1xW1y, θ246.528 deg=5. Solve for the US dyad using equations 5.12.S1xs cosψ().S1x0.342=S1ys sinψ().S1y1.192=A cosγ21 A 0.234=D sinα2D 0.887=B sinγ2B 0.643=Ep21cosδ2.E 1.236=C cosα21 C 0.538=Fp21sinδ2.F 2.138=U1xACS1x.DS1y.E.BCS1y.DS1x.F.2 A.U1x3.180=U1yACS1y.DS1x.F.BCS1x.DS1y.E.2 A.U1y4.439=uU1x2U1y2u 5.461=σatan2 U1xU1y, σ234.381 deg=6. Solve for links 3 and 1 using the vector definitions of V and G.Link 3:V1xz cosφ().s cosψ().V1x1.321=V1yz sinφ().s sinψ().V1y1.636=θ3atan2 V1xV1y, θ3231.086 deg=vV1x2V1y2v 2.103=Link 1: G1xw cosθ().v cosθ3.u cosσ().G1x0.398=2nd Edition, 1999DESIGN OF MACHINERY SOLUTION MANUAL 5-8-3G1yw sinθ().v sinθ3.u sinσ().G1y0.564=θ1atan2 G1xG1y, θ154.796 deg=gG1x2G1y2g 0.690=7. Determine the initial and final values of the input crank with respect to the vector G.θ2iθθ1θ2i301.323 deg=θ2fθ2iβ2θ2f274.323 deg=8. Define the coupler point with respect to point A and the vector V.rpzδpφθ3rp1.075=δp26.686deg=9. Locate the fixed pivots in the global frame using the vector definitions in Figure 5-2.ρ1atan2 P1xP1y, ρ190.000 deg=R1P1x2P1y2R10.000=O2xR1cosρ1.z cosφ().w cosθ().O2x2.441=O2yR1sinρ1.z sinφ().w sinθ().O2y3.811=O4xR1cosρ1.s cosψ().u cosσ().O4x2.838=O4yR1sinρ1.s sinψ().u sinσ().O4y3.247=10. Determine the rotation angle of the fourbar frame with respect to the global frame (angle from the global X axis to the line O2O4.θrotatan2 O4xO2xO4yO2y,θrot54.796deg=11. Determine the Grashof condition.Condition S L,P,Q,()SL S LPQ P Q"Grashof"return SL PQif"non-Grashof"return otherwiseCondition g w,u,v,( ) "Grashof"=12.DESIGN SUMMARYLink 2:w 3.670= θ246.528 deg=Link 3: v 2.103= θ3231.086 deg=Link 4: u 5.461= σ234.381 deg=2nd Edition, 1999DESIGN OF MACHINERY SOLUTION MANUAL 5-8-4Link 1: g 0.690= θ154.796 deg=Coupler: rp1.075=δp26.686deg=Crank angles:θ2i301.323 deg=θ2f274.323 deg=13. Draw the linkage, using the link lengths, fixed pivot positions, and angles above, to verify the design.62.5°2.138V211B1VSP11ZA2A11.236B2P22SY2Z2UW1W2XO41U1GO22nd Edition,


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