DESIGN OF MACHINERY SOLUTION MANUAL 5-6-1! PROBLEM 5-6Statement:See Project P3-20. Define three positions of the dumpster and analytically synthesize a linkage to move through them. The fixed pivots must be located on the existing truck.Solution:See Project P3-20 (p. 142) and Mathcad file P0506.1. This is an open-ended design problem that has many valid solutions. First define the problem more completely than it is stated by deciding on three positions for the dumpster box to move through. The figure below shows one such set of positions (dimensions are in mm).59019992036209412021817226311O2O4PP2P330.3 deg.59.1 deg.0 deg.12. From the figure, the design choices are:P21x590 P21y1202 P31x1999 P31y1817O2x2094 O2y226 O4x2036 O4y311Body angles:θP10 deg.θP230.3 deg.θP359.1 deg.3. The methods of Section 5.8 are used to get a solution for this problem. The solution is sensitive to small changes in the design choices so a trial-and-error approach is warranted.4. Determine the angle changes between precision points from the body angles given.α2θP2θP1α230.300 deg=α3θP3θP1α359.100 deg=5. Using Figure 5-6, determine the magnitudes of R1, R2, and R3 and their x and y components.R1xO2xR1x2.094 103.=R1yO2yR1y226.000=R2xR1xP21xR2x1.504 103.=R2yR1yP21yR2y1.428 103.=2nd Edition, 1999DESIGN OF MACHINERY SOLUTION MANUAL 5-6-2R3xR1xP31xR3x95.000=R3yR1yP31yR3y2.043 103.=R1R1x2R1y2R12.106 103.=R2R2x2R2y2R22.074 103.=R3R3x2R3y2R32.045 103.=6. Using Figure 5-6, determine the angles that R1, R2, and R3 make with the x axis.ζ1atan2 R1xR1y, ζ16.160 deg=ζ2atan2 R2xR2y, ζ243.515 deg=ζ3atan2 R3xR3y, ζ387.338 deg=7. Solve for β2 and β3 using equations 5.34C1R3cosα2ζ3.R2cosα3ζ2.C1495.777=C2R3sinα2ζ3.R2sinα3ζ2.C2212.019=C3R1cosα3ζ1.R3cosζ3.C3786.433=C4R1sinα3ζ1.R3sinζ3.C4130.152=C5R1cosα2ζ1.R2cosζ2.C5189.927=C6R1sinα2ζ1.R2sinζ2.C6176.392=A1C32C42A16.354 105.=A2C3C6.C4C5.A21.140 105.=A3C4C6.C3C5.A31.723 105.=A4C2C3.C1C4.A42.313 105.=A5C4C5.C3C6.A51.140 105.=A6C1C3.C2C4.A63.623 105.=K1A2A4.A3A6.K13.607 1010.=K2A3A4.A5A6.K28.115 1010.=K3A12A22A32A42A622K38.816 1010.=β312 atanK2K12K22K32K1K3.β3172.976 deg=2nd Edition, 1999DESIGN OF MACHINERY SOLUTION MANUAL 5-6-3β322 atanK2K12K22K32K1K3.β3259.100 deg=The second value is the same as α3, so use the first valueβ3β31β21acosA5sinβ3.A3cosβ3.A6A1β2134.802 deg=β22asinA3sinβ3.A2cosβ3.A4A1β2234.802 deg=Since both values are the same,β2β218. Repeat steps 2, 3, and 4 for the right-hand dyad to find γ1 and γ2.R1xO4xR1x2.036 103.=R1yO4yR1y311.000=R2xR1xP21xR2x1.446 103.=R2yR1yP21yR2y1.513 103.=R3xR1xP31xR3x37.000=R3yR1yP31yR3y2.128 103.=R1R1x2R1y2R12.060 103.=R2R2x2R2y2R22.093 103.=R3R3x2R3y2R32.128 103.=9. Using Figure 5-6, determine the angles that R1, R2, and R3 make with the x axis.ζ1atan2 R1xR1y, ζ18.685 deg=ζ2atan2 R2xR2y, ζ246.297 deg=ζ3atan2 R3xR3y, ζ389.004 deg=10. Solve for γ2 and γ3 using equations 5.34C1R3cosα2ζ3.R2cosα3ζ2.C1486.018=C2R3sinα2ζ3.R2sinα3ζ2.C2161.777=C3R1cosα3ζ1.R3cosζ3.C3741.712=C4R1sinα3ζ1.R3sinζ3.C4221.269=C5R1cosα2ζ1.R2cosζ2.C5154.965=C6R1sinα2ζ1.R2sinζ2.C6217.266=2nd Edition, 1999DESIGN OF MACHINERY SOLUTION MANUAL 5-6-4A1C32C42A15.991 105.=A2C3C6.C4C5.A21.269 105.=A3C4C6.C3C5.A31.630 105.=A4C2C3.C1C4.A42.275 105.=A5C4C5.C3C6.A51.269 105.=A6C1C3.C2C4.A63.247 105.=K1A2A4.A3A6.K12.406 1010.=K2A3A4.A5A6.K27.828 1010.=K3A12A22A32A42A622K37.953 1010.=γ312 atanK2K12K22K32K1K3.γ3186.724 deg=γ322 atanK2K12K22K32K1K3.γ3259.100 deg=The second value is the same as α3, so use the first valueγ3γ31γ21acosA5sinγ3.A3cosγ3.A6A1γ2139.743 deg=γ22asinA3sinγ3.A2cosγ3.A4A1γ2239.743 deg=Since both angles are the same,γ2γ2111. Use the method of Section 5.7 to synthesize the linkage. Start by determining the magnitudes of the vectors P21 and P31 and their angles with respect to the X axis.p21P21x2P21y2p211338.994=δ2atan2 P21xP21y, δ2116.144 deg=p31P31x2P31y2p312701.387=δ3atan2 P31xP31y, δ3137.731 deg=2nd Edition, 1999DESIGN OF MACHINERY SOLUTION MANUAL 5-6-512. Evaluate terms in the WZ coefficient matrix and constant vector from equations (5.25) and form the matrix and vector:A cosβ21 B sinβ2C cosα21D sinα2Ep21cosδ2.F cosβ31G sinβ3H cosα31 K sinα3Lp31cosδ3.Mp21sinδ2.Np31sinδ3.AAAFBGBGAFCHDKDKCHCCELMNW1xW1yZ1xZ1yAA1CC.13. The components of the W and Z vectors are:W1x 3.194 103.=W1y 829.763=Z1x 1.100 103.=Z1y 603.763=14. The length of link 2 is:w W1x2W1y2w 3299.543=15. Evaluate terms in the US coefficient matrix and constant vector from equations (5.25) and form the matrix and vector:A' cosγ21 B' sinγ2C cosα21D sinα2Ep21cosδ2.F' cosγ31G' sinγ3H cosα31 K sinα3Lp31cosδ3.Mp21sinδ2.Np31sinδ3.AAA'F'B'G'B'G'A'F'CHDKDKCHCCELMNU1xU1yS1xS1yAA1CC.16. The components of the W and Z vectors are:U1x 1.621 103.=U1y 13.492=S1x 415.016=S1y 297.508=17. The length of link 4 is:u U1x2U1y2u 1621.040=18. Solving for links 3 and 1 from equations 5.2a and 5.2b.V1x Z1x S1x V1x 1.515 103.=2nd Edition, 1999DESIGN OF MACHINERY SOLUTION MANUAL 5-6-6V1y Z1y S1y V1y 901.272=The length of link 3 is: v V1x2V1y2v 1762.404=G1x W1x V1x U1x G1x 58.000=G1y W1y V1y U1y G1y 85.000=The length of link 1 is: g G1x2G1y2g 102.903=19. Check the location of the fixed pivots with respect to the global frame using the calculated vectors W1, Z1, U1, and S1.O2x Z1x W1x O2x 2094.000=O2y Z1y W1y O2y 226.000=O4x S1x U1x O4x 2036.000=O4y S1y U1y O4y 311.000=These check with the design choices shown in the figure above.20. Determine the location of the coupler point with respect to point A and line AB.Distance from A to Pz Z1x2Z1y2z 1254.370=rPzAngle BAP (δp) s S1x2S1y2s 510.637=ψatan2 S1x S1y,()ψ35.635 deg=φatan2 Z1x Z1y,()φ151.228 deg=θ3atan2 z cosφ().s cosψ().z sinφ().s sinψ().,()θ3149.244 deg=δpφθ3δp1.984deg=21.DESIGN SUMMARYLink 1:g 102.9=Link 2: w 3299.5=Link 3: v 1762.4=Link 4: u 1621.0=Coupler point: rP1254.4=δp1.984deg=22.VERIFICATION: The calculated values of g (length of the ground link) and of the coordinates of O2 and O4 give the same values as those on the problem statement, verifying that the calculated values for the other links and the coupler point are correct. The solution is drawn below to show the locations of the moving pivots for the three positions chosen (see next page).23. The design needs to be checked for the presence of toggle positions within its desired range of motion. This design has none, but is close to toggle at position 3.2nd Edition,