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# Dynamics of Machines MECH 324

CSU

GPA 3.87

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This 22 page Class Notes was uploaded by Lionel Hansen on Tuesday September 22, 2015. The Class Notes belongs to MECH 324 at Colorado State University taught by David Alciatore in Fall. Since its upload, it has received 70 views. For similar materials see /class/210249/mech-324-colorado-state-university in Mechanical Engineering at Colorado State University.

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Date Created: 09/22/15

Multiple Dwell Cam Design Example Motion Sgecifioations camshaft m 10002 30 sec rise estartin39se 3039deg 31158 6039deg hrise l39in falli estartifall 120deg Bfan 50deg riseja estartin39seifall 200ng B seifall 130ng hn39seifall 15in Modified Trapezoidal 39 quot Functions from Equations 815819 b 025 c 05 d 25 from Figure 818 2 C 41 a 2 2 C 4888 1 7 8b2 7 d2 7 21E1E 7 2b 1 a 2 igm Mk 2 Gy igg gijgwkgw Tt s ET60 Smod traplte 43 sh h39Ca39 12 B 2 ha92H2m4mwaLik H 2 3 2N3 n2 8 4 2 2 2 2 39 hC 322d b 1 b 7d 7 bNs39 1373 1f 17 1113 2 4 1 dB J Usinq MathCAD39s 39 Numericall instead of Eguations 815819 for v Vmod traplte J3 sh i esmoditrage J3 sh h 6 amoditrape J3 sh 1 Ca39 39S39In 39 b if0elt 3 b 2 Ch fb ltelt17d 1 a2 2 2 2 2 I B J Cah quot 2 1 if lig geg 2 d l3 J 2 3 Example Modified Trapezoidal 39 quot s va Diaqrams 3100deg 6 02deg3 h 1 Smodftraplte s B sh 05 0 0 50 100 9 deg 15 Vmoditrape s B s h arnoditrape s B s h 0 2 0 50 100 9 deg 345 6 Sinqle Dwell Polvnomial Function Equation 826 93 943 95 96 S345693h7h64 j 192B 192B 64h Usin MathCADS mbolic Differentiation co and aste ex ression put cursor next to 9 select from menu SvmbolicsVariableDifferentiate 09y exgression to function definition B B 3 4 5 6 are 1942 1942 449 B B r 2 4 5 e e e e h 192 3 7 768 4 960 5 7 384 6 k B B B B 2 e3 e4 e5 V3456e3h 7 h 192 3 7 768 960 7 384 B B4 B B6 3 4 a345693h 7 h 384 7 2304 38406 7 19206 3 4 5 6 B B B 2 3 384 e e e J3456e3h 7 h 7 4608 11520 7 7680 3 B B B B6 Example 345 6 Polvnomial sva Diaqrams m IOOdeg lad 02deg3 It 1 534549 sh 05 V345595B5h 0 2 l 0 50 100 9 deg 10 a34559s 3 h 0 3910 I 0 50 100 9 deg sva Diagrams for Entire Cam 0 if Odeg S 6 lt estmti se Smoditrape esta i ses seshrise if estautgise g e lt esta irise Brise hrise if 9 sta irise Brise g e lt estai ifaill hrise 7 Smoditrape 7 estartifalls fallsh se if estalyfan S 9 ltesta ifa114r Bfan 0 if estartifall Bfall g e lt estartirise S3456e 7 esta in39seifallsB seifallshriseifall if estartiriseifall S 9 ltesta in39seifa114r B seifall 0 if 330deg S 6 3N 0deg1deg360deg 004 I I I I I I I 003 se 002 001 0 I I I I I I I 0 50 100 150 200 250 300 350 9 deg Maximum 39 quot will occur in risefall seqment 3A esta in39seifallsesta i seifall l39deg estautitiseifall Briseifall 01 0 a3455 9 9 startiriseifalls Briseifan hrise 01 02 I I 200 250 300 9 deg Peak 39 quot amplitude occurs where the ierk is zero 9 270deg initial guess MN emaxgcc r00tj3456e estartiriseifallsBriseifallsh see emaxiacc 265001 deg amax a3456emaxiacc estartiriseifalls Briseifallshn39sen amax 0388 ft 2 3 ft a m amax a 426gtlt 10 5e02 39 21bf 32174 ft 3664 g 2 sec W W m 0062 slug g ma F 2648321bf DESIGN OF MACHINERY SOLUTION MANUAL 12 41 g PROBLEM 124 Statement A threebladed ceiling fan has 1 5 ft by 025 ft equispaced rectangular blades that normally weigh 2 lb each Manufacturing tolerances will cause the blade weight to vary up to plus or minus 5 The mounting accuracy of the blades will vary the location of the CG versus the spin axis by plus or minus 10 of the blades39 diameters Calculate the weight of the largest steel counterweight needed at a 2in radius to statically balance the worstcase blade assembly Given Blade dimensions Length lb 1 l 5ft Width w b 025ft Nominal weight W Imam Z2vlbf Manufacturing tolerances Weight I w 005 CG offset tCG 010 Assumptions 1 The blades are held in place by a bracket such that their base is 6 in from the center of rotation making the tip 24 in from the center Thus the blade sweep diameter is 48 in 2 There is one heavy maximum weight blade at 0 deg and two light minimum weight ones at 120 and 240 deg respectively Thus W1lt1twgtvan0m W12100lbf r11l5vin 6110vdeg W21 twgtvan0m W21900 lbf r2115vin 621120vdeg W31 twgtvan0m W31900 lbf r3l5vin 63240vdeg Solution See Mathcad file P1204 There are two factors to be taken into account the variation in blade weight and the error or eccentricity in the location of the global CG The variation in blade weight about its spin axis will be considered first Resolve the position vectors into xy componenm in the arbitrary coordinate system associated with the freeze frame position of the linkage chosen for analysis R 1x1r1vcmlt6 1 R 1x15oooom R 1y1r1vsine 1 R 1y 0000 in R2x r2vcose 2 R2x 7500om R2y r2vsine 2 R2y 12990vm R 3x 7500 in R 3 l2990 in R 3x r3vcoslte 3 R 3y r3vSinlt6 3 y Solve equation l22c for the massradius product components 39W139R1x W239R2x Ws39R3xgt wax mbe3000vinvlb g W R W R W R mRb FM mRb 0000 invlb y g y Solve equations l22d and l22e for the position angle and massradius product required 0 b1atan2ltmR bx mR by 0 b l80000 deg 39 2 2 quot39Rbb ml be mRby mR bb 3000vinvlb 2nd Edition 1999 DESIGN OF MACHINERY SOLUTION MANUAL 1242 5 Now account for the fact that the blades spin axis can be eccentric from their CG Maximum eccentricity r e 48vinvt CG r e 4800 in mR product due to eccentricity WIW2W3 re mRbe mRbe 283200invlb g 6 Add the two mR producm and divide by the 2in radius specified for the counterweight to find the maximum weight required Rcw20vin mszmRbbmRbe mRb3l320 invlb mRb N mow 1566011 Row Note that 90 of the counterweight is required to balance the eccentricity The manufacturer would be well advised to try to control this variation more tightly 2nd Edition 1999 DESIGN OF MACHINERY SOLUTION MANUAL 581 5 PROBLEM 58 Statement Design a linkage to carry the body in Figure P5l through the two positions P1 andP2 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 ansz with respect toPl P 1x 0o P 1y 0o P2x 1236 P2y 2138 Angles made by the body in positions 1 and 2 0P11210vdeg 0P2Zl475deg Free choices for the WZ dyad z l075 52 270vdeg j 2044vdeg Free choices for the US dyad S l240 7 2 400vdeg w 740vdeg Two argument inverse tangent atan2xy I return 0571 x0 return atanltlt gtgt xgt0 x atanltlt gtgt 7 otherwise x Solution See Figure P5l and Mathcad file P0508 1 Note that this is a twoposition 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 53 will be used here 2 Define the position vectors R1 and R2 and the vector P21 using Figure 51 and equation 51 P Ix P 2x P 21x R12 R22 R2 R1 P21x1236 P 1y P 2y P 21y P 21y 2138 P 2 P 2 2 470 P21 21x 21y P21 3 From the trigonometric relationships given in Figure 51 determine 0L2 and a210p2 0P1 a262500 deg 5 2 atan2ltP 21xP21ygt 5 2 120033 deg 4 Solve for the WZ dyad using equations 58 Z1x1zvcos ZIx0979 Z1y1zvsin ZIy0444 2nd Edition 1999 DESIGN OF MACHINERY SOLUTION MANUAL 582 A cas5 2gt1 A0109 Dsinlta 2 D0887 B Sinlt5 2 B 0454 E p 21vcoslt5 2 E 1236 Ccosa 21 C0538 Fp 21vsinlt5 2 F2138 AvCVZ1xDZIyEgt BltCvzlyDZIxFgt W W 1462 Ix 2VA Ix AvCVZ1y DZ1xFgtBVltCVZ1x DZ1y E W1 W1 3367 y 2A y w W1x21W1y2 w3670 e atan2W1x W1ygt e 246528odeg 5 Solve for the US dyad using equations 512 S 1x1svcos4 S 1x 0342 S 1y S39Sin1 S 1y 1192 A coslty 2 1 A 0234 D Sinlta 2 D 0887 B sinlt7 2 B 0643 E p 21vcoslt5 2 E 1236 Ccoslta 2gt1 C0538 F1p21vsinlt5 2 F2138 AvCVS1xDSIyEgtBvltCVS1y DS1xFgt le 2A le3180 AvCvSIy DVS1xFgtBVltCVS1x DSIy Egt U1 U1 4439 y 2A y 2 2 u U1xU1y u5461 0 amn2U1x U1ygt 0 234381 odeg 6 Solve for links 3 and 1 using the vector definitions ofV and G Link 3 V1x1zvcos Svcosw VIx1321 V1y1zvsin Svsinw VIy1636 e 3 atan2ltV1xV1ygt e 3 231086odeg 2 2 v V1xVIygt v2103 Link 1 01x wvcos6 vvcoslte 3 mom 01x 0398 2nd Edition 1999 DESIGN OF MACHINERY SOLUTION MANUAL 583 G1y1wvsinevvsinlte 3gt uvsino GIy0564 e 1 atan2ltG1xGIygt e 1 54796 deg 2 2 g Glx 0 g0690 7 Determine the initial and final values of the input crank with respect to the vector G 02126 61 021301323 deg 02fi02i15 2 02f 274323 deg 8 Define the coupler point with respect to point A and the vector V rpz 5pi e3 rp 1075 5p26686 deg 9 Locate the fixed pivots in the global frame using the vector definitions in Figure 52 p 1 atan2ltP 1xP 1y p 1 90000 deg 39 2 2 R1 P1xP1y R1o000 O2xIR1vcosltp1gt zvcos wvcose 02x2441 02y1R1vsinp1gt zvsin wvsine 02y3811 04xR1vcosp1 Svcos1 uvcos0 04x2838 04y R Ivsinltp1gt S39Sin1 31110 04y 3247 10 Determine the rotation angle of the fourbar frame with respect to the global frame angle from the global X axis to the line 0204 0 m atan2 0 4x 02x 0 4y 02ygt 0 mt 54796 deg 11 Determine the Grashof condition ConditiorKSLPQ SLe SL PQq P Q return quotGrashofquot SLSPQ return quotnonGrashof otherwise ConditiorKgwu v quotGrashofquot 12 DESIGN SUMNIARY Link 2 w 3670 e 246528odeg Link 3 v 2103 e 3 2310860deg Link 4 u 5461 5 234381 deg 2nd Edition 1999 DESIGN OF MACHINERY Link 1 g 0690 Coupler rp 1075 Crank angles 02 301323 ndeg 0 2f 274323 odeg SOLUTION MANUAL 584 9 154796Ideg 5p 26686 deg 13 Draw the linkage using the link lengths fixed pivot positions and angles above to verify the design 2nd Edition 1999 DESIGN OF MACHINERY SOLUTION MANUAL 8251 g PROBLEM 825 Statement Write a computer program or use an equation solver such as Mathcad or TKSolver to calculate and plot the S v aj diagrams for a 4567 polynomial displacement cam function for any specified values of lift and duration Test it using a lift of 20 mm over an interval of 60 deg at l radsec Enter Lift 11 1 20mm Duration 5 60vdeg Solution See Mathcad file P0825 l The 4567 polynomial is defined in local coordinates by equation 821 Differentiate it to get v a and 362007 4 5 6 14203 1403 B B 6 B e 4 5 s6hv 35H 84 70 B 6 3 ve 3 1403 420 B B 6 3 4 5 21003 8403 B B 2 3 4 840039 420039 B B Plot the displacement velocity acceleration andjerk functions over the lift interval 6 Z0vdeg 05vdeg B 2 16 llv 4203 1680v 52 B 3 19 1lv 8403 5040 e B B B N 59 mm Displacement in 6 3g Cam Rotation Angle deg 2nd Edition 1999 DESIGN OF MACHINERY SOLUTION MANUAL 8252 VELOCITY v 60 40 e a V 393 mm e E 20 0 0 0 0 40 0 0 6 We Cam RotationAngle deg 200 100 E altegt g mm 0 E 8 v 100 200 6 We Cam RotationAngle deg 1000 500 5 19 23quot mm 0 39500 391000 6 3g Cam Rotation Angle deg 2nd Edition 1999 DESIGN OF MACHINERY SOLUTION MANUAL 12 71 g PROBLEM 127 Statement A wheel and tire assembly has been run at 100 rpm on a dynamic balancing machine as shown in Figure l2l 2 p 591 The force measured at the left bearing had a peak of 6 lb at a phase angle of 60 deg with respect to the zero reference angle on the tire The force measured at the right bearing had a peak of 4 lb at a phase angle of 150 deg with respect to the zero reference on the tire The center distance between the two bearings on the machine is 10 in The left edge of the wheel rim is 4 in from the centerline of the closest bearing The wheel is 7in wide at the rim Calculate the size and location with respect to the tire39s zero reference angle of balance weighs needed on each side of the rim to dynamically balance the tire assembly The wheel rim diameter is 15 in Units rpm 2vT3vradvmin1 Given Bearing forces and plane locations with respect to correction planeA Left F1 6vlbf 6 160vdeg 14vin Right F2 4be e 21150vdeg l2 6vin Distance between correction planes B 7 vin Correction weight radii RA 75vin RB 75vin Tire rotational speed a l 00 rpm Solution See Figure l2l 2 and Mathcad file P1207 l Resolve the force vectors into xy componenm with respect to the zero reference angle of the tire FIxF1vcoslte 1gt le 3000 lbf FlyF1vsinlte 1 Fly 5196 lbf F2x F2vcoslt6 2 F2x 3464 lbf F2y F2vsinlte 2 F2y 2000olbf 2 Solve equations l24e for summation of momenm about 0 modified for the bearing forces F vi F vl Ix I 2x 2 mRBx 2 mRBx 16489 mvlb 300 F vi F vl 1y I 2y 2 mRBy 2 WIRBy 16489 mvlb 300 3 Solve equations l22d and l22e for the position angle and massradius product required in plane B Also solve for the weight required at the given radius 03 atan2mRBx mRBygt QB 135000 deg 2 2 mRB mRBx mRBy mRB23319 mvlb quot39RB W3 g W3311 lbf RB 4 Solve equations l24c for forces inx andy directions in plane A 2nd Edition 1999 DESIGN OF MACHINERY SOLUTION MANUAL 12 72 F F Ix 2x mRAx 2 mRBx mRAx 14855 mvlb D F F 1y 2y 7 a WRAy 2 mRBy mRAy727742 ml 3 Solve equations 122d and 122e for the position angle and massradius product required in plane A 4 0A1atan2ltmRAxmRAygt 0A 61832 deg 2 2 mRA mRAx mRAy mRA 31469vmvlb mRAvg W4 W4 420 lbf RA 2nd Edition 1999 210 BoltCasm Devices Sensitive contact gage uses a Variabie eteetricai resistor Typewriter linkage has a Rocking mechanism derived rocking pair to decrease the has a rocking surface instead rocking pair that actuates up from 4bar linkage has con e ect of friction and 1n of a sliding brush to reduce crease the accuracy wear with smooth operation smoother quieter action per or lower case letters with stant spring length which transmits no force to bearing Type Iead fype lever r Sfop E H 39 9 Spring Roll com Keylever ox Sprmg 6 W Drive roll 2 Electrictypewriter mechanism uses roll com for motion ampli cation Here path of pivot point on roll com is curvilinear Roi cams are also employed in iBM electric typewrit ers Fig 2 Here the cam is triggered by touch of the typist s nger to power the type heads The roll is driven by an electric motor at constant speed Cycle begins when the tynist depresses the he iever which rotates the cam into contact with the drixr soil The can is connected 39 39 3 at Dir point A Rotation of drive to 39 i 39 in Sin 6 ts rotate quotte ciocknise caus tne type head facts the platen not shown At end of the cycle the camstrikes stop C and loses contact with drive roll Spring G then returns the cam to its position against the stop K During this time while cam and drive are disengaged the type head continues to approach the platen because of kinetic energy stored in type lever link 6 After type head strikes platen spring H returns the linkage to the home position where link 3 contacts stop DESIGN OF MACHINERY SOLUTION MANUAL 4211 g PROBLEM 421 Statement For the linkage in Figure P48 find its limit toggle positions in terms of the angle of link 02A referenced to the line of centers 0204 when driven from link 02A Then calculate and plot the xy coordinates of coupler pointP between those limim referenced to the line of centers 0204 Given Link lengths p Input 02A a 500vin Coupler AB 1 440vin Rocker 043 c 500vin Ground link d 950vin Coupler point data p 890vin 56vdeg Coordinate transformation angle at l 4vdeg Two argument inverse tangent atan2xy i return 057139 xl0 02 return atanltltzgtgt xgt0 x atanltltzgtgt 7 otherwise x Solution See Figure P48 and Mathcad file P0421 1 Define the coordinate systems The local frame has origin at 02 with the positive x axis going through 04 Let the global frame also have its origin at 02 with the positiveX axis to the right 2 Check the GTashof condition of the linkage ConditionSLP Q SL4 SL PQ4P Q return quotGrashofquot SLSPQ return quotnonG39rashof otherwise Conditionb d a c quotnonGrashof 3 Using equations 433 determine the crank angles relative to the line AD at which links 3 and 4 are in toggle ltagt2ltdgt2 ltbgt2 ltcgt2 bc 1 39 arg arg 1209 2nd ad 1 2 2 2 2 arg2 FW 50 arg2 0283 2001 ad 0 2mggle acosltarg2gt 0 2mggle 7360deg The other toggle angle is the negative of this 2nd Edition 1999 DESIGN OF MACHINERY SOLUTION MANUAL 4212 4 Define one cycle of the input crank between limit positions 6 2 39 0 2togglequot 0 2toggle 2 degquot 0 2toggle 5 Determine the values of the constants needed for finding 93 from equations 4llb and 412 2 2 2 2 K1i K4 i K5 F a b 2111 KIl9000 K421591 K524911 Dlte 2 coslte 2 K I K4vcoslte 2 K5 Elte 2 2vsinlte 2 Flte 2 K I ltK4 1gtvcoslte 2 K5 6 Use equation 413 to find values of 93 for the open circuit 9 39 2 2ltamn2lt2pe 2e 2 159 at 4Dlte Zwe 2 7 Use equations 427 to define the x and ycomponents of the vector RP RPRARPAI RA iavltcoslt6 2 j vsinlt9 2 RPA pvltcoslt6 35gtj vsine 315 RPxlt6 2 avcoslte 2 pvcoslte 3lt6 2 5 RPylt6 2 avsinlte 2 pvsinlte 3lte 2 5 8 Transform the coupler point coordinates in the local frame to the global frame using coordinate transformation equations XPlt6 2 RPxlte 2gtvcosa RPylte 2gtvsina YPlt6 2 RPxe 2 vsinaRPye 2 com 9 Plot the coordinates of the coupler point in the global system COUPLER POINT COORDINA YES Coupler Point Coordinates 9 2 72g CrankAngle deg 2nd Edition 1999 myrva PMS Iquot aNCWE 405W va 0F CAM SURFPK F Cow 1 Dws 4 L 0 1 012 4 a 9 WCWE f w a W I 3 gumvf v M f 0 gt WM 39 M i F16 quot 9 I M SURFKC Mylar BE M M F PermLYquot ll aveN Son x 3 N4 S LEcmeh if j v u T7 REWWf Unbalcum W45 W m THE P FME M i SuMYSELEcrEn A R39E fUI TI l 5510 PMcrlcc m AMA gt Z R Overt We CYCLE or 0 A e 4 36 3 3 PLOT w W643 U Mar cumea Kr A CIJFECK g P Ivemu an 0319 4 360 Jgar Wt FLH FRED Powwng 39 NW W7quot 2e A0 F a uefi w Sac39nor 8 6 541 1 viMS INFO Rama 9 CWW at cm SURFA CE atam F O Lawm 1 TCH Comf CW7M d CUTTER Pk r H CW Aaouub CAM I I TU Evmse CHANGE SIGN OF 4 OR 1 9 gt 3 963327 5 gt 3 All 5 lt RON w M 9 301 C6 39 1 tom r c w Se K i 1 W OZ 7 39 GETSX quot Rme 6 i 14a af ashgr 3 f Ep f bz 3 6 MW F 5quot 59 7 2422 1 RPSMWAB e quotam A m 5 QMQB ea 63 2 RN9 Co694 We Siw 63 a RMg5 Na 0 f Fm W I w w Ma i CUTTEK Q bulf Kata gash 9 Raw Rb lgve 7 l WW X40 W9 79 Rbffeb 0 M59quot 5 V 40 XCM a 379B Sam 4 gram 9 559 m XMLe anes Ma L56ltJ 30 GBM R659

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