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# ELEMENTS OF MECH DESIGN MANE 4030

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This 116 page Class Notes was uploaded by Hugh Wilkinson on Monday October 19, 2015. The Class Notes belongs to MANE 4030 at Rensselaer Polytechnic Institute taught by Lucy Zhang in Fall. Since its upload, it has received 22 views. For similar materials see /class/224906/mane-4030-rensselaer-polytechnic-institute in Mechanical and Aerospace Engineering at Rensselaer Polytechnic Institute.

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EMD Homework Solutions to Problems from Mechanical Design of Machine Elements and Machines by J aok A Collins prepared by Antoinette M Maniatty and Jing Lu 113 This appears to be exactly what some US rms do on a routine basis However if you think it is a solution to the ethical dilemma posed re examine Section 5b of the NSPE Code p 862 It states 77Engineers shall not offer give solicit or receive either directly or indirectly any contribution to in uence the award of a contract by public authority or which Clearly the use of a subcontractor in the proposed manner is indirectly giving the gift The practice is therefore not ethical Note this answer comes from Chapter 1 Reference 8 of the textbook 253 Based on the information given the rating numbers assigned to each of the eight rating factors might be Rating Factor Selected Rating Number RN 1 Accuracy of loads knowledge 0 2 Accuracy of stress calculation 0 3 Accuracy of strength knowledge 1 4 Need to conserve 4 5 Seriousness of failure consequences 3 6 Quality of manufacture 0 7 Operating conditions 0 8 Quality of inspectionmaintenance 1 From 2 85 t 00 l 43001 1 and since t Z 76 from 2 86 254 Based on the information given the rating factors assigned might be the numbers need not match exactly but should be close solution depends to some degree on interpre tation Rating Factor Selected Rating Number RN 1 Accuracy of loads knowledge 1 2 Accuracy of stress calculation 0 3 Accuracy of strength knowledge 0 4 Need to conserve material7 weight7 space7 dollars 3 5 Seriousness of failure consequences 4 6 Quality of manufacture 0 7 Operating conditions 0 8 Quality of inspectionmaintenance 0 total t 0 Using Eqn 2 86 then 2 10 0 nd 1 100 gt 2 256 The failure mode to be investigated is probably yield based on the information given From Table 357 we obtain for Stainless steel alloy AM 3507 at 800 F7 the yield strength is Syp 1197 000 psi Since the safety factor is nd 187 then the design stress must satisfy Syp 7 119000 psi gt 0d 7 0d 71d 18 which means ad lt 66111 psi The predicted stress for the design is calculated using P i 10 0001b 0d 2 A 77 lt 66111psi Thus7 d gt 0439 in 263 The design life for the component is to be 107 cycles The mean and standard deviation of the fatigue strength for the component material at 107 cycles of loading are given as g 687 000 psi and 65 2500 psi The mean and standard deviation of the operating fatigue stress level are given as La 607 000 psi and 6 57 000 psi7 respectively Failure is predicted if the stress is greater than the strength Thus7 we want to know the probability Py S 7 039 gt 0 The mean and standard deviation of random variable y is 0y is 7 0a 68 000 psi 7 60000 psi 8 000 psi fry iamp amp 57590 psi The critical value of y 0 the point when the stress equals the strength and the onset of failure is predicted is X pg 143 standard deviations away from the mean 1 From Table 297 we nd that the reliability is then 9236 264 From Table 297 for the reliability to be 0999997 X 427 The mean and standard deviation of the stress level are given as La 345 MPa and 6 28 Mli a7 respectively The standard deviation of the fatigue strength is given as 65 20 MPa We are interested in ensuring that the probability of y S 7 039 gt 0 is at least 099999 The standard deviation ofy is Q mi 03 344 MPa y si a si345MPa and the mean of y is For a reliability of at least 0999997 we need y to be at least X 427 standard deviations to the right of the limiting value7 y 0 Thus7 y 2 427344 MPa 147 MPa So nally7 using the equation above7 the requires that s Z 492 psi 269 a Using Eqn 2 1017 for a system having 3 components in series7 each with a reliability Of 090 Rs 093 0729 Thus the system reliability would be 729 b Adding a duplicate redundant system7 then we use Eqn 2 102 to obtain the system reliability R5 1 i 1 i 07292 0927 Thus7 the system reliability would be 927 if a redundant system were added This gives a signi cant improvement in reliability about 20 higher c Adding a third redundant system7 following the same procedure as in part b we obtain a reliability of 987 which is a 5 increase from the 2 redundant system case So we see the improvement in reliability diminishes fast as more redundant subsystems are added d Adding redundant subsystems increases cost weight and space 6 10 Since the sheave must rotate in a stable manner on the shaft at relatively high speeds we pick RC5 from Table 64 then check Table 65 in row 197 315 in if approximate 50 mm as 2 inches for hole 18 0 for shaft 25 37 So the hole size 200000001820018 largest 20000020000 smallest for shaft size 20000 0002519975 largest 20000 0003719963 smallest lf take 50mm 19685in then using row 119 197 in for hole 16 0 for shaft 2 3 So the hole size 196850001619701 largest 19685019685 smallest for shaft size 19685 000219665 largest 19685 000319655 smallest 6 11 For a medium drive t we need FN 2 class t Table 66 From Table 67 we nd for a nominal size of 25 in the limits on the sleeve outer diameter and on the housing bore diameter are 25027 i dS 1n sleeve 25020 25000 1 d 7 25012 in hole That is the sleeve outer diameter must be between 25020 and 25027 inches and the hous ing bore diameter must be between 25000 and 25012 inches Note for the shaft we put the larger number on top because in the metal removal process to get the shaft down to the desired diameter we go from larger to smaller For the hole the smaller number is on top for a similar reason during machining the hole goes from a smaller to a larger diameter The limits of interference are 00008 to 00027 in EMD homework solutions1 Due 23 1 13 2 53 2 63 2 69 6 10 2 15 2 18 2 44 113 This appears to be exactly what some US rms do on a routine basis However if you think it is a solution to the ethical dilemma posed re examine Section 5b of the NSPE Code p 862 It states 77Engineers shall not offer give solicit or receive either directly or indirectly any contribution to in uence the award of a contract by public authority or which Clearly the use of a subcontractor in the proposed manner is indirectly giving the gift The practice is therefore not ethical 253 Based on the information given the rating numbers assigned to each of the eight rating factors might be Rating Factor Selected Rating Number RN 1 Accuracy of loads knowledge 0 2 Accuracy of stress calculation 0 3 Accuracy of strength knowledge 1 4 Need to conserve 4 5 Seriousness of failure consequences 3 6 Quality of manufacture 0 7 Operating conditions 0 8 Quality of inspectionmaintenance 1 From 2 85 t 00 1 43001 1 and since It 2 76 from 2 86 263 The design life for the component is to be 107 cycles The mean and standard deviation of the fatigue strength for the component material at 107 cycles of loading are given as g 68000 psi and 65 2500 psi The mean and standard deviation of the operating fatigue stress level are given as t 60 000 psi and if 5 000 psi respectively Failure is 1 predicted if the stress is greater than the strength Thus7 we want to know the probability Py S 7 039 gt 0 The mean and standard deviation of random variable y is y is 7 La 68 000 psi 7 600 psi 8 000 psi fry i 559O psi The critical value of y 0 the point when the stress equals the strength and the onset of failure is predicted is X pg 143 standard deviations away from the mean From Table 297 we nd that the reliability is then 9236 269 a Using Eqn 2 1017 for a system having 3 components in series7 each with a reliability Of 090 R5 093 0729 Thus the system reliability would be 729 b Adding a duplicate redundant system7 then we use Eqn 2 104 to obtain the system reliability R5 1 i 1 i 07292 0927 Thus7 the system reliability would be 927 if a redundant system were added This gives a signi cant improvement in reliability about 20 higher c Adding a third redundant system7 following the same procedure as in part b we obtain a reliability of 987 which is a 5 increase from the 2 redundant system case So we see the improvement in reliability diminishes fast as more redundant subsystems are added d Adding redundant subsystems increases cost7 weight7 and space 610 Since the sheave must rotate in a stable manner on the shaft at relatively high speeds7 we pick R05 from Table 647 then check Table 657 in row 197 315 in if approximate 50 mm as 2 inches7 for hole 187 0 for shaft 257 37 So the hole size 200000001820018 largest 20000020000 smallest for shaft size 20000 0002519975 largest 20000 0003719963 smallest lf take 50mm 19685in7 then using row 119 197 in7 for hole 16 07 for shaft 27 3 So the hole size 196850001619701 largest 19685019685 smallest for shaft size 19685 000219665 largest 19685 000319655 smallest 215 a In the linear elastic regime7 we see that at a load of 16000 lb7 the elongation is 002 in The cross sectional area of the bar tested is A ww 1227 2712 and the length is given as 10 inches Thus7 at a stress of 039 FA 167 000 lb1227 inz 137 038 psi7 the bar sustains a strain of E 6l 002 m10 m 0002 The elastic modulus of the material can then be estimated to be E 06 652 gtlt 106 psi Looking at Table 39 also given on inside cover of the book7 we see that the modulus is closest to magnesium7 of the materials we have to choose from Thus7 the material is probably magnesium b For a load of 8000 lb on a rod of this material with a diameter of 1128 in7 or cross sectional area of A WW 100 0712 the stress will be 039 FA 87 000 psi7 and the strain will be 6 UE 8000652 gtlt 106 0001227 For a 7 foot 84 inch length7 the elongation will be 6 El 0001227 84 in 0103 in7 which is higher than the allowable limit of 0040 inch Thus7 the material for this application should not be approved 218 The temperature difference between members B and E will cause these members to have different lengths with the difference being AL LE 7 LB This difference in lengths will cause the line of sight to be misoriented relative to the desired angle 7 and will result in an error7 AB in radar tracking at a distance of 40000 ft that is related to AL as AL AR 15m 40000 ft Thus7 40000 ftaA920m AR 15 m where A9 50 F here and the temperature range to be considered is between 150 F and 200 F Using Table 38 we nd a For steel members 04 63 gtlt 10 6 F thus AR 168 ft b For aluminum members 044 119 gtlt 10 6 F thus AR 317 ft c For magnesium members ong 160 X 10 6 F thus AR 426 ft 2 44 a 73 dm W L 21X10 x70 N L 00015771 gsypA 9275 gtlt 106 Nm20025 m0013 m a L 821 m t L5 821 m 348 mm 058 hr 30revmm 7T gtlt 025mrev b Maintenance every hr is too frequent not acceptable c Using Table 27 ratio of k of excellent lubrication over k of unlubricated is 10 7 k u 2 gtlt 10 6 72x10 5g gidixio 4 5 gtlt 10 3 km 5 gtlt 10 3 Since service time is proportional to 1k this means a time increase by a factor of 2500 to 50000 meaning 60 days or more Much more acceptable d Select a better combination of materials From Table 27 we see that non metal on metal offers roughly a thousand times increase in wear duration EMD homework solutions2 Due 210 3 5 4 8 4 25 5 6 5 71 35 Using the speci cations described7 the following application requirement table Table 1 based on Tables 31 and 32 in book could be lled in Material data for these evaluation Table 1 Material Speci c Application Requirements and Corresponding Performance In dices Application Requirement Evaluation Index 1 Strengthvolume ratio Ulitmate or yield strength 2 Stiffness Modulus of elasticity 3 Ductility Percent elongation in 2 inches 4 Ability to store energy elastically Energyunit volume at yield 5 Ability to dissipate energy plastically Energyunit volume at rupture 6 Resistance to chemically reactive environment Dimensional loss in environment 7 Cost constraints Costunit weight also machinability indices may be found in Tables 337 397 3107 3117 3127 3147 3187 and 319 Making a short list of candidate materials from each of these tables7 we obtain Table 2 Surveying the lists in Table 27 we see that no material is common to all requirements However7 except for corrosion resistance and ductility7 ultra high strength steel shows high ratings7 and carbon steel also has good ratings Corrosion resistant platings could be used for either Therefore7 it is recommended that plated ultra high strength steel be selected 48 a Here7 it is asking only for the shear stress due to torsion Since this shaft presumably is operating at a constant angular velocity7 the torques resulting from the loads applied to the two elements need to balance We can compute the resulting torque on the larger element T 1200 N i 400 N012 m 96 N m The smaller element then generates an equal and opposite torque7 thus FH006 m 96 N m a FH 1600 N So there is a constant torque of 96 N m applied to the shaft between the two elements This Table 2 Candidate materials by evaluation index able carbon able copper7 able 31 able 311 able able 31 cost gray cast able 31 7 7 able low carbon medium carbon generates a shear stress that is maximum on the shaft surface and is equal to Ta Tg 16T 1696 N m 7 7 764MP T J 312m st 7r004 m3 a b To nd the maximum tensile bending stress7 we need to determine the maximum bending moment Since there are forces in each of the 2 transverse directions7 we need to consider the resulting moments in each plane If we let the x axis be aligned with the shaft7 and then let the y axis be the vertical axis and the z axis be the horizontal axis in the gure on the right in Fig 13487 then we have the following 2 free body diagrams in each plane see Fig 1 From force and moment equilibrium in each plane7 it is straight forward to determine RAy 4173 N7 RBy 71973 N7 RAZ 713867 N7 and R31 718133 N The shear and moment diagrams can then be constructed and are shown in Fig 2 It is clear that the maximum magnitude of the bending moment is at 0 since this is where the bending moment is largest in magnitude in each plane Thus the overall maximum bending moment is Mmam 277342 83462 2896 N m The maximum tensile stress7 which will occur on the surface of the shaft at axial position 800 N 02m 028m 012m y 580 N Ray Vy N A 4173 M N39m 2370 A c 0 2 048 0 6 8346 RAZ 1600 N 1600 N R52 x m Figure 2 4 8 Shear and moment diagrams 5m M5 r I RA R5 Figure 3 4 25 Free body diagram 0 is then MC Mg 32M 322896 N m 77 461MP U I 614m 7rD3 7r004 m3 a 425 A free body diagram is shown in Fig 3 We have only two non trivial equilibrium equations 2 Fy 0 and 2 M2 0 assumingy is downward and z is into the plane and three unknown reactions BA BB and MB thus the reactions cannot be solved just from the equilibrium equations and this problem is statically indeterminate We have one more supporting force or moment than necessary to prevent the shaft from moving Using the singularity function handout I can write an equation for the downward de ection 1 dzv w 2 7E La 7 5s where I dont need to use the singularity function brackets because both terms kick in right at the start of the beam x 0 Integrating twice d1 RA w iEli 7 277 3 0 dz 2 z 695 1 R w 7E11 fsgi x lJrClanCz Note after integrating twice I now have 3 unknowns RA and the two constants of integration Cl and 02 I can solve for these 3 unknowns using the boundary conditions on the de ections and slope 00 0 ML 0 and L 0 where L 5 m Using these boundary conditions in the above equations7 yields 7EIv0 020 RA 3 w 4 7E L 7L 77L OL0 M 6 24 1 d1 7 RA 2 w 3 i 7EIL 7 7L 73L 0170 a RA wL9375N 637ng 8 48 So the de ection and slope equations for the beam are then Ms K12 7ng 2x4 L d1 7 w 2 3 3 i M 79Lx 8x L The maximum de ection is when the slope is zero and 0 lt z lt L For the slope to be zero7 the following must be zero 79m 8x3 L3 0 and the only solution to the above equation on the interval of interest is z 04216L 211 m Now7 we can use the de ection equation Ms to nd the de ection at this location Since the shaft is steel7 we can assume Table 39 E 207 GPa 5000 Nm 48207 gtlt109 Nm26i1012 m4 803 gtlt 10 3 m 803 mm 735 m211 m3 2211m4 5 m3211 m ZLmam Note7 this problem can also be solved using superposition by adding the solutions in Table 41 for cases 10 and 11 56 We see that the shaft is subjected to a combination of torsion and bending The torsion creates a shear stress that is maximum on the surface of the shaft and will be the same everywhere on the shaft surface7 which is T1 T E 2 16T 165007 000 N mm 3979 MPa J d 7Td3 7T40 mm3 The bending moment is due to the transverse loads From symmetry7 the reactions at A and B are just RA 57 000 N and RB 57 000 N The shear and moment diagrams are shown in V2 5000 N 015 030 x m 5000 My Nm x In 750 Figure 4 5 6 Shear and moment diagrams Fig 4 So the maximum bending moment is 750 N m or 750000 N mm7 and occurs at the middle of the shaft x 150 Thus7 the maximum tensile bending stress will be at the middle of the shaft x 150 mm on the bottom surface 2 720 mm7 and is Aiwgc wag 32Ag 732U50000N4nm m 444444444444471193704P U I d 7Td3 7T40 mm3 a The critical location is where this maximum bending stress acts x 150 mm7 y 0 mm7 z 720 mm7 because the shear stress due to the torsion is the same everywhere on the shaft surface A stress element at this critical location is shown in Fig 57 where Tm 739 3979 MPa and Um 11937 MPa Note that we only need to consider 2D because it is a free surface7 thus the normal and shear stresses on the face are zero7 ie Uz 07 739 07 and Tyz 0 The principal stresses can be determined using the Mohr7s circle diagram7 shown in Fig 6 The principal stresses are then Figure 5 56 Stress element at critical location Figure 6 576 Mohr s circle diagram for critical stress element R 1193722 39792 71732 MPa 11937 01 2 R 59685 71732 13142 MPa 02 0 11937 03 T i R 59685 7 71732 71205 MPa and the maximum shear stress is Tm 7173 MPa 571 Using Hooke7s law 1 6000 ex EUm m 20 gtlt 10 4 y 036000 5 6y i am 7W 760 gtlt 10 y 036000 5 61 i am 7m 760 gtlt 10 em 6 ez 20 i 06 i 06 gtlt10 4 80 gtlt 10 5 EMD homework solutions3 Due 217 5 20 4 32 4 35 4 36 520 First7 compute the torque Let H represent the power 67 600 m 7 lbs hp 42 hpgtlt 2 d 39 27720271 7 lbs T0 T3400 revmm W m gt T51 5 T 7785271 7 lb a Using Table 457 case 3 with b a for square cross section 814 7 1667 3 Q 480 a for the hollow square section 125 3 100 3 Q ther 7 Qinne r 7 2713 Note that this gives quite a different result from that obtained if case 5 was chosen thin hollow tube The reason is because for a square or angular cross section7 the assumption of uniform stress in the wall of the hollow tube used in case 5 is not a good assumption So rather than getting 7mm you would get Tam Using 4 42 7 T 7 77852717 lb Wife 3920 39 T Q 019862713 pm From Table 337 Syp 357 000psi Assuming the yield stress in shear is about half this value7 Typ 177 5001052 The existing safety factor is 7 21 7 w i 7 446 n 3 TM 3920 b Using case 3 of Table 457 for square section 16 1 K a4 E i 336 17 225304 for the hollow square 4 4 125 l K Km 7 2253 4 5 l 02032714 1 Using 4 447 also get G 106 gtlt 106psi from Table 3 9 TL 7785271 71b20m 9 7 000724 d 041 KG 0203 m4106 gtlt 106 lipm m 432 A free body diagram is shown in Figure 1 13 kN M Figure 1 Force and moment on section A A From force balance P 13 kN Assume the reaction force and moment act through the centroid of the cross section this is not exactly correct7 but since the centroid is usually quite close to the neutral surface7 it is a good approximation for computing the moment Find out the location of the centroid To The reference point is at the center of curvature7 and n 38 mm7 r0 66 mm Arc ZAJrcj 739 10 gtlt 2510 gtlt18rc 10 gtlt 25 gtlt 38510 gtlt 18gtlt 38109 i To 488605 mm i M 13700090 488605 1805 gtlt 106 N mm Check Table 337 ultimate strength in tension is 345 Mll a7 in compression is 1172 MPa Thus7 easier to fail in tension at the inner surface Using Table 43 case 5 A 10gtlt2510gtlt18430mm2 dA Tih1 To 7 bl 7 bl 7 T 1 H Ti 2 nTih1 3810 25 l 7 10 I 902491 n 38 H n3810 mm A Tn 7476459771771 6 rc7rn1215 mm 01 rn7n9646 mm 00 r0 7 Tn 183541 mm M F 1805 106 9646 13000 gt C i w 7 gemmm2 907MPa eAn A 121543038 430 M 0 F 1805 106 183541 13000 U0 1 C 7 X x 77931MPa eAro A 121543066 430 Comparing with the ultimate tensile strength7 the part is predicted to fail at the inner surface 435 The contact force is just the weight of the steel sphere A181 1020 steel is low carbon steel see Table 337 and its weight density Table 34 is w 7681 kNms Thus7 the contact force is just 4 F wV w gwrg where r is the radius of the sphere Using Eqn 4 677 we nd the size ofthe circular contact area for a sphere on at surface7 with both materials being the same to be Q 33F217EI273 liyz sag 4Fd E w grins 2r 3 27Twr4 1 22 where7 for steel7 V 03 and E 207 GPa Table 39 The maximum interface pressure at i 3 gt4gtlf40 the point of contact is 3F 7 47T LUT3 7 2w13r13 W 2w 2m ltgt23r83 2w 12 pmaz Let7s assume yielding occurs when the maximum shear stress equals half the yield stress The maximum shear stress in terms of 10mm can be found using Fig 416 T771111 For yield to occur Tm ZIP2 For 1020 HR steel Syp 207 MPa Table 33 Thus yield is predicted when 0322w13r13 7 032276810 Nm313r13 V 23 23 2wltl 27 Tmaz SCIZ7 7W 2977 gtlt 108 Nm73r13 7 1035 x 106 Nm2 r0042m42mm So the allegation is true and the sphere need not be all that big to induce yielding at the location of maximum shear stress slightly below the surface If we only considered yielding at the surface then from Fig 4 16 7mm 01pmm Using this in the equations above instead results in a sphere of much larger radius r 138 m to cause yielding at the surface of the sphere 436 a Two spur gears in contact may be approximated as two cylinders in contact Assuming both gears to be steel V1 V2 V 030 E1 E2 E 207 GPa The contact width is b 4F1 V2 4gt180 NW 03932 00235 mm 7TEL 711 207 gtlt 109P0039025 m 00214m 00l2m 2F 7 2180 N W77 195MP p m 7r00235 mm25 mm a 0 Looking at Figure 418 and reading in text7 7mm 0310lem 585 MPa d 075 00176 mm 22 EMD homework solutions4 Due 33 4 19 4 26 4 34 5 4 modi ed 5 84 a 419 The square bar is subjected to bending and torsion while the round one is subjected to only bending Here7 we use subscript 1 to refer to the square bar and subscript 2 to refer to the round bar Let 1 be distance from the end of the square bar to a point in the square bar and x2 be the distance from the end of the round bar to a point on the bar The total strain energy of the bracket is then L1 M2 L1 T2 L2 M2 7 1 2 0 21711 d 0 2KG d 0 2E12 d where M1 Pil M2 Pig T PLZ I i 54 702034539 2 1 7 127 m d4 12 7T7011984239714 64 s 4 16 11 125m 4 4 K i i 34336 7 22533703438m For steel7 E 30 gtlt 106 psi and G 115 X 1061051 Differentiating with respect to the load7 P7 which is at the location and in the direction where we want to compute the de ection results in 6U 1 L1 2 1 L1 2 1 L2 2 yo mO 2Px1dx1mO 2PL2d1mO 2Pz2d2 PL PL PLng 3ELL BEIZ KG 1000 lb10 m3 1000 lb5 m3 330 gtlt 106 ps 020345 M 330 gtlt 106 ps 011984 M 1000 lb10 m5 m Jr115 gtlt 106 ps 03438 m4 00546 m001159m06323 m 013m Note one may also choose to make the cut breaking up the two members at the location where the round bar intersects with the square bar This would change the answer some because instead of integrating the last integral above from 0 to L2 it would become 0 to L2 7 52 Neither is exactly correct but the above is a bit more conservative and includes the de ection where the round bar connects to the square bar due to twisting of the square bar Remember that these equations assume long thin members and the round bar is not so long relative to its diameter so this is a rough approximation in either case 426 A free body diagram is shown in Fig 1 Note the reactions at the supports RA RB and MB depend on the load P which is emphasized in the gure The reactions are whatever they need to be to constrain motion of the beam under the given loading condition The load P does not depend on the reactions Thus when we write the strain energy to nd the reaction RA we do not try to write P in terms RA On the other hand in part b below where we solve for the de ection at the applied load P we do write RA in terms of P yap RMP RBP Figure 1 Free body diagram for 4 26 a To nd the reaction at the support using Castigliano s method we will write the strain energy in terms of RA and then differentiate with respect to RA which gives us the de ection at the location where RA is applied which we know is 0 The beam is only subjected to bending The moment along the beam can be expressed as 2 equations M x RAx forO xSa M2x RAxiPQia foragmgL Note we don t use singularity functions here to express the moment as 1 equation but rather we write it out in separate equations for the different segments of the member Singularity functions complicate the integral so when we use Castigliano s method we don t use them The derivatives of the moments with respect to RA are then 6M1 7 QRA 7 6M2 7 613A x The derivative of the strain energy is then 6U 1 1 6M1 L 6M2 76 0 7 Mid M d am A E 1613A xa 2613A 9 1 a L 7 Ptsz dx Ptsz 7 P2 7 ax dx E 0 a i 1 RIMS 1 RAz3 Pz3 Pazz L 7 E1 3 0 3 3 2 a i i RAL3 7 PL3 PaLZ 7 1 i E 3 3 2 6 Solving for RA in terms of P yields Plugging in the numbers listed in part b7 P 400 lb7 a 4 ft7 and L 10 ft7 we get RA 1728 lb b Putting the relationship for RA in terms of P back into the moment equations we get 3a a3 M1x Plt17 x forO xSa 3a a3 13 3a M2x Plt17 z7Pz7aPm7 zPa foragng Letting c 17 37 and d c 71 2 17 7 37 then we get the moments and derivatives of the moments with respect to P as M1x Pcz 7 661 oz for 0 S x S a 3M M2x PdPa 7 6P2da forangL The de ection at the applied load P in the direction of P is then QM6M1 d LM6M2d 0 1 6P 95 a 2 6P 95 a L Pczx2 dm Pd 12 dz 0 a a 1 L 02x2 dx 8 d2 a2d dx 0 1 3a 3L 629 M 3 0 d 3 2 13 1 3 3 c dLa daa 6U Wm l l l E EN EN Elw Eln Eln 3 21 c EchatcSaB Substituting the numbers given in the problem statement P 400 lb E 30 gtlt 106 psi I 100 in4 1 4 ft 48 in L 10 ft 120 in we obtain 6 226 gtlt10 3in 434 a From the free body diagram below we can nd M V and N in terms of P M PRcos be 7 cos 9 V P sin 6 N P cos 6 b We can nd the de ection of point A using Castigliano7s method by taking the derivative of the strain energy with respect to P for the upper half of the member AC see Fig 13434 in text Then from symmetry we know that the de ection of point D will be the same in the opposite direction thus the total change in separation distance between A and D will just be twice the de ection of point A We are given R 1 in and h 02 in Thus Rh 5 which is less than 10 the rule of thumb I gave you in class for when the simpli ed form of Castigliano7s method could be used with good con dence for computing de ections in curved members We will proceed with the simpli ed form anyway but realize that this Figure 2 Free body diagram for 4 34 is an approximation 1 MaM SAD N 2 bo iapptdo 2 7r E to PR3cos be 7 cos 92 d0 213R1 1 1 1 E Ocos 072cos 0cos0 cos20d0 2PR3 2 i 6 1 i quot m E cos 1506 7 2 cos be s1n0 E 1 sin 20 0 3 21 cos2 be 7 be 2 sin 2 where I used the following trigonometric identities in the above calculations cos26 1 cos 26 and 2cos be sin be sin 2 Now7 if I plug in the rest of the numbers given in the problem P 10 lb7 E 30 gtlt 106 psi7 I 04 m02 m3 2667 gtlt 10427147 and be 10 rad7 I get SAD 00115 in The error in using the simpli ed form is approximately hR2 004 or about 4 If we wanted a more accurate estimate7 we would need to use the form given in class c The graph is shown in below 54 Modi ed from book as follows Compute all the stresses acting on the in nitesimal Elmz DEM 7 7 EIEIEIB 7 7 displacement m BEDS 7 El EIEI7 wall llm lxnc I 5 mm Figure 4 Schematic for Problem 5 4 Referring to Fig 4 at the support where elements A and B are7 the applied load will cause a torque about 7x a bending moment about 72 and a shear force in the 7y direction with magnitudes as follows lTl 700 N200 mm 140000 N mm lel 700 N325 mm 227500 N mm W 700 N Since A and B are elements on free surfaces no contact we know that there are no stresses on the free surfaces and so we only need to consider 2 D planar stress elements at these locations Due to the loading we can immediately determine what the stress states on the elements should look like which is shown in Fig 5 where 017 is the bending stress 7quot is the torsional shear stress and TV is the direct shear stress where the stresses act in the directions the arrows are drawn For the bending stress we see that the member is bending down under the load and thus there will be tension along the axis of the member on top at A and compression on the bottom and no bending stress at the neutral surface y 0 coinciding with B For the torsion induced shear stress it will be the same everywhere on the surface of the member and we get the direction based on the direction the torsion is twisting the member For the direct shear the shear stress will be zero on the top at A and bottom of the member and will be maximum at the center y 0 B and the shear stress acts in the direction of the shearing force Now we can compute the stresses The bent 17 gt12TV x y b b a lt A gt0 B Z x 17 lt film Figure 5 Stress elements for Problem 5 4 member has a hollow cylindrical cross section with the outer diameter d 30 mm and wall thickness of 5 mm thus the inner diameter is d 20 mm On the surface of the member the shear stress due to the torsion will be the same everywhere with magnitude Tc 7 T i 16Td J 014 7 d W e d 16140 000 N mm30 mm W 30 mm4 7 20 mm4 32399 Mpa A normal stress distribution acting along the axis ofthe member through the cross section will result from the bending moment 7 217 7 2y i 7634sz I wed de ed 64227 500 N mmy W 7130Nmm y Um where the minus sign cancels out because the moment M2 is acting in the 72 direction As mentioned before you can see that since the member will bend down the top of the shaft y gt 0 will be in tension and the underside of the member will be in compression y lt 0 so from intuition you might also see why the signs are the way they are Thus at A y d2 we have ab a 7130Nmm315 mm 1070 MPa Finally the maximum shear stress due to V which acts at y 0 coinciding with the stress element at B is computed using Table 44 TV 2K 8V 8700 N A 7Td2 7 d 7T 30 mm2 20 mm 36 MPa So on element A we have Um 1070 MPa and 739 7329 MPa with all other components zero On element B we have Tm 293 MPa with all other components zero To check for yielding let7s rst verify that the material is ductile In Table 310 we nd for Aluminum 2024 T3 e 22 thus it is ductile We can apply the distortion energy theory maximum shear stress theory OK too but would require extra step of nding principal stresses Element A will be the critical location because it has both a high bending stress and torsion shear stress Finding the equivalent uniaxial stress U U 3731 1212 MPa From Table 33 the yield strength of Aluminum 2024 T3 is Syp 345 MPa Thus yield is not predicted The safety factor is then S n 285 Tag 584a For the static conditions given under combined loads of axial tension and torsional shear the critical point will be at the root of the 3 mm radius llet Since 6 lt 05 the material is brittle so we need to use the maximum normal stress theory and we need to apply the full theoretical stress concentration factors Stress concentration factors must be separately determined for the tensile load and the torsional load Using Figure 54b for the axial load with 3 3 006 d 50 and B 112 d 50 The value of Ktp may be read as Ktp 18 Using Figure 54c for the torsional load the value of KtT may be read as KQT 13 Actual stresses at the root of the llet then are U K9 g Mp and Try KtT lt For this stress state we can easily use a Mohr7s circle analysis to determine the principal stresses We nd Um Um 2 2 01 3 Tmy253 MPa 0 0m 0m 2 03 37 T y747MPa 1393 162048 N m M005 mg 1085 MPa Since 01 gt lagl and since brittle materials are stronger in compression than tension we only need to check that 01 lt Sm to avoid failure Since 01 253 MPa lt Sm 414 MPa then failure is not predicted and the arm should be able to support the load EMD homework solutions5 Due 310 5 9 5 28 5 37 5 40 5 42 59 The beam is loaded in bending which produces an axial compressive stress on the top and tensile stress on the bottom Failure due to yielding on the bottom of the beam or by crack propagation from the crack emanating from the lower hole should be checked From Table 52 Syp 1570 MPa K10 62 MPm First we compute the maximum tensile stress on the bottom of the beam 01 and the nominal tensile stress at the hole 00 Note we are taking an average value of the linearly varying nominal stress which is a reasonable approximation since the hole and crack are small also note that for the crack emanating from a hole we use the nominal far eld stress as illustrated in Figure 5 21 in the stress intensity factor K1 calculation ie we compute the stress as though the hole and crack are not there ignoring the stress concentration at the hole which is already accounted for in CI The stresses are then BL RR 225 kN M 225 kN125 m 28125 kN m on central part of beam 28125 kN0125 m 675 MP Ul 004 m025 m312 a U0 28125 kN01 m 540 Mpa 004 m025 m312 From this we see that yielding is not an issue factor of safety guarding against yield is my 1570675 23 Now to check crack propagation First we need to check if we can assume plane strain 62 MPm 2 1570 MP0 gt 00039771 39mm lt 4cm Bmm 25 Therefore plane strain OK Situation is similar to Figure 521 with A 0 a 15 mm B 15 mm Therefore 01 F0 14 Computing K1 K1 OIUhxTl 14540 MPa 7T00015 m 519 MPaxR So the factor of safety guarding against fracture is no 62519 119 which is OK for approving this one time emergency use 528 Given Thumbnail crack with 20 016 in a 005 in like in Fig 522 in member 1 A in Figure P528 At maximum thrust load P 18000 lb Material is AA 70757T6 From Table 52 Syp 75 ksi K16 26 From free body diagram of member B see Fig 1 we nd the tensile force on member A in terms of P P15 m 7 FAsin45 10in 0 7 FA 2121P T H a lo Ru Figure 1 Free body diagram of member B for problem 528 The nominal stress in member A at maximum load is then 7 FA 7 2121P 7 2 7 c7 7 j 7 7 3635 m P 7 654 ks Checking yield rst we nd the safety factor guarding against yield at maximum load is S 75k39 ny i 81 c7 654 km 13915 lt12 So at maximum load the safety factor for yielding is below the allowable Thus we see that to satisfy the 12 safety factor requirement against yielding 0 lt Syp12 625 ksi gt P lt 17190 lb or the thrust should be limited to 955 of capacity Now to check for crack propagation First checking to see if plane strain assumptions hold we compute the minimum thickness for plane strain 2 2 25 25 030mltB0312m Syp 75 ks Thus plane strain assumption is OK The limiting stress intensity factor using our safety factor of 12 is KIC12 2167 Now we compute the stress intensity factor 719y 171y 112 083 and using 120 03125 we nd the aw shape parameter 2 from the graph in Fig 522 to be Q m 145 So the stress intensity factor is 112 K 7 039V 7Ta i 039 lt 5878 ksz 112 Tl74005 m lt 2167 ksz39m 145 This stress is less than the stress we used in our initial guess based on avoiding yield7 039 625 ksi7 Thus7 we have a new USyp 587875 0787 which does not change the aw shape parameter Q from Fig 522 considerably7 so we7ll leave this stress as the nal answer Thus7 P lt 16172 lb7 or the thrust should be limited to 898 capacity in order to ensure a safety factor of 12 guarding against fracture 537 a From Tabel 337 the properties of A181 1020 CD steel are S 421 MPa Using method on P247 S V Sn 421 MPa at N 1 cycle 8 055 211 MPa at N 106 cycles since SM lt 1379 MPa 200 ksi The resulting plot is shown in Figure 2 N jzd Figure 2 S N gure for 5 37 b From the plot7 ng os 211 MPa This is from the estimated gure c From Figure 531 132467 5106 37 ksi 255 MPa This is from actual test data and is about 20 higher than the estimated value Thus7 the estimate is conservative in this case 540 a From Tabel 337 the properties of A181 1060 HR steel are S 987 000 psi7 Syp 54 000 psi Using approximation on P247 5 m 055 49 000 psi Now we apply modifying factors to S Sf kgkwekfks39rkszkrskfrkcrkspkrs where kg 10 no details pick typical value km 10 no weld mentioned kf 10 ignore stress concentration k5 076 machined nish Fig 533 km 09 no details pick typical value kS 10 no details kf 10 no fretting kc 10 no corrosion mentioned k57 10 200 cpm pick typical value k 081 Table 54 for 99 reliability So Sf 0760908149 1m 2715 km This application is for a power plant which presumably runs 24 hoursday and 7 daysweek The cycle rate is 200 cycles per minute Thus in a single week we expect in excess of 2 gtlt 106 cycles So we should design for in nite life lgnoring stress concentration and safety factor and noting that the loading conditions are completely reversed ie Um 0 then the minimum required diameter d is computed as i P778000lbiS 727150 7 A ap4 f 7 p gt d191m 542 The current reliability is 99 and the goal is to get to a reliability of 999 From Ta ble 54 our strength at a reliability of 99 is obtained by multiplying the test data strength at 50 reliability by a factor k99 081 while for a reliability of 999 we would multiply by a factor of k7ggvg 075 Thus we need to reduce our stresses by a factor of 075081 0926 assuming all other factors remain constant EMD homework solutions6 Due 3 24 5 47 5 53 5 54 5 61 modi ed version of 5 84 5 88 547 a Aluminum bar SM 100000 psi Syp 80000 psi SN105 40 000 psi assume fully modi ed but if it were not should apply modifying factors to this value and e 8 in 2 inches thus ductile Axial member subject to cyclic loading me 5000 lb and Fm 10 000 lb First assuming failure due to fatigue at N 105 cycles we compute the diameter Umam Um 10000 lb U I 7 5000 lb A m H L 7 A 1 10000 lb 5 000 lb 7 7500 lb 2 A A 7 A 1 10000 lb 7 5 000 lb 7 2500 lb 2 A A 7 A an 2500 lb 7 aim A 7 5001b SN105 40 000 pm Sb 100000psl39 d2 A 013752712 T d 0418271 Now for this diameter let7s verify the rod does not yield 10 000 lb W 72727 39ltS U 01375 712 p52 3 The maximum stress is below yield so fatigue is the failure mode at N 105 cycles for the given diameter b Consider now me 15 000 lb and me 20 000 lb Following the same procedure as above 1 20000 lb 15000 lb f 17500 lb 5 A A A 1 20000 lb 15000 lb 7 2500 lb 5 lt A 7 A A 7 A lb SN105 40 000 psb u 100000psl d2 A 023752712 T d 0550 m Checking for yield again 207000lb 84 211 39gt S Um 023752712 7 p52 1 Since the maximum stress is above yield the diameter d 0550 in is not acceptable To avoid yield the diameter must be increased to d 0564 in c Although the higher range of stresses had the same alternating stress the increase in the tensile mean stress required an increase in bar diameter to avoid yield and fatigue fracture 553 Table 1 lists the load levels and number of cycles at each load level We want to nd the diameter of the member at which the member will survive just 50 duty cycles ie what is the diameter that will give stress levels such that the number of cycles to failure for each stress level N will satisfy 60000 185 X 106 215 X 105 1 N1 N2 N3 Since the relationship between the number of cycles to failure and the completely reversed Table 1 Stress and cycle counts for 5 53 completely reversed stress level no cycles in 1 duty cycle no cycles in 50 duty cycles UeqicR 71 W 1200 60000 W 37000 185 gtlt 106 W 4300 215 x 105 stress level is very non linear from Figure 531 we need to use trial and error Let7s rst gure out the cross section areas A ofthe member at which it willjust fail at each ofthe load levels to get a rst estimate and also to see if any of the load levels will dominate Looking at Figure 531 for 2024 T4 aluminum the stress level with failure at about 711 60000 cycles 1043978 cycles is about 45 ksi 310 MPa the stress level with failure at about 712 185 gtlt 106 cycles 10627 cycles is about 28 ksi 193 MPa and the stress level with failure at about 713 215 gtlt 105 cycles 1053933 cycles is about 37 ksi 255 MPa The areas associated with each of these stress levels for the given load levels are then 50 000 N A 310 MPa a A1 161 mm2 1 31 000 N A 193 MPa a A2 161 mm2 2 40 000 N A 255 MPa a A3 157 mm2 3 From this we see that with a cross section area of about 160 mmz each of the 3 load levels at 50 duty cycles would use up the entire life ie 3 lives would be used up altogether and none of the load levels appears to dominate as each requires about the same cross sectional area Thus we know A gt 160 mmz Since each of the load levels seems to contribute almost equally let7s now nd the cross sectional area where 13 of the life is used up by the highest load level and try that for our next guess Thus let N1 3711 180000 1053926 cycles The stress associated with this number of cycles to failure is about 37 ksi 255 MPa This gives an area of W 255 MPaa A 196 mm2 The other stress levels and associated number of cycles to failure for this area are 31000 N 7 W 158 MPa 7 23 82 a N2 7 10 40 000 N w 204 MPa 30 ksi N3 1062 158 X 106 mm 60000 185 gtlt106 215 gtlt105 7 065 180000 107 158 gtlt 106 Thus A 196 mm2 is too high and is associated with 65 of the life used up So now we know 160 mm2 lt A lt 196 7717712 Let7s now guess A 180 mmz 50 000 N W 278 MPa 40 52 a N1 1052 158 X 105 mm 31 000 N W 172 MPa 25 ksi N2 1067 501X106 mm 40 000 N W 222 MPa32 Im a N3 106 mm 60 000 185 gtlt106 215 gtlt105 N1 158000 501 X 106 106 N Thus for a cross sectional area of A 180 mmz failure is just predicted at 50 duty cycles 15 mm B be lhe number cl blocks lo allure The dlamelel ls lhen uslng A 7r 2 Rem ow counlmgshown m nguxe 1 Lel 1 6 gtlt 10 1 00 M 5 36x10 Cycle counl labulaled m Table 2 nguxe 1 Cycle Counlmg for P55A Table 2 Ramllow cycle counl lox 5 5A Ncles l Sclullon here assumes plale mall ls 150 mm as xsl slaled m problem ml 300 A mm as stated later ii Solution here takes coef cient in the Paris Law to be 34 gtlt 10 12 rather than 48 gtlt 10 27 which is consistent with equation in Figure P560 for the same material7 and converting USC units to metric First7 we must determine the critical crack length in which this crack will propagate in an unstable manner fracturing the plate when the maximum load is appled We use fracture mechanics Pm 100 000 N mam 556MP U A 12 mm150 mm a K c 2 372 2 BM 25 I 25 7 00102 m 102 mm lt12 mm so plane strain Syp 582 K10 372 MPaxE KI CIUmawWam 01556 MPa 7Tam To compute am we need 01 but 01 from Figure 519 depends on ba7 where b 150 mm and a am at onset of failure So we will iterate Try a 50 mm Then ab 0337 and from Fig 277 0117 03332 0987 resulting in CI 179 Now 372 MPaxR 179556 MPa 7Tam results in am 00445 m 445 mm Which was different from our guess of 50 mm So now update guess by taking midpoint between 50 mm and 445 mm for 1 So let a 4725 mm Then ab 03157 and 011 7 031532 0977 resulting in CI 171 Now 372 MPaxE 171 556 MPa 7Tam results in am 487 mm So the critical crack length is between 4725 mm and 487 mm Lets take it to be am 48 mm7 and at this crack length7 CI 173 At the initial crack size of a0 1 mm7 ab 000677 01 112 To obtain an estimate of the number of cycles to reach the critical crack length7 we integrate the Paris Law da 7 712 3 W 7 34 gtlt 10 AK where dadN is in mcycle and AK is in MPmR AK CIAUxWa Taking CI to be the average between the initial and critical crack sizes7 CI 143 Now compute AU 0mm PM 450 N 025 MPa A 12 mm150 mm AU 0mm 7 0mm 553 MPa So AK 1402E Plugging into the Paris Law d i 937gtlt10 6 3 dN 7 a am0048 3 N a ida 937gtlt10 6 dN 100001 0 72 0048 4 0001 937 gtlt 10 6N N 578 gtlt106 cycles Therefore the crack length is expected to become critical after 578 gtlt 106 cycles At a cycle rate of 5 per second 578 gtlt 1060ycles1s5 cycles1hr36005 321 hr Thus after about an additional 320 hours of service the crack is expected to be at a critical length 5 84 modi ed as follows For the geometry in P5 84 if the material were A181 1060 CD steel subjected to a constant torque of 1800 N m no axial load and a moment uctuating between 0 and 1200 N m for a reliability of 999 assuming all critical surfaces ne ground is the design adequate if an in nite life is desired and if so what is the safety factor A181 1060 CD from Table 310 e 10 thus ductile and from Table 33 S 621 MPa 90 ksi and SW 483 MPa Since we have no fatigue data we will estimate the fatigue limit Using method from p 247 and modifying factors k5 from Fig 533 km typical value k from Table 54 S 055 3105 MPa Sf kyksszS 09090753105 MPa 189 MPa Torque is constant and thus yields a mean shear stress and no alternating stress 7 16T i 161800 N m i Tm i W i 7T03905 m3 i 733 MPa The uctuating moment yields a mean and alternating stress 32Mmm 32 1200 N Umazltn0m m MP8 7Td3 7T005 m3 0 Umln 1 Um Umamn0m Umln MP3 1 Uan0m EltUmamn0m 7 Umln MP3 We apply the fatigue stress concentration factor only to the alternating components7 in this case Jamom Dd 5650 112 and rd 350 0067 from Fig 54 for bending7 Kt 18 The notch radius is 3 mm 012 in from Fig 546 q x 083 Thus7 the fatigue stress concentration factor for the alternating bending stress is Kf qKt 7 1 1 166 0a Kfaamom 812 MPa Now we plug our mean and alternating components into the distortion energy theory DET to nd the equivalent unlamlal mean and alternating stresses Ueqm U 377 4892 37332 1361 MPa 05W 0a 812 MPa Finally7 to get an equivalent completely reversed stress7 we plug our equivalent uniaxial al ternating and mean stresses into the modi ed Goodman equation Us a 812 MPa aeqima 1 7 gm 1 7 m 104 MPa 5 621 u Since Herc lt Sf failure is not predicted and an in nite life is expected The safety factor guarding against fatigue is 71f SfUeq10R 189104 182 We should also verify that there wont be failure by large scale77 yielding at the maximum7 nominal stress levels ie maximum stresses without local stress concentration factors Applying the distortion energy theory U5qmam Ufnadnom 3T72rtamnom MPa lt SitZ7 Thus yielding is also not predicted 588 a See Figure 422 N 1052 m 16 gtlt 105 b In nite NOTE initial tensile prestress leaves notch root with compressive residual stresses it is the compressive residual stress that improves fatigue life c N m 104 Similar to b7 note that the initial compressive prestress leaves tensile residual stress in the notch which reduces fatigue life EMD homework solutions7 Due 331 2 26 2 30 2 34 2 36 6 4 9 14 226 In this case the column end conditions are free xed L5 2L 2H It will buckle about the axis of minimum I 2nd area moment of the cross section From Table A3 we nd for a W6 gtlt 25 wide anged beam A 74 2712 and the minimum I 17 M The minimum safety factor on load is n 2 Thus the maximum height to avoid buckling is 7T2EI i 7T230 gtlt 106 p5217 m4 13mm anm2100001b200001blt L2 7 My Hm 251 m Checking also for yield IdemW PdesignA 2 702 psi This is well below the yield stress for any steel so yielding is not expected to be an issue We see that the maximum allowable height is considerably larger than for the cylindrical pipe of problem 2 25 230 For buckling due to torsion of long thin wire 2 E 2 207x103N 2 M Mm L W 831 N mm 0831 N m mm Checking for yielding 1058 N mm 106 N m Wt 125 mm Comparing above results buckling governs and the failure torque Mf is w SW2w 690 MPa2gtlt lt227Wgt a a Mf 1 0831 N m 234 a Let KE be the kinetic energy just before impact which is the total system energy at that time since there is no potential or strain energy just before impact Let SE be the strain energy at maximum compression which is the total system energy at that instant since there will be no potential or kinetic energy at maximum compression KE l771112EK112 2 2 9 SE k62AlE 6W1 7 Alainm TE KE SE 7 EW 039 7 max 8839 968 103 39 707 n2386m5210mgt mS X W 236 We set the kinetic energy of the tow truck before impact Eb when the tow rope is slack equal to the strain energy Ea when the rope is at maximum extension after impact7 noting that the tow truck has been stopped because the wrecked vehicle doesn7t move Thus7 2 1 2 2 2 Eb7mv Ea k6 a 7m k6 where the rope can be treated as an elastic spring such that F k6 and W mg So we can rewrite the above relationship as 2 2 my a F M 9 k 9 The spring constant of the rope can be found by considering the load de ection relationship for an axial member F E6 AE AE UiZiEEiT a Fde a k7 The maximum stress can then be expressed as 2 9 2 2 U E WEv 22000 N83 x 10 Nm 222 ms 710 Mpa A gLA 981 m527 m026 gtlt 10 3m2 Since 039 710 MPa lt S 1380 Mll a7 I dont expect the rope to break 64 Reviewing the last of con gurational guidelines in 627 the potentially applicable guide lines for the actuating lever of Figure 164 would include 2 Tailor element shape to loading gradient 5 Utilize hollow cylinders and 1 beams to achieve near uniform stress In bending7 l beams a good choice hollow cylinders good in torsion in this case consider l beam shape 6 Provide conforming surfaces at mating interfaces 7 Remove lightly stressed or 77lazy77 material 8 Merge different shapes gradually from one to another 10 Spread loads at joints lncorporating these guidelines to re ne the shape of the actuating lever7 one initial proposal might take the form sketched here Note7 the deeper beam where the moment is higher7 and a hollow cross section to reduce lazy material and achieve a more uniform stress Of course7 other variations are possible SEEUDHAVA Figure 1 6 4 one possible design 914 The nominal size at the t is 1 inch For a class FN4 interference t7 using Table 677 the minimum diametral interference is Amm 0001 in and the maximum diametral interference is Am 00023 in a Since the shaft and hub are both steel and the shaft is solid7 we can use Eqn 9 49 to nd the pressure at the interface between the shaft and hub In this problem7 the inner and outer radii of the hub are a 05 in and b 1 in For steel7 E 30 X 106 psi For the case of loosest t minimum interference7 then 7 EA 1 a2 30 gtlt106psz0001z39n 17 M 7 if 11250 39 p 4a b2 405m 12 p52 3 The hub only sees an internal pressure p and the maximum radial and tangential stresses are located at the inner radius r 1 Using Eqns 9 27 and 9 28 we nd azp 52 39 Urhub b27a2 17 7p7117250p82 azp b 05211 250 psi 12 u 1 7 7 1 18 750 39 W 17gt 52 7 a2 lt f 12 i 05 lt 05 p The shaft only sees an external pressure p Using Eqns 9 32 and 9 33 and noting that for the shaft the outer radius is 05 in which is labeled b in these Eqns and the inner radius is zero solid shaft which is labeled a note that by setting a 0 the dependence on 7 drops out The maximum stresses in the shaft are then Urshaft 10 11 Utshaft 10 711 250 psi b For the condition of tightest t we consider the case of maximum interference Substi tuting Deltamm into the equation above for the pressure we obtain p 25875 psi With the dimensions the same the maximum stress equations just scale by the pressure thus mam 7p 725875 psi 05225875 psi 12 u 1 7 43125 7 b 12 i 05 T 05 W Urshaft 7p 7257 Utshaft 7p 725875 psi c For the maximum torque that can be transmitted across the press t before slippage we need the static friction coef cient for steel on steel Looking at Appendix A1 a typical value is p 011 if the interface is lubricated and p 045 if dry for hardened steel Considering the tightest t the maximum torque that could be transmitted before slippage is Tf 2pp7ra2L 201125875 psi7r05 m215 m 6700 in lb if a lubricated interface and Tf 27 400 in lb if dry EMD homework solutions8 Due 4 7 15 3 15 8 15 9 15 16 153 a Recalling that external meshes are from 15 2 no 3 7 75 7033 71m 711 N2 48 or 712 7033711 70331725 rpm 7575 rpm so since implies CCW rotation 71 n2 575 rpm COW b Since B is a compound shaft 712 n3 7133 Again using 15 2 on N 16 u EEii103933 71m 713 712 N4 48 n4 7033713 7033575 rpm 71917 rpm and With 713 rotating CCW implies CW rotation and nout no 714 1917 rpm CW c Since the power into the system is 1 kW and the ef ciency of each gear mesh is 98 the power out is Hm 0980981 kW 096 kW 960 W The power is the product of the angular velocity in radianssec and the torque We already know the angular velocity of the output shaft so 2 d 39 Hm 960 N ms mem 1917 W m mm Tm mm r61 60 5 Tom 478 N m 158 a The input shaft drives both gears 2 and 6 Gear 2 drives gear 3 which drives is connected to the arm 9 Gear 6 drives gear 4 which is connected to gear 5 and gear 5 is effectively the sun gear for this planetary system So lets rst get the speed of the arm 9 and the sun 5 The direction of the input shaft is speci ed to be CW so we will take CW as positive The input speed is win tag wa 720 rpm Note I usually use the symbol in for angular velocity regardless of the units while the book likes to use in when angular velocity is in rads and 71 when angular velocity is in revrnin or revsec N 21 mg WZ 763720 rpm 7240 710m W9 N 35 M iiNw6 730720 rpm 7840 710m W5 Now for this planetary gear train involving gears 5 7 8 and 10 and arm 9 using 15 4 we have W10 W9 i 1Vst 1 W5 W9 N7N10 wlo 240 rpm 3924 7840 rpm 240 rpm 7 7 7832 a wlo 715 rpm Note the signs in the above equation I consistently use to denote CCW thus since the resulting angular velocity is negative this indicates the output shaft rotates CCW Also note the minus sign on the right hand side of the Eqn For bevel gears the rule of a change of sign with every extermal interaction does not hold like for spur gears If we imagine the arm xed and drive gear 5 CCW we see that gear 10 will move CW thus the direction is reversed and thus the minus sign on the right hand side of Eqn b If we assume 100 ef ciency then lwoutTtmtl m 720 r p lel w m 7 300lb 1n 144001n lb Wont 159 For this planetary gear train the motion can be described by 107 7102 7 N3N4N6 i N3N6 W3 102 7 N4N5N7 i N5N7 If we choose CW as positive then we are given tag 500 rpm and wg 7900 rpm Now we have all the values to put into the equation above except N7 We can determine N7 by realizing that the diameters are proportional to the number of teeth on each gear since they all have the same diametral pitch Nd Thus 1 1 1 1 1 1 1 7N 7N N 7N 7N 7 18 25 7 39 7 21 64 27 2342526 2 22 7 N7 128 Now we can solve for the speed and direction of the annular gear 7 407 7 500 rpm 18 21 7900 rpm 7 500 rpm 39128 Thus 407 394 rpm and since the sign is positive it is rotating CW 1516 a From 15 15 7 i750 N 9 17 N9 75016120 teeth b To convert from diametral pitch Pd 12 teethinch diameter to circular pitch inches circumferencetooth 7T 7T i 7 02618 Pd 12 Tm m p0 c The center to center distance is just the sum of the radii or half the sum of the pitch diameters dpdg i NpNg i 16T120T 567 2 213 212 Tm m C d The base circle radius is just Tb rcos p see Fig 1510 So for the pinion and gear 16T 200 i 0 626 cosltpi212Tmcos i in 120T 77g rgcos p 79 cos p 213d 2 12 Tm Np 77p rpcosltp 213d cos 20 4698 in e From Table 153 for a 16 tooth pinion the maximum number of gear teeth without having interface is 101 Since for this gear set Ng 120 gt 101 interference would be expected EMD homework solutions9 Due 413 15 31 15 38 15 58 8 5 8 9 8 15 1531 a From the power and the angular velocity we can determine the torque transfer Then knowing the radius of the gear we can determine the tangential force transferred which through the pressure angle is related to the radial force on the gear So we determine for forces on gears 2 and 3 as follows W2 T2 N 24 1725 2 d 39 wg 771w if W m w 712043 rad5 N2 36 mm r67 60 s or 12043 rads cw TQWZ rads 6 600 lb 20 hp T3 1 0961 111 110 d2 E 36T 60in Pd 6Tm d3 6121 30 in T2 10961 in lb F 26 m 112 7 36537 lb T3 7 10961 in lb F 33m th 73073 lb FT FttanltpFttan20 172 13298lb F326597 lb 1 A HAAAA O OO VVVVVV A H H Assuming both gears to be straddle mounted between closely spaced bearings the top View of shaft B is shown in Fig 1 b The front View of shaft B may be sketched as shown in Fig 1 c Torsional moment is constant over whole shaft length so if diameter is constant maxi mum torsional shear stress is constant over whole surface of the shaft Because of straddle mounted bearings no signi cant bending moment is generated in the shaft Direct shear or transverse shear is generated in the shaft at the edge of each bearing where it adds to torsional shear Stress concentration probably is a concern at this location as well The conclusion then is7 since bearing reaction forces are larger at gear 3 than at gear 27 and shaft torque is the same7 the governing critical section is adjacent to gear 37 and the governing critical point lies at the shaft surface 139730711 RM 1827 lb on eachbearing R3 3654 lb on each bearing Fa 3654 lb R2v 665 lb on each bearing I33v 133 lb on each bearing Fl 266 lb FIZ 133 lb Figure 1 Problem 15 31 a top view above and b front view below 1538 Following the design procedure suggested in 15117 for a single reduction straight spur gear unit A rst conceptual sketch may be based on the following criteria and spec i cations power transferred is H 50 hp7 71p 5100 rpm7 n9 1700 rpm7 Qv 127 materials for both gears is Grade 2 A181 4620 steel carburized and case hardened to 60 RC Furthermore7 a reliability is to be 99 at a life of 107 pinion revolutions with a design safety factor of nd 13 We also want to minimize Np without undercutting A rst conceptual sketch may be made as follows a Tentatively choose a standard full depth AGMA involute pro le 20 pressure angle tooth system as de ned in Table 151 b Following the guidance of 15117 step 67 a tentative selection will be made of Pd 10 m 1 c From Table 1597 to avoid interference and undercutting7 for areduction ratio of 71png 3 or 317 tentatively select Np 21T Ng 3Np 63T 2 2 so for Pd 10 N17 21T N9 63T d7721 di P Pd ioTm m 9 Pd ioTm 63 in 3 d From the pitch diameters selected in part c7 the center distance is then 21 63 Crprg42in 4 e Frorn 15 167 the face width must satisfy 9 14 7 g b g 5 10271 10271 or 09 in g b g 14 in 6 In addition7 the face width must be large enough to keep the bending and surface stresses from exceeding allowable values So we will start by looking at the bending stress Frorn 15 417 i Fth 05 J KaK39LKmKI 7 The tangential force can be computed from the power transfer and the pitch line velocity ft 7 m 7 2804 fitmm 8 1 V wprp 5100 revmm27r radrev105 m 12 33 000 ft lb 50 hp EV E2804 fitmm 9 E 533511 10 H The other factors may be taken to be as follows assurning higher precision gearing7 which is reasonable with high quality value speci ed Km 10 Table 156 electric rnotor uniforrn7 assurne driven rnachinery uniforrnQ11 Ky 11 Figure 1524 with Qv 12 and V 2804 ftrnin 12 Km 13 Table 157 with precision gears 13 K1 10 one way bending 14 J 034 Table 159 higher precision gears 15 Substituting these values into 22 5885 lb10239n 24 752 lbz n 10 11 13 10 16 ab 034 lt gtlt gtlt gtlt gt b lt gt The bending stress needs to be compared to the bending strength P rorn 15 427 SW YNRgsgbf 17 From Table 15107 for carburized and case hardened steel 60 RC7 Grade 27 S bf 65000 psi Frorn Figure 15287 for 107 cycles7 YN 107 and from Table 15137 for 99 reliablity7 Rg 10 So the bending strength is7 SW 101065 000 pm 657 000 psi 18 In order to satisfy the design criteria on the safety factor then Stb f 65 000 ps b gt 13 19 ab 24 752 lbm d l b 2 05 m 20 new Since this is below the range in Eq 67 the lower limit in Eq 6 is still the lower limit Now we check the surface contact stress From 15 467 E W 7 om lbdeKaKva 21 Noting from Table 39 that for steel rnaterials7 E 30 gtlt 106psi and 1 037 the elastic coef cient OP may be evaluated as 1 Op 229 gtlt 103mm 22 77 30x1 6m The geometry factor 1 may be evaluated from the gear ratio7 where mg a 3 F giving 20 20 3 I wi 01205 23 2 3 1 Using the other factors de ned previously 5885 lb Us 229x10 E W 132 336 lb z39n1395 05f T The surface stress needs to be compared to the surface strength From 15 47 101113 24 53f ZNRGSf 25 From Table 1515 for carburized and case hardened steel 60 RC Grade 2 S f 225000 psi From Table 1513 R9 10 and from Figure 1531 ZN107 10 So the surface strength is 55f 1010225 000 pm 225 000 psi 26 In order to satisfy the design criteria on the safety factor then SS 225 000 39 b new f w 2nd13 27 05f 132336 lbml5 b 2 06 m 23 So the constraint in Eq 6 is still restricting the face width Thus if we stick with the choices we7ve made to this point we should choose a face width of b 09 in To improve the design make it smaller we should repeat these steps choosing a smaller tooth size ie a larger diametral pitch Pd For example we might repeat with Pd 12 Tz n This would lead to dp 175 in and d9 525 in The new bounds on the face width would be 075 m S b S 117 m A face width of 08 in would be found to satisfy the strength design criteria Many other acceptable design con gurations exist too f From part c Np 21T and N9 63T 1558 From speci cations Mounting Gear straddle mounted Pinion overhung Np 15T Ng 60T 71p 900 rpm Pd 6 ltp 20 b 125 in 90 reliability life of 108 cycles and safety factor of nd 25 Material Grade 1 steel through hardened 300 BHN To nd the maximum horsepower we need to nd the maximum pinion torque that can be transmitted through this bevel gearset since the pinion angular velocity is speci ed We will use the AGMA formulas on p 626 First we compute the allowable bending strength and compare against the bending stress Stbf YNRgS bf YN 095 from Fig 1528 R9 118 from Table 1513 S bf 36 000 psi from Fig 1525 Stbf 09511836 000 psi 40 356 psi The bending stress due to the loads is 2Tde 03917 dpr KaKva The pinion diameter is obtained from the diametral pitch and number of teeth N 15T Pd6Tmli gt dp25m d1 dz The following factors are obtained from Figs 1543 1544 1524 and Table 156 J 0235 Km 112 Km 125 assuming uniform drive moderate shock for driven machine I i VIEW Javewp T17 7 E Sln YWp N 15 tan y F a y 1404 see Fig 1541 277 rad 1ft ije 12539 7062539 14040 900 39 I m msin revmmlt my 12 m 5176 fitmm Ky 12 assuming Qv 8 Substituting into 2 then mwm WL2512112 2745 whim 4 03917 For a safety factor of nd 257 then R 2 5 SW 40356psz d 39 ab 2745271 3Tp Tammi 588 111 15 5 Now7 we compute the allowable surface strength and compare against the surface contact stress 89f ZNRgSf ZN 09 from Fig 1531 gf 125700p5239 from Fig 1530 85f 09118125700psz133493ps 6 The contact stress is o QEKKK m Usf bevel a 1 m P 5ng where 3E 3 30 gtlt 106 p52 O m 28054 m l V47r17y2 47007032 W I 0077 Substituting into 7 then 2T 5 2 80544 39 quot 125 12 112 6 6302 lb 39 2395 wT 0f pSZ125m25m20077 x fm P 8 Fmeglemyfemdm sum 5 Hues 1m M 2 5 7 4 Mao 2 WWW 1m4 65 Mb 9 Sm me emum allowable Wane exam mam slmses 91510Nm lhsn um m amid amass hemdmg 5m 5 the memmum e wehls memum pm 5 011m mm 1 ss mu Now m m es thWoo yeam 1 10 375 e A mm dlsgem 5 shown In g 2 mm athhnum we ammm me reactions at the beam gure 2 Problem 875 neelmdy megem EMA mmmmummxzommLounxzsm o a Rs 7 we lt1 2M imamWwamnxzosmo mama 2 Twasanst o 1250 a mummiunn RA 465m A 25 shimmy Badman 5 Zn hammusom RM1856N e no bakes body dmgemend qecdnpulsi reaction mmweomem mesheexend momemd smmayrzendzilplsmsesshnwnmngz Theexxslkndend a mane dlsgems ere shown m Hg A an my gure A dewlem 875 Mus m end 9x131 xeed mg dlsgems h m mam 5 we heve e hemmg mm which due to shell mum heads to enemeung hemmg stress end e mane whmh leg to e mean shAex sues Rat e m eexe m we means in nmplmny But we M gnheakend cheech 191m We wm use equslm u d e 37 Mama 3 MAW neheamgmem es e 2wa 3331 g 7 MeI Ig m 1 NemY26N7m 23mm Mmo a The tmque 15 um Tm 125 m Te 1020 sled emeexedy a m 00 ps1 m We Hum Hg 5 n5 31000123 m We Appxymgewmpnezemmmg fecmxsm 5y mmgyumd nish 1c came diam 1c 09 end amhshxhly em a 1e 0 a me m o s m even aha es mmm could be We 5 o 9o 9o axm MPa 166 MPa Since no information is given about potential stress raisers at B7 we will take be 17 but should verify For a fatigue safety factor of nf 18 d 2234 N m 3125 N510 g 00306 771 306 mm 10 Rounding up to a standard size see Table 637 choose d 35 mm Now check if this diameter is adequate against yield and fatigue including the effect of the axial load Note that the axial load causes a mean axial compressive stress 32234 N m a 5559 MP 11 0y 7r0035 m3 a 02quot 7w 213 MPa 12 7T0035 7n 03an ay z 02quot 5772 MPa 13 16125 N m m mm 1485 MP 14 Tquot 7quot 7r0035 m3 a 5514 5772 MP52 31485 MP52 63 MP5 15 2 y 7 47 16 7 0557mm 63 55H 5559 MP5 17 55H 213 MPa2 31485 MP52 2581 MP5 18 5559 MP5 Ueqicg 595 MPO 393 nf Sf 279 20 qu 595 So choosing d 35 mm more than satis es the design criteria on the yield and fatigue safety factors Note that the axial stress due to the axial load7 02quot is relatively small7 so neglecting it in the rst calculation was quite reasonable If a slightly lower safety factor were allowed nf 167 the allowable diameter could be reduced to 30 mm 89 From Tables 33 and 3107 for annealed A181 1020 steel7 Sn 393 Mll a7 Syp 296 Mli a7 and the material is ductile with a 25 elongation in 2 inches From Fig 5317 S 33000 psi 227 MPa Assuming ground nish k5 097 size effect ksz 097 and a reliability of 90 k 09 since info is not given7 other assumptions could be made Sf 090909227 MPa 165 MPa 1 10 We can compute the torque mean from the transmitted power and angular velocity as N 2 d 39 H 75 kW 757 000 m wT 1725 revmm W W W T 5 rev 60 s Tm 415in lb Assuming the maximum stress will be at the left shoulder with the smaller llet radius7 the moment leading to an alternating stress at this location is P Ma RL02 m 3 02 m 01 3 The bending fatigue stress concentration at this shoulder is determined as follows 2 38 rd 700625 Dd7119 32 32 Kt 185 from Fig 54 19 1 07185 i 1 16 Setting US013 Sfnf 165 p5 13 127 MPa then using Eqn 8 6 393 MPa21701 mP 4 Wm gtlt 106 Nmz 7 415 N431 P 1720 N 5 Ueq70R 127MPa Should also check yield to verify It is easy to compute using equation 8 9 Harm 124 Mll a7 which is well below yield 8 15 a From equilibrium7 the reaction forces can be determined to be RL 225 kN and RR 45 kN The bending moment7 using singularity functions7 is then 22500x 7 367 000 lt s 7 0275 gt 9000 lt s 7 06 gt 1 on 0 g x S 1 m7 where z is the axial position along the shaft in meters7 and M is in units of N m The transverse de ection 1 and slope dvdx are then approximating the shaft with a uniform diameter d2 7 Mm i 1 22 500 36 000 lt 0 275 gt 9000 lt 0 6 gt 2 dzz EI EI z7 9539 x39 d 1 l 7 11 250952 18 000 lt z 0275 gt2 4500 lt x 06 gt2 01 3 dz E1 1 55 3 3750953 6000 lt z 0275 gt3 1500 lt z 06 gt3 0195 02 4 1 0 7 O 0 O 0 5 7J EIlt 2 2 1 51 15595 011 0 01 15595 6 1 55 375053 6000 lt z 0275 gt3 1500 lt z 06 gt3 15595z 7 d 1 d1 5 11 250952 18 000 lt z 0275 gt2 4500 lt z 06 gt2 15595 8 The maximum de ection will be at the location where the slope dvdx is zero From the diagram we can guess that this will be on the interval 0275 lt z lt 06 m For this interval the slope is d1 1 2 2 d7 3 11 2505 18 000z 0275 15595 i 1 6750z2 990095 2921 9 Setting the above equation equal to zero we nd the slope is zero at z 041 m The second area moment for the assumed cross section d 01 m is I 491 gtlt 10 6 m4 and for steel E 207 GPa So the de ection at z 041 m is 11mm 389 gtlt 10 4 m 0389 mm b The slopes at the bearings are then d 9A dl0000153md0088 10 i d 90 61110000934md0054 11 i where recall the y axis is assumed downward EMD homework solutions 10 Due 421 8 16 8 197 18 27 18 6 13 9 816 The bending moment obtained in the previous problem 8 15 may be graphically represented by Fig 1a Now scaling this by EI to obtain 7dzvdz2 MEI gives Fig 1b7 where AB IE0 155 gtlt10 6 m4 IBD 491gtlt10 6 m4 and DE 120 gtlt10 5 m4 integrating once d1 dzv dzv M 6 wdx01 or de7de0701 1 We can do this in Excel by breaking the length of the member up into small sections Ax n1 7 z where 71 indicates a discrete point along the member and using the trapezoidal rule we can solve for the slope offset by a constant 01 along the member with 7w 7 01m 7 7w 7 01 W17 m lt2 So plotted in Fig 1c is dx 70 7 01 We dont know 01 yet because we have no boundary conditions on the slope 0 so we arbitrarily started our integration at 0 for z 0 Integrating again d2 v6dzOgd7dzdx01z02 3 Again7 we apply the trapezoidal rule to carry out this integration So plotted in Fig 1d is 7f f 3272 dx dx 71 7 Ca 7 02 Now we can apply the boundary conditions 00 0 which is already satis ed and thus leads to Cg 0 this factor just shifts the whole curve up or down The factor 01x needs to be such that 01 07 thus7 Clz should be the straight line shown on Fig 1d Subtracting this line from the curve for 7 f f dm dm gives the de ection curve shown in Fig 1e The maximum de ection is then 11mm 05 mm at z 0525 m The maximum de ection is thus a bit higher than the approximation in 8 15 and at a location to the right of that found in the approximation The angular de ections at the bearings can now be determined using the value for 01 206 gtlt 10 3 and Fig 1c 0A 000206 rad 01180 and 00 7000159 rad 70091quot These results are not too far off from the approximation in 8 15 819 a The critical speed for lateral vibration of this shaft may be estimated from Rayleigh7s method From the problem statement7 shaft weight and bearing stiffness effects will be neglected for this estimate To use Rayleigh7s method7 the shaft de ections7 yA and 1 7 a M NW 72812 in man u nna D m n ena nn 7e MEl mm H mm u an BMW 3 n ma n um n nus A a 0 ran u an H mm mm Awe we Av mm Figure 1 Problem 8 16 Numerical integration 1113 will be required Since it is a uniform shaft the shaft de ection can be easily determined using singularity functions First we need to determine the left reaction BL Assuming the bearings do not resist any moment as stated in the problem ZMR BL90 m i 80 lb60m i 1201b20m 0 a BL 80 lb 1 Now7 the bending equation is in units of inches and lbs 77 80x780ltx730gt7120lt770gtl 2 7 7 402740ltz730 gt2 760ltm770 gt2 01 3 HQ 7 37ltz730 gt3 720ltz770 gt3 0102 4 007020 020 7490 7 6680 000 0190 0 7 02 7742 Ms 7 z3 7 g lt x 7 30 gt3 720 lt x 7 70 gt3 7747 222x 5 Noting that for steel E 30 gtlt 106psi and the cross section 2nd area moment is 7 7T2 m4 I 7 078542714 64 Evaluating the de ection at the coupling A and ywheel B then yA 630 7W goof i 7422230 007922271 6 11 670 4W gwof 4 lt403 4 7422270 006262 277 The critical speed is then we 386 27182 80 lb007922ym 120 lb006262 8 80 lb007922 m2 120 lb006262 2712 r61 60 5 7414 d 708 9 Ta 8 2 rad 1 mm rpm Since the shaft rotates at 240 rpm7 this estimate predicts that the shaft rotation is well below the critical speed by a factor of 3 b If the shaft mass is included in the analysis7 a third term must be considered7 for which the shaft weight applied at midspan must be included and the other de ections must be updated to consider the effect of the shaft weight at these locations The density of this low carbon steel can be found in Table 34 The weightlength of the 2 inch diameter steel shaft may be estimated as w 0283lbm37r1m2 08891 lbm 10 and the total shaft weight is WS wL 08891lbz39n90m 800 lb 11 The de ection of the shaft due to its own weight is just the solution for a uniformly distributed load7 which is given in Table 417 case 37 with L 90in vwz L3 7 2L2 x3 24E 08891lbz nz 3 2 3 Wm 2430gtlt106p8 07854m4 90271 729027095 95 mm 1572 gtlt 10 9 n 3z90 m3 7 290 mm 3 12 Evaluating at A7 B7 and the center of the shaft MA vw30002801m 13 MB vw70002091m 14 we vw45003223m 15 From the de ection Eqn 57 we can also compute the de ection at the shaft center due to the weight of the coupling and ywheel yam M45 009210 m 16 Now the overall de ections at the coupling A7 the ywheel B7 and the center of the shaft 0 including the weight of the shaft are 114 007922871 002801 m 01072871 17 133 006262 in 002091 m 008353 in 18 yo 009210 m003223 m 01243871 19 The critical speed is then m 801501072m12015008353m801501243mr we 7 386 ms 2 2 g 80 lb01072 m 120 lb008353 m 80 lb01243 m r61 60 s 7 6068 rads 27T md 1mm 7 579 rpm 21 20 This is quite a bit lower than before so the shaft weight is important in this case Since the shaft rotate at 240 rpm this estimate predicts that the shaft rotation is still well below the critical speed by a factor of 24 c An estimate of the contribution of bearing de ection may be made by calculating bearing de ection at each bearing due to the reactions BL and RR The left reaction neglecting the shaft weight is computed in If we add to this half the shaft weight and solve for the right reaction as well we nd BL 80 lb40 lb 120 lb RR80 lb120 lb80 lbi120 lb 160 lb 22 Using the provided bearing spring rate we can now compute the de ections at the bearings 120 lb 24x10 439 23 M 5 gtlt105lbz39n m l 160 lb yR 32 x 10 4 m 24 5 gtlt105lbz39n Interpolating between these de ections to the locations of the coupling A ywheel B and shaft center 0 leads to additional de ections resulting from the bearing de ections of ybA 267X10 4239n 25 ybB 302gtlt10 4239n 26 we 28gtlt10 4z39n 27 These de ections will be added to yg y and yC to nd a new estimate of the critical speed that includes the effects of both shaft mass and bearing de ection Thus yx 01072 m 0000267 2n 01075 m 28 y 008353 m 0000302 m 008383 m 29 ye 01243271 000028 m 01246271 30 We see that the de ections are hardly different from before7 so we dont expect much of an effect due to the bearing de ections Just to be sure7 we compute the critical speed again 80 lb01075 m 120 lb008383 m 80 lb01246 m m 386 39 2 31 w lt 15 801b01075m2 120 lb008383 712 80 lb01246m2 r61 60 s 6059 d 579 32 m 8 2 rad 1mm rpm So there is effectively no difference when the bearing stiffness is included Thus7 the bearing stiffness is negligible 182 a The constant torque electric motor must supply a torque approximately equal to the average load torque Averaging the load torque to nd the driving torque Td iOZWTI d6 i 8 18kN m 6 kN m 55 kN m See Fig 2 Torque kNm 057 17 157 27 6 6rad 91 2 1 ayes Figure 2 Problem 18 2a and c7 torque vs angular displacement b See Fig 3 c Um is the area shown in Fig 2 Thus7 92 3 Um 7 T 7 Ti d0 17T55 kN m 4125 kN m 127 960 kN m 91 ax 057 131 15 7t 6rad Figure 3 Problem 18 2b7 sketch of angular velocity vs angular displacement d Assuming Cf 027 and given the average angular velocity of 3450 rpm7 the required mass moment of inertia J for the ywheel is then 2 r 2 Um 12 960 N m Jofwgm J02 34503 i mm r61 60 5 J 0496 N m sZ 186 The maximum stresses are at the inside radius of the ywheel which matches the out side radius of the shaft At the inside radius7 the only non zero stress component assuming no interference t is the tangentialhoop stress Evaluating at r a 8 2 2 2 13 2 2 b 7 7 b U 8 pa a 3V a 1 at prz 1 7 Va2 3 Vb2 For ultra high strength steel Table 347 we convert the weight density to mass density 7 76810 Nm3 7 7830 k 3 p 981m 52 97 Also7 for most steel V 03 The angular velocity is w 107 000 rpm 1047 rads Solving for at then noting that 1 kg 77152 1 N 1 at Z7330 kgm31047 rads21i 03003 m2 3 0303 m2 630 MPa For 4340 steel at 425 C7 the yield strength is Table 35 Syp 1283 MPa So the safety factor is then 139 a Using 13 15 1 F P Fi b 55 1 From problem speci cations FZ 67 000 lb7 P 87 000 lb Now7 we compute the member stiffness The problem states that the effective load carrying area of the steel anges and the gasket is 075 sq in If this were not given7 we would estimate the effective load carrying area using 7Td3n 7 304 where dm 2d7 and dub db see discussion on pp 5007 5017 which would give us an effective load carrying area of 133 sq in However7 using the value given7 the spring rate for the two steel ange members together and for the gasket are km AME 1125 gtlt107lbz39n Lstl 210 m AgEg i 075 n253 gtlt 106 psi Lg 00625 m kg 636 gtlt 107 lbm The combined member stiffness is then 1 1 7 1125 gtlt107lbz39n 636 gtlt107lbz39n 71 km 9559 gtlt 10616m To calculate the spring rate of the bolt7 we can either use the approximate formula given in 13 17 or the more precise equation given in 13 18 Lets do both to compare First7 using equation 13 17 7 deEb 7 7T075 m230 gtlt 106 psi 7 4L5 7 42062523971 b 6426 gtlt 10616m Now7 let7s use 13 18 First7 we need to know the length of the threaded region and the length of the unthreaded region in the grip We start by selecting the bolt length L such that L gt Leff H7 where H gt 05db Thus7 L gt 20625 m 0375 m 24375 m Let7s assume a bolt length of L 30 m The portion of the bolt that is threaded is of length Lt 2db 025 m 175m and so the unthreaded length is Lub 3 m 7175271 125 m Thus7 the portion in the effective load carrying portion of the bolt grip which is unthreaded is Lth Leff 7 Lub 20625 m 7125271 08125 m Also7 for a standard 7 16 bolt7 the tensile area see Table 131 is At 03730 2712 Now we can compute the bolt stiffness using 13 18 assuming UNC threads kb 4Lub Lth 1 i 4125 m 08125 m 1 deEb AtEb T 7T075 m230 gtlt 106 p52 03730 in 30 gtlt 106 p52 5991 gtlt 1061bz39n Note that this value is less than that obtained using Eqn 13 17 This is because Eqn 13 18 accounts for the threaded portion of the bolt in the load carrying portion7 which is less stiff than the unthreaded portion Now7 using the more accurate value of kb we nd 5991 gtlt 106 F 8000 lb 6000 lb 9082 11 t b 5991x106 9559 X106 SUSIOH EMD homework solutions11 Due 428 13 11 13 14 13 19 13 20 1311 From problem speci cations Bolt is 13 UNC steel with S 101000psi Syp 85000psi Sf 50000 psi th 3 assuming this is the fatigue stress concentration factor and not the theoretical stress concentration factor and Am 088 0712 a For the case with no pre load all of the uctuation stress is supported by the bolt F1 Pb P The bolt thread at the inner end of the nut is the critical point because of high tensile stress plus stress concentration from the threads plus a smaller root area supporting the load From Table 131 A 01257 0712 Only an axial load is applied Remembering that we only apply the stress concentration factor to the alternating stress here we assume that the value of Sf given is not already modi ed by the fatigue stress concentration factor we obtain 2500 lb maxinom 7 19 I U 01257 m M Umininom i 0 Uninom 05Umamin0m 7 Umminom 97 pm Um 05Ummimm Uminwom 9 944 108239 0a thaan0m 3 9 944 psi 29832 psi First we see that amamnmm lt SW therefore there is no yield Even if we apply the stress concentration factor to the maximum stress it is less than the yield stress so there is no large scale and no local yielding Now checking fatigue 0 7 29832 psi 17 m 717 9344 33090ps St 101000 UeqicR The existing safety factor for this case is then Sf 50 000 5 151 Ueq70R 33 090 77 b From 13 16 the minimum preload F to prevent loss of compression in the housing Fm S 0 is km Figt Pmam Since the load carrying portion of the bolt is entirely unthreaded7 we can use 13 17 to obtain the spring rate for the bolt 7 wdgEb 7 7r05 ln230 gtlt 106 psl 7 5890 gtlt 106 lb 4Leff 4Leff Leff b The clamped members are entirely steel7 and the effective load carrying area is speci ed to be Am 088 m Thus 7 AmEm 7 088 ln230 gtlt 106 psl 7 264 gtlt 107 lb m Leff Leff Leff Solving now for F1 264 107 F12 X 2500 lb 2044 lb 589 gtlt106 264 gtlt107 So the minimum pre load to avoid member separation is 2044 lb c If E 37 0001b7 then Fbmam 589 gtlt106 264 gtlt107 3000 lb 589 106 lt X 2500lb3000 lb3456 lb For these bolt loads7 we now have following similar analysis as in part a 3456lb 27 494 Umam 0 r 7 W 01257 2712 p 3 000 lb mi inom 7 7 I U 012572712 p52 gainom 05Umaminom 7 Umininom 17 814 psi Um 0395ltUmaminom i 0min7n0m 257 U0 tha39ainom 57 Again7 we see that Umakmm lt Sm77 therefore there is no yield even if we applied the stress concentration factor7 we would not predict yield7 so no local yield is predicted either Now checking fatigue7 7 0a 7 54421082 77 297 UeerR 17 m i 1 25680 i 7 1052 Su 7 101000 y r B V y 47 Th v Th 3 lt x gt 51n r 9C C Figure 1 Problem 13 14 The existing safety factor for this case is then 685 n 7 sf 7 50 000 5 7 5qu 7 5838 Note that preloading the joint with a 3000 lb preload7 not only prevents member separation7 but also increases the safety factor by more than a factor of 4 from 151 to 685 This is because by applying the preload7 only a fraction of the uctuating externally applied load kbkb km 01824 or just 18 is carried by the bolt So although the mean stress is increased7 the alternating part is decreased comparing with part a 1314 For the joint con guration of Figure 1313 147 both direct shear and torsion like shear must be considered The directions of the direct shear stress 71 and the torsion like shear stress 7 on each bolt is shown in Fig 1 The shear due to torsion Will be higher on bolts A and 0 because TA To gt TB and thus 73A 730 gt 733 Furthermore7 since the shear stresses are more closely aligned at C than at A7 the overall shear stress on bolt 0 will be highest Thus7 bolt 0 is the most critically loaded bolt a For direct shear7 assuming the shear force P 107 000 lb to be equally distributed over the 3 bolts and the bolts carry the shear stress on the unthreaded part7 then P 107 000 lb4 7 7545 39 Tb 3Ab 37r075 m2 p52 The shear stress acts downward as shown in Fig 1 and is the same on each bolt b For torsion like shear7 we must rst nd the centroid of the joint Since the bolts are all the same7 each with bolt area Ab the centroid is located at 321192 2Ab0 21175271 a 92 167 m 3211 2Ab0 A175 m a g 167 m The torque acting about the centroid due to the load P is then T 10000 lb11m 167 m 1267 700 in lb Now7 we can compute To and 60 To 167m 5 m i 167 m 373 m 167 60 tanil 7 266 5 i 167 To compute Jj we need TB too TB 167 m 167 m 236 m Now7 we compute J for the joint7 77075 m Auriw w 4 2373 m2 236 m 14752714 The shear stress due to torsion at C is then T 126 700 1b 373 39 no LC 7 m x m 3204 1m Jj 1475 m4 The overall shear stress at C is then vector sum 70 77545 7 3204cos266 239 sin266 j 728657 21873 km Thus7 the magnitude of F0 is then Fol 360 165239 c The uniaxial equivalent stress7 using the distortion energy theory and assuming the only stress acting on the critical bolt at C is the shear stress found in part b7 then gag V370 624 km For SAE grade 8 bolts Table 133 Syp 130 ksi Thus the safety factor against yield is d Find the bearing stress The bearing stress is the force that is acting on the surface of the bolt divided by the projected area over which the force is acting That bearing force is the same as the force that is shearing the bolt Thus F 70A 36 000 psiw F 15 904 lb d 075m1m 15 904 lb 0 212 ksz where t is the thickness of the plate 1319 In this case we have 2 parallel welds and a transverse weld This is a fatigue problem so we will need to consider the fatigue stress concentration factors for these types of welds From Table 139 at end of parallel llet weld Kf 27 and at the toe of transverse llet weld Kf 15 The load will be equally carried on all parts of the weld for this loading con guration so failure will occur at the end of the parallel llet weld which has the higher fatigue stress concentration factor The loading is completely reversed and using 13 48 T FW 7 18000 N w WWW 0707st T 07078017 7 150 gtlt 105 Nm Twnomimam f 7150 gtlt105 Nm Tu nomimin s Twm 0 7150gtlt105Nm 404gtlt105Nm m Kneading 20 s s Harm 0 700 X 105 N m Ueqia 7 07101 UeqioR Ueqia The electrode strength for E60 series Table 1313 is Sm 427 MPa Assuming a cast microstructure reasonable since welding involves melting and solidi cation S f 045m 171 MPa p 248 Comparing this to the base metal 1020 steel where S 228 MPa Fig 531 we see that the electrode material is weaker So we will base our calculation on the weaker material for safety Sf kegs 075171 MPa 128 MPa For a design safety factor of 25 then Sf i 128 gtlt106 Nm2s 5qu 7 700 gtlt 105 Nm 5 0014771 14 mm 71d 25 A llet weld size of 14 mm is recommended This is not compatible with the 10 mm plate thickness Either a stronger electrode material is needed or a thicker plate A stronger electrode material is recommended E 90 1320 Direct shear and torsion lead to shear stress The welds are along A B and 13 0 The furthest points from the weld centroid are points A and C7 so these are where the torsion induced shear stress will be highest At both points7 the direct shear stress acts straight down Based on the direction of the torsion7 at A7 the torsion induced shear stress will act up and to the right7 and at C7 it will act down and to the left Thus7 point C will be the critical point as the direct shear and torsion induced shear stress are more closely aligned and will have a higher vector sum The throat is t 070757 and s 0375 in from diagram The weld length is Lw 10 in So the shear stress due to direct shear is P 5000 lb 7 1 886 39 th 07070375 m10m p52 Tu Using the weld table for torsional properties of llet welds Table 13147 we can nd that the centroid of the weld is located 125 in down and 125 in to the left of point B The unit second polar moment of area can also be computed noting that in this case b d 5in 2b 476b4 5123 Ju L 7 5208m3 24b 12 T 50001b525m 26 250 m 7 lb The shear stress due to the torque at G acts down and to the left at an angle of 184quot from the horizontal Tr 267 250 m 71b3953m 7 7 515 39 7 tJu 07070375 m5208 m W F 4in tcos184 2s1n184 5 77713047260p5 F 8300 psz The direction is at an angle of b tan 1 309quot above the leftward horizontal EMD homework solutions 12 Due 55 10 3 10 12 10 3 This is oscillatatory motion the average sliding velocity is calculated by dividing the distance traveled in a single oscillation by the time which is inversely related to the frequency The links rotate b 10 01745 rad both ways ie and for each oscillation at a frequency of 60 oscillationsmin The bearing bore diameter is 125 mm Thus V2 2 rf df 01745 rad00125 m60mm 0131mmz39n 1 The projected bearing pressure P and PV are then W 2000 N P 7 64MPa 2 dL 00125 m0025 m PV 0131mmm64 MPa 0838 3 771771 Checking Table 101 all the calculated values are below the limiting values for nylon so the design is acceptable 10 12 Known information from the problem statement Bearing is 360 journal is to be steel sleeve is to be copper lead alloy L710 37705 1 rd210m 2 W 1000 lb 71 3 000 rpm 50 AAAA co VVVV Using Table 102 for automotive crank shaft copper lead bearing for a shaft diameter under 2 inches we get 0 00014271 c 7 00014 7 7 T 7 00014 C The shaft diameter is 2 inches so the clearance 0 should probably be a little higher but we will use this value for now For SAE 20 oil at an estimated operating temperature of 90 130 F7 from Fig 1037 we nd the lubricant viscosity to be 77 50 X 10 6rehns a Now7 we want to nd rst the minimum lm thickness This is directly related to the eccentricity ratio7 so we will use Fig 1014 Furthermore7 the problem statement mentions that load carrying capability and low friction are both important7 so we know we want to be in the mid range of the gray area We have everything we need to compute S W 1000 lb P E 7 2 m1 m 7 500 W 7 2 2 6 S 1 50 gtlt 10 ps1secy50 revsec 03926 8 c P 00014 500 p52 From Figure 10147 for Ld 05 and S 0267 we nd that we are in the gray region between minimum friction and maximum load7 but closer to maximum load To obtain a better balance between minimum friction and maximum load7 we could increase the clearance c Mid range between minimum friction and maximum load along the curve Ld 057 S m 016 To attain this S7 the clearance should be 2 6 i 5016 50gtlt 10 ps1sec50 revsec 60390018271 c 500 psi For this value of clearance and corresponding S reading across horizontally on Fig 10147 we nd the eccentricity ratio for this design is E 072 The minimum lm thickness is then ho c 7 e 01 7 e 00018m17 072 000050 m 9 b Using 10 317 we obtain limits on the surface roughness7 where we set p1 and p2 to both be 50 since no other information is available ho 2 lej pgRb Rb 000050 39 gt Rj Rb 100uin 10 Using Figure 6117 if the sleeve were reamed7 Rb 63 Min This would mean the journal roughness would be limited to Bi 3 37 Min Choosing the grinding process for the journal with7 Rj 32 Min So the recommended manufacturing processes are 1 Ream the sleeve to 63pm or less 2 Grind the journal to 32pm or less c The power loss is the tangential friction force F times the sliding velocity U The sliding velocity is just U 2mm 27T10 m 50 revsec 314 271560 11 The friction force can be obtained from Fig 109 For E 072 and Ld 05 F1 HT m 95 12 r where F1 FL Now putting the known values into 127 and solving for the friction force F F 00018 95 F 83 lb 13 50 gtlt 106 lb s n2314msec1m T Now solving for the power loss 1hp H FV 83 lb 314 ms 04 hp 14 d The oil ow rate in the bearing clearance space can be determined using Fig 1011 for E 072 and Ld 05 Q 7quotan 5392 15 Q 52 16 Q 0472 7135 17 EMD homework solutions 13 Due 510 11 10 11 1316 13 1110 a By problem speci cation dmmm 2 163 m 1 F 800 lb 2 Fa 0 lb 3 n 1725 rpm 4 R 90 5 Ld 1725 6om7musoo hr 1863 gtlt 108 m 6 mm hr and moderate shock loading exists which from Table 113 for V belts and moderate shock conditions IF 175 The reliability factor associated with 90 reliability is just KR 1 A single row radial ball bearing is to be selected Since there is only a radial load P5 F 800 lb 7 We can now calculate the basic dynamic radial load rating using 11 4 od9oeq 1 mpg g 175800 lb 7996 lb 8 From Table 115 the smallest acceptable bearing with a bore at least 163 inch is bearing No 6309 limiting speed ok Checking static load rating P55 800 lb and from Table 115 C5 7080 lb gt PS 800 lb so the static load rating is also ok So select bearing No 6309 and locally increase the shaft diameter at the bearing site to d 17717 inch with appropriate tolerance b Using Table 115 look for P5IF 800175 1400 S Pf 9 The in nite life requirement is satis ed by bearing No 6040 which has a bore size of 7874in No 6040 has a size which is much bigger than No 6309 and is probably much larger than allowable 1113 First7 calculate bearing reactions at A and B It is straightforward to nd the following reactions at bearing A Fall 2050 lb 1 FaA 2500 lb 2 and at bearing B FTB 4950 lb 3 F03 01b 4 The angular velocity is n 350rprn and the desired reliability is R90 A life of 3 years at 8 hrday7 5 dayswk corresponds to Ld 350 revmm 60 mlnhr8 firday 5dayswk 52wkyr 3 yr 131 gtlt 108 Tab lt5 Using Table 1137 for commercial gearing7 IF 12 Using Table 112 for R 907 KR 10 For bearing A7 using 11 3 and Table 114 for single row roller bearingoz 7E 07 which is for preliminary selection of tapered roller bearings to support a thrust load Xdl 17 Ydl 07 ng 047 and X12 04cotOl Since 04 is not known till bearing is selected7 we will assume a typical value for X12 from Table 117 For bearings with a bore size of at least 1375in7 a typical value is ng 177 and revise later when the bearing is selected and re check P51 2050 lb sz 042 050 lb 172500 lb 5 070 lb P5L lt Pe2 P5A Pe2 5 070 lb 6 Calculating the basic dynarnic radial load using 11 47with a for roller bearings l0d90lreq E 131 X 108 1257070 lb 26 265 lb 7 From Table 1177 tentatively select bearing No 323087 which has a value of Cd 27700lb and X12 174 This is slightly higher than the assumed value7 so lets recalculate to check it is still OK 32 042050 lb1742500 lb5170 lb 7 131 gtlt108 125170 lb 26783 lb 8 Od90lleq which is still below the allowable value Checking the static load rating now From Table 1147 we nd X51 17 Ysl 07 X52 057 and Y52 05ng 0877 for the chosen bearing Then7 Peg1 2 050 lb P592 052050 lb 0872500 lb 3200 lb Pes1 lt Pes2 PBSA P592 3200 lb 9 For bearing No 323087 05 337 700 lb gt 1355M7 so we are well below the static load rating So the recommendation for bearing site A is bearing No 32308 bore 15748 in outside diameter 3543 in width 13878 in Repeat the above for bearing B Since there is no axial load7 P5B 1355 FAB 47 950lb Then at B 131 gtlt 108 12 4 950 lb 25 643 lb 10 LOW lt gtlt gt ltgt MW From Table 1177 we nd that bearing No 32308 is acceptable again with Cd 27700lb and OS 337 700lb So the bearing at B is identical to that at A 1613 Figure 1 Problem 1543a From speci cations A 1 kN 1 a 4 see 2 m 100 mtg 510p a u CW 4 M 0 5 5 b EM at D CCW NbrhNdrFea 0 s Nb 7 w 7 so the brake e selfrenewm smce the mctmh moment ms F In applying the brake From 7 A u 8 1 Wm be selHockmg checkmg this W 0 32 gt 0 9 so the brake e mt selfrlackmg c Taking the drum as a free body summing moments about the center of drum rotation Tf Ru N 10 From 7 Faa W 3111 N 11 b 7 ud 30 i 0515 So 10 gives Tf 015 m053111 N 2333 N 771 12 d Summing forces horizontally R MN 0 13 or R iuN 7053111 N 71556 N 14 Summing forces vertically R N i F 0 15 or BmFaiN1000N73lllN72111N 16 e To bring the moving system to a stop the energy dissipated Ed is just the torque angular force times the angular stopping distance 27139 rad Ed Tf6 2333 N m100 rev lt rev 146 600 N 771 1466 kJ 17 f To make the brake self locking 2 7 2 60 cm 18 Meme 1 Ma nee body dmgem EMD Homewozk Soluuons w Peobxens from Mecfmnwal 17mm 0 Mom Hmmts am Macmesby 1m A Collan pxepaxed by Antoinette M MemeLLy and 1111 Lu 1 shown In g 1 where we assumed lhsl me heexmgel 5 supmne mmmmd es xx end 23 A001 ddmum Wm he meme end usmg our mngu emy fumm hemem mm desse we can Me for the downwexd demeame m 5 Mz 53 sum 3 lt2 7 so gt We 1 1 dc mm m use esmeuemy mm m the st mm because Mus mm 1m In ng Oeend xenmxymnemememgeemeheemem 2 mm mm punmg wesmgulsnlyfuncnmlmns a me meem mememe 2 A0 me these an m In new en the em ex the beam end um heve he Impact meet me ehoveequsums We 5 emeazeweeem 5 em eaaaeum lt 27 so gt3Cncz Nexu new we houndsxy mum at me two heexmsuppmeem 0 end we to obtain the two constants of integration Cl and 02 7EIU0 020 78717430 7889303Cl30 0 a 01 8000 So we have nally 1 vz E 889953 i 889 lt z i 30 gt3 7800095 The maximum de ection will occur7 either between the bearing supports or at the overhung end Let7s rst compute the de ection at the overhung end Since the shaft is steel7 E 30 gtlt 106 psi and I D 07854 in Evaluating at the overhung end then gives 040 0010 in Between the supports 0 lt z lt 307 the maximum de ection will occur when 2 267952 7 8000 0 Note7 for z lt 30 the singularity function term lt 730 gt 07 thus7 we don7t need to include it here We can now nd the location of the maximum de ection between the supports by solving the above equation7 and we nd it is at z 173 in Now solving for the de ection at this location7 we nd 0173 70004 in So the magnitude of the maximum de ection between the supports is much less than that at the overhung end In fact7 if we want to avoid large de ections7 we should avoid having shafts overhung like this one 417 Placing a dummy load Q at the free end of the beam where the de ection is to be determined7 the bending moment at any point along the beam is see Figure 2 MABz Pa 7 g Q 7 L MBicW QW L P100b Q L25 mo A B c Figure 2 4 1 7 The stored energy is L a M343 L M54 U d d 0 2E 95 2E1 95 Using Castigliano7s theorem we differentiate U with respect to Q to nd the de ection 60 at C in the direction of Q 6U 1 aMAB 0 M434 SeacEIUo MHlt 6Q dzMB C 6Q 6 Next differentiating the MAB and MBC with respect to Q gives aMAiB i 5MBic 6Q 6Q L So the de ection equation becomes iMWimmiwdw Qz 7 Lx 7 L ds 3 Setting the dummy load Q 0 results in L P 7 L Sci O z7 7Ld L P 3 3L L2 607 277 i d Ei0 95 295 2 95 A 2 01 or 6 77 37ng2 Liz 75PL3 c E1 3 4 2 0 48E For steel material E 30 gtlt 106psz39 Evaluating7 5 100 lb 10 39 3 60 417 gtlt 10 32397101 4830 gtlt 10 p52T 419 The square bar is subjected to bending and torsion while the round one is subjected to only bending Here7 we use subscript 1 to refer to the square bar and subscript 2 to refer to the round bar Let 1 be distance from the end of the square bar to a point in the square bar and x2 be the distance from the end of the round bar to a point on the bar The total strain energy of the bracket is then L1 M2 L1 T2 L2 M2 1 2 7 d 7 d 7 d 0 2E11 1 1 0 2KG 1 1 0 2EIZ 952 where M1 P l M2 Pig T PLZ I i 54 702034539 2 1 7 127 m d4 12 L011984m4 64 4 16 11 12539 4 K f i336i J 2253303438m4 2 3 12 2 For steel7 E 30 gtlt 106 psi and G 115 gtlt 106psz39 Differentiating with respect to the load7 P7 which is at the location and in the direction where we want to compute the de ection results in 6U 1 L1 2 1 L1 2 1 L2 2 077 2Pd 7 2PLd 7 2Pd y 6P 2E110 9 1 2KGO 2 lam0 9 2 952 PL PL PL1L 3E1l 3E12 KG 10001110239713 1000115239713 3 30 gtlt 106 p5 020345 m4 330 x 106 psz011984 W 1000 lb10 m5 m 115 gtlt 106 ps 03438 m4 00546m001159m06323071 013m Note one may also choose to make the cut breaking up the two members at the location where the round bar intersects with the square bar This would change the answer some because instead of integrating the last integral above from 0 to L2 it would become 0 to L2 7 52 Neither is exactly correct but the above is a bit more conservative and includes the de ection where the round bar connects to the square bar due to twisting of the square bar Remember that these equations assume long thin members and the round bar is not so long relative to its diameter so this is a rough approximation in either case 420 A free body diagram of the shaft with the gear on it is shown in Figure 3 below where l7ve assumed that the axial force is supported at bearing B the problem fails to indicate which bearing supports the axial force but normally in a situation like this the bearing sitting against the gear will support the axial thrust load We can determine the reactions RAy R3 and R31 using moment and force equilibrium as follows 016 EMA RBy06miP016m0 4 R3 P 06 016 211 0 a RAy7RByiWP 2F 0 a BM P The shear and moment diagrams for the shaft are shown in Figure 4 Neglecting strain energy in the gear this is reasonable since the gear is large so the average stress and strain will be relatively small compared to that due to bending in the shaft the strain energy in 50 mm D 600 mm 160 mm 1 Figure 3 Free body diagram for 4 20 the system in terms of the applied load P is 06 M2 7 0 2E1 1 where 016 M RAy i Ps By Castigliano7s theorem then the de ection at the location of the applied load P in the direction of P is i 6U i 0396 M 6M 6 7 7 7 776 6P 0 E 6P 9 where 6M 7 701695 6P 06 Thus 06 1 0396 016 2 2 1 016 2 3 5 P95 d m P 1 lt0162P06m3 E 06 0 3 For the steel shaft E 20 GPa 207 gtlt 109Nm2 The load is given as P 2400 N For the solid round shaft 7 l 4 l I 7 64d 64 So the de ection is then 6 194 gtlt 10 4 m 0194 mm 005 m4 V N x 111 702667 P Figure 4 Shear and moment diagrams for 4 20 426 A free body diagram is shown in Fig 5 Note the reactions at the supports RA RB and MB depend on the load P which is emphasized in the gure The reactions are whatever they need to be to constrain motion of the beam under the given loading condition The load P does not depend on the reactions Thus when we write the strain energy to nd the reaction RA we do not try to write P in terms RA On the other hand in part b below where we solve for the de ection at the applied load P we do write RA in terms of P gtMBP RMP RBP Figure 5 Free body diagram for 4 26 a To nd the reaction at the support using Castigliano s method we will write the strain energy in terms of RA and then differentiate with respect to RA which gives us the de ection at the location where RA is applied which we know is 0 The beam is only subjected to bending The moment along the beam can be expressed as 2 equations M1lt M2lt V l 7 RIM for0 a RAziPQia foragng V l l Note7 we dont use singularity functions here to express the moment as 1 equation7 but rather we write it out in separate equations for the different segments of the member Singularity functions complicate the integral7 so when we use Castigliano7s method7 we don7t use them The derivatives of the moments with respect to RA are then 6M1 7 QRA 7 6M2 7 QRA i The derivative of the strain energy is then 6U 1 1 6M1 L 6M2 7 5 0 7 M d M d am A E 1613A xa 2613A 9 7 1 a 2 L 2 2 i RAz dza RAz P ax dm i i RA3Q Puffy Paxz L i E 3 0 3 3 2 a i 1 RAL3 PL3 PaLZ Pa3 E1 3 3 2 6 Solving for RA in terms of P yields 3a a3 p 1 7 7 7 RA 2L 2L3 Plugging in the numbers listed in part b7 P 400 lb7 a 4 ft7 and L 10 ft7 we get RA 1728 lb b Putting the relationship for RA in terms of P back into the moment equations we get 3a a3 7 7 lt lt M x Plt1 2L2L3 for07ia 3 3 3 3 M2x Plt1727P27aPizPa foragng Letting c 1 7 3 and d c 71 2L 7 3 then we get the moments and derivatives of the moments with respect to P as M105 M205 3M1 7 lt lt Pcz 6P 0x for07ia 3M PdPa 6P2da foragng The de ection at the applied load P in the direction of P is then 61 Substituting the numbers given in the problem statement P 400 lb E 30 gtlt 106 psi I 100 in4 1 4 ft 48 in L 10 ft 120 in we obtain 6 226 gtlt10 3in 427 We can use Castigliano7s method to solve this problem by noting that if the beam touches the support at B there will be a reaction force R3 at this location acting in the direction opposite the 5 mm de ection However rst we might want to check if the beam will de ect under the prescribed loading 5 mm at B and touch the support Figure 6 shows a free body diagram of the beam and if the beam does not touch the support at B the force RB will be zero and this case will be identical to that shown in Table 41 Case 3 The de ection at B may be computed then as note I use 1 to denote the de ection in the direction of y rather than y so as not to confuse the coordinate direction with the de ection JLU 53P 1 6M L 6M M1 1dx M2 2cm 0 P a L i dLa3 7 daa3 1 El 6 6P Pczxzdera Pdxa2 dz 1 L czmzdxa da2ddz 3 WM E1 3 0 d 3 a P E P dL 13 i c3a3 wa 7 37 2 3 v7 24EIL 2La a where w 5000 Nm 1 2 m L 8 m E 207 gtlt 109 Pa steel and 1 1 Ems m002 m3 2 gtlt10 8 m4 Substituting into the equation for 1 we nd 1 459 m This is a huge de ection and clearly beyond what these small deformation formulas can predict accurately Clearly7 this problem has not been well de ned because the beam is inadequate to support the given load At any rate7 we will proceed7 noting that the beam will de nitely touch the support at B under the prescribed loading Now7 if we want to solve for the reaction at B7 we need to write the Figure 6 Free body diagram for 4 27 strain energy in terms of RB First7 lets get RA in terms of RB Summing the moments about 0 1 1 b 2M0 RALiiwL2RBb0 a RA wLifRB This beam is only subjected to transverse loads7 and since it is long and thin7 the strain energy will be dominated by that due to the bending moment The bending moment7 in terms of RB can be expressed as 1 1 1 b M1 iRAx7wx2iwx277sziRB for0 a 2 2 2 L 1 2 1 b 1 2 M2 7R4anwa 7RBxia7 wLZRBx wx 7RB7a 1 iiszlt1iggtRBalt1iggt foragng where we used our expression above for RA in terms of RB and also b L 7 a to obtain the above relations The de ection at B in the direction of RB7 as it is drawn in Figure 67 is then using Castigliano7s method 6U 1 1 6M1 L 6M2 6 77 Mid M d B 6R3 E1 1633 a 26133 95 where 6M1 b 633 33 6M 6R Thus 7 1 wb 3 wb 2 13312 2 L 2 z 2 waL z 2 63 i 0 7x L2 x dx a R36 1 2 z 1 dx i wba4 wbag RBb2a3 RBa2b3 7 waLg Lag 6L2 7 80L 3612 E 8L 6 3L2 3L2 24 24L 7 1 wa 2 3 3 RBaZb2 EI24 2La a L 3L Solving the above for R3 results in wL 3 3 2 363LEI RBL 0 2LaW and putting in the numbers from above and noting that 63 70005m note the minus sign because 63 is in the opposite direction of RB we nally nd RB 31 66667 N7345 N 31663 N Note because the beam is too thin for the load making I too small we dont see much effect from the second term above Since the above equation is symbolic it will be easy to use when the design is changed or to help in redesigning the member changing the cross section or the lengths etc 440 a De ection of a cantilever spring subjected to a load at its end is Table 41 case 8 FL3 100 lb16 m3 W 330 gtlt 106 p5 T122 n05 n3 b De ection of the spring with a spring rate of k57 300 lbin is 02184271 F 100 lb 5 k5 3001bm 03333 m c De ection of cantilever and spring when spring is placed under the end of the cantilever In this case the cantilever will carry part of the applied force the spring will carry the remainder of the force7 and both will de ect the same amount Fb F9 d F1 6041 1b FbL3 Fs 7 d F1001bF Fs 3E kw an H Fb16z39n3 Fs 330 gtlt 106 ps f 2 m05 my 3001bm 15259FS 7 F5 FS 25259FS 100 lb 3059 lb 3959 lb F 6041 lb 6 1 3001bm 0132 m EMD Homework Solutions to Problems from Mechanical Design of Machine Elements and Machines by J aok A Collins prepared by Antoinette M Maniatty and Jing Lu 566 To utilize Mohr7s Circle to solve problem 5 65 the sketch of gure 1a is typical of all critical points The critical points are located along the extreme bers farthest from the neutral surface of bending due to M1 since M produces critical points over the entire surface The X y plane of Figure 1a may be redrawn as shown in Figure 1b Mohr7s circle may next be constructed as shown in Figure 1c by plotting points A and B as shown passing a circle through them and nding principal stresses 01 and 03 semi graphically The second principal stress 02 0 is plotted at the origin From Figure 1c Figure 1 5 66 Using 4 33 and using 4 5 Then Ur 6 1723 l2739m2y 0392 0 i 7 Um Um 0393 O R17237 i27m2y 1 71mg 716M 1 J 7rd3 0 7 i 32M7 F I 7rd3 7 16Mb 16Mb 2 16M 27 16 2 2 7 7rd3 7Td3 7Td3 i leJr Mb Mt 0 l6 lei VMb2Mt2l The principal shearing stresses are just the radii of each of the circles 0397039 8 Mllzsb7g7MWWM 2 039 70 16 imig wmhm 039 70 8 lTsl ll leVMb2Mt2l 568 a Special points on the cross section of the bar referring to Figure 2 Shear stress caused by torque reaches rnax value at the outer surface of the bar and Shear stress caused by F1 reach max value near neutral axis so point B and D are of special interest However since the shear stress due to torsion and that due to F1 both act in the semedlrectmel 1717 s dyeew In em 0119115 Name sues byhmdmgmanemxeedq mexvelueellhempendh lumpmn39s m h A end Cexe ax Spams mm Firmware emwem A end 0 A 1 male mch smoe n hes mug stress So we Wm mm Points A end D gure 2 568 amrsecnm dmmm h m sues mm cf m mmnmel emsz ea mm A end D ere shown In gure a mug me these exe 27D stress as on m Surface cf mm we could also st chew the 271 stress damn shown m The stress wed w max mam 91 mm A 5 7 7 7 50 ooomitb15 m 1 1 7m o m 6A 51w mused w Wane an mm surface 5 Haw 2 7 7 1 macaw V J so Sheexmusedby Hummus ng 7 Jr so x How 11 amomeA 39 E 4 lTM 7 4 ar 03 mm a sea sues mm m 17an A Ian and 17an s um a Figure 4 5 68 2 D stress elements for point A left and point B right Now we know the stresses acting on the 2 D elements in Figure 4 Using a Mohr7s circle analysis it is easy to show that the principal stresses at A are 01 26434p5239 0392 0 03 7757105239 and at B 01 31124psz39 0392 0 03 731124 psi c From the principal stresses it is easy to compute the maximum shear stress for point A max 7 mm 26435 7571 Tmaz U 2 U 2 17 003 p52 and for point D 7 31124 31124 Tm 7 2 31124 psz 582 From Table 33 S 38ksi Syp 27ksi From Table 310 e 5 in 2 inch gage length therefore this material is on the border between brittle and ductile so to be conservative we will treat as brittle stress concentration effects important apply full There are three stress concentration spots For the hole in the plate the hole is near the center of the plate the nominal stress there is zero even with the stress concentration of the hole the stress near the hole will still be near zero7 so we dont need to consider the hole That leaves the 2 llets Here hl 125in7 hg 15in7 hg 45in For the llet of r 0125inrh1 017 haI711 127 from Figure 57a Kn 17 6M 6 850 m 7 lb MC 039 Kai Kn I W 2 29 594 psz gt Syp 271m 01875 m125 m So the stress at the llet of r 0125 in exceeds yield7 but does not exceed the ultimate strength Furthermore7 the nominal stress7 without the stress concentration factor is only 17408 psi7 which is well below yield7 so large scale yield is not predicted7 and local fracture is also not predicted Now consider the llet of r 015in7 rh2 017 h3h2 37 from Figure 57a Ktz 19 MC 6M 6 850 mill UKtliKt1 7 1 22969 39ltS I W 01875m15m2 7 p52 3 So local yielding only at the llet of r 0125 in is predicted7 but complete failure fracture is not expected since the stress at this location is still well below the ultimate strength of the material 583 The cyclically loaded machine part shown in Figure P5 83 has three potential critical points one at the i diameter hole7 one at the 025 radius llet7 and one at the 018 radius llet Since the part is subjected to cyclic pure bending7 and the hole is at the neutral axis of bending7 the normal stress is near zero Even with stress concentration7 the actual stress at the hole will therefore be near zero7 so the hole may be ignored as a critical point Comparing the two llets7 it may be observed that see Figure 57a for the 025 radius llet7 the ratio Hh is smaller and the ratio rh is larger than for the 018 radius llet7 so at the 018 radius the stress concentration factor is larger and the nominal bending stress is also larger Clearly then7 the governing critical point is at the 018 llet If it were not clear7 for any reason it would be simple enough to calculate the actual stress7 am at each potential critical point and select the largest as the governing critical point For the 018 radius llet then7 r i 018 H 200 h 7 7011 7122 h 164 164 So7 from Figure 57a7 Kt 17 From Figure 5467 under cyclic bending loads7 for steel with S 57 0001052 and for llet radius r 018m q z 08 So using 5 92 Kf qKt 71 1 0817 i11 156 The nominal alternating bending stress at the root of the llet is 7 MC 7 6M 7 64000 m 7 lb nomii 23795 39 U I W 0375 m164 m2 p Since the bending moment is the only applied load and is completely reversed there is only a single alternating stress component due to bending and no mean stress components The alternating bending stress including the fatigue stress concentration effect is then an Kfamm 156 23 795 p52 37120 psz Then 05W 37 120 psz 05147 0 Ueq70R 37120 psz and from Figure 531 the estimated life would be N m 106 cycles EMD Homework Solutions to Problems from Mechanical Design of Machine Elements and Machines by Jack A Collins prepared by Antoinette M Maniatty and Jing Lu 537 a From Tabel 337 the properties of AlSl 1020 CD steel are S 421 MPa Using method on P247 Slv Sn 421 MPa at N 1 cycle 8 055 211 MPa at N 106 cycles since SM lt 1379 MPa 200 ksi The resulting plot is shown in Figure 1 N yed Figure 1 S N gure for 5 37 b From the plot7 ng os 211 MPa This is from the estimated gure c From Figure 531 P2467 5106 37 ksi 255 MPa This is from actual test data and is about 20 higher than the estimated value Thus7 the estimate is conservative in this case 540 a From Tabel 337 the properties of AlSl 1060 HR steel are S 987 000 psi7 Syp 547 000 psi Using approximation on P247 5 z 055 497 000 psi Now we apply modifying factors to S Sf kgkwekfks39rkszkrskfrkcrkspk39rs where kg 10 no details7 pick typical value km 10 no weld mentioned kf 10 ignore stress concentration k5 076 machined nish Fig 533 km 09 no details7 pick typical value k7S 10 no details ka 10 no fretting km 10 no corrosion mentioned ksp 10 200 cpm7 pick typical value k7 081 Table 547 for 99 reliability So Sf 0760908149 1m 2715 km This application is for a power plant7 which presumably runs 24 hoursday and 7 daysweek The cycle rate is 200 cycles per minute Thus7 in a single week7 we expect in excess of 2 gtlt 106 cycles So we should design for in nite life lgnoring stress concentration and safety factor7 and noting that the loading conditions are completely reversed ie Um 07 then the minimum required diameter d is computed as 7 P778000lbiS 727150 7 A ap4 f 7 p gt d191m 541 a From Tabel 337 the properties of Ti 6Al 4V are S 1507 000psz397 Syp 1287 000psz39 Using method to estimate 6 va SM 150000psz39 at N 1 cycle S 0558 82500105239 at N 106 cycles where we pick 055 as the midrange value Sf kgkwekfks39rkszkrskfrkcrkspkrs 2 and then we pick values for those factors kg 10 no details pick typical value km 08 weld but no details kf 10 ignore stress concentration k5 068 machined nish note graph is for steel may not appropriate ksz 09 no details pick typical value kS 10 no details kf 10 no fretting kc 10 no corrosion mentioned k57 10 no details pick typical value k 069 Table 23 high reliability required So Sf 1008100680910101010069825ks 2787ksz 51 blue 3 5 10 m 4 m 1 gt3 0 I 2 3 39 I VEu 39 Figure 2 S N gure for 2 40 Now from S N curve see Figure 2 SN 510g10N Sut b 720361m SN 72036log10N150 atN105 SN 4821050 P 50000 27 7N S 7 P 7 50000 aN Os 48200 5 102m Ta 542 The current reliability is 99 and the goal is to get to a reliability of 999 From Ta ble 54 our strength at a reliability of 99 is obtained by multiplying the test data strength at 50 reliability by a factor k99 081 while for a reliability of 999 we would multiply by a factor of k7ggvg 075 Thus we need to reduce our stresses by a factor of 075081 0926 assuming all other factors remain constant 543 Aluminum 2024 T47 SM 469Mll a7 Syp 324Mll a7 e 20 therefore7 ductile Axial member subject to cyclic loading First compute maximum and minimum stress 20 000 N am 7 1852 MPa 18mm X 6mm 72 000 N 71852 MPa 18mm gtlt 6mm Umin Since am lt Syp no yielding Now check fatigue Compute equivalent completely reversed stress Um 05 1852 71852 MPa 8334 MPa 0a 05 1852 1852 MPa 10186 MPa 0a 7 10186 MPa UeerR 7 m 1 7 8334 SH 469 1239 MPa 1797 ksz Now compute the long life fatigue strength at N 5 gtlt 108 From Fig 5317 ng 20ksi at N 5 X 108 Now de ne relevant modifying factors based on information provided7 which isnt much7 so mostly take typical values Modifying factors not equal to one are ks 07 km 09 Long life fatigue strength at N 5 gtlt 108 is then SN5X103 070920 ksz39 126 ksz lt Ueqicg Therefore failure is before N 5 gtlt 108 To estimate fatigue life7 estimate va by dividing Ueq70R by the modifying factors since S 7 N curve not linear7 dif cult to estimate modi ed curve7 so adjust Ueq70R instead 1797 km 2852 k 39 N 0709 52 From Fig 5317 logN 637 thus7 N m 2 gtlt 106 cycles 545 From Table 33 for Ti 6Al 4V SM 1034 MPa and SW 883 MPa and from Table 310 e 10 at room temperature thus ductile a Since the loading is cyclic with a non zero mean stress positive mean stress on the tensile side of the beam fatigue is a potential failure mode and yielding is a potential failure mode b The critical point will be at the center of the beam on the bottom It is at the center because the bending moment will be highest and thus the bending stress will be highest at the center and it is on the bottom of the beam because this is where the tensile stress due to bending is highest on the top of the beam the stress will be equal in magnitude but compressive which is less likely to cause fatigue failure c The bending moment at the center of the beam will be 1 4 M P025 m The bending stresses at the center bottom of the beam is 7 E 7 Mh2 7 7 6P025 m I 01312 bhz 005 m01m2 and the maximum and minimum bending stresses are then 0W 7 6270 000 N025 m 7 810 Mpa 005 m 01 771 690 000 N025 m mm 270 MP U 005 m01m2 a Then the alternating and mean stresses are 1 0a 507mm 7 0mm 270 MPa 1 Um 507mm 0mm 540 MPa The equivalent completely reversed stress is a 270 MP qu U a 565 MPa am 540 1 st 1 m For titanium alloys we can estimate the fatigue limit using the guidelines found on P248 using a middle value for the range listed 8 0555 0551034 MPa 569 MPa at N 106 cycles No information is given about surface nish reliability etc so we will use typical values for modifying factors The modifying factors not equal to one are k5 07 T l 01 I L a E 9 N o nguxe 3 Cycle Ccuhtmg for P55A 1m 09 k 9 assumes 90 xeheblhty Thus the letlgue lmut ls estuheted to b S 0 70 90 9569 MPe 323 at us the equweleht compleleh reversed stress Week gt 5 so lh mte hle s not expected We ceh estuhete the s N curve by essummg aslralghl hhe between 5 et 10S cycles end su et1 cycle on asemlrlogscele then S blogNA etN1 Sublog1AA etN10 s39tylog10 s u H17 5 71186MPalogN103A MPe 71186MP9 Equetmg the ebcve to as allows us to estlmete the cycles to allure s Week 565 Mm 7110 6 We logm 103A Mpe a N s 9 000 232125 Note thls was essumlhg 90 lelleblhty thus et N 9000 cycles e 10 plcbeblhty cl allure 15 estlmeted 554 Rem ow couhtmg shown m nguxe 3 Cycle count tabulated m Table 1 Let B be Table 1 Rain ow cycle count for 554 Umam Umin Um U11 Ueqi CR Number ksi ksi ksi ksi ksi the number of blocks to failure 1 Bf6gtlt103 1 Bf 6 gtlt103 100 hrs 560 First7 we must determine the critical crack length in which this crack will propagate in an unstable manner fracturing the plate when the maximum load is appled We use fracture mechanics PM 160000 lb am 5 32 000 p52 A 05 m X 10 m K c 2 33 800 2 EM 25 S 25 84500 04 m lt 05 m so plane strain K10 337 800 psz Vz KI Cgrmav Mme 018 333 psi Mme To compute am we need 01 but 01 from Figure 519 depends on ab7 where b 10 m and a am at onset of failure So we will iterate Let7s initially assume that 1 ltlt b7 so 11 is small and 01 w 11 Now 33800 psz xz39 11327 000 p82 7Tam results in am 0293071 Now we go back7 using this value of a and update 01 11 00293 and 0117 0029332 117 resulting in CI 115 Now 33800 psz xz 115 327 000 psi Mme results in am 0269071 This is close enough to the previous value so we will take the critical crack size to be ac 0269 in and at this crack length 01 115 We also need to make sure that we dont have large scale yielding when the crack is at this length The average stress over the remaining cross section excluding the crack is 160 000 lb 32 885 39ltS 10m40269m05m p52 3 Urne Thus large scale yielding is not an issue Fracture will occur rst To obtain an estimate of the number of cycles before fracture we integrate the Paris Law da 18 10 19 AK 3 cm X where AK OIAU39xTl39a At the initial crack size of a 0075 in ab 00075 01 113 Taking CI to be the average between the initial and nal crack sizes 01 114 and A0 0mm 7 0mm 32 000 psi So AK 64 659E Plugging into the Paris Law d 4 87 gtlt 10 5 3 7 a dN am0269 3 N 1 5 da 487x 105 dN 100075 0 ml 1 i2 aw 7 a0 2 42 0269 7 0075 345 487 gtlt 10 5N N 70800 cycles 561 Notes Solution here assumes plate width is 150 mm as rst stated in problem not 300 mm as stated later ii Solution here takes coef cient in the Paris Law to be 34 gtlt 10 12 rather than 48 gtlt 10 27 which is consistent with equation in Figure P560 for the same material and converting USC units to metric First we must determine the critical crack length in which this crack will propagate in an unstable manner fracturing the plate when the maximum load is appled We use fracture mechanics PW 100000 N W 556 MP U A 12 mm150 mm a K c 2 372 2 EM 25 71 25 7 00102 m 102 mm lt12 mm so plane strain W 582 K10 372 MPaxR KI OijaFTl am 01556 MPa 7Tam To compute am we need 01 but 01 from Figure 519 depends on ba7 where b 150 mm and a am at onset of failure So we will iterate Try a 50 mm Then ab 0337 and from Fig 277 0117 03332 0987 resulting in CI 179 Now 372 MPmm 179 556 MPa mam results in am 00445 m 445 mm Which was different from our guess of 50 mm So now update guess by taking rnidpoint between 50 mm and 445 mm for 1 So let a 4725 mm Then ab 03157 and 011 7 031532 097 resulting in CI 171 Now 372 MPaxR 171556 MPa 7ram results in am 487 mm So the critical crack length is between 4725 mm and 487 mm Lets take it to be am 48 rnrn7 and at this crack length7 CI 173 At the initial crack size of a0 1 rnrn7 ab 000677 01 112 To obtain an estimate of the number of cycles to reach the critical crack length7 we integrate the Paris Law da 7 712 3 W 7 34 gtlt 10 AK where dadN is in mcycle and AK is in MPmR AK OIAUxTl39a Taking CI to be the average between the initial and critical crack sizes7 CI 143 Now cornpute AU Pmm 450 N 0mm 025 MPa A 12 mm150 mm AU 0mm 7 0mm 553 MPa So7 AK 1402E Plugging into the Paris Law d i 937x10 6ag dN am0048 3 N a ida 937gtlt10 6 dN a 0 00001 72 0048 7 0001 937 gtlt 10 6N N 578 gtlt106 cycles Therefore7 the crack length is expected to become critical after 578 gtlt 106 cycles At a cycle rate of 5 per second7 578 X 1060yclesls5 cycleslhr36005 321 hr Thus7 after about an additional 320 hours of service7 the crack is expected to be at a critical length

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