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# ELEMENTS OF MECH DESIGN MANE 4030

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This 84 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 Antoinette Maniatty in Fall. Since its upload, it has received 164 views. For similar materials see /class/224914/mane-4030-rensselaer-polytechnic-institute in Mechanical and Aerospace Engineering at Rensselaer Polytechnic Institute.

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Date Created: 10/19/15

MANE 4030 EMD Homework Solutions 2 Due 27 1 2 16 The potential failure modes are force and temperature induced de ection and yielding The relevant material properties for A181 304 Stainless steel the from tables in Chapter 3 are Syp 241 MPa Table 337 w 7871 kNm3 Table 347 Oz 173 gtlt 1076 C Table 387 and E 193 GPa Table 39 First7 let7s check yield At the top of the support rod7 where it is xed7 the stress in the rod will be highest because the rod must support both the 30kN test package and its own weight The stress at this location will be F wV 30000 N 787 710 Nm37r001m25 039 7 A 7T00l 7n 9549 039 MPa 959 MPa lt Syp 241 MPa Thus7 yielding is not a problem Now7 let7s compute the force and temperature induced de ection Note that the force on the rod is linearly varying from just the weight of the test package at the lower end to the weight of the test package plus the weight of the rod at the top end7 as each location in the rod must carry the weight of the test package plus the portion of the rod below it Thus7 the stress and resulting strain are also linearly varying The force induced de ection will be the average strain along the length of the rod due to the average stress7 0a times the length of the rod 957 Ta 7 6 7 8quot at 7 f aAe L 7 193 000 173 gtlt 107680 5 m 000247 m 000692 m 00094 m 94 mm Since 94 mm gt 8 mm7 failure is predicted7 and the design should not be approved E0 The wear depth is d kmes At any location on the brake pad7 the sliding distance of the contacting brake rotor will be LS 27WN7 where r is the radial position and N is the number of revolutions Thus7 the sliding distance is proportional to the radial position 7 For the wear depth to be uniform ie constant7 the nominal contact pressure will then need to vary inversely with r Thus7 c Pm r where c is a constant 3 3 5 Using the speci cations described7 the following application requirement table Table 1 based on Tables 31 and 32 in book could be lled in 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 Material data for these evaluation 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 4 a At A7 we can see that there will be shear stress due to torsion and bending stress To compute the stress components7 we need to nd the torque and bending moment acting at A The torque acting on the shaft at this location is T 600 N m To compute the bending moment7 we need to make a cut at A and nd the moment that balances the forces and moments on the remainder of the shaft Before we do that7 we need to nd the force P and the reactions at the bearings7 using equilibrium Balancing the torque moment about z allows us to nd P7 T 7 P015 m 600 N m i P015 m 0 e P 4000 N Next7 balancing the moments about 2 we assume the ywheel is right against the Table 2 Candidate materials by evaluation index able carbon able copper7 able 31 able 311 able able 31 cost gray cast able 31 Zinc 7 able low carbon medium carbon bearing7 so that no moment is generated by its weight and the weight of the ywheel is entirely supported by the left bearing at the origin7 713025 mR0y05 m 74000 N025 mR0y05 m 0 a Boy 2000 N Since there is no force acting in the z direction on this system7 R01 0 Also7 we will assume that the bearings only support a normal load7 so there are only force reactions7 no moment reactions at the bearings ie they prevent de ections of the shaft in y and 2 but they dont prevent rotation of the shaft about y and Now we know all the forces acting on the crankshaft to the right of A Thus we can make a free body diagram of this section and nd the bending moment in the shaft at A Fig 1 In addition to the torque due to P7 there is a moment about an axis parallel to 2 at A Balancing the moments about an axis parallel to the z axis at A yields Mj i 4000 N015 m 2000 N04 m 0 a MZA i200 N m Since the sign is negative7 the moment is acting in the direction opposite to the way it is drawn in the free body diagram This will generate a compressive stress on the Figure 1 Free body diagram of crank element at A along the a direction as shown on the stress element in Fig 2 If we look at the direction the torque is acting to the left of A balanced by that due to the force P to the right of A we can determine the direction of the resulting shear stress which is also shown on the stress element in Fig 2 We can see that the shear stress Tm is acting in the negative direction it is acting in the z direction on the surface whose unit normal is along 3 and in the direction on the surface whose unit normal is along z wX u e A 6 u x I 2 Figure 2 Stress element at A looking down from the top Note the stresses are shown in the direction in which they are acting ie compressive negative axial stress 03 and negative shear stress Tm The stresses at A due to the torsion and bending moment are then A 036 M2 c 200 N rn00175 m 47 51MPa I 50035 m4 P fig rXE3 07117 Pp pr wwrr lt5 7 51 L Ll3976lquot7ll39l 0x tYE Figure 3 Mohr s circle diagram for stress element shown in Fig 2 E i 600 N m00175 m 7127 MP J g20035 m4 a Tm b A Mohr s circle is shown in Fig 3 This case is different from what we ve done in class because it is with respect to the a z axes instead of the a y axes Note that in this coordinate frame the 3 axis is 90 counterclockwise from the z axis As such making the analog to the a y axes that was discussed in class here the z axis is like the 3 axis in the a y frame and the 3 axis is like the y axis in the a y frame Thus the points we plot on the Mohr s circle are 02 739m and 033 Tm This is important so that the rotations on the Mohr s circle will be in the same direction as in physical space Another way of thinking of it that may be easier to remember is that we plot clockwise shear stresses as positive and counterclockwise sheer stresses as negative First look at the face with normal along z in Fig 2 on this face we have a 0 and 739 7127 MPa acting clockwise about the center of element A and so we plot the point 0 7127 Now look at the face with normal along as on this face we have a 4751 MPa and 739 7127 MPa acting counterclockwise about the center of element A so we plot the point 4751 7127 This gives us the diameter of the Mohr s circle From the Mohr s circle diagram we can compute the principal stresses 1 a E 4751 MPa 2376 MPa R 23762 7127 7513 MPa a1aaR5l37llPa 030a R 9889MPa 020 V quot7quot IL UthY39bu V Figure 4 Stress elements oriented with the c principal stresses and d with the maximum shear stress Note that 02 is zero because the stresses on the shaft surface at A axis are zero ie 0y my Tyz 0 We can also get the angle of rotation to the principal stress directions gz p from this diagram From geometry we see that 7127 tan 2ng 23 76 gt gz p 358 c Figure 4 c shows the principal stress element oriented with respect to the a and z axes Note that we rotate the stress element in the same direction as on the Mohr s circle ie note that 2ng is going clockwise on the Mohr s circle taking us from the stress state in the a z coordinates to the principal stresses so we rotate the the stress element by gz p in the clockwise direction We can also see on the Mohr s circle that the point 02 739m rotates to 01 Thus this face on the element rotated to the principal stresses will have 01 acting on it d We can compute the maximum shear stress also from the Mohr s circle diagram It is the radius of the largest circle In this case 7mm R 7513 MPa The normal stress at this location is ac 2376 MPa The rotation to the element with maximum shear stress can be computed from 2376 tan 2 s W sz 92 Note on the Mohr s circle diagram 2ng 2 90 ie the principal stresses and maximum shear stress are always 90 apart on the Mohr s circle and thus p s 45 6 800 N Al 0 D B A l x x 02m 028m 012m 02m 012m RAy 580 N Ray RAZ 1600 N 1600 N R52 Figure 5 4 8 Free bo y diagrams F gure 4 shows an element oriented wit sr onding normal stress h the m ximum shear stress and corre 4 8 a Here it is asking only for the shear stress due to torsion Since this shaft presumably s operating at a constant angular velocity th torque e lting from the loads applied t the two elements need to balance We can om I e the resulting to ue on the larger T 1200 N i 400 N 012 m 96 N m The smaller element then generates an equal and opposite torque thus FH006 m 96 N m FH 1600 N So there is a constant torque of 96 N m applied to the shaft between the two elements This generates a shear stress that is maximum on the shaft surface and is equal to D Ta T3 7 16T 1696 N m 7 7 764MP T J 312m 703 No04 m3 a b To nd the maximum tensile bending stress we need to determine the maximum bending moment Since there are forces in each of the 2 transverse directions we need to consider the resulting moments in each plane If we let the z axis be aligned with the shaft 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 1348 then we have the following 2 free body diagrams in each plane see Fig 5 From force and moment equilibrium in each plane it is straight forward to determine RAy 4173 N RBy 71973 N RAZ 713867 N and R31 718133 N The shear and moment diagrams can then be constructed and are shown in Fig 6 It is clear that the maximum magnitude of the bending moment Figure 6 4 8 Shear and moment diagrams is at 0 since this is where the bending moment is largest in magnitude in each plane Thus7 the overall maximum bending moment is Mmam 277342 8346 2896 N m The maximum tensile stress7 which will occur on the surface of the shaft at axial position 0 is then 7 MC Mg 7 32M 7 322896 N m I 374m st 7r004 m3 461 MPa EMD Homework Solutions for HW1 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 229 For 1020 annealed steel Syp 43km First check yielding then we will check buckling 7 10000 lb 720k lt 431m 723 9k 7 052712 52 18 39 52 so the link is acceptable based on yielding Then buckling Check the two possible cases of buckling in case one it buckles around axis a a treat as xed at each end in case two it buckles around c c treat as pinned at each end The critical buckling loads in each case are HE i 7r230 gtlt 106 www P i 30 843 lb 1 L3 0520 271 7 i in 3 P i WZEIiWigo 843lb 2 7 L3 7 20 m T 7 Note it is only by chance that the buckling load is the same in each direction in this case This is not generally true The safety factor is ig g 3084 gt 18 so the link is acceptable 231 a For a cantilever beam with point load at the end K 4013 Table 22 in the buckling criterion A safety factor n 2 is required Steel with properties E 30 gtlt 106 psi G 115 gtlt 106 psi and SW 45000 psi First computing the maximum load allowed to avoid yield S M P 14 39 2 39 Uallowable 22500 psz gt 76 n I EO125 m 4 m Pmam emiing Now computing the maximum load allowed to avoid buckling For the geometry given J5 dt33 4 m0125 27033 2604 gtlt1073 m4 1 4 m0125 m312 6510 x10 4m4 P i 130 i K GJ5EIy 7 4013115 gtlt 1062604 gtlt 107330 gtlt 1066510 gtlt 1074 mazlmckllng 7 n 7 2L2 7 2476 lb b Thus failure is governed by the allowable load to avoid buckling and the maximum allowable load is 247 lb 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 t Z 76 from 2 86 1071 100 nd1 18 254 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 rnaterial7 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 n 1 10 0 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 SW 7 119000 psi gt Ud T Ud 71d which means ad lt 66111 psi The predicted stress for the design is calculated using P i 10 0001b 7 lt 66 111 ad A W 7 ps1 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 My g 7 La 687 000 psi 7 607 000 psi 87 000 psi fry qamp 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 M70 7 8000 A i 143 standard deviations away from the mean 7y 5590 From Table 297 we nd that the reliability is then 9236 1 F 9 MANE 4030 EMD Homework Solutions 1 Due 131 1 13 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 77 Engineers 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 2 54 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 interpretation 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 material weight space 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 2 2 64 From Table 29 for the reliability to be 099999 X 427 The mean and standard deviation of the stress level are given as t 345 MPa and 6 28 MPa 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 r 9 standard deviation of y is and the mean of y is y si a si345MPa For a reliability of at least 099999 we need y to be at least X 427 standard deviations to the right ofthe limiting value y 0 Thus y 2 427 344 MPa 147 MPa So nally using the equation above the requires that s Z 492 psi 2 69 a Using Eqn 2 101 for a system having 3 components in series each with a relia bility of 090 Rs 093 0729 Thus the system reliability would be 729 b Adding a duplicate redundant system then we use Eqn 2 102 to obtain the system reliability R5 1 i 1 4 07292 0927 Thus 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 system 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 11 For a medium drive t we need EN 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 dS in sleeve 25020 25000 i d 25012 in hole That is the sleeve outer diameter must be between 25020 and 25027 inches and the housing bore diameter must be between 25000 and 25012 inches Note for the shaft we put the larger number on top7 because in the metal removal process to get the shaft down to the desired diameter7 we go from larger to smaller For the hole7 the smaller number is on top for a similar reason7 during machining7 the hole goes from a smaller to a larger diameter The limits of interference are 00008 to 00027 in MAME 4030 EMD Homework Solutions 3 Due 214 1 a The Mohr s circle is shown in Fig 1 From the circle7 the principal stresses are computed as 1 0a 525 50 MPa 375 MPa R x 50 3752 152 1953 MPa 01 0a R 5703 MPa 02 0a R 1797 MPa 03 0 The maximum shear stress is the radius of the largest circle Thus7 1 1 7mm 01 03 55703 0 MPa 2851 MPa 2 mm 15539QZ l m z 1 495x61 task 7 5111 39 1 15149 7 4i6mm 7 93937 Figure 1 Mohr s circle diagram Stress element diagram shown in upper right b From the Mohr s circle7 we can nd the rotation angle gz p to the principal stresses 15 251 50 375 gt 15 tan2 p Fig 2 shows the stress element oriented relative to the principal stresses with the principal stresses 01 and 02 acting on it c The local strain state can be found using Hooke s Law7 Eqns 5 15 5 20 in your book The material is steel7 so the elastic properties are Table 39 E 207 g 1150 5 l 1 3 0 1 T l 71 y 67 K 1 6 X Figure 2 Stress element oriented with the principal stresses GPa G 79 GPa and V 030 1 2 4 1 5 e 2070005 0350 83x 0 1 2 2 1 4 a 207 00050 035 05x 0 1 Z A 2 1 4 e 20000 0 03550 09x 0 15 1 1 4 79000 90X 0 sz 0 Vyz 0 2 The torque moment about the shaft axis as due to the 56 kN load applied on gear 4 must be equal and opposite to that resulting from F A applied to gear 3 Thus T FAcos 20 025 m 56 kNcos 20 01875 m 9867 N m gt FA 42 kN Fig 3 shows free body diagrams of the shaft in the a y and a z planes The reaction forces at the bearings O and B can be computed from equilibrium They are Roy 25421N RBy 5107 N R02 27513N and R32 65771N From the forces on the shaft in each plane we can construct the shear and moment diagrams in each plane which are also shown in Fig 3 The torque between A and C is constant and is T 9867 N m We will have high stresses just to the right of A where we have the full torque and the bending moment at A and at B where we also have the full torque acting as well as the bending moment at B We need to determine which bending moment is higher and that will be the 7 W930 Roz Rm 626mm A 5 ol l A B l Ll 0L K m W m m 2 0qu 046 My J 1 jl Ra 36mm r s 143660 y Y 2 V l N3 1942 3 52523 0 A a 4 L 055 q 005 r 040 I W WM FL g a X M3 39 41qu mw 39 i 1539 M 1 M Wm 1 6i M A 6 x o a g A e C l l 4513 I quot ZIUHH Figure 3 Free body diagrams of shaft in a y and a z planes and associated shear and moment diagrams critical location Computing the overall bending moment at each location MA 139822 151322 20603 N m MB 76612 210492 2240 N m Thus the maximum bending moment occurs at B so this is the critical location The stresses acting at this location are aXial stress 03 due to bending and shear stress 739 due to torsion The stresses at B are MBC i MBd2 32MB 322240 N m as 7 845 MP 0 I g 4d4 d3 003 m3 a Tc Td2 16T 169867 N m 186MP T J g gd4 d3 003 m3 a Fig 4 shows the stresses acting on the critical stress element at B 3 4 13 A free body diagram is shown in Fig 5 where we assumed that the bearing at B supports the thrust load problem doesn t specify how this is supported and how 3 Figure 4 Stress element at critical location B M 4P 1600 111113 A R B f 39 BE 1539 Pmlb R a R b I I 1 Figure 5 4 13 Free body diagram it is supported has no effect on the solution From force and moment equilibrium the reactions may be computed as Ry 533 lb minus sign indicates this force acts downward RyB 533 lb and Rf 400 lb Now letting the y aXis be in the downward direction so that a downward deflection will be positive and using our singularity function handout from class we can write for the downward deflection 11 CPU El 533 533 lt a 30 gt a Note 1 I don t bother to use a singularity function for the rst term because this term kicks in from a 0 and I m only consider a over the length of the beam and 2 I don t bother putting the singularity function terms for the force and moment at a 40 in because these only kick in right at the end of the beam and thus have no impact Integrating the above equations twice do E1d 2672 267 lt a 30 gt2 01 1 EIv 8893 889 lt a 30 gt3 019 02 Next I apply the boundary conditions at the two bearing supports 110 0 and 7 030 0 to obtain the two constants of integration Cl and 02 7EIv0 020 7EIv30 7889303Cl300 a 018000 So we have nally ii 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 FZD 07854 in4 Evaluating at the overhung end then gives M40 0010 in Between the supports 0 lt z lt 307 the maximum de ection will occur when d1 1 77 267 278000 0 dz E lt x Note7 for m lt 30 the singularity function term lt z 7 30 gt 07 thus7 we dont 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 M173 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 4 31 a The force acting on link CD at D from the leaf spring should be half the force acting on the leaf spring7 since it is symmetrical7 or 9 kN7 up The link CD is pinned at each end7 so the force acting on the link must be along the axis of the link So the force acting at C on the curved member must be 9 kN 9 than225 373 kN Cy Cm Now we want to nd the loads acting on section A 7 B which we assume to be the critical section We know that there will be forces in z and y and a bending moment at this location due to the forces at 0 found above To nd the bending moment we must know the location of the axis about which the moment is acting Technically this is the neutral axis however the neutral axis is close to the centroidal axis so the taking the moment about the centroid of the section would be a good approximation Here we will use the neutral axis So we need to nd the location of the neutral axis rst From the geometry we can nd the inner radius n outer radius To and radius to the centroid of the section To n38mm r076mm rc57mm The neutral axis is at radial position 7quot and is found from using Table 43 r i blnr70 n f 7 7 7 bh h T bln7ln7 38 mm548224 mm 111 Note that when l compute Tn I keep several signi cant gures This is because later I will need to nd 6 r0 7 Tn and To is close to Tn lf 1 round off To and Tn I will not have an accurate calculation of 6 So now I can calculate the forces and moments on section A 7 B are P 7373 kN Py 79 kN M 9 000 N0025 00548224 m 7 3 730 N0057 m 5058 N m Note this is slightly lower than what I would have gotten if I took the moment about the centroid axis M 9 000 N0025 0038 0019 m 7 3730 N0057 m 5254 N m So taking the moment about the centroid is also a valid approximation and is typically conservative as it will overestimate the moment Since the axial load induces a compressive stress across the whole cross section and the bending moment induces a compressive load on the inner radius the stress will be highest in magnitude on the inner surface of the bend The shear stress due to the shear load will be zero at this location and will be maximum at the center of the cross section but this is 9 expected to be relatively small compared to the dominant bending stress Computing the stress then at the inner radius r0 7 Tn 218 mm Ci Tn 7 n 1682 mm Py MCZ39 HA 7 X T eAn 97 000N 5058 N m001682 m 7 0025 m0038 m 7 000218 m0025 m0038 m0038 m 71176 MPa Now7 we compute the stress on the outer radius7 which will be the highest tensile stress on the section in case the material of interest is weaker in tension co r0 7 Tn 2118 mm Py M00 U3 7 K eAro 97 000N 5058 N m002118 m 0025 m0038 m 000218 m0025 m0038 m0076 m 586 MPa b For ASTM A 48 Class 50 gray cast iron7 the strength is higher in compression SM 1172 MPa than in tension Sm 345 MPa Computing the safety factor at A7 we obtain 71A 3 100 and at B7 we obtain 71 345 59 586 The non circular shaft transmits pure torque The torque T in this application is found by equating the power to the torque times the angular velocity H N 2 d 39 60 kW 60 000 im Tu T1500 revmm H W s r61 60 s 7 T 3820 N m From Table 457 critical points are located at the midpoint of each side of the square7 as shown Fig 6 Next7 based on 4 42 Q T 382 N 7m 240 gtlt 106 Nm2 7 Q 1592 gtlt 106 m3 Tmaz Figule s Czo rsection of intezest Usmg the expzesslon for Q fzom Table 4 5 fox the squaxe section b a MM 8 i 1667a2 1592 gtlt10 s mz Q3a18b48a a985 gtlt 10393m9 85mm Each side of the squaxe section IS 2 so 2a2x985mm197mm Thm the sides of the squaxe should be at least 20 mm r r quot 0 mc l ova 00010 vanMM T aM cess M3 7th t m A wire rope is to be selected for use on an elevator transporting people to the 201 h floor of a building A limiting load capacity is listed in the elevator however there is no way to verify that people will follow this guideline The wire rope manufacturer has substantial test data available and it shows moderate strength variability The analysis of the stress induced in the rope is mostly straight forward except at the connection points of the wire rope where the stress concentration can vary The state requires annual inspections with a more stringent inspection every 5 years What design safety factor should be chosen 2J5 Rehabikih Rquot w ijwqh P sw ogt 073 W3 95 39 i 1 5quot S i 1 It RMK S famn 9 9m m as m homgemsm ammmgm39 t 1 303 3mm apl W quotw rm Wok amp 3M 2 gm 0W di lau w ms M33 0 W Ts 25 togwnoV mA cm A ADM ia TJ VNWZS BJF 0 S 5 6 5y 39 4 an 0F g Mai 74 J S6 5 mifw viaow a mow MM CICFSM GD 0 m R P gum 01 HQ JM r 0 3 lt1 Q mama w V Mi m MW 6 QMW V w M A Va 2 20 3 0 A mayo VarCcJolas uwb UJLWH M M1 0 Ht hH 0mg 2 is A 2 Rainquot Q XP w Ju xx 4 3 it I an V 12s w 2 o kg Sammmg ac m WMAOUSMUW We w W 39F 6x30 42 H m woo Wk m Mix Tow 2 61 vMQs L Moi V39 39 x 3 w j 330quot gag41X 3 0 V32 0 35 0 3 A supplier of 4340 steel has shipped enough material to fabricate 100 fatizguencritical tension links for an aircraft application As required in the purchase contract the supplier has conducted uniaxial fatigue tests on random specimens drawn from the lot of material and has certified that the mean fatigue strength corresponding to a life of 106 cycles is 68000 psi the standard deviation on strength corresponding to a life of 106 cycles is 3200 psi and that the distribution of strength at a life of 106 cycles is normal The tension link has a cylindrical crosssection with a diameter of 05 inch with a tolerance of i 0002 inch From dynamic simulations it has been determined that the expected force amplitude on this component is 10000 lb with a standard deviation of 1500 lb You are to estimate how many tension links from the lot of 100 are expected to fail when operated for 106 cycles Reliability Example a To determine the stress we need the force and cross sectional area The force is not too accurate with a standard deviation of 15 of the mean The cross sectional area is related to the square of the diameter which is pretty accurate Note that assuming a natural tolerance the standard deviation in the diameter is 1 3 the tolerance thus the standard deviation is 6667 X 10 or 013 of the mean Even if we look at the variability in the cross sectional area we see that the standard deviation is only about 027 which is sn39iall compared to the deviation in the force Thus the deviation in the crosseseetional area is relatively negligible Neglecting the variability in the deviation in the crossesectional area then the mean and standard deviation of the amplitude of the applied stress is 10700012 quot 74025271 1500 lb A0 m 7639 39 a 7r025 in2 pm 2 50 930 psi HIS 55 quotET 2 E Q o 3 Figure 1 Sketch of probability distributions of stress and strength of tensile member 0 Now we find the mean and standard deviation of the random variable y m S 0 y 45 g 2 68 000 psi 50930 psi 2 17 070 psi a 2 Meg a 3 200 p502 7 639 psi 2 282 psi d I need to nd the parameter X to use in Table 29 The way I like to think of this is that I need to know how many standard deviations are between the mean and the value I m interested in which in this case is 0 ie I want to know the probability 3 lt 0 or Py lt 0 y r lailule vegion en 43 o 10 quotyes 30 3943 50 vikpsi Figure 2 Sketch of probability distributions for random variable associated with reliability7 y So then 0 17 070 X 2 206 So there are 206 standard deviations between the mean and 0 I look up X z 206 since the i iormalized distribution is symmetric about X z 0 it doesn t matter if it is plus or minus I just need to remember now I ve reflected the distribution about the mean since I changed the sign In Table 29 I nd the FX 09803 Now F X is the integral from oo to X on the standard normal distribution I m interested in the region in the tail which is on the other side of X and is 1 FX 000197 197 This is the probability of failure so out of 100 parts 2 are expected to fail A h4 6 0 quot391 100 l 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