Shigly . design

591
Chapter 1 D B G F F acc A E f f 1 1 cr C Impending motion to left F cr Consider force F at G, reactions at B and D. Extend lines of action for fully-developed fric- tion DE and BE to find the point of concurrency at E for impending motion to the left. The critical angle is θ cr . Resolve force F into components F acc and F cr . F acc is related to mass and acceleration. Pin accelerates to left for any angle 0 <θ<θ cr . When θ>θ cr , no magnitude of F will move the pin. D B G F F acc A E E f f 1 1 C d Impending motion to right F cr cr Consider force F at G, reactions at A and C. Extend lines of action for fully-developed fric- tion AE and CE to find the point of concurrency at E for impending motion to the left. The critical angle is θ cr . Resolve force F into components F acc and F cr . F acc is related to mass and acceleration. Pin accelerates to right for any angle 0 cr . When θ cr , no mag- nitude of F will move the pin. The intent of the question is to get the student to draw and understand the free body in order to recognize what it teaches. The graphic approach accomplishes this quickly. It is im- portant to point out that this understanding enables a mathematical model to be constructed, and that there are two of them. This is the simplest problem in mechanical engineering. Using it is a good way to begin a course. What is the role of pin diameter d ? Yes, changing the sense of F changes the response. Problems 1-1 through 1-4 are for student research. 1-5

description

mechanic field

Transcript of Shigly . design

Chapter 1DBGFFaccAEf f1 1crCImpending motion to leftFcrConsider force F at G, reactions at B and D. Extend lines of action for fully-developed fric-tion DE and BE to nd the point of concurrency at E for impending motion to the left. Thecritical angle is cr. Resolve force F into components Faccand Fcr. Faccis related to mass andacceleration. Pin accelerates to left for any angle 0 < < cr. When > cr, no magnitudeof F will move the pin.DBGFFaccAE Ef f1 1CdImpending motion to rightFcrcrConsider force F

at G, reactions at A and C. Extend lines of action for fully-developed fric-tion AE

and CE

to nd the point of concurrency at E

for impending motion to the left. Thecritical angle is

cr. Resolve force F

into components F

accand F

cr. F

accis related to massand acceleration. Pin accelerates to right for any angle 0 <

<

cr. When

>

cr, no mag-nitude of F

will move the pin.The intent of the question is to get the student to draw and understand the free body inorder to recognize what it teaches. The graphic approach accomplishes this quickly. It is im-portant to point out that this understanding enables a mathematical model to be constructed,and that there are two of them.This is the simplest problem in mechanical engineering. Using it is a good way to begin acourse.What is the role of pin diameter d?Yes, changing the sense of F changes the response.Problems 1-1 through 1-4 are for student research.1-5shi20396_ch01.qxd 6/5/03 12:11 PM Page 1

2 Solutions Manual Instructors Solution Manual to Accompany Mechanical Engineering Design1-6 (a)

Fy = F f N cos + N sin = 0 (1)

Fx = f N sin + N cos Tr = 0F = N(sin f cos ) Ans.T = Nr( f sin +cos )CombiningT = Fr1 + f tan tan f = KFr Ans. (2)(b) If T detent self-locking tan f = 0 cr = tan1f Ans.(Friction is fully developed.)Check: If F = 10 lbf, f = 0.20, = 45, r = 2 inN =100.20 cos 45+sin 45 = 17.68 lbfTr = 17.28(0.20 sin 45+cos 45) = 15 lbff N = 0.20(17.28) = 3.54 lbfcr = tan1f = tan1(0.20) = 11.3111.31 < < 901-7(a) F = F0+k(0) = F0T1 = F0r Ans.(b) When teeth are about to clearF = F0+kx2From Prob. 1-6T2 = Fr f tan +1tan fT2 = r( F0+kx2)( f tan +1)tan f Ans.1-8Given, F = 10 +2.5x lbf, r = 2 in, h = 0.2 in, = 60, f = 0.25, xi = 0, xf = 0.2Fi = 10 lbf; Ff = 10 +2.5(0.2) = 10.5 lbf Ans.xyFf NN Trshi20396_ch01.qxd 6/5/03 12:11 PM Page 2

Chapter 1 3From Eq. (1) of Prob. 1-6N = Ff cos +sin Ni =100.25 cos 60+sin 60 = 13.49 lbf Ans.Nf =10.51013.49 = 14.17 lbf Ans.From Eq. (2) of Prob. 1-6K =1 + f tan tan f =1 +0.25 tan 60tan 600.25 = 0.967 Ans.Ti = 0.967(10)(2) = 19.33 lbf inTf = 0.967(10.5)(2) = 20.31 lbf in1-9(a) Point vehiclesQ =carshour = vx =42.1v v20.324Seek stationary point maximumdQdv= 0 =42.1 2v0.324 v* = 21.05 mphQ* =42.1(21.05) 21.0520.324 = 1367.6 cars/h Ans.(b)Q = vx +l =

0.324v(42.1) v2 + lv

1Maximize Q with l = 10/5280 miv Q22.18 1221.43122.19 1221.43322.20 1221.435 22.21 1221.43522.22 1221.434% loss of throughput 1368 12211221 = 12% Ans.xl2l2vxvshi20396_ch01.qxd 6/5/03 12:11 PM Page 3

4 Solutions Manual Instructors Solution Manual to Accompany Mechanical Engineering Design(c) % increase in speed 22.2 21.0521.05 = 5.5%Modest change in optimal speed Ans.1-10 This and the following problem may be the students rst experience with a gure of merit. Formulate fom to reect larger gure of merit for larger merit. Use a maximization optimization algorithm. When one gets into computer implementa-tion and answers are not known, minimizing instead of maximizing is the largest errorone can make.

FV = F1sin W = 0

FH = F1cos F2 = 0From whichF1 = W/sin F2 = W cos /sin fom = S = (volume).= (l1A1+l2A2)A1 = F1S = WS sin , l2 = l1cos A2 =

F2S

= W cos S sin fom =

l2cos WS sin +l2W cos S sin

= Wl2S

1 +cos2cos sin

Set leading constant to unityfom0 20 5.8630 4.0440 3.2245 3.0050 2.8754.736 2.82860 2.886Check second derivative to see if a maximum, minimum, or point of inection has beenfound. Or, evaluate fom on either side of *.* = 54.736 Ans.fom* = 2.828Alternative:dd

1 +cos2cos sin

= 0And solve resulting tran-scendental for *.shi20396_ch01.qxd 6/5/03 12:11 PM Page 4

Chapter 1 51-11(a) x1+ x2 = X1+e1+ X2+e2error = e = (x1+ x2) ( X1+ X2)= e1+e2 Ans.(b) x1 x2 = X1+e1( X2+e2)e = (x1 x2) ( X1 X2) = e1e2 Ans.(c) x1x2 = ( X1+e1)( X2+e2)e = x1x2 X1X2 = X1e2+ X2e1+e1e2.= X1e2+ X2e1 = X1X2

e1X1+ e2X2

Ans.(d) x1x2= X1+e1X2+e2= X1X2

1 +e1/X11 +e2/X2

1 + e2X2

1.= 1 e2X2and

1 + e1X1

1 e2X2

.= 1 + e1X1 e2X2e = x1x2 X1X2.= X1X2

e1X1 e2X2

Ans.1-12(a) x1 =5 = 2.236 067 977 5X1 = 2.23 3-correct digitsx2 =6 = 2.449 487 742 78X2 = 2.44 3-correct digitsx1+ x2 =5 +6 = 4.685 557 720 28e1 = x1 X1 =5 2.23 = 0.006 067 977 5e2 = x2 X2 =6 2.44 = 0.009 489 742 78e = e1+e2 =5 2.23 +6 2.44 = 0.015 557 720 28Sum = x1+ x2 = X1+ X2+e= 2.23 +2.44 +0.015 557 720 28= 4.685 557 720 28 (Checks) Ans.(b) X1 = 2.24, X2 = 2.45e1 =5 2.24 = 0.003 932 022 50e2 =6 2.45 = 0.000 510 257 22e = e1+e2 = 0.004 442 279 72Sum = X1+ X2+e= 2.24 +2.45 +(0.004 442 279 72)= 4.685 557 720 28 Ans.shi20396_ch01.qxd 6/5/03 12:11 PM Page 5

6 Solutions Manual Instructors Solution Manual to Accompany Mechanical Engineering Design1-13(a) = 20(6.89) = 137.8 MPa(b) F = 350(4.45) = 1558 N = 1.558 kN(c) M= 1200 lbf in (0.113) = 135.6 N m(d) A = 2.4(645) = 1548 mm2(e) I = 17.4 in4(2.54)4= 724.2 cm4(f) A = 3.6(1.610)2= 9.332 km2(g) E = 21(1000)(6.89) = 144.69(103) MPa = 144.7 GPa(h) v = 45 mi/h (1.61) = 72.45 km/h(i) V = 60 in3(2.54)3= 983.2 cm3= 0.983 liter1-14(a) l = 1.5/0.305 = 4.918 ft = 59.02 in(b) = 600/6.89 = 86.96 kpsi(c) p = 160/6.89 = 23.22 psi(d) Z =1.84(105)/(25.4)3= 11.23 in3(e) w = 38.1/175 = 0.218 lbf/in(f) = 0.05/25.4 = 0.00197 in(g) v = 6.12/0.0051 = 1200 ft /min(h) = 0.0021 in/in(i) V =30/(0.254)3= 1831 in31-15(a) =20015.3 = 13.1 MPa(b) =42(103)6(102)2 = 70(106) N/m2= 70 MPa(c) y =1200(800)3(103)33(207)(6.4)(109)(102)4 = 1.546(102) m = 15.5 mm(d) =1100(250)(103)79.3(/32)(25)4(109)(103)4 = 9.043(102) rad = 5.181-16(a) =60020(6) = 5 MPa(b) I =1128(24)3= 9216 mm4(c) I = 64324(101)4= 5.147 cm4(d) =16(16)(253)(103)3 = 5.215(106) N/m2= 5.215 MPashi20396_ch01.qxd 6/5/03 12:11 PM Page 6

Chapter 1 71-17(a) =120(103)(/4)(202) = 382 MPa(b) =32(800)(800)(103)(32)3(103)3 = 198.9(106) N/m2= 198.9 MPa(c) Z = 32(36)(364264) = 3334 mm3(d) k =(1.6)4(79.3)(103)4(109)8(19.2)3(32)(103)3 = 286.8 N/mshi20396_ch01.qxd 6/5/03 12:11 PM Page 7

(b) f/(Nx) = f/(69 10) = f/690Eq. (2-9) x = 848069 = 122.9 kcyclesEq. (2-10) sx =_1 104 600 84802/6969 1_1/2= 30.3 kcycles Ans.x f f x f x2f/(Nx)60 2 120 7200 0.002970 1 70 4900 0.001580 3 240 19 200 0.004390 5 450 40 500 0.0072100 8 800 80 000 0.0116110 12 1320 145 200 0.0174120 6 720 86 400 0.0087130 10 1300 169 000 0.0145140 8 1120 156 800 0.0116150 5 750 112 500 0.0174160 2 320 51 200 0.0029170 3 510 86 700 0.0043180 2 360 64 800 0.0029190 1 190 36 100 0.0015200 0 0 0 0210 1 210 44 100 0.001569 8480 1 104 600Chapter 22-1(a)060 210 190 200 180 170 160 150 140 130 120 110 100 90 80 7024681012shi20396_ch02.qxd 7/21/03 3:28 PM Page 8

Chapter 2 92-2 Data represents a 7-class histogram with N = 197.2-3Form a table: x = 454858 = 78.4 kpsisx =_359 088 45482/5858 1_1/2= 6.57 kpsiFrom Eq. (2-14)f (x) = 16.572exp_12_x 78.46.57_2_x f f x f x264 2 128 819268 6 408 27 74472 6 432 31 10476 9 684 51 98480 19 1520 121 60084 10 840 70 56088 4 352 30 97692 2 184 16 92858 4548 359 088x f f x f x2174 6 1044 181 656182 9 1638 298 116190 44 8360 1 588 400198 67 13 266 2 626 688206 53 10 918 2 249 108214 12 2568 549 552220 6 1320 290 400197 39 114 7 789 900 x = 39 114197 = 198.55 kpsi Ans.sx =_7 783 900 39 1142/197197 1_1/2= 9.55 kpsi Ans.shi20396_ch02.qxd 7/21/03 3:28 PM Page 9

10 Solutions Manual Instructors Solution Manual to Accompany Mechanical Engineering Design2-4 (a)y f f y f y2y f/(Nw) f (y) g(y)5.625 1 5.625 31.64063 5.625 0.072 727 0.001 262 0.000 2955.875 0 0 0 5.875 0 0.008 586 0.004 0886.125 0 0 0 6.125 0 0.042 038 0.031 1946.375 3 19.125 121.9219 6.375 0.218 182 0.148 106 0.140 2626.625 3 19.875 131.6719 6.625 0.218 182 0.375 493 0.393 6676.875 6 41.25 283.5938 6.875 0.436 364 0.685 057 0.725 0027.125 14 99.75 710.7188 7.125 1.018 182 0.899 389 0.915 1287.375 15 110.625 815.8594 7.375 1.090 909 0.849 697 0.822 4627.625 10 76.25 581.4063 7.625 0.727 273 0.577 665 0.544 2517.875 2 15.75 124.0313 7.875 0.145 455 0.282 608 0.273 1388.125 1 8.125 66.01563 8.125 0.072 727 0.099 492 0.1067255 396.375 2866.859For a normal distribution, y = 396.375/55 = 7.207, sy =_2866.859 (396.3752/55)55 1_1/2= 0.4358f ( y) = 10.43582exp_12_x 7.2070.4358_2_For a lognormal distribution, x = ln 7.206 818 ln1 +0.060 4742= 1.9732, sx = ln1 +0.060 4742= 0.0604g( y) = 1x(0.0604)(2) exp_12_ln x 1.97320.0604_2_(b) Histogram00.20.40.60.811.25.63 5.88 6.13 6.38 6.63 6.88log N7.13 7.38 7.63 7.88 8.13DataNLNfshi20396_ch02.qxd 7/21/03 3:28 PM Page 10

Chapter 2 112-5 Distribution is uniform in interval 0.5000 to 0.5008 in, range numbers are a = 0.5000,b = 0.5008 in.(a) Eq. (2-22) x = a +b2 = 0.5000 +0.50082 = 0.5004Eq. (2-23) x = b a23 = 0.5008 0.500023 = 0.000 231(b) PDF from Eq. (2-20)f (x) =_1250 0.5000 x 0.5008 in0 otherwise(c) CDF from Eq. (2-21)F(x) =___0 x < 0.5000(x 0.5)/0.0008 0.5000 x 0.50081 x > 0.5008If all smaller diameters are removed by inspection, a = 0.5002, b = 0.5008x = 0.5002 +0.50082 = 0.5005 in x = 0.5008 0.500223 = 0.000 173 inf (x) =_1666.7 0.5002 x 0.50080 otherwiseF(x) =___0 x < 0.50021666.7(x 0.5002) 0.5002 x 0.50081 x > 0.50082-6 Dimensions produced are due to tool dulling and wear. When parts are mixed, the distributionis uniform. From Eqs. (2-22) and (2-23),a = x 3s = 0.6241 3(0.000 581) = 0.6231 inb = x +3s = 0.6241 +3(0.000 581) = 0.6251 inWe suspect the dimension was 0.6230.625 in Ans.shi20396_ch02.qxd 7/21/03 3:28 PM Page 11

12 Solutions Manual Instructors Solution Manual to Accompany Mechanical Engineering Design2-7 F(x) = 0.555x 33 mm(a) Since F(x) is linear, the distribution is uniform at x = aF(a) = 0 = 0.555(a) 33a = 59.46 mm. Therefore, at x = bF(b) = 1 = 0.555b 33b = 61.26 mm. Therefore,F(x) =___0 x < 59.46 mm0.555x 33 59.46 x 61.26 mm1 x > 61.26 mmThe PDF is dF/dx, thus the range numbers are:f (x) =_0.555 59.46 x 61.26 mm0 otherwise Ans.From the range numbers,x = 59.46 +61.262 = 60.36 mm Ans. x = 61.26 59.4623 = 0.520 mm Ans.1(b) is an uncorrelated quotient F = 3600 lbf, A = 0.112 in2CF = 300/3600 = 0.083 33, CA = 0.001/0.112 = 0.008 929From Table 2-6, for = FA= 36000.112 = 32 143 psi Ans. = 32 143_(0.083332+0.0089292)(1 +0.0089292)_1/2= 2694 psi Ans.C = 2694/32 143 = 0.0838 Ans.Since F and A are lognormal, division is closed and is lognormal too. = LN(32 143, 2694) psi Ans.shi20396_ch02.qxd 7/21/03 3:28 PM Page 12

Chapter 2 132-8 Cramers rulea1 =y x2xy x3x x2x2x3= yx3xyx2xx3(x2)2 Ans.a2 =x yx2xyx x2x2x3= xxy yx2xx3(x2)2 Ans.0.0500.050.10.150.20.250.30 0.2 0.4 0.6 0.8 1DataRegressionxyx y x2x3xy0 0.01 0 0 00.2 0.15 0.04 0.008 0.0300.4 0.25 0.16 0.064 0.1000.6 0.25 0.36 0.216 0.1500.8 0.17 0.64 0.512 0.1361.0 0.01 1.00 1.000 0.0103.0 0.82 2.20 1.800 0.406a1 = 1.040 714 a2 = 1.046 43 Ans.Data Regressionx y y0 0.01 00.2 0.15 0.166 2860.4 0.25 0.248 8570.6 0.25 0.247 7140.8 0.17 0.162 8571.0 0.01 0.005 71shi20396_ch02.qxd 7/21/03 3:28 PM Page 13

14 Solutions Manual Instructors Solution Manual to Accompany Mechanical Engineering Design2-90204060801001201400 100 200 SuSe300 400DataRegressionData RegressionSu S

e S

e S2u SuS

e0 20.3567560 30 39.08078 3600 180064 48 40.32905 4096 307265 29.5 40.64112 4225 1917.582 45 45.94626 6724 3690101 51 51.87554 10201 5151119 50 57.49275 14161 5950120 48 57.80481 14400 5760130 67 60.92548 16900 8710134 60 62.17375 17956 8040145 64 65.60649 21025 9280180 84 76.52884 32400 15120195 78 81.20985 38025 15210205 96 84.33052 42025 19680207 87 84.95466 42849 18009210 87 85.89086 44100 18270213 75 86.82706 45369 15975225 99 90.57187 50625 22275225 87 90.57187 50625 19575227 116 91.196 51529 26332230 105 92.1322 52900 24150238 109 94.62874 56644 25942242 106 95.87701 58564 25652265 105 103.0546 70225 27825280 96 107.7356 78400 26880295 99 112.4166 87025 29205325 114 121.7786 105625 37050325 117 121.7786 105625 38025355 122 131.1406 126025 433105462 2274.5 1251868 501855.5m = 0.312067 b = 20.35675 Ans.shi20396_ch02.qxd 7/21/03 3:28 PM Page 14

Chapter 2 152-10E =

_y a0a2x2_2Ea0= 2

_y a0a2x2_= 0

y na0a2

x2= 0

y = na0+a2

x2Ea2= 2

_y a0a2x2_(2x) = 0

xy = a0

x +a2

x3Ans.Cramers rulea0 =y x2xy x3 n x2x x3= x3y x2xynx3xx2a2 = n yx xy n x2x x3= nxy xynx3xx2a0 = 800 000(56) 12 000(2400)4(800 000) 200(12 000) = 20a2 = 4(2400) 200(56)4(800 000) 200(12 000) = 0.002DataRegression051015yx20250 20 40 60 80 100Data Regressionx y y x2x3xy20 19 19.2 400 8000 38040 17 16.8 1600 64000 68060 13 12.8 3600 216000 78080 7 7.2 6400 512000 560200 56 12000 800000 2400shi20396_ch02.qxd 7/21/03 3:28 PM Page 15

16 Solutions Manual Instructors Solution Manual to Accompany Mechanical Engineering Design2-11Data Regressionx y y x2y2xy x x (x x)20.2 7.1 7.931803 0.04 50.41 1.42 0.633333 0.4011111110.4 10.3 9.884918 0.16 106.09 4.12 0.433333 0.1877777780.6 12.1 11.838032 0.36 146.41 7.26 0.233333 0.0544444440.8 13.8 13.791147 0.64 190.44 11.04 0.033333 0.0011111111 16.2 15.744262 1.00 262.44 16.20 0.166666 0.0277777782 25.2 25.509836 4.00 635.04 50.40 1.166666 1.3611111115 84.7 6.2 1390.83 90.44 0 2.033333333 m = k = 6(90.44) 5(84.7)6(6.2) (5)2 = 9.7656b = Fi = 84.7 9.7656(5)6 = 5.9787(a) x = 56; y = 84.76 = 14.117Eq. (2-37)syx =_1390.83 5.9787(84.7) 9.7656(90.44)6 2= 0.556Eq. (2-36)sb = 0.556_16 + (5/6)22.0333 = 0.3964 lbfFi = (5.9787, 0.3964) lbf Ans.Fx 0510152025300 1 0.5 1.5 2 2.5DataRegressionshi20396_ch02.qxd 7/21/03 3:28 PM Page 16

Chapter 2 17(b) Eq. (2-35)s m = 0.5562.0333 = 0.3899 lbf/ink = (9.7656, 0.3899) lbf/in Ans.2-12 The expression = /l is of the form x/y. Now = (0.0015, 0.000 092) in, unspecieddistribution; l = (2.000, 0.0081) in, unspecied distribution;Cx = 0.000 092/0.0015 = 0.0613Cy = 0.0081/2.000 = 0.000 75From Table 2-6, = 0.0015/2.000 = 0.000 75

= 0.000 75_0.06132+0.004 0521 +0.004 052_1/2= 4.607(105) = 0.000 046We can predict and

but not the distribution of .2-13 = E = (0.0005, 0.000 034) distribution unspecied; E = (29.5, 0.885) Mpsi, distributionunspecied;Cx = 0.000 034/0.0005 = 0.068,Cy = 0.0885/29.5 = 0.030 is of the form x, yTable 2-6 = E = 0.0005(29.5)106= 14 750 psi = 14 750(0.0682+0.0302+0.0682+0.0302)1/2= 1096.7 psiC = 1096.7/14 750 = 0.074 352-14 = FlAEF = (14.7, 1.3) kip, A = (0.226, 0.003) in2, l = (1.5, 0.004) in, E = (29.5, 0.885) Mpsi dis-tributions unspecied.CF = 1.3/14.7 = 0.0884; CA = 0.003/0.226 = 0.0133; Cl = 0.004/1.5 = 0.00267;CE = 0.885/29.5 = 0.03Mean of : = FlAE = Fl_1A__1E_shi20396_ch02.qxd 7/21/03 3:28 PM Page 17

18 Solutions Manual Instructors Solution Manual to Accompany Mechanical Engineering DesignFrom Table 2-6, = Fl(1/ A)(1/ E) = 14 700(1.5)10.226129.5(106)= 0.003 31 in Ans.For the standard deviation, using the rst-order terms in Table 2-6, .=FlA E_C2F +C2l +C2A+C2E_1/2= _C2F +C2l +C2A+C2E_1/2 = 0.003 31(0.08842+0.002672+0.01332+0.032)1/2= 0.000 313 in Ans.COVC = 0.000 313/0.003 31 = 0.0945 Ans.Force COV dominates. There is no distributional information on .2-15 M = (15000, 1350) lbf in, distribution unspecied; d = (2.00, 0.005) in distributionunspecied. = 32Md3 , CM = 135015 000 = 0.09, Cd = 0.0052.00 = 0.0025 is of the form x/y, Table 2-6.Mean: = 32 M

d3.= 32 M d3 = 32(15 000)(23)= 19 099 psi Ans.Standard Deviation: = __C2M +C2d3___1 +C2d3__1/2From Table 2-6, Cd3.= 3Cd = 3(0.0025) = 0.0075 = __C2M +(3Cd)2__(1 +(3Cd))2_1/2= 19 099[(0.092+0.00752)/(1 +0.00752)]1/2= 1725 psi Ans.COV:C = 172519 099 = 0.0903 Ans.Stress COV dominates. No information of distribution of .shi20396_ch02.qxd 7/21/03 3:28 PM Page 18

Chapter 2 192-16Fraction discarded is +. The area under the PDF was unity. Having discarded +fraction, the ordinates to the truncated PDF are multiplied by a.a = 11 ( +)New PDF, g(x) , is given byg(x) =_ f (x)/[1 ( +)] x1 x x20 otherwiseMore formal proof: g(x) has the property1 =_ x2x1g(x) dx = a_ x2x1f (x) dx1 = a__ f (x) dx _ x10f (x) dx _ x2f (x) dx_1 = a {1 F(x1) [1 F(x2)]}a = 1F(x2) F(x1) = 1(1 ) = 11 ( +)2-17(a) d = U[0.748, 0.751]d = 0.751 +0.7482 = 0.7495 in d = 0.751 0.74823 = 0.000 866 inf (x) = 1b a = 10.751 0.748 = 333.3 in1F(x) = x 0.7480.751 0.748 = 333.3(x 0.748)x1f (x)xx2 shi20396_ch02.qxd 7/21/03 3:28 PM Page 19

20 Solutions Manual Instructors Solution Manual to Accompany Mechanical Engineering Design(b) F(x1) = F(0.748) = 0F(x2) = (0.750 0.748)333.3 = 0.6667If g(x) is truncated, PDF becomesg(x) = f (x)F(x2) F(x1) = 333.30.6667 0 = 500 in1x = a

+b

2 = 0.748 +0.7502 = 0.749 in x = b

a

23 = 0.750 0.74823 = 0.000 577 in2-18 From Table A-10, 8.1% corresponds to z1 = 1.4 and 5.5% corresponds to z2 = +1.6.k1 = + z1 k2 = + z2 From which = z2k1 z1k2z2 z1= 1.6(9) (1.4)111.6 (1.4)= 9.933 = k2k1z2 z1= 11 91.6 (1.4) = 0.6667The original density function isf (k) = 10.66672exp_12_k 9.9330.6667_2_ Ans.2-19 From Prob. 2-1, = 122.9 kcycles and = 30.3 kcycles.z10 = x10 = x10122.930.3x10 = 122.9 +30.3z10From Table A-10, for 10 percent failure, z10 = 1.282x10 = 122.9 +30.3(1.282)= 84.1 kcycles Ans.0.748g(x) 500xf (x) 333.30.749 0.750 0.751shi20396_ch02.qxd 7/21/03 3:28 PM Page 20

Chapter 2 212-20x f f x f x2x f/(Nw) f (x)60 2 120 7200 60 0.002899 0.00039970 1 70 4900 70 0.001449 0.00120680 3 240 19200 80 0.004348 0.00300990 5 450 40500 90 0.007246 0.006204100 8 800 80000 100 0.011594 0.010567110 12 1320 145200 110 0.017391 0.014871120 6 720 86400 120 0.008696 0.017292130 10 1300 169000 130 0.014493 0.016612140 8 1120 156800 140 0.011594 0.013185150 5 750 112500 150 0.007246 0.008647160 2 320 51200 160 0.002899 0.004685170 3 510 86700 170 0.004348 0.002097180 2 360 64800 180 0.002899 0.000776190 1 190 36100 190 0.001449 0.000237200 0 0 0 200 0 5.98E-05210 1 210 44100 210 0.001449 1.25E-0569 8480 x = 122.8986 sx = 22.88719x f/(Nw) f (x) x f/(Nw) f (x)55 0 0.000214 145 0.011594 0.01093555 0.002899 0.000214 145 0.007246 0.01093565 0.002899 0.000711 155 0.007246 0.00651865 0.001449 0.000711 155 0.002899 0.00651875 0.001449 0.001951 165 0.002899 0.0032175 0.004348 0.001951 165 0.004348 0.0032185 0.004348 0.004425 175 0.004348 0.00130685 0.007246 0.004425 175 0.002899 0.00130695 0.007246 0.008292 185 0.002899 0.00043995 0.011594 0.008292 185 0.001449 0.000439105 0.011594 0.012839 195 0.001449 0.000122105 0.017391 0.012839 195 0 0.000122115 0.017391 0.016423 205 0 2.8E-05115 0.008696 0.016423 205 0.001499 2.8E-05125 0.008696 0.017357 215 0.001499 5.31E-06125 0.014493 0.017357 215 0 5.31E-06135 0.014493 0.015157135 0.011594 0.015157shi20396_ch02.qxd 7/21/03 3:28 PM Page 21

22 Solutions Manual Instructors Solution Manual to Accompany Mechanical Engineering Design2-21x f f x f x2f/(Nw) f (x)174 6 1044 181656 0.003807 0.001642182 9 1638 298116 0.005711 0.009485190 44 8360 1588400 0.027919 0.027742198 67 13266 2626668 0.042513 0.041068206 53 10918 2249108 0.033629 0.030773214 12 2568 549552 0.007614 0.011671222 6 1332 295704 0.003807 0.0022411386 197 39126 7789204 x = 198.6091 sx = 9.695071x f/(Nw) f (x)170 0 0.000529170 0.003807 0.000529178 0.003807 0.004297178 0.005711 0.004297186 0.005711 0.017663186 0.027919 0.017663194 0.027919 0.036752194 0.042513 0.036752202 0.042513 0.038708202 0.033629 0.038708210 0.033629 0.020635210 0.007614 0.020635218 0.007614 0.005568218 0.003807 0.005568226 0.003807 0.00076226 0 0.00076DataPDF00.0050.010.0150.020.0250.030.0350.040.045150 170 190 210x230fHistogramPDF00.0020.0040.0060.0080.010.0120.0140.0160.018fx0.020 50 100 150 200 250shi20396_ch02.qxd 7/21/03 3:28 PM Page 22

Chapter 2 232-22x f f x f x2f/(Nw) f (x)64 2 128 8192 0.008621 0.0054868 6 408 27744 0.025862 0.01729972 6 432 31104 0.025862 0.03770576 9 684 51984 0.038793 0.05674280 19 1520 121600 0.081897 0.05895984 10 840 70560 0.043103 0.04229888 4 352 30976 0.017241 0.02095292 2 184 16928 0.008621 0.007165624 58 4548 359088 x = 78.41379 sx = 6.572229x f/(Nw) f (x) x f/(Nw) f (x)62 0 0.002684 82 0.081897 0.05230562 0.008621 0.002684 82 0.043103 0.05230566 0.008621 0.010197 86 0.043103 0.0311866 0.025862 0.010197 86 0.017241 0.0311870 0.025862 0.026749 90 0.017241 0.01283370 0.025862 0.026749 90 0.008621 0.01283374 0.025862 0.048446 94 0.008621 0.00364774 0.038793 0.048446 94 0 0.00364778 0.038793 0.06058178 0.081897 0.0605812-23 = 4 Pd2 = 4(40)(12) = 50.93 kpsi = 4 Pd2 = 4(8.5)(12) = 10.82 kpsi sy = 5.9 kpsiDataPDFx 00.010.020.030.040.050.060.070.080.0960 70 80 90 100fshi20396_ch02.qxd 7/21/03 3:28 PM Page 23

24 Solutions Manual Instructors Solution Manual to Accompany Mechanical Engineering DesignFor no yield, m = Sy 0z = m m m= 0 m m= m mm = Sy = 27.47 kpsi, m =_ 2 + 2Sy_1/2= 12.32 kpsiz = 27.4712.32 = 2.230From Table A-10, pf = 0.0129R = 1 pf = 1 0.0129 = 0.987 Ans.2-24 For a lognormal distribution,Eq. (2-18) y = ln x ln_1 +C2xEq. (2-19) y =_ln_1 +C2x_From Prob. (2-23)m = Sy = xy =_ln Syln_1 +C2Sy__ln ln_1 +C2_= ln_Sy _1 +C21 +C2Sy_ y =_ln_1 +C2Sy_+ln_1 +C2__1/2=_ln__1 +C2Sy__1 +C2__z = = ln_Sy _1 +C21 +C2Sy__ln__1 +C2Sy__1 +C2__ = 4 Pd2 = 4(30)(12) = 38.197 kpsi = 4 Pd2 = 4(5.1)(12) = 6.494 kpsiC = 6.49438.197 = 0.1700CSy = 3.8149.6 = 0.076 810mshi20396_ch02.qxd 7/21/03 3:28 PM Page 24

Chapter 2 25z = ln__ 49.638.197_ 1 +0.17021 +0.076 812___ln_(1 +0.076 812)(1 +0.1702)_ = 1.470From Table A-10pf = 0.0708R = 1 pf = 0.929 Ans.2-25(a) a = 1.000 0.001 inb = 2.000 0.003 inc = 3.000 0.005 ind = 6.020 0.006 in w = d a b c = 6.020 1 2 3 = 0.020 intw =

tall = 0.001 +0.003 +0.005 +0.006= 0.015 inw = 0.020 0.015 in Ans.(b) w = 0.020 w =_ 2all =__0.0013_2+_0.0033_2+_0.0053_2+_0.0063_2= 0.004 86 0.005 in (uniform)w = 0.020 0.005 in Ans.2-26V +V = (a +a)(b +b)(c +c)V +V = abc +bca +acb +abc + small higher order termsVV.= aa + bb + cc Ans.V = ab c = 1.25(1.875)(2.75) = 6.4453 in3VV = 0.0011.250 + 0.0021.875 + 0.0032.750 = 0.00296V = VVV = 0.00296(6.4453) = 0.0191 in3Lower range number:V V = 6.4453 0.0191 = 6.4262 in3Ans.Upper range number:V +V = 6.4453 +0.0191 = 6.4644 in3Ans.shi20396_ch02.qxd 7/21/03 3:28 PM Page 25

26 Solutions Manual Instructors Solution Manual to Accompany Mechanical Engineering Design2-27(a)wmax = 0.014 in, wmin = 0.004 in w = (0.014 +0.004)/2 = 0.009 inw = 0.009 0.005 in w =

x

y = a b c0.009 = a 0.042 1.000 a = 1.051 intw =

tall0.005 = ta+0.002 +0.002ta = 0.005 0.002 0.002 = 0.001 ina = 1.051 0.001 in Ans.(b) w =_

2all =_ 2a + 2b + 2c 2a = 2w 2b 2c=_0.0053_2_0.0023_2_0.0023_2 2a = 5.667(106) a =_5.667(106) = 0.00238 in a = 1.051 in, a = 0.00238 in Ans.2-28 Choose 15 mm as basic size, D, d. Table 2-8: t is designated as 15H7/h6. From Table A-11, the tolerance grades are D = 0.018 mm and d = 0.011 mm.Hole: Eq. (2-38)Dmax = D +D = 15 +0.018 = 15.018 mm Ans.Dmin = D = 15.000 mm Ans.Shaft: From Table A-12, fundamental deviation F = 0. From Eq. (2-39)dmax = d +F = 15.000 +0 = 15.000 mm Ans.dmin = d +Rd = 15.000 +0 0.011 = 14.989 mm Ans.2-29 Choose 45 mm as basic size. Table 2-8 designates t as 45H7/s6. From Table A-11, thetolerance grades are D = 0.025 mm and d = 0.016 mmHole: Eq. (2-38)Dmax = D +D = 45.000 +0.025 = 45.025 mm Ans.Dmin = D = 45.000 mm Ans.ac bwshi20396_ch02.qxd 7/21/03 3:28 PM Page 26

Chapter 2 27Shaft: From Table A-12, fundamental deviation F = +0.043 mm. From Eq. (2-40)dmin = d +F = 45.000 +0.043 = 45.043 mm Ans.dmax = d +F +d = 45.000 +0.043 +0.016 = 45.059 mm Ans.2-30 Choose 50 mm as basic size. From Table 2-8 t is 50H7/g6. From Table A-11, the tolerancegrades are D = 0.025 mm and d = 0.016 mm.Hole:Dmax = D +D = 50 +0.025 = 50.025 mm Ans.Dmin = D = 50.000 mm Ans.Shaft: From Table A-12 fundamental deviation = 0.009 mmdmax = d +F = 50.000 +(0.009) = 49.991 mm Ans.dmin = d +F d= 50.000 +(0.009) 0.016= 49.975 mm2-31 Choose the basic size as 1.000 in. From Table 2-8, for 1.0 in, the t is H8/f7. FromTable A-13, the tolerance grades are D = 0.0013 in and d = 0.0008 in.Hole: Dmax = D +(D)hole = 1.000 +0.0013 = 1.0013 in Ans.Dmin = D = 1.0000 in Ans.Shaft: From Table A-14: Fundamental deviation = 0.0008 indmax = d +F = 1.0000 +(0.0008) = 0.9992 in Ans.dmin = d +F d = 1.0000 +(0.0008) 0.0008 = 0.9984 in Ans.Alternatively,dmin = dmaxd = 0.9992 0.0008 = 0.9984 in. Ans.2-32Do = W+Di +WDo = W + Di + W= 0.139 +3.734 +0.139 = 4.012 intDo =

tall = 0.004 +0.028 +0.004= 0.036 inDo = 4.012 0.036 in Ans.DoW DiWshi20396_ch02.qxd 7/21/03 3:28 PM Page 27

28 Solutions Manual Instructors Solution Manual to Accompany Mechanical Engineering Design2-33Do = Di +2WDo = Di +2 W = 208.92 +2(5.33)= 219.58 mmtDo =

allt = tDi +2tw= 1.30 +2(0.13) = 1.56 mmDo = 219.58 1.56 mm Ans.2-34Do = Di +2WDo = Di +2 W = 3.734 +2(0.139)= 4.012 mmtDo =_

allt2=_t2Do +(2 tw)2_1/2= [0.0282+(2)2(0.004)2]1/2= 0.029 inDo = 4.012 0.029 in Ans.2-35Do = Di +2WDo = Di +2 W = 208.92 +2(5.33)= 219.58 mmtDo =_

allt2= [1.302+(2)2(0.13)2]1/2= 1.33 mmDo = 219.58 1.33 mm Ans.2-36(a) w = F W w = F W = 0.106 0.139= 0.033 intw =

allt = 0.003 +0.004tw = 0.007 inwmax = w +tw = 0.033 +0.007 = 0.026 inwmin = w tw = 0.033 0.007 = 0.040 inThe minimum squeeze is 0.026 in. Ans.wWFshi20396_ch02.qxd 7/21/03 3:28 PM Page 28

Chapter 2 29(b)Y = 3.992 0.020 inDo+w Y = 0w = Y Do w = Y Do = 3.992 4.012 = 0.020 intw =

allt = tY +tDo = 0.020 +0.036 = 0.056 inw = 0.020 0.056 inwmax = 0.036 inwmin = 0.076 in

O-ring is more likely compressed than free prior to assembly of theend plate.2-37(a) Figure denes w as gap.The O-ring is squeezed at least 0.75 mm.(b)From the gure, the stochastic equation is:Do+w = Yor, w = Y Do w = Y Do = 218.48 219.58 = 1.10 mmtw =

allt = tY +tDo = 1.10 +0.34 = 1.44 mmwmax = w +tw = 1.10 +1.44 = 0.34 mmwmin = w tw = 1.10 1.44 = 2.54 mmThe O-ring is more likely to be circumferentially compressed than free prior to as-sembly of the end plate.Ymax = Do = 219.58 mmYmin = max[0.99 Do, Do1.52]= max[0.99(219.58, 219.58 1.52)]= 217.38 mmY = 218.48 1.10 mmYDo ww = F W w = F W= 4.32 5.33 = 1.01 mmtw =

allt = tF +tW = 0.13 +0.13 = 0.26 mmwmax = w +tw = 1.01 +0.26 = 0.75 mmwmin = w tw = 1.01 0.26 = 1.27 mmwWFYmax = Do = 4.012 inYmin = max[0.99 Do, Do0.06]= max[3.9719, 3.952] = 3.972 inYDo wshi20396_ch02.qxd 8/6/03 11:07 AM Page 29

30 Solutions Manual Instructors Solution Manual to Accompany Mechanical Engineering Design2-38wmax = 0.020 in, wmin = 0.040 in w = 12(0.020 +(0.040)) = 0.030 intw = 12(0.020 (0.040)) = 0.010 inb = 0.750 0.001 inc = 0.120 0.005 ind = 0.875 0.001 in w = a b c d0.030 = a 0.875 0.120 0.750 a = 0.875 +0.120 +0.750 0.030 a = 1.715 inAbsolute:tw =

allt = 0.010 = ta+0.001 +0.005 +0.001ta = 0.010 0.001 0.005 0.001= 0.003 ina = 1.715 0.003 in Ans.Statistical: For a normal distribution of dimensionst2w =

allt2= t2a +t2b +t2c +t2dta =_t2wt2b t2c t2d_1/2= (0.01020.00120.00520.0012)1/2= 0.0085a = 1.715 0.0085 in Ans.2-39x n nx nx293 19 1767 164 31195 25 2375 225 62597 38 3685 357 54299 17 1683 166 617101 12 1212 122 412103 10 1030 106 090105 5 525 55 125107 4 428 45 796109 4 436 47 524111 2 222 24 624136 13364 1315 704 x = 13 364/136 = 98.26 kpsisx =_1 315 704 13 3642/136135_1/2= 4.30 kpsib cwdashi20396_ch02.qxd 7/21/03 3:28 PM Page 30

Chapter 2 31Under normal hypothesis,z0.01 = (x0.0198.26)/4.30x0.01 = 98.26 +4.30z0.01= 98.26 +4.30(2.3267)= 88.26 .= 88.3 kpsi Ans.2-40 From Prob. 2-39, x = 98.26 kpsi, and x = 4.30 kpsi.Cx = x/x = 4.30/98.26 = 0.043 76From Eqs. (2-18) and (2-19),y = ln(98.26) 0.043 762/2 = 4.587 y =_ln(1 +0.043 762) = 0.043 74For a yield strength exceeded by 99% of the population,z0.01 = (ln x0.01y)/ y ln x0.01 = y+ yz0.01From Table A-10, for 1% failure, z0.01 = 2.326. Thus,ln x0.01 = 4.587 +0.043 74(2.326) = 4.485x0.01 = 88.7 kpsi Ans.The normal PDF is given by Eq. (2-14) asf (x) = 14.302exp_12_x 98.264.30_2_For the lognormal distribution, from Eq. (2-17), dening g(x),g(x) = 1x(0.043 74)2exp_12_ln x 4.5870.043 74_2_x (kpsi) f/(Nw) f (x) g(x) x (kpsi) f/(Nw) f (x) g(x)92 0.00000 0.03215 0.03263 102 0.03676 0.06356 0.0613492 0.06985 0.03215 0.03263 104 0.03676 0.03806 0.0370894 0.06985 0.05680 0.05890 104 0.01838 0.03806 0.0370894 0.09191 0.05680 0.05890 106 0.01838 0.01836 0.0186996 0.09191 0.08081 0.08308 106 0.01471 0.01836 0.0186996 0.13971 0.08081 0.08308 108 0.01471 0.00713 0.0079398 0.13971 0.09261 0.09297 108 0.01471 0.00713 0.0079398 0.06250 0.09261 0.09297 110 0.01471 0.00223 0.00286100 0.06250 0.08548 0.08367 110 0.00735 0.00223 0.00286100 0.04412 0.08548 0.08367 112 0.00735 0.00056 0.00089102 0.04412 0.06356 0.06134 112 0.00000 0.00056 0.00089Note: rows are repeated to draw histogramshi20396_ch02.qxd 7/21/03 3:28 PM Page 31

32 Solutions Manual Instructors Solution Manual to Accompany Mechanical Engineering DesignThe normal and lognormal are almost the same. However the data is quite skewed andperhaps a Weibull distribution should be explored. For a method of establishing theWeibull parameters see Shigley, J. E., and C. R. Mischke, Mechanical Engineering Design,McGraw-Hill, 5th ed., 1989, Sec. 4-12.2-41 Let x = (S

f e)104x0 = 79 kpsi, = 86.2 kpsi, b = 2.6Eq. (2-28) x = x0+( x0)(1 +1/b) x = 79 +(86.2 79)(1 +1/2.6)= 79 +7.2 (1.38)From Table A-34, (1.38) = 0.88854 x = 79 +7.2(0.888 54) = 85.4 kpsi Ans.Eq. (2-29) x = ( x0)[(1 +2/b) 2(1 +1/b)]1/2= (86.2 79)[(1 +2/2.6) 2(1 +1/2.6)]1/2= 7.2[0.923 76 0.888 542]1/2= 2.64 kpsi Ans.Cx = x x = 2.6485.4 = 0.031 Ans.2-42x = Sutx0 = 27.7, = 46.2, b = 4.38x = 27.7 +(46.2 27.7)(1 +1/4.38)= 27.7 +18.5 (1.23)= 27.7 +18.5(0.910 75)= 44.55 kpsi Ans.f (x)g(x)Histogram00.020.040.060.080.10.120.140.1690 92 94 96 98 100 102 104 106 108x (kpsi)Probability density110 112shi20396_ch02.qxd 7/21/03 3:28 PM Page 32

Chapter 2 33 x = (46.2 27.7)[(1 +2/4.38) 2(1 +1/4.38)]1/2= 18.5[(1.46) 2(1.23)]1/2= 18.5[0.8856 0.910 752]1/2= 4.38 kpsi Ans.Cx = 4.3844.55 = 0.098 Ans.From the Weibull survival equationR = exp__x x0 x0_b_= 1 pR40 = exp__x40 x0 x0_b_= 1 p40= exp__ 40 27.746.2 27.7_4.38_= 0.846p40 = 1 R40 = 1 0.846 = 0.154 = 15.4% Ans.2-43x = Sutx0 = 151.9, = 193.6, b = 8x = 151.9 +(193.6 151.9)(1 +1/8)= 151.9 +41.7 (1.125)= 151.9 +41.7(0.941 76)= 191.2 kpsi Ans. x = (193.6 151.9)[(1 +2/8) 2(1 +1/8)]1/2= 41.7[(1.25) 2(1.125)]1/2= 41.7[0.906 40 0.941 762]1/2= 5.82 kpsi Ans.Cx = 5.82191.2 = 0.0302-44x = Sutx0 = 47.6, = 125.6, b = 11.84 x = 47.6 +(125.6 47.6)(1 +1/11.84) x = 47.6 +78 (1.08)= 47.6 +78(0.959 73) = 122.5 kpsi x = (125.6 47.6)[(1 +2/11.84) 2(1 +1/11.84)]1/2= 78[(1.08) 2(1.17)]1/2= 78(0.959 73 0.936 702)1/2= 22.4 kpsishi20396_ch02.qxd 7/21/03 3:28 PM Page 33

34 Solutions Manual Instructors Solution Manual to Accompany Mechanical Engineering DesignFrom Prob. 2-42p = 1 exp__x x0 0_b_= 1 exp__ 100 47.6125.6 47.6_11.84_= 0.0090 Ans.y = Syy0 = 64.1, = 81.0, b = 3.77 y = 64.1 +(81.0 64.1)(1 +1/3.77)= 64.1 +16.9 (1.27)= 64.1 +16.9(0.902 50)= 79.35 kpsiy = (81 64.1)[(1 +2/3.77) (1 +1/3.77)]1/2y = 16.9[(0.887 57) 0.902 502]1/2= 4.57 kpsip = 1 exp__y y0 y0_3.77_p = 1 exp__70 64.181 64.1_3.77_= 0.019 Ans.2-45 x = Sut = W[122.3, 134.6, 3.64] kpsi, p(x > 120) = 1 = 100% since x0 > 120 kpsip(x > 133) = exp__ 133 122.3134.6 122.3_3.64_= 0.548 = 54.8% Ans.2-46 Using Eqs. (2-28) and (2-29) and Table A-34,n = n0+( n0)(1 +1/b) = 36.9 +(133.6 36.9)(1 +1/2.66) =122.85 kcycles n = ( n0)[(1 +2/b) 2(1 +1/b)] = 34.79 kcyclesFor the Weibull density function, Eq. (2-27), fW(n) = 2.66133.6 36.9_ n 36.9133.6 36.9_2.661exp__ n 36.9133.6 36.9_2.66_For the lognormal distribution, Eqs. (2-18) and (2-19) give,y = ln(122.85) (34.79/122.85)2/2 = 4.771 y =_[1 +(34.79/122.85)2] = 0.2778shi20396_ch02.qxd 7/21/03 3:28 PM Page 34

Chapter 2 35From Eq. (2-17), the lognormal PDF isfLN(n) = 10.2778 n2exp_12_ln n 4.7710.2778_2_We form a table of densities fW(n) and fLN(n) and plot.n (kcycles) fW(n) fLN(n)40 9.1E-05 1.82E-0550 0.000991 0.00024160 0.002498 0.00123370 0.004380 0.00350180 0.006401 0.00673990 0.008301 0.009913100 0.009822 0.012022110 0.010750 0.012644120 0.010965 0.011947130 0.010459 0.010399140 0.009346 0.008492150 0.007827 0.006597160 0.006139 0.004926170 0.004507 0.003564180 0.003092 0.002515190 0.001979 0.001739200 0.001180 0.001184210 0.000654 0.000795220 0.000336 0.000529The Weibull L10 life comes from Eq. (2-26) with a reliability of R = 0.90. Thus,n0.10 = 36.9 +(133 36.9)[ln(1/0.90)]1/2.66= 78.1 kcycles Ans.f (n)n, kcycles00.0040.0020.006 0.0080.0100.0120.0140 100 50 150 200LNW250shi20396_ch02.qxd 7/21/03 3:28 PM Page 35

36 Solutions Manual Instructors Solution Manual to Accompany Mechanical Engineering DesignThe lognormal L10 life comes from the denition of the z variable. That is,ln n0 = y+ yz or n0 = exp(y+ yz)From Table A-10, for R = 0.90, z = 1.282. Thus,n0 = exp[4.771 +0.2778(1.282)] = 82.7 kcycles Ans.2-47 Form a tablex g(x)i L(105) fi fix(105) fix2(1010) (105)1 3.05 3 9.15 27.9075 0.05572 3.55 7 24.85 88.2175 0.14743 4.05 11 44.55 180.4275 0.25144 4.55 16 72.80 331.24 0.31685 5.05 21 106.05 535.5525 0.32166 5.55 13 72.15 400.4325 0.27897 6.05 13 78.65 475.8325 0.21518 6.55 6 39.30 257.415 0.15179 7.05 2 14.10 99.405 0.100010 7.55 0 0 0 0.062511 8.05 4 32.20 259.21 0.037512 8.55 3 25.65 219.3075 0.021813 9.05 0 0 0 0.012414 9.55 0 0 0 0.006915 10.05 1 10.05 101.0025 0.0038100 529.50 2975.95 x = 529.5(105)/100 = 5.295(105) cycles Ans.sx =_2975.95(1010) [529.5(105)]2/100100 1_1/2= 1.319(105) cycles Ans.Cx = s/ x = 1.319/5.295 = 0.249y = ln 5.295(105) 0.2492/2 = 13.149 y =_ln(1 +0.2492) = 0.245g(x) = 1x y2exp_12_ln x y y_2_g(x) = 1.628x exp_12_ln x 13.1490.245_2_shi20396_ch02.qxd 7/21/03 3:28 PM Page 36

Chapter 2 372-48x = Su = W[70.3, 84.4, 2.01]Eq. (2-28) x = 70.3 +(84.4 70.3)(1 +1/2.01)= 70.3 +(84.4 70.3)(1.498)= 70.3 +(84.4 70.3)0.886 17= 82.8 kpsi Ans.Eq. (2-29) x = (84.4 70.3)[(1 +2/2.01) 2(1 +1/2.01)]1/2 x = 14.1[0.997 91 0.886 172]1/2= 6.502 kpsiCx = 6.50282.8 = 0.079 Ans.2-49 Take the Weibull equation for the standard deviation x = ( x0)[(1 +2/b) 2(1 +1/b)]1/2and the mean equation solved for x x0 x x0 = ( x0)(1 +1/b)Dividing the rst by the second, x x x0= [(1 +2/b) 2(1 +1/b)]1/2(1 +1/b)4.249 33.8 =_ (1 +2/b)

2(1 +1/b) 1 =R = 0.276300.10.20.30.40.5105 g(x)x, cyclesSuperposedhistogramand PDF3.05(105) 10.05(105)shi20396_ch02.qxd 7/21/03 3:28 PM Page 37

38 Solutions Manual Instructors Solution Manual to Accompany Mechanical Engineering DesignMake a table and solve for b iterativelyb .= 4.068 Using MathCad Ans. = x0+ x x0(1 +1/b) = 33.8 + 49 33.8(1 +1/4.068)= 49.8 kpsi Ans.2-50x = Sy = W[34.7, 39, 2.93] kpsi x = 34.7 +(39 34.7)(1 +1/2.93)= 34.7 +4.3(1.34)= 34.7 +4.3(0.892 22) = 38.5 kpsi x = (39 34.7)[(1 +2/2.93) 2(1 +1/2.93)]1/2= 4.3[(1.68) 2(1.34)]1/2= 4.3[0.905 00 0.892 222]1/2= 1.42 kpsi Ans.Cx = 1.42/38.5 = 0.037 Ans.2-51x (Mrev) f f x f x21 11 11 112 22 44 883 38 114 3424 57 228 9125 31 155 7756 19 114 6847 15 105 7358 12 96 7689 11 99 89110 9 90 90011 7 77 84712 5 60 720Sum 78 237 1193 7673x = 1193(106)/237 = 5.034(106) cycles x =_7673(1012) [1193(106)]2/237237 1 = 2.658(106) cyclesCx = 2.658/5.034 = 0.528b 1 + 2/b 1 + 1/b (1 +2/b) (1 +1/b)3 1.67 1.33 0.90330 0.89338 0.3634 1.5 1.25 0.88623 0.90640 0.2804.1 1.49 1.24 0.88595 0.90852 0.271shi20396_ch02.qxd 7/21/03 3:28 PM Page 38

Chapter 2 39From Eqs. (2-18) and (2-19),y = ln[5.034(106)] 0.5282/2 = 15.292 y =_ln(1 +0.5282) = 0.496From Eq. (2-17), dening g(x),g (x) = 1x(0.496)2exp_12_ln x 15.2920.496_2_x (Mrev) f /(Nw) g(x) (106)0.5 0.00000 0.000110.5 0.04641 0.000111.5 0.04641 0.052041.5 0.09283 0.052042.5 0.09283 0.169922.5 0.16034 0.169923.5 0.16034 0.207543.5 0.24051 0.207544.5 0.24051 0.178484.5 0.13080 0.178485.5 0.13080 0.131585.5 0.08017 0.131586.5 0.08017 0.090116.5 0.06329 0.090117.5 0.06329 0.059537.5 0.05063 0.059538.5 0.05063 0.038698.5 0.04641 0.038699.5 0.04641 0.025019.5 0.03797 0.0250110.5 0.03797 0.0161810.5 0.02954 0.0161811.5 0.02954 0.0105111.5 0.02110 0.0105112.5 0.02110 0.0068712.5 0.00000 0.00687z = ln x y y ln x = y+ yz = 15.292 +0.496zL10 life, where 10% of bearings fail, from Table A-10, z = 1.282. Thus,ln x = 15.292 +0.496(1.282) = 14.66 x = 2.32 106rev Ans.HistogramPDFx, Mrevg(x)(106)00.050.10.150.20.250 2 4 6 8 10 12shi20396_ch02.qxd 7/21/03 3:28 PM Page 39

3-1 From Table A-20Sut = 470 MPa (68 kpsi), Sy = 390 MPa (57 kpsi) Ans.3-2 From Table A-20Sut = 620 MPa (90 kpsi), Sy = 340 MPa (49.5 kpsi) Ans.3-3 Comparison of yield strengths:Sut of G10500 HR is 620470 = 1.32 times larger than SAE1020 CD Ans.Syt of SAE1020 CD is 390340 = 1.15 times larger than G10500 HR Ans.From Table A-20, the ductilities (reduction in areas) show,SAE1020 CD is 4035 = 1.14 times larger than G10500 Ans. The stiffness values of these materials are identical Ans.Table A-20 Table A-5Sut Sy Ductility Stiffness MPa (kpsi) MPa (kpsi) R% GPa (Mpsi)SAE1020 CD 470(68) 390 (57) 40 207(30)UNS10500 HR 620(90) 340(495) 35 207(30)3-4 From Table A-211040 Q&T Sy = 593 (86) MPa (kpsi) at 205C (400F) Ans.3-5 From Table A-211040 Q&T R = 65% at 650C (1200F) Ans.3-6 Using Table A-5, the specic strengths are:UNS G10350 HR steel: SyW = 39.5(103)0.282 = 1.40(105) in Ans.2024 T4 aluminum: SyW = 43(103)0.098 = 4.39(105) in Ans.Ti-6Al-4V titanium: SyW = 140(103)0.16 = 8.75(105) in Ans.ASTM 30 gray cast iron has no yield strength. Ans.Chapter 3shi20396_ch03.qxd 8/18/03 10:18 AM Page 40

Chapter 3 413-7 The specic moduli are:UNS G10350 HR steel: EW = 30(106)0.282 = 1.06(108) in Ans.2024 T4 aluminum: EW = 10.3(106)0.098 = 1.05(108) in Ans.Ti-6Al-4V titanium: EW = 16.5(106)0.16 = 1.03(108) in Ans.Gray cast iron: EW = 14.5(106)0.26 = 5.58(107) in Ans.3-8 2G(1 +) = E = E 2G2GFrom Table A-5Steel: = 30 2(11.5)2(11.5) = 0.304 Ans.Aluminum: = 10.4 2(3.90)2(3.90) = 0.333 Ans.Beryllium copper: = 18 2(7)2(7) = 0.286 Ans.Gray cast iron: = 14.5 2(6)2(6) = 0.208 Ans.3-9 0100 0.0020.10.0040.20.0060.30.0080.40.0100.50.0120.60.0140.70.0160.8(Lower curve)(Upper curve)20304050Stress PA0 kpsiStrain, 607080EYUSu 85.5 kpsi Ans.E 900.003 30 000 kpsi Ans.Sy 45.5 kpsi Ans.R (100) 45.8% Ans.A0 AFA0 0.1987 0.10770.1987 ll0l l0l0ll0 1AA0 1shi20396_ch03.qxd 8/18/03 10:18 AM Page 41

42 Solutions Manual Instructors Solution Manual to Accompany Mechanical Engineering Design3-10 To plot true vs. , the following equations are applied to the data.A0 = (0.503)24 = 0.1987 in2Eq. (3-4) = ln ll0for 0 L 0.0028 in = ln A0A for L > 0.0028 intrue = PAThe results are summarized in the table below and plotted on the next page.The last 5 points of data are used to plot log vs log The curve t gives m = 0.2306log 0 = 5.1852 0 = 153.2 kpsiAns.For 20% cold work, Eq. (3-10) and Eq. (3-13) give,A = A0(1 W) = 0.1987(1 0.2) = 0.1590 in2 = ln A0A = ln 0.19870.1590 = 0.2231Eq. (3-14):S

y = 0m= 153.2(0.2231)0.2306= 108.4 kpsi Ans.Eq. (3-15), with Su = 85.5 kpsi from Prob. 3-9,S

u = Su1 W = 85.51 0.2 = 106.9 kpsi Ans.P L A true log log true0 0 0.198 713 0 01000 0.0004 0.198 713 0.0002 5032.388 3.69901 3.7017742000 0.0006 0.198 713 0.0003 10064.78 3.52294 4.0028043000 0.0010 0.198 713 0.0005 15097.17 3.30114 4.1788954000 0.0013 0.198 713 0.00065 20129.55 3.18723 4.3038347000 0.0023 0.198 713 0.001149 35226.72 2.93955 4.5468728400 0.0028 0.198 713 0.001399 42272.06 2.85418 4.6260538800 0.0036 0.1984 0.001575 44354.84 2.80261 4.6469419200 0.0089 0.1978 0.004604 46511.63 2.33685 4.6675629100 0.1963 0.012216 46357.62 1.91305 4.66612113200 0.1924 0.032284 68607.07 1.49101 4.83636915200 0.1875 0.058082 81066.67 1.23596 4.90884217000 0.1563 0.240083 108765.2 0.61964 5.0364916400 0.1307 0.418956 125478.2 0.37783 5.09856814800 0.1077 0.612511 137418.8 0.21289 5.138046shi20396_ch03.qxd 8/18/03 10:18 AM Page 42

Chapter 3 433-11 Tangent modulus at = 0 isE0 = .= 5000 00.2(103) 0 = 25(106) psiAt = 20 kpsiE20.= (26 19)(103)(1.5 1)(103) = 14.0(106) psi Ans.(103) (kpsi)0 00.20 50.44 100.80 161.0 191.5 262.0 322.8 403.4 464.0 495.0 543-12 From Prob. 2-8, for y = a1x +a2x2a1 = yx3xyx2xx3(x2)2 a2 = xxy yx2xx3(x2)2log log y 0.2306x 5.18524.84.955.15.21.6 1.4 1.2 1 0.8 0.6 0.4 0.2 0truetrue (psi)0200004000060000800001000001200001400001600000 0.1 0.2 0.3 0.4 0.5 0.6 0.7 (103)(Sy)0.001 35 kpsi Ans. (kpsi)01020304050600 1 2 3 4 5shi20396_ch03.qxd 8/18/03 10:18 AM Page 43

44 Solutions Manual Instructors Solution Manual to Accompany Mechanical Engineering DesignLet x represent (103) and y represent (kpsi),x y x2x3xy0 0 0 0 00.2 5 0.04 0.008 1.00.44 10 0.1936 0.085184 4.40.80 16 0.64 0.512 12.81.0 19 1.00 1.000 19.01.5 26 2.25 3.375 39.02.0 32 4.00 8.000 64.02.8 40 7.84 21.952 112.03.4 46 11.56 39.304 156.44.0 49 16.00 64.000 196.05.0 54 25.00 125.000 270.0 = 21.14 297 68.5236 263.2362 874.6Substituting,a1 = 297(263.2362) 874.6(68.5236)21.14(263.2362) (68.5236)2 = 20.993 67a2 = 21.14(874.6) 297(68.5236)21.14(263.2362) (68.5236)2 = 2.142 42The tangent modulus isdydx = dd= 20.993 67 2(2.142 42)x = 20.993 67 4.284 83xAt = 0, E0 = 20.99 Mpsi Ans.At = 20 kpsi20 = 20.993 67x 2.142 42x2x = 1.069, 8.73Taking the rst root, = 1.069 and the tangent modulus isE20 = 20.993 67 4.284 83(1.069) = 16.41 Mpsi Ans.Determine the equation for the 0.1 percent offset liney = 20.99x +b at y = 0, x = 1 b = 20.99y = 20.99x 20.99 = 20.993 67x 2.142 42x22.142 42x220.99 = 0 x = 3.130(Sy)0.001 = 20.99(3.13) 2.142(3.13)2= 44.7 kpsi Ans.3-13 Since |o| = |i|

ln R +hR + N

=

ln RR + N

=

ln R + NR

R +hR + N = R + NR( R + N)2= R( R +h)From which, N2+2RN Rh = 0shi20396_ch03.qxd 8/18/03 10:18 AM Page 44

Chapter 3 45The roots are: N = R

1

1 + hR

1/2

The + sign being signicant,N = R

1 + hR

1/21

Ans.Substitute for N ino = ln R +hR + NGives 0 = lnR +hR + R

1 + hR

1/2 R= ln

1 + hR

1/2Ans.These constitute a useful pair of equations in cold-forming situations, allowing the surfacestrains to be found so that cold-working strength enhancement can be estimated.3-14 = 16Td3 = 16T(12.5)3106(103)3 = 2.6076T MPa =

180

rL =

180

(12.5)350 = 6.2333(104)For G, take the rst 10 data points for the linear part of the curve. (103) (MPa)T (deg.) (103) (MPa) x y x2xy0 0 0 0 0 0 0 07.7 0.38 0.236865 20.07852 0.236865 20.07852 0.056105 4.755915.3 0.80 0.498664 39.89628 0.498664 39.89628 0.248666 19.894823.0 1.24 0.772929 59.9748 0.772929 59.9748 0.597420 46.356330.7 1.64 1.022261 80.05332 1.022261 80.05332 1.045018 81.835438.3 2.01 1.252893 99.87108 1.252893 99.87108 1.569742 125.127846.0 2.40 1.495992 119.9496 1.495992 119.9496 2.237992 179.443653.7 2.85 1.776491 140.0281 1.776491 140.0281 3.155918 248.758661.4 3.25 2.025823 160.1066 2.025823 160.1066 4.103957 324.347669.0 3.80 2.368654 179.9244 2.368654 179.9244 5.610522 426.178676.7 4.50 2.804985 200.0029 = 11.45057 899.8828 18.62534 1456.698680.0 5.10 3.178983 208.60885.0 6.48 4.039178 221.64690.0 8.01 4.992873 234.68495.0 9.58 5.971501 247.722100.0 11.18 6.968829 260.76shi20396_ch03.qxd 8/18/03 10:18 AM Page 45

46 Solutions Manual Instructors Solution Manual to Accompany Mechanical Engineering Designy = mx +b, = y, = x where m is the shear modulus G,m = Nxy xyNx2(x)2 = 77.3 MPa103 = 77.3 GPa Ans.b = y mxN = 1.462 MPaFrom curve Sys.= 200 MPa Ans.Note since is not uniform, the offset yield does not apply, so we are using the elasticlimit as an approximation.3-15 x f f x f x238.5 2 77.0 2964.5039.5 9 355.5 14 042.2540.5 30 1215.0 49 207.5041.5 65 2697.5 111 946.3042.5 101 4292.5 182 431.3043.5 112 4872.0 211 932.0044.5 90 4005.0 178 222.5045.5 54 2457.0 111793.5046.5 25 1162.5 54 056.2547.5 9 427.5 20 306.2548.5 2 97.0 4 704.5049.5 1 49.5 2 450.25 = 528.0 500 21708.0 944 057.00 x = 21 708/500 = 43.416, x =

944 057 (21 7082/500)500 1 = 1.7808Cx = 1.7808/43.416 = 0.041 02, y = ln 43.416 ln(1 +0.041 022) = 3.7691 (103) (MPa)0501001502002503000 1 2 3 4 5 6 7shi20396_ch03.qxd 8/18/03 10:18 AM Page 46

Chapter 3 47 y =

ln(1 +0.041 022) = 0.0410,g(x) = 1x(0.0410)2exp

12

ln x 3.76910.0410

2

x f /( Nw) g(x) x f /( Nw) g(x)38 0 0.001488 45 0.180 0.14226838 0.004 0.001488 45 0.108 0.14226839 0.004 0.009057 46 0.108 0.07381439 0.018 0.009057 46 0.050 0.07381440 0.018 0.035793 47 0.050 0.02941040 0.060 0.035793 47 0.018 0.02941041 0.060 0.094704 48 0.018 0.00915241 0.130 0.094704 48 0.004 0.00915242 0.130 0.172538 49 0.004 0.00225942 0.202 0.172538 49 0.002 0.00225943 0.202 0.222074 50 0.002 0.00044943 0.224 0.222074 50 0 0.00044944 0.224 0.20674844 0.180 0.206748Sy = LN(43.42, 1.781) kpsi Ans.3-16 From Table A-22AISI 1212 Sy = 28.0 kpsi , f = 106 kpsi, Sut = 61.5 kpsi0 = 110 kpsi, m = 0.24, f = 0.85From Eq. (3-12) u = m = 0.24Eq. (3-10) A0A

i= 11 W = 11 0.2 = 1.25Eq. (3-13) i = ln 1.25 = 0.2231 i < uxf (x)00.050.10.150.20.2535 40 45 50HistogramPDFshi20396_ch03.qxd 8/18/03 10:18 AM Page 47

48 Solutions Manual Instructors Solution Manual to Accompany Mechanical Engineering DesignEq. (3-14) S

y = 0mi = 110(0.2231)0.24= 76.7 kpsi Ans.Eq. (3-15) S

u = Su1 W = 61.51 0.2 = 76.9 kpsi Ans.3-17 For HB = 250,Eq. (3-17) Su= 0.495 (250) = 124 kpsi= 3.41 (250) = 853 MPa Ans.3-18 For the data given,

HB = 2530

H2B = 640 226HB = 253010 = 253 HB =

640 226 (2530)2/109 = 3.887Eq. (3-17)Su = 0.495(253) = 125.2 kpsi Ans. su = 0.495(3.887) = 1.92 kpsi Ans.3-19 From Prob. 3-18, HB = 253 and HB = 3.887Eq. (3-18)Su = 0.23(253) 12.5 = 45.7 kpsi Ans. su = 0.23(3.887) = 0.894 kpsi Ans.3-20 (a) uR.= 45.522(30) = 34.5 in lbf/in3Ans.(b)P L A A0/A 1 = P/A00 0 0 01000 0.0004 0.0002 5032.392000 0.0006 0.0003 10064.783000 0.0010 0.0005 15097.174000 0.0013 0.000 65 20129.557000 0.0023 0.00115 35226.728400 0.0028 0.0014 42272.068800 0.0036 0.0018 44285.029200 0.0089 0.004 45 46297.979100 0.1963 0.012 291 0.012 291 45794.7313 200 0.1924 0.032 811 0.032 811 66427.5315 200 0.1875 0.059 802 0.059 802 76492.3017 000 0.1563 0.271 355 0.271 355 85550.6016 400 0.1307 0.520 373 0.520 373 82531.1714 800 0.1077 0.845059 0.845 059 74479.35shi20396_ch03.qxd 8/18/03 10:18 AM Page 48

Chapter 3 49uT.=5

i =1Ai = 12(43 000)(0.001 5) +45 000(0.004 45 0.001 5)+12(45 000 +76 500)(0.059 8 0.004 45)+81 000(0.4 0.059 8) +80 000(0.845 0.4).= 66.7(103)in lbf/in3Ans.02000010000300004000050000600007000080000900000 0.2 0.4 0.6 0.8A3A4 A5Last 6 data pointsFirst 9 data points0A1 A2 15000100005000200002500030000350004000045000500000 0.002 0.001 0.003 0.004 0.00502000010000300004000050000600007000080000900000 0.2 0.4All data points0.6 0.8shi20396_ch03.qxd 8/18/03 10:18 AM Page 49

Text Eq. refers to Mechanical Engineering Design, 7th edition text by Joseph Edward Shigley, Charles R. Mischke and Richard G. Budynas; equations and figures with the prefix T refer to the present tutorial. MECHANICAL ENGINEERING DESIGN TUTORIAL 4 15: PRESSURE VESSEL DESIGN PRESSURE VESSEL DESIGN MODELS FOR CYLINDERS: 1. Thick-walled Cylinders 2. Thin-walled Cylinders THICK-WALL THEORY Thick-wall theory is developed from the Theory of Elasticity which yields the state of stress as a continuous function of radius over the pressure vessel wall. The state of stress is defined relative to a convenient cylindrical coordinate system: 1. t Tangential Stress 2. r Radial Stress 3. l Longitudinal Stress Stresses in a cylindrical pressure vessel depend upon the ratio of the inner radius to the outer radius ( /o ir r ) rather than the size of the cylinder. Principal Stresses (1 2 3, , ) 1. Determined without computation of Mohrs Circle; 2. Equivalent to cylindrical stresses ( , ,t r l ) Applicable for any wall thickness-to-radius ratio. Cylinder under Pressure Consider a cylinder, with capped ends, subjected to an internal pressure, pi, and an external pressure, po, FIGURE T4-15-1 oporirip lrt rlt

Shigley, Mischke & Budynas Machine Design Tutorial 415: Pressure Vessel Design 2/10 The cylinder geometry is defined by the inside radius, ,ir the outside radius, ,or and the cylinder length, l. In general, the stresses in the cylindrical pressure vessel ( , ,t r l ) can be computed at any radial coordinate value, r, within the wall thickness bounded by ir and ,or and will be characterized by the ratio of radii, / .o ir r = These cylindrical stresses represent the principal stresses and can be computed directly using Eq. 4-50 and 4-52. Thus we do not need to use Mohrs circle to assess the principal stresses. Tangential Stress: 2 22 2 2 2 2/ ) (i oi o o i o o i itr rr p p r r r p r p = for i or r r (Text Eq. 4-50) Radial Stress: 2 22 2 2 2 2/ ) (i oi o o i o o i irr rr p p r r r p r p + = for i or r r (Text Eq. 4-50) Longitudinal Stress: Applicable to cases where the cylinder carries the longitudinal load, such as capped ends. Only valid far away from end caps where bending, nonlinearities and stress concentrations are not significant. 2 22 2i oo o i ilr rr p r p= for i or r r (Modified Text Eq. 4-52) Two Mechanical Design Cases 1. Internal Pressure Only ( 0 =op ) 2. External Pressure Only ( 0 =ip ) Design Case 1: Internal Pressure Only Only one case to consider the critical section which exists at ir r = . Substituting 0 =op into Eqs. (4-50) and incorporating / ,o ir r = the largest value of each stress component is found at the inner surface: 2 2 2,max2 2 21( )1o it i t i i i tio ir rr r p p p Cr r + += = = = = (T-1)

Shigley, Mischke & Budynas Machine Design Tutorial 415: Pressure Vessel Design 3/10 where 2 2 22 2 211o itio ir rCr r+ += = is a function of cylinder geometry only. i r i rp r r = = =max ,) ( Natural Boundary Condition (T-2) Longitudinal stress depends upon end conditions: i lip C Capped Ends (T-3a) l = 0 Uncapped Ends (T-3b) where 211liC=. Design Case 2: External Pressure Only The critical section is identified by considering the state of stress at two points on the cylinder: r = ri and r = ro. Substituting pi = 0 into Text Eqs. (4-50) for each case: r = ri 0 ) ( = =i rr r Natural Boundary Condition (T-4a) ( )2 2,max2 2 22 21ot i t o o o too irr r p p p Cr r = = = = = (T-4b) where, 2 22 2 22 21otoo irCr r= = . r = ro o r o rp r r = = =max ,) ( Natural Boundary Condition (T-5a) 2 2 22 2 21( )1o it o o o o tio ir rr r p p p Cr r + += = = = (T-5b) Longitudinal stress for a closed cylinder now depends upon external pressure and radius while that of an open-ended cylinder remains zero: o lop C Capped Ends (T-6a) l = 0 Uncapped Ends (T-6b)

Shigley, Mischke & Budynas Machine Design Tutorial 415: Pressure Vessel Design 4/10 where 221loC =. Example T4.15.1: Thick-wall Cylinder Analysis Problem Statement: Consider a cylinder subjected to an external pressure of 150 MPa and an internal pressure of zero. The cylinder has a 25 mm ID and a 50 mm OD, respectively. Assume the cylinder is capped. Find: 1. the state of stress (r, t, l ) at the inner and outer cylinder surfaces; 2. the Mohrs Circle plot for the inside and outside cylinder surfaces; 3. the critical section based upon the estimate of max. Solution Methodology: Since we have an external pressure case, we need to compute the state of stress ( ,r ,t l ) at both the inside and outside radius in order to determine the critical section. 1. As the cylinder is closed and exposed to external pressure only, Eq. (T-6a) may be applied to calculate the longitudinal stress developed. This result represents the average stress across the wall of the pressure vessel and thus may be used for both the inner and outer radii analyses. 2. Assess the radial and tangential stresses using Eqs. (T-4) and (T-5) for the inner and outer radii, respectively. 3. Assess the principal stresses for the inner and outer radii based upon the magnitudes of ( ,r ,t l ) at each radius. 4. Use the principal stresses to calculate the maximum shear stress at each radius. 5. Draw Mohrs Circle for both states of stress and determine which provides the critical section. Solution: 1. Longitudinal Stress Calculation: OD 50 mm ID 25mm25mm; 12.5mm2 2 2 2o ir r = = = = = = Compute the radius ratio, 25 mm2.012.5 mmoirr = = =

Shigley, Mischke & Budynas Machine Design Tutorial 415: Pressure Vessel Design 5/10 Then, 2 22 2222(2)1 (2) 1( ) ( ) ( 150MPa)(1.3333 mm )1lol i l o o o loCr r r r p p C = = = = = = = = = 21.3333 mmMPa 200 = == =l 2. Radial & Tangential Stress Calculations: Inner Radius (r = ri) 2 22 22,max 2 22 2(2)1 (2) 12( ) ( 150MPa)(2.6667)toot i t o o too iCrr r p p Cr r = = = = = = = = 2.6667 400 MPat i (r r ) Compressive = = 0 p for Condition Boundary Naturali = = = 0 ) r (r i r Outer Radius (r = ro) 2 22 22 2,min 2 21 (2) 11 (2) 1( ) ( 150MPa)(1.6667)tio it o t o o tio iCr rr r p p Cr r + += = = += = = = = 1.6667 e Compressiv MPa 250 = == = = == = ) r (r o t Condition Boundary Natural MPa 150 = == = = == = = == =o i rp ) r (r 3. Define Principal Stresses: Inner Radius (r = ri ) Outer Radius (r = ro ) MPa 400MPa 200MPa 0321 = = = == =tlr MPa 250MPa 200MPa 150321 = = = = = =tlr 4. Maximum Shear Stress Calculations: Inner Radius (r = ri ) 1 3max0 ( 400)( )2 2ir r = = = = 200 MPa

Shigley, Mischke & Budynas Machine Design Tutorial 415: Pressure Vessel Design 6/10 Outer Radius (r = ro ) 1 3max( 150) ( 250)( )2 2or r = = = = 50 MPa 5. Mohrs Circles: Inner Radius (r = ri ) Outer Radius (r = ro ) Critical Section ! Radius Inside at is Section Critical r ri = == = = == = MPa 200 ) (max 1150 MPa = 3 = -250 MPa 2 = -200 MPamax = 50 MPa 2 = -200 MPa3 = -400 MPa 10 MPa = max = 200 MPa FIGURE T4-15-2 FIGURE T4-15-3

Shigley, Mischke & Budynas Machine Design Tutorial 415: Pressure Vessel Design 7/10 THIN-WALL THEORY Thin-wall theory is developed from a Strength of Materials solution which yields the state of stress as an average over the pressure vessel wall. Use restricted by wall thickness-to-radius ratio: 1According to theory, Thin-wall Theory is justified for20tr 1In practice, typically use a less conservative rule, 10tr State of Stress Definition: 1. Hoop Stress, t , assumed to be uniform across wall thickness. 2. Radial Stress is insignificant compared to tangential stress, thus, 0.r 3. Longitudinal Stress, l SExists for cylinders with capped ends; SAssumed to be uniformly distributed across wall thickness; SThis approximation for the longitudinal stress is only valid far away from the end-caps. 4. These cylindrical stresses ( , , )t r l are principal stresses ( , , )t r l which can be determined without computation of Mohrs circle plot. Analysis of Cylinder Section 1 t FV FHoop FHoop Pressure Acting over Projected Vertical Areadi FIGURE T4-15-4

Shigley, Mischke & Budynas Machine Design Tutorial 415: Pressure Vessel Design 8/10 The internal pressure exerts a vertical force, FV, on the cylinder wall which is balanced by the tangential hoop stress, FHoop. t pd F F Ft t A Fpd d p pA Ft i Hoop V yt t stressed t Hoopi i proj V = = == = == = =2 2 0)} 1 )( {()} 1 )( {( Solving for the tangential stress, Hoop Stress2itpdt = (Text Eq. 4-53) Comparison of state of stress for cylinder under internal pressure verses external pressure: Internal Pressure Only 20(Text Eq. 4-55)4 2itri tlpdHoop StresstBy DefinitionpdCapped Caset=== = External Pressure Only 204 2otro tlpdHoop StresstBy DefinitionpdCapped Caset=== = Example T4.15.2: Thin-wall Theory Applied to Cylinder Analysis Problem Statement: Repeat Example T1.1 using the Thin-wall Theory (po = 150 MPa, pi = 0, ID = 25 mm, OD = 50 mm). Find: The percent difference of the maximum shear stress estimates found using the Thick-wall and Thin-wall Theories.

Shigley, Mischke & Budynas Machine Design Tutorial 415: Pressure Vessel Design 9/10 Solution Methodology: 1. Check t/r ratio to determine if Thin-wall Theory is applicable. 2. Use the Thin-wall Theory to compute the state of stress 3. Identify the principal stresses based upon the stress magnitudes. 4. Use the principal stresses to assess the maximum shear stress. 5. Calculate the percent difference between the maximum shear stresses derived using the Thick-wall and Thin-wall Theories. Solution: 1. Check t/r Ratio: 10120121mm 25mm 5 . 12orrt = = The application of Thin-wall Theory to estimate the stress state of this cylinder is thus not justified. 2. Compute stresses using the Thin-wall Theory to compare with Thick-wall theory estimates. definition by Stress Radial b.mm) 2(12.5) mm 50 )( MPa 150 (2wall) across uniform stress, (average Stress Hoop a.r0MPa 300= ===td po ot c. Longitudinal Stress (average stress, uniform across wall)4 2o o tlp dt = = = 150 MPa 3. Identify Principal Stresses in terms of Average Stresses: MPa 300MPa 150MPa 0321 = = = == =tlr 4. Maximum Shear Stress Calculation: MPa 1502) MPa 300 ( 023 1max + = == 5. Percent Difference between Thin- and Thick-wall Estimates for the Critical Section:

Shigley, Mischke & Budynas Machine Design Tutorial 415: Pressure Vessel Design 10/10 max,Thin max,Thickmax,Thick% 100%( 150) ( 200)(100%)( 200)Difference = + += = +25% Thin -wall estimate is 25% low!

Text Eq. refers to Mechanical Engineering Design, 7th edition text by Joseph Edward Shigley, Charles R. Mischke and Richard G. Budynas; equations with the prefix T refer to the present tutorial. MECHANICAL ENGINEERING DESIGN TUTORIAL 4-17: PRESS AND SHRINK FITS APPLICATION OF THICK-WALL CYLINDRICAL PRESSURE VESSEL THEORY RELATING TO STRESSES DEVELOPED FROM INTERFERENCE FITS: 1. Design application which uses the cylindrical pressure vessel Thick-Wall Theory. 2. Stresses develop between cylinders due to the contact pressure generated by an interference fit. The interference fit is achieved by pressing a larger inside member into the smaller opening of an outside member. In the specific case of a shaft press fit into the hub of a gear, the outside diameter (OD) of the shaft is slightly larger than the inside hole diameter (ID) of the hub. The diametral difference between the shaft OD and the ID of the hub hole is referred to as the interference fit. The radial deformation required by the interference fit causes an interfacial pressure, p, to develop at the nominal radius, at r = R. Consequently, radial and tangential stresses, r and t , are produced. Assuming uncapped ends ( 0),l = a biaxial state of stress exists for which two non-zero principal stresses must be considered. From the cylindrical pressure vessel theory, the radial and tangential stresses represent principal stresses. The length of the outer member is assumed to be equal to the length of the inner member. (b) Cross-section of cylinders showing internal outside radius larger than external inside radius by a small amount of . (a) End view of inner and outer members, press fit together. riR ro R roriInner Member Outer Member FIGURE T4-17-1 Interference fit of two cylinders of finite length and equal lengths.

Shigley, Mischke & Budynas Machine Design Tutorial 4-17: Press and Shrink Fits 2/11 3. Referring to Fig. T4-17-1, the geometric features of the cylindrical parts are defined as: the inside radius of the inner cylindernominal radius of internal outside radius and external inside radius after assemblyoutside radius of the outer cylinderradial interferenceiorRr==== INSIDE CYLINDER Inner member experiences an external pressure, po = p, resulting in compressive tangential and radial stresses. Thick-Wall Theory may be applied with ro = R: 2 22 2( ) (Text Eq. 4-57)( )it i r R o itir i r R oR rp pCR rp p== += = = = OUTSIDE CYLINDER Outer member only experiences internal pressure, pi = p, resulting in tensile tangential stress and compressive radial stress. Thick-Wall Theory is, as always, applicable with ri = R: 2 22 2( ) (Text Eq. 4-58)( )ot o r R i otor i r R ir Rp pCr Rp p== += = = = DEFINITION OF INTERFACIAL PRESSURE We presently have two equations and three unknowns for both the inside and outside cylinder analyses. A third equation which relates the contact pressure and the interference can be derived by examining the deformation of the members. Deflection Equation The total radial interference may be defined as: o i + =total where,

Shigley, Mischke & Budynas Machine Design Tutorial 4-17: Press and Shrink Fits 3/11 decrease in radius of inner cylinder increase in radius of holeio== The deformation may also be expressed as: total i opRK pRK = + (Modified Text Eq. 4-59) where the outside member constant, Ko, is defined as, [ ]2 22 21 1oo o o oo o or RK CE E r R += + = + Using the radius ratio form defined for the cylindrical pressure vessel formulation, /o or R = , we can define 2211oooC += Note that the Co term is a function of geometry only, while the member constant term Ko is a function of both geometry and material parameters. Similarly for the inside member constant Ki, [ ]2 22 21 1ii i i ii i iR rK CE E R r += = with /i iR r = , Ci is defined as: 2211iiiC += For the case of a solid shaft, ri = 0, /i iR r = = and Ci = 1. We can now solve for the deformation for a given class of interference fits, pR K Ki o] [total + = Or, rearranging, the contact pressure, p, may be expressed as a function of the interference without assumptions regarding material property values: R1total+=i oK Kp (Modified Text Eq. 4-60)

Text Eq. refers to Mechanical Engineering Design, 7th edition text by Joseph Edward Shigley, Charles R. Mischke and Richard G. Budynas; equations with the prefix T refer to the present tutorial. Example T4.17.1: Shaft & Hub Shrink Fit Problem Statement: A carbon-steel gear hub having a nominal hole diameter of 1 inch is to be shrink-fitted to a carbon-steel shaft using a class FN4 fit. The hub has a nominal thickness of inch. Find: 1. the maximum tangential and radial stresses in the hub and shaft when the loosest fit is obtained; 2. the same as (a), except using the tightest fit as a condition. Solution Methodology: 1. Using Table A-1-2 of this document, identify the diametral size ranges for the shaft and the hole based on an ANSI US Customary Standard class FN4 fit which is discussed below. 2. Calculate the interferences for the loosest and tightest fits. 3. Compute the interfacial pressure for the loosest and tightest fits. 4. For each fit, calculate the radial and tangential stresses for both the hub and the shaft. Schematic: Solution: 1. Shaft and Hub Size Ranges: This problem specifies a class FN4 fit. The class FN4 fit is a specification defined by the American National Standards Institute (ANSI), American Standard Limits for Cylindrical Parts ANSI B4.1-1978. The FN4 fit is a Force fit suitable for parts which can be highly stressed or for shrink fits where the heavy pressing forces required are impractical. Excerpts from the ANSI B4.1-1978 standard are provided in Table A-1 of this tutorial. From Table A-1-2, for a nominal shaft/hole diameter of 1 in., the appropriate size range is 0.951.19 in. The allowable tolerances, in thousandths of an inch, are: ShaftHub

Shigley, Mischke & Budynas Machine Design Tutorial 4-17: Press and Shrink Fits 5/11 Largest Tolerance Smallest Tolerance (10-3 in.) (10-3 in.) Hub hole +0.8 0.0 Shaft +2.3 +1.8 The shaft and hub size ranges for a nominal 1 in. diameter are: Largest Diameter Smallest Diameter (in.) (in.) Hub hole 1.0008 1.0000 Shaft 1.0023 1.0018 2. Diametral and Radial Interference Calculations: Loosest Fit Tightest Fit Hub hole 1.0008 1.0000 Shaft 1.0018 1.0023 0.0010 0.0023 The radial interferences are therefore = 0.0005 in. (loosest fit) and = 0.001 15 in. (tightest fit). The interference is taken as a positive number by convention. Consequently, the radial interferences for the two cases are: = 0.0005 in. (loosest fit) = 0.001 15 in. (tightest fit) 3. Interfacial Pressure The interface pressure can be computed from Modified Eq. (4-60): R1total+=i oK Kp For this problem, the geometric features have been defined as: 1 in.0.5 in. 1.0 in. 0.0 0.5 in.2 2 2o id dr t r R = + = + = = = = Thus, for the hub (outside member),

Shigley, Mischke & Budynas Machine Design Tutorial 4-17: Press and Shrink Fits 6/11 [ ]( )[ ]2 22 286(1 in.)2(0.5 in.)1 (2) 1 51.66671 (2) 1 31 11.667 0.292 6.5289 10 (1/psi)30 10 psioooooo o oorRCK CE = = =+ += = = = = + = + = Similarly for the shaft (inside member) iiRr = = [ ]( )[ ]228611.011 11.0 0.292 2.360 10 (1/psi)30 10 psiiiii i iiCK CE += == = = For the loosest fit case: loosest fitloosest fit 8 81 1 0.0005R 6.5289 10 2.360 10 0.5o ipK K = = + + = 11 250 psi (loosest fit) For the tightest fit case: tightest fittightest fit 8 81 1 0.00115R 6.5289 10 2.360 10 0.5o ipK K = = + + = 25 875 psi (tightest fit) 4. Radial & Tangential Stress Calculations for Shaft and Hub: Loosest Fit (p = 11 250 psi) Hub: r (r R) p = = = 11 250 psi 2 22 2( ) (11250psi)(1.6667)ot oor Rr R p pCr R += = = = =18 750 psi Shaft: r (r R) p = = = 11 250 psi 2 22 2( ) ( 11250psi)(1)it iiR rr R p pCR r += = = = = 11 250 psi Tightest Fit (p = 25 875 psi)

Shigley, Mischke & Budynas Machine Design Tutorial 4-17: Press and Shrink Fits 7/11 Hub: r (r R) p = = = 25 875 psi 2 22 2( ) (25875psi)(1.6667)ot oor Rr R p pCr R += = = = =43 125 psi Shaft: r (r R) p = = = 25 875 psi 2 22 2( ) ( 25 875psi)(1)it iiR rr R p pCR r += = = = = 25 875 psi Thus the shaft has equal radial and tangential stress for each tightness condition, whereas, the hubs tangential stress is consistently higher than its radial stress for both the loosest and the tightest condition. Note: Since the shaft length is greater than the hub length, which is typical in practice, this design case violates one of the assumptions of the interfacial pressure development. In this case, there would be an increase in the interfacial pressure at each end of the hub. This condition of increased interfacial pressure would typically be accounted for by applying a stress concentration factor, Kt, for stresses calculated at points at the end of the hub such as, actual , tangential actual , radialt t tr t rKK ==

Shigley, Mischke & Budynas Machine Design Tutorial 4-17: Press and Shrink Fits 8/11 Table A-1 LIMITS AND FITS FOR CYLINDRICAL PARTS The limits shown in the accompanying tabulations are in thousandths of an inch. The size ranges include all sizes over the smallest size in the range, up to and including the largest size in the range. The letter symbols are defined as follows: RC Running and sliding fits are intended to provide a similar running performance, with suitable lubrication allowance, throughout the range of sizes. The clearance for the first two classes, used chiefly as slide fits, increases more slowly with diameter than the other classes, so that accurate location is maintained even at the expense of free relative motion. RC1 Close sliding fits are intended for the accurate location of parts which must assemble without perceptible play. RC2 Sliding fits are intended for accurate location but with greater maximum clearance than class RC1. Parts made to this fit move and turn easily, but are not intended to run freely, and in the larger sizes may seize with small temperature changes. RC3 Precision running fits are about the closest fits which can be expected to run freely and are intended for precision work at slow speeds and light journal pressures, but are not suitable where appreciable temperature differences are likely to be encountered. RC4 Close running fits are intended chiefly for running fits on accurate machinery with moderate surface speeds and journal pressures, where accurate location and minimum play is desired. RC5RC6 Medium running fits are intended for higher running speeds, heavy journal pressures, or both. RC7 Free running fits are intended for use where accuracy is not essential or where large temperature variations are likely to be encountered, or under both of these conditions. RC8RC9 Loose running fits are intended for use where wide commercial tolerances may be necessary, together with an allowance, on the external member. L Locational fits are fits intended to determine only the location of the mating parts; they may provide rigid or accurate location, as with interference fits, or provide some freedom of location, as with clearance fits. Accordingly, they are divided into three groups: clearance fits, transition fits, and interference fits. LC Locational clearance fits are intended for parts which are normally stationary but which can be freely assembled or disassembled. They run from snug fits for parts requiring accuracy of location, through the medium clearance fits for parts such as ball, race, and housing, to the looser fastener fits where freedom of assembly is of prime importance. LT Locational transition fits are a compromise between clearance and interference fits, for application where accuracy of location is important, but either a small amount of clearance or interference is permissible. LN Locational interference fits are used where accuracy of location is of prime importance and for parts requiring rigidity and alignment with no special requirements for bore pressure. Such fits are not intended for parts designed to transmit frictional loads form one part to another by virtue of the tightness of fit, since these conditions are covered by force fits. FN Force and shrink fits constitute a special type of interference fit, normally characterized by maintenance of constant bore pressure throughout the range of sizes. The interference therefore varies almost directly with diameter, and the difference between its minimum and maximum value is small so as to maintain the resulting pressures within reasonable limits. FN1 Light drive fits are those requiring light assembly pressures and producing more or less permanent assemblies. They are suitable for thin sections or long fits or in cast-iron external members. FN2 Medium drive fits are suitable for ordinary steel parts or for shrink fits on light sections. They are about the tightest fits that can be used with high-grade cast-iron external members. FN3 Heavy drive fits are suitable for heavier steel parts or for shrink fits in medium sections. FN4FN5 Force fits are suitable for parts which can be highly stressed or for shrink fits where the heavy pressing forces required are impractical. Extracted from American Standard Limits for Cylindrical Parts ANSI B4.1-1978, with the permission of the publishers, The American Society of Mechanical Engineers, United Engineering Center, 345 East 47th Street, New York 10017. Limit dimensions are tabulated in this standard for nominal sizes up to and including 200 in. An SI version is also available.

Shigley, Mischke & Budynas Machine Design Tutorial 4-17: Press and Shrink Fits 9/11 Table A-1-1 RUNNING AND SLIDING FITS Diameter Size Range (in.) Class 0.00 - 0.12 0.12 - 0.24 0.24 - 0.40 0.40 - 0.71 RC1 Hole +0.20 -0.00 +0.20 -0.00 +0.25 -0.00 +0.30 -0.00 Shaft +0.10 -0.25 -0.15 -0.30 -0.20 -0.35 -0.25 -0.45RC2 Hole +0.25 -0.00 +0.30 -0.00 +0.40 -0.00 +0.40 -0.00 Shaft -0.10 -0.30 -0.15 -0.35 -0.20 -0.45 -0.25 -0.55RC3 Hole +0.40 -0.00 +0.50 -0.00 +0.60 -0.00 +0.70 -0.00 Shaft -0.30 -0.55 -0.40 -0.70 -0.50 -0.90 -0.60 -1.00RC4 Hole +0.60 -0.00 +0.70 -0.00 +0.90 -0.00 +1.00 -0.00 Shaft -0.30 -0.70 -0.40 -0.90 -0.50 -1.10 -0.60 -1.30RC5 Hole +0.60 -0.00 +0.70 -0.00 +0.90 -0.00 +1.00 -0.00 Shaft -0.60 -1.00 -0.80 -1.30 -1.00 -1.60 -1.20 -1.90RC6 Hole +1.00 -0.00 +1.20 -0.00 +1.40 -0.00 +1.60 -0.00 Shaft -0.60 -1.20 -0.80 -1.50 -1.00 -1.90 -1.20 -2.20RC7 Hole +1.00 -0.00 +1.20 -0.00 +1.40 -0.00 +1.60 -0.00 Shaft -1.00 -1.60 -1.20 -1.90 -1.60 -2.50 -2.00 -3.00RC8 Hole +1.60 -0.00 +1.80 -0.00 +2.20 -0.00 +2.80 -0.00 Shaft -2.50 -3.50 -2.80 -4.00 -3.00 -4.40 -3.50 -5.10RC9 Hole +2.50 -0.00 +3.00 -0.00 +3.50 -0.00 +4.00 -0.00 Shaft -4.00 -5.60 -4.50 -6.00 -5.00 -7.20 -6.00 -8.80 Diameter Size Range (in.) Class 0.71 - 1.19 1.19 - 1.97 1.97 - 3 .15 3.15 - 4.73 RC1 Hole +0.40 -0.00 +0.40 -0.00 +0.50 -0.00 +0.60 -0.00 Shaft -0.30 -0.55 -0.40 -0.70 -0.40 -0.70 -0.50 -0.90RC2 Hole +0.50 -0.00 +0.60 -0.00 +0.70 -0.00 +0.90 -0.00 Shaft -0.30 -0.70 -0.40 -0.80 -0.40 -0.90 -0.50 -1.10RC3 Hole +0.80 -0.00 +1.00 -0.00 +1.20 -0.00 +1.40 -0.00 Shaft -0.80 -1.30 -1.00 -1.60 -1.20 -1.90 -1.40 -2.30RC4 Hole +1.20 -0.00 +1.60 -0.00 +1.80 -0.00 +2.20 -0.00 Shaft -0.80 -1.60 -1.00 -2.00 -1.20 -2.40 -1.40 -2.80RC5 Hole +1.20 -0.00 +1.60 -0.00 +1.80 -0.00 +2.20 -0.00 Shaft -1.60 -2.40 -2.00 -3.00 -2.50 -3.70 -3.00 -4.40RC6 Hole +2.00 -0.00 +2.50 -0.00 +3.00 -0.00 +3.50 -0.00 Shaft -1.60 -2.80 -2.00 -3.60 -2.50 -4.30 -3.00 -5.20RC7 Hole +2.00 -0.00 +2.50 -0.00 +3.00 -0.00 +3.50 -0.00 Shaft -2.50 -3.70 -3.00 -4.60 -4.00 -5.80 -5.00 -7.20RC8 Hole +3.50 -0.00 +4.00 -0.00 +4.50 -0.00 +5.00 -0.00 Shaft -4.50 -6.50 -5.00 -7.50 -6.00 -9.00 -7.00 -10.50RC9 Hole +5.00 -0.00 +6.00 -0.00 +7.00 -0.00 +9.00 -0.00 Shaft -7.00 -10.50 -8.00 -12.00 -9.00 -13.50 -10.00 -15.00

Shigley, Mischke & Budynas Machine Design Tutorial 4-17: Press and Shrink Fits 10/11 Table A-1-2 FORCE AND SHRINK FITS Diameter Size Range (in.) Class 0.00 - 0.12 0.12 - 0.24 0.24 - 0.40 0.40 - 0.56 FN1 Hole +0.25 -0.00 +0.30 -0.00 +0.40 -0.00 +0.40 -0.00 Shaft +0.50 +0.30 +0.60 +0.40 +0.75 +0.50 +0.80 +0.50FN2 Hole +0.40 -0.00 +0.50 -0.00 +0.60 -0.00 +0.70 -0.00 Shaft +0.85 +0.60 +1.00 +0.70 +1.40 +1.00 +1.60 +1.20FN3 Hole Shaft FN4 Hole +0.40 -0.00 +0.50 -0.00 +0.60 -0.00 +0.70 -0.00 Shaft +0.95 +0.70 +1.20 +0.90 +1.60 +1.20 +1.80 +1.40FN5 Hole +0.60 -0.00 +0.70 -0.00 +0.90 -0.00 +1.00 -0.00 Shaft +1.30 +0.90 +1.70 +1.20 +2.00 +1.40 +2.30 +1.60 Diameter Size Range (in.) Class 0.56 - 0.71 0.71 - 0.95 0.95 - 1.19 1.19 - 1.58 FN1 Hole +0.40 -0.00 +0.50 -0.00 +0.50 -0.00 +0.60 -0.00 Shaft +0.90 +0.60 +1.10 +0.70 +1.20 +0.80 +1.30 +0.90FN2 Hole +0.70 -0.00 +0.80 -0.00 +0.80 -0.00 +1.00 -0.00 Shaft +1.60 +1.20 +1.90 +1.40 +1.90 +1.40 +2.40 +1.80FN3 Hole +0.80 -0.00 +1.00 -0.00 Shaft +2.10 +1.60 +2.60 +2.00FN4 Hole +0.70 -0.00 +0.80 -0.00 +0.80 -0.00 +1.00 -0.00 Shaft +1.80 +1.40 +2.10 +1.60 +2.30 +1.80 +3.10 +2.50FN5 Hole +1.00 -0.00 +1.20 -0.00 +1.20 -0.00 +1.60 -0.00 Shaft +2.50 +1.80 +3.00 +2.20 +3.30 +2.50 +4.00 +3.00 Diameter Size Range (in.) Class 1.58 - 1.97 1.97 - 2.56 2.56 - 3 .15 3.15 - 3.94 FN1 Hole +0.60 -0.00 +0.70 -0.00 +0.70 -0.00 +0.90 -0.00 Shaft +1.40 +1.00 +1.80 +1.30 +1.90 +1.40 +2.40 +1.80FN2 Hole +1.00 -0.00 +1.20 -0.00 +1.20 -0.00 +1.40 -0.00 Shaft +2.40 +1.80 +2.70 +2.00 +2.90 +2.20 +3.70 +2.80FN3 Hole +1.00 -0.00 +1.20 -0.00 +1.20 -0.00 +1.40 -0.00 Shaft +2.80 +2.20 +3.20 +2.50 +3.70 +3.00 +4.40 +3.50FN4 Hole +1.00 -0.00 +1.20 -0.00 +1.20 -0.00 +1.40 -0.00 Shaft +3.40 +2.80 +4.20 +3.50 +4.70 +4.00 +5.90 +5.00FN5 Hole +1.60 -0.00 +1.80 -0.00 +1.80 -0.00 +2.20 -0.00 Shaft +5.00 +4.00 +6.20 +5.00 +7.20 +6.00 +8.40 +7.00

Shigley, Mischke & Budynas Machine Design Tutorial 4-17: Press and Shrink Fits 11/11 Table A-1-2 FORCE AND SHRINK FITS (CONTINUED) Diameter Size Range (in.) Class 3.94 - 4.73 4.73 - 5.52 5.52 - 6.30 6.30 - 7.09 FN1 Hole +0.90 -0.00 +1.00 -0.00 +1.00 -0.00 +1.00 -0.00 Shaft +2.60 +2.00 +2.90 +2.20 +3.20 +2.50 +3.50 +2.80FN2 Hole +1.40 -0.00 +1.60 -0.00 +1.60 -0.00 +1.60 -0.00 Shaft +3.90 +3.00 +4.50 +3.50 +5.00 +4.00 +5.50 +4.50FN3 Hole +1.40 -0.00 +1.60 -0.00 +1.60 -0.00 +1.60 -0.00 Shaft +4.90 +4.00 +6.00 +5.00 +6.00 +5.00 +7.00 +6.00FN4 Hole +1.40 -0.00 +1.60 -0.00 +1.60 -0.00 +1.60 -0.00 Shaft +6.90 +6.00 +8.00 +7.00 +8.00 +7.00 +9.00 +8.00FN5 Hole +2.20 -0.00 +2.50 -0.00 +2.50 -0.00 +2.50 -0.00 Shaft +9.40 +8.00 +11.60 +10.00 +13.60 +12.00 +13.60 +12.00

Text. refers to Mechanical Engineering Design, 7th edition text by Joseph Edward Shigley, Charles R. Mischke, and Richard G. Budynas; equations and figures with the prefix T refer to the present tutorial. MECHANICAL ENGINEERING DESIGN TUTORIAL 4-20: HERTZ CONTACT STRESSES CHARACTERISTICS OF CONTACT STRESSES 1. Represent compressive stresses developed from surface pressures between two curved bodies pressed together; 2. Possess an area of contact. The initial point contact (spheres) or line contact (cylinders) become area contacts, as a result of the force pressing the bodies against each other; 3. Constitute the principal stresses of a triaxial (three dimensional) state of stress; 4. Cause the development of a critical section below the surface of the body; 5. Failure typically results in flaking or pitting on the bodies surfaces. TWO DESIGN CASES Two design cases will be considered, 1. Sphere Sphere Contact (Point Contact Circular Contact Area) 2. Cylinder Cylinder Contact (Line Contact Rectangular Contact Area) SPHERE SPHERE CONTACT TEXT FIGURE 4-42 Two Spheres in Contactz x yFFz x y FF2a1d2d(b) Contact stress has an elliptical distribution across contact over zone of diameter 2a. (a) Two spheres held in contact by force F.

Shigley, Mischke & Budynas Machine Design Tutorial 4-20: Hertz Contact Stresses 2/10 Consider two solid elastic spheres held in contact by a force F such that their point of contact expands into a circular area of radius a, given as: 31/ 32 21 1 2 21 21 21 21 2(Modified Text Eq. 4-72)(1 ) / (1 ) / 3where 8 (1/ ) (1/ ) applied force, Poisson's ratios for spheres 1 and 2, elastic modulii for spheres 1 and 2, diameteaaa K FE EKd dFE Ed d = + = + ==== rs of spheres 1 and 2 This general expression for the contact radius can be applied to two additional common cases: 1. Sphere in contact with a plane 2( ); d = 2. Sphere in contact with an internal spherical surface or cup 2( ). d d = Returning to the sphere-sphere case, the maximum contact pressure, maxp , occurs at the center point of the contact area. max23(Text Eq. 4-73)2Fpa = State of Stress The state of stress is computed based on the following mechanics: 1. Two planes of symmetry in loading and geometry dictates that ;x y = 2. The dominant stress occurs on the axis of loading: max;z = 3. The principal stresses are y x = = =2 1 and z =3 given 1 2 3, ; 4. Compressive loading leads to , , and x y z being compressive stresses. Calculation of Principal Stresses ( )1max21 21 11 tan (1 ) (Modified Text Eq. 4-74)2 1x aaayp = + + = = =

Shigley, Mischke & Budynas Machine Design Tutorial 4-20: Hertz Contact Stresses 3/10 max32(Modified Text Eq. 4-75)1zap = =+ where / nondimensional depth below the surface Poisson's ratio for the sphere examined (1 or 2)az a = == Mohrs Circle Plotting the principal stresses on a Mohrs circle plot results in: one circle, defined by 1 2 = , shrinking to a point; and two circles, defined by 1 3, and 2 3, , plotted on top of each other. The maximum shear stress, max , for the plot is calculated as: 1 3max(Modified Text Eq. 4-76)2 2 2y zx z = = = If the maximum shear stress,max , and principal stresses, 1 2 3, , and , are plotted as a function of maximum pressure, maxp , below the surface contact point, the plot of Fig. 4-43 is generated. This plot, based on a Poissons ratio of 3 . 0 = , reveals that a critical section exists on the load axis, approximately 0.48a below the sphere surface. Many authorities theorize that this maximum shear stress is responsible for the surface fatigue failure of such contacting elements; a crack, originating at the point of maximum shear, progresses to the surface where lubricant pressure wedges a chip loose and thus creates surface pitting. TEXT FIGURE 4-43: Magnitude of the stress components below the surface as a function of maximum pressure of contacting spheres.

Shigley, Mischke & Budynas Machine Design Tutorial 4-20: Hertz Contact Stresses 4/10 CYLINDERCYLINDER CONTACT Consider two solid elastic cylinders held in contact by forces F uniformly distributed along the cylinder length l. The resulting pressure causes the line of contact to become a rectangular contact zone of half-width b given as: 1/ 22 21 1 2 21 21 21 21 2 (Modified Text Eq. 4-77)(1 ) / (1 ) / 2where (1/ ) (1/ ) applied force, Poisson's ratios for cylinders 1 and 2, elastic modulii for cylinders 1 and 2, dibbb K FE EKl d dFE Ed d = + = + ====1 2ameters of spheres 1 and 2 length of cylinders 1 and 2 ( assumed) l l l = = (b) Contact stress has an elliptical distribution across contact zone of width 2b. (a) Two right circular cylinders held in contact by forces F uniformly distributed along cylinder length l.z xy FFz xyFF2b 1d2dlTEXT FIGURE 4-44 Two Cylinders in Contact

Shigley, Mischke & Budynas Machine Design Tutorial 4-20: Hertz Contact Stresses 5/10 This expression for the contact half-width, b, is general and can be used for two additional cases which are frequently encountered: 1. Cylinder in contact with a plane, e.g. a rail 2( ); d = 2. Cylinder in contact with an internal cylindrical surface, for example the race of a roller bearing 2( ). d d = The maximum contact pressure between the cylinders acts along a longitudinal line at the center of the rectangular contact area, and is computed as: max2(Text Eq. 4-78)Fpbl = State of Stress The state of stress is computed based on the following mechanics: 1. One plane of symmetry in loading and geometry dictates that ;x y 2. The dominant stress occurs along the axis of loading: max;z = 3. The principal stresses are equal to , , and x y z with z =3; 4. Compressive loading leads to , , and x y z being compressive stresses. Calculation of Principal Stresses and Maximum Shear Stress 3 z max21y2max2y max21(Modified Text Eq. 4-81)1 for 0 0.436 for 0.436where,2 1 (Modified Text Eq. 4-79)1 22 (Modified Text Eq. 4-1bx bbx b bbbbppp = = + = = + + = + 80)/bz b =

Shigley, Mischke & Budynas Machine Design Tutorial 4-20: Hertz Contact Stresses 6/10 1/ 3max1/ 3The maximum shear stress is thus given as:( ) / 2 for 0 0.436( ) / 2 for 0.436z x bz y b = = = When these equations are plotted as a function of maximum contact pressure up to a distance 3b below the surface contact point, the plot of Fig. 4-45 is generated. Based on a Poissons ratio of 0.3, this plot reveals that max attains a maxima for / 0.786bz b = = and 0.3pmax. Example T4.20.1: Problem Statement: A 6-in-diameter cast-iron wheel, 2 in wide, rolls on a flat steel surface carrying a 800 lbf load. Find: 1. The Hertzian stresses 1/3, , and x y z in the cast iron wheel at the cr