Ch 5 and 6 Practice problems

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Page 1 of 19 Some Practice Material from Ch 5 and 6 (10 th Edition) This is not a practice exam. This is just study material. Also be sure to study your past homework and quizzes. For best results, try to solve these problems before looking at the solutions. Some of these problems are simple and the exam may have more involved problems 5-2, 3 ,4 (all repetitious) 5-7,8,9,10 (all part of one table) 5-14,15,17 (all part of one table) 5-64 based on 5-63 6-2,4,5 6-10 6-14,15,17 6-26,27,31 CHAPTER 5 5-2 SOLUTION S y = 350 MPa. a b c d e MSS 3.5 1.75 3.5 2.33 3.5 DE 3.5 2.02 4.04 2.65 4.04 (a) MSS: 1 3 350 100 MPa, 0 3.5 . 100 0 n Ans σ σ = = ⇒ = = − DE: 2 2 1/2 350 3.5 . [100 (100)(100) 100 ] n Ans = = − + (b) MSS: 1 3 350 100 , 100 MPa 1.75 . 100 ( 100) n Ans σ σ = = − ⇒ = = − − DE: ( ) 1/2 2 2 350 2.02 . 100 (100)( 100) 100 n Ans = =   − − + −  

Page 2 of 19 (c) MSS: 1 3 350 100 MPa, 0 3.5 . 100 0 n Ans σ σ = = ⇒ = = − DE: 1/2 2 2 350 4.04 . 100 (100)(50) 50 n Ans = =   − +   (d) MSS: 1 3 350 100, 50 MPa 2.33 . 100 ( 50) n Ans σ σ = = − ⇒ = = − − DE: ( ) 1/2 2 2 350 2.65 . 100 (100)( 50) 50 n Ans = =   − − + −   ( e ) MSS: 1 3 350 0, 100 MPa 3.5 . 0 ( 100) n Ans σ σ = = − ⇒ = = − − DE: ( ) ( ) 1/2 2 2 350 4.04 . 50 ( 50)( 100) 100 n Ans = =   − − − − + −   5-3 SOLUTION From Table A-20, S y = 37.5 kpsi a b c d e MSS 1.5 1.25 1.33 1.16 0.96 DE 1.72 1.44 1.42 1.33 1.06 (a) MSS: 1 3 37.5 25 kpsi, 0 1.5 . 25 0 n Ans σ σ = = ⇒ = = − DE: 1/2 2 2 37.5 1.72 . 25 (25)(15) 15 n Ans = =   − +   (b) MSS: 1 3 37.5 15 kpsi, 15 1.25 . 15 ( 15) n Ans σ σ = = − ⇒ = = − − DE: ( ) 1/2 2 2 37.5 1.44 . 15 (15)( 15) 15 n Ans = =   − − + −   (c) 2 2 20 20 , ( 10) 24.1, 4.1 kpsi 2 2 A B σ σ   = ± + − = −     1 2 3 24.1, 0, 4.1 kpsi σ σ σ = = = − MSS: 37.5 1.33 . 24.1 ( 4.1) n Ans = = − −

Page 3 of 19 DE: ( ) 1/2 2 2 37.5 20 3 10 26.5 kpsi 1.42 . 26.5 n Ans σ   ′ = + − = ⇒ = =   (d) 2 2 12 15 12 15 , ( 9) 17.7, 14.7 kpsi 2 2 A B σ σ − + − −   = ± + − = −     1 2 3 17.7, 0, 14.7 kpsi σ σ σ = = = − MSS: 37.5 1.16 . 17.7 ( 14.7) n Ans = = − − DE: 1/2 2 2 2 ( 12) ( 12)(15) 15 3( 9) 28.1 kpsi σ ′   = − − − + + − =   37.5 1.33 . 28.1 n Ans = = (e) ( ) 2 2 24 24 24 24 , 15 9, 39 kpsi 2 2 A B σ σ − − − +   = ± + − =− −     1 2 3 0, 9, 39 kpsi σ σ σ = = − = − MSS: ( ) 37.5 0.96 . 0 39 n Ans = = − − DE: ( ) ( )( ) ( ) ( ) 1/2 2 2 2 24 24 24 24 3 15 35.4 kpsi σ   ′ = − − − − + − + − =   37.5 1.06 . 35.4 n Ans = = 5-4 SOLUTION From Table A-20, S y = 47 kpsi. a b c d e MSS 1.57 0.78 1.57 1.57 0.78 DE 1.57 0.90 1.81 1.81 0.84 (a) MSS: 1 3 47 30 kpsi, 0 1.57 . 30 0 n Ans σ σ = = ⇒ = = − DE: 2 2 1/2 47 1.57 . [30 (30)(30) 30 ] n Ans = = − + (b) MSS: 1 3 47 30 , 30 kpsi 0.78 . 30 ( 30) n Ans σ σ = = − ⇒ = = − − DE: ( ) 1/2 2 2 47 0.90 . 30 (30)( 30) 30 n Ans = =   − − + −  

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Page 4 of 19 (c) MSS: 1 3 47 30 kpsi, 0 1.57 . 30 0 n Ans σ σ = = ⇒ = = − DE: 1/2 2 2 47 1.81 . 30 (30)(15) 15 n Ans = =   − +   (d) MSS: 1 3 47 0, 30 kpsi 1.57 . 0 ( 30) n Ans σ σ = = − ⇒ = = − − DE: ( ) ( ) 1/2 2 2 47 1.81 . 30 ( 30)( 15) 15 n Ans = =   − − − − + −   ( e ) MSS: 1 3 47 10, 50 kpsi 0.78 . 10 ( 50) n Ans σ σ = = − ⇒ = = − − DE: ( ) 1/2 2 2 47 0.84 . 50 ( 50)(10) 10 n Ans = =   − − − +   5-7 SOLUTION S y = 295 MPa, σ A = 75 MPa, σ B = − 35 MPa, ( a ) ( ) ( ) ( ) 1/2 1/2 2 2 2 2 295 3.03 . 75 75 35 35 y A A B B S n Ans σ σ σ σ = = =   − + − − + −   ( b ) Note: The drawing in this manual may not be to the scale of original drawing. The measurements were taken from the original drawing.

Page 5 of 19 2.50" 3.01 . 0.83" OB n Ans OA = = = 5-8 SOLUTION S y = 295 MPa, σ A = 30 MPa, σ B = − 100 MPa, ( a ) ( ) ( ) ( ) 1/2 1/2 2 2 2 2 295 2.50 . 30 30 100 100 y A A B B S n Ans σ σ σ σ = = =   − + − − + −   ( b ) Note: The drawing in this manual may not be to the scale of original drawing. The measurements were taken from the original drawing. 2.50" 3.01 . 0.83" OB n Ans OA = = = 5-9 SOLUTION S y = 295 MPa, 2 2 100 100 , ( 25) 105.9, 5.9 MPa 2 2 A B σ σ   = ± + − = −    

Page 6 of 19 ( a ) ( ) ( ) ( ) 1/2 1/2 2 2 2 2 295 2.71 . 105.9 105.9 5.9 5.9 y A A B B S n Ans σ σ σ σ = = =   − + − − + −   ( b ) Note: The drawing in this manual may not be to the scale of original drawing. The measurements were taken from the original drawing. 2.87" 2.71 . 1.06" OB n Ans OA = = = _____________________________________________________________________________ 5-10 SOLUTION S y = 295 MPa, 2 2 30 65 30 65 , 40 3.8, 91.2 MPa 2 2 A B σ σ − − − +   = ± + = − −     ( a ) ( ) ( ) ( )( ) ( ) 1/2 1/2 2 2 2 2 295 3.30 . 3.8 3.8 91.2 91.2 y A A B B S n Ans σ σ σ σ = = =   − + − − − − + −   ( b ) Note: The drawing in this manual may not be to the scale of original drawing. The measurements were taken from the original drawing. 3.00" 3.33 . 0.90" OB n Ans OA = = =

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Page 7 of 19 5-14 SOLUTION Since ε f > 0.05, and S yt ≠ S yc , the Coulomb-Mohr theory for ductile materials will be used. ( a ) From Eq. (5-26), 1 1 3 1 150 50 1.23 . 235 285 yt yc n Ans S S σ σ − −   −   = − = − =           ( b ) Plots for Problems 5-14 to 5-18 are found here . Note: The drawing in this manual may not be to the scale of original drawing. The measurements were taken from the original drawing. 1.94" 1.23 . 1.58" OB n Ans OA = = = ______________________________________________________________________________ 5-15 SOLUTION ( a ) From Eq. (5-26), 1 1 3 1 50 150 1.35 . 235 285 yt yc n Ans S S σ σ − −   −   = − = − =           2.14" 1.35 . 1.58" OD n Ans OC = = = ______________________________________________________________________________

Page 8 of 19 5-16 SOLUTION 2 2 125 125 , ( 75) 160, 35 kpsi 2 2 A B σ σ   = ± + − = −     ( a ) From Eq. (5-26), 1 1 3 1 160 35 1.24 . 235 285 yt yc n Ans S S σ σ − −   −   = − = − =           2.04" 1.24 . 1.64" OF n Ans OE = = = ______________________________________________________________________________ 5-17 SOLUTION 2 2 80 125 80 125 , 50 47.7, 157.3 kpsi 2 2 A B σ σ − − − +   = ± + = − −     ( a ) From Eq. (5-26), 1 1 3 1 0 157.3 1.81 . 235 285 yt yc n Ans S S σ σ − −   −   = − = − =           2.99" 1.82 . 1.64" OH n Ans OG = = = ______________________________________________________________________________ 5-18 SOLUTION 2 2 125 80 125 80 , ( 75) 180.8, 24.2 kpsi 2 2 A B σ σ + −   = ± + − =     ( a ) From Eq. (5-26), 1 1 3 1 180.8 0 1.30 . 235 285 yt yc n Ans S S σ σ − −     = − = − =           2.37" 1.30 . 1.83" OJ n Ans OI = = =

Page 9 of 19 5-64 SOLUTION As in Prob. 5-63, we will assume this to be a statics problem. Since the proportions are unchanged, the bearing reactions will be the same as in Prob. 5-63 and the bending moment will still be a maximum at point C . Thus xy plane: 127(3) 381 lbf in C M = = xz plane: 328(3) 984 lbf in C M = = So 1/2 2 2 max (381) (984) 1055 lbf in M   = + = ⋅   3 3 3 3 32 2(1055) 10 746 10.75 psi kpsi C x M d d d d σ π π 3 = = = Since the torsional stress is unchanged, 3 5.09 kpsi xz d τ = For combined bending and torsion, the maximum shear stress is found from 1/2 1/2 2 2 2 2 max 3 3 3 10.75 5.09 7.40 kpsi 2 2 x xz d d d σ τ τ           = + = + =                         Using the MSS theory, as was used in Prob. 5-63, gives ( ) max 3 7.40 67 0.762 in . 2 2 2 y S d Ans n d τ = = = ⇒ =

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Page 10 of 19

Page 11 of 19 CHAPTER 6 6-2 SOLUTION (a) Table A-20: S ut = 80 kpsi Eq. (6-8): 0.5(80) 40 kpsi . e S Ans ′ = = (b) Table A-20: S ut = 90 kpsi Eq. (6-8): 0.5(90) 45 kpsi . e S Ans ′ = = (c) Aluminum has no endurance limit. Ans. (d) Eq. (6-8): S ut > 200 kpsi, 100 kpsi . e S Ans ′ = 6-4 SOLUTION rev 1600 MPa, 900 MPa ut S σ = = Fig. 6-18: S ut = 1600 MPa = 232 kpsi. Off the graph, so estimate f = 0.77. Eq. (6-8): S ut > 1400 MPa, so S e = 700 MPa Eq. (6-14): [ ] 2 2 0.77(1600) ( ) 2168.3 MPa 700 ut e f S a S = = = Eq. (6-15): 1 1 0.77(1600) log log 0.081838 3 3 700 ut e f S b S     = − = − = −         Eq. (6-16): 1 1/ 0.081838 rev 900 46 400 cycles . 2168.3 b N Ans a σ −     = = =         ______________________________________________________________________________ 6-5 SOLUTION 230 kpsi, 150 000 cycles ut S N = = Fig. 6-18, point is off the graph, so estimate: f = 0.77 Eq. (6-8): S ut > 200 kpsi, so 100 kpsi e e S S ′ = =

Page 12 of 19 Eq. (6-14): [ ] 2 2 0.77(230) ( ) 313.6 kpsi 100 ut e f S a S = = = Eq. (6-15): 1 1 0.77(230) log log 0.08274 3 3 100 ut e f S b S     = − = − = −         Eq. (6-13): 0.08274 313.6(150 000) 117.0 kpsi . b f S aN Ans − = = = 6-10 SOLUTION d = 1.5 in, S ut = 110 kpsi Eq. (6-8): 0.5(110) 55 kpsi e S ′ = = Table 6-2: a = 2.70, b = − 0.265 Eq. (6-19): 0.265 2.70(110) 0.777 b a ut k aS − = = = Eq. (6-20): k b = 0.879 d − 0.107 = 0.879(1.5) − 0.107 =0.842 Eq. (6-18): S e = k a k b e S ′ = 0.777(0.842)(55) = 36.0 kpsi Ans. 6-14 SOLUTION Given: w =2.5 in, t = 3/8 in, d = 0.5 in, n d = 2. From Table A-20, for AISI 1020 CD, S ut = 68 kpsi and S y = 57 kpsi. Eq. (6-8): 0.5(68) 34 kpsi e S ′ = = Table 6-2: 0.265 2.70(68) 0.88 a k − = = Eq. (6-21): k b = 1 (axial loading) Eq. (6-26): k c = 0.85 Eq. (6-18): S e = 0.88(1)(0.85)(34) = 25.4 kpsi Table A-15-1: / 0.5 / 2.5 0.2, 2.5 t d K = = w Get the notch sensitivity either from Fig. 6-20, or from the curve-fit Eqs. (6-34) and (6-35a). The relatively large radius is off the graph of Fig. 6-20, so we will assume the curves continue according to the same trend and use the equations to estimate the notch sensitivity. ( ) ( ) ( ) ( ) ( )( ) 2 3 5 8 3 0.246 3.08 10 68 1.51 10 68 2.67 10 68 0.09799 a − − − = − + − = 1 1 0.836 0.09799 1 1 0.25 q a r = = = + +

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Page 13 of 19 Eq. (6-32): 1 ( 1) 1 0.836(2.5 1) 2.25 f t K q K = + − = + − = 2.25 = 3 (3/ 8)(2.5 0.5) a a a f a F F K F A σ = = − Since a finite life was not mentioned, we’ll assume infinite life is desired, so the completely reversed stress must stay below the endurance limit. 25.4 2 3 e f a a S n F σ = = 4.23 kips . a F Ans = What is the percent error if you had used Kt unmodified, rather than Kf? Note that this problem did not use the Sf = aN b equation. Hopefully, it is clear why. This is a good example of the different ways we need to think about fatigue. 6-15 SOLUTION Given: max min 2 in, 1.8 in, 0.1 in, 25 000 lbf in, 0. D d r M M = = = = ⋅ = From Table A-20, for AISI 1095 HR, S ut = 120 kpsi and S y = 66 kpsi. Eq. (6-8): ( ) 0.5 0.5 120 60 kpsi e ut S S ′ = = Eq. (6-19): 0.265 2.70(120) 0.76 b a ut k aS − = = = Eq. (6-24): e 0.370 0.370(1.8) 0.666 in d d = = Eq. (6-20): 0.107 0.107 0.879 0.879(0.666) 0.92 b e k d − − = = = Eq. (6-26): 1 c k = Eq. (6-18): (0.76)(0.92)(1)(60) 42.0 kpsi e a b c e S k k k S ′ = = = Fig. A-15-14: / 2 /1.8 1.11, / 0.1/1.8 0.056 D d r d = = = = 2.1 t K ⇒ = Get the notch sensitivity either from Fig. 6-20, or from the curve-fit Eqs . (6-34) and (6-35a). Using the equations, ( ) ( ) ( ) ( ) ( )( ) 2 3 5 8 3 0.246 3.08 10 120 1.51 10 120 2.67 10 120 0.04770 a − − − = − + − = 1 1 0.87 0.04770 1 1 0.1 q a r = = = + + Eq. (6-32): 1 ( 1) 1 0.87(2.1 1) 1.96 f t K q K = + − = + − = 4 4 4 ( / 64) ( / 64)(1.8) 0.5153 in I d π π = = =

Page 14 of 19 max min 25 000(1.8 / 2) 43 664 psi 43.7 kpsi 0.5153 0 Mc I σ σ = = = = = Eq. (6-36): ( ) ( ) max min 43.7 0 1.96 42.8 kpsi 2 2 m f K σ σ σ + + = = = ( ) ( ) max min 43.7 0 1.96 42.8 kpsi 2 2 a f K σ σ σ − − = = = Eq. (6-46): 1 42.8 42.8 42.0 120 a m f e ut n S S σ σ = + = + 0.73 . f n Ans = Plot the fatigue diagram and Goodman line that correspond to the above solution. A factor of safety less than unity indicates a finite life. OK, but how do you find a value for the finite life, N? (Hint: the book almost doesn’t explain this, but you can find it in the details of Example 6-12.) – Also check out video on ‘constant fatigue life’ Check for yielding. It is not necessary to include the stress concentration for static yielding of a ductile material. max 66 1.51 . 43.7 y y S n Ans σ = = = ______________________________________________________________________________

Page 15 of 19 6-17 SOLUTION From a free-body diagram analysis, the bearing reaction forces are found to be R A = 2000 lbf and R B = 1500 lbf. The shear-force and bending-moment diagrams are shown. The critical location will be at the shoulder fillet between the 1-5/8 in and the 1-7/8 in diameters, where the bending moment is large, the diameter is smaller, and the stress concentration exists. M = 16 000 – 500 (2.5) = 14 750 lbf ∙ in With a rotating shaft, the bending stress will be completely reversed. rev 4 14 750(1.625 / 2) 35.0 kpsi ( / 64)(1.625) Mc I σ π = = = This stress is far below the yield strength of 71 kpsi, so yielding is not predicted. Fig. A-15-9: r/d = 0.0625/1.625 = 0.04, D/d = 1.875/1.625 = 1.15, K t =1.95 Get the notch sensitivity either from Fig. 6-20, or from the curve-fit Eqs . (6-34) and (6-35a). Using the equations, ( ) ( ) ( ) ( ) ( ) ( ) 2 3 3 5 8 0.246 3.08 10 85 1.51 10 85 2.67 10 85 0.07690 a − − − = − + − = 1 1 0.76 0.07690 1 1 0.0625 q a r = = = + + . Eq. (6-32): 1 ( 1) 1 0.76(1.95 1) 1.72 f t K q K = + − = + − = Eq. (6-8): ' 0.5 0.5(85) 42.5 kpsi e ut S S = = Eq. (6-19): 0.265 2.70(85) 0.832 b a ut k aS − = = = Eq. (6-20): 0.107 0.107 0.879 0.879(1.625) 0.835 b k d − − = = = Eq. (6-26): 1 c k = Eq. (6-18): ' (0.832)(0.835)(1)(42.5) 29.5 kpsi e a b c e S k k k S = = = ( ) rev 29.5 0.49 . 1.72 35.0 e f f S n Ans K σ = = = Infinite life is not predicted. Use the S-N diagram to estimate the life. Fig. 6-18: f = 0.867

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Page 16 of 19 ( ) [ ] 2 2 0.867(85) Eq. (6-14): 184.1 29.5 1 1 0.867(85) Eq. (6-15): log log 0.1325 3 3 29.5 ut e ut e f S a S f S b S = = =     = − = − = −         1 1 0.1325 rev (1.72)(35.0) Eq. (6-16): 4611 cycles 184.1 b f K N a σ −     = = =         N = 4600 cycles Ans. Make sure you can draw the S-N diagram that corresponds to the above answer. How does the above answer change if σ rev = 50 MPa? Hint: quite a bit, and you need to use another equation (i.e. Low cycle fatigue). 6-26 SOLUTION max min 590 MPa, 490 MPa, 28 kN, 12 kN ut y S S F F = = = = Check for yielding 2 max max 28 000 147.4 N/mm 147.4 MPa 10(25 6) F A σ = = = = − max 490 3.32 . 147.4 y y S n Ans σ = = = Determine the fatigue factor of safety based on infinite life. From Prob. 6-25 : 208.6 MPa, 2.2 e f S K = =

Page 17 of 19 ( ) max min 28 000 12 000 2.2 92.63 MPa 2 2(10)(25 6) a f F F K A σ − − = = = − max min 28 000 12 000 2.2 231.6 MPa 2 2(10)(25 6) m f F F K A σ   + + = = =   −   Modified Goodman criteria: 1 92.63 231.6 208.6 590 a m f e ut n S S σ σ = + = + 1.20 . f n Ans = Gerber criteria: 2 2 2 1 1 1 2 ut a m e f m e ut a S S n S S σ σ σ σ         = − + +             2 2 1 590 92.63 2(231.6)(208.6) 1 1 2 231.6 208.6 590(92.63)         = − + +             1.49 . f n Ans = ASME-Elliptic criteria: 2 2 2 2 1 1 ( / ) ( / ) (92.63/ 208.6) (231.6 / 490) f a e m y n S S σ σ = = + + = 1.54 Ans . The results are consistent with Fig. 6-27, where for a mean stress that is about half of the yield strength, the Modified Goodman line should predict failure significantly before the other two. 6-27 SOLUTION 590 MPa, 490 MPa ut y S S = = (a) max min 28 kN, 0 kN F F = = Check for yielding 2 max max 28 000 147.4 N/mm 147.4 MPa 10(25 6) F A σ = = = = − max 490 3.32 . 147.4 y y S n Ans σ = = =

Page 18 of 19 From Prob. 6-25: 208.6 MPa, 2.2 e f S K = = max min max min 28 000 0 2.2 162.1 MPa 2 2(10)(25 6) 28 000 0 2.2 162.1 MPa 2 2(10)(25 6) a f m f F F K A F F K A σ σ − − = = = −   + + = = =   −   For the modified Goodman criteria, 1 162.1 162.1 208.6 590 a m f e ut n S S σ σ = + = + 0.95 . f n Ans = Since infinite life is not predicted, estimate a life from the S-N diagram. First, find an equivalent completely reversed stress. For the modified Goodman criteria, see Ex. 6-12. rev 162.1 223.5 MPa 1 ( / ) 1 (162.1/ 590) a m ut S σ σ σ = = = − − Fig. 6-18: f = 0.87 Eq. (6-14): [ ] 2 2 0.87(590) ( ) 1263 208.6 ut e f S a S = = = Eq. (6-15): 1 1 0.87(590) log log 0.1304 3 3 208.6 ut e f S b S     = − = − = −         Eq. (6-16): 1 1/ 0.1304 rev 223.5 586 000 cycles . 1263 b N Ans a σ −     = = =         (b) max min 28 kN, 12 kN F F = = The maximum load is the same as in part (a), so max 147.4 MPa σ = 3.32 . y n Ans = Factor of safety based on infinite life: max min max min 28 000 12 000 2.2 92.63 MPa 2 2(10)(25 6) 28 000 12 000 2.2 231.6 MPa 2 2(10)(25 6) a f m f F F K A F F K A σ σ − − = = = −   + + = = =   −   1 92.63 231.6 208.6 590 a m f e ut n S S σ σ = + = + 1.20 . f n Ans = (c) max min 12 kN, 28 kN F F = = −

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Page 19 of 19 The compressive load is the largest, so check it for yielding. min min 28 000 147.4 MPa 10(25 6) F A σ − = = = − − min 490 3.32 . 147.4 yc y S n Ans σ − = = = − Factor of safety based on infinite life: ( ) ( ) max min max min 12 000 28 000 2.2 231.6 MPa 2 2(10)(25 6) 12 000 28 000 2.2 92.63 MPa 2 2(10)(25 6) a f m f F F K A F F K A σ σ − − − = = = −   + − + = = = −   −   For σ m < 0, 208.6 0.90 . 231.6 e f a S n Ans σ = = = Since infinite life is not predicted, estimate a life from the S-N diagram. For a negative mean stress (COMPRESSION) , we shall assume the equivalent completely reversed stress is the same as the actual alternating stress. Get a and b from part (a). Eq. (6-16): 1 1/ 0.1304 rev 231.6 446 000 cycles . 1263 b N Ans a σ −     = = =         A properly drawn fatigue diagram helps to make sense of the above statements.