Chapter (a) (b) f/(n x) = f/(69 10) = f/690

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1 Chapter -1 (a) (b) f/(n x) = f/(69 10) = f/690 x f fx fx f/(n x) Eq. (-9) x = 8480 = 1.9 kcycles 69 [ /69 Eq. (-10) s x = 69 1 = 30.3 kcycles ] 1/

2 Chapter 9 - Data represents a 7-class histogram with N = 197. x f fx fx x = = kpsi 197 [ /197 s x = = 9.55 kpsi ] 1/ -3 Form a table: x f fx fx From Eq. (-14) x = 4548 = 78.4 kpsi 58 [ ] 1/ /58 s x = = 6.57 kpsi 58 1 f (x) = [ π exp 1 ( ) ] x

3 10 Solutions Manual Instructor s Solution Manual to Accompany Mechanical Engineering Design -4 (a) y f fy fy y f/(nw) f (y) g(y) For a normal distribution, ( ( ) 1/ /55) ȳ = /55 = 7.07, s y = = [ 1 f (y) = π exp 1 ( ) ] x For a lognormal distribution, x = ln ln = 1.973, s x = ln = [ 1 g(y) = x(0.0604)( π) exp 1 ( ) ] ln x (b) Histogram f 1. 1 Data N LN log N

4 Chapter 11-5 Distribution is uniform in interval to in, range numbers are a = , b = in. (a) Eq. (-) µ x = a + b = = Eq. (-3) σ x = b a 3 (b) PDF from Eq. (-0) = = (c) CDF from Eq. (-1) { x in f (x) = 0 otherwise 0 x < F(x) = (x 0.5)/ x x > If all smaller diameters are removed by inspection, a = 0.500, b = µ x = = in ˆσ x = = in 3 { x f (x) = 0 otherwise 0 x < F(x) = (x 0.500) x x > Dimensions produced are due to tool dulling and wear. When parts are mixed, the distribution is uniform. From Eqs. (-) and (-3), a = µ x 3s = ( ) = in b = µ x + 3s = ( ) = in We suspect the dimension was 0.63 in 0.65

5 1 1 Solutions Manual Instructor s Solution Manual to Accompany Mechanical Engineering Design -7 F(x) = 0.555x 33 mm (a) Since F(x) is linear, the distribution is uniform at x = a F(a) = 0 = 0.555(a) 33 a = mm. Therefore, at x = b F(b) = 1 = 0.555b 33 b = 61.6 mm. Therefore, 0 x < mm F(x) = 0.555x x 61.6 mm 1 x > 61.6 mm The PDF is df/dx, thus the range numbers are: { x 61.6 mm f (x) = 0 otherwise From the range numbers, µ x = = mm ˆσ x = = 0.50 mm (b) σ is an uncorrelated quotient F = 3600 lbf, Ā = 0.11 in C F = 300/3600 = , C A = 0.001/0.11 = From Table -6, for σ σ = µ F = 3600 µ A 0.11 = psi [ ( ] 1/ ) ˆσ σ = = 694 psi ( ) C σ = 694/3 143 = Since F and A are lognormal, division is closed and σ is lognormal too. σ = LN(3143, 694) psi

6 Chapter 13-8 Cramer s rule a 1 = a = y x xy x 3 x x x x 3 x y x xy x x x x 3 = y x3 xy x x x 3 ( x ) = x xy y x x x 3 ( x ) x y x x 3 xy a 1 = a = Data Regression x y y y Data Regression x

7 14 Solutions Manual Instructor s Solution Manual to Accompany Mechanical Engineering Design -9 Data Regression S u S e S e S u S u S e m = b = S e Data Regression S u

8 Chapter E = ( y a0 a x ) E a 0 = ( y a0 a x ) = 0 y na0 a x = 0 y = na 0 + a x E a = ( y a0 a x ) (x) = 0 xy = a 0 x + a x 3 Cramer s rule a 0 = a = y x xy x 3 n = x x x 3 x n xy y n = x x x 3 x 3 y x xy n x 3 x x n xy x y n x 3 x x Data Regression x y y x x 3 xy a 0 = a = (56) 1 000(400) 4( ) 00(1 000) = 0 4(400) 00(56) 4( ) 00(1 000) = 0.00 y 5 Data Regression x

9 16 Solutions Manual Instructor s Solution Manual to Accompany Mechanical Engineering Design -11 Data Regression x y y x y xy x x (x x) ˆm = k = 6(90.44) 5(84.7) 6(6.) (5) = ˆb = F i = (5) 6 = F 30 Data Regression x (a) x = 5 6 ; ȳ = Eq. (-37) = Eq. (-36) (84.7) (90.44) s yx = 6 = s ˆb = (5/6) = lbf.0333 F i = (5.9787, ) lbf

10 Chapter 17 (b) Eq. (-35) s ˆm = = lbf/in k = (9.7656, ) lbf/in -1 The expression ɛ = δ/l is of the form x/y. Now δ = (0.0015, ) in, unspecified distribution; l = (.000, ) in, unspecified distribution; C x = / = C y = /.000 = From Table -6, ɛ = /.000 = [ ˆσ ɛ = = 4.607(10 5 ) = We can predict ɛ and ˆσ ɛ but not the distribution of ɛ. -13 σ = ɛe ɛ = (0.0005, ) distribution unspecified; E = (9.5, 0.885) Mpsi, distribution unspecified; C x = / = 0.068, C y = /9.5 = σ is of the form x, y Table -6 σ = ɛē = (9.5)10 6 = psi ˆσ σ = ( ) 1/ = psi C σ = / = ] 1/ -14 δ = Fl AE F = (14.7, 1.3) kip, A = (0.6, 0.003) in, l = (1.5, 0.004) in, E = (9.5, 0.885) Mpsi distributions unspecified. C F = 1.3/14.7 = ; C A = 0.003/0.6 = ; C l = 0.004/1.5 = ; C E = 0.885/9.5 = 0.03 Mean of δ: δ = Fl AE = Fl ( 1 A )( ) 1 E

11 18 Solutions Manual Instructor s Solution Manual to Accompany Mechanical Engineering Design From Table -6, δ = F l(1/ā)(1/ē) 1 1 δ = (1.5) (10 6 ) = in For the standard deviation, using the first-order terms in Table -6,. F l ( ˆσ δ = C F + Cl + C A ĀĒ + ) 1/ C E = δ ( CF + C l + C A + 1/ E) C COV ˆσ δ = ( ) 1/ = in C δ = / = Force COV dominates. There is no distributional information on δ. -15 M = (15 000, 1350) lbf in, distribution unspecified; d = (.00, 0.005) in distribution unspecified. σ = 3M πd 3, C M = = 0.09, C d = = σ is of the form x/y, Table -6. Mean: σ = 3 M π d 3 Standard Deviation:. 3 M 3(15 000) = = π d 3 π( 3 ) = psi )] 1/ ˆσ σ = σ [( CM + )/( C d 1 + C 3 d 3. From Table -6, C d 3 = 3C d = 3(0.005) = ˆσ σ = σ [( CM + (3C d) )/ (1 + (3C d )) ] 1/ COV: = [( )/( )] 1/ = 175 psi C σ = 175 = Stress COV dominates. No information of distribution of σ.

12 Chapter f(x) x 1 x x Fraction discarded is α + β. The area under the PDF was unity. Having discarded α + β fraction, the ordinates to the truncated PDF are multiplied by a. New PDF, g(x), is given by g(x) = a = More formal proof: g(x) has the property 1 1 (α + β) { f (x)/[1 (α + β)] x1 x x 0 otherwise 1 = x x 1 [ 1 = a g(x) dx = a f (x) dx x x 1 x1 0 f (x) dx f (x) dx x ] f (x) dx 1 = a {1 F(x 1 ) [1 F(x )]} a = 1 F(x ) F(x 1 ) = 1 (1 β) α = 1 1 (α + β) -17 (a) d = U[0.748, 0.751] µ d = = in ˆσ d = = in 3 f (x) = 1 b a = = 333.3in 1 F(x) = x = 333.3(x 0.748)

13 0 Solutions Manual Instructor s Solution Manual to Accompany Mechanical Engineering Design (b) F(x 1 ) = F(0.748) = 0 If g(x) is truncated, PDF becomes F(x ) = ( )333.3 = g(x) 500 f(x) x g(x) = f (x) F(x ) F(x 1 ) = = 500 in µ x = a + b = ˆσ x = b a = = in = in -18 From Table A-10, 8.1% corresponds to z 1 = 1.4 and 5.5% corresponds to z =+1.6. From which k 1 = µ + z 1 ˆσ k = µ + z ˆσ µ = z k 1 z 1 k z z 1 = = 1.6(9) ( 1.4) ( 1.4) ˆσ = k k 1 z z 1 = ( 1.4) = The original density function is [ 1 f (k) = π exp 1 ( ) ] k From Prob. -1, µ = 1.9 kcycles and ˆσ = 30.3 kcycles. z 10 = x 10 µ ˆσ = x x 10 = z 10 From Table A-10, for 10 percent failure, z 10 = 1.8 x 10 = ( 1.8) = 84.1 kcycles

14 Chapter 1-0 x f fx fx x f/(nw) f(x) E E x = s x = x f/(nw) f (x) x f/(nw) f (x) E E E E

15 Solutions Manual Instructor s Solution Manual to Accompany Mechanical Engineering Design f Histogram PDF x -1 x f fx fx f/(nw) f (x) x = s x = x f/(nw) f (x) f Data PDF x

16 Chapter 3 - x f fx fx f/(nw) f (x) x = s x = x f/(nw) f(x) x f/(nw) f(x) f Data PDF x -3 σ = 4 P πd = 4(40) π(1 ) ˆσ σ = 4 ˆσ P πd = 4(8.5) π(1 ) ˆσ sy = 5.9 kpsi = kpsi = 10.8 kpsi

17 4 Solutions Manual Instructor s Solution Manual to Accompany Mechanical Engineering Design For no yield, m = S y σ 0 z = m µ m = 0 µ m ˆσ m ˆσ m µ m = S y σ = 7.47 kpsi, ˆσ m = = µ m ˆσ m ( ˆσ σ +ˆσ S y ) 1/ = 1.3 kpsi z = =.30 From Table A-10, p f = R = 1 p f = = m -4 For a lognormal distribution, Eq. (-18) µ y = ln µ x ln 1 + Cx Eq. (-19) ˆσ y = ln ( ) 1 + Cx From Prob. (-3) µ m = S y σ = µ x ( ) µ y = ln S y ln 1 + C Sy [ ] S y 1 + C = ln σ σ 1 + CS y [ ) ˆσ y = ln (1 + CS y + ln ( 1 + Cσ ) ] 1/ ) (1 ) = ln [(1 ] + C Sy + C σ ( ) S y 1 + C ln σ σ 1 + CS z = = µˆσ y ) (1 ln [(1 + C ) ] Sy + C σ σ = 4 P πd = 4(30) = kpsi π(1 ) ˆσ σ = 4 ˆσ P πd = 4(5.1) = kpsi π(1 ) C σ = = C Sy = 3.81 = ( ) ln σ ln 1 + Cσ

18 Chapter 5 ln z = ln [ ( )( ) ] = From Table A-10 p f = R = 1 p f = (a) a = ± in b =.000 ± in c = ± in d = 6.00 ± in w = d a b c = = 0.00 in t w = t all = = in w = 0.00 ± in (b) w = 0.00 (0.001 ) ( ) ( ) ( ) ˆσ w = ˆσ all = = in (uniform) w = 0.00 ± in -6 V + V = (a + a)(b + b)(c + c) V + V = abc + bc a + ac b + ab c + small higher order terms V. = a V a + b b + c c V =ā b c = 1.5(1.875)(.75) = in 3 V V = = V = V V V = (6.4453) = in 3 Lower range number: V V = = 6.46 in 3 Upper range number: V + V = = in 3

19 6 Solutions Manual Instructor s Solution Manual to Accompany Mechanical Engineering Design -7 (a) w a c b w max = in, w min = in w = ( )/ = in w = ± in w = x ȳ =ā b c =ā ā = in t w = t all = t a t a = = in a = ± in (b) ˆσ w = ˆσ all = ˆσ a +ˆσ b +ˆσ c ˆσ a =ˆσ w ˆσ b ˆσ c ( ) = 3 ˆσ a = 5.667(10 6 ) ( ) ˆσ a = 5.667(10 6 ) = in ( ) ā = in, ˆσ a = in -8 Choose 15 mm as basic size, D, d. Table -8: fit is designated as 15H7/h6. From Table A-11, the tolerance grades are D = mm and d = mm. Hole: Eq. (-38) D max = D + D = = mm D min = D = mm Shaft: From Table A-1, fundamental deviation δ F = 0. From Eq. (-39) d max = d + δ F = = mm d min = d + δ R d = = mm -9 Choose 45 mm as basic size. Table -8 designates fit as 45H7/s6. From Table A-11, the tolerance grades are D = 0.05 mm and d = mm Hole: Eq. (-38) D max = D + D = = mm D min = D = mm

20 Chapter 7 Shaft: From Table A-1, fundamental deviation δ F = mm. From Eq. (-40) d min = d + δ F = = mm d max = d + δ F + d = = mm -30 Choose 50 mm as basic size. From Table -8 fit is 50H7/g6. From Table A-11, the tolerance grades are D = 0.05 mm and d = mm. Hole: D max = D + D = = mm D min = D = mm Shaft: From Table A-1 fundamental deviation = mm d max = d + δ F = ( 0.009) = mm d min = d + δ F d = ( 0.009) = mm -31 Choose the basic size as in. From Table -8, for 1.0 in, the fit is H8/f7. From Table A-13, the tolerance grades are D = in and d = in. Hole: D max = D + ( D) hole = = in D min = D = in Shaft: From Table A-14: Fundamental deviation = in d max = d + δ F = ( ) = in d min = d + δ F d = ( ) = in Alternatively, d min = d max d = = in. -3 W D i W D o D o = W + D i + W D o = W + D i + W = = 4.01 in t Do = t all = = in D o = 4.01 ± in

21 8 Solutions Manual Instructor s Solution Manual to Accompany Mechanical Engineering Design -33 D o = D i + W D o = D i + W = (5.33) = mm t Do = all t = t Di + t w = (0.13) = 1.56 mm D o = ± 1.56 mm -34 D o = D i + W D o = D i + W = (0.139) = 4.01 mm t Do = t = [ td o + ( t w ) ] 1/ all = [ () (0.004) ] 1/ = 0.09 in D o = 4.01 ± 0.09 in -35 D o = D i + W D o = D i + W = (5.33) = mm t Do = t = [ () (0.13) ] 1/ all = 1.33 mm D o = ± 1.33 mm -36 (a) F W w w = F W w = F W = = in t w = all t = t w = in w max = w + t w = = 0.06 in w min = w t w = = in The minimum squeeze is 0.06 in.

22 Chapter 9 (b) D o Y w Y max = D o = 4.01 in Y min = max[0.99 D o, D o 0.06] D o + w Y = 0 = max[3.9719, 3.95] = 3.97 in Y = 3.99 ± 0.00 in w = Y D o w = Ȳ D o = = 0.00 in t w = all t = t Y + t Do = = in w = 0.00 ± in } w max = in O-ring is more likely compressed than free prior to assembly of the w min = in end plate. -37 (a) (b) Figure defines w as gap. F The O-ring is squeezed at least 0.75 mm. From the figure, the stochastic equation is: or, W w D o Y w = F W w = F W = = 1.01 mm t w = t = t F + t W = = 0.6 mm all w max = w + t w = = 0.75 mm w min = w t w = = 1.7 mm w Y max = D o = mm Y min = max[0.99 D o, D o 1.5] = max[0.99(19.58, )] = mm Y = ± 1.10 mm D o + w = Y w = Y D o w = Ȳ D o = = 1.10 mm t w = t = t Y + t Do = = 1.44 mm all w max = w + t w = = 0.34 mm w min = w t w = =.54 mm The O-ring is more likely to be circumferentially compressed than free prior to assembly of the end plate.

23 30 Solutions Manual Instructor s Solution Manual to Accompany Mechanical Engineering Design -38 b c w d w max = 0.00 in, w min = in w = 1 ( ( 0.040)) = in a t w = 1 ( 0.00 ( 0.040)) = in Absolute: t w = all b = ± in c = 0.10 ± in d = ± in w =ā b c d =ā ā = ā = in t = = t a t a = = in a = ± in Statistical: For a normal distribution of dimensions t w = all t = t a + t b + t c + t d t a = ( tw t b t c ) 1/ t d = ( ) 1/ = a = ± in -39 x n nx nx x = /136 = 98.6 kpsi ( ) 1/ /136 s x = = 4.30 kpsi 135

24 Chapter 31 Under normal hypothesis, z 0.01 = (x )/4.30 x 0.01 = z 0.01 = (.367) = 88.6 = kpsi -40 From Prob. -39, µ x = 98.6 kpsi, and ˆσ x = 4.30 kpsi. From Eqs. (-18) and (-19), C x =ˆσ x /µ x = 4.30/98.6 = µ y = ln(98.6) / = ˆσ y = ln( ) = For a yield strength exceeded by 99% of the population, z 0.01 = (ln x 0.01 µ y )/ ˆσ y ln x 0.01 = µ y +ˆσ y z 0.01 From Table A-10, for 1% failure, z 0.01 =.36. Thus, ln x 0.01 = (.36) = x 0.01 = 88.7 kpsi The normal PDF is given by Eq. (-14) as [ 1 f (x) = 4.30 π exp 1 ( ) ] x For the lognormal distribution, from Eq. (-17), defining g(x), [ 1 g(x) = x( ) π exp 1 ( ) ] ln x x (kpsi) f/(nw) f (x) g (x) x (kpsi) f/(nw) f (x) g (x) Note: rows are repeated to draw histogram

25 3 Solutions Manual Instructor s Solution Manual to Accompany Mechanical Engineering Design Histogram f(x) g(x) Probability density x (kpsi) The normal and lognormal are almost the same. However the data is quite skewed and perhaps a Weibull distribution should be explored. For a method of establishing the Weibull parameters see Shigley, J. E., and C. R. Mischke, Mechanical Engineering Design, McGraw-Hill, 5th ed., 1989, Sec Let x = (S fe) x 0 = 79 kpsi, θ = 86. kpsi, b =.6 Eq. (-8) x = x 0 + (θ x 0 )Ɣ(1 + 1/b) x = 79 + (86. 79)Ɣ(1 + 1/.6) = Ɣ(1.38) From Table A-34, Ɣ(1.38) = Eq. (-9) x = ( ) = 85.4 kpsi ˆσ x = (θ x 0 )[Ɣ(1 + /b) Ɣ (1 + 1/b)] 1/ = (86. 79)[Ɣ(1 + /.6) Ɣ (1 + 1/.6)] 1/ = 7.[ ] 1/ =.64 kpsi C x = ˆσ x x =.64 = x = S ut x 0 = 7.7, θ = 46., b = 4.38 µ x = ( )Ɣ(1 + 1/4.38) = Ɣ(1.3) = ( ) = kpsi

26 Chapter 33 ˆσ x = ( )[Ɣ(1 + /4.38) Ɣ (1 + 1/4.38)] 1/ = 18.5[Ɣ(1.46) Ɣ (1.3)] 1/ = 18.5[ ] 1/ = 4.38 kpsi C x = 4.38 = From the Weibull survival equation [ ( ) ] x b x0 R = exp = 1 p θ x 0 [ ( ) ] x40 x b 0 R 40 = exp = 1 p 40 θ x 0 [ ( ) ] = exp = p 40 = 1 R 40 = = = 15.4% -43 x = S ut x 0 = 151.9, θ = 193.6, b = 8 µ x = ( )Ɣ(1 + 1/8) = Ɣ(1.15) = ( ) = 191. kpsi ˆσ x = ( )[Ɣ(1 + /8) Ɣ (1 + 1/8)] 1/ = 41.7[Ɣ(1.5) Ɣ (1.15)] 1/ = 41.7[ ] 1/ = 5.8 kpsi C x = = x = S ut x 0 = 47.6, θ = 15.6, b = x = ( )Ɣ(1 + 1/11.84) x = Ɣ(1.08) = ( ) = 1.5 kpsi ˆσ x = ( )[Ɣ(1 + /11.84) Ɣ (1 + 1/11.84)] 1/ = 78[Ɣ(1.08) Ɣ (1.17)] 1/ = 78( ) 1/ =.4 kpsi

27 34 Solutions Manual Instructor s Solution Manual to Accompany Mechanical Engineering Design From Prob. -4 [ ( ) ] x b x0 p = 1 exp θ θ 0 [ ( ) ] = 1 exp = y = S y y 0 = 64.1, θ = 81.0, b = 3.77 ȳ = ( )Ɣ(1 + 1/3.77) = Ɣ(1.7) = ( ) = kpsi σ y = ( )[Ɣ(1 + /3.77) Ɣ(1 + 1/3.77)] 1/ σ y = 16.9[( ) ] 1/ = 4.57 kpsi [ ( ) ] y 3.77 y0 p = 1 exp θ y 0 [ ( ) ] p = 1 exp = x = S ut = W[1.3, 134.6, 3.64] kpsi, p(x > 10) = 1 = 100% since x 0 > 10 kpsi [ ( ) ] p(x > 133) = exp = = 54.8% -46 Using Eqs. (-8) and (-9) and Table A-34, µ n = n 0 + (θ n 0 )Ɣ(1 + 1/b) = ( )Ɣ(1 + 1/.66) = 1.85 kcycles ˆσ n = (θ n 0 )[Ɣ(1 + /b) Ɣ (1 + 1/b)] = kcycles For the Weibull density function, Eq. (-7), ( ) [.66 n f W (n) = exp For the lognormal distribution, Eqs. (-18) and (-19) give, µ y = ln(1.85) (34.79/1.85) / = ˆσ y = [1 + (34.79/1.85) ] = ( n 36.9 ) ]

28 Chapter 35 From Eq. (-17), the lognormal PDF is [ 1 f LN (n) = n π exp 1 We form a table of densities f W (n) and f LN (n) and plot. ( ) ] ln n n (kcycles) f W (n) f LN (n) E E f(n) LN W n, kcycles 50 The Weibull L10 life comes from Eq. (-6) with a reliability of R = Thus, n 0.10 = ( )[ln(1/0.90)] 1/.66 = 78.1kcycles

29 36 Solutions Manual Instructor s Solution Manual to Accompany Mechanical Engineering Design The lognormal L10 life comes from the definition of the z variable. That is, ln n 0 = µ y +ˆσ y z or n 0 = exp(µ y +ˆσ y z) From Table A-10, for R = 0.90, z = 1.8. Thus, n 0 = exp[ ( 1.8)] = 8.7kcycles -47 Form a table x g(x) i L(10 5 ) f i f i x(10 5 ) f i x (10 10 ) (10 5 ) x = 59.5(10 5 )/100 = 5.95(10 5 )cycles [ (10 10 ) [59.5(10 5 )] /100 s x = = 1.319(10 5 )cycles C x = s/ x = 1.319/5.95 = 0.49 µ y = ln 5.95(10 5 ) 0.49 / = ˆσ y = ln( ) = 0.45 [ 1 g(x) = exp 1 ( ) ] ln x µy x ˆσ y π ˆσ y g(x) = 1.68 x [ exp 1 ( ) ] ln x ] 1/

30 Chapter g(x) Superposed histogram and PDF (10 5 ) 10.05(10 5 ) x, cycles -48 x = S u = W[70.3, 84.4,.01] Eq. (-8) µ x = ( )Ɣ(1 + 1/.01) = ( )Ɣ(1.498) = ( ) = 8.8 kpsi Eq. (-9) ˆσ x = ( )[Ɣ(1 + /.01) Ɣ (1 + 1/.01)] 1/ ˆσ x = 14.1[ ] 1/ = 6.50 kpsi C x = 6.50 = Take the Weibull equation for the standard deviation and the mean equation solved for x x 0 Dividing the first by the second, ˆσ x = (θ x 0 )[Ɣ(1 + /b) Ɣ (1 + 1/b)] 1/ x x 0 = (θ x 0 )Ɣ(1 + 1/b) ˆσ x = [Ɣ(1 + /b) Ɣ (1 + 1/b)] 1/ x x 0 Ɣ(1 + 1/b) = Ɣ(1 + /b) Ɣ (1 + 1/b) 1 = R = 0.763

31 38 Solutions Manual Instructor s Solution Manual to Accompany Mechanical Engineering Design Make a table and solve for b iteratively b 1 + /b 1 + 1/b Ɣ(1 + /b) Ɣ(1 + 1/b) b. = Using MathCad θ = x 0 + x x 0 Ɣ(1 + 1/b) = 49.8 kpsi = Ɣ(1 + 1/4.068) -50 x = S y = W[34.7, 39,.93] kpsi x = ( )Ɣ(1 + 1/.93) = Ɣ(1.34) = (0.89 ) = 38.5 kpsi ˆσ x = ( )[Ɣ(1 + /.93) Ɣ (1 + 1/.93)] 1/ = 4.3[Ɣ(1.68) Ɣ (1.34)] 1/ = 4.3[ ] 1/ = 1.4 kpsi C x = 1.4/38.5 = x (Mrev) f fx fx Sum µ x = 1193(10 6 )/37 = 5.034(10 6 ) cycles 7673(10 ˆσ x = 1 ) [1193(10 6 )] /37 =.658(10 6 ) cycles 37 1 C x =.658/5.034 = 0.58

32 Chapter 39 From Eqs. (-18) and (-19), µ y = ln[5.034(10 6 )] 0.58 / = 15.9 ˆσ y = ln( ) = From Eq. (-17), defining g(x), [ 1 g (x) = x(0.496) π exp 1 ( ) ] ln x x (Mrev) f/(nw) g(x) (10 6 ) g(x)(10 6 ) Histogram PDF x, Mrev z = ln x µ y ˆσ y ln x = µ y +ˆσ y z = z L 10 life, where 10% of bearings fail, from Table A-10, z = 1.8. Thus, ln x = ( 1.8) = x = rev

Chapter (a) (b) f/(n x) = f/(69 10) = f/690

Chapter (a) (b) f/(n x) = f/(69 10) = f/690 Chapter -1 (a) 1 1 8 6 4 6 7 8 9 1 11 1 13 14 15 16 17 18 19 1 (b) f/(n x) = f/(69 1) = f/69 x f fx fx f/(n x) 6 1 7.9 7 1 7 49.15 8 3 4 19.43 9 5 45 4 5.7 1 8 8 8.116 11 1 13 145.174 1 6 7 86 4.87 13