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

Transcription

1 Chapter -1 (a) (b) f/(n x) = f/(69 1) = f/69 x f fx fx f/(n x) Eq. (-9) x = 848 = 1.9 kcycles 69 [ /69 Eq. (-1) s x = 69 1 = 3.3 kcycles ] 1/

2 Solutions Manual Instructor s Solution Manual to Accompany Mechanical Engineering Design - 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 Chapter 3-4 (a) y f fy fy y f/(nw) f (y) g(y) For a normal distribution, ( ( ) 1/ /55) ȳ = /55 = 7.7, s y = = [ 1 f (y) =.4358 π exp 1 ( ) ] x For a lognormal distribution, x = ln ln = 1.973, s x = ln =.64 [ 1 g(y) = x(.64)( π) exp 1 ( ) ] ln x (b) Histogram f 1. 1 Data N LN log N

4 4 Solutions Manual Instructor s Solution Manual to Accompany Mechanical Engineering Design -5 Distribution is uniform in interval.5 to.58 in, range numbers are a =.5, b =.58 in. (a) Eq. (-) µ x = a + b = =.54 Eq. (-3) (b) PDF from Eq. (-) σ x = b a 3 = =. 31 (c) CDF from Eq. (-1) { 15.5 x.58 in f (x) = otherwise x <.5 F(x) = (x.5)/.8.5 x.58 1 x >.58 If all smaller diameters are removed by inspection, a =.5, b =.58 µ x = ˆσ x = f (x) = F(x) = =.55 in.58.5 =. 173 in 3 { x.58 otherwise x < (x.5).5 x.58 1 x >.58-6 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 =.641 3(. 581) =.631 in b = µ x + 3s = (. 581) =.651 in We suspect the dimension was.63 in.65

5 1 Chapter 5-7 F(x) =.555x 33 mm (a) Since F(x) is linear, the distribution is uniform at x = a F(a) = =.555(a) 33 a = mm. Therefore, at x = b F(b) = 1 =.555b 33 b = 61.6 mm. Therefore, x < mm F(x) =.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) = otherwise From the range numbers, µ x = ˆσ x = = 6.36 mm =.5 mm (b) σ is an uncorrelated quotient F = 36 lbf, Ā =.11 in C F = 3/36 =.83 33, C A =.1/.11 =.8 99 From Table -6, for σ σ = µ F = 36 µ A.11 = psi [ ( ] 1/ ) ˆσ σ = = 694 psi ( ) C σ = 694/3 143 =.838 Since F and A are lognormal, division is closed and σ is lognormal too. σ = LN(3143, 694) psi

6 6 Solutions Manual Instructor s Solution Manual to Accompany Mechanical Engineering Design -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.3.5 Data Regression x

7 Chapter 7-9 Data Regression S u S e S e S u S u S e m =.3167 b = S e 14 1 Data Regression S u

8 8 Solutions Manual Instructor s Solution Manual to Accompany Mechanical Engineering Design -1 E = ( y a a x ) E a = ( y a a x ) = y na a x = y = na + a x E a = ( y a a x ) (x) = xy = a x + a x 3 Cramer s rule a = 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 = a = 8 (56) 1 (4) 4(8 ) (1 ) = 4(4) (56) 4(8 ) (1 ) =. y 5 Data Regression x

9 Chapter 9-11 Data Regression x y y x y xy x x (x x) ˆm = k = 6(9.44) 5(84.7) 6(6.) (5) = ˆb = F i = (5) 6 = F 3 Data Regression x (a) x = 5 6 ; ȳ = Eq. (-37) = Eq. (-36) (84.7) (9.44) s yx = 6 = s ˆb = (5/6) =.3964 lbf.333 F i = (5.9787,.3964) lbf

10 1 Solutions Manual Instructor s Solution Manual to Accompany Mechanical Engineering Design (b) Eq. (-35) s ˆm = =.3899 lbf/in k = (9.7656,.3899) lbf/in -1 The expression ɛ = δ/l is of the form x/y. Now δ = (.15,. 9) in, unspecified distribution; l = (.,.81) in, unspecified distribution; C x =. 9/.15 =.613 C y =.81/. =. 75 From Table -6, ɛ =.15/. =. 75 [ ˆσ ɛ = = 4.67(1 5 ) =. 46 We can predict ɛ and ˆσ ɛ but not the distribution of ɛ. -13 σ = ɛe ɛ = (.5,. 34) distribution unspecified; E = (9.5,.885) Mpsi, distribution unspecified; C x =. 34/.5 =.68, σ is of the form x, y Table -6 C y =.885/9.5 =.3 σ = ɛē =.5(9.5)1 6 = psi ] 1/ ˆσ σ = 14 75( ) 1/ = psi C σ = 196.7/14 75 = δ = Fl AE F = (14.7, 1.3) kip, A = (.6,.3) in, l = (1.5,.4) in, E = (9.5,.885) Mpsi distributions unspecified. C F = 1.3/14.7 =.884; C A =.3/.6 =.133; C l =.4/1.5 =.67; C E =.885/9.5 =.3 Mean of δ: δ = Fl ( )( ) 1 1 AE = Fl A E

11 Chapter 11 From Table -6, δ = F l(1/ā)(1/ē) 1 1 δ = 14 7(1.5).6 9.5(1 6 ) =.3 31 in For the standard deviation, using the first-order terms in Table -6, COV. F l ( ˆσ δ = C F + Cl + C A ĀĒ + ) 1/ C E = δ ( CF + C l + C A + 1/ E) C ˆσ δ =.3 31( ) 1/ =. 313 in C δ =. 313/.3 31 =.945 Force COV dominates. There is no distributional information on δ. -15 M = (15, 135) lbf in, distribution unspecified; d = (.,.5) in distribution unspecified. σ = 3M πd 3, C M = =.9, C d =.5. =.5 σ is of the form x/y, Table -6. Mean: Standard Deviation: σ = 3 M π d 3. 3 M 3(15 ) = = 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(.5) =.75 COV: ˆσ σ = σ [( C M + (3C d) )/ (1 + (3C d )) ] 1/ = 19 99[( )/( )] 1/ = 175 psi C σ = 175 = Stress COV dominates. No information of distribution of σ.

12 1 Solutions Manual Instructor s Solution Manual to Accompany Mechanical Engineering Design -16 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 otherwise 1 = x x 1 [ 1 = a g(x) dx = a f (x) dx x x 1 x1 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[.748,.751] µ d = =.7495 in ˆσ d = =. 866 in 3 f (x) = 1 b a = = 333.3in 1 F(x) = x.748 = 333.3(x.748)

13 Chapter 13 (b) F(x 1 ) = F(.748) = If g(x) is truncated, PDF becomes F(x ) = ( )333.3 = g(x) 5 f(x) x g(x) = f (x) F(x ) F(x 1 ) = = 5 in µ x = a + b = ˆσ x = b a = =.749 in =. 577 in -18 From Table A-1, 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) =.6667 The original density function is [ 1 f (k) =.6667 π exp 1 ( ) ] k From Prob. -1, µ = 1.9 kcycles and ˆσ = 3.3 kcycles. z 1 = x 1 µ ˆσ = x x 1 = z 1 From Table A-1, for 1 percent failure, z 1 = 1.8 x 1 = ( 1.8) = 84.1 kcycles

14 14 Solutions Manual Instructor s Solution Manual to Accompany Mechanical Engineering Design - 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 Chapter 15 f..18 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 16 Solutions Manual Instructor s Solution Manual to Accompany Mechanical Engineering Design - x f fx fx f/(nw) f (x) x = s x = x f/(nw) f(x) x f/(nw) f(x) f.9.8 Data PDF x -3 σ = 4 P πd = 4(4) π(1 ) ˆσ σ = 4 ˆσ P πd = 4(8.5) π(1 ) ˆσ sy = 5.9 kpsi = 5.93 kpsi = 1.8 kpsi

17 Chapter 17 For no yield, m = S y σ z = m µ m = 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 = =.3 From Table A-1, p f =.19 R = 1 p f = 1.19 =.987 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(3) = kpsi π(1 ) ˆσ σ = 4 ˆσ P πd = 4(5.1) = kpsi π(1 ) C σ = =.17 C Sy = 3.81 = ( ) ln σ ln 1 + Cσ

18 18 Solutions Manual Instructor s Solution Manual to Accompany Mechanical Engineering Design ln z = ln [ ( )( ) ] = 1.47 From Table A-1 p f =.78 R = 1 p f =.99-5 x n nx nx x = /136 = 98.6 kpsi ( ) 1/ /136 s x = = 4.3 kpsi 135 Under normal hypothesis, z.1 = (x )/4.3 x.1 = z.1 = (.367) = 88.6 = kpsi -6 From Prob. -5, µ x = 98.6 kpsi, and ˆσ x = 4.3 kpsi. C x =ˆσ x /µ x = 4.3/98.6 = From Eqs. (-18) and (-19), µ y = ln(98.6) / = ˆσ y = ln( ) = For a yield strength exceeded by 99% of the population, z.1 = (ln x.1 µ y )/ ˆσ y ln x.1 = µ y +ˆσ y z.1

19 Chapter 19 From Table A-1, for 1% failure, z.1 =.36. Thus, ln x.1 = (.36) = x.1 = 88.7 kpsi The normal PDF is given by Eq. (-14) as [ 1 f (x) = 4.3 π exp 1 ( ) ] x For the lognormal distribution, from Eq. (-17), defining g(x), [ 1 g(x) = x(.43 74) 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 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

20 Solutions Manual Instructor s Solution Manual to Accompany Mechanical Engineering Design Weibull parameters see Shigley, J. E., and C. R. Mischke, Mechanical Engineering Design, McGraw-Hill, 5th ed., 1989, Sec Let x = (S fe) x = 79 kpsi, θ = 86. kpsi, b =.6 Eq. (-8) x = x + (θ x )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 )[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 = 7.7, θ = 46., b = 4.38 µ x = ( )Ɣ(1 + 1/4.38) = Ɣ(1.3) = (.91 75) = kpsi 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 x R = exp = 1 p θ x

21 Chapter 1 [ ( ) ] x4 x b R 4 = exp = 1 p 4 θ x = exp [ ( ) ] = p 4 = 1 R 4 = =.154 = 15.4% -9 x = S ut x = 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 = =.3-3 x = S ut x = 47.6, θ = 15.6, b = From Prob. -8 x = ( )Ɣ(1 + 1/11.84) x = Ɣ(1.8) = ( ) = 1.5 kpsi ˆσ x = ( )[Ɣ(1 + /11.84) Ɣ (1 + 1/11.84)] 1/ = 78[Ɣ(1.8) Ɣ (1.17)] 1/ = 78( ) 1/ =.4 kpsi [ ( ) ] x b x p = 1 exp θ θ [ ( ) ] = 1 exp =.9

22 Solutions Manual Instructor s Solution Manual to Accompany Mechanical Engineering Design y = S y y = 64.1, θ = 81., b = 3.77 ȳ = ( )Ɣ(1 + 1/3.77) = Ɣ(1.7) = (.9 5) = kpsi σ y = ( )[Ɣ(1 + /3.77) Ɣ(1 + 1/3.77)] 1/ σ y = 16.9[( ).9 5 ] 1/ = 4.57 kpsi [ ( ) ] y 3.77 y p = 1 exp θ y [ ( ) ] p = 1 exp = x = S ut = W[1.3, 134.6, 3.64] kpsi, p(x > 1) = 1 = 1% since x > 1 kpsi [ ( ) ] p(x > 133) = exp =.548 = 54.8% -3 Using Eqs. (-8) and (-9) and Table A-34, µ n = n + (θ n )Ɣ(1 + 1/b) = ( )Ɣ(1 + 1/.66) = 1.85 kcycles ˆσ n = (θ n )[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) ] =.778 From Eq. (-17), the lognormal PDF is [ 1 f LN (n) =.778 n π exp 1 We form a table of densities f W (n) and f LN (n) and plot. ( n 36.9 ) ] ( ) ] ln n

23 Chapter 3 n (kcycles) f W (n) f LN (n) 4 9.1E-5 1.8E f(n).14.1 LN W n, kcycles 5 The Weibull L1 life comes from Eq. (-6) with a reliability of R =.9. Thus, n.1 = ( )[ln(1/.9)] 1/.66 = 78.1kcycles The lognormal L1 life comes from the definition of the z variable. That is, ln n = µ y +ˆσ y z or n = exp(µ y +ˆσ y z) From Table A-1, for R =.9, z = 1.8. Thus, n = exp[ ( 1.8)] = 8.7kcycles

24 4 Solutions Manual Instructor s Solution Manual to Accompany Mechanical Engineering Design -33 Form a table x g(x) i L(1 5 ) f i f i x(1 5 ) f i x (1 1 ) (1 5 ) x = 59.5(1 5 )/1 = 5.95(1 5 )cycles [ (1 1 ) [59.5(1 5 )] /1 s x = 1 1 = 1.319(1 5 )cycles C x = s/ x = 1.319/5.95 =.49 µ y = ln 5.95(1 5 ).49 / = ˆσ y = ln( ) =.45 [ 1 g(x) = exp 1 ( ) ] ln x µy x ˆσ y π ˆσ y g(x) = 1.68 [ exp 1 ( ) ] ln x x.45 ] 1/

25 Chapter g(x).5.4 Superposed histogram and PDF (1 5 ) 1.5(1 5 ) x, cycles -34 x = S u = W[7.3, 84.4,.1] Eq. (-8) µ x = ( )Ɣ(1 + 1/.1) = ( )Ɣ(1.498) = ( ) = 8.8 kpsi Eq. (-9) ˆσ x = ( )[Ɣ(1 + /.1) Ɣ (1 + 1/.1)] 1/ ˆσ x = 14.1[ ] 1/ = 6.5 kpsi C x = = Take the Weibull equation for the standard deviation and the mean equation solved for x x Dividing the first by the second, ˆσ x = (θ x )[Ɣ(1 + /b) Ɣ (1 + 1/b)] 1/ x x = (θ x )Ɣ(1 + 1/b) ˆσ x = [Ɣ(1 + /b) Ɣ (1 + 1/b)] 1/ x x Ɣ(1 + 1/b) = Ɣ(1 + /b) Ɣ (1 + 1/b) 1 = R =.763

26 6 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. = 4.68 Using MathCad θ = x + x x Ɣ(1 + 1/b) = 49.8 kpsi = Ɣ(1 + 1/4.68) -36 x = S y = W[34.7, 39,.93] kpsi x = ( )Ɣ(1 + 1/.93) = Ɣ(1.34) = (.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(1 6 )/37 = 5.34(1 6 ) cycles 7673(1 ˆσ x = 1 ) [1193(1 6 )] /37 =.658(1 6 ) cycles 37 1 C x =.658/5.34 =.58

27 Chapter 7 From Eqs. (-18) and (-19), µ y = ln[5.34(1 6 )].58 / = 15.9 ˆσ y = ln( ) =.496 From Eq. (-17), defining g(x), [ 1 g (x) = x(.496) π exp 1 ( ) ] ln x x (Mrev) f/(nw) g(x) (1 6 ) g(x)(1 6 ) Histogram PDF x, Mrev z = ln x µ y ˆσ y ln x = µ y +ˆσ y z = z L 1 life, where 1% of bearings fail, from Table A-1, z = 1.8. Thus, ln x = ( 1.8) = x = rev

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