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Shigley's Mechanical Engineering Design 10th Edition Solution Manual, Lecture notes of Mechanical Systems Design

Composed by Michael A. Latcha, PhD. for ME 486, Oakland University (OU)

Typology: Lecture notes

2020/2021

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Shigley’s MED, 10
th
edition Chapter 1 Solutions, Page 1/12
Chapter 1
Problems 1-1 through 1-6 are for student research. No standard solutions are provided.
1-7 From Fig. 1-2, cost of grinding to ± 0.0005 in is 270%. Cost of turning to ± 0.003 in is
60%.
Relative cost of grinding vs. turning = 270/60 = 4.5 times Ans.
______________________________________________________________________________
1-8 C
A
= C
B
,
10 + 0.8 P = 60 + 0.8 P 0.005 P
2
P
2
= 50/0.005 P = 100 parts Ans.
______________________________________________________________________________
1-9 Max. load = 1.10 P
Min. area = (0.95)
2
A
Min. strength = 0.85 S
To offset the absolute uncertainties, the design factor, from Eq. (1-1) should be
( )
2
1.10
1.43 .
0.85 0.95
d
n Ans
= =
______________________________________________________________________________
1-10 (a) X
1
+ X
2
:
( ) ( )
1 2 1 1 2 2
1 2 1 2
1 2
error
.
x x X e X e
e x x X X
e e Ans
+ = + + +
= = + +
= +
(b) X
1
X
2
:
(
)
( ) ( )
1 2 1 1 2 2
1 2 1 2 1 2
.
x x X e X e
= + +
= =
(c) X
1
X
2
:
(
)
(
)
1 2 1 1 2 2
1 2 1 2 1 2 2 1 1 2
1 2
1 2 2 1 1 2
1 2
.
x x X e X e
e x x X X X e X e e e
e e
X e X e X X Ans
X X
= + +
= = + +
+ = +
pf3
pf4
pf5
pf8
pf9
pfa
pfd
pfe
pff
pf12
pf13
pf14
pf15
pf16
pf17
pf18
pf19
pf1a
pf1b
pf1c
pf1d
pf1e
pf1f
pf20
pf21
pf22
pf23
pf24
pf25
pf26
pf27
pf28
pf29
pf2a
pf2b
pf2c
pf2d
pf2e
pf2f
pf30
pf31
pf32
pf33
pf34
pf35
pf36
pf37
pf38
pf39
pf3a
pf3b
pf3c
pf3d
pf3e
pf3f
pf40
pf41
pf42
pf43
pf44
pf45
pf46
pf47
pf48
pf49
pf4a
pf4b
pf4c
pf4d
pf4e
pf4f
pf50
pf51
pf52
pf53
pf54
pf55
pf56
pf57
pf58
pf59
pf5a
pf5b
pf5c
pf5d
pf5e
pf5f
pf60
pf61
pf62
pf63
pf64

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Chapter 1

Problems 1-1 through 1-6 are for student research. No standard solutions are provided.

1-7 From Fig. 1-2, cost of grinding to ± 0.0005 in is 270%. Cost of turning to ± 0.003 in is 60%. Relative cost of grinding vs. turning = 270/60 = 4.5 times Ans.


1-8 CA = CB ,

10 + 0.8 P = 60 + 0.8 P − 0.005 P^2

P^2 = 50/0.005 ⇒ P = 100 parts Ans.


1-9 Max. load = 1.10 P Min. area = (0.95)^2 A Min. strength = 0.85 S To offset the absolute uncertainties, the design factor, from Eq. (1-1) should be

2

n d = = Ans


1-10 ( a ) X 1 + X 2 :

1 2 1 1 2 2 1 2 1 2 1 2

error .

x x X e X e e x x X X e e Ans

( b ) X 1 − X 2 :

1 2 1 1 2 2 1 2 1 2 1 2.

x x X e X e e x x X X e e Ans

( c ) X 1 X 2 :

1 2 (^1 1 )(^2 2 )

1 2 1 2 1 2 2 1 1 2 1 2 1 2 2 1 1 2 1 2

x x X e X e e x x X X X e X e e e e e X e X e X X Ans X X

( d ) X 1 / X 2 : 1 1 1 1 1 1 2 2 2 2 2 2 1 2 2 1 1 1 2 1 2 2 2 2 2 1 2 1 2 1 1 1 1 2 2 2 2 1 2

1 1 then 1 1 1 1

Thus,.

x X e X e X x X e X e X

e e e X e e e e X X e X X X X X x X X e e e Ans x X X X X

+ ^ + 
 +^  ≈^ −^   ≈^  +^  −^ ≈^ +^ −
______________________________________________________________________________

1-11 ( a ) x 1 = 7 = 2.645 751 311 1 X 1 = 2.64 (3 correct digits) x 2 = 8 = 2.828 427 124 7 X 2 = 2.82 (3 correct digits) x 1 + x 2 = 5.474 178 435 8 e 1 = x 1 − X 1 = 0.005 751 311 1 e 2 = x 2 − X 2 = 0.008 427 124 7 e = e 1 + e 2 = 0.014 178 435 8 Sum = x 1 + x 2 = X 1 + X 2 + e = 2.64 + 2.82 + 0.014 178 435 8 = 5.474 178 435 8 Checks ( b ) X 1 = 2.65, X 2 = 2.83 (3 digit significant numbers) e 1 = x 1 − X 1 = − 0.004 248 688 9 e 2 = x 2 − X 2 = − 0.001 572 875 3 e = e 1 + e 2 = − 0.005 821 564 2 Sum = x 1 + x 2 = X 1 + X 2 + e = 2.65 +2.83 − 0.001 572 875 3 = 5.474 178 435 8 Checks


( ) (^ )

3 3

1.006 in. d^ 2.

S

d Ans n d

σ π

Table A-17: d = 1 14 in Ans.

Factor of safety:

( )

3

3

S

n Ans σ π

______________________________________________________________________________

2 + 1 + 3 + 5 + 8 + 12 = 31 (the data is not perfectly normal)


1-

x f f x f x^2 174 6 1044 181656 182 9 1638 298116 190 44 8360 1588400 198 67 13266 2626668 206 53 10918 2249108 214 12 2568 549552 222 6 1332 295704 Σ 197 39126 7789204

Eq. (1-6) 1

198.61 kpsi 197

k i i i

x f x N (^) =

= (^) ∑ = =

Eq. (1-7)

(^2 2) 1 2 2 1 7 789 204^ 197(198.61)^ 9.68 kpsi. 1 197 1

k i i i x

f x N x s Ans N

=

______________________________________________________________________________

1-15 L = 122.9 kcycles and sL =30.3 kcycles

Eq. (1-5) 10 10 10

x L

x x L x z s

μ σ

Thus, x 10 = 122.9 + 30.3 z 10 = L 10

From Table A-10, for 10 percent failure, z 10 = −1.282. Thus,

L 10 = 122.9 + 30.3(−1.282) = 84.1 kcycles Ans.


x f fx fx^2 93 19 1767 164331 95 25 2375 225625 97 38 3686 357542 99 17 1683 166617 101 12 1212 122412 103 10 1030 106090 105 5 525 55125 107 4 428 45796 109 4 436 47524 111 2 222 24642 Σ 136 13364 1315704

Eq. (1-6) 1

13 364 / 136 98.26471 = 98.26 kpsi

k i i i

x f x N (^) =

= (^) ∑ = =

Eq. (1-7)

(^2 2) 1 2 2 1 1 315 704^ 136(98.26471)^ 4.30 kpsi 1 136 1

k i i i x

f x N x s N

=

Note , for accuracy in the calculation given above, x needs to be of more significant figures than the rounded value.

For a normal distribution, from Eq. (1-5), and a yield strength exceeded by 99 percent ( R = 0.99, pf = 0.01), 0.01 0.

x x x

x x x x z s

μ σ

Solving for the yield strength gives

x 0.01 = 98.26 + 4.30 z 0.

From Table A-10, z 0.01 = − 2.326. Thus

x 0.01 = 98.26 + 4.30(− 2.326) = 88.3 kpsi Ans. ______________________________________________________________________________

1-17 Eq. (1-9): R = 1

n i i

R

=

∏ = 0.98(0.96)0.94 = 0. Overall reliability = 88 percent Ans.


Eq. (1-11):

(^2 2 2 2 2 )

d d S

n z n C C σ

Interpolating Table A-10, 1.61 0. 1.6127 Φ ⇒ Φ = 0. 1.62 0.

R = 1 − 0.0534 = 0.9466 Ans.

2 2

1.020 in. / ( / 4) 4 95.

y y y y

S S d S (^) Pn n d Ans P d P S

π σ π π π

( b ) n =1.

2 2 2

z

R = 1 − 0.00015645 = 0.9998 Ans.

1.140 in. y^ 95.

Pn d Ans π S π

______________________________________________________________________________

1-20 μ σ (^) max = σ (^) max = σ (^) a + σ b = 90 + 383 =473 MPa

From footnote 9, p. 25 of text,

max^ (^ )

2 2 1/ 2 2 2 1/ 2 σ^ ˆ (^) σ = σˆ^ σ a + σˆ^ σ b = (8.4 + 22.3 ) =23.83 MPa

max max max max max

C σ σ σ σ

σ σ μ σ

y y y y

S S S S y

C
S

σ σ μ

max

n S^ y Ans σ

Eq. (1-11): ( )

(^2 2 2 2 2 )

d d S

n z n C C σ

From Table A-10, Φ(− 1.635) = 0.

R = 1 − 0.05105 = 0.94895 = 94.9 percent Ans.


1-21 a = 1.500 ± 0.001 in b = 2.000 ± 0.003 in c = 3.000 ± 0.004 in d = 6.520 ± 0.010 in ( a ) w = dabc = 6.520 − 1.5 − 2 − 3 = 0.020 in t (^) w = (^) ∑ t all= 0.001 + 0.003 + 0.004 +0.010 = 0. w = 0.020 ± 0.018 in Ans.

( b ) From part (a), w min = 0.002 in. Thus, must add 0.008 in to d. Therefore,

d = 6.520 + 0.008 = 6.528 in Ans.


1-22 V = xyz , and x = a ± ∆ a , y = b ± ∆ b , z = c ± ∆ c ,

V = abc

V ( a a ) ( b b )( c c )

abc bc a ac b ab c a b c b c a c a b a b c

The higher order terms in ∆ are negligible. Thus,

Vbca + acb + abc

and,.

V bc a ac b ab c a b c a b c Ans V abc a b c a b c

For the numerical values given, V = 1.500 1.875 3.000( ) =8.4375 in^3

1-26 From O-Rings, Inc. (oringsusa.com), Di = 34.52 ± 0.30 mm, d = 3.53 ± 0.10 mm

Do = Di + 2 d = 34.52 + 2 3.53 ( ) =41.58 mm

tD (^) o = (^) ∑ t all = 0.30 + 2 0.10 ( )=0.50 mm

Do = 41.58 ± 0.50 mm Ans.


1-27 From O-Rings, Inc. (oringsusa.com), Di = 5.237 ± 0.035 in, d = 0.103 ± 0.003 in

Do = Di + 2 d = 5.237 + 2 0.103 ( ) =5.443 in

tD (^) o = (^) ∑ t all = 0.035 + 2 0.003 ( ) =0.041 in

Do = 5.443 ± 0.041 in Ans.


1-28 From O-Rings, Inc. (oringsusa.com), Di = 1.100 ± 0.012 in, d = 0.210 ± 0.005 in

Do = Di + 2 d = 1.100 + 2 0.210 ( ) =1.520 in

tD (^) o = (^) ∑ t all = 0.012 + 2 0.005 ( ) =0.022 in

Do = 1.520 ± 0.022 in Ans.


1-29 From Table A-2,

( a ) σ = 150/6.89 = 21.8 kpsi Ans.

( b ) F = 2 /4.45 = 0.449 kip = 449 lbf Ans.

( c ) M = 150/0.113 = 1330 lbf ⋅ in = 1.33 kip ⋅ in Ans.

( d ) A = 1500/ 25.4^2 = 2.33 in^2 Ans.

( e ) I = 750/2.54^4 = 18.0 in^4 Ans.

( f ) E = 145/6.89 = 21.0 Mpsi Ans.

( g ) v = 75/1.61 = 46.6 mi/h Ans.

( h ) V = 1000/946 = 1.06 qt Ans.


1-30 From Table A-2,

( a ) l = 5(0.305) = 1.53 m Ans.

( b ) σ = 90(6.89) = 620 MPa Ans.

( c ) p = 25(6.89) = 172 kPa Ans.

( d ) Z =12(16.4) = 197 cm^3 Ans.

( e ) w = 0.208(175) = 36.4 N/m Ans.

( f ) δ = 0.001 89(25.4) = 0.048 0 mm Ans.

( g ) v = 1 200(0.0051) = 6.12 m/s Ans.

( h ) ∫ = 0.002 15(1) = 0.002 15 mm/mm Ans.

( i ) V = 1830(25.4^3 ) = 30.0 (10^6 ) mm^3 Ans.


1- ( a ) σ = M /Z = 1770/0.934 = 1895 psi = 1.90 kpsi Ans.

( b ) σ = F /A = 9440/23.8 = 397 psi Ans.

( c ) y =Fl^3 / 3 EI = 270(31.5)^3 /[3(30)10^6 (0.154)] = 0.609 in Ans.

( d ) θ = Tl /GJ = 9 740(9.85)/[11.3(10^6 )( π /32)1.00^4 ] = 8.648(10−^2 ) rad = 4.95° Ans.


1- ( a ) σ =F / wt = 1000/[25(5)] = 8 MPa Ans.

( b ) I = bh^3 /12 = 10(25)^3 /12 = 13.0(10^3 ) mm^4 Ans.

( c ) I = π d^4 /64 = π (25.4)^4 /64 = 20.4(10^3 ) mm^4 Ans.

( d ) τ = 16 T / π d^3 = 16(25)10^3 /[ π (12.7)^3 ] = 62.2 MPa Ans.


1-

Chapter 2

2-1 From Tables A-20, A-21, A-22, and A-24c, ( a ) UNS G10200 HR: Sut = 380 (55) MPa (kpsi), Syt = 210 (30) MPa (kpsi) Ans. ( b ) SAE 1050 CD: Sut = 690 (100) MPa (kpsi), Syt = 580 (84) MPa (kpsi) Ans. ( c ) AISI 1141 Q&T at 540°C (1000°F): Sut = 896 (130) MPa (kpsi), Syt = 765 (111) MPa (kpsi) Ans. ( d ) 2024-T4: Sut = 446 (64.8) MPa (kpsi), Syt = 296 (43.0) MPa (kpsi) Ans. ( e ) Ti-6Al-4V annealed: Sut = 900 (130) MPa (kpsi), Syt = 830 (120) MPa (kpsi) Ans.


2-2 ( a ) Maximize yield strength: Q&T at 425°C (800°F) Ans.

( b ) Maximize elongation: Q&T at 650°C (1200°F) Ans.


2-3 Conversion of kN/m^3 to kg/ m^3 multiply by 1(10^3 ) / 9.81 = 102 AISI 1018 CD steel: Tables A-20 and A-

370 10 3 47.4 kN m/kg. 76.5 102

S (^) y Ans ρ

= = ⋅

2011-T6 aluminum: Tables A-22 and A-

169 10^3 62.3 kN m/kg. 26.6 102

S (^) y Ans ρ

= = ⋅

Ti-6Al-4V titanium: Tables A-24c and A-

830 10^3 187 kN m/kg. 43.4 102

S (^) y Ans ρ

= = ⋅

ASTM No. 40 cast iron: Tables A-24a and A-5.Does not have a yield strength. Using the ultimate strength in tension

42.5 6.89 103 40.7 kN m/kg 70.6 102

S ut (^) Ans ρ

= = ⋅


2- AISI 1018 CD steel: Table A-

6 30.0 10 (^) 106 10 (^6) in.

E

Ans γ

2011-T6 aluminum: Table A-

6 10.4 10 (^) 106 10 (^6) in.

E

Ans γ

Ti-6Al-6V titanium: Table A-

6 16.5 10 (^) 103 10 (^6) in.

E

Ans γ

No. 40 cast iron: Table A-

6 14.5 10 (^) 55.8 10 (^6) in.

E

Ans γ

______________________________________________________________________________
E G

G v E v G

Using values for E and G from Table A-5,

Steel:

v Ans

The percent difference from the value in Table A-5 is

0.304 0. 0.0411 4.11 percent.

Ans

Aluminum:

v Ans

The percent difference from the value in Table A-5 is 0 percent Ans.

Beryllium copper:

v Ans

The percent difference from the value in Table A-5 is

0.286 0. 0.00351 0.351 percent.

Ans

Gray cast iron:

v Ans

The percent difference from the value in Table A-5 is

0.208 0. 0.0142 1.42 percent.

Ans

______________________________________________________________________________

2-6 ( a ) A 0 = π (0.503)^2 /4 = 0.1987 in^2 , σ = Pi / A 0

( a ) Linear range

( b ) Offset yield

( c ) Complete range

y = 3.05E+07x - 1.06E+

0

10000

20000

30000

40000

50000

0.000 0.001 0.001 0.

Stress (psi)

Strain

Series Linear (Series1)

0

5000

10000

15000

20000

25000

30000

35000

40000

45000

50000

0.000 0.002 0.004 0.006 0.008 0.010 0.012 0.

Stress (psi)

Strain

Y

0

10000

20000

30000

40000

50000

60000

70000

80000

90000

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.

Stress (psi)

Strain

U

( c ) The material is ductile since there is a large amount of deformation beyond yield.

( d ) The closest material to the values of Sy, Sut, and R is SAE 1045 HR with Sy = 45 kpsi, Sut = 82 kpsi, and R = 40 %. Ans.


2-7 To plot σ (^) true vs. ε, the following equations are applied to the data.

true

P
A

σ =

Eq. (2-4)

0 0

ln for 0 0.0028 in (0 8 400 lbf )

ln for 0.0028 in ( 8 400 lbf )

l l P l A l P A

ε

ε

where

2 2 0

0.1987 in 4

A

The results are summarized in the table below and plotted on the next page. The last 5 points of data are used to plot log σ vs log ε

The curve fit gives m = 0. log σ 0 = 5.1852 ⇒ σ 0 = 153.2 kpsi Ans.

For 20% cold work, Eq. (2-14) and Eq. (2-17) give,

A = A 0 (1 – W ) = 0.1987 (1 – 0.2) = 0.1590 in^2

0

0

ln ln 0.

Eq. (2-18): 153.2(0.2231) 108.4 kpsi. Eq. (2-19), with 85.6 from Prob. 2-6,

107 kpsi. 1 1 0.

m y u u u

A
A

S Ans S S S Ans W

ε

′ σ ε

______________________________________________________________________________

2-8 Tangent modulus at σ = 0 is

6 3

25 10 psi 0.2 10 0

E

σ −

ò (^) −

Ans.

At σ = 20 kpsi

Shigley’s MED, 10th^ edition

20

E

∫ (10-^3 ) σ (kpsi) 0 0 0.20 5 0.44 10 0.80 16 1.0 19 1.5 26 2.0 32 2.8 40 3.4 46 4.0 49 5.0 54

______________________________________________________________________________
2-9 W = 0.20,

( a ) Before cold working: kpsi, σ 0 = 90.0 kpsi, m After cold working: Eq. ( Eq. (2-14), 0 i^1 1 0.

A
A W

Eq. (2-17), ε i = ln = ln1.25 = 0.223< ε u

Eq. (2-18), S (^) y ′^ = σ ε i = = Ans

Eq. (2-19), u 1 1 0. S ′^ = = = Ans − −

( b ) Before:

u y

S
S

Lost most of its ductility


2-10 W = 0.20, ( a ) Before cold working: AISI 1212 HR steel. Table A σ 0 = 110 kpsi, m = 0.24, After cold working: Eq. (

Chapter 2

3 6 3

14.0 10 psi 1.5 1 10 −

≈ = Ans.

______________________________________________________________________________

) Before cold working: Annealed AISI 1018 steel. Table A-22, Sy = 0.25, ∫ f = 1. After cold working: Eq. (2-16), ∫ u = m = 0. 1 1

A 1 W 1 0.

ln 0 ln1.25 0. i u i

A
A

ε = = = < ε

0 90 0.223^ 61.8 kpsi^.

m S (^) y = σ ε i = = Ans 93% increase

61.9 kpsi. 1 1 0.

S S^ u Ans W

25% increase

= = After:

u y

S
S

ductility.


) Before cold working: AISI 1212 HR steel. Table A-22, Sy = 28 kpsi, 0.24, ∫ f = 0. q. (2-16), ∫ u = m = 0.

Chapter 2 Solutions, Page 8/

______________________________________________________________________________

y = 32 kpsi,^ Su = 49.

93% increase Ans.

25% increase Ans.

Ans.

______________________________________________________________________________

= 28 kpsi, Su = 61.5 kpsi,