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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
−
+ ^ +
+^ ≈^ −^ ≈^ +^ −^ ≈^ +^ −
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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
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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
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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 = d − a − b − c = 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,
∆ V ≈ bc ∆ a + ac ∆ b + ab ∆ c
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,
