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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
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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
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
Chapter 1 Solutions - Rev. B, Page 1/
( 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 1 2 1
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 (^) 2
2
e
X X e X X X X
x X X e e e Ans x 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
0.799 in. d 2.
S
d A n d
ns
Table A-17: d = 7 8
in Ans.
Factor of safety:
3
3 7 8
S
n A
ns
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1-13 Eq. (1-5): R =
1
n
i i
R
Overall reliability = 88 percent Ans.
Chapter 1 Solutions - Rev. B, Page 2/
w max = 0.05 in, w min = 0.004 in
0.05 0. 0.027 in 2
w =
Thus, w = 0.05 0.027 = 0.023 in, and then, w = 0.027 0.023 in.
1.569 in
a b c
a
a
w =
tw = 0.023 = t all ^ t a^ + 0.002 + 0.005^ ^ ta^ = 0.016 in
Thus, a = 1.569 0.016 in Ans.
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1-17 Do Di 2 d 3.734 2 0.139 4.012 in
tD (^) o (^) t all 0.028 2 0.004 0.036 in
Do = 4.012 0.036 in Ans.
1-18 From O-Rings, Inc. (oringsusa.com), Di = 9.19 0.13 mm, d = 2.62 0.08 mm
Do Di 2 d 9.19 2 2.62 14.43 mm
tD (^) o (^) t all 0.13 2 0.08 0.29 mm
Do = 14.43 0.29 mm Ans.
1-19 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.
Chapter 1 Solutions - Rev. B, Page 4/
1-20 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.
1-21 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.
1-22 From Table A-2,
( a ) = 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.
2 = 2.33 in
2 Ans.
( e ) I = 750/2.
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.
1-23 From Table A-2,
( a ) l = 5(0.305) = 1.53 m Ans.
( b ) = 90(6.89) = 620 MPa Ans.
( c ) p = 25(6.89) = 172 kPa Ans.
Chapter 1 Solutions - Rev. B, Page 5/
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 540C (1000F): 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 425C (800F) Ans.
( b )Maximize elongation: Q&T at 650C (1200F) Ans.
2-3 Conversion of kN/m
3 to kg/ m
3 multiply by 1(
3 ) / 9.81 = 102
AISI 1018 CD steel: Tables A-20 and A-
3 370 10 47.4 kN m/kg. 76.5 102
S (^) y Ans
2011-T6 aluminum: Tables A-22 and A-
3 169 10 62.3 kN m/kg. 26.6 102
S (^) y Ans
Ti-6Al-4V titanium: Tables A-24c and A-
3 830 10 187 kN m/kg. 43.4 102
S (^) y Ans
ASTM No. 40 cast iron: Tables A-24a and A-5.Does not have a yield strength. Using the
ultimate strength in tension
3 42.5 6.89 10 40.7 kN m/kg 70.6 102
S ut (^) Ans
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AISI 1018 CD steel: Table A-
6 6
106 10 in.
E
Ans
2011-T6 aluminum: Table A-
6 10.4 10 (^6) 106 10 in.
E
Ans
Ti-6Al-6V titanium: Table A-
6 6
103 10 in.
E
Ans
No. 40 cast iron: Table A-
6 14.5 10 (^6) 55.8 10 in.
E
Ans
______________________________________________________________________________
E G
G v E v G
From Table A-
Steel:
v A
ns
Aluminum:
v A
ns
Beryllium copper:
v A
ns
Gray cast iron:
v A
ns
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2-6 ( a ) A 0 = (0.503)
2
/4, = Pi / A 0
For data in elastic range, = l / l 0 = l / 2
For data in plastic range, 0 0 0 0 0
l l l l A
l l l A
On the next two pages, the data and plots are presented. Figure ( a ) shows the linear part of
the curve from data points 1-7. Figure ( b ) shows data points 1-12. Figure ( c ) shows the
complete range. Note : The exact value of A 0 is used without rounding off.
( b ) From Fig. ( a ) the slope of the line from a linear regression is E = 30.5 Mpsi Ans.
From Fig. ( b ) the equation for the dotted offset line is found to be
6
The equation for the line between data points 8 and 9 is
5
( b ) Offset yield
( c ) Complete range
( 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 true vs. , the following equations are applied to the data.
true
P
A
Eq. (2-4)
0
0
ln for 0 0.0028 in
ln for 0.0028 in
l l l A l A
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 vs log
The curve fit gives m = 0.
log 0 = 5.1852 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 A
S
S
S Ans W
ns
P L A true log log true
3 6 (^20 )
14.0 10 psi 1.5 1 10
E
Ans.
)^ ^ (kpsi)
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2-9 W = 0.20,
( a ) Before cold working: Annealed AISI 1018 steel. Table A-22, Sy = 32 kpsi, Su = 49.
kpsi, 0 = 90.0 kpsi, m = 0.25, f = 1.
After cold working: Eq. (2-16), u = m = 0.
Eq. (2-14),
i^1 1 0.
A
A W
Eq. (2-17), (^) i ln 0 ln1.25 0.223 u i
A
A
Eq. (2-18), S 93% increase Ans.
0 90 0.223^ 61.8 kpsi^.
m
y^ i
Ans
Eq. (2-19),
61.9 kpsi. 1 1 0.
u u
S
S A
W
ns 25% increase Ans.
( b ) Before:
u
y
S
S
After:
u
y
S
S
Ans.
Lost most of its ductility
2-10 W = 0.20,
( a ) Before cold working: AISI 1212 HR steel. Table A-22, Sy = 28 kpsi, Su = 61.5 kpsi,
0 = 110 kpsi, m = 0.24, f = 0.
After cold working: Eq. (2-16), u = m = 0.
Eq. (2-14),
i^1 1 0.
A
A W
Eq. (2-17), (^) i ln 0 ln1.25 0.223 u i
A
A
Eq. (2-18), 174% increase Ans.
0 110 0.223^ 76.7 kpsi^.
m
S y i A
ns
Eq. (2-19),
76.9 kpsi. 1 1 0.
u u
S
S A
W
ns 25% increase Ans.
( b ) Before:
u
y
S
S
After:
u
y
S
S
Ans.
Lost most of its ductility
2-11 W = 0.20,
( a ) Before cold working: 2024-T4 aluminum alloy. Table A-22, Sy = 43.0 kpsi, Su =
64.8 kpsi, 0 = 100 kpsi, m = 0.15, f = 0.
After cold working: Eq. (2-16), u = m = 0.
Eq. (2-14), 0
i^1 1 0.
A
A W
Eq. (2-17), (^) i ln 0 ln1.25 0. i
A
A
f Material fractures. Ans.
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2-12 For HB = 275, Eq. (2-21), Su = 3.4(275) = 935 MPa Ans.
2-13 Gray cast iron, HB = 200.
Eq. (2-22), Su = 0.23(200) 12.5 = 33.5 kpsi Ans.
From Table A-24, this is probably ASTM No. 30 Gray cast iron Ans.
2-14 Eq. (2-21), 0.5 HB = 100 HB = 200 Ans.
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41.4 kpsi. 10
u u
S
S A
N
ns
Eq. (20-8),
10 2 2 2 1 17152.63^ 10 41.^
1.. u 1 9
u u i S
S NS
s A N
ns
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2-17 ( a )
2 45.5 (^3) 34.5 in lbf / in. 2(30)
u (^) R Ans
( b )
P L A A 0 / A – 1^ ^ = P / A 0
From the figures on the next page,
5
1
3 3
66.7 10 in lbf/in.
T i i
u A
Ans
A-20, perform a Brinell hardness test, then use Eq. (2-21) to estimate an ultimate strength
of. Assuming the material is hot-rolled due to the
rough surface finish, appropriate choices from Table A-20 would be one of the higher carbon steels, such as hot-rolled AISI 1050, 1060, or 1080. Ans.
Su 0.5 HB 0.5(200) 100 kpsi
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2-22 First, try to find the broad category of material (such as in Table A-5). Visual, magnetic,
and scratch tests are fast and inexpensive, so should all be done. Results from these three favor a softer, non-ferrous material like aluminum. If it is desired to confirm this, either a
weight or bending test could be done to check density or modulus of elasticity. The
weight test is faster. From the measured weight of 2.90 lbf, the unit weight is determined
to be
3 3 2
2.9 lbf 0.103 lbf/in 0.10 lbf/in [ (1 in) / 4](36 in)
W
Al
w
which agrees reasonably well with the unit weight of 0.098 lbf/in
3 reported in Table A-
for aluminum. No other materials come close to this unit weight, so the material is likely
aluminum. Ans.
2-23 First, try to find the broad category of material (such as in Table A-5). Visual, magnetic,
and scratch tests are fast and inexpensive, so should all be done. Results from these three
favor a softer, non-ferrous copper-based material such as copper, brass, or bronze. To further distinguish the material, either a weight or bending test could be done to check
density or modulus of elasticity. The weight test is faster. From the measured weight of
9 lbf, the unit weight is determined to be
3 3 2
9.0 lbf 0.318 lbf/in 0.32 lbf/in [ (1 in) / 4](36 in)
W
Al
w
which agrees reasonably well with the unit weight of 0.322 lbf/in
3 reported in Table A-
for copper. Brass is not far off (0.309 lbf/in
3 ), so the deflection test could be used to gain
additional insight. From the measured deflection and utilizing the deflection equation for an end-loaded cantilever beam from Table A-9, Young’s modulus is determined to be
3 3
4
17.7 Mpsi (^3 3) (1) 64 (17 / 32)
Fl E
Iy
which agrees better with the modulus for copper (17.2 Mpsi) than with brass (15.4 Mpsi).
The conclusion is that the material is likely copper. Ans.
2-24 and 2-25 These problems are for student research. No standard solutions are provided.
2-26 For strength, = F/A = S A = F/S
For mass, m = Al = ( F/S ) l
Thus, f 3 ( M ) = /S , and maximize S/ ( = 1)
In Fig. (2-19), draw lines parallel to S/
From the list of materials given, both aluminum alloy and high carbon heat treated
steel are good candidates, having greater potential than tungsten carbide or polycarbonate. The higher strength aluminum alloys have a slightly greater potential. Other factors, such
as cost or availability, may dictate which to choose. Ans.
2-27 For stiffness, k = AE/l A = kl/E
For mass, m = Al = ( kl/E ) l = kl
2
/E
Thus, f 3 ( M ) = /E , and maximize E/ ( = 1)
In Fig. (2-16), draw lines parallel to E/
