Answers to Selected Problems Appendix
B
1013
B–1 Chapter 1
1–8
10 ≤ F ≤ 10.5 lbf
,
13.5 ≤ N ≤ 14.2 lbf,
K =
0.967,
19.3 ≤ T ≤ 20.3 lbf · in
1–12 (a)
e
1
= 0.006 067 977
,
e
2
= 0.009 489 743
,
e = 0.015 557 720
, (b)
e
1
=−0.003 932 023
,
e
2
=−0.000 510 257
,
e =−0.004 442 280

¯
F
i
= 5.979 lbf
,
s
Fi
= 0.396 lbf
;
(b)
¯
k = 9.766
lbf/in,
s
k
= 0.390
lbf/in
2–19
L
10
= 84.1kcycles
2–23
R = 0.987
2–25 (a)
w = 0.020 ± 0.015 in
, (b)
w = 0.020 ±
0.005 in
2–32
¯

,
area reduction = 45.8 percent
3–11
(S
y
)
0.001
.
= 35 kpsi, E
σ=0
= 25 Mpsi,
E
σ=20 kpsi
= 14 Mpsi
3–14
G = 77.3GPa, S
ys
.
= 200 MPa
3–18
¯
S
ut
= 125.2 kpsi, ˆσ
S
ut
= 1.9 kpsi
3–20 (a)
u
R

71.43x − 14
1
− 60x − 18
1
lbf · in
4–6 (a)
M
max
= 253 lbf · in,
(b)
(a/l)
∗
= 0.207,
M
∗
= 214 lbf · in
4–8 (a)
σ
1
= 14,σ
2
= 4,σ
3
= 0, 2θ = 53.1
◦
cw;
(b)
σ
1
= 18.6,σ

ν = 0.292,
2
=−0.000 099 1,d =
−0.000 049 6 in
4–18
σ
1
= 30 MPa,σ
2
= 10 MPa,σ
3
=−20 MPa,
τ
max
= 25 MPa
4–22 (a)
M
max
= 21 600 kip · in,
(b)
x
max
= 523
in
from left or right supports
4–23 (a)
σ
A
= 42 kpsi,σ
B

max
= 59.90 lbf · in,θ= 0.0960 rad,
k
t
= 624 lbf · in/rad
4–43
d
C
= 45 mm
4–47
σ
max
= 11.79 kpsi,τ
max
= 7.05 kpsi
4–53
p
i
= 639 psi
4–57
(σ
r
)
max
= 3656 psi
4–65
δ
max
= 0.038 mm,δ
min

=−15.8 kpsi
shi20361_app_B.qxd 6/11/03 12:32 PM Page 1013
1014 Mechanical Engineering Design
4–77
σ
i
= 71.3 kpsi,σ
o
=−34.2 kpsi
4–81
p
max
= 399F
1/3
MPa,σ
max
= 399F
1/3
MPa,
τ
max
= 120F
1/3
MPa
B–5 Chapter 5
5–1 (a)
k = (1/k
1
+ 1/k
2

y =−0.908 in
5–15
y
left
=−0.0506 in, y
right
=−0.0506 in
,
y
midspan
= 0.0190 in
5–18
y
max
=−0.0130 in
5–20
z
A
= 0.0368 in, z
B
= 0.00430 in
5–26 Use
d = 1
3
8
in
5–30
y
B
= 0.0459 in

c
=−13.3 kpsi
5–56
R
O
= 3.89 kip, R
C
= 1.11 kip,
both in
same direction
5–59
σ
BE
= 140 MPa,σ
DF
= 71.2MPa,
y
B
=−0.670 mm, y
C
=−2.27 mm,
y
D
=−0.341 mm
5–64
δ
A
= (π + 4)PR
3
/(4EI), δ

DE:
n = 2.41,
(c) MSS:
n = 4.17,
DE:
n = 4.81
6–3 (a) MSS:
n = 2.17,
DE:
n = 2.50,
(b) MSS:
n = 1.45,
DE:
n = 1.56,
(c) MSS:
n = 1.52,
DE:
n = 1.65,
(c) MSS:
n = 1.27,
DE:
n = 1.50
6–9 (a) DE:
σ

= 12.29 kpsi, n = 3.42
6–10 (a) DCM:
σ
1
= 90 kpsi,σ

3
8
in
6–25
d = 18 mm
6–32 (a)
δ = 0.0005 in, p = 3516
psi,
(σ
t
)
i
=−5860 psi,(σ
r
)
i
=−3516 psi,
(σ
t
)
o
=−9142 psi,(σ
r
)
o
=−3516 psi
6–35
n
o
= 2.81, n

−0.0851
kpsi, 10
3
≤
N ≤ 10
6
cycles
7–6
S
e
= 243 MPa
7–10
S

e
= 221.8MPa
,
k
a
= 0.899
,
k
b
= 1
,
k
c
= 0.85
,
S

= 1.32
7–17
n
y
= 5.06,
(a)
n
f
= 2.44
, (b)
n
f
= 2.55
7–23 At the fillet
n
f
= 1.70
7–24 (a)
T = 3.42 N · m,
(b)
T = 4.21 N · m,
(c)
n
y
= 1.91
7–27 (a)
P
all
= 16.1kN, n
y

LN(1, 0.15) kpsi,
z =−2.373
,
R = 0.991
B–8 Chapter 8
8–1 (a) Thread depth 2.5 mm, thread width 2.5 mm,
d
m
= 22.5mm, d
r
= 20
mm,
l = p = 5
mm
8–4
T
R
= 16.23 N · m, T
L
= 6.62 N · m, e = 0.294
8–8
T = 16.5 lbf · in, d
m
= 0.5417 in,
l = 0.1667 in,
sec
α = 1.033,
T = 0.0696F
,
T

l
d
= 0.500 in, l
t
= 0.625 in
8–15 (a)
A
d
= 0.442 in
2
,
A
tube
= 0.552 in
2
,
k
b
= 1.02(10
6
) lbf/in, k
m
= 1.27(10
6
) lbf/in,
C = 0.445
, (b)
F
i
= 11 810 lbf

n = 3.26
. Bolt bearing:
n = 5.99
.
Member bearing:
n = 3.71
. Member tension:
n = 5.36
8–48
F = 2.22 kN
8–50 Bearing on bolt,
n = 9.58;
shear of bolt,
n = 5.79;
bearing on members,
n = 5.63;
bending of members,
n = 2.95
B–9 Chapter 9
9–1
F = 17.7 kip
9–3
F = 11.3 kip
9–5 (a)
τ

= 1.13F kpsi,τ

x
= τ

4
in
9–20
τ
max
= 18
kpsi
9–22
n = 3.57
B–10 Chapter 10
10–3 (a)
L
0
= 5.17
in, (b)
F
S
sy
= 45.2
lbf,
(c)
k = 11.55
lbf/in, (d)
(L
0
)
cr
= 5.89
in, guide
spring

p = 10
mm,
L
s
= 44.2
mm,
N
a
= 12 turns
,
(b)
k = 1080
N/m, (c)
F
s
= 81.9
N, (d)
τ
s
=
271
MPa
10–29 (a)
L
0
= 16.12
in, (b)
τ
i
= 14.95

C
10
= 18.59
kN,
02–30 mm deep-groove ball bearing,
R = 0.919
11–4
R = R
1
R
2
= 0.927(0.942) = 0.873
, goal not met
11–8
x
D
= 180
,
C
10
= 57.0
kN
11–11
C
10
= 8.88
kN
11–13
R
O

/c = 0.595
,
rf/c = 5.8
,
Q/(rcNl) = 3.98
,
Q
s
/Q = 0.5
,
h
0
= 0.000 446
in,
H = 0.0134
Btu/s,
Q = 0.0274
in
3
/s,
Q
s
= 0.0137
in
3
/s
12–3 SAE 10:
h
0
= 0.000 275

3
/s,
Q
s
= 793
mm
3
/s
12–11
T
av
= 65
◦
C
,
h
0
= 0.0272 mm, H = 45.2W,
Q
s
= 1712
mm
3
/s
12–20 15.2 mPa
·
s
shi20361_app_B.qxd 6/11/03 12:32 PM Page 1015
1016 Mechanical Engineering Design
B–13 Chapter 13

d
2b
= 8.77
in,
p
b
= 0.984
in
13–5
d
P
= 2.333
in,
d
G
= 5.333
in,
γ = 23.63
◦
,
 = 66.37
◦
,
A
0
= 2.911
in,
F = 0.873
in
13–8 (a) 13, (b) 15, (c) 18, (d) 16

mm
13–15
e = 4/51
,
n
d
= 47.06
rev/min cw
13–22
n
A
= 68.57
rev/min cw
13–27
n
b
/n
a
= 11/36
same sense
13–33
F
A
= 71.5 i + 53.4 j + 350.5 k lbf,
F
B
=−149.5 i − 590.4 k lbf
13–40
F
C

kW
(pinion and gear wear)
14–18
W
t
= 1356
lbf,
H = 34.1
hp (pinion bending);
W
t
= 1720
lbf,
H = 43.3
hp (gear bending),
W
t
= 265 lbf; H = 6.67
hp (pinion and gear wear)
14–22
W
t
= 775
lbf,
H = 19.5
hp (pinion bending);
W
t
= 300
lbf,

t
P
= 464
lbf,
H
3
= 11.0
hp,
W
t
G
= 531
lbf,
H
4
= 12.6
hp
15–8 Pinion core 300 Bhn, case, 373 Bhn; gear core
339 Bhn, case, 345 Bhn
15–9 All four
W
t
= 690
lbf
15–11 Pinion core 180 Bhn, case, 266 Bhn; gear core,
180 Bhn, case, 266 Bhn
B–16 Chapter 16
16–1 (a) Right shoe:
p
a

y
= 171
lbf,
R = 215
lbf
16–3 LH shoe:
T = 161.4
N
·
m,
p
a
= 610
kPa; RH
shoe:
T = 59.0
N
·
m,
p
a
= 222.8
kPa,
T
total
=
220.4 N
·
m
16–5

(b)
R = 901
lbf; (c)
p|
θ=0
= 70
psi,
p|
θ=270
◦
= 27.3
psi
16–17 (a)
F = 1885
lbf,
T = 7125
lbf
·
in;
(c) torque capacity exhibits a stationary point
maximum
16–18 (a)
d
∗
= D/
√
3
; (b)
d
∗

= 0.08
,
t = 5.30
in
16–26 (b)
I
e
= I
M
+ I
P
+ n
2
I
P
+ I
L
/n
2
;
(c)
I
e
= 10 + 1 + 10
2
(1) + 100/10
2
= 112
16–27 (c)
n

= 2.5
hp,
n
fs
= 1.0;
(c) 0.151 in
shi20361_app_B.qxd 6/11/03 12:32 PM Page 1016
Answers to Selected Problems 1017
17–3 A-3 polyamide belt,
b = 6
in,
F
c
= 77.4
lbf,
T =
10 946 lbf
·
in,
F
1
= 573.7
lbf,
F
2
= 117.6
lbf,
F
i
= 268.3

hp,
n
fs
= 1.1
17–7
R
x
= (F
1
+ F
2
){1 − 0.5[(D − d)/(2C)]
2
},
R
y
= (F
1
− F
2
)(D − d)/(2C)
. From Ex. 17–2,
R
y
= 1214.4
lbf,
R
x
= 34.6 lbf
17–14 With

H
a
= 7.91
hp; (b)
C = 18
in; (c)
T =
1164
lbf
·
in,
F = 744
lbf
17–27 Four-strand No. 60 chain,
N
1
= 17
teeth,
N
2
= 84
teeth, rounded
L/p = 134
,
n
fs
= 1.17
, life
15 000
h (pre-extreme)

d = 24 mm, D = 32 mm, r = 1.6mm
18–20 (a) Static:
d = 1.526
in; (b) DE-Gerber:
d = 1.929 in;
ASME-elliptic:
d = 1.927 in;
MSS-Soderberg:
d = 1.932 in;
DE-Goodman:
d = 2.008 in
18–24 Slope:
y

m
= y

; deflection:
y
m
= sy = y/2
;
moment:
M
m
= s
3
M = M/8;
Force:
F