Chapter 11
11-1
For the deep-groove 02-series ball bearing with R = 0.90, the design life x D , in multiples
of rating life, is
L
60D nD 60 25000 350
xD D
525 Ans.
LR
L10
106
The design radial load is
FD 1.2 2.5 3.0 kN
1/3
Eq. (11-6):
525
C10 3.0
1/1.4
Chapter 9
Figure for Probs.
9-1 to 9-4
9-1
Given, b = 50 mm, d = 50 mm, h = 5 mm, allow = 140 MPa.
F = 0.707 hl allow = 0.707(5)[2(50)](140)(103) = 49.5 kN Ans.
_
9-2
Given, b = 2 in, d = 2 in, h = 5/16 in, allow = 25 kpsi.
F = 0.707 hl allow = 0.707(5/16
Chapter 7
7-1
(a) DE-Gerber, Eq. (7-10):
A 4 K f M a 3 K fsTa 4 (2.2)(70) 3 (1.8)(45) 338.4 N m
2
2
2
2
B 4 K f M m 3 K fsTm 4 (2.2)(55) 3 (1.8)(35) 265.5 N m
2
2
2
6
8(2)(338.4) 2(265.5) 210 10
d
1 1
6
6
210 10 338.4 700 10
3
d = 25.85 (10 ) m = 25.8
Chapter 6
Eq. (2-21):
Eq. (6-8):
Table 6-2:
Eq. (6-19):
Sut 3.4 H B 3.4(300) 1020 MPa
Se 0.5Sut 0.5(1020) 510 MPa
a 1.58, b 0.085
b
ka aSut 1.58(1020) 0.085 0.877
Eq. (6-20):
6-1
kb 1.24d 0.107 1.24(10) 0.107 0.969
Se ka kb Se (0.877)(0.969)(510) 433 MPa
Chapter 4
For a torsion bar, k T = T/ = Fl/, and so = Fl/k T . For a cantilever, k l = F/ , = F/k l . For
the assembly, k = F/y, or, y = F/k = l +
Thus
F Fl 2 F
y
k
kT kl
Solving for k
kk
1
k 2
2l T
Ans.
l
1 kl l kT
kT kl
_
4-1
For a torsion bar, k T =
Chapter 2
2-1
From Tables A-20, A-21, A-22, and A-24c,
(a) UNS G10200 HR: S ut = 380 (55) MPa (kpsi), S yt = 210 (30) Mpa (kpsi) Ans.
(b) SAE 1050 CD: S ut = 690 (100) MPa (kpsi), S yt = 580 (84) Mpa (kpsi) Ans.
(c) AISI 1141 Q&T at 540C (1000F): S ut = 8
Chapter 1
Problems 1-1 through 1-6 are for student research. No standard solutions are provided.
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-
Chapter 17
17-1
Given: F-1 Polyamide, b = 6 in, d = 2 in with n = 1750 rev/min, H nom = 2 hp, C = 9(12) =
108 in, velocity ratio = 0.5, K s = 1.25, n d = 1
V = d n / 12 = (2)(1750) / 12 = 916.3 ft/min
Eq. (17-1):
D = d / vel ratio = 2 / 0.5 = 4 in
42
Dd
Chapter 16
16-1 Given: r = 300/2 = 150 mm, a = R = 125 mm, b = 40 mm, f = 0.28, F = 2.2 kN, 1 = 0, 2 = 120, and a = 90. From which, sin a = sin90 = 1. Eq. (16-2):
Mf 0.28 pa (0.040)(0.150) 120 0 sin (0.150 0.125 cos ) d 1 2.993 10 4 pa N m
Eq. (16-3):
MN
Chapter 14
d
14-1
N
22
3.667 in
P
6
Table 14-2:
Y = 0.331
dn (3.667)(1200)
Eq. (13-34): V
1152 ft/min
12
12
1200 1152
Eq. (14-4b): K v
1.96
1200
H
15
Eq. (13-35) : W t 33 000
33 000
429.7 lbf
V
1152
K vW t P 1.96(429.7)(6)
7633 psi 7.63 kpsi Ans.
Chapter 15
15-1
Given: Uncrowned, through-hardened 300 Brinell core and case, Grade 1, N C =
109 rev of pinion at R = 0.999, N P = 20 teeth, N G = 60 teeth, Q v = 6, P d = 6
teeth/in, shaft angle = 90, n p = 900 rev/min, J P = 0.249 and J G = 0.216 (Fig.
Chapter 13
d P 17 / 8 2.125 in
N
1120
dG 2 d P
2.125 4.375 in
N3
544
13-1
NG PdG 8 4.375 35 teeth
Ans.
C 2.125 4.375 / 2 3.25 in
Ans.
_
nG 1600 15 / 60 400 rev/min
p m 3 mm Ans.
13-2
Ans.
C 3 15 60 2 112.5 mm Ans.
_
NG 16 4 64 teeth
13-3
Ans.
dG NG m 64
Chapter 12
12-1
Given: d max = 25 mm, b min = 25.03 mm, l/d = 1/2, W = 1.2 kN, = 55 mPas, and N =
1100 rev/min.
b d max 25.03 25
0.015 mm
cmin min
2
2
r 25/2 = 12.5 mm
r/c = 12.5/0.015 = 833.3
N = 1100/60 = 18.33 rev/s
P = W/ (ld) = 1200/ [12.5(25)] = 3.
Chapter 10
10-1
From Eqs. (10-4) and (10-5)
KW K B
4C 1 0.615 4C 2
4C 4
C
4C 3
Plot 100(K W K B )/ K W vs. C for 4 C 12 obtaining
We see the maximum and minimum occur at C = 4 and 12 respectively where
Maximum = 1.36 % Ans.,
and Minimum = 0.743 % Ans.
_
Chapter 8
Note to the Instructor for Probs. 8-41 to 8-44. These problems, as well as many others in this
chapter are best implemented using a spreadsheet.
8-1
(a) Thread depth= 2.5 mm Ans.
Width = 2.5 mm Ans.
d m = 25 - 1.25 - 1.25 = 22.5 mm
d r = 25 - 5