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8. Phillips Screw Company, www.phillips-screw.com
Developer of the Phillips'" screwdriver Manufacturer of
related fasteners for the aerospace, automotive,
construction, and industrial markets.
9. SPS Technologies, Inc. www.spstech.coin/unbrako
Manufac

502
Chapter 11 Keys, Couplings, and Seals
FIGURE 11-5
Details for proposed
design of key and
keyseats
Top view of keyseat
in shaft
a
0.30 in
1.00 in
I.SO in
1.7.3 in-
T
i/:xi/2X IV: key
Ring groove
0.25 in = HI2
Shaft section at keyseat
Hub section at k

618
Chapter 14 Rolling Contact Bearings
FIGURE 14-14
Locknut and
lockwasher for
retaining bearings
(SKF USA, Inc.,
Norristown, PA)
Bearing
Lockwasher
roove
shaft
X Thread for
locknut
B + L ->r T
Shaft
Locknut
Lockwasher
(c)
on the bearings. It is desirabl

316
Chapter 8 Kinematics of Gears
O
Center distance (C): The distance from the center of the pinion to the center ofthe
gear; the sum of the pitch radii of two gears in mesh. That is, because radius =
diameter/2,
Center Distance
C = Da/2 + Dp/2 = cfw_Da +

Section 9-1 Section Title
443
3. American Gear Manufacturers Association, Standard
1106A97. Tooth Proportions for Pla.stic Gears. Washington, DC: American Gear Manufacturers Association.
6. American Gear Manufacturers Association. Standard
2001-C95. Funda

708
Chapter 17 Linear Motion Elements
phenomenon of critical speed at which the screw would tend to vibrate or whiri about its
axis, possibly reaching dangerous amplitudes. Therefore, it is recommended that the operating speed of the screw be below 0.80 t

Section 8-13 Devising Gear Trains
355
FIGURE 8-31
Triple-reduction gear
train
->
C
F
4
Output
shaft
B
/)
1
Input
shaft
3
A
E
Factors of 320
One method is to divide by the smallest prime numbers that will divide evenly into the given
number, typically 2, 3

Problems
259
.S.O in
Machine
a
L(Kld
Platform
Wall
4x4x1/2
Steel tube
18.5 ft
lO.-SOft
Floor
FIGURE P6-33
terial has a yield strength of 36 ksi. The total load on the
platform is 55 000 lb. uniformly distributed.
34. Compute the allowable axial load on a

Machine Frames, Bolted
Connections, and Welded Joints
The Big Picture
You Are the Designer
20-1
Objectives of This Chapter
20-2
Machine Frames and Structures
20-3
Eccentrically Loaded Bolted Joints
20-4
Welded Joints
773

Section 5-11 Design Examples
201
real situations, you must ensure that each design decision is compatible with the totality ofthe design.
Design Example 5-1
Solution
Objective
A large electrical transformer is to be suspended from a roof truss of a buildi

536
Chapter 12 Shaft Design
(12-2)
W, = T/cfw_D/2)
Tangential Force
where P = power being transmitted in hp
n = rotational speed in rpm
T torque on the gear in lb in
D = pitch diameter of the gear in inches
The angle between the total force and the tangen

Section 12-7 Shaft Design Example
551
ufacturer's catalog will specify the minimum acceptable diameter to the right of
the bearing to provide a suitable shoulder against which to seat the bearing.
Point C: Point C is the location of gear C with a well-rou

Section 9-16 Power-Transmitting Capacity
429
transmitting capacity. In this analysis, it is assumed that the operating temperature of
the gears and their lubricants is 250F and that gears are produced with the appropriate
surface finish.
Bending
We start

Chapter 17 Linear Motion Elements
698
You Are the Designer
You are a member of a plant engineering team for a large
steel processing plant. One of the furnaces in the plant in
which steel is heated prior to final heat treatment is installed beneath the fl

456
Chapter 10 Helical Gears, bevel uears, ana wormgearing
FIGURE 10-5
Geometry factor (J) for
15 normal pressure
angle
Generating rack
Value for J is for an element of indicated
numbers of teeth and a 75-tooth mate.
E
o
Helix angle, i|;
(a) Geometry fact

490
Chapter 10 Helical Gears, Bevel Gears, and Wormgearing
gear, efficiency, input speed, input power, and stress on
the gear teeth. If the worm is hardened steel and the
gear is chilled bronze, evaluate the rated load, and determine whether the design is

174
Chapter 5 Design for Different Types of Loading
3. Use Figure 5-8 to estimate the modified endurance strength, s.
4. Apply a material factor, C^, from
Wrought steel: C, = 1.00
Cast steel:
C, = 0.80
Powdered steel: C, = 0.76
the following list.
Malleab

682
Chapter 16 Plain Surface Bearings
Comment
A qualitative evaluation of the result would require more knowledge about the applicafion.
But note that a coefficient of friction of 0.0042 is quite low. It is likely that a machine requiring such a large sha

Section 19-4 Stresses and Deflection for Helical Compression Springs
745
where 6 = angle of twist in radians
T = applied torque
L = length of the wire
G = modulus of elasticity ofthe material in shear
J = polar moment of inertia of the wire
Again, for con

Section 15-5 General Layout and Design Details of the Reducer
645
ing ofthe shaft are used to produce a clearance to facilitate assembly. When the parts return
to normal temperatures, the final fit is produced.
We now show the determination of the limit d

402
O l laf^LCI S
FIGURE 9-23
External spur pinion
geometry factor. /, for
standard center
distances. All curves
area for the lowest
point of single-tooth
contact on the pinion.
(Extracted from AGMA
Standard 218.01,
Rating the Pitting
Resistance and
Bendi

Section 11-7 Couplings
515
FIGURE 11-20
Bellows coupling. The
inherent flexibility of
the bellows
accommodates the
misalignment.
(Stock Drive Products,
New Hyde Park, NY)
FIGURE 11-21
PARA-FLEX
coupling. Using an
elastomeric element
permits misalignment
a

The
Big
Picture
Helical Gears, Bevel Gears, an(d Wormgearing
Discussion Map
The geometry of helical
gears, bevel gears, and
wormgearing was described
in Chapter 8.
D The principles of stress
analysis of gears were
discussed in Chapter 9 for
spur gears. M

382
Chapter 9 Spur Gear Design
Eff ective case depth is defined as depth of case
wh ich has a minimum hardness of 50 HRC.
Tol al case depth to core cartron is approximat ely
1 ,3 X effective case ucpiii
^o
S^*
/if = 0.119 935 xP<i-0.861 05
15
1NJonmal cas

CONJUGATE FRACTIONS
15
Step 3. Add the respective numerators and denominators of the conjugates 01 and 11 to
create a new conjugate e/f between 0 and 1: e/f = (a + c)/(b + d) = (0 + 1)/(1 + 1) = 12.
Step 4. Since 0.4749693 lies between 01 and 12, e/f must

INTEREST
25
S = P ( 1 + i)n
P = S (1 + i)
i = (S P)
n
1n
1
n = ( log S log P ) log ( 1 + i )
Example:At 10 per cent interest compounded annually for 10 years, a principal amount
P of $1000 becomes a sum S of
S = 1000 ( 1 + 10 100 ) 10 = $2 ,593.74
If a su

PROGRESSIONS
To Find
l
n
a
Formulas for Geometrical Progression
Given
Use Equation
l
a = -n
r
rn 1
( r 1 )S
a = -r
S
rn 1
l
r
S
a = lr ( r 1 )S
l
n
S
a ( S a )n 1 = l ( S 1 )n 1
a
n
r
l = ar n 1
a
r
S
1
l = - [ a + ( r 1 )S ]
r
a
n
S
l ( S l)n 1 = a ( S a

INVESTMENTS
29
as the effects of income taxes, salvage values, expected life, uncertainty about the required
rate of return, changes in the cost of borrowed funds, and others.
A tabulation of annual costs of alternative plans A, B, C, etc., is a good way

COORDINATE SYSTEMS
39
b
where = a direction, the angle whose tangent is b a, thus = atan - ; and, r = a
a
2
2
magnitude = a + b .
A complex number can be plotted on a real-imaginary coordinate system known as the
complex plane. The figure below illustrate