Galling Failures in Pin Joints
by
Greg A. Radighieri
B.S. in Mechanical Engineering, Texas A&M University, 2000
Submitted to the Department of Mechanical Engineering
in partial fulfillment of the requirements for the degree of
Master of Science in Mechanical Engineering
at the
MASSACHUSETTS INSTITUTE OF TECHNOLOGY
May 2002
@ Massachusetts Institute of Technology 2002. All rights reserved.
Author
Department of Mecfanical Engineering
May 18, 2002
Certified by
Samir Nayfeh
Associate Professor, Mechanical Engineering Department
Thesis Supervisor
Accepted by
Ain A. Sonin
Chairman, Department Committee on Graduate Students
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2
Galling Failures in Pin Joints
by
Greg A. Radighieri
Submitted to the Department of Mechanical Engineering
on May 18, 2002, in partial fulfillment of the
requirements for the degree of
Master of Science in Mechanical Engineering
Abstract
A pin joint is defined as an assembly of a pin and a bushing, under sliding contact,
that transfers loads between two components while allowing rotation. Under heavy
loads and low sliding velocities, pin joints are considered to be in boundary lubrication. Failure of heavily loaded pin joints is yet to be fully understood and is often
generalized as galling failure. Galling is really the end result of failure.
A summary review of important parameters involved in boundary lubrication and
galling is included. At the end of this review is a section hypothesizing how pin joint
failures (and galling in general) may occur.
To gain insight on boundary lubrication and the galling phenomenon, a test machine was designed and built to force "galling" failure in pin joints. Initial tests results
are given.
Thesis Supervisor: Samir Nayfeh
Title: Associate Professor, Mechanical Engineering Department
3
4
Acknowledgments
I have many people to thank, for it takes a lot of teamwork to get things done. I
should first thank my advisor for handing me this project. I could still be looking
for a project if it weren't for him. He also provided me lots of insight into machines,
machine dynamics, and machine design. He knows his stuff forwards and backwards,
and he serves as a fantastic resource. He takes very good care of his students, including
me. Thanks, Samir.
Many thanks to this project's corporate sponsor. I hope my work will bring
rewards to both the sponsor and MIT. My reward is the knowledge and the experience
I've gained.
5
6
Contents
1
Introduction
1.1
2
3
Pin Joint Analysis
2.1 Pin Joint Description . . . . . . . . . . . . . . . .
. . . . . . . . .
2.2 Dimensional Analysis (Scaling)
2.2.1 Fluid Mechanics Example . . . . . . . . .
2.2.2 Pin Jo it . . . . . . . . . . . . . . . . . .
2.3 Macroscopic Deformation.....
. . . . . . . . .
. . . . . . . . .
2.3.1 Beam Bending......
2.3.2 Torsion . . . . . . . . . . . . . . . . . . .
2.3.3 Transverse Shear . . . . . . . . . . . . . .
. . . . . . . . .
2.3.4 Combined Stress.....
. . . . . . . . .
2.4 Microscopic Deformation.....
2.4.1 Radial Gap Geometry
. . . . . . . . .
2.4.2 Surface Parameters .
. . . . . . . . .
2.4.3 Undeformed Contact
. . . . . . . . .
2.4.4 Contact Mechanics .
. . . . . . . . .
2.5 Tribological Considerations.
. . . . . . . . .
.
2.5.1 Lubrication . . . ....
. . . . . . . . .
2.5.2 Friction . . . . . . . . . .
. . . . . . . . .
2.5.3 Frictional Heating .....
. . . . . . . . .
2.5.4 Sliding Wear . . . . . . . .
. . . . . . . . .
2.6 Failure Theories . . . . . . . . . .
. . . . . . . . .
2.6.1 List of Parameters .
2.6.2 Discussion of Possible Faili ure Mechanisms
Design of a Pin Joint Testing
3.1 Motivation . . . . . . . . . .
3.2 Basic Requirements . . . . .
3.3 Basic Design Rules . . . . .
3.3.1
3.4
17
17
Motivation . . . . . . . . . . . . .
Machine
. . . . . .
. . . . . .
. . . . . .
Rules of Thumb . . . . . . . . .
3.3.2 Kinematic Constraint . . . . . .
3.3.3 Iteration . . . . . . . . . . . . .
Normal Load Design . . . . . . . . . .
7
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19
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25
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35
35
39
40
41
43
43
44
47
47
47
47
47
48
48
48
3.5
3.6
3.7
3.4.1 Everything Turns Into Rubber
3.4.2 Hydraulics . . . . . . . . . . .
3.4.3 Threaded Fasteners and Lever Arms
Torque and Rotational Speed Design
3.5.1 Taper Analysis . . . . . . ..
Machine Elements . . . . . .
3.6.1 Bearings . . . . . . .
3.6.2 Bearing Ring . . . .
3.6.3 Support Base . . . .
3.6.4 Bushing Housing
3.6.5 Pin Housing.....
3.6.6 Torque Transmission
3.6.7 Torque Support . . .
3.6.8 Sensors . . . . . . . .
3.6.9 Test Stand . . . . . .
3.6.10 Seals . . . . . . . . .
Final Design . . . . . . . . .
3.7.1 Relative Motion . . .
3.7.2 Bearings . . . . . . .
3.7.3 Bearing Ring . . . .
3.7.4 Support Base .
3.7.5 Bushing Housing
3.7.6 Pin Housing.....
3.7.7 Torque Transmission
3.7.8 Torque Support . . .
3.7.9 Sensors . . . . . . . .
3.7.10 Seals . . . . . . . . .
3.7.11 Test Stand ......
3.7.12 Assembled Test Mach ine
4
Pin Joint Designs
4.1 Undercut Bushing ..........
5
Test Results
5.1 Initial Tests . . . . . . . . . . . .
5.1.1 Standard Pin Joint . . . .
5.1.2 Undercut Bushing . . . . .
5.1.3 Discussion and Conclusion
5.1.4 Recommendations . . . . .
48
49
49
53
53
55
55
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56
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60
61
65
66
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68
70
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73
73
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75
75
75
76
79
79
A Four Point Beam Bending Analysis
81
B Centroids of Circular Sections
83
C Bolt Calculations
85
8
D Differential Screw Analysis
87
E Taper Analysis
89
F Drawings
91
G Testing Procedure
125
9
---hIiiIII-~iI~M~hEH~I~Iih~iil
iihII
no -MIN
I~lIhI~i~~~i IE
lii
II
EhhhHIlIM
'I
List of Figures
2-1
2-2
2-3
2-4
2-5
2-6
2-7
Pin Joint Schematic. . . . . . . . . . . . . . . . . . . . . . . . . . . .
Four Point Pin Loading. . . . . . . . . . . . . . . . . . . . . . . . . .
Four Point Pin Loading, Dimensions Added . . . . . . . . . . . . . .
Pin deflection versus distance for different loads. . . . . . . . . . . . .
Distributed Pin Loading . . . . . . . . . . . . . . . . . . . . . . . . .
Pin Cross Section . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
von Mises Effective Stress for Applied Load P = 100, 000lbf, Applied
Torque T = 2240N . m . . . . . . . . . . . . . . . . . . . . . . . . . .
2-8 von Mises Effective Stress for Applied Load P = 100, 000lbf. Applied
Torque T = 2240,3240,4240, 5240N - m (left to right). Maximum von
Mises Stress (left to right): 230,241,255,272MPa .. . . . . . . . . . . .
2-9 Radial Gap Schematic . . . . . . . . . . . . . . . . . . . . . . . . . .
2-10 Gap Ratio 2
as a function of the Pin/Bushing Ratio R.........
.
R
.
2-11 Gap Ratio I
R as a function of the Pin/Bushing Ratio R............
2-12 Pressure versus Length of Line Contact.... .. .. .. .. .. .. .. .
3-16
Bolt Force for an Applied Torque T = 80lbf - ft . . . .
Bolt Factor of Safety for Applied Torque T = 80lbf - ft
Total Force for a 10-Bolt Array, Applied Torque T = 80
Bolt Optimization . . . . . . . . . . . . . . .
.
.
Force Amplification versus Taper Angle . . .
.
Koyo Spherical Bearing . . . . . . . . . . . .
Pin Joint with Bushing Housing and Clamps
.
.
Cross section of bearing ring assembly . . .
.
Section cut of bearing ring assembly . . . . .
Exploded view of bearing ring assembly . . .
.
Torque Support . . . . . . . . . . . . . . . .
.
.
Seal design (original) . . . . . . . . . . . . .
.
Seal design (revised) . . . . . . . . . . . . .
.
Test Machine, Front View . . . . . . . . . .
.
Test Machine, Top View . . . . . . . . . . .
Test Machine, Front View . . . . . . . . . .
.
4-1
Undercut Bushing (after testing failure)
3-1
3-2
3-3
3-4
3-5
3-6
3-7
3-8
3-9
3-10
3-11
3-12
3-13
3-14
3-15
11
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lbf
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- ft
. .
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19
23
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27
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34
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51
51
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58
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60
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62
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68
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70
70
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71
74
5-1
5-2
5-3
5-4
Test Results for Standard Pin Joint. (a) Normal Load versus Time;
(b) Third Load Cell versus Time (c) Temperature versus Time . . . .
Calculated (a) Reaction Torque and (b) Friction Coefficient for Standard Pin Joint . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Test Results for Undercut Bushing. (a) Normal Load versus Time; (b)
Third Load Cell versus Time (c) Temperature versus Time . . . . . .
Calculated (a) Reaction Torque and (b) Friction Coefficient for Undercut Bushing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
A-I Four Point Pin Loading, Dimensions Added
76
77
77
78
. . . . . . . . . . . . . .
81
B-i Centroid Schematic . . . . . . . . . . . . . . . . . . . . . . . . . . . .
83
D-1 Differential Screw Schematic
. . . . . . . . . . . . . . . . . . . . . .
D-2 Free body diagram with respect to lower wedge . . . . . . . . . . . .
D-3 Free body diagram with respect to upper wedge . . . . . . . . . . . .
87
87
87
E-1 Taper Free Body Diagram . . . . . . . . . . . . . . . . . . . . . . . .
89
F-1
F-2
F-3
F-4
F-5
F-6
F-7
F-8
F-9
F-10
F-11
F-12
F-13
F-14
F-15
F-16
F-17
F-18
F-19
F-20
F-21
F-22
F-23
F-24
F-25
F-26
F-27
Bearing Ring, Page I . . . . . . . . . . . . .
Bearing Ring, Page 2 . . . . . . . . . . . . .
Bearing Ring, Page 3 . . . . . . . . . . . . .
Bearing Ring, Page 4 . . . . . . . . . . . . .
Bushing Housing, Page 1 . . . . . . . . . . .
Bushing Housing, Page 2 . . . . . . . . . . .
Bushing Housing, Page 3 . . . . . . . . . . .
Bushing Housing, Page 4 . . . . . . . . . . .
Bushing Clamp, Page 1 . . . . . . . . . . . .
Bushing Clamp, Page 2 . . . . . . . . . . . .
Bushing Clamp, Page 3 . . . . . . . . . . . .
Bushing Housing.Mount, Page 1 . . . . . . .
Bushing Housing Mount, Page 2 . . . . . . .
Pin Connector, Page 1 . . . . . . . . . . . .
Pin Connector, Page 2 . . . . . . . . . . . .
Pin Connector, Page 3 . . . . . . . . . . . .
Pin Clamp . . . . . . . . . . . . . . . . . . .
Pin Support . . . . . . . . . . . . . . . . . .
"C" Piece, Page 1 . . . . . . . . . . . . . . .
"C" Piece, Page 2 . . . . . . . . . . . . . . .
"C" Piece, Page 3 . . . . . . . . . . . . . . .
Connector between Load Cell and "C" Piece
Support for Third Load Cell . . . . . . . . .
Support Base, Page 1 . . . . . . . . . . . . .
Support Base, Page 2 . . . . . . . . . . . . .
Nut Preload Strip . . . . . . . . . . . . . . .
Nut Preload Strip, Slotted . . . . . . . . . .
12
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92
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Il
112
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114
115
116
117
118
F-28
F-29
F-30
F-31
F-32
Nut Preload Strip Removal Block
Test Stand, Page 1 ....................
Test Stand, Page 2 . . . . . . . .
Test Stand, Page 3 . . . . . . . .
Test Stand, Page 4 . . . . . . . .
13
. . . . . . . . . .
.....
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119
120
121
122
123
I
U
U
14
List of Tables
2.1
2.2
2.3
2.4
2.5
Pin Joint Clearance at Bushing Midsection
Pin Joint Clearance at Bushing Edge . . .
Kinematic viscosities for 80W-90 oil. . . .
Friction Coefficient Ranges for Lubrication
Film Parameter for Lubrication Regimes .
.
.
.
.
.
20
20
35
37
37
A .1
Pin Param eters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
82
C.1
Factor of Safety Calculations for Applied Torque = 80lbf - ft . . . . .
86
15
. . . . .
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Regimes
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16
Chapter 1
Introduction
1.1
Motivation
A pin joint is defined as an assembly of a pin and a bushing, under sliding contact,
that transfers loads between two components while allowing rotation. Under heavy
loads and low sliding velocities, pin joints are considered to be in boundary lubrication. Failure of heavily loaded pin joints is yet to be fully understood and is often
generalized as galling failure.
There is no established definition for galling, but Budinski's definition seems the
most direct [2, p. 34]:
e Galling - Damage to one or both members in a solid-solid sliding system caused
by macroscopic plastic deformation of the apparent area of contact, leading to
the formation of surface excrescences that interfere with sliding.
Blaming pin joint failures on galling turns out to be an oversimplification. Galling
is really the end result of failure. The circumstances leading up to galling must be
carefully considered. Boundary lubrication, a pre-galling condition for bearings under
high loads, is a state in which the lubricant can no longer sustain elastohydrodynamic
lubrication (EHL). In other words, surface asperities are in contact with each other.
Boundary lubrication is difficult to analyze and still requires further research. Some
key questions include:
* What are the surface interactions?
* How high are the contact stresses and how are they reduced?
* What is the ratio of the real contact area compared to the apparent contact
area?
* How much wear debris is formed?
17
.
How does the lubricant interact?
* What kind of lubricant should be used?
The beginning of this report includes a summary review of important parameters
involved in boundary lubrication and galling. At the end of this review is a section
hypothesizing how pin joint failures (and galling in general) may occur.
To gain insight on boundary lubrication and the galling phenomenon, a test machine was designed and built to force "galling" failure in pin joints. Initial tests results
are given.
18
Chapter 2
Pin Joint Analysis
2.1
Pin Joint Description
L
D
d
u
A
SECTION A-A
BUSHING
Figure 2-1: Pin Joint Schematic
A diagram showing the overall dimensions of a pin joint is shown in Figure 2-1.
To more accurately describe the pin joint geometry, the clearance between the pin
and the bushing should be considered. For the actual pin joint studied, maximum
and minimum clearances for the bushing midsection and edge are given in Tables 2.1
and 2.2.
2.2
Dimensional Analysis (Scaling)
In analyzing a system, dimensional analysis can prove a very powerful tool for resolving the key parameters. It can answer questions like
" Which ratios are important?
" How can I simplify this?
19
Table 2.1: Pin Joint Clearance at Bushing Midsection
Bushing minimum diameter:
Pin maximum diameter:
67.88 - 0.25
66.725 + 0.025
=
=
Minimum Clearance:
67.63mm
66.75mm
00.88mm
Bushing maximum diameter:
Pin minimum diameter:
67.88 + 0.25
66.725 - 0.050
68.13mm
66.675mm
Maximum Clearance:
1.455mm
Table 2.2: Pin Joint Clearance at Bushing Edge
Bushing minimum diameter:
Pin maximum diameter:
67.18 - 0.13
66.725 + 0.025
=
=
67.05mm
66.75mm
00.30mm
Minimum Clearance:
Bushing maximum diameter:
Pin minimum diameter:
67.18 + 0.13
66.725 - 0.050
Maximum Clearance:
67.31mm
66.675mm
00.64mm
9 Can I scale the pin joint so that I can make a smaller test machine?
Dimensional analysis can greatly simplify a problem.
2.2.1* Fluid Mechanics Example
A classic example in fluid mechanics is the problem of drag force on a smooth sphere
in a uniform stream. The force F would appear to be a function of the following
variables: sphere diameter D, fluid velocity V, fluid viscosity p, and fluid density p.
F = f(D, V, p, p)
To test the relationships between these variables, one variable must be
the others must be varied, one at a time. To test 10 values for each
total number of combinations is 104 . That means there must be 104
must be a better way. There is: dimensional analysis. By looking at the
units involved, it can be shown that a functional relationship exists
nondimensional parameters:
(p7D)
pV2D2
20
fixed and all
variable, the
tests! There
fundamental
between two
This relationship reduces the necessary tests from
of 3 orders of magnitude.
2.2.2
104 to 10
[3, p. 298-299], a reduction
Pin Joint
As in the fluid mechanics example, it is desirable to find a fundamental set of nondimensional parameters to represent the forces and/or kinematics involved with the pin
joint. For testing purposes, it would be favorable if the pin joint forces or geometry
(or both) could be scaled (another advantage of dimensional analysis), so that tests
could be performed faster and cheaper.
Assuming that the deformation is linearly elastic, the stress distribution depends
on the following: stress o, force F, elastic modulus E, Poisson's ratio v, bushing
diameter D, pin diameter d, and pin length L.
a = f (F,E, v, D, d, L)
Poisson's ratio is dimensionless; hence, we can set it aside for now.
The fundamental units are mass M, length L, and time t. Reducing the key pin
joint parameters to fundamental units:
: J-
F:
E :
d:
L:
L
L
Creating the first Buckingham Pi equation,
0
Lt
11 = Fad'o- = ML
(MLa
(L)
(
Dt2
t2
= M 0 L t0
and the system of equations,
M:
t:
a+1=0
a+b-1=0
-2a-2=0
we can solve for the unknowns using least squares.
[a ]E[-,l
ad2
=1l
F
[bJ[2j
_
a
(F)
This is simply the equation for stress: force divided by area. There are several lengths
to consider. In this case, the area is most sensibly represented by length times the
bushing diameter; hence, the resulting equation becomes
01F
SDL
21
Performing the same analysis for the elastic modulus yields the same result.
E
2
F
(DL
For the pin length, we derive a simple aspect ratio.
n3 = -
d
H4
=
-
D
Since the Poisson's ratio is dimensionless and because it is important in solid mechanics, the final set of nondimensional parameters is 1
(E-) d d~v
(AT)
\DLI
1Coming up with this set of nondimensional parameters systematically is rather
tricky. It turns
out that choosing the number of repeating variables to be the same as the number of fundamental
units yields a matrix with dependent columns. Without including the bushing diameter or Poisson's ratio, the rank of the matrix is two instead of three. This results in three rather than two
nondimensional parameters.
22
2.3
2.3.1
Macroscopic Deformation
Beam Bending
Analyzing the bending in the pin will give an order of magnitude approximation for
the deflections and stresses in the pin joint. The simplest free body diagram that
models pin bending is the four point bending model.
I
I
E
I
I
i
Figure 2-2: Four Point Pin Loading
Four Point Bending Analysis
Consider the overall load to be 2P and the pin to be solid. This simplifies the analysis.
The free body diagram then appears as in Figure 2-3.
P
P
Li
L2
P
P
Figure 2-3: Four Point Pin Loading, Dimensions Added
Assuming that the flexural rigidity of the beam is uniform along its length, the
deflection y is governed by
Ed2y =
EI
M(x)
(2.1)
dzX2
where E is the elastic modulus, I is the bending inertia, M is the moment, and x is
the distance along the beam.
The boundary conditions are
{
x=a
x=a
x=a +
y=o
dx1= daX 2
L
x=a +
x~a+ Li
23
dX2
y=O
d
2
dX3
Where
a =
L2 -
L1
2
From analysis (Appendix A), the beam deflection is as shown in Figure 2-4.
0.2
0.15
0.1
-I
0.05
0
-
-.
0
0.05
0.1
- - -
--
0.15
--.
100,0001bf
75,0001bf
50,0001bf
- - - 25,0001bf
2,OOb
0.2
50
0
200
150
100
300
250
Distance (mm)
Figure 2-4: Pin deflection versus distance for different loads.
The maximum deflection for a full load turns out to be 0.24 mm at the center of
the pin. One valid question is whether or not this is greater than the clearance in
the pin joint. The answer is no. As you can see from Table 2.1, the clearance is 0.88
mm, well over twice the predicted deflection at full load.
At maximum load, the bending stiffness is
P
_
lbf
N
222 500 N
6
- = '
-= 930x1006 - =5.6x10
3
m
0.24x10- m
3
The bending stress can be calculated.
01-=
Mc
in
(2.2)
where M is the moment, c is the pin radius, and I is the area inertia. The maximum
moment for a 100,0001bf load is ~ 6200 N -m; hence, the maximum bending stress is
o- = 220 MPa
(2.3)
Considering a distributed load would be more accurate and would look something
like Figure 2-5.
24
ad
E 'M
U9
Lb tt
I
Figure 2-5: Distributed Pin Loading
2.3.2
Torsion
Next, we need to estimate the pin joint torque. With a normal load of 100,000 lbf
and a friction coefficient of 0.15, the required torque can be estimated:
T =
(2.4)
pNr
=
(0.15)(100, 000 lbf)(0.11 ft) = 1650 lbf - ft
-
2240 N-m
The shear stress due to torsion is
Tr
T =
(2.5)
-
where r is the radius and J is the polar inertia of the pin. The polar inertia is
- c)
j=
where c, is the outer radius and ci is the inner radius.
From Equations 2.4 and 2.5, the maximum shear stress caused by torsion is estimated:
(2240 N - m)(0.033 m)
(2.6)
- 4
= 39.6 MPa
T =
(0.033 M)
Another parameter of interest is the torsional stiffness, defined as
T
T
2.3.3
_
JG
#L
(1.86x10-6 m)(75x109
0.24 m
;-)
(2.7)
5.82A
5
N-m
radian
(2.8)
Transverse Shear
Transverse shear stress can be calculated as
VQ
it
25
(2.9)
where V is the internal shear force, I is the inertia, and t is the thickness at the point
of concern. The variable Q is defined as
JydA = 9A'
Q=
where A' is a partial cross-sectional area and 9 is the distance from the pin center
axis to the centroid of A'. From analysis (Appendix B), the distance is
[r 2 x _
c 2 X]
-
A'
[r 2 x
r
2
(9
-
-
-c
-
2
x]
sin 0 cos 0)
(2.10)
The highest transverse shear is located on the bending axis of the pin. For a full load,
the transverse shear is
VQ
37
(I
-
2.3.4
)t
(222, 500 N)(1.35x10- 2 m)(1.58x10(7.98m10 7 m4 )(0.066 m)
90.1 MPa
3
M2 )
Combined Stress
Cubic Stress Equation
Generally, Mohr's Circle is used to find combined stresses, but Mohr's circle cannot
be used for cases in which there are no shear-free faces; hence, the roots of the cubic
stress equation must be calculated [7, p. 30].
63__(e,
0Y + U-2)or + (or.'-y + o-zo-z + oyo-(UXowoz + 2 TxyTyzTzx - ax2
TX_
y
2
77 z - Tz X ) 0
TT 2
-0
(2.11)
At point A (Figure 2-6), the roots are
A,1
=
220 MPa
CA,2
=
CA,3
=
39.7 MPa
-39.7 MPa
(2.12)
26
Ir
I
A
I
B
I|
Q
Section C-C
C
Figure 2-6: Pin Cross Section
At point B, the roots are
UB,1
96.3 MPa
OB,2
0 MPa
OTB,3
-96.3 MPa
(2.13)
Distortion-Energy / von-Mises-Hencky Theory
The distortion-energy theory predicts that yielding will occur whenever the distortion energy in a unit volume equals the distortion energy in the same volume when
uniaxially stressed to the yield strength [7, p. 245].
[(
--
2
) + (a2
-
a 3 ) 2 + (a
-
0 1)2
2
The effective stresses at points A and B are
=
230 MPa
('7B =
167 MPa
(TA
(2.14)
Stress Maps and Contours
Up until now, the calculated stresses have only covered the worst cases, but we are
also interested in the overall picture. How do the stresses combine? Are there any
stress concentrations or gradients to worry about?
By iterating (and careful MatLab programming), the stresses at every point in
the pin cross section can be calculated. The bending, shear, and torsional stresses
are easy to visualize by themselves, but the von Mises stresses are not as trivial.
27
Figure 2-7: von Mises Effective Stress for Applied Load P
Torque T = 2240N . m
100,000lbf, Applied
As it can be seen from Figure 2-72, hourglass-shaped contours appear in the von
Mises stress distribution. Since this is rather peculiar, examine the stress distribution
for increasing torque, as shown in Figure 2-8.
401
40
4
20 40 08
801
801
f)O
60
40
40:
20
~
20 40608W
X) 40 60 0
20
20)40'W00
Figure 2-8: von Mises Effective Stress for Applied Load P = 100,000lbf. Applied
Torque T = 2240,3240,4240, 5240N - m (left to right). Maximum von Mises Stress
(left to right): 230,241,255,272MPa.
As it can be seen from Figure 2-8, the contour lines become more and more circular
as the torsion stress increases.
2
The numbers on the coordinate axes in the figure do not correlate to von Mises stress. They
relate only to the graphing program.
28
2.4
2.4.1
Microscopic Deformation
Radial Gap Geometry
Pin joints naturally have radial clearance between the pin and the bushing. A mathematical description of this geometry is useful for analyzing contact stress.
I
Radal Gap
R
0
r
r
I
g
Figure 2-9: Radial Gap Schematic
From Figure 2-9, it can be shown that
A = R -r
ARr
= -
(2.15)
#
(2.16)
(2.17)
#=
2
- (a + 0)
I
sin/
sinO0
A
r
Combining Equations 2.16 and 2.18, we obtain
2
2
q$-2O
Now, combining 2.20 with 2.19, we have
29
# +0
(2.18)
(2.19)
(2.20)
(2.21)
0 + arcsin A sin 0
1r
(2.22)
I
The radial gap can then be calculated:
g., = Rsin0- rsin#
gy
=
Rcos0-(rcos#+A)
=
Rcos-(rcos#+R-r)
(2.23)
(2.24)
(2.25)
Using the Pythagorean Theorem,
(2.26)
g +g2
g =
This mathematical description becomes more useful if it can be non-dimensionalized.
Selecting the bushing radius R as the key parameter, the gap becomes a function of
the pin and bushing radii, -. The corresponding equations are
#
0 + arcsin[(-
R
-
r
1) sin 0]
= sinO - 7sin#
R
(2.27)
(2.28)
RR
R
9= V(L)2 + (
R
R
)2
(2.30)
Figure 2-10 shows the gap ratio (Equation 2.30) as a function of the bushing half
angle, 0.
2.4.2
Surface Parameters
Though contacting surfaces are generally viewed as smooth, they are rough on the
microscopic scale. Two parameters are generally used to characterize a surface: centerline average roughness (Ra) and root-mean-square roughness (Rq)[1, p. 316]:
1N
Ra =
Rq =
(
Izil
(2.31)
z2)
( 2.32)
where zi is the height from the centerline and N is the number of measurements.
30
0 06
.
r
0.05 ----- 0.956
-.-.- 0.961
--
0.967
0.972
0.978
0.983
-
0.989
---
i
5 0.04
-
--
>
CD
0.994-
-1.000
0.03 -
a0.02
0.01
0
10
0
40
30
20
60
50
70
80
90
Bushing Half Angle (degrees)
of the Pin/Bushing Ratio
as a function
R
Figure 2-10: Gap Ratio
2.4.3
L
R
Undeformed Contact
Assume that any gap less than or equal to the surface roughness of the pin or the
bushing can be considered to be in contact. With a guess that the surface roughness
will be from 0.1 pm to 1 pm [1, p. 317], the gap ratio becomes
{
3x10
6
R=0.1 pm
R=1 pm
3x10- 5
Since this is a much smaller scale than the scale shown in Figure 2-10, an enlarged
view will be needed. (Figure 2-11) From Figure 2-11, the corresponding contact angles
are
{0 2O0.60 , R=0.1 pm
, R=1 pm
0
20
Hence, the radial contact length (s = r(20)) is
{
{
0.69 mm
2.3 mm
,
,
R=0.1 pm
R=1 pm
Taking the bushing length, the order of the undeformed stresses can be calculated.
.445,000
N
m)i
(0.00239m)(0.24 m)
2.7 GPa
= 800 MPa
, R=0.1 pm
, R=1 pm
The undeformed stress calculations indicate that smoother, contacting surfaces
will experience higher stresses, attributed to the smaller effective contact area. This
31
9
8
7
-6
Cu
x 10-6
_r
0.950
0.956
. 0.961
0.967
0.972
- 0.978
... 0.983
0.989
0.994
--- 1.000
-
C
5
M
C1
Ca
CD4
CL3
cc
2
O0' -
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Bushing Half Angle (degrees)
Figure 2-11: Gap Ratio I as a function of the Pin/Bushing Ratio
j
may partially explain why polished, contacting surfaces tend to gall more readily.
From Section 2.4.4, the estimated deformed contact is about 13mm and the maximum
pressure is 170 MPa
.
3
1t is not clear whether the 13 mm is the arc length or the cord of the arc. The cord is assumed.
32
2.4.4
Contact Mechanics
Hertzian Contact
Contact between the pin and the bushing can be considered line contact.
defined an effective elastic modulus and effective radius 4 as
E* =
+ 12
F
R=
Hertz
(2.33)
1-
I
(2.34)
I
With load per unit length P, the semi-contact width can be calculated:
(4PR)2
a = rE*
(2.35)
(.5
The maximum contact pressure can also be written as
Pmax =
(PE*
N
----7rR
2P
-
(2.36)
7ra
Assuming that both the bushing and the pin are steel:
110 GPa
2(1 - (0.3)2)
200 GPa
R
R 0.033
P=
4(1.85x10 6
ar(110x10
n
0.0335 m1
44,000 NN
0.24 m
N)(2.21
9
N)
m)
2(1.85x10
Pmax -
= 1.85x10 6 m
2
6
2.21 mn
N)
'
7(0.00688 m)
0.00688 m= 6.88 mm
=
171 MPa
Figure 2-12 shows that as the length of contact decreases, the maximum pressure
increases dramatically.
4
The second term in the formula is negative because the surfaces are conformal rather than
non-conformal.
33
1800
1600
-
-
445 kN
334 kN
223 kN
1400-
0-
1200
1000
-
E 800 -
~600400-
20
50
0
150
100
200
250
Line Contact Length (mm)
Figure 2-12: Pressure versus Length of Line Contact
Non-Hertzian Contact
In reality, the contact seen in a pin joint does not qualify as "Hertzian".
explains it beautifully:
Johnson
... smooth non-conforming surfaces in contact were defined: the initial
separation between such surfaces in the contact region can be represented to an adequate approximation by a second-order polynomial. Nonconforming surfaces can therefore be chracterised completely by their radii
of curvature at the point of first contact. However when the undeformed
profiles conform rather closely to each other a different description of their
initial separation may be necessary. Conforming surfaces in contact frequently depart in another way from the conditions in which the Hertz
theory applies. Under the application of load the size of the contact area
grows rapidly and may become comparable with the significant dimensions of the contacting bodies themselves. A pin in a hole with a small
clearance is an obvious case in point. When the are of contact occupies
an appreciable fraction of the circumference of the hole neither the pin
nor the hole can be regarded as an elastic half-space so that the Hertz
treatment is invalid [4, p. 114-115].
34
2.5
Tribological Considerations
The mechanics and lubrication of a pin joint are surprisingly difficult to model. The
interactions at contacting surfaces must be understood. This describes the focus
of tribology. Tribology is a combination of fluid mechanics, solid mechanics, heat
transfer, thermodynamics, chemistry and more. Tribology, in short, is the study
of friction, lubrication, and wear. Machine elements are bound to have contacting
surfaces, and for a well-designed machine, the likely failures are located at these
contacts. Why? The interactions between surfaces are not yet fully understood.
2.5.1
Lubrication
Lubrication is critical for any contacting surfaces in relative motion. It provides a
thin layer of separation between contacting solids, effectively reducing friction and
wear. Secondary benefits include wear particle removal and surface cooling. The
most important lubricant parameter is viscosity. For a lubricant to be effective it
must be viscous enough to maintain a lubricant film under operating conditions but
should be as fluid as possible to remove heat and to avoid power loss due to viscous
drag [1, p. 319].
Viscosity
The absolute or dynamic viscosity is defined as the ratio of the shear stress to the
shear strain rate in the fluid [1, p. 320.
F/A
7 =
(2.37)
v/h
where F is the shearing force, A is the shear area and v is velocity at a height h.
Sometimes, it is more useful to describe a lubricant by its kinematic viscosity 77k,
77
k
p
where p is the density. The oil type used in the pin joint is 80W-90. Some kinematic
viscosities are listed in Table 2.3.
Table 2.3: Kinematic viscosities for 80W-90 oil.
SAE Grade
'nk©4 0'C
%A@100C
(W)
8OW-90
1.4x 10-4
35
(2)
1
.42x10-b
Pressure-Viscosity Relationships
The most well-known pressure-viscosity relationship was formulated by Barus:
(2.38)
= 77e"i
-,
where 77 is the viscosity at pressure p, r7 is the atmospheric
pressure-viscosity coefficient, and p is the pressure of concern.
equation is not very accurate for pressures above 0.5GPa [8, p.
provides an order of magnitude estimate.
Stachowiak describes the problem facing the determination
analytical pressure-viscosity relationship:
viscosity, a is the
Unfortunately, this
17], but it at least
of a fundamental,
One of the problems associated with available formulae is that they only
allow the accurate calculation of pressure-viscosity coefficients at low shear
rates. In many engineering applications, especially in heavily loaded contacts however, the lubricant operates under very high shear rates, and the
precise values of the pressure-viscosity coefficient are needed for the evaluation of the minimum film thickness. An accurate value of this coefficient
can be determined experimentally...
If an accurate analytical formula could be developed it... would open up
the possibilities of modifying the chemical make up of the lubricant in
order to achieve the desired pressure-viscosity coefficient for specific applications [8, p. 19].
Viscosity-Temperature Effects
In general, liquid lubricant viscosity decreases with temperature. One important
property is the rate of decrease. Viscosity can decease by two orders of magnitude in
a 100 0C range.
Lubrication Regimes and Film Parameter
There are several lubrication regimes. In order or increasing severity, they are:
1. Hydrodynamic Lubrication
2. Elastohydrodynamic Lubrication (EHL)
3. Partial Lubrication
4. Boundary Lubrication
Each regime can be associated with a certain friction coefficient range. These
ranges are listed in Table 2.4 [1, p. 314].
Another way to classify the lubrication regime is by the film parameter, A.
36
Table 2.4: Friction Coefficient Ranges for Lubrication Regimes
Friction
Lubrication
Coefficient
Regime
0.0006 to 0.002
Hydrodynamic
0.002 to 0.06
Elastohydrodynamic
Boundary
0.06 to 0.2
Unlubricated
;> 0.5
hmin
(Rq2a +
(2.39)
)2
where hmin is the minimum film thickness, and Rqa and Rqa are the RMS surface
roughness' of the two contacting surfaces. A general relation has been established
comparing the lubrication regime to the minimum film thickness and to surface roughness, as shown in Table 2.5[1, p. 318]5.
Table 2.5: Film Parameter for
Lubrication
Regime
Hydrodynamic
Elastohydrodynamic
Partial
Boundary
Lubrication Regimes
Film
Parameter
5 < A < 100
3 <A < 10
T< A < 5
A < 1
Boundary and Extreme Pressure Lubrication
As tests have indicated, the pin joint operating regime classifies as boundary lubrication (compare friction coefficients from Chapter 5 with those in Table 2.4)6.
Stachowiak describes this as "extreme pressure lubrication", a mechanism that
takes place in lubricated contacts in which loads and speeds are high enough to
result in high transient friction temperatures sufficient to cause desorption of available
adsorption lubricants [8, p. 388].
Adsorption Lubrication
For contact pressures of up to 1 GPa and surface temperatures between 100 and
150 0C, the lubrication mechanism is generally known as adsorption lubrication. This
5
Hamrock mentions partial lubrication [1] as a phase in which the asperities have just begun to
contact.
6
See Stachowiak's Engineering Tribology [8] for much more information on lubricants and lubrication regimes (especially the chapter on Boundary and Extreme Pressure.)
37
differs from EHL because a mono-molecular layer separates the contacting surfaces,
and this layer is so thin that the mechanics of the asperity contact are identical to
those of dry surfaces in contact [8, p. 360 ]7.
Adsorption describes the affinity of the lubricant or lubricant additive to the contacting surfaces, specifically the worn surfaces. Polarity is key for proper adsorption.
Ideally, a lubricant molecule will strongly adhere to the metallic surface while its
exposed end repels other molecules; hence, the adhering action keeps the molecule
from being removed, and the repulsive action causes the shear strength of the sliding
interface to be weak (and thus yields low friction). Organic polar molecules such as
fatty acids perform this role well.
Under dynamic conditions, loss of lubricant is poorly understood, but Stachowiak
provides some insight:
The rate-limiting step in the formation of an adsorbate film under sliding
conditions is believed to be re-adsorption, and a minimum concentration
of fatty acid is required for this process to occur within the time available
between successive sliding contacts [8, p. 371].
Another factor that is tough to account for the surface temperature. As the
temperature increases, the lubricant and/or lubricant additives will tend to desorb,
or be removed from the sliding surfaces.
7
This is great news, for the tribological equations of frictional heating and of wear are based upon
dry contact.
38
2.5.2
Friction
The coefficient of friction can be divided into a few basic friction mechanisms [9, p.73:
" Asperity deformation,
Pd
* Plowing by wear particles and hard surface asperities, pp
" Adhesion between flat surfaces, ya
Suh dismisses asperity deformation when considering dynamics [9, p.78]:
In a dynamic situation where the surfaces become smooth, most of the
normal load is carried by the entrapped wear particles and the flat contacts. Therefore, the actual contribution of the asperity deformation to
the frictional force is expected to be a small fraction of the estimated value
in a dynamic situation. Before the onset of sliding between two surfaces,
Ad largely controls the static coefficient of friction.
Friction by Plowing
Given a wear particle of radius r and width of penetration into a surface (w), the
coefficient of friction for plowing is [9, p.81]
W
[2-
2r
_W
_)~2r
P =
H _W
sin--
)2
1
-2
_
In short, this equation shows that p, increases strongly with increasing w/2r. The
plowing coefficient can range from 0 to 1 and is commonly found to be around 0.2.
Friction by Adhesion
Adhesion describes the attractive force between two flat contacting surfaces.
nearly flat surfaces, the adhesive friction coefficient can be defined as [9, p.79]
A"1
+ 1! + cos-'
where pa varies from 0 to 0.39 as
f
f
f
+ sin(cos-
1
f)
changes from 0 to 1.
39
For
2.5.3
Frictional Heating
Friction plays a very important role in virtually every machine. Friction generates
heat, and heat can greatly alter machine performance8 . In this case, how is friction
involved with galling failure? Pin joints are generally hot during observed failures,
but is the frictional heating causing the failure, or is something else raising the friction
which then induces failure? This is an unresolved question, discussed in the Failure
Theories Section (Section 2.6).
In general, the heat generated (q) between two contacting solids in relative motion
can be described as a function of the normal force F, relative velocity v, nominal
contact area An, and the friction coefficient p.
pFv
q An
Bulk Temperature
Assuming uniform heat flow, the bulk surface temperature T can be calculated from
Tb - To = pb
(2.40)
11
12b
where T. is the sink temperature, k1 and k 2 are thermal conductivities, and
12b are lengths.
11b
and
Flash Temperature
At the microscopic or asperity level, temperatures can be much higher than the bulk
surface temperature. The equation is similar to that of T, but in this case the real
contact area Ar is used along with the effective sink temperature T'. The temperature
at the asperity level is known as the "flash" temperature Tf. It is given by
T Fv
T - Tb'= A,
(2.41)
1
&+E
11f
k221
12f
where 11f and l2f are lengths.
8
For more information on frictional heating, see Kong and Ashby's Friction - Heating Maps and
Their Applications [5].
40
2.5.4
Sliding Wear
-
Wear refers to the loss of small pieces of a surface in relative motion with another
surface. Though this loss can be very small and even difficult to locate, it is very
important when considering any two surfaces in sliding contact.
The most importantfactor in sliding wear is the distance travelled.
Time and speed affect the heat transfer dynamics. Since the distance travelled is
key, the term wear rate refers to the volume of wear debris per unit distance:
V
Delamination Wear
Delamination wear describes the formation of subsurface cracks and the ultimate
creation of wear particles during solid-on-solid sliding. In Tribophysics, Nam Suh
describes the events which lead to loose wear sheet formation [9, p. 199-200 ]:
1. When two surfaces come into contact, normal and tangential loads
are transmitted through the contact points. Asperities of the softer
surface are easily deformed and fractured by the repeated loading action, forming small wear particles. Hard asperities are also removed
but at slower rates. A relatively smooth surface is generated initially,
either when these asperities are deformed or when they are removed.
2. The surface traction exerted by the harder asperities at the contact
points induces incremental plastic deformation per cycle of loading,
accumulating with repeated loading. The increment of permanent
deformation remaining after given cyclic loading is small compared
with the total plastic deformation that occurs during that cycle, since
the direction of shear deformation reverses during a given cycle and
the magnitude of elastic unloading is comparable to plastic strain.
3. As the subsurface deformation continues, cracks are nucleated below
the surface. Crack nucleation very near the surface cannot occur, due
to the triaxial state of compressive loading which exists just below
the contact region.
4. Once cracks are present (either by crack nucleation or from preexisting voids and cracks), further loading and deformation causes
the cracks to extend and propagate, eventually joining neighboring
cracks. The cracks tend to propagate parallel to the surface at a
depth governed by material properties and the state of loading.
5. When the cracks finally shear to the surface, long and thin wear
sheets delaminate. The thickness of the wear sheet is determined by
the location of subsurface crack growth, which is affected by the normal and tangential loads at the surface. The wear rate is controlled
by the crack nucleation rate or the crack propagation rate, whichever
is slower.
41
Adhesive Wear
Two flat contacting surfaces will have an attractive force for each other. While loaded,
if the shearing force required to separate the two surfaces is greater than the shearing
strength of a subsurface just below one of the contacting surfaces, then a break will
occur at the subsurface. The fundamental law of adhesion theory is
V
KF
x
3H
where K is the wear coefficient, F is the applied load, and H is the hardness of the
surface being worn [6, p.1401-141]. The real contact area can be approximated by
F
H
Ar
hence
V
KAr
x
3
Also, the volume of wear debris can be viewed as the average depth h worn away
times the apparent contact area A,.
V = hAa
hence
h
A)
(Kx
3 (Aa
The ratio A,/Aa is very useful. As the ratio increases the wear certainly increases9 .
9
See Lim and Ashby's Wear-Mechanism Maps for an in-depth discussion of wear
42
2.6
Failure Theories
As mentioned earlier, it is unclear which mechanism initiates galling failures. Below,
are summarized many different possible mechanisms or symptoms of failure along
with important pin joint and lubricant properties.
2.6.1
e
List of Parameters
Deformation
1. Macroscopic
(a) Bending stress too high
- Function of normal load
- Function of length and area inertia
(b) Torsion too high
-
Function of rotational resistance (function of normal load and joint
friction)
-
Function of length and polar inertia
(c) Shear too high
- Function of normal load
- Function of length , shear flow, and area inertia
(d) von Mises stress too high
- Function of bending stress, torsion, and shear
(e) Deflection at pin center contacts bushing
- Function of normal load
- Function of area inertia, and elastic modulus
2. Microscopic
(a) Contact pressure too high
- Function of normal load
- Function of contact width, length, surface hardness, and effective
modulus
* Tribology
1. Frictional Heating
(a) Bulk temperature too high
- Function of normal load and relative velocity
- Function of friction coefficient, nominal contact area, length, and
thermal conductivity
- Function of sink temperature
(b) Flash temperature too high
43
- Function of normal load and relative velocity
- Function of friction coefficient, real contact area, length, and thermal conductivity
- Function of effective sink temperature
2. Lubrication
(a) Lubricant "desorbs"
- Function of temperature
- Function of lubricant adsorption at pin and bushing surfaces
(b) Lubricant fractures under extreme pressure
- Function of contact pressure
(c) Lubricant is squeezed out of contact patch
- Function of stress gradient
- Function of pin joint clearance
(d) Lubricant bulk temperature too high
-
Function of lubricant viscosity
- Function of input power (normal load in conjunction with the rotational speed)
- Function of sink temperature
(e) Lubricant vaporizes
- Function of temperature
-
Function of pressure
3. Wear
(a) Oxidation
(b) Corrosion
(c) Delamination
(d) Adhesion
2.6.2
Discussion of Possible Failure Mechanisms
Stresses seen from analysis of macroscopic deformation of the pin joint are relatively
tame compared to those found in the microscopic analysis. It is fairly clear that the
most important factors are:
1. Lubricant Adsorption
2. Lubricant Squeeze-Out
3. Lubricant Load Capability
4. Lubricant Thermal Conductivity
5. Contact Pressure
44
6. Shearing Force
7. Relative Surface Hardness
8. Rotational Speed (Friction Heating)
9. Distance Travelled
10. Pin
/
Bushing Relative Stiffness
11. Pin
/
Bushing Thermal Conductivities
Given a normal load, constant rotational speed, and a lubricant, what are the
ways in which a pin joint could fail?
Assume that the lubrication regime is boundary lubrication. Otherwise, the pin
joint would be likely to last a long time (and wouldn't be considered much of a
problem). Given boundary lubrication, it can be assumed asperities are contacting.
First, take a look at lubricant adsorption. Does the lubricant have an adsorption
affinity for the metal surfaces? If not, a lubricant film will not form on the metal
surfaces, and the pin joint will perform poorly.
Now look at the static normal load. How much of the load is taken by the lubricant,
and how much of the load is taken by the contacting asperities? This ratio is very
important. If much of the load is taken by the adsorbed lubricant film, then wear
and shearing force are kept to a minimum. This is the ideal case for boundary
lubrication and can be viewed as almost elastohydrodynamic lubrication. From the
contact pressure, does the bulk lubricant squeeze out? This most likely occurs; hence,
how much lubricant squeezes out? Does the lubricant return to the contact location
if the normal load is removed? Automotive oils, for example, would likely return (a
self-help feature), whereas grease may not return to the contact patcho.
Now include rotation. What is the friction force? This shearing force really is the
culprit when it comes to failure (but it is the end result). All efforts must attempt
to somehow reduce it. Methods would include better adsorption properties, better
squeeze-out properties, lower contact pressures and better transfer of heat away from
the contacting surfaces.
With rotation comes frictional heating. What are the bulk temperatures of the pin
and the bushing? What are the surface temperatures? If the local surface temperature
becomes too high, the lubricant film will tend to desorb from the pin and bushing
surfaces. As the film desorbs, the asperities will be forced to take more of the load.
The shearing force will climb as asperity adhesion and plowing increase. The effective
film thickness will decrease, but if the bulk temperature remains low enough, readsorption of the lubricant can occur (keeping the film thickness relatively constant).
Re-adsorption is also a function of the pin joint range of motion. Following a point
on either the pin or the bushing at the contact patch, it will periodically move away
10 This is a tough trade-off between load capacity and the ability to re-wet the pin joint contact
surfaces
45
from and then toward the contact patch. If this range of motion is large enough and
the bulk temperature is low enough, the lubricant will have a chance to re-adsorb.
As the bulk temperature rises, the lubricant is less likely to re-adsorb. Consequently, asperity adhesion and plowing will continually grow. The effective film
thickness will decrease; thus, the real contact area will increase. Large wear particles will begin to break from the surface (this can be attributed to a combination of
plowing and delamination). If the particles are from a hard surface, then the plowing
wear will increase (possibly dramatically). As a side note, if the relative hardness of
pin joint is greater than roughly 4 (one surface 4 times harder than the other), the
softer wear particles may continually polish the contacting surfaces; thus, the sliding
surfaces last longer ". This is not necessarily a cure-all rule of thumb. It happens
to work well in applications like commutators for alternators (speed more of a factor
than normal load).
As large wear particles are removed, clean surfaces underneath are exposed. These
clean surfaces are susceptible to adhesion. Adhesion on a greater scale, in such a
case, can greatly increase the necessary friction force to rotate the pin joint. The real
contact area will approach a significant percentage of the apparent contact area. Consequently, if the loading and temperatures remain high (normal load and rotational
speed are not lessened), this combination of lubricant desorption, plowing wear, and
adhesive wear will escalate until the pin joint fails. Keep in mind this can happen
within a short span of time.
"Here, volume of wear debris would likely become the limiting factor
46
Chapter 3
Design of a Pin Joint Testing
Machine
3.1
Motivation
A test machine can be used to seek greater understanding of the fundamentals of
galling and boundary lubrication. Testing machines have been largely unsuccessful in
detecting the onset of galling. Current machines designed to test pin joints for heavy
machinery are generally very large, and test setup time is long. In addition, most of
these machines cannot properly measure the resistive torque in a loaded pin joint.
The following design is for a relatively small test machine with a short setup time
to test full size pin joints '. In addition, setup time for the design is significantly
reduced.
3.2
Basic Requirements
There are three basic requirements on the pin joint test machine:
1. Normal load capability: 100,000 lbf
2. Torque: sufficient to rotate pin and/or bushing
3. Rotational speed: 20 rpm
We can assume that the torque and rotational speed requirements are coupled. These
three requirements were evaluated to simulate field operations.
3.3
3.3.1
Basic Design Rules
Rules of Thumb
Below is a simple list of powerful design rules:
1
"Small" is the size of a desk. Full size pin joints are like those found in the tracks of a bulldozer.
47
e Functional independence
" Kinematic constraint
" Minimize gradients
" Minimize load paths
" Design for manufacture 2
* Design for self-help 3
" Design for fail-safe
3.3.2
Kinematic Constraint
Simply put, "kinematic constraint" implies a design should neither over- nor underconstrain the part or parts in question. In three dimensions, there are six degrees
of freedom, and for any three dimensional part, the designer should attempt to use
exactly six constraints. In a nutshell, that is "kinematic constraint".
3.3.3
Iteration
Another important aspect of design is iteration. A great designer requires very few
iterations to come up with a final design. In any case, the designer must create a
conceptual design and cycle it through design criteria and functional requirements
to create a real, applicable design. Ideally, several candidate designs are considered.
Each one should be as different as possible from the rest. The whole idea of iteration
is to pass into each new stage without having to step backward. Each new stage
should be faster than the previous one. It's analogous to spinning on ice skates. With
your arms held out (initial stages), the rotational speed is slow, and as you pull your
arms in (progressing stagEs), the rotational speed greatly increases.
3.4
3.4.1
Normal Load Design
Everything Turns Into Rubber
A normal load of 100,0001bf is a rather significant load; hence, designs to incorporate
such a load should be very thorough. At such a load, everything literally begins
to act like rubber. The deformations are on a much smaller scale for metal, but the
deformations are definitely significant for design purposes. In designing, it seems to be
2
1f it can't be manufactured, why design it? This is a VERY important concept. A designer
with machining and casting experience can realize designs faster and better than those without the
experience.
3
An example of self-help is to design a bearing housing that allows thermal expansion of a shaft.
This could also be called "robustness"
48
of great importance to actually visualize the metal pieces deforming in an exaggerated
manner...like rubber.
With this in mind, it's now worth mentioning the 100,0001bf load path. Everything
in that load path must be closely examined because all parts in the path will be
susceptible to failure 4
3.4.2
Hydraulics
The first possibility that comes to mind is hydraulics. Hydraulics can easily produce
such a force, given a sufficient hydraulic power supply. Let's get a feel for the size of
the required hydraulic cylinder. McMaster-Carr sells a 55ton-capacity cylinder with
the basic dimensions of 5" OD x 14" average height (with a 3" ram). The cost is
$625. A manually-operated hydraulic power supply costs $2004300, and an electric
supply costs at least $1000. The determining factor may depend upon the dynamics
of the loading. Does the load need to be applied quickly? Can the load be static or
does it need to be cycled?
3.4.3
Threaded Fasteners and Lever Arms
Another normal load possibility is by lever action. This class includes threaded fasteners and lever arms. The feasibility of a differential screw was first examined5 .
Differential Screw Analysis
The hope in a differential screw is that the small axial motion might result in a
higher axial force. There is a well-known equation that relates the applied torque to
the resulting bolt force:
T = KFd
F=
Kd
(3.1)
(3.2)
where F is the applied force, d is the bolt mean diameter, and K is a constant resulting
from the friction coefficient and sines and cosines of a force analysis. In general 6,
K ~ 0.2
With a differential screw, we hope the effective K will be lower. From analysis (Appendix D), the effective K is
p cos A2 - sin A2
y cos A + sin A
K::=:[co ~ 1 lsn 1 +
A
i
]
cos A, - p si~n A, cos A2 + p sin A2
4
(3.3)
The "minimize load path" design rule is thus very important in this case.
A differential screw is a screw with two different threads on the same shaft. The basic idea is
to use two threads that are slightly different in pitch so that one turn of the shaft produces a small
axial motion (because of the "differential" in the threading).
6
ForK , 0.2, the friction coefficient pg 0.15
5
49
For a coefficient of friction
[
=
0.15 and A, = 2.1 deg, A2 = 1.05 deg,
K = 0.32
This is larger than K for a single bolt. Next, check the differential in the angle. For A,
and A2 closer together, K is still greater than K for a single bolt; hence, no mechanical
advantage is gained by the differential screw.
Now, let's take a close look at regular bolts.
Threaded Fastener Analysis
Equation 3.2 is the fundamental equation.
T
Kd
F =T
We can assume K will remain fairly constant; hence, we have two key parameters to
adjust: the applied torque and the bolt diameter. The -applied torque is limited by
the person applying it. Using a torque wrench, an estimate of maximum applied bolt
torque was measured:
T = 80lbf . ft
It seems like the analysis is already done, but we have not accounted for the fact
that bolts of the same diameter have different grades of strength 7 . We need to get
a feel for bolt force and load capacity. Knowing that the required normal force is
100,0001bf, let's consider bolt sizes M8 to M20. Using Equation 3.2, the resulting
forces for an applied torque T = 80lbf - ft are shown in Figure 3-1
Multiplying the area with the strength, the load capacity of each bolt is calculated
(Appendix C). Taking the ratio of the strength to the stress induced by the applied
bolt force gives the factor of safety (Figurer 3-2).
The results are somewhat surprising. The problem turns into an optimization.
If the bolt is too large, not enough torque can be applied. If the bolt is too small,
the force is sufficiently high but the stresses are too high and the bolt will likely fail.
Throw into the mix the fact that multiple bolts will be used and the problem becomes
interesting8 .
Looking at the forces produced by 80lbf - ft, it is determined to use an array of
10 bolts (Figure 3-3). In such a case, the optimization gives only two choices: M1O
or M12. The M10.would result in a net force of 120,0001bf, but the factor of safety is
uncomfortably close to 1; hence, a grade M12 is the final bolt selection (Figure 3-4).
7
Also, you could argue that the torque could be applied by another mechanism or by a longer
wrench. This is true, but two themes of this design will be simplicity and economics. At some level,
things must be done by hand. Assume torque is applied by hand (and wrench!).
8
Perhaps this all seems trivial to you, but I had never thought bolt selection could be so
challenging.
50
181814 -
1210 I
B
a420
L
'
6
8
10
12
14
16
Bok Diameter (mm)
18
20
22
Figure 3-1: Bolt Force for an Applied Torque T = 80lbf - ft
5.0S
0
IL
4.5
4.0 -
U
3.5
-U
S
3.0-
.3
-
-
2.5 -
-a-Grade 4.6
-- Grude 4.8
--
Grade
5.8
-.- Grade 9.0
-+-Grade 10.9
-Grade 12.9
-Minimum
Faclor of Safety
2.0U,
0
0
S
IL
1.51.00.5
dr
-
0.0
6
8
10
14
12
Bolt Diwneter (mrn)
16
18
Figure 3-2: Bolt Factor of Safety for Applied Torque T = 80lbf - ft
51
200,000
180.000
-
140,000 U.
120.000
-
100.000 80,000 -
80,000 S40,00020,000
0
-
'
*
i
16
14
Bolt Diameter (min)
18
20
i
.
.
.
6
8
10
12
'
22
Figure 3-3: Total Force for a 10-Bolt Array, Applied Torque T = 80 lbf - ft
200,000
2.0
Acceptable Strength
and Applied Force
0
1.5
-
/
-
150,000 A
Z/
/
.a
175,000
0
125,000 U.
-
M
0
~
100,000
0
LL
LA.
#00-Grade
-Grade
0.5
-75,000
4.5
-Grade 4.8
/
5.8
50,000
- A
10.9
AP
- -Grade
Grade 9.8
-A "Grade 12.9
-25,000
-enimum Factor of Safety
Total Soft Force
U.0
8
9
10
11
12
13
.
.
14
15
Bolt Diameter (mm)
Figure 3-4: Bolt Optimization
52
0
16
+o
3.5
Torque and Rotational Speed Design
Since there are both torque and rotational speed requirements, rotational power becomes an issue:
P = Tw
where T is the torque and w is the rotational speed. From Section 2.3.2, we know
T = 1650 lbf - ft; hence,
P = (1650 lbf - ft)(2
3.5.1
lbf - ft_
rad
-= 6 hp
) = 3300
s
(3.4)
s
Taper Analysis
In some manner, the pin and the bushing must be affixed to another part of the
machine. In general, this is not a problem, but in this case, the torque is very high
(16501bf -ft). To add to the complexity, this is accomplished by press fits in the real
case, but in our case, we want to be able to disassemble the pin and bushing so that
the machine parts can be used again. One way to do this is with tapers. The normal
force produced for an applied clamping force F is (Appendix E)
F
F
FN =
2(3.5)
sin0 + 2p cosO - p 2 sin 0
Figure 3-5 shows the force magnification for a given taper angle.
12
-
10
-Mua0.06
Mu=0.1
-
-MusO.2
I.
8
2-
0
0
1
2
3
5
4
6
7
8
9
10
Taper Ange (degrees)
Figure 3-5: Force Amplification versus Taper Angle
Holding the pin in place will constitute the limiting case because the moment arm
is smaller and thus requies a higher normal force. The normal force N required to
hold the pin can be approximated as
53
N
T
Pr
1650 lbf - ft
= 75, 000 lbf
(0.2)(0.11 ft)
(3.6)
where T is the applied torque and r is pin radius. For the bushing, the required force
is
N
T
-pIr
1650 lbf - ft
(0.2) (0.18 ft)
=46, 000 lbf
(3.7)
where r is the bushing outer radius.
From Figure 3-5, we can see that a 5 degree taper gives an amplification of 2
to 6x, depending upon the friction coefficient. From the fundamental force-torque
relationship for bolts (Equations 3.1), we know we can easily apply several thousand
pounds of force with a single bolt; hence, the taper fit is a viable option as a substitute
for the press fit. Also, the taper is not permanent like the press fit. Now the torque
and disassembly requirements are met.
54
3.6
3.6.1
Machine Elements
Bearings
Roller bearings fall into a few basic categories:
" Ball
" Cylindrical
* Taper
" Spherical
Each one has distinct advantages and disadvantages. For smooth motion and very
little radial slop, ball bearings are a good choice. But for high loads, the ball is
fundamentally a bad choice. This can be understood through contact mechanics.
The contact for each ball is a point' (or a very small circular patch).
The next three choices have good load capabilities, for they use cylindrical rollers
rather than balls. The roller contact is basically a line (or a thin rectangle)' 0 . Cylindrical bearings are good for instances in which a ball bearing does not provide enough
load capability and there is very little or no axial misalignment. Tapers bearings are
used in many machines. They must be preloaded, but that gives the designer the
ability to adjust stiffness. They can be used in a back-to-back orientation for good
bending stiffness". Again, there must be very little axial misalignment. The spherical bearing allows for significant axial misalignment (several degrees misalignment)
while maintaining high load capabilities.
3.6.2
Bearing Ring
With a rolling-element bearing, a housing will be necessary to either apply or support
loads of up to 100,000 lbf. Keep in mind that the loading may be either static or
dynamic. The ring should have high stiffness.
3.6.3
Support Base
A support base will be needed to support loads from the bolt array. Considering the
bolt array has high stiffness, the support base should also have high stiffness. A low
stiffness means the support base would deflect enough to reduce the axial loading
provided by the bolt array.
9
By Hertz contact mechanics, a point will produce an infinitely high stress. Just think of the
contact area. A point has no area.
10 By Hertz contact, a line also produces an infinitely high stress, but this cause for worry quickly
disappears because any finite thickness in the line provides a sufficient contact area.
"Bending of a shaft is implied.
55
3.6.4
Bushing Housing
Since the bushing outside diameter does not fit any standard bearings, a bushing
housing will be used to transfer applied loads to the bushing. This seems inevitable
since, again, it is desired to apply the loading as predictably as possible.
At first, it seems that the options for loading are endless, but they can be reduced
by imagining the machine. Is it feasible to apply the normal load to the bushing and
the torque to the pin? Yes, this is possible but not wise. The smaller diameter on
the pin (compared to the bushing) means the taper clamping force must be higher
(Section 3.5.1). And because this connection is to be cycled on the order of millions
of cycles, that would indeed be unwise.
What about applying the normal load to the pin? This is also possible, but recall
that the normal load will be applied with an array of 10 bolts. There would definitely
be spatial constraints since the pin length extending out from the bushing is just a
few inches. In addition, this would require two pieces to handle high loads. It may
be easier to design one piece to take the whole load.
3.6.5
Pin Housing
The pin ends will also need a housing to transfer the 100,0001bf load. One design
effort will be to make the load path closed. In other words, the load transferred to
the pin housing is somehow transferred to the bushing housing in a return path. This
means the test stand upon which the machine rests will not be forced to endure very
high loading. Another design effort will be to minimize the length of the load path.
The longer the path the more chances there are for machine elements to fail.
3.6.6
Torque Transmission
Since rotational speed is a requirement, some kind of power transmission element
.must provide the applied torque. This could be accomplished by an electric motor
with heavy duty gearing or by hydraulics. There are other possibilities, but these two
stand as the most obvious and applicable.
3.6.7
Torque Support
Up to this point, only torque transfer has been mentioned, but there must exist a
structure to support the applied torque. In cases of machine testing failure, this
structure could become very important. It may be needed to prevent catastrophic
failure (breaking the test machine instead of the pin joint).
3.6.8
Sensors
Normal Load and Torque
Sensors will be necessary to detect the normal load and the torque applied to the pin
and/or bushing. The measurement device or process will be dictated by instrument
56
precision and by allowable sensor travel. Assume the normal load travel to be small
(on the order of 1mm or less) and the precision to be within 5 percent of the max
load.
Measuring the normal load is relatively easy. A load cell (or cells) could be used
for the normal load. What about the torque? Torque could be measured either by
torque transducers or by properly placed linear load cells. It seems that the major
constraints for sensors will be spatial. Sensor placement within the machine will likely
become important, and from my experience this is generally the case 2 .
Temperature
Since frictional heating will play an important role in galling failure, a temperature
measuring device (or devices) would add significant information toward the indication
of the failure.
Acceleration
Measuring the vibratory acceleration within a pin joint may become helpful in detecting the beginning of galling failure.
3.6.9
Test Stand
With what could potentially become a massive structure, a well-designed test stand
may also become an important part of the overall design. Generally, a high stiffness
is desired for the foundations of a machine. For cases in which vibration becomes a
concern, damping elements may also be desirable.
3.6.10
Seals
The seals are very important and fundamental to the test machine design. In the
actual bulldozer pin joint, the seals are permanently preloaded onto the bushing face.
This is accomplished by pressing part of the track link onto the pin end.
Ideally, the seal forms an airtight (at least oil-tight) enclosure between the pin
and bushing, effectively preventing oil leakage. On the test machine, the seals must
be removed and reassembled; thus, there must be a repeatable method for preloading
the seals. The question of how much preload to apply is debatable.
12I feel compelled to note that allowing for the wiring of electrical devices (such as sensors) is
rather important when space considerations are a premium. I ran into this situation working for
Telerobotics at the Jet Propulsion Laboratory in Pasadena, California
57
3.7
3.7.1
Final Design
Relative Motion
The pin is to be kept stationary while the bushing is rotated. Tapers will be used
to hold the pin stationary. Next, the design needs to allow bushing rotation while
applying a 100,000 lbf load.
3.7.2
Bearings
Although taper bearings were first selected, spherical bearings were chosen for the
final design. This way, the normal load will not wreak havoc if there is any axial
misalignment. The bearing is made by Koyo. The basic specs are: 170 mm O.D., 260
mm O.D., 140,000 lbf static load capacity.
Figure 3-6: Koyo Spherical Bearing
3.7.3
Bearing Ring
The bearing ring turned out to be a difficult design. Possibilities were to use
" Two pieces, bolted together
" Two pieces, welded together
" One piece, casted
For welding, the required weld area would roughly be
58
A F
a
A--
_
100, 000 lbf
12, 000 psi
8.3 in 2
where 12,000 psi is an estimated weld strength with a built-in factor of safety
a thick weld of 1/4", the required lengths for a rectangle of sides 11 and 12 is
(11 + 12)
A
2t
13.
For
8.3 in 2
= 16.7 in
.
(2)(1 in)
This is feasible, but the disadvantage to welding can be the loss of precision. In
this case, precision is mandatory. Though it is possible to keep the precision with
careful manufacturing methods, it easier with the other two bearing ring possibilities.
Also, the large volume of weld material required can be costly. Last, the welds may
withstand the normal load, but the welded pieces may experience very high shear
stresses near the weld.
Using two pieces bolted together is also feasible. This would involve a circular
piece for the roller bearing and a plate to hold the bolt array. It turns out that the
bolts should have no problem handling the loads, but the plate deflection and stress
levels are high.
For these reasons stated above, the bearing ring is to be a casting (Figures 3-8,
3-9, and 3-10). For high strength and low weight, Zinc Aluminum was selected. As
cast, the material should have a yield strength of 50-55ksi.
Gussets (Ribs)
Gussets are to be designed into the bearing ring. Gussets can greatly increase the
strength and the stiffness of the bearing ring. They help distribute the load and,
thus, reduce stress concentrations and overall stress levels.
Flexures
As mentioned earlier, the spherical bearing will be implemented to allow axial misalignment of the pin joint in the test machine. As a result, there should be a method
to control this misalignment. Keep in mind the races of a spherical bearing can
completely misalign by 90 degrees.
Flexures will be used to restrain the bearing ring within the plane. Flexures
operate elastically and thus do not dissipate energy. This is a great attribute for a
testing machine, for energy losses can be narrowed down to the energy losses in the
pin if the testing machine were "elastic" in all respects. Otherwise, the energy losses
add, and results are less determinate.
Two flexures are to be on the diagonal across the bearing ring plate, providing a
support structure for bearing ring misalignment. It had been originally proposed that
the torque support also support the pin joint misalignment, but this idea had several
cons. First, the load path was long and crooked. Second, the torque support and the
13
This number could be raised by stress relieving the welds
59
axial misalignment were coupled. Introducing flexures de-coupled the functions and
greatly reduced the load path.
Retaining Rings
Two internal retaining rings will keep the spherical bearing inside the bearing ring
(Figures 3-8, 3-9, and 3-10). A bearing ring shoulder could have been used in place of
one of the retaining rings, but the manufacturing cost would have been significantly
higher than the cost of another retaining ring.
3.7.4
Support Base
The support base turned out to be a rather simple design. It is a 2" thick slab of
6061-T6 Aluminum to handle the bending load caused by the bolt array. One end
is cut out to allow the reciprocating lever to move in closer to the pin joint (Figures
3-8, 3-9, and 3-10).
Nut Preload
Since it requires two hands to apply enough torque with a torque wrench to achieve
the desired bolt preload, a design was incorporated to clamp the nuts into the support
base. This way, only one person is needed to tension the bolts.
3.7.5
Bushing Housing
Budiing Housng
Pin
Figure 3-7: Pin Joint with Bushing Housing and Clamps
The bushing'housing mount includes a shoulder to provide positive location of the
spherical bearing. The bearing will be press fit up to that shoulder (Figures 3-8, 3-9,
and 3-10).
60
Bushing Housing Mount
The bushing housing will not only support the normal load but also transmit the
necessary torque. A bushing housing mount is to be attached at one end. This
mount will support a lever arm attached to the torque transmission element (in this
case an electric motor).
Bushing Taper Clamps
To keep the bushing from slipping with respect to the bushing housing, tapered clamps
at each end of the housing will be used 1.
3.7.6
Pin Housing
A pin housing will support the normal load at each end of the pin.
Pin Taper Clamps
Similar to the -bushing housing, tapered pin clamps will keep the pin from slipping
with respect to the pin housing (Figures 3-8, 3-9, and 3-10).
4
See taper analysis in Appendix E
61
Retaining Ring
Bearing Ring
Bushing Taper Clamp
Bushing
Pin Taper Block
Bushing Housing
Pin Taper ClmpA
~tL
-
Pin
0
-
Load CelI
Spherical Bearing
Pin / Load Cell
Support
Support Base
Bolt Array
Figure 3-8: Cross section of bearing ring assembly
62
Bearing Ring
Retaining Ring
Bushing
Bushing Housing
Seal
Bushing Taper Clamp
Pin
Pin Taper Block
Pin Taper Clamp
Fo
0
Load Cell
0
C>
Pin / Load Cell
Support
Figure 3-9: Section cut of bearing ring assembly
63
2
3
1) Pin Taper Clamp
2) Pin Taper Block
3) Bushing Taper Clamp
4) Seal
5) Pin
6) Bushing
7) Bushing Housing
8) Retaining Ring
9) Spherical Bearing
10) Bearing Ring
10
6
9
F
-8
2
Figure 3-10: Exploded view of bearing ring assembly
64
3.7.7
Torque Transmission
An electric motor will transmit the torque required to rotate the pin with up to a
100,000 lbf load. To rotate the pin at 20 rpm, a 7.5 horsepower motor was selected.
Reciprocating Lever Arm
In order to rotate the pin joint cyclically, a reciprocating lever arm will be incorporated. A disk affixed to the gearbox output shaft will provide the reciprocating
motion. The design is very similar to the classic crank-slider mechanism, but in this
case it would be more aptly called a crank-partial-rotation mechanism. Instead of a
cyclical linear motion, there will be a cyclical partial rotation of the bushing.
65
3.7.8
Torque Support
"C" Piece
o0
00E
0
0
O
0
0::
0
i
I
C Piece
E
Axis of Rotation
T-ension / Compression
Load Cell
Load Cell Support
Figure 3-11: Torque Support
Shown in Figure 3-11, the torque support is a u-shaped piece attached to the pin
housings" . The midsection of the "U" will connect to a load cell to measure and
support the resistive torque in the pin joint.
3.7.9
Sensors
Normal Load
Two load cells will support the normal load. Each load cell will be placed under the
pin housing and will transmit the load through a pin support to the support base.
The selected load cells are capable of 100,000 lbf each and are made by Sensotec.
Torque
A third load cell will measure and convert the resistive pin joint torque into a linear
force. The force will be transmitted through a load cell support to the support base.
5
1 1t's
initial name was the "C" piece.
66
The load cell is designed by Sensotec for 10,000 lbf of either tension or compression.
Thermocouple
At the moment, only one thermocouple is used to measure the pin joint bulk temperature. As a recommendation, at least one more should be incorporated so that
non-uniform heating can be detected.
Accelerometer
An accelerometer has not yet been incorporated into the machine, but it is strongly
recommended to help catch the very beginning of galling failure. An accelerometer
should be able to detect failure before the sounds caused by galling are audible.
67
3.7.10
Seals
Unfortunately, a repeatable method for preloading the seals had not been designed
before most of the machine had already been built, but fortunately, a good solution
arose. Simply using a block with two holes, the bushing clamps can be used as a
foundation to preload the seals. With holes set on the same bolt diameter as the
threaded holes on the bushing clamp, long shank M8 bolts are passed through the
block to preload the pin connector and the seal assembly.
detailseall
Bushing
Seal (Red Piece)
A
Seal (Black Piece)--
Pin Taper Block
DETAIL A
Figure 3-12: Seal design (original)
68
1R.
M
seala~new
etailedl
Seal (Red Piece)
0 Ring Spacer
Bushing
0 Ring Retainer
Seal (Black Piece)
0 Ring
Pin
DETAIL A
Figure 3-13: Seal design (revised)
69
3.7.11
Test Stand
The test stand is a two piece design. The top piece will contain all the critical welds
and holes necessary for the test machine. The bottom piece is simply for vertical
support. This way, the overall cost is reduced. The original design was one piece,
but it was more expensive because more critical welds were required. Also, the load
path was significantly longer. Last, the loading is cyclical; hence, the test stand will
need heat treating to stress relieve the welds, and heat treating the whole test stand
would have been more expensive.
3.7.12
Assembled Test Machine
Figure 3-14: Test Machine, Front View
$.Nw*jvsx
,
rP~ C~1~~fl
~Cb~u~
Figure 3-15: Test Machine, Top View
70
Figure 3-16: Test Machine, Front View
71
72
Chapter 4
Pin Joint Designs
4.1
Undercut Bushing
The undercut bushing (Figure 4-1) is a good possibility for reducing the expected
stress concentrations. With a lower stress concentration, the pin and bushing surfaces
are less likely to yield and ultimately fail.
The undercut is simply a recess cut into the face of each end of the bushing.
This cut effectively turns each bushing end into a flexure. This is a great idea for
reducing stress while maintaining or increasing component life, for a flexure will trade
deflection for stress and will return to its original position without dissipating energy.
73
Figure 4-1: Undercut Bushing (after testing failure)
74
Chapter 5
Test Results
With the test machine completed, it is now possible to test pin joints and new pin joint
designs. Before reporting test results, it is very important to establish an assembly
procedure and a. testing procedure. After testing several pin joints, a basic assembly
procedure has been established. It is outlined in Appendix G.
A formal testing procedure has not yet been established, for the procedure is still
under debate. Once a testing procedure is established, it should be kept constant.
This ensures that comparison among pin joint tests will provide useful information.
5.1
Initial Tests
During initial tests, data was taken from the three load cells and from a thermocouple.
The thermocouple was placed through one of the pin clamp slits so that it rested
against the pin and the pin connector.
The third load cell data can be used to calculate the reaction torque and subsequently the friction in the joint. The reaction torque is simply the load cell reading
multiplied by the moment arm. The friction coefficient it is a function of the torque
T, normal force N, and the moment arm r.
T
=
Nr
T
P
-Nr
P
T
(F1+F
2
)r
where normal force is the sum of the two normal load cell readings, F and F 2 .
5.1.1
Standard Pin Joint
After establishing a fairly constant test procedure, the standard pin joint was tested.
Testing lasted for about 15 minutes total, but the actual pin joint rotation lasted for
about 10 minutes at about 10,000 lbf. Load cell and temperature data are shown in
Figure 5-1.
The end of testing coincides with about 600 seconds on the plots; hence, the data
shown after such time does not carry any significance.
75
(a)
x 104
1.20
LL
0
1 -1/
I-
100
200
300
400
500
600
(b)
200
-
.D
20*0
-J
-200
~-400
II
100
200
300
400
500
600
100
200
300
Time (s)
400
500
600
IC)
Figure 5-1: Test Results for Standard Pin Joint. (a) Normal Load versus Time; (b)
Third Load Cell versus Time (c) Temperature versus Time
The calculated reaction torque and friction coefficients are shown in Figure 5-2.
Note: the mean from the third load cell data was subtracted in order to calculate the
torque and thus the friction coefficient.
From both plots (a) and (b), it is evident when failure has begun. The third load
cell amplitude increases ( 275 lbf-ft at around 560 seconds). This is equivalent to the
friction coefficient surpassing about 0.13 to 0.14 (Plot b).
5.1.2
Undercut Bushing
The results for the initial undercut bushing test are shown in Figure 5-3.
The test lasted significantly longer. The pin joint endured 15 minutes at 10,000
lbs, 15 minutes at 20,000 lbs, and about 5 minutes at 30,000 lbs (Figure 5-3, Plot a).
The calculated reaction torque and friction coefficients are shown in Figure 5-4.
The reaction torque steadily increased during the 30,0001bf test until scraping
sounds were heard. At 20,0001bf (from 800 to 1700 seconds), the pin joint displayed
no evidence of failure.
76
(a)
300
0-
100
r-20050
100
200
150
250
0
350
400
450
500
550
300
350
Time (s)
400
450
500
550
300
Time (s)
0
0
(b)
0.2 -
0.1
-
100
50
150
200
250
0
Figure
5-2: Calculated (a) Reaction Torque and (b) Friction Coefficient for Standard
Pin Joint
L
0.0
(a)
X104
5
3
-
0
2.5 -2-~
I
I
,
I
.
I
01.5-
200
0
500-
o
400
600
800
1000
(b)
1200
1400
1600
1800
2000
70-I
0)
-0
-100
-
0
200
400
600
800
1000
(C)
1200
1400
1600
1800
2000
200
400
600
800
1000
Time (s)
1200
1400
1600
1800
2000
L70
60 50 -
$40 5300
Figure 5-3: Test Results for Undercut Bushing. (a) Normal Load versus Time; (b)
Third Load Cell versus Time (c) Temperature versus Time
77
(a)
60000
0
200
400
600
80
1000
1200
1400
1600
1800
2000
1200
1400
1600
1800
2000
Time (s)
(b)
0.-
0
0
0
200
400
600
800
1000
Time (s)
Figure 5-4: Calculated (a) Reaction Torque and (b) Friction Coefficient for Undercut
Bushing
78
5.1.3
Discussion and Conclusion
After initial testing, results indicate the undercut bushing may increase the service
life of a pin joint. Note this is by no means a definitive conclusion. There must be
more data to support the hypothesis.
5.1.4
Recommendations
Before running more tests, a test procedure should be established with reference to the
method of applying lubricant. For example, a few questions to be answered include:
" How much lubricant should be applied?
" If the pin joint leaks, should there be a continuous lubricant supply?
- Should the supply be constant flow or constant pressure?
Another thermocouple should be added to the setup. This way, asymmetric pin
joint heating could be detected. Also, it is strongly recommended that an accelerometer be incorporated. It would be very useful in detecting the onset of galling. It
is my guess that the accelerometer can detect the onset of galling much better than
the current method. Currently, galling is judged by listening for scraping sounds
emanating from the pin joint.
79
80
Appendix A
Four Point Beam Bending Analysis
P
P
L1.I
I.
L2P
P
Figure A-1: Four Point Pin Loading, Dimensions Added
L2 - L1
2
EI
=
EIkdxE ==
Px2 + C1
Ely
-- '
El dy
=
EI dx2
EIkdX
d3
Ely
=
-
Pa
=
3
-PL
2
Px=
3____
6
+C 3
+C 3 x 2 + C4
-
Px
1X1+C
Pax2
-
Ely
EI d
Px1
1
-2Pa
PL1 x 3 - 2Pax3 +C5
X2
ELA~
-Pax+3 5 X3 ±+C
2
6
Boundary conditions are
81
(1)
(2)
x=a
y=O
x=a
(3)
x=a +
d=d
dx
dX2
1
(4)
(5)
x=a + L1
x=a + L,
2
d=0
y=O
dy= d
Boundary condition (1):
Pa3
0
+C 1 a +C2
6
6
Boundary condition (3):
C3
2
= -Pa
_
PaL1
2
Boundary Condition (2):
Pa
2
_
Now,
C2
2
PaL1
2
can be found:
Pa3
Pa2 L 1
2
3
We can use either boundary condition (1) or (4) to find C4. For simplicity, take
(1):
Pa3 ± Pa2 L 1
2
2
Now, we can evaluate C5 and C 6 , but we know the deflection curve will be symmetric; hence, the analysis is finished. Simply reproduce the curve from x
0 to
x = a + -L and mirror it.
x,:
Ely
=Px'
6
Ely2=
X2
El
=
-a
2
3
2
Pa x1
_ PaLix, + Pa
2
a
2
2
2
3
PaLix 2 +Pa
2
±2±
+Pa
3
2
2
+Pa
L
2
2
Table A.1: Pin Parameters
Pin Diameter
r
66mm
Bushing Length L, =
240mm
Pin Length
L2=
295mm
9.
3x10E- 7m 4
r =
I =
Inertia
Elastic Modulus E =
200xlO9GPa
82
L,
Appendix B
Centroids of Circular Sections
y
y
0
EMMONS
C
0r
x
Figure B-1: Centroid Schematic
7-4I
AJ_X 1fri2
2
(r2_
-[2r
2A
2
[rec
x
22
2)3 -
-
C2XI
2c2
3
-
Check formula with centroid for a hemisphere. Let c
83
(B.1)
- ca]
=
0,
2
y
2[
3 _r 3 _
-
2
2r 3
lIr2
3
-
c2r]
4r
(B.2)
This matches the centroid for a hemisphere. To generalize, we need the area of
the arc section shown in Figure B-1.
1
Aa=-bh
2
where
h =rcos6
b = 2r sin 9
Combining,
A, = r 2 sin 0 cos 0
Last,
Acone = r 2o
Thus,
A = r 2 (O - sin 0 cos 0)
The final equation for
g
is now
_
[r2 _
2
c 2 X]
r (0 - sinGcos9)
84
Appendix C
Bolt Calculations
PL
BF
FS
where FS is the factor of safety, PL is the proof load, and BF is the calculated
bolt force.
LE
85
Table C.1: Factor of Safety Calculations for Applied Torque = 80lbf - ft
Diameter
Tensile Area
(mm)
(mm2)
10
12
14
16
36.6
80
84.3
115
157
Bolt Force
(kN)
61.8
10 x Bolt Force
(lbf)
152,000
54.3
122,000
45.2
102,000
38.8
87,000
33.9
76,000
86
Grade
Factor of Safety
4.8
Proof Load
(kN)
8.2
11.3
5.8
13.9
0.20
8.8
13.9
0.20
9.8
23.8
0.35
10.9
12.9
4.5
30.4
38.1
13.*1
0.45
0.56
0.24
4.8
18.0
0.33
5.8
22.0
0.41
8.8
9.8
10.9
12.9
22.0
37.7
48.1
56.3
0.41
0.69
0.89
1.04
4.5
4.8
5.8
8.8
9.8
T97.
26.1
32.0
32.0
54.8
0.42
0.58
0.71
0.71
1.21
10.9
70.0
1.55
12.9
81.8
1.81
4.5
25.W
0.67
4.8
5.8
35.7
43.7
0.92
1.13
8.8
9.8
10.9
43.7
74.8
95.5
1.13
1.93
2.46
12.9
4.5
112
35.3
2.89
1.04
4.8
5.8
8.8
9.8
10.9
48.7
59.7
94.2
102
130
1.44
1.76
2.78
3.01
3.83
12.9
152
4.48
5
0.12
0.17
Appendix D
Differential Screw Analysis
FF
-
Figure D-1: Differential Screw Schematic
P
F
N,
Figure D-2: Free body diagram with respect to lower wedge
N
2
F
P
Figure D-3: Free body diagram with respect to upper wedge
EFx = 0:
EFy =0:
F - fi cos A,
-
f 2 cos A 2 - N 1 sinA1 + N 2 sin A2 = 0
N 1 cosA, - N 2 cosA2 EFx2 = 0 :
f sin A, - f 2 sin A2
f 2 cos A2 = N 2 sin A2
87
= 0
(D.1)
(D.2)
(D.3)
EF.2
0:
P = N 2 cos A2 + f 2sinA 2
(D.A)
fi = pNi
(D.5)
pN 2
(D.6)
f2 =
Understand that Equations D.5 and D.6 are true in the limiting case in which the
objects of concern are at the point of slipping.
Simplifying EquationD.2,
N
1
A2
= N 2 cos A2+ 1 sin
cos A, - ft sin A,
(D.7)
and Equation D.4,
N 2 =P
cos A2 +
sin A2
(D.8)
we can combine to get
Ni =P
1
cos A, - p sin A,
(D.9)
Simplifying Equation D.1
F = N 1(p cos A + sin A) + N 2 (p cos A2 - sin A2 )
(D.10)
and combining with Equation D.9, we get
F
Ppcos A + sinA
cos A - 1 sin A
+ p cos A 2 -sin A 2
cos A2 + p sin A 2
(D.11)
Now, we need to validate the analysis. Is this equation valid? Does it make sense?
Well, if the friction coefficient were zero, we would be left with
F = P[tan A, - tan A2 ]
(D.12)
This result is similar to the result you would get for an analysis of a regular
threaded fastener. The only difference is the tan A2 term.
88
Appendix E
Taper Analysis
In some manner, the pin and the bushing must be affixed to another part of the
machine. In general, this is not a problem, but in this case, the torque is very high
(1650lbf -f t). To add to the complexity, this is accomplished by press fits in the real
case. But in our case, we want to be able to disassemble the pin and bushing so that
the machine parts can be used again. -One way to do this is with tapers.
FN
F
F
0
F
N
Figure E-1: Taper Free Body Diagram
EF :
F=Ff - FcsO - FNsin
(E.1)
N = FNCOSO - Ftsin6
(E.2)
EFy:
Ft= pFN
(E.3)
Ff = pN
(E.4)
Simplifying Equation E.A
- FNsinO
F-pFNcosO
fF
Ff
-
(E.5)
FN(pCosO + sinG)
and E.2
N =
=
FNcosO - pFNsirO
FN(cosO - psinO)
89
(E.6)
we combine to get
F
=pFN(COSO - psinO)
=
FN(sin0
-
FN(,pcosO + sin0)
(E.7)
2
+ 2p cos 0 - P sin)
F
sin0+2pcosO-p
90
2
sinO
(E.8)
Appendix F
Drawings
91
U
-
Is
ti
\-
Ir
ra
2
II
3a
g
V7
*1&
;,
71
g' I
- ,
U.,
413'1
4kV
~
---
-
r
I~A
tit
i-A
--4a 1-11.
-------
-I
,--.1 1
Figure F-1: Bearing Ring, Page 1
92
a
0
v
j
j1
L~
I
II
C4
I
:1
~
I&t
.00
00
I--
'I
b
U
0 0
0
0
-I
11
jI&
a
4
f-I
4
III
ii
I
I
I
U
4
Vp
Figure F-2: Bearing Ring, Page 2
93
I)
tr
I
I
0804sm
A
1;
.ti :4
"
*4
pl
16,
111111
IM
5
I
#A
4 1
C
p
Figure F-3: Bearing Ring, Page 3
94
MO
'.
Avwwwm
**
44
4
FiueF4CerigRnPg
95
4
Li
-J
U
4n
Iis
0
NA
IN
A
--
F
II
4&
Ii
a-
C
7
ASA
&
Se~ft
I
'A
1!
Figure F-5: Bushing Housing, Page 1
96
U
A
4
40
-11
yJ.
ii
*
-
A
iI
\\<
II
4,,h
4
I11
II~i
I'4-rt
VI
-V7
-4
Uod
l
-J
-
I;
a
Figure F-6: Bushing Housing, Page 2
97
40
-ti
£
5
L*
<
L.
0
F
--
4-
E
V
s*
Figure F-7: Bushing Housing, Page 3
98
U
'I
fff,
alli
...
.......
Fiur
F-:BsigHosnPg
99
til
I
'i
OC
I'
a
I
if
LU
Esa
5
71/
/
:0:
N
i!
C
10
U
Figure F-9: Bushing Clamp, Page 1
100
t'3)
(D
0m
71
CD
olq
A
I
C
D
&
-.'
t
e#
14"1%
W
ow nmea B
40-
aOia
DETA A
SCALE2. 1
R$HT SIDE VIEW
YI
A
IDnoS at A
I-
7
6
UY
ISOMETRC VIEW 2
BOTTOM VIEW
'A
'fu
C
s'*3I-
SIDING; CLAWP
DETAIL C
$CALE:2 1
oi
SCALE 1:2
SECTION 846
~
mesntaasss
2
9.
04C
t
Nw
*
:ft
II I
I''
'.4
C
0
U
I
Uj
<0
0
4
ii
-a
A.
a
~
-&3
~1
k
~3;
jfk i~
0
U
Figure F-11: Bushing Clamp, Page 3
102
1I
..
su*Vwiuworti
V.W4s
!rt.
-
4-4-
z
OW
&4w(jIS PW&VW pa4 dwQ
toi
Z:1 3lVOSf
slumN
IW
w
4
IVVil
O,
tumam
onmm2m
l* ".0
w~ I" "SPIN*
--A
t
0
fI*IWDt
I,
0
A
I',
'tft..
-
Art
*ta.
~-
s
-U-
4
.n.
P
a
.*
W4 tOOjiAiU
heat. a
iRD UV.
4
p
,.
AlVWVfltdijkL.
a
as to ms it."Avo
'ID
O
C
bCO
w
a
U
t
-
e4
01,
Figure F-13: Bushing Housing Mount, Page 2
104
USI
I
Kl'v
IL
141
ItI
If
_
TZ
4'
ii
WOO
I'
IN
Figure F-14: Pin Connector, Page 1
105
1V
II f
106
-ci
0
CDq
0
,!
CD
oq
A
0
t
W
a
roml3 c
ot"N54
r
a
7
6
-----
-, -A
ISOMAtRIC VIEW
NS VWID th 0a44*.,t W4It
0 wof Riq w.*te
1
6
-V
A
-
P#4C~*4H8IOA
t~'p
w1
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Appendix G
Testing Procedure
1. Cleaning
(a) Clean pin and bushing surfaces
(b) Clean all taper surfaces (bushing housing, pin clamps, bushing clamps, pin
connector)
2. Lubrication
(a) Grease seal spacers
(b) Grease 0 rings
(c) Grease red lip of seal
(d) Grease spherical bearing rollers. The most important rollers are the ones
located at the top of the ring near the eye (used for moving the housing).
3. Assembly
(a) Flexures to Bearing Housing
i. Fasten the flexures to the bearing housing but not to the support base.
Do not tighten them yet. Note: You may have to lay the bearing
housing on it's side to do this.
ii. Slide the bearing housing on the support base in between the pin supports. The housing should be slightly angled and partly overhanging
the support base (because of the flexures on one end). This will make
assembly easier for the "C" piece.
iii. Place 100,0001bf load cells in each support pin's locating hole.
(b) Bushing into Bushing Housing
i. Insert bushing into bushing housing
ii. Insert bushing clamps but do not clamp them yet
iii. Center bushing inside bushing housing
iv. Begin to hand tighten bushing clamps. Make sure the bushing remains
centered while tightening.
125
v. Tighten bushing clamps with wrench (again, make sure the bushing
remains centered). Make sure to tighten bolts uniformly to prevent
the clamp taper from prematurely grasping the bushing housing or the
bushing.
(c) Bushing Housing Mount
i. Fasten the bushing housing mount (curved piece with 4 radial holes
and 3 axial holes) to the bushing housing.
ii. Tighten bushing housing mount
(d) Seal Preloading
i. Completely assemble and tighten one pin connector to the "C" Piece.
Note: The threaded holes located on the taper face should be facing
away from the rest of the "C" piece.
ii. Insert pin into bushing
iii. Insert seal, black lip facing the bushing, on each side.
iv. Insert one plain (thin) seal spacer on each side. Insert shouldered
spacer at each end. Shoulder should face away from the bushing.
v. Insert one 0 ring on each side.
vi. Slide one side of the pin almost completely into the bushing. The 0
ring should be visible.
vii. Place the fastened pin connector's large bore hole over the barely visible side of the pin. Note: You may have to remove a few bolts from
the bushing clamp on the opposite side of the bushing to get this to
happen. Also, you may need to tighten the bushing clamps more.
viii. Carefully slide the pin into the pin connector. Note that the 0 ring
will likely get stuck on the lip of the connector. If so, stop pushing the
pin. Push the caught portion of the 0 ring back toward the bushing
while pushing the pin from the opposite side of the bushing.
ix. If the bearing housing is positioned correctly, you should be able to
rest the "C" Piece on the two 100,0001bf load cells.
x. Center the pin , and place the other pin connector on the pin. Be
careful, it is loose and should be immediately held on with the preload
blocks.
xi. Attach the two preload blocks. This is simply a block of Aluminum
with two through holes. Press one block face against the face of a pin
connector, aligning the block holes with those of the nearest bushing
clamp. Pass two long M8 bolts through the block holes to the two
threaded holes on the bushing clamp. Note: It may help to have one
person maneuver the "C" piece while attaching the preload blocks.
xii. Lightly preload each block. Note: For the loose pin connector, try
centering the connector (with respect to the pin axis) with one hand
while lightly preloading with the other hand. This will make the next
part easier.
126
xiii. The goal now is to align the holes on the loose pin connector. Increase
the preload. Alternate blocks while doing so. Meanwhile, check the
hole alignment. Once the holes are aligned, insert 4 M8's, and tighten
them completely. Make sure all 8 M8's on the "C" piece are firmly
tightened.
xiv. Remove preload blocks.
(e) Cut-Head Bolts
i. Position bushing housing mount such that it is near its linkage with
the electric motor.
ii. More than likely, one side of the "C" piece will be closer to the corresponding bushing clamp than the other. If so, the bolt heads close
to the "C" piece should be replaced with cut-head bolts. Note that
the bushing housing will rotate will rotate through about 30degrees
total. In this rotation, there should be at most three bolt heads (on
the crowded side) that may interfere with the rotation. Replace these
bolts with cut-head bolts. I simply used a grinder to shave some thickness off the heads.
(f) Placing Assembly on Load Cells
i. Make sure third load cell support is affixed to support base. Make
sure third load cell is hand tightened into this support. Using THREE
people, place the assembly onto the load cells. Use two people, one on
each side of the assembly, and a steel pipe through the bearing housing
eye to lift the assembly. The third person should guide hold onto the
"C" piece with both hands, guiding the large bore "C" piece onto the
third load cell. The bottoms of the pin connectors should contact the
100,0001bf load cells.
(g) Bolt Array
i. Align the bolt array holes on the bearing housing with the support
base holes. Place the blue M12's into all ten holes, partially fastening
them to the nuts on the underside of the support base.
(h) Attach Fork Link to Bushing Housing Mount
i. Guided by dowel pins, place two spacers onto the fork link.
ii. Align the dowel pins with the corresponding holes on the bushing
housing mount. This can be done by rotating the bushing housing
with the bushing housing mount while holding the fork link.
iii. Place two spacing blocks (steel block with 3 clearance holes) on the
opposite side of the fork link, and clamp them firmly with three M10's.
(i) Fastening and Positioning
i. Place the flexure pieces on the support base. Each side should include
one aluminum block with two through holes. There are also two shim
pieces. Level the bearing housing and place the shims where necessary.
Usually, only one is needed.
127
ii. Place a flexure cap (slightly smaller than the flexure block) over each
flexure blade and tighten firmly.
iii. Fasten and firmly tighten a washer and a nut on the third load cell.
iv. Using a torque wrench, tighten the bolt array as desired. Note that
the preload distance is small and easy to unload; hence, I recommend
you tighten each group of 5 and then the other group. Repeat this
process one more time to ensure uniform preloading.
128
Bibliography
[1] Bo Jacobson Bernard J. Hamrock and Steven R. Schmid. Fundamentals of Machine Elements. McGraw-Hill, 1999.
[2] Kenneth G. Budinski. Surface Engineeringfor Wear Resistance. Prentice-Hall,
1988.
[3]
Robert W. Fox and Alan T. McDonald. Introduction to Fluid Mechanics. John
Wiley and Sons, Inc., fifth edition, 1998.
[4] K. L. Johnson. Contact Mechanics. Cambridge University Press, 1985.
[5] H.S. Kong and M.F. Ashby. Friction - heating maps and their applications. MRS
Bulletin, pages 41-47, October 1991.
[6] Ernest Rabinowicz. Friction and Wear of Materials. John Wiley and Sons, Inc.,
1965.
[7] Joseph E. Shigley and Charles R. Mischke.
McGraw-Hill, fifth edition, 1989.
Mechanical Engineering Design.
[8] Gwidon W. Stachowiak and Andrew W. Batchelor.
Butterworth- Heinemann, second edition, 2001.
[9] Nam P. Suh. Tribophysics. Prentice-Hall, 1986.
129
Engineering Tribology.
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