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 MASSACHUSEU ITTU TE OF TECHNOLOGY OCT 2

LIBRARIES BARKER Room 14-0551 77 Massachusetts Avenue Cambridge, MA 02139 Ph: 617.253.2800 MITLiaries Email: [email protected] Document Services http://libraries.mit.edu/docs

DISCLAIMER OF QUALITY

Due to the condition of the original material, there are unavoidable flaws in this reproduction. We have made every effort possible to provide you with the best copy available. If you are dissatisfied with this product and find it unusable, please contact Document Services as soon as possible.

Thank you.

The images contained in this document are of the best quality available. 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 lubrica- tion. 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 ma- chine 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 17 1.1 M otivation ...... 17

2 Pin Joint Analysis 19 2.1 Pin Joint Description ...... 19 2.2 Dimensional Analysis (Scaling) ...... 19 2.2.1 Fluid Mechanics Example ...... 20 2.2.2 Pin Jo it ...... 21 2.3 Macroscopic Deformation...... 23 2.3.1 Beam Bending...... 23 2.3.2 Torsion ...... 25 2.3.3 Transverse Shear ...... 25 2.3.4 Combined Stress...... 26 2.4 Microscopic Deformation...... 29 2.4.1 Radial Gap Geometry ...... 29 2.4.2 Surface Parameters ...... 30 2.4.3 Undeformed Contact ...... 31 2.4.4 Contact Mechanics ...... 33 2.5 Tribological Considerations. 35 ...... 2.5.1 Lubrication ...... 35 ...... 2.5.2 Friction ...... 39 ...... 2.5.3 Frictional Heating ..... 40 ...... 2.5.4 Sliding Wear ...... 41 ...... 2.6 Failure Theories ...... 43 ...... 2.6.1 List of Parameters . 43 2.6.2 Discussion of Possible Faili ure Mechanisms . 44

3 Design of a Pin Joint Testing Machine 47 3.1 M otivation ...... 47 3.2 Basic Requirements ...... 47 3.3 Basic Design Rules ...... 47 3.3.1 Rules of Thumb ...... 47 3.3.2 Kinematic Constraint ...... 48 3.3.3 Iteration ...... 48 3.4 Normal Load Design ...... 48

7 3.4.1 Everything Turns Into Rubber 48 3.4.2 Hydraulics ...... 49 3.4.3 Threaded Fasteners and Lever Arms 49 3.5 Torque and Rotational Speed Design 53 3.5.1 Taper Analysis ...... 53 3.6 Machine Elements ...... 55 3.6.1 Bearings ...... 55 3.6.2 Bearing Ring . ... 55 3.6.3 Support Base . ... 55 3.6.4 Bushing Housing 56 3.6.5 Pin Housing..... 56 3.6.6 Torque Transmission 56 3.6.7 Torque Support .. . 56 3.6.8 Sensors ...... 56 3.6.9 Test Stand ...... 57 3.6.10 Seals ...... 57 3.7 Final Design ...... 58 3.7.1 Relative Motion . .. 58 3.7.2 Bearings ...... 58 3.7.3 Bearing Ring . ... 58 3.7.4 Support Base . 60 3.7.5 Bushing Housing 60 3.7.6 Pin Housing..... 61 3.7.7 Torque Transmission 65 3.7.8 Torque Support .. . 66 3.7.9 Sensors ...... 66 3.7.10 Seals ...... 68 3.7.11 Test Stand ...... 70 3.7.12 Assembled Test Mach ine 70

4 Pin Joint Designs 73 4.1 Undercut Bushing ...... 73

5 Test Results 75 5.1 Initial Tests ...... 75 5.1.1 Standard Pin Joint ..... 75 5.1.2 Undercut Bushing ...... 76 5.1.3 Discussion and Conclusion . 79 5.1.4 Recommendations ...... 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 41I1dI1iI1I, ill liii oI Iii ---hIiiIII-~iI~M~hEH~I~Iih~iil iihII I~lIhI~i~~~i IE lii II EhhhHIlIM 'I

no -MIN List of Figures

2-1 Pin Joint Schematic...... 19 2-2 Four Point Pin Loading...... 23 2-3 Four Point Pin Loading, Dimensions Added ...... 23 2-4 Pin deflection versus distance for different loads...... 24 2-5 Distributed Pin Loading ...... 25 2-6 Pin Cross Section ...... 27 2-7 von Mises Effective Stress for Applied Load P = 100, 000lbf, Applied Torque T = 2240N . m ...... 28 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 ...... 28 2-9 Radial Gap Schematic ...... 29

2-10 Gap Ratio R2 as a function of the Pin/Bushing Ratio R...... 31 2-11 Gap Ratio RI as a function of the Pin/Bushing Ratio R...... 32 2-12 Pressure versus Length of Line Contact...... 34

3-1 Bolt Force for an Applied Torque T = 80lbf - ft ...... 51 3-2 Bolt Factor of Safety for Applied Torque T = 80lbf - ft .... . 51 3-3 Total Force for a 10-Bolt Array, Applied Torque T = 80 lbf - ft 52 3-4 Bolt Optimization ...... 52 3-5 Force Amplification versus Taper Angle ...... 53 3-6 Koyo Spherical Bearing ...... 58 3-7 Pin Joint with Bushing Housing and Clamps ...... 60 3-8 Cross section of bearing ring assembly ...... 62 3-9 Section cut of bearing ring assembly ...... 63 3-10 Exploded view of bearing ring assembly ...... 64 3-11 Torque Support ...... 66 3-12 Seal design (original) ...... 68 3-13 Seal design (revised) ...... 69 3-14 Test Machine, Front View ...... 70 3-15 Test Machine, Top View ...... 70 3-16 Test Machine, Front View ...... 71

4-1 Undercut Bushing (after testing failure) .. 74

11 5-1 Test Results for Standard Pin Joint. (a) Normal Load versus Time; (b) Third Load Cell versus Time (c) Temperature versus Time ... . 76 5-2 Calculated (a) Reaction Torque and (b) Friction Coefficient for Stan- dard Pin Joint ...... 77 5-3 Test Results for Undercut Bushing. (a) Normal Load versus Time; (b) Third Load Cell versus Time (c) Temperature versus Time ...... 77 5-4 Calculated (a) Reaction Torque and (b) Friction Coefficient for Under- cut B ushing ...... 78

A-I Four Point Pin Loading, Dimensions Added ...... 81

B-i Centroid Schematic ...... 83

D-1 Differential Screw Schematic ...... 87 D-2 Free body diagram with respect to lower wedge ...... 87 D-3 Free body diagram with respect to upper wedge ...... 87

E-1 Taper Free Body Diagram ...... 89

F-1 Bearing Ring, Page I ...... 92 F-2 Bearing Ring, Page 2 ...... 93 F-3 Bearing Ring, Page 3 ...... 94 F-4 Bearing Ring, Page 4 ...... 95 F-5 Bushing Housing, Page 1 ...... 96 F-6 Bushing Housing, Page 2 ...... 97 F-7 Bushing Housing, Page 3 ...... 98 F-8 Bushing Housing, Page 4 ...... 99 F-9 Bushing Clamp, Page 1 ...... 100 F-10 Bushing Clamp, Page 2 ...... 101 F-11 Bushing Clamp, Page 3 ...... 102 F-12 Bushing Housing.Mount, Page 1 ...... 103 F-13 Bushing Housing Mount, Page 2 ...... 104 F-14 Pin Connector, Page 1 ...... 105 F-15 Pin Connector, Page 2 ...... 106 F-16 Pin Connector, Page 3 ...... 107 F-17 Pin Clamp ...... 108 F-18 Pin Support ...... 109 F-19 "C" Piece, Page 1 ...... 110 F-20 "C" Piece, Page 2 ...... Il F-21 "C" Piece, Page 3 ...... 112 F-22 Connector between Load Cell and "C" Piece ...... 113 F-23 Support for Third Load Cell ...... 114 F-24 Support Base, Page 1 ...... 115 F-25 Support Base, Page 2 ...... 116 F-26 Nut Preload Strip ...... 117 F-27 Nut Preload Strip, Slotted ...... 118

12 F-28 Nut Preload Strip Removal Block ...... 119 F-29 Test Stand, Page 1 ...... 120 F-30 Test Stand, Page 2 ...... 121 F-31 Test Stand, Page 3 ...... 122 F-32 Test Stand, Page 4 ...... 123

13 I U U

14 List of Tables

2.1 Pin Joint Clearance at Bushing Midsection ...... 20 2.2 Pin Joint Clearance at Bushing Edge ...... 20 2.3 Kinematic viscosities for 80W-90 oil...... 35 2.4 Friction Coefficient Ranges for Lubrication Regimes ...... 37 2.5 Film Parameter for Lubrication Regimes ...... 37

A .1 Pin Param eters ...... 82

C.1 Factor of Safety Calculations for Applied Torque = 80lbf - ft ..... 86

15 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 lubrica- tion. 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 ma- chine 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 resolv- ing 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: 67.88 - 0.25 = 67.63mm Pin maximum diameter: 66.725 + 0.025 = 66.75mm

Minimum Clearance: 00.88mm

Bushing maximum diameter: 67.88 + 0.25 68.13mm Pin minimum diameter: 66.725 - 0.050 66.675mm

Maximum Clearance: 1.455mm

Table 2.2: Pin Joint Clearance at Bushing Edge

Bushing minimum diameter: 67.18 - 0.13 = 67.05mm Pin maximum diameter: 66.725 + 0.025 = 66.75mm

Minimum Clearance: 00.30mm

Bushing maximum diameter: 67.18 + 0.13 67.31mm Pin minimum diameter: 66.725 - 0.050 66.675mm

Maximum Clearance: 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 fixed and all the others must be varied, one at a time. To test 10 values for each variable, the total number of combinations is 104 . That means there must be 104 tests! There must be a better way. There is: dimensional analysis. By looking at the fundamental units involved, it can be shown that a functional relationship exists between two nondimensional parameters:

pV2D2 (p7D)

20 This relationship reduces the necessary tests from 104 to 10 [3, p. 298-299], a reduction of 3 orders of magnitude.

2.2.2 Pin Joint As in the fluid mechanics example, it is desirable to find a fundamental set of nondi- mensional 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,

11 = Fad'o- = ML 0Lt

(MLa (L) 0 ( = M L t0 t 2 Dt2 and the system of equations,

M: a+1=0 a+b-1=0 t: -2a-2=0 we can solve for the unknowns using least squares.

2 [a ]E[-,l = 1l ad _ a [bJ[2j F (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 0- 1F 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.

n 3 = d-

H = - 4 D Since the Poisson's ratio is dimensionless and because it is important in solid me- chanics, 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 Pois- son's ratio, the rank of the matrix is two instead of three. This results in three rather than two nondimensional parameters.

22 2.3 Macroscopic Deformation

2.3.1 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 y=o x=a = dx1 daX2 x=a + L dX2 x=a + y=O x~a + Li d dX { 2 3 23 Where

L2 - L1 a = 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 - - - -- 100,0001bf 75,0001bf 0.15 -- 50,0001bf - - - 25,0001bf -. 2,OOb 0.2

0 50 100 150 200 250 300 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 _ 222 500 N N lbf - = ' -= 930x1006 - =5.6x106 - 3 0.24x10- 3 m m in The bending stress can be calculated.

Mc 01-= (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 E ' M ad I U9 Lb tt 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 = pNr (2.4) = (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

j= - c) 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 esti- mated: (2240 N - m)(0.033 m) T = - 4 = 39.6 MPa (2.6) (0.033 M) Another parameter of interest is the torsional stiffness, defined as

T JG (2.7) #L

6 9 5 T _ (1.86x10- m)(75x10 ;-) 5.82A N-m (2.8) 0.24 m radian

2.3.3 Transverse Shear

Transverse shear stress can be calculated as

VQ (2.9) it

25 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

Q = JydA = 9A' 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

2 2 [r x _ - c X] A' 2 2 [r x - --c x] 2 r (9 - 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 )t 2 (222, 500 N)(1.35x10- m)(1.58x10- 3 M2 ) (7.98m10 7 m4 )(0.066 m)

- 90.1 MPa

2.3.4 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- TX_ y 77 z - Tz X ) 0 2 2 -(UXowoz + TxyTyzTzx - ax2 2 TT -0 (2.11)

At point A (Figure 2-6), the roots are

A,1 = 220 MPa CA,2 = 39.7 MPa CA,3 = -39.7 MPa (2.12)

26 Ir I A B Q I| I 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 distor- tion energy in a unit volume equals the distortion energy in the same volume when uniaxially stressed to the yield strength [7, p. 245].

2 2 - ) 2 [( -- 2 ) + (a2 - a 3) + (a 0 1

The effective stresses at points A and B are

(TA = 230 MPa

('7B = 167 MPa (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 100,000lbf, Applied Torque T = 2240N . m

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.

801 801 401 f)O 60 40 40 40: 4 20 ~ 20 20 40 08 X) 40 60 0 20 40608W 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.

2The 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 Microscopic Deformation

2.4.1 Radial Gap Geometry Pin joints naturally have radial clearance between the pin and the bushing. A math- ematical description of this geometry is useful for analyzing contact stress.

Radal Gap I 0

R r r

I g

Figure 2-9: Radial Gap Schematic

From Figure 2-9, it can be shown that

A = R -r (2.15) ARr

= - # (2.16)

(2.17)

#= - (a + 0) (2.18) 2 I sin/ sinO0 (2.19) A r Combining Equations 2.16 and 2.18, we obtain

2 2 # +0 q$-2O (2.20) (2.21)

Now, combining 2.20 with 2.19, we have

29 0 + arcsin A sin 0 (2.22) 1r I The radial gap can then be calculated:

g., = Rsin0- rsin# (2.23)

gy = Rcos0-(rcos#+A) = Rcos-(rcos#+R-r) (2.24) (2.25)

Using the Pythagorean Theorem,

g = g +g2 (2.26) 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 R # 0 + arcsin[(- - 1) sin 0] (2.27) r

= sinO - 7sin# (2.28) R RR

9= V(L)2 + ( )2 (2.30) R R R 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: cen- terline average roughness (Ra) and root-mean-square roughness (Rq)[1, p. 316]:

1N Ra = Izil (2.31)

Rq = ( z2) ( 2.32)

where zi is the height from the centerline and N is the number of measurements.

30 0 0 6 . r

0.05 ------0.956 -.-.- 0.961 i --- 0.967 5 0.04 - 0.972 -- 0.978 > -- 0.983 - 0.989 0.994- CD -1.000 0.03 -

a0.02

0.01

0 0 10 20 30 40 50 60 70 80 90 Bushing Half Angle (degrees)

Figure 2-10: Gap Ratio as a functionR of the Pin/Bushing Ratio LR

2.4.3 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 3x10- 5 R=1 pm 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 0 20 , R=1 pm Hence, the radial contact length (s = r(20)) is

{ 0.69 mm , R=0.1 pm 2.3 mm , R=1 pm

Taking the bushing length, the order of the undeformed stresses can be calculated.

{ .445,000 N m)i 2.7 GPa , R=0.1 pm

(0.00239m)(0.24 m) = 800 MPa , 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 x 10-6 9

8 0.950 0.956 . 0.961 0.967 7 - 0.972 - 0.978 _r... 0.983 -6 - 0.989 Cu - 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. Hertz defined an effective elastic modulus and effective radius 4 as

E* = + 12 (2.33)

F 1- R = I I (2.34)

With load per unit length P, the semi-contact width can be calculated:

(4PR)2 (2.35) a = rE* (.5

The maximum contact pressure can also be written as (PE*N 2P Pmax = ------7rR 7ra (2.36)

Assuming that both the bushing and the pin are steel:

2(1 - (0.3)2) 110 GPa 200 GPa

R 2.21 mn R 0.033 n 0.0335 m1

44,000 NN P = = 1.85x10 6 - 0.24 m m

6 4(1.85x10 N)(2.21 m) 2 ar(110x10 9 N) 0.00688 m= 6.88 mm

6 2(1.85x10 N) Pmax - ' = 171 MPa 7(0.00688 m)

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-

1200 0-

- 1000 -

E 800 -

- ~600-

400-

20

0 50 100 150 200 250 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". Johnson explains it beautifully:

... smooth non-conforming surfaces in contact were defined: the initial separation between such surfaces in the contact region can be repre- sented to an adequate approximation by a second-order polynomial. Non- conforming 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 fre- quently 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 dimen- sions 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©40'C %A@100C (W) (2) 8OW-90 1.4x 10-4 1 .42x10-b

35 Pressure-Viscosity Relationships

The most well-known pressure-viscosity relationship was formulated by Barus:

-,= 77e"i (2.38) where 77 is the viscosity at pressure p, r7 is the atmospheric viscosity, a is the pressure-viscosity coefficient, and p is the pressure of concern. Unfortunately, this equation is not very accurate for pressures above 0.5GPa [8, p. 17], but it at least provides an order of magnitude estimate. Stachowiak describes the problem facing the determination of a fundamental, analytical pressure-viscosity relationship:

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 con- tacts however, the lubricant operates under very high shear rates, and the precise values of the pressure-viscosity coefficient are needed for the evalu- ation 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 ap- plications [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 Lubrication Friction Regime Coefficient Hydrodynamic 0.0006 to 0.002 Elastohydrodynamic 0.002 to 0.06 Boundary 0.06 to 0.2 Unlubricated ;> 0.5

hmin (2.39) (Rq2a + )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 rough- ness, as shown in Table 2.5[1, p. 318]5.

Table 2.5: Film Parameter for Lubrication Regimes Lubrication Film Regime Parameter Hydrodynamic 5 < A < 100 Elastohydrodynamic 3

Boundary and Extreme Pressure Lubrication

As tests have indicated, the pin joint operating regime classifies as boundary lubri- cation (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

5Hamrock 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 lubri- cation 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 con- tacting 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 con- tacts. 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]

_)~2r W [2-- )2 -2 1 P = sin-- H _W 2r _W _ 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. For nearly flat surfaces, the adhesive friction coefficient can be defined as [9, p.79] f A"1 + 1! + cos-' f + sin(cos- 1 f) where pa varies from 0 to 0.39 as f changes from 0 to 1.

39 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 k2 are thermal conductivities, and 11b and 12b are lengths.

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 1 (2.41) T - Tb'= A, &+E k221 11f 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 important factor 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 ac- tion, 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 pre- existing 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 nor- mal 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 Ar H 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 (KxA) 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 List of Parameters e 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 ther- mal 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 ro- tational 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, re- adsorption 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

10This 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. Con- sequently, asperity adhesion and plowing will continually grow. The effective film thickness will decrease; thus, the real contact area will increase. Large wear parti- cles 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. Con- sequently, 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 Basic Design Rules

3.3.1 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 under- constrain 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 Normal Load Design

3.4.1 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 fas- teners 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 (3.1)

F = (3.2) Kd 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 (Ap- pendix D), the effective K is

y cos A + sin A p cos A - sin A K::=:[co ~ 1 lsn 1 + A 2 i 2] (3.3) cos A, - p si~n A, cos A2 + p sin A2

4The "minimize load path" design rule is thus very important in this case. 5 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). 6ForK , 0.2, the friction coefficient pg 0.15

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 F =T Kd 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 strength7 . 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!). 8Perhaps this all seems trivial to you, but I had never thought bolt selection could be so challenging.

50 18-

18-

14 -

12-

10 -

I B a-

4-

2-

0 ' L 6 8 10 12 14 16 18 20 22 Bok Diameter (mm)

Figure 3-1: Bolt Force for an Applied Torque T = 80lbf - ft

5.0-

4.5 - -a-Grade 4.6 S -- Grude 4.8 0 4.0 - -- Grade 5.8 IL -.- Grade 9.0 U 3.5 - -+-Grade 10.9 -U S 3.0- -Grade 12.9 .3 -Minimum Faclor of Safety 2.5 -

2.0-

1.5- U, 0 1.0- dr 0 S 0.5 - IL 0.0 6 8 10 12 14 16 18 Bolt Diwneter (mrn)

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 - S40,000-

20,000 - 0 i . . . ' ' * i 6 8 10 12 14 16 18 20 22 Bolt Diameter (min)

Figure 3-3: Total Force for a 10-Bolt Array, Applied Torque T = 80 lbf - ft

2.0 200,000 Acceptable Strength and Applied Force / 175,000

0 1.5 - 150,000 A - Z/ 125,000 U.0 / - M 0 .a ~ 100,000 0 LL #00-Grade 4.5 -75,000 / -Grade 4.8 -Grade 5.8 50,000 LA. 0.5 +o - APA- -Grade Grade 9.810.9 - -A "Grade 12.9 -25,000 -enimum Factor of Safety Total Soft Force U.0 . . 0 8 9 10 11 12 13 14 15 16 Bolt Diameter (mm)

Figure 3-4: Bolt Optimization

52 3.5 Torque and Rotational Speed Design

Since there are both torque and rotational speed requirements, rotational power be- comes 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,

rad lbf - ft_ P = (1650 lbf -ft)(2 ) = 3300 -= 6 hp (3.4) s s

3.5.1 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 FN = F 22(3.5) sin0 + 2p cosO - p 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 4 5 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 T 1650 lbf -ft N = 75, 000 lbf (3.6) Pr (0.2)(0.11 ft) where T is the applied torque and r is pin radius. For the bushing, the required force is

T 1650 lbf - ft N -- =46, 000 lbf (3.7) pIr (0.2) (0.18 ft) 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 Machine Elements

3.6.1 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 . Cylin- drical 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 spher- ical 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 de- tecting 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 Final Design

3.7.1 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 _ 100, 000 lbf A-- 8.3 in2 a 12, 000 psi where 12,000 psi is an estimated weld strength with a built-in factor of safety 13. For a thick weld of 1/4", the required lengths for a rectangle of sides 11 and 12 is A 8.3 in 2 . = 16.7 in (11 + 12) 2t- (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 mis- alignment 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

13This 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

- Pin

~tL 0

- Load CelI

Spherical Bearing Pin / Load Cell Support

Bolt Array Support Base

Figure 3-8: Cross section of bearing ring assembly

62 Bearing Ring Retaining Ring

Bushing Housing Bushing Seal

Bushing Taper Clamp Pin

Pin Taper Block

Pin Taper Clamp Fo 0

Load Cell 0

Pin / Load Cell C > 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 10 4) Seal 5) Pin 6) Bushing 7) Bushing Housing 8) Retaining Ring 9) Spherical Bearing 10) Bearing Ring

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 incor- porated. 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

E C Piece

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 tem- perature. 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 A Seal (Red Piece)

Seal (Black Piece)-- Pin Taper Block

DETAIL A

Figure 3-12: Seal design (original)

68 M 1R.

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 , r- P~ 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 subse- quently 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 P T -Nr T P (F1+F 2 )r where normal force is the sum of the two normal load cell readings, F and F2 .

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 x 104 (a)

1.2- 0 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

IC)

100 200 300 400 500 600 Time (s)

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-200-

50 100 150 200 250 300 350 400 450 500 550

0 Time (s) 0 0 (b)

0.2 -

0.1 -

50 100 150 200 250 300 350 400 450 500 550 Time (s) 0

Figure 5-2: Calculated (a) Reaction Torque and (b) Friction Coefficient for Standard Pin Joint

L 0.0 X104 (a) 5 3 - 0 2.5 -- 2- -~ I , I I I . 01.5-

0 200 400 600 800 1000 1200 1400 1600 1800 2000 (b) 500- o 70-I

0)

-0 -100

- 0 200 400 600 800 1000 1200 1400 1600 1800 2000 (C) L70 60 - 50 - $40 - 530- 0 200 400 600 800 1000 1200 1400 1600 1800 2000 Time (s)

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)

600-

00

0 200 400 600 80 1000 1200 1400 1600 1800 2000 Time (s)

(b)

0.-

0 0

0 200 400 600 800 1000 1200 1400 1600 1800 2000 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 accelerom- eter 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 I. L1.I

L2- P P

Figure A-1: Four Point Pin Loading, Dimensions Added

L2 - L1 2

EI = Px1

EIkdxE = Px2 + C1 Ely -- ' 1X1+C

El dy = Pa

EI - Pax dx2 2 +C 3

Ely =- +C 3x 2 + C4

EI d = Px 3 -PL 1 -2Pa

EIk d - 2 PL1x 3 - 2Pax +C5 dX3 3 Ely Px=3____ X2 6 ELA~2 -Pax+3 5 X3 ±+C6

Boundary conditions are

81 (1) x=a y=O (2) x=a d=d dx1 dX2 (3) x=a + 2 d=0

(4) x=a + L1 y=O (5) x=a + L, dy= d Boundary condition (1):

Pa3 0 6 +C a +C2 6 1 Boundary condition (3):

2 C3 = -Pa _ PaL 2 1 Boundary Condition (2):

2 _ Pa PaL1 2 2

Now, C2 can be found:

3 2 Pa Pa L 1 3 2 We can use either boundary condition (1) or (4) to find C4. For simplicity, take (1):

3 2 Pa ± Pa L1 2 2

Now, we can evaluate C5 and C6 , but we know the deflection curve will be sym- metric; hence, the analysis is finished. Simply reproduce the curve from x 0 to x = a + -L and mirror it.

2 3 2 =Px' Pa x _ PaLix, + Pa +Pa L x,: Ely 6 2 1 2 3 2 3 2 Ely2= -a 2 PaLix 2 +Pa +Pa L, X2 El= 2 a 2 2 ±2± 2

Table A.1: Pin Parameters Pin Diameter r 66mm Bushing Length L, = 240mm Pin Length L2= 295mm 9 . 7 4 Inertia I = r = 3x10E- m Elastic Modulus E = 200xlO9GPa

82 Appendix B

Centroids of Circular Sections

y

y 0 C EMMONS

0r x

Figure B-1: Centroid Schematic

7-4IAJ_X 1fri22 22

(r2_ 2)3 - C2XI -[2r 2 x - - 2c2 2A 3 [rec -- ca] (B.1)

Check formula with centroid for a hemisphere. Let c = 0,

83 2 3 _r 3 _ y 2 [ - - c2r] 2 2r3 lIr2 3 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,

2 A = r (O - sin 0 cos 0) The final equation for g is now

2 _ [r2 _ c X] r2 (0 - sinG cos9)

84 Appendix C

Bolt Calculations

PL FS BF 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 Bolt Force 10 x Bolt Force Grade Proof Load Factor of Safety (mm) (mm2) (kN) (lbf) (kN) 36.6 61.8 152,000 5 8.2 0.12 4.8 11.3 0.17 5.8 13.9 0.20 8.8 13.9 0.20 9.8 23.8 0.35 10.9 30.4 0.45 12.9 38.1 0.56 10 80 54.3 122,000 4.5 13.*1 0.24 4.8 18.0 0.33 5.8 22.0 0.41 8.8 22.0 0.41 9.8 37.7 0.69 10.9 48.1 0.89 12.9 56.3 1.04 12 84.3 45.2 102,000 4.5 T97. 0.42 4.8 26.1 0.58 5.8 32.0 0.71 8.8 32.0 0.71 9.8 54.8 1.21 10.9 70.0 1.55 12.9 81.8 1.81 14 115 38.8 87,000 4.5 25.W 0.67 4.8 35.7 0.92 5.8 43.7 1.13 8.8 43.7 1.13 9.8 74.8 1.93 10.9 95.5 2.46 12.9 112 2.89 16 157 33.9 76,000 4.5 35.3 1.04 4.8 48.7 1.44 5.8 59.7 1.76 8.8 94.2 2.78 9.8 102 3.01 10.9 130 3.83 12.9 152 4.48

86 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: F - fi cos A, - f 2 cos A 2 - N 1 sinA1 + N 2 sin A2 = 0 (D.1)

EFy =0: N 1 cosA, - N 2 cosA 2 - f sin A, - f 2 sin A2 = 0 (D.2)

EFx2 = 0 : f 2 cos A2 = N 2 sin A2 (D.3)

87 EF.2 0: P = N 2 cos A2 + f 2sin A2 (D.A) fi = pNi (D.5)

f2 = pN 2 (D.6) 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,

cos A2+ 1 sin A2 N 1 = N 2 (D.7) cos A, - ft sin A,

and Equation D.4,

N 2 =P (D.8) cos A2 + sin A2 we can combine to get

Ni =P 1 (D.9) cos A, - p sin A,

Simplifying Equation D.1

F = N 1(p cos A + sin A) + N2 (p cos A2 - sin A2 ) (D.10)

and combining with Equation D.9, we get

-sin A F Ppcos A + sinA + p cos A2 2 (D.11) cos A - 1 sin A cos A2 + p sin A2 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) EFy: N = FNCOSO - Ftsin6 (E.2) Ft= pFN (E.3) Ff = pN (E.4) Simplifying Equation E.A

F-pFNcosOfF - FNsinO (E.5)

Ff - FN(pCosO + sinG) and E.2

N = FNcosO - pFNsirO (E.6) = FN(cosO - psinO)

89 we combine to get

F =pFN(COSO - psinO) - FN(,pcosO + sin0) (E.7) = FN(sin 0 + 2p cos 0 - P2 sin)

F sin0+2pcosO-p 2sinO (E.8)

90 Appendix F

Drawings

91 - U Is ti \- - ra Ir 2 II 3a g V7 *1& ;,

71

g' I

- , U., 413'1 4kV

~ -

--- r I~A tit i-A 0

a ------4a 1 -11. -I ,--.1 1

Figure F-1: Bearing Ring, Page 1

92 j v

j1 L~ I

I :1 U C4 ~ 4 II I&t ii I I

00 .00 f-I

4

I--

'I b U I 0 0 0 0 -I 11 jI& III a Vp 4

Figure F-2: Bearing Ring, Page 2

93 I)

tr I I

0804sm " A 1; *4 .ti :4

16, pl

111111IM

5 #A I

4 1

p C

Figure F-3: Bearing Ring, Page 3

94 MO '. Avwwwm

**

44

4

FiueF4CerigRnPg

95 4

Li

-J U 4n 0 Iis II 4& NA IN F

--A Ii

a- C

7

ASA & Se~ft

'A I

1!

Figure F-5: Bushing Housing, Page 1

96 A U 40 4

yJ. -11 * ii - A \\< ii I

40

-ti £ II 5

4,,h

I4 11 VI -V7 -4 II~i Uod l -J Irt - I;'4-

a

Figure F-6: Bushing Housing, Page 2

97 L* <

L. E

F 0 -- 4- V

s*

Figure F-7: Bushing Housing, Page 3

98 'I U

fff, alli

......

Fiur F-:BsigHosnPg

99 til I

'i

OC

I' a I if

LU

Esa

5

71/ / :0: N

Ci! 10

U

Figure F-9: Bushing Clamp, Page 1

100 6 2

ow nmea B IDnoS at A aOia SCALE1:2

D A UY C

7

14"1%-.' t e#W olq & YI

CD I- C

71 0m

'fu SECTION846 I 40- BOTTOMVIEW ~ R$HTSIDE VIEW

(D t'3)

DETAILC $CALE:21 A

DETA mesntaasss s'*3I- A SIDING; CLAWP SCALE2.1 'A ISOMETRCVIEW 2 oi 9.

04C t * Nw :ft II I '.4 I''

C

0 I U Uj <0

4 0 ii -a A.a

k -&3~1 ~

~3; jfk i~

0 U

Figure F-11: Bushing Clamp, Page 3

102 t 0 4 1I fI*IWDt

w tumam

4 IVVilO, .. su*Vwiuworti l* ".0 w~I" "SPIN* &4w(jIS PW&VW pa4dwQ V.W4-s - 4-4- OW onmm2m toi !rt.

0 C

--A 'ID

- ~- -U- I', bCO

I,

A 4

IW

a

Z:13lVOSf heat. a a as to ms it."Avo O 'tft.. *ta. iRD UV. W4 tOOjiAiU AlVWVfltdijkL. slumN Art s .n. P .* p ,. z 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 a 6 6 roml3 c NS VWID th 0a44*.,t W4It t W ot"N54 r 0 wof iqR w.*te SCALE12

0 ------V

oq

CD

-ci

0 ,!

CDq0 ISOMAtRICVIEW 1 ASNERICVIW 2

&tmactwwth hslNvlv4 1.docb~g~ - t~'p w1 A A P#4C~*4H8IOA

A A

a 7 -, -A ci 9 C ::;i~

I It :6 4 4, 'I ii S I t ee II3RS S a, I-.

N

C,

.0

TUTTtft- 11

Ii 4 I J-

Figure F-17: Pin Clamp

108 C 4 2 I;, A> in ~, ...... h:. ; II:. 'A "u H

It *~ -9,

10 'I 'bit ~ I t a ;.7*~ ~ ~ r K. *

I

______I '-1 I?

4 -R -U I! if K *3 < tad j jIi .jI I;i *

a U

Figure F-18: Pin Support

109 I

Q) 0 L1. 'I 4 A III'

It Ii P T4 Iow 0; A>~n~7jfj I

ze 4.) I I 9 1W' II j0

W

rUr v4

In Bi ap

0 V

Figure F-19: "C" Piece, Page 1

110 LLJ SQ V) 0 i'

F 4!LiaSR

LU

1* [ £ :1

/ 9

13

Figure F-20: "C" Piece, Page 2

111 w

4

LLA 01a

I'Ii I g I I, I 'I I k :y NI FI FR.II

'U

(/-

0

4

-t III sly

a

Figure F-21: "C" Piece, Page 3

112 I a

Piif

T.

"lit

113

3, I

L + - I

Figure F-22: Connector between Load Cell and "C" Piece

113 it 0

TT

Figure F-23: Support for Third Load Cell

114 p

S '-ii -'1 U 110 p i r Y "Y's as r 0s i I I, ~ I t St It ii r3 .1 +~1 al -±4 I 2 rI :4 I * * & S -~ 14 4 9tz

0

0, 0 40

I -

I*

rt 1-00 P 4t

Figure F-24: Support Base, Page 1

115 C.4

41 IIt

rjr

-0 . -

a z

Figure F-25: Support Base, Page 2

116 m

U 'ii

iiy~i

fYI: I Ijr , iij I

v

I I 7 4 011I p

- £

Figure F-26: Nut Preload Strip

117 0. v :11 ItJ I It sAsufinfl. ~rC~FIII9 11<

n.

* 0 ! tIt

~jfi-T Ii if

jl t I V 4lb U IS U ~ "2 I- - ',i&-" UT '. I J L ~!i I 'p

III .------I "4 - C

Figure F-27: Nut Preload Strip, Slotted

118 M

4

N i

rOil ' ! e S at I

SI I

I

I '4 lhlet 1:1I;-Sue ~ ill

Figure F-28: Nut Preload Strip Removal Block

119 __q

0 0 b Ii ~ I ,7rl 'IIi I UO

A ___I Lu 4

Ai ...... ~ -ci 1 0 :1; ~

a in - I 3

3- '1Ii

.1 4 I U-

~u2

w h

a

Figure F-29: Test Stand, Page 1

120 U 3-C -- Wa- -v-i-- i-i WAI@I If Ii Iii III II j0~ F 4 *2

4J~ fl___ IU

2-

if-v whow"Awkw- 91j a-- r 41

'0 - 4-TI 9 1- 44- 01V -t r ,~ I C 2* Lg 'V a - 14-! -OO ~1 ci F II -C, I a

Figure F-30: Test Stand, Page 2

121 4 U

S I II I ~1' aI I 5 .1 0! +1W N 0~ ji! I e tr j

~1 n 9 1

'S a a -m

.01 * I -- - -0- - I I I I

4 2ev I o ~ U a 4) S I I SR

1 g

K''

S r

a LI

Figure F-31: Test Stand, Page 3

122 oj I

INN

vlo

Fiur F-27etStnPg/

123 124 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 sup- ports. 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 vis- ible 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 cor- responding 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 thick- ness 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 Ma- chine Elements. McGraw-Hill, 1999.

[2] Kenneth G. Budinski. Surface Engineering for 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. Mechanical Engineering Design. McGraw-Hill, fifth edition, 1989.

[8] Gwidon W. Stachowiak and Andrew W. Batchelor. Engineering Tribology. Butterworth- Heinemann, second edition, 2001.

[9] Nam P. Suh. Tribophysics. Prentice-Hall, 1986.

129 L1L120 -(c