A Critical Analysis of Pin Bending Behavior Using Finite Element
Analysis
by
Edward Kwon
An Engineering Project Submitted to the Graduate
Faculty of Rensselaer Polytechnic Institute
in Partial Fulfillment of the
Requirements for the degree of
MASTER OF ENGINEERING
Major Subject: MECHANICAL ENGINEERING
Approved:
_________________________________________
Ernesto Gutierrez-Miravete, Project Adviser
Rensselaer Polytechnic Institute
Hartford, CT
December, 2013
© Copyright 2013
by
Edward Kwon
All Rights Reserved
ii
CONTENTS
A Critical Analysis of Pin Bending Behavior Using Finite Element Analysis .................. i
LIST OF FIGURES .......................................................................................................... iv
LIST OF TABLES ............................................................................................................. v
LIST OF SYMBOLS ........................................................................................................ vi
KEYWORDS/GLOSSARY ............................................................................................ vii
ABSTRACT ................................................................................................................... viii
1. INTRODUCTION/BACKGROUND .......................................................................... 1
2. THEORY/METHODOLOGY ..................................................................................... 2
3. RESULTS AND DISCUSSION ................................................................................ 11
4. CONCLUSION.......................................................................................................... 20
REFERENCES ................................................................................................................ 22
APPENDICES ................................................................................................................. 23
iii
LIST OF FIGURES
Figure 1 – Finite Element Model Assembly ...................................................................... 2
Figure 2 – Load Summary ................................................................................................. 3
Figure 3 – Boundary Condition Summary......................................................................... 4
Figure 4 – Free Body Diagram of Uniform Load Distribution on Pin [3] ........................ 6
Figure 5 – Free Body Diagram of Triangular Load Distribution on Pin ........................... 7
Figure 6 – Contact Stress Plot Showing a Triangular Load Distribution on Pin ............. 12
Figure 7 – Contact Stress Plots Showing Clevis Lug Thickness vs. Contact Length ..... 13
Figure 8 – Contact Stress along Topside of Pin for Varying Clevis Lug Thicknesses ... 14
Figure 9 – Contact Stresses along Topside of Pin vs. Clevis Lug Thickness .................. 15
Figure 10 – Comparison of Maximum Bending Stress Results with Varying Clevis Lug
Thicknesses ...................................................................................................................... 16
Figure 11 – Comparison of Bending and Shear Stress Results for Varying Pin Diameters
......................................................................................................................................... 18
iv
LIST OF TABLES
Table 1 – Shear and Bending Stress Equations ................................................................. 9
Table 2 – Finite Element Model Input Values................................................................. 11
v
LIST OF SYMBOLS
Term
Units
Definition
A
in2
Area
b1
in
Length of pin reacted by each clevis lug
b2
in
Length of pin reacted by ½ the tang lug
c
in
Distance from neutral axis to outer fiber
d
in
Diameter of pin
dcrit
in
Critical pin diameter
F
lbf
Force applied to tang
g
in
Gap between tang and clevis face
I
in4
Moment of inertia of pin
M
lbf·in
Maximum bending moment
r
in
Radius of pin
t1
in
Clevis lug thickness
t2
in
Tang lug thickness
σb
psi
Maximum bending stress
σs
psi
Shear stress
vi
KEYWORDS/GLOSSARY
Primary Stress
Exists to keep the structure in equilibrium and can lead to catastrophic failure.
These stresses are not self-limiting, that is, small-scale yielding will not result in stress
redistribution.
Secondary Stress
Stresses due to structural discontinuities, boundary conditions, load application
proximity, etc. These stresses are self-limiting, that is, small-scale yielding will result in
stress redistribution, effectively removing the peak.
Stress Linearization
Technique used when post processing finite element analysis results in order to
separate the primary stresses (consisting of membrane and bending stresses) from the
secondary stresses which are self-limiting.
vii
ABSTRACT
The most common failure method for pins is shear failure. However, there have
been cases of pins failing despite being adequately sized for shear. Many people fail to
take into account pin bending as a legitimate failure mode. There have been some studies
done in the past to try to come up with a theoretical equation for the maximum bending
stress in a pin in double shear. In order to come up with this equation, the pin was
assumed to see a uniform load distribution.
In this project, a new equation for the maximum pin bending stress is developed
based on assuming a triangular load distribution across the pin. This triangular load
distribution assumption is validated by studying finite element analysis contact stress
plots. Maximum bending stresses are calculated using both the old equation from past
studies and the newly derived equation based on this triangular load distribution
assumption. By comparing these calculated maximum bending stresses against finite
element analysis results, the new equation derived in this project is determined to be
more accurate. Based on this new maximum pin bending stress equation, a critical pin
diameter equation is developed and validated through finite element analysis. For pin
sizes smaller than this critical pin diameter, pin failure is expected to occur due to
bending. For pin sizes larger than this critical pin diameter, shear failure is expected to
occur.
viii
1. INTRODUCTION/BACKGROUND
The purpose of this project will be to develop an equation for the maximum pin
bending stress for a double shear joint and then to validate this equation through an
ABAQUS finite element model. A relationship between the pin diameter and the failure
method will also be developed and validated through the finite element model. This
relationship will help engineers to better understand when pin bending should be
considered over shear failure.
Prior studies have been done looking at the various failure modes of a double
shear joint. Bending stresses in the pin were often ignored in the past based on the
assumptions of a close tolerance fit of the pin in the female mating part and a small gap
between the adjacent lugs [1]. Studies that have looked at pin bending as a failure
mechanism have noted the difficulties of quantifying the load distribution on the pin,
thus making it difficult to calculate the bending moment acting on the pin [2]. The most
common assumption made is to assume a uniform load distribution acting on the pin
across the entire lug thickness [2] [3] but this fails to take into account the stiffness of
the pin and the thickness of the lugs. It is safe to assume that a uniform load distribution
won’t act on the pin across the entire lug thickness of an infinitely thick lug. Therefore,
it is necessary to develop a more accurate maximum bending stress equation for the pin
taking into account the flexibility of the pin and the impacts of varying lug thicknesses.
With the advent of finite element analysis software, it has become easier to
develop and validate a more accurate pin bending stress equation than what was used in
the past. Using ABAQUS 12.0 finite element analysis software, a new maximum pin
bending stress equation will be developed and validated. By running multiple finite
element analyses with different model dimensions, the effects of pin diameter and lug
thickness on the maximum bending stress will be determined as well.
1
2. THEORY/METHODOLOGY
An ABAQUS model was created of a clevis connection with a pin in double
shear. The model consists of three parts—the clevis, the clevis pin, and the tang. The
parts were meshed with hex elements and were assembled as shown in Figure 1.
Tang
Clevis Pin
Clevis
Figure 1 – Finite Element Model Assembly
An initial displacement step was created to initialize contact. A surface-tosurface contact was created with the clevis and tang pin holes chosen as the master
surfaces and the pin chosen as the slave surface. A downward displacement of 0.012”
was applied to the clevis and an upward displacement of 0.012” was applied to the tang
in order to establish contact with the pin. A load step was then created to apply a 1000
pound load to the upper surface of the tang in the upward direction with the base of the
clevis fixed. The displacements from the initial step were deactivated for this step since
contact had already been established. Two additional boundary conditions were created
at the pin and the tang. The pin boundary condition was created at the two outer pin
surfaces to prevent axial displacement and rotation of the pin. The tang boundary
2
condition was created on the tang side faces to prevent displacement along the pin axis
and to prevent twisting of the tang. See Figure 2 and Figure 3 for a summary of the loads
and boundary conditions created for this finite element model.
Figure 2 – Load Summary
3
Figure 3 – Boundary Condition Summary
4
Maximum pin shear and bending stresses will be obtained from the results of this
ABAQUS finite element analysis. The default stress plot in ABAQUS is set to show von
Mises stresses. In order to get the maximum shear and bending stresses, S11 and S12
stress component plots will first be viewed to determine where along the pin the
maximum bending and shear stress occurs. Note that 1 refers to the x-axis which in this
case is the pin axis and that 2 refers to the y-axis which is the vertical axis. Thus, S11 is
the normal stress along the pin axis and S12 is the vertical shear stress. The maximum
bending stress is expected to be at the center of the pin and the maximum shear stress is
expected to be along the shear plane where the clevis and tang contact the pin. Once the
maximum bending and shear stress locations are determined, a section cut will be taken
at those locations with the free body plot turned on. This free body plot will display the
average or “linearized” forces and moments acting on that section cut. The linearized
vertical force will be taken to be the shear force while the moment along the z-axis
(perpendicular to the pin axis) will be taken to be the bending moment. This stress
linearization process is a critical tool for post processing finite element analysis results in
order to differentiate the primary stresses from the secondary stresses. The shear and
bending stresses calculated from this linearized shear force and bending moment will be
compared to the shear and pin bending equations developed throughout the rest of this
section (Equation 1, Equation 6, Equation 9, and Equation 11).
The shear stress equation for a pin in double shear is simply:
𝜎𝑠 =
𝐹
𝐹
2𝐹
=
= 2
2
𝐴 2𝜋𝑟
𝜋𝑑
Equation 1
A pin bending equation is derived in Reference [3] assuming a uniform
distributed load as shown in Figure 4.
5
Figure 4 – Free Body Diagram of Uniform Load Distribution on Pin [3]
Based on this load distribution, the maximum pin bending moment is calculated as:
𝑀=
𝐹 𝑏1 𝑏2
( + + 𝑔)
2 2
2
Equation 2
Further assuming that the load in each lug is uniformly distributed across the lug
thickness (b1 = t1 and 2b2 = t2) results in the following equation:
𝑀=
𝐹 𝑡1 𝑡2
( + + 𝑔)
2 2 4
Equation 3
Given the moment of inertia of the pin is
𝐼=
𝜋𝑑 4
64
6
Equation 4
and the distance from the neutral axis to the outer fiber is
𝑐=
𝑑
2
Equation 5
the maximum bending stress of the pin can be calculated as:
𝜎𝑏 =
𝑀𝑐 4𝐹(2𝑡1 + 𝑡2 + 4𝑔)
=
𝐼
𝜋𝑑 3
Equation 6
The pin bending equations derived in Reference [3] assume a uniformly
distributed load. However, a triangular load distribution as shown in Figure 5 may be a
better representation of the load profile on the pin based on the deflection of the pin as
the load is applied. Note that in Figure 5(a), 𝒕𝟏 ≤
t1. In Figure 5(b), 𝒕𝟏 >
𝒕𝟐
𝟐
𝒕𝟐
𝟐
and thus b1 is assumed to be equal to
and b1 is assumed to be no greater than b2.
(a)
(b)
Figure 5 – Free Body Diagram of Triangular Load Distribution on Pin
7
Based on this triangular load distribution, the maximum pin bending moment is
calculated as:
𝑀=
𝐹 𝑏1 𝑏2
( + + 𝑔)
2 3
3
Equation 7
𝑡
For the case where 𝑡1 ≤ 22, Equation 7 becomes:
𝑀=
𝐹 𝑡1 𝑡2
( + + 𝑔)
2 3 6
Equation 8
and the maximum bending stress of the pin can be calculated as:
𝜎𝑏 =
8𝐹(2𝑡1 + 𝑡2 + 6𝑔)
3𝜋𝑑 3
Equation 9
𝑡
For the case where 𝑡1 > 22, Equation 7 becomes:
𝑀=
𝐹 𝑡2
( + 𝑔)
2 3
Equation 10
and the maximum bending stress of the pin can be calculated as:
𝜎𝑏 =
16𝐹(𝑡2 + 3𝑔)
3𝜋𝑑 3
Equation 11
As previously discussed, ABAQUS shear and bending stresses will be compared
to shear and bending stresses calculated using Equation 1, Equation 6, Equation 9, and
Equation 11 (see Table 1). In addition, the uniform and triangular load distribution
assumptions will be tested by studying the contact stresses of the finite element model.
Based on the contact stress plot, a new equation will be developed if appropriate.
8
Table 1 – Shear and Bending Stress Equations
Equation 1 – Shear Stress
𝜎𝑠 =
Equation 6 – Bending Stress (Uniform Load)
𝜎𝑏 =
4𝐹(2𝑡1 + 𝑡2 + 4𝑔)
𝜋𝑑 3
𝜎𝑏 =
8𝐹(2𝑡1 + 𝑡2 + 6𝑔)
3𝜋𝑑 3
𝑡
Equation 9 – Bending Stress (Triangular Load, 𝑡1 ≤ 22 )
2𝐹
𝜋𝑑2
𝑡
Equation 11 – Bending Stress (Triangular Load, 𝑡1 > 22 )
𝜎𝑏 =
16𝐹(𝑡2 + 3𝑔)
3𝜋𝑑 3
Based on Table 1, an equation can be developed for the critical pin diameter at
which bending stresses overtake shear stresses. Since the shear yield strength is 0.577
times the tensile yield strength, the critical pin diameter can be calculated by letting 𝜎𝑠 =
0.577 × 𝜎𝑏 . Assuming a uniform load distribution across the lug thickness, the critical
pin diameter is calculated to be:
2𝐹
𝜋𝑑𝑐𝑟𝑖𝑡
2
= 0.577 ×
4𝐹(2𝑡1 + 𝑡2 + 4𝑔)
𝜋𝑑𝑐𝑟𝑖𝑡 3
𝑑𝑐𝑟𝑖𝑡 = 0.577 × 2(2𝑡1 + 𝑡2 + 4𝑔)
Equation 12
Assuming a triangular load distribution, the critical pin diameter is calculated to be:
2𝐹
𝜋𝑑𝑐𝑟𝑖𝑡
2
= 0.577 ×
8𝐹(2𝑡1 + 𝑡2 + 6𝑔)
3𝜋𝑑𝑐𝑟𝑖𝑡 3
4
𝑑𝑐𝑟𝑖𝑡 = 0.577 × (2𝑡1 + 𝑡2 + 6𝑔)
3
9
Equation 13
𝑡
for the case where 𝑡1 ≤ 22 , and
2𝐹
𝜋𝑑𝑐𝑟𝑖𝑡
2
= 0.577 ×
16𝐹(𝑡2 + 3𝑔)
3𝜋𝑑𝑐𝑟𝑖𝑡 3
8
𝑑𝑐𝑟𝑖𝑡 = 0.577 × (𝑡2 + 3𝑔)
3
Equation 14
𝑡
for the case where 𝑡1 > 2 .
2
These critical pin diameter equations will be tested by varying the pin diameter in
the finite element model and comparing the resulting bending and shear stresses. Pin
failure due to bending should be more of a concern than shear failure for pin diameters
smaller than the critical pin diameter calculated above.
10
3. RESULTS AND DISCUSSION
A total of 15 finite element models of the clevis connection were created for this
project (see Table 2). The first 9 models were used for the maximum bending stress
study in order to compare the uniform and triangular load distribution equations with the
ABAQUS results. A range of clevis lug thicknesses were used for this study in order to
test the assumption made that the length of pin reacted by the clevis lug would be no
greater than the length of pin reacted by half of the tang lug. The next 6 models were
used for the critical pin diameter study in order to validate the critical pin diameter
equation.
Table 2 – Finite Element Model Input Values
Dimensions (in.)
Study
Maximum
Bending
Stress
Study with
Varying
Clevis Lug
Thicknesses
Critical Pin
Diameter
Study
Model
Diameter
1
2
3
4
5
6
7
8
9
10
11
12
13
14
Pin
0.480
0.480
0.480
0.480
0.480
0.480
0.480
0.480
0.480
0.480
1.000
1.500
1.731
3.000
Clevis
0.500
0.500
0.500
0.500
0.500
0.500
0.500
0.500
0.500
0.500
1.020
1.520
1.751
3.020
Tang
0.500
0.500
0.500
0.500
0.500
0.500
0.500
0.500
0.500
0.500
1.020
1.520
1.751
3.020
15
4.000
4.020
4.020
11
Gap
Lug Thickness
Load
(lbf)
0.125
0.125
0.125
0.125
0.125
0.125
0.125
0.125
0.125
0.125
0.125
0.125
0.125
0.125
Clevis
0.200
0.250
0.300
0.350
0.400
0.450
0.500
0.550
0.600
0.400
0.400
0.400
0.400
0.400
Tang
0.750
0.750
0.750
0.750
0.750
0.750
0.750
0.750
0.750
0.750
0.750
0.750
0.750
0.750
1000
1000
1000
1000
1000
1000
1000
1000
1000
1000
1000
1000
1000
1000
0.125
0.400
0.750
1000
The first finite element analysis was performed on Model 5 of Table 2 and the
resulting contact stress plot was studied to determine if the uniform load distribution
assumed in Reference [3] and shown in Figure 4 or if the triangular load distribution
shown in Figure 5 was more accurate. Reviewing the contact stress plots shown in
Figure 6, it can be seen that the triangular load distribution is a more accurate
assumption. Therefore, it can be predicted that the bending stress equations derived from
this triangular load distribution assumption (Equation 9 and Equation 11) will be more
accurate than the equation derived in Reference [3] (Equation 6). Note that for the
ABAQUS plots shown in Figure 6, CPRESS refers to contact pressure or contact stress
and has units of psi.
Figure 6 – Contact Stress Plot Showing a Triangular Load Distribution on Pin
12
The assumption was also made that the length of pin reacted by the clevis lug (b1
in Figure 5) was not going to be any greater than the length of pin reacted by half of the
tang lug (b2 in Figure 5) or 𝒃𝟏 ≤ 𝒃𝟐 . This assumption was put to the test by running the
finite element analysis with varying clevis lug thicknesses. With the tang lug thickness at
0.750” thick, the finite element analysis was performed for clevis lug thicknesses
ranging from 0.200” to 0.600” thick at every 0.050” increment. See Figure 7 for contact
stress plots on the pins with a clevis lug thickness of 0.200”, 0.400”, and 0.600”. The 2nd
row of Figure 7 shows contact stresses for the top side of the pin where the clevis comes
into contact while the 3rd row shows contact stresses at the bottom of the pin where the
tang contacts. Note that the actual contact stress values are left out since this figure is
merely intended to help visualize the load distribution. See Figure 8 and Figure 9 for
actual contact stress values.
Figure 7 – Contact Stress Plots Showing Clevis Lug Thickness vs. Contact Length
13
To accurately assess the load distribution on the pin, the contact stress values
from the contour plots shown in row 2 of Figure 7 were plotted as a line along the length
of the pin in Figure 8. By doing so, it can be seen that the load distribution is indeed
triangular in nature. Note that “Stress” refers to contact stress and has units of psi while
“True distance along path” refers to the distance along the pin and has units of inches.
Also note that unlike in Figure 7, contact stresses for all the different clevis lug thickness
models were plotted for completeness, rather than just for the 0.200”, 0.400”, and 0.600”
clevis lug thickness models. It can also be seen that the maximum length of pin reacted
by each clevis lug is roughly 0.45 inch. Compared to the 0.75 inch thick tang lug, the
maximum length of pin reacted by each clevis lug is roughly 1.2 times the length of pin
𝑡
reacted by half of the tang lug (𝑏1 ≤ ~1.2 × 22 ). See Figure 9 for an overlay of all the
ABAQUS plots in Excel with the appropriate labels and units listed.
Figure 8 – Contact Stress along Topside of Pin for Varying Clevis Lug Thicknesses
14
Topside Pin Contact Stress Plot for Various Clevis Lug Thicknesses
40000
Clevis Lug
35000
Thickness (in.)
Contact Stress (psi)
30000
0.200
0.250
0.300
0.350
0.400
0.450
0.500
0.550
0.600
25000
20000
15000
10000
5000
0
-1.25
-1
-0.75
-0.5
-0.25
0
0.25
0.5
0.75
1
1.25
Distance Along Pin Centered at 0 (in.)
Figure 9 – Contact Stresses along Topside of Pin vs. Clevis Lug Thickness
Based on this result, a new maximum bending stress equation was developed.
𝑡
Assuming 𝑏1 ≤ ~1.2 × 22 , the maximum pin bending moment (Equation 7) for the case
𝑡
where 𝑡1 > 1.2 × 22, becomes:
𝑀=
𝐹 2.2𝑡2
(
+ 𝑔)
2
6
Equation 15
and the maximum bending stress of the pin can be calculated as:
𝜎𝑏 =
8𝐹(2.2𝑡2 + 6𝑔)
3𝜋𝑑 3
15
Equation 16
A plot of maximum bending stresses versus clevis lug thicknesses was created to
compare the actual bending stress results from the finite element analysis with the
calculated bending stresses based on either a uniform load distribution across the entire
𝑡
lug thickness as assumed in Reference [3], a triangular load distribution with 𝑏1 ≤ 22, or
𝑡
a triangular load distribution with 𝑏1 ≤ ~1.2 × 22. See Figure 10 below.
Maximum Bending Stress vs. Clevis Lug Thickness
30000
ABAQUS
Results
28000
Bending Stress (psi)
26000
Triangular Load
Distribution b1 < t2/2
24000
22000
Triangular Load
Distribution b1 < 1.2*t2/2
20000
18000
Uniform Load
Distribution
16000
14000
0.000
0.100
0.200
0.300
0.400
0.500
0.600
0.700
Clevis Lug Thickness (in.)
Figure 10 – Comparison of Maximum Bending Stress Results with Varying Clevis
Lug Thicknesses
Figure 10 reveals the uniform load distribution assumption made in Reference [3]
to be more conservative but less accurate than the triangular load distribution
assumptions. Between the two triangular load distribution assumptions, the original
assumption of 𝑏1 ≤
𝑡2
2
looks to be slightly more accurate while the assumption of 𝑏1 ≤
16
~1.2 ×
𝑡2
2
based on the ABAQUS contact stress plots is slightly more conservative at the
higher clevis lug thickness ranges.
Since the triangular load distribution with 𝑏1 ≤
𝑡2
2
has been established to be the
most accurate of the three load distribution assumptions, the critical pin diameter will be
assumed to be the ones previously derived in the theory/methodology section (Equation
13 and Equation 14).
If the pin diameter is smaller than the critical pin diameter, the bending stress
term outweighs the shear stress term and thus, pin failure due to bending becomes more
of a concern. If the pin diameter is greater than the critical pin diameter, the opposite is
true and shear stress governs. To validate this calculated critical pin diameter, actual
bending and shear stresses from finite element analyses were compared against the
calculated stresses for varying pin diameters (see Figure 11). Six finite element models
were used for this validation with constant clevis lug thicknesses of t 1 = 0.400 inch,
constant tang lug thicknesses of t2 = 0.750 inch, constant gaps of g = 0.125 inch, and
varying pin diameters of 0.48 inch, 1 inch, 1.5 inches, 1.731 inches, 3 inches, and 4
inches. Note that the 0.48 inch pin diameter model was already previously analyzed as
part of the clevis thickness study and the 1.731 inch pin diameter model was created
based on the critical pin diameter calculation on the next page.
17
Bending Stress vs. Shear Stress for Varying Pin Diameters
20000
18000
16000
Stress (psi)
14000
12000
Bending - ABAQUS
10000
Bending - Calculated
8000
Shear - ABAQUS
6000
Shear - Calculated
4000
2000
0
0
1
2
3
4
5
Pin Diameter (in.)
Figure 11 – Comparison of Bending and Shear Stress Results for Varying Pin
Diameters
It can be seen from Figure 11 that the bending stress is much larger than the
shear stress for very small pin diameters. As the pin diameter increases, the bending
stress decreases exponentially. The fact that the ABAQUS bending and shear stresses are
roughly the same as the calculated bending and shear stresses for varying pin diameters
further validates the shear and bending equations. The critical pin diameter equation is
validated by comparing the ABAQUS bending and shear stresses at the calculated
critical pin diameter:
8
8
𝑑𝑐𝑟𝑖𝑡 = 0.577 × (𝑡2 + 3𝑔) = 0.577 × (0.750 + 3 × 0.125) = 1.731 𝑖𝑛𝑐ℎ𝑒𝑠
3
3
18
With the pin diameter at 1.731 inches, the ABAQUS bending stress result is 469 psi and
the ABAQUS shear stress result is 212 psi. This shear stress is 0.45 times the bending
stress, which is close to the expected value of 0.577 considering how closely the
“ABAQUS” plot lines follow the “calculated” plot lines in Figure 11.
19
4. CONCLUSION
The purpose of this project was to develop an equation for the maximum bending
stress of a pin in double shear and then to validate this equation through finite element
analysis. A secondary purpose was to determine, and validate through finite element
analysis, a relationship between the pin size and the shear and bending stresses in order
to better understand when pin bending should be considered over shear failure.
In order to derive an equation for the maximum pin bending stress, the load
profile on the pin had to be determined. The assumption of a triangular load distribution
on the pin was made and validated through finite element analysis. Based on this load
distribution, maximum pin bending stress equations of
𝜎𝑏 =
8𝐹(2𝑡1 + 𝑡2 + 6𝑔)
3𝜋𝑑 3
for the case where the clevis lug thickness is less than or equal to half of the tang lug
𝑡
thickness (𝑡1 ≤ 22 ) and
𝜎𝑏 =
16𝐹(𝑡2 + 3𝑔)
3𝜋𝑑 3
for the case where the clevis lug thickness is greater than half of the tang lug thickness
𝑡
(𝑡1 > 22 ) were derived and validated using ABAQUS finite element analysis software.
A relationship was also determined between the pin size and the shear and
bending stresses. Critical pin diameter equations of
4
𝑑𝑐𝑟𝑖𝑡 = 0.577 × (2𝑡1 + 𝑡2 + 6𝑔)
3
for the case where 𝑡1 ≤
𝑡2
2
and
20
8
𝑑𝑐𝑟𝑖𝑡 = 0.577 × (𝑡2 + 3𝑔)
3
for the case where 𝑡1 >
𝑡2
2
were derived and validated through an ABAQUS finite
element analysis. Failure due to pin bending was shown to be a concern over shear
failure for pin sizes below this critical pin diameter.
The results of this study are summarized below:

Verification of triangular load distribution on pin

Derivation of maximum pin bending stress equation

Validation of maximum pin bending stress equation

Derivation of critical pin diameter equation

Validation of critical pin diameter equation
21
REFERENCES
[1]
Samuel, Andrew and John Weir. Introduction to Engineering Design.
Butterworth-Heinemann, 1999.
[2]
Richards, Keith L. Design Engineer’s Handbook. CRC Press: Taylor & Francis
Group, 2013.
[3]
Maddux, G.E., Leon A. Vorst, F. Joseph Giessler, and Terence Moritz. Stress
Analysis Manual. Dayton: Technology Incorporated, 1969.
22
APPENDICES
23