The Lug Analysis calculator allows for analysis of lifting lugs under axial, transverse, or oblique loading. This calculator follows the Air Force Method as documented in the Stress Analysis Manual of the Air Force Flight Dynamics Laboratory (FDL). See the instructions within the documentation for more details on performing this analysis. See the reference section for details on the methodology and the equations used.
Options:
Inputs
Input the details for the lug and hit 'Submit' to calculate results:
Applied Forces:
Pin Inputs:
Female (Outer) Lug Inputs:
The joint can optionally include an outer set of female lugs. If included, results will be calculated for female lugs and double shear joint strength.
Additional Pin Inputs:
Since female lugs are included, pin strength calculations will be performed and so we need more information about the pin:
Calculate results:
Display Units
Display results in:
Results
The results of the lug analysis are detailed below. Refer to the lug analysis reference section for details on how these results were derived.
Results Summary
Summary tables of results are shown below. These tables give the factors of safety for the lugs and pin, as applicable. Any factors of safety of at least 1 are shown in green, and below 1 are shown in red. It is up to the discretion of the engineer to determine the appropriate factor of safety to use in design.
Male Lug
Total Applied Force  P_{app}  
Lug Strength for Oblique Loading  P_{ult}  
Factor of Safety  FS 
Female Lug
These are the strength results of the female lug, independent of the rest of the joint:
Total Applied Force  P_{app}  
Lug Strength for Oblique Loading  P_{ult}  
Factor of Safety  FS 
Double Shear Joint Strength
This is the overall strength of the joint, accounting for the male lug, the two female lugs, and the pin. The bearing areas between the pin and the lugs are shown in the figure to the right.
Total Applied Force  P_{app}  
Overall Ultimate Load (Joint Strength)  P_{ult}  
Factor of Safety  FS 
Female Lug
There is no female lug in this joint, so these results cannot be calculated.
Double Shear Joint Strength
There is no female lug in this joint, so these results cannot be calculated.
See full result details on the other tabs.
Component Properties
This section details the properties of the lug components.
Materials List
A list of all materials used in the joint is provided below:
Elastic Modulus, E 
Percent Elongation, eL 
Ultimate Strain, ε_{u} 
Tensile Yield Strength, S_{ty} 
Tensile Ultimate Strength, S_{tu} 
Shear Yield Strength, S_{sy} 
Shear Ultimate Strength, S_{su} 
Compressive Yield Strength, S_{cy} 
Compressive Ultimate Strength, S_{cu} 

NOTE: For ductile materials, compressive strength is approximately equal to tensile strength. For brittle materials, compressive strength is higher than tensile strength. In this analysis it is conservatively assumed that compressive strength is equal to tensile strength.
NOTE: The ultimate strain for a material is calculated based on the percent elongation, \(eL\), the elastic modulus, \(E\), and the ultimate tensile strength, \(S_{tu}\):
$$ \varepsilon_u = eL + { S_{tu} \over E } $$Applied Forces
P_{ax}  =  axial force  
P_{tr}  =  transverse force  
P_{app}  =  total applied force  
α  =  angle 
NOTE: There are 2 female lugs that share the applied load, so the load on a single female lug is half of that on the male lug.
Joint Properties
The joint consists of a single male lug, two female lugs, and a pin.
g  =  gap between lugs (for bending arm) 
Pin Properties
Material: 
D_{P}  =  diameter  
A  =  area  
S  =  section modulus  
k_{b.P}  =  plastic bending coefficient 
Lug Dimensions
Lug dimensions are as shown in the figure below. Dimensions for the male lug and the female lugs (if applicable) are given in the following sections.
Male Lug Properties
Material:  
Bushing Material: 
w  =  width  
t_{2}  =  thickness  
D  =  hole diameter  
e  =  edge distance (hole center to edge of lug)  
a  =  distance (edge of hole to edge of lug)  
h_{eff}  =  effective edge distance 
Female Lug Properties
Material  
Bushing Material: 
w  =  width  
t_{1}  =  thickness  
D  =  hole diameter  
e  =  edge distance (hole center to edge of lug)  
a  =  distance (edge of hole to edge of lug)  
h_{eff}  =  effective edge distance 
Male Lug Results
Lug strengths in the axial and transverse directions are first calculated. The lug strength in the applied force direction (i.e. oblique loading) is then calculated, along with a Factor of Safety.
The table below gives a summary of the results. Full detailed calculations are provided in the following sections.
Lug Strength for Pure Axial Loading  P_{u.L.B}  
Lug Strength for Pure Transverse Loading  P_{tru.L.B}  
Lug Strength for Oblique Loading  P_{ult}  
Total Applied Force  P_{app}  
Factor of Safety  FS 
NOTE: There are 2 female lugs that share the applied load, so the load on a single female lug is half of that on the male lug.
Axial Loading
This section calculates the strength of the lug with a pure axial load applied.
Bearing Strength Under Axial Load
The ultimate bearing load accounts for bearing, shearout, and hoop tension. Before calculating the ultimate bearing load, the axial load coefficient and bearing efficiency factor must be determined.
The axial load coefficient is determined from the figure below. This coefficient is only valid for \( D/t \le 5 \), which in this case. From the plot, the coefficient is:
\( K_{axial} = \)  () 
Reference Values
D/t  =  
e/D  =  
D  =  
e  =  
t  = 
NOTE: This plot was generated based on the Air Force Manual, Figure 92.
The bearing efficiency factor is determined from the figure below. This coefficient is only valid for \( D/t \gt 5 \), which in this case. From the plot, the coefficient is:
\( K_{bear} = \)  () 
Reference Values
D/t  =  
e/D  =  
D  =  
e  =  
t  = 
NOTE: This plot was generated based on the Air Force Manual, Figure 93. The dotted lines in this figure were generated by finding the surrounding D/t curves, and the solid blue line was generated by interpolating between the surrounding curves.
Since \( D/t \le 5 \), the axial load coefficient is used:
\( K = K_{axial} = \) 
Since \( D/t \gt 5 \), the bearing efficiency factor is used:
\( K = K_{bear} = \) 
The ultimate bearing load is calculated as:
$$ P_{bru.L} = K \cdot \min{(S_{tu}, 1.304 S_{ty})} \cdot D t \cdot \left \begin{array}{ll} {a \over D} & \text{if } e/D \lt 1.5 \\ 1 & \text{otherwise} \end{array} \right. = $$ 
Reference Values
S_{ty}  =  
S_{tu}  =  
D  =  
a  =  
e  =  
t  = 
Bushing Bearing Strength Under Axial Load
The ultimate bushing load is calculated as:
$$ P_{u.B} = 1.304 S_{cy.B} D_p t = $$ 
NOTE: There was no bushing specified for this lug, so the bushing bearing strength was calculated using the compressive yield strength of the lug body material, .
Reference Values
Bushing:
S_{cy.B}  =  
D_{P}  =  
t  = 
NetSection Strength Under Axial Load
The netsection ultimate load accounts for tension failure across the net section. It depends on the net tension stress coefficient determined from the plot below:
\( K_n = \) 
Reference Values
D/w  =  
S_{ty} / S_{tu}  =  
S_{tu} / (E⋅ε_{u})  = 
NOTE: This plot was generated based on the Air Force Manual, Figure 94 (b), (c), and (d). The dotted lines in this figure were generated by interpolating within each of the three Air Force Manual figures based on the ratio S_{tu} / (E⋅ε_{u}). The solid line was generated by interpolating between the figures based on the ratio S_{ty}/S_{tu}.
The netsection ultimate load is calculated as:
$$ P_{nu.L} = K_n \cdot \min{(S_{tu}, 1.304 S_{ty})} \cdot (w  D) \cdot t = $$ 
Reference Values
S_{ty}  =  
S_{tu}  =  
D  =  
w  =  
t  = 
Design Strength Under Axial Load
The design ultimate load for an axially loaded lug is the minimum of the ultimate bearing load, the ultimate bushing load, and the ultimate netsection load:
$$ P_{u.L.B} = \min{( P_{bru.L}, P_{u.B}, P_{nu.L} )} = $$ 
Reference Values
P_{bru.L}  =  
P_{u.B}  =  
P_{nu.L}  = 
Transverse Loading
This section calculates the strength of the lug with a pure transverse load applied.
Lug Strength Under Transverse Load
The ultimate transverse load is dependent on the transverse ultimate and yield load coefficients determined from the plot below:
\( K_{try} = \)  transverse yield load coefficient  
\( K_{tru} = \)  transverse ultimate load coefficient 
Reference Values
h_{eff} / D  =  
h_{eff}  =  
D  = 
NOTE: This plot was generated based on the Air Force Manual, Figure 98.
The ultimate transverse load is calculated as:
$$ P_{tru.L} = \left \begin{array}{ll} K_{tru} S_{tu} D t & \text{if } S_{tu} \le 1.304 S_{ty} \\ 1.304 K_{try} S_{ty} D t & \text{otherwise} \end{array} \right. = $$ 
Reference Values
S_{ty}  =  
S_{tu}  =  
D  =  
t  = 
Bushing Bearing Strength Under Transverse Load
The bearing strength for the bushing in a transversely loaded lug is the same as for an axially loaded lug:
$$ P_{tru.B} = P_{u.B} = $$ 
Design Strength Under Transverse Load
The design ultimate load for a transversely loaded lug is the minimum of the ultimate lug load and the ultimate bushing load:
$$ P_{tru.L.B} = \min{( P_{tru.L}, P_{tru.B} )} = $$ 
Reference Values
P_{tru.L}  =  
P_{tru.B}  = 
Oblique Loading
This section calculates the strength of the lug under oblique loading (combined axial and transverse loading). Note that pure axial loading and pure transverse loading are special cases of oblique loading where only one load component is nonzero.
Applied Forces
The forces applied to the lug are:
\( P_{ax} = \)  axial force  
\( P_{tr} = \)  transverse force  
\( P_{app} = \)  total applied force  
\( \alpha = \)  applied force angle 
NOTE: There are 2 female lugs that share the applied load, so the load on a single female lug is half of that on the male lug.
Lug Strength for Pure Axial and Transverse Loading
The lug strengths for pure loading in the axial and transverse directions are:
\( P_{u.L.B} = \)  lug strength for pure axial loading  
\( P_{tru.L.B} = \)  lug strength for pure transverse loading 
Lug Strength for Oblique Loading
Lug strength for oblique loading is defined by an interaction equation that accounts for the interaction between the axial and transverse loading and strength. The interaction equation for lug strength is:
$$ \left({ P_{ax.ult} \over P_{u.L.B} }\right)^{1.6} + \left({ P_{tr.ult} \over P_{tru.L.B} }\right)^{1.6} = 1 $$The failure locus of the plot to the left is defined by the interaction equation above. The axes of the plot are load ratios, which relate the applied load to the lug strength.
The current point on the plot is defined by the load ratios in the axial and transverse directions:
$$ R_{ax} = { P_{ax} \over P_{u.L.B} } = $$  axial load ratio  
$$ R_{tr} = { P_{tr} \over P_{tru.L.B} } = $$  transverse load ratio 
The load line in the plot extends from the origin through the current point and intersects with the failure locus. The critical point is the point at which the load line intersects the failure locus. This point gives the load ratios at which the lug fails.
$$ R_{ax.ult} = { P_{ax.ult} \over P_{u.L.B} } = $$  ultimate axial load ratio  
$$ R_{tr.ult} = { P_{tr.ult} \over P_{tru.L.B} } = $$  ultimate transverse load ratio 
The lug strength components for oblique loading in the applied direction are:
$$ P_{ax.ult} = \left({ 1 \over \left({ 1 \over P_{u.L.B} }\right)^{1.6} + \left({ \tan{(\alpha)} \over P_{tru.L.B} }\right)^{1.6} }\right)^{0.625} = $$  axial strength component  
$$ P_{tr.ult} = P_{ax.ult} \cdot \tan{(\alpha)} = $$  transverse strength component 
Reference Values
P_{u.L.B}  =  
P_{tru.L.B}  =  
α  = 
The lug strength for oblique loading in the applied direction is:
$$ P_{ult} = \sqrt{ P_{ax.ult}^2 + P_{tr.ult}^2 } = $$ 
Reference Values
P_{ax.ult}  =  
P_{tr.ult}  = 
Reference Values
P_{ult}  =  
P_{app}  = 
Female Lug Results
Lug strengths in the axial and transverse directions are first calculated. The lug strength in the applied force direction (i.e. oblique loading) is then calculated, along with a Factor of Safety.
The table below gives a summary of the results. Full detailed calculations are provided in the following sections.
Lug Strength for Pure Axial Loading  P_{u.L.B}  
Lug Strength for Pure Transverse Loading  P_{tru.L.B}  
Lug Strength for Oblique Loading  P_{ult}  
Total Applied Force  P_{app}  
Factor of Safety  FS 
NOTE: There are 2 female lugs that share the applied load, so the load on a single female lug is half of that on the male lug.
Axial Loading
This section calculates the strength of the lug with a pure axial load applied.
Bearing Strength Under Axial Load
The ultimate bearing load accounts for bearing, shearout, and hoop tension. Before calculating the ultimate bearing load, the axial load coefficient and bearing efficiency factor must be determined.
The axial load coefficient is determined from the figure below. This coefficient is only valid for \( D/t \le 5 \), which in this case. From the plot, the coefficient is:
\( K_{axial} = \)  () 
Reference Values
D/t  =  
e/D  =  
D  =  
e  =  
t  = 
NOTE: This plot was generated based on the Air Force Manual, Figure 92.
The bearing efficiency factor is determined from the figure below. This coefficient is only valid for \( D/t \gt 5 \), which in this case. From the plot, the coefficient is:
\( K_{bear} = \)  () 
Reference Values
D/t  =  
e/D  =  
D  =  
e  =  
t  = 
NOTE: This plot was generated based on the Air Force Manual, Figure 93. The dotted lines in this figure were generated by finding the surrounding D/t curves, and the solid blue line was generated by interpolating between the surrounding curves.
Since \( D/t \le 5 \), the axial load coefficient is used:
\( K = K_{axial} = \) 
Since \( D/t \gt 5 \), the bearing efficiency factor is used:
\( K = K_{bear} = \) 
The ultimate bearing load is calculated as:
$$ P_{bru.L} = K \cdot \min{(S_{tu}, 1.304 S_{ty})} \cdot D t \cdot \left \begin{array}{ll} {a \over D} & \text{if } e/D \lt 1.5 \\ 1 & \text{otherwise} \end{array} \right. = $$ 
Reference Values
S_{ty}  =  
S_{tu}  =  
D  =  
a  =  
e  =  
t  = 
Bushing Bearing Strength Under Axial Load
The ultimate bushing load is calculated as:
$$ P_{u.B} = 1.304 S_{cy.B} D_p t = $$ 
NOTE: There was no bushing specified for this lug, so the bushing bearing strength was calculated using the compressive yield strength of the lug body material, .
Reference Values
Bushing:
S_{cy.B}  =  
D_{P}  =  
t  = 
NetSection Strength Under Axial Load
The netsection ultimate load accounts for tension failure across the net section. It depends on the net tension stress coefficient determined from the plot below:
\( K_n = \) 
Reference Values
D/w  =  
S_{ty} / S_{tu}  =  
S_{tu} / (E⋅ε_{u})  = 
NOTE: This plot was generated based on the Air Force Manual, Figure 94 (b), (c), and (d). The dotted lines in this figure were generated by interpolating within each of the three Air Force Manual figures based on the ratio S_{tu} / (E⋅ε_{u}). The solid line was generated by interpolating between the figures based on the ratio S_{ty}/S_{tu}.
The netsection ultimate load is calculated as:
$$ P_{nu.L} = K_n \cdot \min{(S_{tu}, 1.304 S_{ty})} \cdot (w  D) \cdot t = $$ 
Reference Values
S_{ty}  =  
S_{tu}  =  
D  =  
w  =  
t  = 
Design Strength Under Axial Load
The design ultimate load for an axially loaded lug is the minimum of the ultimate bearing load, the ultimate bushing load, and the ultimate netsection load:
$$ P_{u.L.B} = \min{( P_{bru.L}, P_{u.B}, P_{nu.L} )} = $$ 
Reference Values
P_{bru.L}  =  
P_{u.B}  =  
P_{nu.L}  = 
Transverse Loading
This section calculates the strength of the lug with a pure transverse load applied.
Lug Strength Under Transverse Load
The ultimate transverse load is dependent on the transverse ultimate and yield load coefficients determined from the plot below:
\( K_{try} = \)  transverse yield load coefficient  
\( K_{tru} = \)  transverse ultimate load coefficient 
Reference Values
h_{eff} / D  =  
h_{eff}  =  
D  = 
NOTE: This plot was generated based on the Air Force Manual, Figure 98.
The ultimate transverse load is calculated as:
$$ P_{tru.L} = \left \begin{array}{ll} K_{tru} S_{tu} D t & \text{if } S_{tu} \le 1.304 S_{ty} \\ 1.304 K_{try} S_{ty} D t & \text{otherwise} \end{array} \right. = $$ 
Reference Values
S_{ty}  =  
S_{tu}  =  
D  =  
t  = 
Bushing Bearing Strength Under Transverse Load
The bearing strength for the bushing in a transversely loaded lug is the same as for an axially loaded lug:
$$ P_{tru.B} = P_{u.B} = $$ 
Design Strength Under Transverse Load
The design ultimate load for a transversely loaded lug is the minimum of the ultimate lug load and the ultimate bushing load:
$$ P_{tru.L.B} = \min{( P_{tru.L}, P_{tru.B} )} = $$ 
Reference Values
P_{tru.L}  =  
P_{tru.B}  = 
Oblique Loading
This section calculates the strength of the lug under oblique loading (combined axial and transverse loading). Note that pure axial loading and pure transverse loading are special cases of oblique loading where only one load component is nonzero.
Applied Forces
The forces applied to the lug are:
\( P_{ax} = \)  axial force  
\( P_{tr} = \)  transverse force  
\( P_{app} = \)  total applied force  
\( \alpha = \)  applied force angle 
NOTE: There are 2 female lugs that share the applied load, so the load on a single female lug is half of that on the male lug.
Lug Strength for Pure Axial and Transverse Loading
The lug strengths for pure loading in the axial and transverse directions are:
\( P_{u.L.B} = \)  lug strength for pure axial loading  
\( P_{tru.L.B} = \)  lug strength for pure transverse loading 
Lug Strength for Oblique Loading
Lug strength for oblique loading is defined by an interaction equation that accounts for the interaction between the axial and transverse loading and strength. The interaction equation for lug strength is:
$$ \left({ P_{ax.ult} \over P_{u.L.B} }\right)^{1.6} + \left({ P_{tr.ult} \over P_{tru.L.B} }\right)^{1.6} = 1 $$The failure locus of the plot to the left is defined by the interaction equation above. The axes of the plot are load ratios, which relate the applied load to the lug strength.
The current point on the plot is defined by the load ratios in the axial and transverse directions:
$$ R_{ax} = { P_{ax} \over P_{u.L.B} } = $$  axial load ratio  
$$ R_{tr} = { P_{tr} \over P_{tru.L.B} } = $$  transverse load ratio 
The load line in the plot extends from the origin through the current point and intersects with the failure locus. The critical point is the point at which the load line intersects the failure locus. This point gives the load ratios at which the lug fails.
$$ R_{ax.ult} = { P_{ax.ult} \over P_{u.L.B} } = $$  ultimate axial load ratio  
$$ R_{tr.ult} = { P_{tr.ult} \over P_{tru.L.B} } = $$  ultimate transverse load ratio 
The lug strength components for oblique loading in the applied direction are:
$$ P_{ax.ult} = \left({ 1 \over \left({ 1 \over P_{u.L.B} }\right)^{1.6} + \left({ \tan{(\alpha)} \over P_{tru.L.B} }\right)^{1.6} }\right)^{0.625} = $$  axial strength component  
$$ P_{tr.ult} = P_{ax.ult} \cdot \tan{(\alpha)} = $$  transverse strength component 
Reference Values
P_{u.L.B}  =  
P_{tru.L.B}  =  
α  = 
The lug strength for oblique loading in the applied direction is:
$$ P_{ult} = \sqrt{ P_{ax.ult}^2 + P_{tr.ult}^2 } = $$ 
Reference Values
P_{ax.ult}  =  
P_{tr.ult}  = 
Reference Values
P_{ult}  =  
P_{app}  = 
There is no female lug in this joint, so these results cannot be calculated.
Double Shear Joint Strength
In this section, the double shear joint strength is calculated which gives the overall strength of the joint accounting for the interactions between the lugs and the pin. The table below gives a summary of the results. Full detailed calculations are provided in the following sections.
Overall Ultimate Load (Joint Strength)  P_{ult}  
Total Applied Force  P_{app}  
Factor of Safety  FS 
Nominal Joint Strength
The strengths of the male and female lugs, calculated in the previous sections, are:
$$ P_{ult.M} = $$  male lug strength  
$$ P_{ult.F} = $$  female lug strength 
The nominal joint strength, ignoring the effects of pin bending, is:
\( P_{u.J.nom} = \min{( 2 \cdot P_{ult.F}, P_{ult.M} )} = \)  nominal joint strength 
Nominal Pin Strength
The nominal shear strength and bending strength of the pin are:
$$ P_{us.P} = 2 \left( { \pi \over 4 } D_p^2 \right) S_{su.P} = $$  pin ultimate shear load  
$$ P_{ub.P} = { \pi D_P^3 \cdot k_{b.P} \cdot S_{tu.P} \over 16 \left( { t_1 \over 2 } + { t_2 \over 4 } + g \right) } = $$  pin ultimate bending load 
Reference Values
S_{tu.P}  =  
S_{su.P}  =  
D_{P}  =  
k_{b.P}  =  
t_{1}  =  
t_{2}  =  
g  = 
Pin Strong or Weak in Bending?
For the pin to be considered strong in bending, the pin ultimate bending load \( P_{ub.P} \) must be greater than either the pin ultimate shear load \( P_{us.P} \) or the nominal ultimate joint load \( P_{u.J.nom} \). One of the conditions below must hold true for the pin to be strong in bending:
\( P_{ub.P} \ge P_{u.J.nom} \) or \( P_{ub.P} \ge P_{us.P} \) 
At least one of the conditions above does hold true, so the pin is strong in bending.
The conditions above do not hold true, so the pin is weak in bending.
Reference Values
P_{ub.P}  =  
P_{us.P}  =  
P_{u.J.nom}  = 
Strong Pin
For a strong pin, the the pin ultimate bending load is calculated assuming that the load distributes evenly over the full width of the lugs (i.e. same as the nominal value):
$$ P_{ub.P} = { \pi D_P^3 \cdot k_{b.P} \cdot S_{tu.P} \over 16 \left( { t_1 \over 2 } + { t_2 \over 4 } + g \right) } = $$  pin ultimate bending load 
Reference Values
S_{tu.P}  =  
D_{P}  =  
k_{b.P}  =  
t_{1}  =  
t_{2}  =  
g  = 
For a strong pin, the pin bending does not affect the joint strength and the ultimate joint load is equal to the nominal ultimate joint load:
\( P_{u.J} = P_{u.J.nom} = \)  ultimate joint load 
Weak Pin
For a weak pin, the load will not be distributed evenly over the lug widths and will instead concentrate toward the outer portions of the lugs at the shear planes. A "balanced design" is found that accounts for the reduced bearing areas between the pin and the lugs. The bearing widths for the male and female lugs are calculated such that the male lug strength, the female lug strength, and the pin bending strength are all equal.
The bearing widths for the male and female lugs are the widths over which the pin loads are supported:
\( b_1 = \)  bearing width over female lug  
\( b_2 = \)  bearing width over male lug 
Reference Values
t_{1}  =  
t_{2}  = 
Because the bearing widths are smaller than the full lug thicknesses, the moment arm is reduced and the pin ultimate bending load is increased:
$$ P_{ub.P} = { \pi D_P^3 \cdot k_{b.P} \cdot S_{tu.P} \over 16 \left( { b_1 \over 2 } + { b_2 \over 2 } + g \right) } = $$  pin ultimate bending load 
Reference Values
S_{tu.P}  =  
D_{P}  =  
k_{b.P}  =  
g  = 
In the balanced design, the ultimate joint load is equal to the pin ultimate bending load:
\( P_{u.J} = P_{ub.P} = \)  ultimate joint load 
Note that the same joint strength would be achieved if the width of each female lug was reduced to \(b_1 = \) and the width of the male lug reduced to \(2 b_2 = \) .
Overall Ultimate Load
The overall ultimate load accounts for both the ultimate joint load and the ultimate pin shear load:
\( P_{ult} = \min{( P_{u.J}, P_{us.P} )} = \)  overall ultimate load 
Reference Values
P_{u.J}  =  
P_{us.P}  = 
The overall ultimate load must be compared to the applied force to determine the factor of safety:
\( P_{app} = \)  total applied force 
The Factor of Safety (FS) is:
$$ FS = { P_{ult} \over P_{app} } = $$ 

Reference Values
P_{ult}  =  
P_{app}  = 
There is no female lug in this joint, so these results cannot be calculated.
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