# Lifting Lug Design – USA Air Force Method

## 1. Introduction

Lug or Padeye is basically a plate with a hole in it where the hole is used to connect a clevis pin, a chain, a hook or a rope, with the objective to lift an equipment. This article will only cover the design of lifting lug according to the USA Air Force Method. It’s highly advised that you read and fully understand this article before using our lifting lug design spreadsheet.

The Air Force Method is widely used in the industries all around the world and is described in the Stress Analysis Manual of the Air Force Flight Dynamics Laboratory (FDL), which you can be downloaded by free in the Defense Technical Information Center website. This method is similar to those presented in the Analysis and Design of Flight Vehicle Structures from E. F. Bruhn (this book appears in the reference page) and Analysis of Lugs and Shear Pins from M. A. Melcon and F. M. Hoblit.

The method described in the book presents static strength analysis procedures for uniformly loaded lugs and bushings, for double shear joints, and for single shear joints, subject to axial, transverse, or oblique loading. It considers materials having ultimate elongation of at least 5% in any directon in the plane of the lug, but there’s part with modifications which can be used with materials having less than 5% of elongation. In addition, a short section on the stress due to press fit bushing is also presented.

The first thing that the designer will notice is that this method heavily relies on curves generated by empirical data (results from tests and interpolation) and the second thing is that it is somewhat more complex compared to other lug analysis methods (ASME BTH for example), as it takes in account the bearing, shear-out and hoop tension (combined in a single failure mode) in the lug. The interaction between the lug and the pin (double shear joint) is also considered.

## 2. Uniform Axial Load

The axially loaded lugs in tension must be checked for bearing strength and for net-section strength, which we are considering as failure modes. The bearing strength of the lug shown in the Figure 1, depends largely on the interaction between bearing, shear-out, and hoop-tension stresses in the part of the lug ahead of the pin (right side of the section 1-1).

Figure 1 – Lugs Loaded in Tension

Nomenclature shown in the Figure 1:

$P$ = Load
$w_T$ = Width of the “lug neck”
$w$ = Lug width
$D$ = Lug hole diameter
$D_P$ = Pin diameter
$a$ = Distance from the edge of the hole to the edge of the lug
$e$ = Edge distance
$t$ = Lug thickness

### 2.1. Lug Bearing Strength

The bearing stresses and loads for a lug failure involving bearing, shear-out, or hoop tension can be determined using the equations below, with an allowable load coefficient ($
K$
) determined from the charts Figure 2 and 3. For values of ${e/D lt 1,5}$, lug failures are likely to involve shear-out or hoop-tension, and for values of $e/D ge 1,5$, the bearing is likely to be critical. Actual lug failures may involve more than one failure mode, but such interaction effects are accounted for in the values of $K$.

The lug ultimate bearing stress ($F_{bruL}$) can be found using the Equation 1 or 2:

$normalsize color{black}{F_{bruL} = K { a over D } F_{tux} quad textrm{if} quad {e over D} lt 1,5 tag{1}}$
$normalsize color{black}{F_{bruL} = K F_{tux} quad textrm{if} quad {e over D} ge 1,5 tag{2}}$

Where:

$F_{tux}$ = Cross grain tensile ultimate stress of lug material

The lug yield bearing stress ($F_{bryL}$) can be found using Equation 3 or 4:

$normalsize color{black}{F_{bryL} = K {a over D} F_{tyx} quad textrm{if} quad {e over D} lt 1,5 tag{3}}$
$normalsize color{black}{F_{bryL} = K F_{tyx} quad textrm{if} quad {e over D} ge 1,5 tag{4}}$

Where:

$F_{tyx}$ = Cross grain tensile yield stress of lug material

The value of $K$ can be found using the chart in the Figure 2, if $D/t le 5$, or Figure 3 if ${D/t gt 5}$.

Figure 2 – Allowable Uniform Axial Load Coefficient for D/t less than or equal to 5

Figure 3 – Bearing Effeciency Factors of Lugs, Aluminium Alloys, and Alloy Steek with Fty < 160 KSi

The allowable lug ultimate bearing ($P_{bruL}$) for lug failure in bearing, shear-out, or hoop tension is:

$normalsize color{black}{P_{bruL} = F_{bruL} D t quad textrm{if} quad F_{tux} le 1,304 F_{tyx} tag{5}}$
$normalsize color{black}{P_{bruL} = 1,304 F_{bryL} D t quad textrm{if} quad F_{tux} gt 1,304 F_{tyx} tag{6}}$

The Equation 7 and 8 can only be used if the load is uniformly distributed across the lug thickness. If the pin is too flexible and bends excessively, the load on the lug will tend to peak up near the shear faces and possibly cause premature failure of the lug. The bending strength of the pin can be checked using the equations in the section “Double Shear Joint Strength Under Uniform Axial Load“.

### 2.2. Lug Net-Section Strength

The allowable lug net-section tensile ultimate stress ($F_{nuL}$) on section 1-1 in Figure 4 is affected by the ability of the lug material to yield and thereby relieve the stress concentration at the edge of the hole, it can be calculated using the Equation 7.

$normalsize color{black}{F_{nuL} = K_{n} F_{tu} tag{7}}$

Where:

$K_{n}$ = net-section stress coefficient, which is obtained from the graph shown in Figure 5, 6 or 7. These graphs are function of the ultimate and yield stress and strains of the lug material in the direction of the applied load.

Figure 4 – Net Tension Stress Coefficient

Figure 5 – Net Tension Stress Coefficient Fty/Ftu = 1,0

Figure 6 – Net Tension Stress Coefficient Fty/Ftu = 0,8

Figure 7 – Net Tension Stress Coefficient Fty/Ftu = 0,6

The allowable lug net-section tensile yield stress ($F_{nyL}$) can be found using the Equation 11:

$normalsize color{black}{F_{nyL} = K_{n} F_{ty} tag{8}}$

The allowable lug net-section ultimate load ($P_{nuL}$) can be found using the Equation 9 or 10:

$normalsize color{black}{P_{nuL} = F_{nuL} (w - D) t quad textrm{if} quad F_{tu} le 1,304 F_{ty} tag{9}}$
$normalsize color{black}{P_{nuL} = 1,304 F_{nyL} (w - D) t quad textrm{if} quad F_{tu} gt 1,304 F_{ty} tag{10}}$

### 2.3. Lug Design Strength

The allowable design ultimate load for the lug ($P_{uL}$) is the lower of the values obtained from Equations 5, 6, 9 and 10:

$normalsize color{black}{P_{uL} le P_{bruL} quad textrm{or} quad P_{nuL} tag{11}}$

### 2.4. Bushing Bearing Strength

The allowable bearing yield stress for bushings ($F_{bryB}$) is restricted to the compressive yield stress ($F_{cyB}$) of the bushing material, unless higher values are substantianed by tests.

The allowable bearing ultimate stress for bushings ($F_{bruB}$) can be foud using Equation 12:

$normalsize color{black}{F_{bruB} = 1,304 F_{cyB} tag{12}}$

The allowable bushing ultimate load ($P_{uB}$) can be found using Equation 13:

$normalsize color{black}{P_{uB} = 1,304 F_{cyB} D_{P} t tag{13}}$

This assumes that the bushing extends through the full thickness of the lug.

### 2.5. Combined Lug-Bushing Design Strength

The allowable lug-bushing ultimate load ($P_{uLB}$) is the lower of the loads obtained from Equations 11 and 13.

$normalsize color{black}{P_{uLB} le P_{uL} quad textrm{or} quad P_{uB} tag{14}}$

### 2.6. Double Shear Joint Strength

The strength of a joint such as the one shown in Figure 8 depends on the lug-bushing ultimate strength ($P_{uLB}$) and on the pin shear and pin bendind strengths.

Figure 8 – Double Shear Lug Joint

#### 2.6.1. Lug-Bushing Design Strength

The allowable lug-bushing ultimate load ($P_{uLB}$) for the joint is computed, using Equation 14. For the symmetrical joint shown in the figure, Equation 14 is used to calculate the ultimate load for the outer lugs and bushings ($2P_{ULB1}$) and the ultimate load for the inner lug and bushing ($P_{uLB2}$}. The allowable value of $P_{uLB}$ for the joint is the lower of these two values.

$normalsize color{black}{P_{uLB} le 2P_{uLB1} quad textrm{or} quad P_{uLB2} tag{15}}$

#### 2.6.2. Pin Shear Strength

The pin ultimate shear load ($P_{usp}$) for the symmetrical joint shown in Figure 8 is the double shear strength of the pin:

$normalsize color{black}{P_{usp} = 1,571 {D_p}^2 F_{sup} tag{16}}$

Where $F_{sup}$ is the ultimate shear stress of the pin material.

#### 2.6.3. Pin Bending Strength

Although actual pin bending failures are infrequent, excessive pin deflections can cause the load in the lugs to peak up near the shear planes instead of being uniformly distributed across the lug thickness, thereby leading to premature lug or bushing failures at loads less than those predicted by Equation 15. At the same time, however, the concentration of load near the lug shear planes reduces the bending arm and therefore, the bending moment in the pin, making the pin less critical in bending. The following procedure is used in determining the pin ultimate bending load.

Assume that the load in each lug is uniformly distributed across the lug thickness ($b_1 = t_1$, and $2b_2 = t_2$). For the symmetrical joint shown in Figure 8, the resulting maximum pin bending is:

$normalsize color{black}{M_{maxP} = Big({P over 2}Big) Big({t_1 over 2} + {t_2 over 4 } + gBig) tag{17}}$

The ultimate failing moment for the pin is:

$normalsize color{black}{M_{uP} = 0,0982 cdot k_{bP} {D_P}^3 F_{tuP} tag{18}}$

Where $normalsize color{black}{k_{bP}}$ is the plastic bending coefficient for the pin. The value of $normalsize color{black}{k_{bP}}$ varies from 1.0 for a perfectly elastic pin to 1.7 for a perfectly plastic pin, with a value of 1.56 for pins made from reasonably ductile materials (more than 5% elongation).

The pin ultimate bending load ($normalsize color{black}{P_{ubP}}$) is, therefore:

$normalsize color{black}{P_{ubP} = {({0.1963 cdot k_{bP} {D_P}^3 F_{tuP}}) over Big({t_1 over 2} + {t_2 over 4 } + gBig)} tag{19}}$

If $normalsize color{black}{P_{ubP}}$ is equal to or greater than either $normalsize color{black}{P_{uLB}}$ (Equation 15) or $normalsize color{black}{P_{uSP}}$ (Equation 16), then the pin is a relatively strong pin that is not critical in bending, and no further pin bending calculations are required. The allowable load for the joing ($normalsize color{black}{P_{all}}$ can be determined by going directly to Equation 26.

If $normalsize color{black}{P_{ubP}}$ (Equation 19) is less than both $normalsize color{black}{P_{uLB}}$ (Equation 15) and $normalsize color{black}{P_{uSP}}$ (Equation 16), the pin is considered a relatively weak pin, critical in bending. However, such a pin may deflect sufficiently under load to shift the CG of the bearing loads toward the shear faces of the lugs, resulting in a decreased pin bending moment and a increased value of $normalsize color{black}{P_{ubP}}$. These shifted loads are assumed to be uniformly distributed over widths $normalsize color{black}{b_1}$ and $normalsize color{black}{2b_2}$, respectively, as shwon in Figure 8. The portions of the lugs and bushings not included in $normalsize color{black}{b_1}$ and $normalsize color{black}{2b_2}$ are considered ineffective. The new increased value of pin ultimate bending load is:

$normalsize color{black}{P_{ubP} = {({0.1963 cdot k_{bP} {D_P}^3 F_{tuP}}) over Big({b_1 over 2} + {b_2 over 2 } + gBig)} tag{20}}$

The maximum allowable value of $normalsize color{black}{P_{ubP}}$ is reached when $normalsize color{black}{b_1}$ and $normalsize color{black}{b_2}$ are suficiently reduced so that $normalsize color{black}{P_{ubP}}$ (Equation 19) is equal to $normalsize color{black}{P_{uLR}}$ (Equation 15), provided that $normalsize color{black}{b_1}$ and $normalsize color{black}{2b_2}$ are substuted for $normalsize color{black}{t_1}$ and $normalsize color{black}{t_2}$, respectively. At this point we have a balanced design where the joint is equally critical in pin-bending failure or lug-bushing failure.

The following equations give the “balanced design” pin ultimate bending load ($normalsize color{black}{P_{ubPmax}}$) and effective bearing width ($normalsize color{black}{b_1min}$ and $normalsize color{black}{2b_2min}$):

$normalsize color{black}{P_{ubPmax} = {2 cdot C sqrt{{P_{ubP} over C} cdot Big({t_1 over 2} + {t_2 over 4} + gBig) + g^2} - 2 cdot C g} tag{21}}$

Where:

$normalsize color{black}{C = {{P_{uLB1} P_{uLB2}} over {P_{uLB1} t_2 + P_{uLB2} t_1}} tag{22}}$

The value of $normalsize color{black}{P_{ubP}}$ on the right hand side of Equation 20 and the values of $normalsize color{black}{P_{uLB1}}$ amd $normalsize color{black}{P_{uLB2}}$ in the expression for $normalsize color{black}{C}$ are based on the assumption that the full thicknesses of the lugs are effective and have already been calculated (Equations 14 and 19).

If the inner lug strength is equal to the total strength of the two outer lugs ($normalsize color{black}{P_{uLB2} = 2P_{uLB1}}$), and if $normalsize color{black}{g =
0}$
, then:

$normalsize color{black}{P_{ubPmax} = {sqrt{P_{ubP} P_{uLB2}}} tag{23}}$

The “balanced design” effective bearing widths are:

$normalsize color{black}{b_{1min} = {{P_{ubPmax} {t_1}} over {2P_{uLB1}}} tag{24}}$

$normalsize color{black}{2b_{2min} = {{P_{ubPmax} {t_2}} over {2P_{uLB2}}} tag{25}}$

Where $normalsize color{black}{P_{ubPmax}}$ is obtained from Equation 20 and $normalsize color{black}{P_{uLB1}}$ and $normalsize color{black}{P_{uLB2}}$ are the previously calculated values based on the full thicknesses of the lugs. Since any lug thicknesses greater than $normalsize color{black}{b_{1min}}$ or $normalsize color{black}{b_{2min}}$ are not considered effective static strength design would have $normalsize color{black}{t_1 = b_{1min}}$ and $normalsize color{black}{t_2 = 2b_{2min}}$.

The allowable joint ultimate load ($normalsize color{black}{P_{all}}$ for the double-shear joint is obtained as follows:

If $normalsize color{black}{P_{ubP}}$ (Equation 19) is greater than either $normalsize color{black}{P_{uLB}}$ (Equation 15) or $normalsize color{black}{P_{uSP}}$ (Equation 16), then $normalsize color{black}{P_{all}}$ is lower of the values of $normalsize color{black}{P_{uLB}}$ or $normalsize color{black}{P_{uSP}}$.

$normalsize color{black}{P_{all} le {P_{ULB}} textrm{or} P_{uSP} tag{26}}$

If $normalsize color{black}{P_{ubP}}$ (Equation 19) is less than both $normalsize color{black}{P_{uLB}}$ and $normalsize color{black}{P_{uSP}}$, then $normalsize color{black}{P_{all}}$ is the lower of the values of $normalsize color{black}{P_{uSP}}$ and $normalsize color{black}{P_{ubPmax}}$.

$normalsize color{black}{P_{all} le {P_{uSP}} textrm{or} P_{ubPmax} tag{27}}$

#### 2.6.4. Lug Tang Strength

If Equation 26 has been used to determine the joint allowable load, then we have a condition where the load in the lugs and tangs is assumed unifromly distributed. The allowable stress in the tangs is $normalsize color{black}{F_{tur}}$. The lug tang strength ($normalsize color{black}{P_{r}}$) is the lower of the following values:

$normalsize color{black}{P_{r} = {2F_{tur} W_{r1} t_1} tag{28}}$

$normalsize color{black}{P_{r} = {2F_{tur2} W_{r2} t_2} tag{29}}$

If Equation 27 was used to determine the joint allowable load, the tangs of the outer lugs should be checked for the combined axial and bending stresses resulting from the eccentric application of the bearing loads. Assuming that the lug thickness remains constant beyond the pin, a load ($normalsize color{black}{P/2}$) applied over the width $normalsize color{black}{b_1}$ in each outer lug will produce the following bending moment in the tangs:

$normalsize color{black}{M_1 = {P over 2} Big({{t_1 - b_1} over 2}Big) tag{30}}$

A simple, but generally conservative approximation to the maximum combined stress in the outer lug tangs is:

$normalsize color{black}{F_{tr1} = {P over {2W_{r1} t_1}} + {6M_1 over {k_{br} W_{r1} t_{1}^2}} tag{31}}$

Where $normalsize color{black}{k_{br}}$, the plastic bending coefficient for a lug tang of retangular cross-section, varies from 1.0 for a perfectly elastic tang to 1.5 for a perfectly plastic tang, with a value of 1.4 representative of a rectangular cross sections with materials of reasonable ductility (more than 5% elongation). The allowable value of $normalsize color{blac}{F_{tr1}}$ is $normalsize color{black}{F_{tur1}}$. The lug tang strength is the lower of the following values:

$normalsize color{black}{P_r = {{2F_{tur1} W_{t1} t_1} over {1 + {3 over k_{br}} Big({1 - {b_{1min} over t_1}}Big)}} tag{32}}$

$normalsize color{black}{P_r = {F_{tur2} W_{t2} t_2}} tag{33}$

Where $normalsize color{black}{b_{1min}}$ is given by Equation 24.

## 3. Single-Shear Joint Strength

In single-shear joints, lug and pin bending are more critical than in double-shear joints. The amount of bending can be significantly affected by bolt clamping. In the cases considered in Figure 9, no bolt clamping is assumed, and the bending moment in the pin is resisted by socket action in the lugs.

Figure 9 – Single Shear Lug Joint

In Figure 9 a representative single-shear joint is shown, with centrally applied loads ($normalsize color{black}{P}$) in each lug, and bending moments ($normalsize color{black}{M}$ and $normalsize color{black}{M_1}$) that keep the system in equilibrium. (Assuming that there is no gap between the lugs, $normalsize color{black}{M + M_1 = P (t + t_1) / 2}$). The individual values of $normalsize color{black}{M}$ and $normalsize color{black}{M_1}$ are determined from the loading of the lugs as modified by the deflection, if any, of the lugs, according to the principles of mechanics.

The strength analysis procedure outlined below applies to either the lug. The joint strength is determined by the lowest of the margins of safety calculated for the different failure modes defined by Equations 34 through 40.

### 3.1. Lug Bearing Strength

The bearing stress distribution between lug and bushing is assumed to be similar to the stress distribution that would be obteined in a rectangular cross section of width ($normalsize color{black}{D}$) and depth ($normalsize color{black}{t}$), subjected to a load ($normalsize color{black}{P}$) and moment ($normalsize color{black}{M}$). At ultimate load the maximum lug bearing stress ($normalsize color{black}{F_{brmaxL}}$) is approximated by:

$normalsize color{black}{F_{brmaxL} = {P over Dt} + {6M over {k_{brL} Dt^2}} tag{34}}$

Where $normalsize color{black}{k_{brL}}$ is a plastic bearing coefficient for the lug material, and is assumed to be the same as the plastic bending coefficient ($normalsize color{black}{k_{bL}}$) for a rectangular section.

The allowable ultimate value of $normalsize color{black}{F_{brmaxL}}$ is either $normalsize color{black}{F_{bruL}}$ (Equations 1, 2) or $normalsize color{black}{1.304F_{bryL}}$ (Equations 3, 4), whichever is lower.

### 3.2. Net-Section Strength

At ultimate load the nominal value of the ouver fiber tensile stress in the lug net-section is approximated by:

$normalsize color{black}{F_{tmax} = {P over {(w-D)t}} + {6M over {k_{bL} (w-D)t^2}} tag{35}}$

Where $normalsize color{black}{k_{bL}}$ is the plastic bending coefficient for the lug net-section.

The allowable ultimate value of $normalsize color{black}{F_{tmax}}$ is $normalsize color{black}{F_{nuL}}$ (Equation 11) or $normalsize color{black}{1.304F_{nyL}}$ (Equation 11), whichever is lower.

### 3.3. Bushing Strength

The bearing stress distribution between bushing and pin is assumed to be similar to that between the lug and bushing. At ultimate bushing load the maximum bushing bearing stress is approximated by:

$normalsize color{black}{F_{brmaxB} = {P over {D_P t}} + {6M over {k_{brL} D_P t^2}} tag{36}}$

Where $normalsize color{black}{k_{brL}}$, the plastic bearing coefficient, is assumed the same as the palstic bending coefficient ($normalsize color{black}{k_{bL}}$) for a rectangular section.

The allowable ultimate value of $normalsize color{black}{F_{brmaxB}}$ is $normalsize color{black}{1.304F_{cyB}}$, where $normalsize color{black}{F_{cyB}}$ is the bushing material compressive yield strength.

### 3.4. Pin Shear Strength

The maximum value of pin shear can occur either within the lug or at the common shear face of the two lugs, depending upon the value of $normalsize color{black}{M/Pt}$. At the lug ultimate load the maximum pin shear stress $normalsize color{black}{F_{smaxP}}$ is approximated by:

If $normalsize color{black}{M/Pt le 2/3}$:

$normalsize color{black}{F_{smaxP} = {1.273P over {{D_p}^2}} tag{37}}$

If $normalsize color{black}{M/Pt > 2/3}$:

$normalsize color{black}{F_{smaxP} = {1.273P over {D_p}^2} {{sqrt{Big({2M over Pt}Big)^2 + 1} - 1} over {{2M over Pt} + 1 - sqrt{Big({2M over Pt}Big)^2 + 1}}} tag{38}}$

Equation 37 defines the case where the maximum pin shear is obtained at the common shear face of the lugs, and Equation 38 defines the case where the maximum pin shear occurs away from the shear face.

The allowable ultimate value of $normalsize color{black}{F_{bmaxP}}$ is $normalsize color{black}{F_{suP}}$, the ultimate shear stress of the pin material.

### 3.5. Pin Bending Strength

The maximum pin bending moment can occur within the lug or at the common shear faces of the two lugs, depending on the value of $normalsize color{black}{M/Pt}$. At the lug ultimate load the maximum pin bending stress ($normalsize color{black}{F_{bmaxP}}$) is approximated by:

If $normalsize color{black}{M/Pt le 3/8}$:

$normalsize color{black}{F_{smaxP} = {10.19M over {k_{bP} {D_P}^3}} {Big({Pt over 2M} -1 Big)} tag{39}}$

If $normalsize color{black}{M/Pt > 3/8}$:

$normalsize color{black}{F_{smaxP} = {10.19M over {k_{bP} {D_P}^3}} {{Big(sqrt{({2M over Pt})^2 +1} -1 Big)} over {2M over Pt}} tag{40}}$

Where $normalsize color{black}{k_{bP}}$ is the plastic bending coefficient for the pin.

Equation 39 defines the case where the maximum pin bending moment is obtained at the common shear face of lugs, and Equation 40 defines the case where the maximum pin bending moment occurs away from the shear face, where the pin shear is zero.

The allowable ultimate value of $normalsize color{black}{F_{bmaxP}}$ is $normalsize color{black}{F_{tuP}}$, the ultimate tensile stress of the pin material.

## 4. Transverse Load

Transversely loaded lugs and bushings are checked in the same general manner as axially loaded lugs. The transversely loaded lug, however, is a more redundant structure than an axially loaded lug, and it has a more complicated failure mode. Figure 10 illustrate the different lug dimensions that are critical in determining the lug strength.

Figure 10 – Schematic of Lugs Under Transverse Loads

### 4.1. Lug Strength

The lug ultimate bearing stress ($normalsize color{black}{F_{bruL}}$) is:

$normalsize color{black}{F_{bruL} = {F_{tru}} {F_{tux}} tag{41}}$

Where $normalsize color{black}{K_{tru}}$, the transverse ultimate load coefficient, is obtained from Figure 11 as a function of the “effective” edge distance ($normalsize color{black}{h_{av}}$):

$normalsize color{black}{h_{av} = {6 over {{3 over h_1} + {1 over h_2} + {1 over h_3} + {1 over h_4}}} tag{42}}$

Figure 11 – Transverse Ultimate and Yield Load Coefficient

The effective edge distance can be found by using the nomograph in Figure 12. The nomograph is used by first connecting the $normalsize color{black}{h_1}$ and $normalsize color{black}{h_2}$ lines at the appropriate value of $normalsize color{black}{h_1}$ and $normalsize color{black}{h_2}$. The intersection with line A is noted. Next connect the $normalsize color{black}{h_3}$ and $normalsize color{black}{h_4}$ lines similarly, and note the B line intersection. Connecting the A and B line intersection gives the value of $normalsize color{black}{h_{av}}$ to be read at the intersection with the $normalsize color{black}{h_{av}}$ line. The different edge distances ($normalsize color{black}{h_1}$, $normalsize color{black}{h_2}$, $normalsize color{black}{h_3}$, $normalsize color{black}{h_4}$) indicate different critical regions in the lug, $normalsize color{black}{h_1}$ being the most critical. The distance $normalsize color{black}{h_3}$ is the smallest distance from the hole to the edge of the lug. If the lug is a concentric lug with parallel sides, $normalsize color{black}{h_{av}/D}$ can be obtained directly from Figure 13 for any value of $normalsize color{black}{e/D}$. In concentric lugs, $normalsize color{black}{h_1 = h_4}$ and $normalsize color{black}{h_2 = h_3}$.

The lug yield bearing stress ($normalsize color{black}{F_{bryL}}$) is:

$normalsize color{black}{F_{bryL} = {K_{try}} {F_{tyx}} tag{43}}$

Where $normalsize color{black}{K_{try}}$, the transverse yield load coefficient, is obtained from Figure 12.

The allowable lug transverse ultimate load ($normalsize color{black}{P_{truL}}$) is:

If $normalsize color{black}{F_{tux} leq 1.304F_{tyx}}$:

$normalsize color{black}{P_{truL} = F_{brL} Dt tag{44}}$

If $normalsize color{black}{F_{tux} > 1.304F_{tyx}}$:

$normalsize color{black}{P_{truL} = 1.304F_{bryL} Dt tag{45}}$

Where $normalsize color{black}{F_{bruL}}$ and $normalsize color{black}{F_{bryL}}$ are obtained from Equations 42 and 43.

Figure 12 – Nomograph for Effective Edge Distance

Figure 13 – Effective Edge Distance

If the lug is not of constant thickness, then $normalsize color{black}{A_{av}/A_{br}}$ is substituted for $normalsize color{black}{h_{av}/D}$ on the horizontal scale of the graph in Figure 11, where $normalsize color{black}{A_{br}}$ is the lug bearing area, and:

$normalsize color{black}{A_{av} = {6 over {3/A_{1} + 1/A_{2} + 1/A_{3} + 1/A_{4}}} tag{46}}$

$normalsize color{black}{A_{1}}$, $normalsize color{black}{A_{2}}$, $normalsize color{black}{A_{3}}$ and $normalsize color{black}{A_{4}}$ are the areas of the sections defined by $normalsize color{black}{h_{1}}$, $normalsize color{black}{h_{2}}$, $normalsize color{black}{h_{3}}$ and $normalsize color{black}{h_{4}}$, respectively.

The values of $normalsize color{black}{K_{tru}}$ and $normalsize color{black}{K_{try}}$ corresponding to $normalsize color{black}{A_{av}/A_{br}}$ are then obtained from the graph in Figure 11 and the allowable bearing stresses are obtained aas before from Equation 42 and 43.

### 4.2. Bushing Strength

The allowable bearing stress on the bushing is the same as that for the bushing in an axially loaded lug and is given by Equation 11. The allowable bushing ultimate load ($normalsize color{black}{P_{truB}}$) is equal to $normalsize color{black}{P_{uB}}$ (Equation 9).

### 4.3. Double Shear Joints

The strength calculations needed for double shear joint strength analysis are basically the same as those needed for axially loaded. Equations 15 through 27 can be used; however, the maximum lug bearing stresses at ultimate and yield loads must not exceed those given by Equations 42 and 43.

### 4.4. Single Shear Joints

The previous discussion on double shear joint applies to single shear joint strength analysis except the equations to be used are now Equations 34 through 40.

## 5. Oblique Load

The analysis procedures used to check the strength of axially loaded lugs and of transversely loaded lugs are combined to analyze obliquely loaded lugs such as the one shwon in Figure 14. These procedures apply only if $normalsize color{black}{alpha}$ does not exceed 90°.

Figure 14 – Obliquely Loaded Lug

### 5.1. Lug Strength

The obliquely applied load ($normalsize color{black}{P_{alpha}}$) os resolved into an axial component ($normalsize color{black}{P = P_{alpha} cos{alpha}}$) and a transverse component ($normalsize color{black}{P_{tr} = P_{alpha} sin{alpha}}$). The allowable ultimate value of $normalsize color{black}{P_{alpha}}$ is $normalsize color{black}{P_{alpha L}}$ and its axial and transverse components satisfy the following equation:

$normalsize color{black}{Big({P over P_{uL}}Big)^{1.6} + Big({P_{tr} over P_{truL}}Big)^{1.6} = 1 tag{47}}$

Where $normalsize color{black}{P_{uL}}$ is the strength of an axially loaded lug (Equation 8) and $normalsize color{black}{P_{truL}}$ is the strength of a transversely loaded lug (Equations 44 and 45). The allowable load curve defined by Equation 16 is plotted on the graph in Figure 15.

Figure 15 – Allowable Load Curve

For any given value of $normalsize color{black}{alpha}$ the allowable load ($normalsize color{black}{P_{alpha L}}$ for a lug can be determined from the graph shown in Figure 15 by drawing a line from the origin with a slope equal to ($normalsize color{black}{P_{uL}/P_{truL}}$. The intersection of this line with the allowable load curve (point 1 on the graph) indicates the allowable value of $normalsize color{black}{P/P_{uL}}$ and $normalsize color{black}{P_{tr}/P_{truL}}$, from which the axial and transverse components, P and $normalsize color{black}{P_{tr}}$ of the allowable load can be readily obtained.

### 5.2. Bushing Strength

The bushing strength calculations are identical to those for axial loading (Equations 9 and 10).

### 5.3. Double Shear Joints

The strength calculations are basically the same as those for an axially loaded joint except that the maximum lug bearing stress at ultimate load must not exceed $normalsize color{black}{P_{alpha L}/Dt}$, where $normalsize color{black}{P_{alpha L}}$ is defined by Equation 47. Use Equations 15 through 27.

### 5.4. Single Shear Joints

The previous discussion on double shear joints applies to single shear joint strength analysis except the equations to be used are now Equations 34 through 40.

## 6. Multiple Shear and Single Shear Connections

Lug-pin combinations having the geometry indicated in Figure 16 should be analyzed according to the following cirteria:

1. The load carried by each lug should be determined by distributing the total applied load $normalsize color{black}{b}$ among the lugs as indicated in Figure 16, $normalsize color{black}{P}$ being obtained in Figure 17. This distribution is based on the assumption of plastic behavior (at ultimate load) of the lugs and elastic bending of the pin, and gives approximately zero bending deflection of the pin.
2. The maximum shear load on the pin is given in Table XX.
3. The maximum bending moment in the pin is given by the formulae:

$normalsize color{black}{M = {{P_{1} b} over 2}}$

Where $normalsize color{black}{b}$ is given in Figure 17.

Figure 16 – Schematic of Multiple Shear Joint in Tension

Figure 17 – Values of b

## 7. Axially Loaded Lug Design

This section presents procedures for the optimized design of lugs, bushings and pin in a symmetrical, double-shear joint, such as shown in Figure 8, subjected to a static axial load (P). One design procedure applies to the case where the pin is critical in shear, the other to the case where the pin is critical in bending. A method is given to help determine which mode of pin failure is more likely, so that the appropriate design procedure will be used.

Portions of the design procedures may be useful in obtaining efficient design for joints other than symmetrical, double-shear joints.

### 7.1. Pin Failure

An indication of whether the pin in an optimized joint design is more likely to fail in shear or in bending can be obtained from the value of R (Equation 48). If R is less than 1.0, the pin is likely to fail in shear and the design procedure for joints with pins critical in shear should be used to get an optimized design. If R is greater than 1.0, the pin is likely to be critical in bending and the design procedures for joints with pins critical in bending should be used.

$normalsize color{black}{R = {{pi F_{sup}} over {k_{bt} F_{tup}}} Big({F_{sup} over F_{brall1}} + {F_{sup} over F_{brall2}} Big) tag{48}}$

Where $normalsize color{black}{F_{sup}}$ and $normalsize color{black}{F_{tup}}$ are the ultimate shear and ultimate tension stresses for the pin material, $normalsize color{black}{k_{bp}}$ is the plastic bending coefficient for the pin, and $normalsize color{black}{F_{brall1}}$ and $normalsize color{black}{F_{brall2}}$ are allowable bearing stresses in the female and male lugs. The value of $normalsize color{black}{F_{brall1}}$ can be approximated by the lowest of the following three values:

$normalsize color{black}{K F_{tux1} {D over D_p}}$; $normalsize color{black}{K F_{tyx1} {D over D_p}}$; $normalsize color{black}{1.304 F_{cyB1}}$

Where $normalsize color{black}{F_{tux1}}$ and $normalsize color{black}{F_{tyx1}}$ are the cross-grain tensile ultimate and tensile yield stress for female lugs, $normalsize color{black}{F_{cyB1}}$ is the compressive yield stress of the bushings in the female lugs, and $normalsize color{black}{K}$ is obtained from Figure 18. Assume $normalsize color{black}{D = D_p}$ if a better estimate cannot be made. $normalsize color{black}{F_{brall2}}$ is approximated in a similar manner.

Figure 18 – Allowable Bearing Coefficient

#### 7.1.1. Failure in the Shearing Mode

1. Pin and Bushing Diameter.The minimum allowable diameter for a pin in double shear is:$normalsize color{black}{D_p = 0.798 sqrt{P over F_{sup}} tag{49}}$

The outside diameter of the bushing is $normalsize color{black}{D = D_p + 2 t_B}$, where $normalsize color{black}{t_B}$ is the bushing wall thickness.

2. Edge Distance Ratio (e/D). The value of $normalsize color{black}{e/D}$ that will minimize the combined lug and pin wiehgt is obtained from Figure 19 for the case where lug bearing failure and pin shear failure occur simultaneously. The lug is assumed not critical in net tension, and the bushing is assumed not critical in bearing.

Figure 19 – Edge Distance Ratio Pin Critical in Shear

Figure 20 – Edge Distance Ratio Pin Critical in Bearing

The curves in Figure 20 apply specifically to concentric lugs ($normalsize color{black}{a = e - D/2}$, and $normalsize color{black}{w = 2e}$), but they can be used for reasonably similar lugs.

1. Allowable Loads. The allowable loads for the different failure modes (lug bearing failure, lug net-tension failure, and bushing failure) are determined from Equations 5, 6, 9, 10 and 13 in terms of the (unknown) lug thickness. The lowest of these loads is critical.
2. Lug Thicknesses. The required male and female lug thicknesses are determined by equating the applied load in each lug to the critical failure load for the lug.
3. Pin Bending. To prevent bending failure of the pin before lug or bushing failure occurs in a uniformly loaded symmetrical double-shear joint, the required pin diameter is:$normalsize color{black}{D_p = sqrt[3]{{{2.55P} over {k_{bp} F_{tup}}} Big({t_1 + {t_2 over 2} + 2g}Big)} tag{50}}$Where $normalsize color{black}{k_{bp}}$ is the plastic bending coefficient for the pin. If the value of $normalsize color{black}{D_p}$ from Equation 50 is greater than that from Equation 49, the joint must be redesigned because the pin is critical in bending.
4. Reduced Edge Distance. If allowable bushing load (Equation 13) is less than the allowable lug load (Equations 5 and 6), a reduced value of $normalsize color{black}{e}$, obtained by using the curve shown in Figure 21 for optimum $normalsize color{black}{e/D}$, will give a ligther joint in which lug bearing failure and bushing bearing failure will occur simultaneously. The previously calculated pin diameter and lug thicknesses are unchanged.
5. Reduced Lug Width. If the lug net-tension strength (Equation 9 and 10) exceeds the bearing strength (Equation 5 and 6), the net-section width can be reduced by the ratio of the bearing strength to the net-tension strength.

#### 7.1.2. Failure in the Bending Mode

1. Pin and Bushing Diameters (First Approximation). A first approximation to the optimum pin diameter is shwon in Equation 51.

$normalsize color{black}{D_p = sqrt[4]{{1.273 over k_{bp}} Big({P over F_{tup}}Big)^2 Big({{F_{tup} over F_{tall1}} + {F_{tup} over F_{tall2}} Big)}} tag{51}}$

Where $normalsize color{black}{F_{tall1}}$ is either $normalsize color{black}{F_{tux1}}$ or $normalsize color{black}{1.304 F_{tyx1}}$, whichever is smaller; and $normalsize color{black}{F_{tall2}}$ is either $normalsize color{black}{F_{tux2}}$ or $normalsize color{black}{1.304 F_{Ftyx2}}$, whichever is smaller. This approximation becomes more accurate when there are no bushings and when there is no gap between the lugs.

The first approximation to the outside diameter of the bushing is $normalsize color{black}{D = D_p + 2 t_B}$.

Figure 21 – Edge Distance Ratio

1. Edge Distance Ratio (e/D). The value of $normalsize color{black}{e/D}$ that will minimize the combined lug and pin weight is obtained from Figure 20 for the case of symmetrical double-shear joints in which lug bearing failure and pin bending failure occur simultaneously. The lug is assumed not critical in tension and the bushing is assumed not critical in bearing.The curves apply specifically to concentric lugs ($normalsize color{black}{a = e - D/2}$ and $normalsize color{black}{w = 2e}$), but can be used for reasonably similar lugs.
2. Allowable Loads (Frist Approximation). The allowable loads for the different failure modes (lug bearing failuyre, lug net-tension failure, and bushing failure) are determined from Equations 5, 6, 9, 10 and 13, in terms of the (unknown) lug thickness. The lowest of these loads is critical.
3. Lug Thicknesses (First Approximation). The frist approximation to the required male and female lug thicknesses are determined by equating the applied load in each lug to the lowest allowable load for the lug.
4. Pin Diameter (Second Approximation). The second approximation to the pin diameter is obtained by substuting the frist approximation lug thicknesses into Equation 50.
5. Final Pin and Bushing Diameters and Lug Thicknesses. The final optimum pin diameter is very closely approximated by:$normalsize color{black}{D_{popt} = 1/3 D_p + 2/3 D_p tag{52}}$

An average value, however, is generally sufficient. If the final optimum value is not a standard pin diameter, choose the next larger standard pin and bushing.

The final lug thicknesses corresponding to the standard pin and bushing are then determined.

6. Pin Shear. The pin is checked for shear strength (Equation 49).
7. Reduced Edge Distance. If the bushing bearing strength (Equation 13) is less than the lug bearing strength (Equations 5 and 6), a reduced value of $normalsize color{black}{e/D}$, obtained from the curve in Figure 21, will give a lighter joint. The pin diameter and lug thicknesses are unchanged.
8. Reduced Lug Width. If the lug net-tension strength (Equations 9 and 10) exceeds the lug bearing strength (Equations 5 and 6), the net-section width can be reduced by the ratio of the bearing strength to the net-tension strength.

## 8. Analysis of Lugs with Less Than 5% Elongation

The procedures given for determining the static strength of lugs apply to lugs made from materials which have ultimate elongations, $normalsize color{black}{varepsilon_u}$, of at least 5% in all directions in the plane of the lug. This section describes procedures for calculating reductions in strength for lugs made from materials which do not meet the elongation requirement. In addtion to using these procedures, special consideration must be given to possible further loss in strength resulting from material defects wyhen the short transverse gain direction of the lug material is in the plane of the lug. The analysis procedures for lugs made from materials without defects but with kess than 5% elongation are as follows.

### 8.1. Axially Loaded Lug

#### 8.1.1. Bearing Strength

1. Determine $normalsize color{black}{F_{ty} F_{tu}}$ and $normalsize color{black}{varepsilon_y varepsilon_u}$, using values of $normalsize color{black}{F_{ty}}$, $normalsize color{black}{F_{tu}}$, $normalsize color{black}{varepsilon_y}$ and $normalsize color{black}{varepsilon_u}$ that correspond to the minimum value of $normalsize color{black}{varepsilon_u}$ in the plane of the lug.
2. Determine the value of $normalsize color{black}{B}$, the ductility factor, from the graph shown in Figure 22.
3. Determine a second value of $normalsize color{black}{B}$ (denoted by $normalsize color{black}{B_{0.05}}$) for the same values of $normalsize color{black}{F_{ty}}$, $normalsize color{black}{F_{tu}}$, and $normalsize color{black}{varepsilon_y}$ as before, but with $normalsize color{black}{varepsilon_u = 0.05}$.
4. Multiply the bearing stress and bearing load allowables given by Equations 1 through 6 by $normalsize color{black}{B/B_{0.05}}$ to obtain the corrected allowables.

Figure 22 – Ductility Factor

#### 8.1.2. Net-Section Strength

The procedure for determining net-section allowables is the same for all values of $normalsize color{black}{varepsilon_u}$. The graphs in Figures 5, 6 and 7 are used to obtain a value of $normalsize color{black}{K_n}$ which is substituted in Equations 7 and 8. If the grain direction of the material is known, the values of $normalsize color{black}{F_{ty}}$, $normalsize color{black}{F_{tu}}$, and $normalsize color{black}{varepsilon_u}$ used in entering the graphs should correspond to the grain direction parallel to the load. Otherwise, use values corresponding to the minimum value of $normalsize color{black}{varepsilon_u}$ in the plane of the lug.

#### 8.1.3. Strength of Lug Tangs

The plastic bending coefficient for a rectangular cross section can be approximated by $normalsize color{black}{k_{bL} = 1.5 B}$, where $normalsize color{black}{B}$ is obtained from Figure 22, in which $normalsize color{black}{y}$ and $normalsize color{black}{u}$ are the yield and ultimate strains of the lug tang material in the direction of loading. The maximum allowable value of $normalsize color{black}{k_{bL}}$ for a rectangle is 1.4.

#### 8.1.4. Lug-Bushing Strength / Single-Shear Joint

The values of $normalsize color{black}{k_{brL}}$ and $normalsize color{black}{k_{bL}}$ for rectangular cross sections are approximated by $normalsize color{black}{1.5 B}$, where $normalsize color{black}{B}$ is determined from the graph as described in Figure 22. The maximum allowable values of $normalsize color{black}{k_{brL}}$ and $normalsize color{black}{k_{bL}}$ are 1.4.

### 8.2. Transversely Loaded Lugs

#### 8.2.1. Bearing Strength

The same procedure as that for the bearing strength of axially loaded lugs is used.

1. Determine $normalsize color{black}{B}$ and $normalsize color{black}{B_{0.05}}$ as described for axially loaded lugs, where $normalsize color{black}{B}$ corresponds to the minimum value of $normalsize color{black}{varepsilon_u}$ in the plane of the lug.
2. Multiply the bearing stress and bearing load allowables given by Equations 42 through 45 by $normalsize color{black}{B/B_{0.05}}$ to obtain the corrected allowables.

## 9. Stresses Due to Press Fit Bushings

Pressure between a lug and bushing assembly having negative clearance can be determined from consideration of the radial displacements. After assembly, the in crease in inner radius (lug) plus the decrease in outer radius of the bushing equals the difference between the radii of the bushing and ring before assembly:

$normalsize color{black}{delta = u_{ring} - u_{bushing} tag{53}}$

Where:

$normalsize color{black}{delta}$ = Difference between outer radius of bushing and inner radius of the ring.

$normalsize color{black}{u}$ = Radial displacement, positive away from the axis of ring or bushing.

Radial displacement at the inner surface of a ring subjected to internal pressure $normalsize color{black}{p}$ is:

$normalsize color{black}{u = {D_p over E_{ring}} Big[{{C^2 + D^2} over {C^2 - D^2}} + mu_{ring}Big]tag{54}}$

Raial displacement at the outer surface of a bushing subjected to external pressure $normalsize color{black}{p}$ is:

$normalsize color{black}{u = - {B_p over E_{bush}}Big[{{B^2 + A^2} over {B^2 - A^2}} - mu_{bush}Big]tag{55}}$

Where:

$normalsize color{black}{A}$ = Inner radius of bushing
$normalsize color{black}{B}$ = Outer radius of bushing
$normalsize color{black}{C}$ = Outer radius of ring (lug)
$normalsize color{black}{D}$ = Inner radius of ring (lug)
$normalsize color{black}{E}$ = Modulus of elasticity
$normalsize color{black}{mu}$ = Poisson’s ratio

Substitute Equations 54 and 55 into Equation 53 and solve for $normalsize color{black}{p}$:

$normalsize color{black}{p = {delta over {{D over E_{ring}} Big({{C^2 + D^2} over {C^2 - D^2}} + mu_{ring}}Big) + {B over E_{bish}} Big({{B^2 + A^2} over {B^2 - A^2}} - mu_{bush}Big)} tag{56}}$

Maximum radial and tangential stresses for a ring subjected to internal pressure occur at the inner surface of the ring (lug):

$normalsize color{black}{F_r = -p tag{57}}$

$normalsize color{black}{F_t = p Big[{{C^2 + D^2} over {C^2 - D^2}}Big] tag{58}}$

Positive sign indicates tension. The maximum shear stress at this point is:

$normalsize color{black}{F_s = {{F_t - F_r} over 2} tag{59}}$

The maximum radial stress for a bushing subjected to external pressure occurs at the outer surface of the bushing is:

$normalsize color{black}{F_r = -p tag{60}}$

The maximum tangential stress for a bushing subjected to external pressure occurs at the inner surface of the bushing is:

$normalsize color{black}{F_t = {{2_{pb}^2 over {B^2 = A^2}}} tag{61}}$

The allowable press fit stress may be based on:

1. Stress Corrosion. The maximum allowable press fit stress in magnesium alloys should not exceed 8000 psi. For all aluminum alloys the maximum press fit stress should not exceed $normalsize color{black}{0.50 F_{ty}}$.
2. Static Fatigue. Static fadigue is the brittle fracture of metals under sustained loading, and in steel may result from several different phenomena, the most familiar of which is hydrogen embrittlement. Steel parts heat treated above 200 ksi, which by nature of their function or other considerations are exposed to hydrogen embrittlement, should be designed to an allowable press fit stress of $normalsize color{black}{25% F_{tu}}$.
3. Ultimate Strength. Ultimate strength cannot be exceeded, but is not usually critical in a press fit application.
4. Fatigue Life. The hoop tension stresses resulting from the press fir of a bushing in a lug will reduce the stress range for oscillating loads, thereby improving fatigue life.

The presence of hard brittle coatings in holes that contain a press fit bushing or bearing can cause premature failure by cracking of the coating or by high press fit stresses caused by build-up of coating. Therefore, Hardcoat or HAE coatings should not be used in holes that will subsequently contain a press fit bushing or bearing.

Figures 23 and 24 permit determining the tangential stress, $normalsize color{black}{F_r}$, for bushings pressed into aluminum rings. Figure 23 presents data for general steel bushings, and Figure 24 presents data for the NAS 75 class bushings. Figure 25 gives limits for maximum interference fits for steel bushings in magnesium alloy rings.

Figure 23 – Tangential Stresses for Pressed Steel Bushings in Aluminum Rings

Figure 24 – Tangential Stresses for Pressed NAS 75 Bushings

Figure 25 – Maximum Interference Fits of Steel Bushings in Magnesium Alloy Rings