Lug Analysis
This page provides the chapter on lug analysis from the "Stress Analysis Manual," Air Force Flight Dynamics Laboratory, October 1986.
Other related chapters from the Air Force "Stress Analysis Manual" can be seen to the right.
 Lug Analysis
 Lug Fatigue Analysis
9.1 Introduction to Lug Analysis
Lugs are connectortype elements widely used as structural supports for pin connections. In the past, the lug strength was overdesigned since weight and size requirements were for the most part unrestricted. However, the refinement of these requirements have necessitated conservative methods of design.
This section presents static strength analysis procedures for uniformly loaded lugs and bushings, for double shear joints, and for single shear joints, subjected to axial, transverse, or oblique loading. Also listed is a section which applies to lugs made from materials having ultimate elongations of at least 5% in any direction in the plane of the lug. Modifications for lugs with less than 5% elongation are also presented. In addition, a short section on the stresses due to press fit bushings is presented.
9.2 Lug Analysis Nomenclature
F_{bru.L}  =  Lug ultimate bearing stress 
F_{bry.L}  =  Lug yield bearing stress 
F_{tux}  =  Cross grain tensile ultimate stress of lug material 
F_{tyx}  =  Cross grain tensile yield stress of lug material 
F_{bru}  =  Allowable ultimate bearing stress, MHBS 
F_{bry}  =  Allowable yield bearing stress, MHBS 
F_{tu}  =  Ultimate tensile stress 
F_{nu.L}  =  Allowable lug netsection tensile ultimate stress 
F_{ny.L}  =  Allowable lug netsection tensile yield stress 
F_{bry.B}  =  Allowable bearing yield stress for bushings 
F_{bru.B}  =  Allowable bearing ultimate stress for bushings 
F_{cy.B}  =  Bushing compressive yield stress 
F_{su.P}  =  Ultimate shear stress of the pin material 
F_{tu.P}  =  Pin ultimate tensile stress 
F_{tu.T}  =  Allowable ultimate tang stress 
F_{br.max.L}  =  Maximum lug bearing stress 
F_{br.max.B}  =  Maximum bushing bearing stress 
F_{s.max.P}  =  Maximum pin shear stress 
F_{b.max.P}  =  Maximum pin bending stress 
P_{bru.L}  =  Allowable lug ultimate bearing load 
P_{nu.L}  =  Allowable lug netsection ultimate load 
P_{u.B}  =  Allowable bushing ultimate load 
P_{u.L}  =  Allowable design ultimate load 
P_{u.L.B}  =  Allowable lugbushing ultimate load 
P_{us.P}  =  Pin ultimate shear load 
P_{ub.P}  =  Pin ultimate bending load 
P_{ub.P.max}  =  "Balanced design" pin ultimate bending load 
P_{all}  =  Allowable joint ultimate load 
P_{T}  =  Lug tang strength 
P_{tru.L}  =  Allowable lug transverse ultimate load 
P_{tru.B}  =  Allowable bushing transverse ulti.Inate load 
K_{n}  =  Nettension stress coefficient 
K_{b.P}  =  Plastic bending coefficient for pin 
K_{b.T}  =  Plastic bending coefficient for tang 
K_{br.L}  =  Plastic bearing coefficient for lug 
K_{b.L}  =  Plastic bending coefficient for lug 
K_{tru}  =  Transverse ultimate load coefficient 
K_{try}  =  Transverse yield load coefficient 
M_{max.P}  =  Maximum pin bending moment 
M_{u.P}  =  Ultimate pin failing moment 
A  =  Area, in^{2} 
a  =  Distance from edge of hole to edge of lug, inches 
B  =  Ductility factor for lugs with less than 5% elongation 
b  =  Effective bearing width, inches 
D  =  Hole diameter of pin diameter, inches 
E  =  Modulus of elasticity, psi 
e  =  Edge distance, inches 
f  =  Stress, psi 
g  =  Gap between lugs, inches 
h_{1}..h_{4}  =  Edge distances in transversely loaded lug, inches 
h_{av}  =  Effective edge distance in transversely loaded lug 
K  =  Allowable stress (or load) coefficient 
M  =  Bending moment, in.lbs. 
P  =  Load, lbs. 
t_{B}  =  Bushing wall thickness, inches 
t  =  Lug thickness, inches 
w  =  Lug width, inches 
α  =  Angle of load to axial direction, degrees 
ϵ  =  Strain, inches/inch 
ρ  =  Density, lbs/in^{3} 
9.3 Lug and Bushing Strength Under Uniform Axial Load
Axially loaded lugs in tension must be checked for bearing strength and for netsection strength. The bearing strength of a lug loaded in tension, as shown in Figure 91, depends largely on the interaction between bearing, shearout, and hooptension stresses in the part of the lug ahead of the pin. The netsection of the lug through the pin must be checked against nettension failure. In addition, the lug and bushing must be checked to ensure that the deformations at design yield load are not excessive.
9.3.1 Lug Bearing Strength Under Uniform Axial Load
The bearing stresses and loads for lug failure involving bearing, sheartearout, or hoop tension in the region forward of the netsection in Figure 91 are determined from the equations below, with an allowable load coefficient (K) determined from Figures 92 and 93. For values of e/D less than 1.5, lug failures are likely to involve shearout or hoop tension; and for values of e/D greater than 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_{bru.L}) is
The graph in Figure 92 applies only to cases where D/t is 5 or less, which covers most of the cases. If D/t is greater than 5, there is a reduction in strength which can be approximated by the curves in Figure 93. The lug yield bearing stress (F_{bry.L}) is
The allowable lug ultimate bearing load (P_{bru.L}) for lug failure in bearing, shearout, or hoop tension is
P_{bru.L}/Dt should not exceed either F_{bru} or 1.304F_{bry}, where F_{bru} and F_{bry} are the allowable ultimate and yield bearing stresses for the lug material for e/D = 2.0, as given in MILHDBK5 or other applicable specification.
Equations (93a) and (93b) apply only 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.
A procedure to check the pin bending strength in order to prevent premature lug failure is given in Section 9.4 entitled "Double Shear Joint Strength Under Uniform Axial Load."
9.3.2 Lug NetSection Strength Under Uniform Axial Load
The allowable lug netsection tensile ultimate stress (F_{nu.L}) on Section 11 in Figure 94a is affected by the ability of the lug material to yield and thereby relieve the stress concentration at the edge of the hole.
K_{n}, the nettension stress coefficient, is obtained from the graphs shown in Figure 94 as a function of the ultimate and yield stress and strains of the lug material in the direction of the applied load. The ultimate strain (ϵ_{u}) can be obtained from MILHDBK5.
The allowable lug netsection tensile yield stress (F_{ny.L}) is
The allowable lug netsection ultimate load (P_{nu.L}) is
9.3.3 Lug Design Strength Under Uniform Axial Load
The allowable design ultimate load for the lug (P_{u.L}) is the lower of the values obtained from Equations (93) and (96).
$$ P_{u.L} \le $$  $$ P_{bru.L} $$  (Equations (93a) and (93b)), or 
$$ P_{nu.L} $$  (Equations (96a) and (96b)) 
9.3.4 Bushing Bearing Strength Under Uniform Axial Load
The allowable bearing yield stress for bushings (F_{bry.B}) is restricted to the compressive yield stress (F_{cy.B}) of the bushing material, unless higher values are substantiated by tests.
The allowable bearing ultimate stress for bushings (F_{bru.B}) is
The allowable bushing ultimate load (P_{u.B}) is
This assumes that the bushing extends through the full thickness of the lug.
9.3.5 Combined LugBushing Design Strength Under Uniform Axial Load
The allowable lugbushing ultimate load (P_{u.L.B}) is the lower of the loads obtained from Equations (97) and (99).
$$ P_{u.L.B} \le $$  $$ P_{u.L} $$  (Equation (97)), or 
$$ P_{u.B} $$  (Equation (99)) 
9.4 Double Shear Joint Strength Under Uniform Axial Load
The strength of a joint such as the one shown in Figure 95 depends on the lugbushing ultimate strength (P_{u.L.B}) and on the pin shear and pin bending strengths.
9.4.1 LugBushing Design Strength for Double Shear Joints Under Uniform Axial Load
The allowable lugbushing ultimate load (P_{u.L.B}) for the joint is computed, using Equation (910). For the symmetrical joint shown in the figure, Equation (910) is used to calculate the ultimate load for the outer lugs and bushings (2P_{u.L.B.1}) and the ultimate load for the inner lug and bushing (P_{u.L.B.2}). The allowable value of P_{u.L.B} for the joint is the lower of these two values.
$$ P_{u.L.B} \le $$  $$ 2 P_{u.L.B.1} $$  (Equation (910)), or 
$$ P_{u.L.B.2} $$  (Equation (910)) 
9.4.2 Pin Shear Strength for Double Shear Joints Under Uniform Axial Load
The pin ultimate shear load (P_{us.P}) for the symmetrical joint shown in Figure 95 is the double shear strength of the pin:
where F_{su.P} is the ultimate shear stress of the pin material.
9.4.3 Pin Bending Strength for Double Shear Joints Under Uniform Axial Load
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 (911). 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 95, the resulting maximum pin bending moment is
The ultimate failing moment for the pin is
where k_{b.P} is the plastic bending coefficient for the pin. The value of k_{b.P} 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 (P_{ub.P}) is, therefore,
If P_{ub.P} is equal to or greater than either P_{u.L.B} (Equation (911)) or P_{us.P} (Equation (912)), 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 joint (P_{all}) can be determined by going directly to Equation (919a).
If P_{ub.P} (Equation (915)) is less than both P_{u.L.B} (Equation (911)) and P_{us.P} (Equation (912)), the pin is considered a relatively weak pin, critical in bending. However, such a pin may deflect sufficiently under load to shift the c.g. of the bearing loads toward the shear faces of the lugs, resulting in a decreased pin bending moment and an increased value of P_{ub.P}. These shifted loads are assumed to be uniformly distributed over widths b_{1} and 2b_{2}, which are less than t_{1} and t_{2}, respectively, as shown in Figure 95. The portions of the lugs and bushings not included in b_{1} and 2b_{2} are considered ineffective. The new increased value of pin ultimate bending load is
The maximum allowable value of P_{ub.P} is reached when b_{1} and b_{2} are sufficiently reduced so that P_{ub.P} (Equation (915a)) is equal to P_{u.L.B} (Equation (911)), provided that b_{1} and 2b_{2} are substituted for t_{1} and t_{2}, respectively. At this point we have a balanced design where the joint is equally critical in pinbending failure or lugbushing failure.
The following equations give the "balanced design" pin ultimate bending load (P_{ub.P.max}) and effective bearing widths (b_{1.min} and 2b_{2.min}):
where
$$ C = { P_{u.L.B.1} ~P_{u.L.B.2} \over P_{u.L.B.1} ~t_2 + P_{u.L.B.2} ~t_1 } $$The value of P_{ub.P} on the right hand side of Equation (916) and the values of P_{u.L.B.1} and P_{u.L.B.2} in the expression for C are based on the assumption that the full thicknesses of the lugs are effective and have already been calculated. (Equations (910) and (915)).
If the inner lug strength is equal to the total strength of the two outer lugs (P_{u.L.B.2} = 2P_{u.L.B.1}), and if g = 0, then
The "balanced design" effective bearing widths are
where P_{ub.P.max} is obtained from Equation (916) and P_{u.L.B.1} and P_{u.L.B.2} are the previously calculated values based on the full thicknesses of the lugs. Since any lug thicknesses greater than b_{1.min} or b_{2.min} are not considered effective, an efficient static strength design would have t_{1} = b_{1.min} and t_{2} = 2b_{2.min}.
The allowable joint ultimate load (P_{all}) for the doubleshear joint is obtained as follows:
If P_{ub.P} (Equation (915) is greater than either P_{u.L.B} (Equation (911)) or P_{us.P} (Equation. (912)), then P_{all} is the lower of the values P_{u.L.B} or P_{us.P}.
$$ P_{all} \le $$  $$ P_{u.L.B} $$  (Equation (911)), or 
$$ P_{us.P} $$  (Equation (912)) 
If P_{ub.P} (Equation (915)) is less than both P_{u.L.B} and P_{us.P}, then P_{all} is the lower of the values of P_{us.P} and P_{ub.P.max}.
$$ P_{all} \le $$  $$ P_{us.P} $$  (Equation (912)), or 
$$ P_{ub.P.max} $$  (Equation (916)) 
9.4.4 Lug Tang Strength for Double Shear Joints Under Uniform Axial Load
If Equation (919a) has been used to determine the joint allowable load, then we have a condition where the load in the lugs and tangs is assumed uniformly distributed. The allowable stress in the tangs is F_{tu.T}. The lug tang strength (P_{T}) is the lower of the following values.
If Equation (919b) 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 (P/2) applied over the width b_{1} in each outer lug will produce the following bending moment in the tangs:
$$ M_1 = {P \over 2} \left({ t_1  b_1 \over 2 }\right) $$A simple, but generally conservative, approximation to the maximum combined stress in the outer lug tangs is
where k_{b.T}, the plastic bending coefficient for a lug tang of rectangular 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 rectangular cross sections with materials of reasonable ductility (more than 5% elongation). The allowable value of F_{t.T.1} is F_{tu.T.1}. The lug tang strength is the lower of the following values:
where b_{1.min} is given by Equation (918a).
9.5 SingleShear Joint Strength Under Uniform Axial Load
In singleshear joints, lug and pin bending are more critical than in doubleshear joints. The amount of bending can be significantly affected by bolt clamping. In the cases considered in Figure 96, no bolt clamping is assumed, and the bending moment in the pin is resisted by socket action in the lugs.
In Figure 96 a representative singleshear joint is shown, with centrally applied loads (P) in each lug, and bending moments (M and M_{1}) that keep the system in equilibrium. (Assuming that there is no gap between the lugs, M + M_{1} = P(t + t_{1})/2). The individual values of M and 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 lug. The joint strength is determined by the lowest of the margins of safety calculated for the different failure modes defined by Equations (923) through (927).
9.5.1 Lug Bearing Strength for Single Shear Joints Under Uniform Axial Loads
The bearing stress distribution between lug and bushing is assumed to be similar to the stress distribution that would be obtained in a rectangular cross section of width (D) and depth (t), subjected to a load (P) and moment (M). At ultimate load the maximum lug bearing stress (F_{br.max.L}) is approximated by
where k_{br.L} is a plastic bearing coefficient for the lug material, and is assumed to be the same as the plastic bending coefficient (k_{b.L}) for a rectangular section.
The allowable ultimate value of F_{br.max.L} is either F_{bru.L} (Equations (91a) (91b)) or 1.304F_{bry.L} (Equations (92a) (92b)), whichever is lower.
9.5.2 Lug NetSection Strength for Single Shear Joints Under Uniform Axial Load
At ultimate load the nominal value of the outer fiber tensile stress in the lug netsection is approximated by
where k_{b.L} is the plastic bending coefficient for the lug netsection.
The allowable ultimate value of F_{t.max} is F_{nu.L} (Equation (94)) or 1.304F_{ny.L} (Equation (95)), whichever is lower.
9.5.3 Bushing Strength for Single Shear Joints Under Uniform Axial Load
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
where k_{br.L}, the plastic bearing coefficient, is assumed the same as the plastic bending coefficient (k_{b.L}) for a rectangular section.
The allowable ultimate value of F_{br.max.B} is 1.304F_{cy.B}, where F_{cy.B} is the bushing material compressive yield strength.
9.5.4 Pin Shear Strength for Single Shear Joints Under Uniform Axial Load
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 M/Pt. At the lug ultimate load the maximum pin shear stress (F_{s.max.P}) is approximated by
Equation (926a) defines the case where the maximum pin shear is obtained at the common shear face of the lugs, and Equation (926b) defines the case where the maximum pin shear occurs away from the shear face.
The allowable ultimate value of F_{b.max.P} is F_{su.P}, the ultimate shear stress of the pin material.
9.5.5 Pin Bending Strength for Single Shear Joints Under Uniform Axial Load
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 M/Pt. At the lug ultimate load the maximum pin bending stress (F_{b.max.P}) is approximated by
where k_{bP} is the plastic bending coefficient for the pin.
Equation (927a) defines the case where the maximum pin bending moment is obtained at the common shear face of the lugs, and Equation (927b) 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 F_{b.max.P} is F_{tu.P}, the ultimate tensile stress of the pin material.
9.6 Example of Uniform Axially Loaded Lug Analysis
Determine the static strength of an axially loaded, double shear joint, such as shown in Section 9.4, with dimensions and material properties given in Table 91.
Female Lugs, 1  Male Lug, 2  Bushings, 1 and 2  Pin  

Material  2024T351 Plate  7075T651 Plate  Al. Bronze  4130 Steel 
F_{tu}  64,000 psi (Xgrain) 
77,000 psi (Xgrain) 
110,000 psi  125,000 psi 
F_{ty}  40,000 psi (Xgrain) 
66,000 psi (Xgrain) 
60,000 psi  103,000 psi 
F_{cy}  60,000 psi  
F_{su}  82,000 psi  
E  10.5 x 10^{6} psi  10.3 x 10^{6}  29 x 10^{6} psi  
ϵ_{u}  0.12  0.06  
D or D_{P}  D = 1.00 in  D = 1.00 in  D_{P} = 0.75 in, D = 1.00 in 
D_{P} = 0.75 in 
e  1.25 in  1.5 in  
a  0.75 in  1.00 in  
w = W_{T}  2.50 in  3.00 in  
t  0.50 in  0.75 in  0.50 and 0.75 in  
g  0.10 in 
(1) Female Lugs and Bushings
F_{tux} = 64,000 psi; 1.304F_{tyx} = 1.304 × 40000 = 52,160 psi.
a) Lug Bearing Strength (Equations (92a) and (93b))
\( {e_1 \over D} = {1.25 \over 1.00} = 1.25 \); therefore K_{1} = 1.46 (from Figure 92)
P_{bru.L.1} = 1.304 × 1.46 × 0.75 × 40000 × 1.00 × 0.50 = 28600 lbs.
b) Lug NetSection Tension Strength (Equations (95) and (96b))
\( {D \over w_1} = {1.00 \over 2.50} = 0.40 \); \( {F_{ty} \over F_{tu}} = {40000 \over 64000} = 0.625 \);
\( {F_{tu} \over E \varepsilon_u} = {64000 \over 10.5 \times 10^6 \times 0.12} = 0.051 \); therefore, K_{n.1} = 0.74 (by interpolation from Figure 94)
P_{nu.L} = 2 × 1.304 × 0.74 × 40000 × (2.5  1.0) × 0.5 = 57,898 lbs.
c) Lug Design Strength (Equation (97))
P_{u.L.1} = P_{bru.L.1} = 28600 lbs.
d) Bushing Bearing Strength (Equation (99))
P_{u.B.1} = 1.304 × 60000 × 0.75 × 0.50 = 29300 lbs.
e) Combined LugBushing Design Strength (Equation (910))
P_{u.L.B.1} = P_{u.L.1} = 28600 lbs.
(2) Male Lugs and Bushings
F_{tux} = 77,000 psi; 1.304F_{tyx} = 1.304 × 66000 = 86,100 psi.
a) Lug Bearing Strength (Equations (91b) and (93a))
\( {e_2 \over D} = {1.50 \over 1.00} = 1.50 \); therefore K_{2} = 1.33 (from Figure 92)
P_{bru.L.2} = 1.33 × 77000 × 1.00 × 0.75 = 77000 lbs.
b) Lug NetSection Tension Strength (Equations (94) and (96a))
\( {D \over w_2} = {1.00 \over 3.00} = 0.333 \); \( {F_{ty} \over F_{tu}} = {66000 \over 77000} = 0.857 \);
\( {F_{tu} \over E \varepsilon_u} = {77000 \over 10.3 \times 10^6 \times 0.06} = 0.125 \); therefore, K_{n.2} = 0.87 (by interpolation from Figure 94)
P_{nu.L} = 1.304 × 87 × 66000 × (3.0  1.0) × 0.75 = 112,313 lbs.
c) Lug Design Strength (Equation (97))
P_{u.L.2} = P_{bru.L.2} = 77000 lbs.
d) Bushing Bearing Strength (Equation (99))
P_{u.B.2} = 1.304 × 60000 × 0.75 × 0.75 = 44000 lbs.
e) Combined LugBushing Design Strength (Equation (910))
P_{u.L.B.2} = P_{u.B.2} = 44000 lbs.
(3) Joint Analysis
a) LugBushing Strength (Equation (911))
P_{u.L.B} = P_{u.L.B.2} = 44000 lbs.
b) Pin Shear Strength (Equation (912))
P_{us.P} = 1.571 × (0.75)^{2} × 82000 = 72400 lbs.
c) Pin Bending Strength (Equation (915))
The pin ultimate bending load, assuming uniform bearing across the lugs, is
Since P_{ub.P} is less than both P_{u.L.B} and P_{us.P}, the pin is a relatively weak pin which deflects sufficiently under load to shift the bearing loads toward the shear faces of the lugs. The new value of pin bending strength is, then,
(from Equation (916)) where
Therefore, P_{ub.P.max} = 2 × 29000 × (0.754  0.10) = 37900 lbs.
The "balanced design" effective bearing widths are
\( b_{1.min} = {37900 \times 0.50 \over 2 \times 28600} = 0.331 ~\text{in} \) (from Equation(918a))
\( 2b_{2.min} = {37900 \times 0.75 \over 44000} = 0.646 ~\text{in} \) (from Equation(918b))
Therefore, the same value of P_{ub.P.max} would be obtained if the thickness of each female lug was reduced to 0.331 inches and the thickness of the male lug reduced to 0.646 inches.
d) Joint Strength (Equation (919b))
The final allowable load for the joint, exclusive of the lug tangs, is
P_{all} = P_{ub.P.max} = 37900 lbs.
(4) Lug Tang Analysis
$$ P_T = { 2 \times 64000 \times 2.50 \times 0.50 \over 1 + {3 \over 1.4} \times \left(0.1  {0.331 \over 0.500}\right) } = 92700 ~\text{lbs} $$

(from Equation (922a)) 
or
P_{T} = 77000 × 3.00 × 0.75 = 173300 lbs (from Equation (922b))
Therefore, the lug tangs are not critical and the allowable joint load remains at 37900 pounds.
9.7 Lug and Bushing Strength Under 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 97 illustrates the different lug dimensions that are critical in determining the lug strength.
9.7.1 Lug Strength Under Transverse Load
The lug ultimate bearing stress (F_{bru.L}) is
where K_{tru}, the transverse ultimate load coefficient, is obtained from Figure 98 as a function of the "effective" edge distance (h_{av}):
$$ h_{av} = { 6 \over 3/h_1 + 1/h_2 + 1/h_3 + 1/h_4 } $$The effective edge distance can be found by using the nomograph in Figure 99. The nomograph is used by first connecting the h_{1} and h_{2} lines at the appropriate value of h_{1} and h_{2}. The intersection with line A is noted. Next connect the h_{3} and h_{4} lines similarly, and note the B line intersection. Connecting the A and B line intersection gives the value of h_{av} to be read at the intersection with the h_{av} line. The different edge distances (h_{1}, h_{2}, h_{3}, h_{4}) indicate different critical regions in the lug, h_{1} being the most critical. The distance 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, h_{av}/D can be obtained directly from Figure 910 for any value of e/D. In concentric lugs, h_{1} = h_{4} and h_{2} = h_{3}.
The lug yield bearing stress (F_{bry.L}) is
where K_{try}, the transverse yield load coefficient, is obtained from Figure 98.
The allowable lug transverse ultimate load (P_{tru.L}) is
where F_{bru.L} and F_{bry.L} are obtained from Equations (928) and (929).
If the lug is not of constant thickness, then A_{av}/A_{br} is substituted for h_{av}/D on the horizontal scale of the graph in Figure 98, where A_{br} is the lug bearing area, and
$$ A_{av} = { 6 \over 3/A_1 + 1/A_2 + 1/A_3 + 1/A_4 } $$A_{1}, A_{2}, A_{3}, and A_{4} are the areas of the sections defined by h_{1}, h_{2}, h_{3}, and h_{4}, respectively.
The values of K_{tru} and K_{try} corresponding to A_{av}/A_{br} are then obtained from the graph in Figure 98 and the allowable bearing stresses are obtained as before from Equations (928) and (929).
9.7.2 Bushing Strength Under Transverse Load
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 (98). The allowable bushing ultimate load (P_{tru.B}) is equal to P_{u.B} (Equation (99)).
9.8 Double Shear Joints Under Transverse Load
The strength calculations needed for double shear joint strength analysis are basically the same as those needed for axially loaded. Equations (911) through (919) can be used; however, the maximum lug bearing stresses at ultimate and yield loads must not exceed those given by Equations (928) and (929).
9.9 Single Shear Joints Under Transverse Load
The previous discussion on double shear joint applies to single shear joint strength analysis except the equations to be used are now Equations (923) through (927).
9.10 Lug and Bushing Strength Under 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 shown in Figure 911. These procedures apply only if α does not exceed 90°.
9.10.1 Lug Strength Under Oblique Load
The obliquely applied load (P_{α}) is resolved into an axial component (P = P_{α} cosα) and a transverse component (P_{tr} = P_{α} sinα). The allowable ultimate value of P_{α} is P_{α.L} and its axial and transverse components satisfy the following equation:
where P_{u.L} is the strength of an axially loaded lug (Equation (97)) and P_{tru.L} is the strength of a transversely loaded lug (Equations (930a), (930b)). The allowable load curve defined by Equation (931) is plotted on the graph in Figure 912.
For any given value of α the allowable load (P_{α.L}) for a lug can be determined from the graph shown in Figure 912 by drawing a line from the origin with a slope equal to (P_{u.L}/P_{tru.L}). The intersection of this line with the allowable load curve (point 1 on the graph) indicates the allowable values of P/P_{L} and P_{tr}/P_{tru.L}, from which the axial and transverse components, P and P_{tr} of the allowable load can be readily obtained.
9.10.2 Bushing Strength Under Oblique Load
The bushing strength calculations are identical to those for axial loading (Equations (98) and (99)).
9.11 Double Shear Joints Under Oblique Load
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 P_{α.L}/Dt, where P_{α.L} is defined by Equation (931). Use Equations (911) through (919)).
9.12 Single Shear Joints Under Oblique Load
The previous discussion on double shear joints applies to single shear joint strength analysis except the equations to be used are now Equations (923) through (927).
9.13 Multiple Shear and Single Shear Connections
Lugpin combinations having the geometry indicated in Figure 913 should be analyzed according to the following criteria:
 The load carried by each lug should be determined by distributing the total applied load P among the lugs as indicated in Figure 913, b being obtained in Table 92. 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.
 The maximum shear load on the pin is given in Table 92.
 The maximum bending moment in the pin is given by the formulae
$$ M = {P_1 b \over 2} $$  where b is given in Table 92. 
Total number of lugs including both sides 
C  Pin Shear  b 

5  0.35  0.50P_{1}  $$ 0.28 ~{ t' + t'' \over 2 } $$ 
7  0.40  0.53P_{1}  $$ 0.33 ~{ t' + t'' \over 2 } $$ 
9  0.43  0.54P_{1}  $$ 0.37 ~{ t' + t'' \over 2 } $$ 
11  0.44  0.54P_{1}  $$ 0.39 ~{ t' + t'' \over 2 } $$ 
∞  0.50  0.50P_{1}  $$ 0.50 ~{ t' + t'' \over 2 } $$ 
9.14 Axially Loaded Lug Design
This section presents procedures for the optimized design of lugs, bushings and pin in a symmetrical, doubleshear joint, such as shown in Figure 95, 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 designs for joints other than symmetrical, doubleshear joints.
9.14.1 Axial Lug Design for 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 (932)). 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.
where F_{su.P} and F_{tu.P} are the ultimate shear and ultimate tension stresses for the pin material, k_{b.P} is the plastic bending coefficient for the pin, and F_{br.all.1} and F_{br.all.2} are allowable bearing stresses in the female and male lugs. The value of F_{br.all.1} can be approximated by the lowest of the following three values:
K F_{tux.1} D/D_{P}; 1.304K F_{tyx.1} D/D_{P}; 1.304F_{cy.B.1}
where F_{tux.1} and F_{tyx.1} are the crossgrain tensile ultimate and tensile yield stress for female lugs, F_{cy.B.1} is the compressive yield stress of the bushings in the female lugs, and K is obtained from Figure 914. Assume D = D_{P} if a better estimate cannot be made. F_{br.all.2} is approximated in a similar manner.
9.14.1.1 Axial Lug Design for Pin Failure in the Shearing Mode
Pin and Bushing Diameter
The minimum allowable diameter for a pin in double shear is
The outside diameter of the bushing is D = D_{P} + 2t_{B} where t_{B} is the bushing wall thickness.
Edge Distance Ratio (e/D)
The value of e/D that will minimize the combined lug and pin weight is obtained from Figure (915)(a) 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.
The curves in Figure 915 apply specifically to concentric lugs (a = e  D/2, and w = 2e), but they can be used for reasonably similar lugs.
Allowable Loads
The allowable loads for the different failure modes (lug bearing failure, lug nettension failure, and bushing failure) are determined from Equations (93), (96), and (99) in terms of the (unknown) lug thickness. The lowest of these loads is critical.
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.
Pin Bending
To prevent bending failure of the pin before lug or bushing failure occurs in a uniformly loaded symmetrical doubleshear joint, the required pin diameter is
where k_{b.P} is the plastic bending coefficient for the pin. If the value of D_{P} from Equation (934) is greater than that from Equation (933), the joint must be redesigned because the pin is critical in bending.
Reduced Edge Distance
If the allowable bushing load (Equation (99)) is less than the allowable lug load (Equation (93)), a reduced value of e, obtained by using the curve shown in Figure 916 for optimum e/D, will give a lighter joint in which lug bearing failure and bushing bearing failure will occur simultaneously. The previously calculated pin diameter and lug thicknesses are unchanged.
Reduced Lug Width
If the lug nettension strength (Equation (96)) exceeds the bearing strength (Equation (93)), the netsection width can be reduced by the ratio of the bearing strength to the nettension strength.
9.14.1.2 Axial Lug Design for Pin Failure in the Bending Mode
Pin and Bushing Diameters (First Approximation)
A first approximation to the optimum pin diameter is shown in Equation (935).
where F_{t.all.1} is either F_{tux.1} or 1.304F_{tyx.1} whichever is smaller; and F_{t.all.2} is either F_{tux.2} or 1.304F_{tyx.2} whichever is smaller. This approximation becomes more accurate when there are no bushings and when there is no gap between lugs.
The first approximation to the outside diameter of the bushing is D = D_{P} + 2t_{B}.
Edge Distance Ratio (e/D)
The value of e/D that will minimize the combined lug and pin weight is obtained from Figure (915)(b) for the case of symmetrical doubleshear 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 (a = e  D/2, and w = 2e), but can be used for reasonably similar lugs.
Allowable Loads (First Approximation)
The allowable loads for the different failure modes (lug bearing failure, lug nettension failure, and bushing failure) are determined from Equations (93), (96), and (99), in terms of the (unknown) lug thickness. The lowest of these loads is critical.
Lug Thicknesses (First Approximation)
The first 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.
Pin Diameter (Second Approximation)
The second approximation to the pin diameter is obtained by substituting the first approximation lug thicknesses into Equation (934).
Final Pin and Bushing Diameters and Lug Thicknesses
The final optimum pin diameter is very closely approximated by
D_{P.opt} = 1/3 D_{P} (Equation (935)) + 2/3 D_{P} (Equation (934))
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.
Pin Shear
The pin is checked for shear strength (Equation (933)).
Reduced Edge Distance
If the bushing bearing strength (Equation (99)) is less than the lug bearing strength (Equation (93)), a reduced value of e/D, obtained from the curve in Figure 916, will give a lighter joint. The pin diameter and lug thicknesses are unchanged.
Reduced Lug Width
If the lug nettension strength (Equation (96)) exceeds the lug bearing strength (Equation (93)), the netsection width can be reduced by the ratio of the bearing strength to the nettension strength.
9.14.1.3 Example of Axially Loaded Lug Design
Using the same materials for the lug, bushing and pin as mentioned in Section 9.6, and assuming the same allowable static load of 37900 pounds, a symmetrical doubleshear joint will be designed to carry this load. A 0.10inch gap is again assumed between the lugs. The bushing wall thickness is assumed to be 1/8 inch.
The lug will first be assumed to be concentric (a = e  D/2, and w = 2e) but the final minimum weight design will not necessarily be concentric.
Pin Failure Mode (Equation (932))
The pin is first checked to determine whether it will be critical in shear or bending, using Equation (932). Assuming D = D_{P} as a first approximation, determine F_{br.all.1} and F_{br.all.2}, using the graph in Figure 914 to determine K.
KF_{tux.1} = 1.02 × 64000 = 65300 psi; 1.304KF_{tyx.1} = 1.304 × 1.02 × 40000 = 53100 psi
1.304F_{cy.B.1} = 1.304 × 60000 = 78200 psi; therefore, F_{br.all.1} = 53100 psi
KF_{tux.2} = 1.02 × 77000 = 78500 psi; 1.304KF_{tyx.2} = 1.304 × 1.02 × 66000 = 53100 psi
1.304F_{cy.B.2} = 1.304 × 60000 = 78200 psi; therefore, F_{br.all.2} = 78200 psi
Therefore,
$$ R = { \pi \times 82000 \over 1.56 \times 125000 } \times \left( {82000 \over 53100} + {82000 \over 78200} \right) = 3.4 $$

(Equation (932)) 
Therefore, the design procedure for pins critical in bending applies.
Pin and Bushing Diameters  First Approximation (Equation (935))
D = 0.741 + 2 × 0.125 = 0.991 in
Edge Distance Ration (e/D)
The optimum value of e/D for both male and female lugs is 1.24 (Figure 915 (b)). Therefore a/D is 0.74 and w/D is 2.48 for a concentric lug (therefore, w = 2.46 in).
Allowable Loads  Female Lugs and Bushings (First Approximation)
(a) Lug Bearing Strength (Equations (92a) and (936))
P_{bru.L.1} = 1.304 × 1.46 × 0.74 × 40000 × 0.991t_{1} = 55900t_{1} lbs
where K = 1.46 is obtained from Figure 92 for e/D = 1.24
(b) Lug NetSection Tension Strength (Equations (95) and (96b))
K_{n.1} = 0.74 (obtained by interpolation from the graphs shown in Figure 94) for D/w = 0.403; F_{ty}/F_{tu} = 0.625; F_{tu}/Eϵ_{u} = 0.051
P_{nu.L.1} = 1.304 × 40000 × (2.46 − 0.991) t_{1} = 56600 t_{1} lbs
(c) Bushing Bearing Strength (Equation (99))
P_{u.B.1} = 1.304 × 60000 × 0.741 t_{1} = 58000 t_{1} lbs
Allowable Loads  Male Lug and Bushing (First Approximation)
(a) Lug Bearing Strength (Equations (9la) and (93a))
P_{bru.L.2} = 1.46 × 0.74 × 77000 × 0.991 t_{2} = 82500 t_{2} lbs
(b) Lug NetSection Tension Strength (Equations (94) and (96a))
K_{n.2} = 0.88 (obtained by interpolation from the graphs shown in Figure 94) for D/w_{2} = 0.403; F_{ty}/F_{tu} = 0.857; F_{tu}/Eϵ_{u} = 0.125
P_{nu.L.2} = 0.88 × 77000 × (2.46 − 0.991) t_{2} = 99500 t_{2} lbs
(c) Bushing Bearing Strength (Equation (99))
P_{u.B.2} = 1.304 × 60000 × 0.741 t_{2} = 58000 t_{2} lbs
Lug Thicknesses (First Approximation)
\( t_1 = {37900 \over 2 \times 55900} \) = 0.339 in; \( t_2 = {37900 \over 58000} \) = 0.654 in
Pin Diameter  Second Approximation (Equation (934))
D = 0.7555 + 2 × 0.125 = 1.0005 in
Final Pin and Bushing Diameter (Equation (936))
D_{P.opt} = 0.741/2 + 0.755/2 = 0.748 in (Use 0.750 inch pin)
D = 0.750 + 2 × 0.125 = 1.000 in
Pin Shear (Equation (933))
$$ D_P = 0.798 \sqrt{ 37900 \over 82000 } = 0.541 ~\text{in} $$Therefore, the pin is not critical in shear.
Final Lug Thicknesses
t_{1} = 0.339 × 0.991/1.000 = 0.336 in
t_{2} = 0.654 × 0.741/0.750 = 0.646 in
Reduced Edge Distance
The lug tension strength (Equation (93)) exceeds the bushing strength (Equation (99)) for the male lug. Therefore, a reduced e/D can be obtained for the male lug shown in Figure 916.
Therefore, e/D = 0.97 (male lug)
Reduced Lug Width
The lug netsection tension strength (Equation (96)) exceeds the bearing strength (Equation (93)) for both the male and female lugs. Therefore, the widths can be reduced as follows:
Final Dimensions
D_{P} = 0.750 in; D = 1.000 in;
t_{1} = 0.336 in; e_{1} = 1.24 in; w_{1} = 2.46 in;
t_{2} = 0.646 in; e_{2} = 0.97 in; w_{2} = 2.23 in;
Since w_{2} is larger than 2e_{2}, the final male lug is not concentric.
9.15 Analysis of Lugs with Less Than 5 PCT Elongation
The procedures given through Section 914 for determining the static strength of lugs apply to lugs made from materials which have ultimate elongations, ϵ_{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 addition to using these procedures, special consideration must be given to possible further loss in strength resulting from material defects when 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 less than 5% elongation are as follows:
9.15.1 Bearing Strength of Axially Loaded Lugs with Less Than 5 PCT Elongation
 Determine F_{ty}/F_{tu} and ϵ_{y}/ϵ_{u}, using values of F_{ty}, F_{tu}, ϵ_{y} and ϵ_{u} that correspond to the minimum value of ϵ_{u} in the plane of the lug.
 Determine the value of B, the ductility factor, from the graph shown in Figure 917.
 Determine a second value of B (denoted by B_{0.05}) for the same values of F_{ty}, F_{tu}, and ϵ_{y} as before, but with ϵ_{u} = 0.05.
 Multiply the bearing stress and bearing load allowables given by Equations (91a) through (93b) by B/B_{0.05} to obtain the corrected allowables.
9.15.2 NetSection Strength of Axially Loaded Lugs with Less Than 5 PCT Elongation
The procedure for determining netsection allowables is the same for all values of ϵ_{u}. The graphs in Figure 94 are used to obtain a value of K_{n} which is susbstituted in Equations (94) and (95). If the grain direction of the material is known, the values of F_{ty}, F_{tu}, and ϵ_{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 ϵ_{u} in the plane of the lug.
9.15.3 Strength of Lug Tangs in Axially Loaded Lugs with Less Than 5 PCT Elongation
The plastic bending coefficient for a rectangular cross section can be approximated by k_{b.L} = 1.5B, where B is obtained from Figure 917, in which y and u are the yield and ultimate strains of the lug tang material in the direction of loading. The maximum allowable value of k_{b.L} for a rectangle is 1.4.
9.15.4 LugBushing Strength in AxiallyLoaded SingleShear Joint with Less Than 5 PCT Elongation
The values of k_{br.L} and k_{b.L} for rectangular cross sections are approximated by 1.5B, where B is determined from the graph as described in Figure 917. The maximum allowable values of k_{br.L} and k_{b.L} are 1.4.
9.15.5 Bearing Strength of Transversely Loaded Lugs with Less Than 5% Elongation
(Equations (928) through (930b) in Section 9.7.1)
The same procedure as that for the bearing strength of axially loaded lugs is used.
 Determine B and B_{0.05} as described for axially loaded lugs, where B corresponds to the minimum value of ϵ_{u} in the plane of the lug.
 Multiply the bearing stress and bearing load allowables given by Equations (928) through (930b) by B/B_{0.05} to obtain the corrected allowables.
9.16 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 increase in inner radius of the ring (lug) plus the decrease in outer radius of the bushing equals the difference between the radii of the bushing and ring before assembly:
where
δ = Difference between outer radius of bushing and inner radius of the ring.
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 p is
Radial displacement at the outer surface of a bushing subjected to external pressure p is
where
A = Inner radius of bushing  D = Inner radius of ring (lug) 
B = Outer radius of bushing  E = Modulus of elasticity 
C = Outer radius of ring (lug)  μ = Poisson's ratio 
Substitute Equations (937) and (938) into Equation (936) and solve for p:
Maximum radial and tangential stresses for a ring subjected to internal pressure occur at the inner surface of the ring (lug).
$$ F_r = p $$  $$ F_t = p \left[{ C^2 + D^2 \over C^2  D^2 }\right] $$ 
Positive sign indicates tension. The maximum shear stress at this point is
$$ F_s = {F_t  F_r \over 2} $$The maximum radial stress for a bushing subjected to external pressure occurs at the outer surface of the bushing is
$$ F_r = p $$The maximum tangential stress for a bushing subjected to external pressure occurs at the inner surface of the bushing is
$$ F_t = { 2_{PB}^2 \over B^2 = A^2 } $$The allowable press fit stress may be based on:
 Stress Corrosion. The maximum allowable press fit stress in magnesium alloys should not exceed 8000 psi. For all alumimun alloys the maximum press fit stress should not exceed 0.50F_{ty}.
 Static Fatigue. Static fatigue 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 25% F_{tu}.
 Ultimate Strength. Ultimate strength cannot be exceeded, but is not usually critical in a press fit application.
 Fatigue Life. The hoop tension stresses resulting from the press fit 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 buildup of coating. Therefore, Hardcoat or HAE coatings should not be used in holes that will subsequently contain a press fit bushing or bearing.
Figures 918 and 919 permit determining the tangential stress, F_{T}, for bushings pressed into aluminum rings. Figure 918 presents data for general steel bushings, and Figure 919 presents data for the NAS 75 class bushings. Figure 920 gives limits for maximum interference fits for steel bushings in magnesium alloy rings.