Lug Analysis Method: US Air Force AFFDL-TR-69-42 Chapter 9
Outputs
Geometry
Results
Results Summary
The tables below summarize the results, listing the lug and pin safety factors wherever they apply. A factor of safety of 1 or more (a margin of safety of 0 or more) is highlighted green, and anything lower is highlighted red. Choosing a suitable design factor of safety remains the responsibility of the engineer.
Male Lug
| Total Applied Force | Papp | |
| Lug Strength for Oblique Loading | Pult | |
| Factor of Safety | FS | |
| Margin of Safety | MS |
Female Lug
Strength results for a single female lug, considered on its own apart from the joint:
| Total Applied Force | Papp | |
| Lug Strength for Oblique Loading | Pult | |
| Factor of Safety | FS | |
| Margin of Safety | MS |
Double Shear Joint Strength
This is the strength of the complete joint, combining the male lug, both female lugs, and the pin. The figure at right marks where the pin presses on each lug.
| Total Applied Force | Papp | |
| Overall Ultimate Load (Joint Strength) | Pult | |
| Factor of Safety | FS | |
| Margin of Safety | MS |
Female Lug
These results require female lugs, which are not part of this joint.
Double Shear Joint Strength
These results require female lugs, which are not part of this joint.
The remaining tabs contain the complete calculations.
Component Properties
This section lists the properties of each joint component.
Materials List
The materials used in this joint are listed below:
| Elastic Modulus, E |
Percent Elongation, eL |
Ultimate Strain, εu |
Tensile Yield Strength, Sty |
Tensile Ultimate Strength, Stu |
Shear Yield Strength, Ssy |
Shear Ultimate Strength, Ssu |
Compressive Yield Strength, Scy |
Compressive Ultimate Strength, Scu |
Al Grain Cutoff (Fig. 9-3) |
|---|
Note: Ductile materials have compressive strengths close to their tensile strengths, while brittle materials are stronger in compression than in tension. As a conservative simplification, this analysis treats compressive strength as equal to tensile strength.
Note: Each material's ultimate strain is found from its percent elongation, eL, its elastic modulus, E, and its tensile ultimate strength, Stu:
Applied Forces
| Pax | = | axial force | |
| Ptr | = | transverse force | |
| Papp | = | total applied force | |
| α | = | angle |
Note: The applied load is split between the two female lugs, so each female lug carries half of the load on the male lug.
Joint Properties
The joint is made up of one male lug, a pair of female lugs, and a pin.
| g | = | clearance between lugs (adds to the bending arm) |
Pin Properties
| Material: |
| DP | = | diameter | |
| A | = | area | |
| S | = | section modulus | |
| kb.P | = | plastic bending coefficient |
Male Lug Properties
| Material: | |
| Bushing Material: |
| w | = | width | |
| t2 | = | thickness | |
| D | = | hole diameter | |
| e | = | edge distance (from hole center to lug end) | |
| a | = | distance (from hole edge to lug end) | |
| α | = | upper side angle | |
| β | = | lower side angle (toward transverse load) | |
| et | = | hole offset from centerline (+ toward β side) | |
| W | = | width across the hole centerline | |
| h0, h2 | = | sections above / below the hole | |
| h4, h1 | = | sections at the 45° points, upper / lower | |
| h3 | = | minimum section toward the lug end | |
| heff | = | effective edge distance |
Female Lug Properties
| Material: | |
| Bushing Material: |
| w | = | width | |
| t1 | = | thickness | |
| D | = | hole diameter | |
| e | = | edge distance (from hole center to lug end) | |
| a | = | distance (from hole edge to lug end) | |
| α | = | upper side angle | |
| β | = | lower side angle (toward transverse load) | |
| et | = | hole offset from centerline (+ toward β side) | |
| W | = | width across the hole centerline | |
| h0, h2 | = | sections above / below the hole | |
| h4, h1 | = | sections at the 45° points, upper / lower | |
| h3 | = | minimum section toward the lug end | |
| heff | = | effective edge distance |
Male Lug Results
Female Lug Results
These results require female lugs, which are not part of this joint.
Double Shear Joint Strength
This section finds the double shear strength of the joint, which is the overall joint capacity once the interaction between the lugs and the pin is included. The table below summarizes the outcome, and the sections that follow show every step of the calculation.
| Overall Ultimate Load (Joint Strength) | Pult | |
| Total Applied Force | Papp | |
| Factor of Safety | FS | |
| Margin of Safety | MS |
Nominal Joint Strength
From the earlier lug calculations, the male and female lug strengths are:
| Pult.M = | male lug strength | |
| Pult.F = | female lug strength |
Setting aside any pin bending effects, the nominal joint strength is:
| Pu.J.nom = min( 2·Pult.F , Pult.M ) = | nominal joint strength |
Nominal Pin Strength
The pin's nominal shear and bending strengths are:
| Pus.P = 2(π4Dp2)Ssu.P = | pin ultimate shear load | |
| Pub.P = πDP3 · kb.P · Stu.P16(t12 + t24 + g) = | pin ultimate bending load |
Calculated values
| Stu.P | = | |
| Ssu.P | = | |
| DP | = | |
| kb.P | = | |
| t1 | = | |
| t2 | = | |
| g | = |
Pin Strong or Weak in Bending?
A pin counts as strong in bending when its ultimate bending load Pub.P is at least as large as either its shear capacity Pus.P or the nominal joint capacity Pu.J.nom. The pin is strong in bending if either of these conditions is met:
| Pub.P ≥ Pu.J.nom | or | Pub.P ≥ Pus.P |
One or more of the conditions above is met, meaning the pin is strong in bending.
Neither condition above is met, meaning the pin is weak in bending.
Calculated values
| Pub.P | = | |
| Pus.P | = | |
| Pu.J.nom | = |
Strong Pin
With a strong pin, the ultimate bending load assumes the load spreads uniformly across the full lug thicknesses (the same as the nominal value):
| Pub.P = πDP3 · kb.P · Stu.P16(t12 + t24 + g) = | pin ultimate bending load |
Because the pin is strong, bending does not reduce the joint strength, so the ultimate joint load equals the nominal ultimate joint load:
| Pu.J = Pu.J.nom = | ultimate joint load |
Calculated values
| Stu.P | = | |
| DP | = | |
| kb.P | = | |
| t1 | = | |
| t2 | = | |
| g | = |
Weak Pin
When the pin is weak, it bends enough that the load no longer spreads evenly across the lugs and instead gathers near the shear planes at the lug faces. A "balanced design" is then found using the reduced contact widths between the pin and the lugs. The male and female bearing widths are chosen so the male lug, the female lugs, and the pin in bending all reach the same strength.
Following AFFDL-TR-69-42 §9.4.3, the lug strengths are assumed proportional to the effective bearing widths. The "balanced design" pin ultimate bending load is (Eq. 9-16):
| C = Pult.F · Pult.MPult.F t2 + Pult.M t1 = | |
| Pub.P.max = 2C√Pub.PC(t12 + t24 + g) + g2 − 2Cg = |
The balanced value exceeds the nominal lug strength, which would need bearing widths larger than the lugs. The joint is therefore limited to the nominal joint strength, Pu.J.nom.
The bearing widths are the portions of the male and female lugs that actually carry the pin load (Eq. 9-18a, 9-18b):
| b1 = Pub.P.max t12 Pult.F = | bearing width over female lug | |
| 2b2 = Pub.P.max t2Pult.M = | bearing width over male lug (b2 = per side) |
Calculated values
| Pult.F | = | |
| Pult.M | = | |
| Pub.P | = | |
| t1 | = | |
| t2 | = | |
| g | = |
Check: narrower bearing widths shorten the moment arm, and recomputing the bending capacity with Eq. 9-15a returns the balanced value:
| Pub.P = πDP3 · kb.P · Stu.P16(b12 + b22 + g) = | pin ultimate bending load |
Calculated values
| Stu.P | = | |
| DP | = | |
| kb.P | = | |
| g | = |
For the balanced design, the ultimate joint load is the pin ultimate bending load:
| Pu.J = Pub.P = | ultimate joint load |
The joint would reach this same strength if each female lug were thinned to b1 = and the male lug were thinned to 2b2 = .
Overall Ultimate Load
The overall ultimate load is governed by the lower of the ultimate joint load and the pin ultimate shear load:
| Pult = min( Pu.J , Pus.P ) = | overall ultimate load |
Calculated values
| Pu.J | = | |
| Pus.P | = |
Comparing the overall ultimate load with the applied force gives the factor of safety:
| Papp = | total applied force |
The resulting Margin of Safety (MS) and Factor of Safety (FS) are:
| MS = FS − 1 = | FS = PultFF · LF · Papp = |
Calculated values
| Pult | = | |
| Papp | = |
These results require female lugs, which are not part of this joint.
Lug Optimizer
The Lug Optimizer maps the margin of safety over a range of two lug dimensions, shows what governs, and finds the lightest design that still meets the target margin. Designs whose ratios fall outside the design charts (Figures 9-2, 9-3, 9-4 and 9-8) are not calculated. It starts from the current inputs; every other value is held as entered.
NOTE: Lug section geometry and design-chart coefficients (derived from Air Force Manual figures) adhere to the LugCalc reference implementation. This software provides a theoretical analysis; consequently, all designs require empirical testing to ensure structural validation.
The lug strength is found first for purely axial and purely transverse loads. Those results are then combined into the strength along the applied load direction (oblique loading) and a Factor of Safety.
The results are summarized in the table below, and the sections that follow show each calculation in full.
| Lug Strength for Pure Axial Loading | Pu.L.B | |
| Lug Strength for Pure Transverse Loading | Ptru.L.B | |
| Lug Strength for Oblique Loading | Pult | |
| Total Applied Force | Papp | |
| Factor of Safety | FS | |
| Margin of Safety | MS | |
Note: The applied load is split between the two female lugs, so each female lug carries half of the load on the male lug.
Axial loading
Here the lug strength is found for a load acting purely along the lug axis.
Bearing (shear-out, hoop tension)GovernsFig. 9-2 / 9-3
The ultimate bearing load covers bearing, shear-out, and hoop tension failures. It depends on the axial load coefficient and the bearing efficiency factor, which are found first.
The figure below gives the axial load coefficient. It is valid only for D/t ≤ 5, which here. Reading from the plot, the coefficient is:
| Kaxial = | () |
Calculated values
| D/t | = | |
| e/D | = | |
| D | = | |
| e | = | |
| t | = | |
Note: Plot built from the Air Force Manual (AFFDL-TR-69-42), Figure 9-2.
The figure below gives the bearing efficiency factor. It is valid only for D/t > 5, which here. Reading from the plot, the coefficient is:
| Kbear = | () |
Calculated values
| D/t | = | |
| e/D | = | |
| D | = | |
| e | = | |
| t | = | |
Note: Plot built from the Air Force Manual (AFFDL-TR-69-42), Figure 9-3, using LugCalc's polynomial fit of each curve, held constant past each fit's valid range. Dashed lines show the neighboring D/t curves; the solid blue line interpolates between them. When the lug material has an aluminum grain cutoff, curve (A) or (B) from the figure note is drawn in red and limits the factor.
Warning: D/t = exceeds 5. The LugCalc reference treats lug solutions with D/t > 5 as invalid. The Air Force Manual approximates the strength reduction for thin lugs with Figure 9-3, which is used below. Consider adjusting the bore diameter or thickness to achieve D/t ≤ 5.
Since D/t ≤ 5, the calculation uses the axial load coefficient:
| K = Kaxial = |
Since D/t > 5, the calculation uses the bearing efficiency factor:
| K = Kbear = |
The ultimate bearing load is then:
| Pbru.L = K · min(Stu, 1.304 Sty) · D t · aDif e/D < 1.51otherwise = Pbru.L = K · min(Stu, 1.304 Sty) · D t = |
Calculated values
| Sty | = | |
| Stu | = | |
| D | = | |
| a | = | |
| e | = | |
| t | = |
Bushing bearingGovernsEq. 9-9
Bushing Bearing Strength Under Axial Load
The ultimate bushing load is then:
| Pu.B = 1.304 Scy.B Dp t = |
Note: No pin is included in this analysis, so the pin is assumed to fill the hole: Dp = D.
Note: This lug has no bushing selected; its bushing bearing strength therefore uses the compressive yield strength of the lug's own material, .
Calculated values
Bushing:
| Scy.B | = | |
| DP | = | |
| t | = |
Net-section tensionGovernsFig. 9-4
The net-section ultimate load covers tensile failure through the section beside the hole, across the width W at the hole centerline (equal to w for straight sides). It uses the net tension stress coefficient read from the plot below:
| Kn = |
Calculated values
| D/W | = | |
| Sty / Stu | = | |
| Stu / (E⋅εu) | = |
Note: Plot built from the Air Force Manual (AFFDL-TR-69-42), Figure 9-4 (b), (c), and (d), drawn for Sty/Stu = 1.0, 0.8 and 0.6. Dashed lines interpolate within each figure using the ratio Stu / (E⋅εu). The solid line interpolates between the figures using Sty/Stu, which is limited to the range 0.6 to 1.0.
The net-section ultimate load is then:
| Pnu.L = Kn · min(Stu , 1.304 Sty) · (W − D) · t = |
Calculated values
| Sty | = | |
| Stu | = | |
| D | = | |
| W | = | |
| t | = |
Design strength, axialResult
For axial loading, the design ultimate load is the smallest of the ultimate bearing, ultimate bushing, and ultimate net-section loads:
| Pu.L.B = min( Pbru.L , Pu.B , Pnu.L ) = |
Calculated values
| Pbru.L | = | |
| Pu.B | = | |
| Pnu.L | = |
Transverse loading
Here the lug strength is found for a load acting purely across the lug axis.
Lug strength, transverseGovernsFig. 9-8
The ultimate transverse load uses the transverse ultimate and yield load coefficients, both read from the plot below:
| Ktry = | transverse yield load coefficient | |
| Ktru = | transverse ultimate load coefficient |
Calculated values
| heff / D | = | |
| heff | = | |
| D | = | |
Note: Plot built from the Air Force Manual (AFFDL-TR-69-42), Figure 9-8.
The ultimate transverse load is then:
| Ptru.L = Ktru Stu D tif Stu ≤ 1.304 Sty1.304 Ktry Sty D totherwise = |
Calculated values
| Sty | = | |
| Stu | = | |
| D | = | |
| t | = |
Bushing bearing, transverseGovernsEq. 9-9
Under transverse load, the bushing bearing strength is the same value found for axial load:
| Ptru.B = Pu.B = |
Design strength, transverseResult
For transverse loading, the design ultimate load is the lesser of the ultimate lug and ultimate bushing loads:
| Ptru.L.B = min( Ptru.L , Ptru.B ) = |
Calculated values
| Ptru.L | = | |
| Ptru.B | = |
Oblique loading
Here the lug strength is found for oblique loading, where axial and transverse loads act together. Purely axial and purely transverse loads are simply oblique cases in which one of the two load components is zero.
Applied forces and pure strengths
Applied Forces
The lug carries these applied forces:
| Pax | = | axial force | |
| Ptr | = | transverse force | |
| Papp | = | total applied force | |
| α | = | applied force angle |
Note: The applied load is split between the two female lugs, so each female lug carries half of the load on the male lug.
Lug Strength for Pure Axial and Transverse Loading
From the preceding sections, the purely axial and purely transverse lug strengths are:
| Pu.L.B | = | strength under purely axial load | |
| Ptru.L.B | = | strength under purely transverse load |
Oblique strength, margin and factor of safetyResult
The oblique lug strength comes from an interaction equation that combines the axial and transverse loads with their matching strengths. The interaction equation is:
The failure locus in the plot at right traces the interaction equation above. Both axes are load ratios, each comparing an applied load with the matching lug strength.
The current point in the plot is set by the axial and transverse load ratios:
| Rax = PaxPu.L.B = | axial load ratio | |
| Rtr = PtrPtru.L.B = | transverse load ratio |
The load line runs from the origin through the current point until it meets the failure locus. That meeting point is the critical point, and its coordinates give the load ratios that would cause the lug to fail.
| Rax.ult = Pax.ultPu.L.B = | ultimate axial load ratio | |
| Rtr.ult = Ptr.ultPtru.L.B = | ultimate transverse load ratio |
Along the applied load direction, the oblique strength components are:
| Pax.ult = (1(1Pu.L.B)1.6 + (tan(α)Ptru.L.B)1.6)0.625 = | axial strength component | |
| Ptr.ult = Pax.ult · tan(α) = | transverse strength component |
Calculated values
| Pu.L.B | = | |
| Ptru.L.B | = | |
| α | = |
The oblique lug strength along the applied load direction is therefore:
| Pult = √Pax.ult2 + Ptr.ult2 = |
Calculated values
| Pax.ult | = | |
| Ptr.ult | = |
The resulting Margin of Safety (MS) and Factor of Safety (FS) are:
| MS = FS − 1 = | FS = PultFF · LF · Papp = |
Calculated values
| Pult | = | |
| Papp | = |
Values must include units. Shear and compressive strengths are optional; if left blank, Ssy = 0.577 Sty, Ssu = 0.577 Stu, Scy = Sty, Scu = Stu.