Calculate e/d, W/d, d/t, Abr = dt, At = (W – d)/t (see Figures 3.5 and 3.6 for nomenclature).
Determine Kbr using the coordinates e/d and d/t.
The ultimate load for shear-bearing failure, P′bru, is
P′bru = Kbr Ftu Abr (3.1)
Ultimate load for tension failure:
From Figure 3.6 using W/d, obtain Kt for the lug candidate material.
The ultimate load for tension failure P′tu is
P′tu = Kt Ftu At (3.2)
where Ftu = ultimate tensile strength of lug material.
d
dh
W
W/2
Bushing e
t P
FIGURE 3.5 Detail of lug with bushing.
Load for yielding of the lug.
From Figure 3.8 with coordinates of e/d to obtain Kbry. The yield load, P′y, is
Py′ = Kbry Abr Fty (for Kbry, see Figure 3.8) (3.3) Load for yielding of the bushing in bearing (if used).
Pbry′ = 1.85 Fcy Abrb (3.4)
where Fcy = compressive yield stress of the bushing material.
For the shear pin bending stress see Section 3.7.
Curve 1.
4130, 4140, 4340, and 8630 steel.
2014-T6 and 7075-T6 plate ≤ 0.5 in (12.7 mm) (L, T).
7075-T6 bar and extrusion (L).
2014-T6 hand-forging billet ≤ 144 in2 (92903 mm2) (L).
2014-T6 and 7075-T6 die forgings (L).
W F
e
d Load
01.00 1.50 2.00 2.50 3.00
W/d
3.50 4.00 4.50 5.00
0.10 0.20 0.30 Kt
0.40 0.50 0.60 0.70 0.80 0.90 1.00
7
6
4 5
3 2
1
8
FIGURE 3.6 Effect of grain direction subject to axial load. L, T, N indicates the grain direction “F” in sketch; L = Longitudinal; T = Long transverse; N = Short transverse (normal).
Curve 2.
2014-T6 and 7075-T6 plate > 0.5 in, ≤ 1.0 in (>12.7 mm, ≤25.4 mm).
7075-T6 extrusion (T, N).
7075-T6 hand-forged billet ≤ 36 in2 (23226 mm2) (L).
2014-T6 hand-forged billet > 144 in2 (92903 mm2) (L).
2014-T6 hand-forged billet ≤ 36 in2 (23226 mm2) (T).
17-4 PH 17-7 PH-THD.
Curve 3.
2024-T6 plate (L, T).
2024-T4 and 2024-T42 extrusion (L, T, N).
Curve 4.
2024-T4 plate (L, T).
2024-T3 plate (L, T).
2014-T6 and 7075-T6 plate > 1.0 in (25.4 mm).
2024-T4 bar (L, T).
7075-T6 hand-forged billet > 36 in2 (23226 mm2) (L).
7075-T6 hand-forged billet ≤ 16 in2 (10323 mm2) (T).
Curve 5.
195T6, 220T4, and 356T6 aluminium alloy casting.
7075-T6 hand-forged billet > 16 in2 (10323 mm2) (T).
2014-T6 hand-forged billet > 36 in2 (10323 mm2) (T).
Curve 6.
Aluminium alloy plate, bar, hand forged billet and die forging (N).
Note: for die forgings, N direction exists only at the parting plane.
7075-T6 bar (T).
Curve 7.
18-8 stainless steel, annealed.
Curve 8.
18-8 stainless steel, full hard.
Note: for 1/4, 1/2, and 3/4 hard interpolate between curves 7 and 8.
In Figure 3.7, curve A is a cutoff to be used for all aluminum alloy hand-forged billets when the long-transverse grain direction has the general direction G. Curve B is to be used for all aluminum alloy plate, bar, and hand-forged billets when the short-transverse grain direction has the general direction G, and for die forgings when the lug contains the parting plane in a direction approxi-mately normal to the direction G.
In addition to the limitations provided by curves A and B, in no event shall a Kbr greater than 2.00 be used for lugs made from 12.7 mm thick or thicker aluminum alloy plate, bar, or hand-forged billets. (See Figure 3.8.)
0 0.600.801.001.201.401.601.802.002.20 e/d
2.60
2.40 2.803.003.203.403.60 4.00 30 20 2515 10 9 8
d/t
7 6 5 4 3 2
3.80 0.20
0.40 0.60 0.80 1.00 0.20 1.40 1.60 1.80 2.00 2.20 2.60 2.40 2.80 3.00
P d
G
e
A
B t
Kbr Efficiency Factor for Shear Bearing
FIGURE 3.7 Shear efficiency factor Kbr.
0 0 1.00 2.00 3.00 4.00
e/d 0.50
1.00 1.50 2.00 2.50
Kbry
FIGURE 3.8 Factors for calculating yield axial loads attributable to shear bearing of lug and pin.
3.4 ANALYSIS OF LUGS WITH TRANSVERSE LOADING: ALLOWABLE LOADS In the case where the loading being applied to the lug is at α = 90°, as in Figure 3.9, the loading is treated as transverse. In order to determine the ultimate and yield loads for the lug, the shape of the lug must be taken into account. This is accomplished by the use of a shape parameter given by
Shape parameter = AAavbr
where
Abr = the bearing area = dt
Avr = the weighted average area given by:
A
A A A A
vr=
+
+
+
3 1 6 1 1
1 2 3 4
A1, A2, A3, and A4 are areas of the lug sections indicated in Figure 3.9.
Obtain the areas A1, A2, A3, and A4 as follows: A1, A2, A3, and A4 are measured on the planes indicated in Figure 3.9(a) (perpendicular to the axial center line), except that in a necked lug (Figure 3.9(c)). A1 and A4 should be measured perpendicular to the local centerline.
A3 is the least area on any radial section around the hole. Since the choice of areas and the method of averaging has been substantiated only for the lugs of the shapes shown in Figure 3.10, thought should always be given to ensure that the areas A1, A2, A3, and A4 adequately reflect the strength of the lug.
A4
A1 A2
A3 A4
A1 A2
45° A3 45°
A4
A A2
A3
Note:
Stress concentration at neck not covered by this analysis.
(a) (b) (c)
FIGURE 3.9 Transverse areas calculations.
FIGURE 3.10 Substantiated lug shapes.
Obtain the weighted average of these areas using A
A A A A
vr=
+
+
+
3 1 6 1 1
1 2 3 4
Compute:
Abr=dt and A, Aavbr
Determine the allowable ultimate load:
P = K A F ′ P = Allowable ultima
tru tru br tux
tru tte load as determined for the transverse looad.
K = Efficiency factor for transversetru load (ultimate) (see Figure 3.11).
A = Prbr oojected bearing area.
F = Ultimate tensiltux ee strength of the lug material across the ggrain.
where:
Yield load Py′ of the lug:
Obtain Ktry from Figure 3.11.
Py′ = Ktry Abr Ftyx where:
Py′ = Allowable yield load on the lug
Ktry = Efficiency factor for transverse load (yield); see Figure 3.11 Abr = Projected bearing area
Ftyx = Tensile yield stress of lug material across the grain Determine the allowable yield-bearing load on the bushing using
Pbry = 1.85 FcyAbrb where:
Pbry = Allowable yield-bearing load on bushing Fcy = Compressive yield stress of bushing material
Abrb = The smaller of the bearing areas of bushing on pin or bushing on lug (the latter may be smaller as a result of external chamfer on the bushing)
In Figure 3.11 all curves are for Ktru except for the one noted as Ktry.
Note that the curve for 125,000 HT steel agrees closely with test data. Curves for other materials have been obtained by the best available means of correcting for material properties and may pos-sibly be very conservative in some places.
In no case should the ultimate transverse load be taken as less than that which can be carried by cantilever beam action of the portion of the lug under the load (Figure 3.12). The load that can be carried by cantilever beam action is indicated very approximately by curve A in Figure 3.7; should Ktru be below curve A, separate calculations, as a cantilever beam, are warranted.
1.7
2014-T6 and 7075-T6 plate > 25 mm (1.0 ins).
7075-T6 hand forged billet ≤ 0.0103 m2 (16 ins2).
2024-T6 plate
2024-T4 and 2024-T42 extrusion.
2014-T6 and 7075-T6 plate > 12.7 mm (0.5 ins), ≤25 mm (1.0 ins).
7075-T6 extrusion.
2014-T6 hand forged billet ≤ 0.023 m2 (36 ins2).
2014-T6 and 7075-T6 die forging.
2014-T6 and 7075-T6 plate ≤ 12.7 mm (0.5 ins) Curve 1 aluminum and steel
H.T. = 180 ksi
2024-T3-T4 plate t ≤ 0.5 in
356-T6
Curve A-approximate cantilever strength, if Ktru is below this curve,
see Fig. 7.5.10.
2024-T3 or -T4 plate > 0.5 in
2024-T4 bar
FIGURE 3.11 Efficiency factor for transverse load Ktru.
P
FIGURE 3.12 Lug as a cantilever carrying load.
Investigate pin bending as for axial load (see Section 3.6) with the following modifications:
Take (Pu′) min = Ptru′. In the equation:
r = e – d2
t use for the e – d2
term the edge distance at a = 90°