Tools › Structures › Joints › Lug Strength › Example
AFFDL-TR-69-42 Section 9.6: double-shear lug joint analysed lug by lug (Table 9-1)
| Tool | Lug Strength |
| Source | Maddux, G. E., Vorst, L. A., Giessler, F. J., and Moritz, T. (1969). Stress Analysis Manual. AFFDL-TR-69-42, Air Force Flight Dynamics Laboratory, Wright-Patterson Air Force Base, OH. Section 9.6, Table 9-1. |
| Level | Professional |
Problem
A double-shear joint of Figure 9-5 carries an axial limit load of 20,000 lb: two female aluminum lugs (D = 1.0 in, w = 2.5 in, e = 1.25 in, t = 0.5 in, F_tu = 64 ksi, F_ty = 40 ksi) and a male lug (D = 1.0 in, w = 3.0 in, e = 1.5 in, t = 0.75 in, F_tu = 77 ksi, F_ty = 66 ksi), each with a steel bushing (F_cy = 60 ksi), on a 0.75 in steel pin (F_su = 82 ksi, F_tu = 125 ksi) with a 0.10 in gap. Find the bearing, net-section and bushing strengths of each lug, the pin shear and bending strengths with the balanced-design reduction, the tang strengths and the allowable joint load.
Inputs to Enter
- Limit load P = 20,000 lb axial (α = 0), fitting factor FF 1.15, FS 1.5 (P_U = 34,500 lb), FS_yld 1.15
- Lug 1 (female): D 1.0, w 2.5, e 1.25, t 0.5, w_T 2.5 in; F_tu = F_tux 64 ksi, F_ty = F_tyx 40 ksi, E 10.5 MSI, ε_u 0.12; bushing F_cyb 60 ksi
- Lug 2 (male): D 1.0, w 3.0, e 1.5, t 0.75, w_T 3.0 in; F_tu = F_tux 77 ksi, F_ty = F_tyx 66 ksi, E 10.3 MSI, ε_u 0.06; bushing F_cyb 60 ksi
- Pin D_p 0.75 in, F_sup 82 ksi, F_tup 125 ksi, k_bp 1.56, gap g 0.10 in, k_bT 1.4; joint type double shear
Answers
- Lug 1: K = 1.46 (Fig. 9-2 at e/D 1.25), P_bruL = 28,600 lb (9-3b); K_n = 0.74 (Fig. 9-4), P_uB = 29,300 lb
- Lug 2: K = 1.33 (Fig. 9-2 at e/D 1.5), P_bruL = 77,000 lb (9-3a); K_n = 0.87, P_uB = P_uLB2 = 44,000 lb
- Joint P_uLB = 44,000 lb (9-11); pin shear P_usp = 72,400 lb (9-12); pin bending P_ubp = 30,100 lb (9-15)
- Balanced design: C = 29,000 lb/in, P_ubpmax = 37,900 lb (9-16), b_1min = 0.331 in (9-18a), 2 b_2min = 0.646 in (9-18b)
- Allowable joint load P_all = 37,900 lb (9-19b); tangs 92,700 lb (9-22a) and 173,300 lb (9-22b)
- Applied ultimate P_U = 1.15 × 1.5 × 20,000 = 34,500 lb
Notes
The tool's default inputs are the manual's example; the page opens in Simple mode (one lug), so switch to Advanced to see the whole joint. The manual's printed values are reproduced within the test tolerances (0.3 % on the loads, 1–3 % on the K_n read from the digitized Figure 9-4). The manual gives no margins; the tool prints M.S. = allowable / P_U − 1.