Question

A support fitting is machined from 7075- T651 aluminum alloy plate, and is attached to a beam using 6 steel bolts as shown. The external loads are applied through a two-lug clevis configuration, and it is assumed that that each lug supports 60% of the applied external load. A fitting factor of 1.15 is to be applied to the lugs, and the bolts have a maximum allowable shear load of 1900 lbf. All dimensions are in inches. (a) Find the resultant force on each bolt in the connection. (b) Determine the design factor of safety for the lugs as well as the most highly loaded bolt in the connection. (c) Determine the design factor of safety for bearing at the most highly loaded bolt hole in the fitting.

2. A support fitting is machined from 7075- T651 aluminum alloy plate, and is attached to a beam using 6 steel bolts as shown. The external 07R loads are applied through a two-lug clevis configuration, and it is assumed that that each lug supports 60% of the applied external load. A fitting factor of 1.15 is to be applied to the lugs, and the bolts have a maximum allowable shear load of 1900 lbf. All dimensions are in inches 1.17 R-1150 Ibf 10 723 DIA .10 R. 2750 TbE 3.08 61 Find the resultant force on each bolt in the connection Determine the design factor of safety for the lugs as well as the most highly loaded bolt in the connection (a) .61 (b) 1.50 1875 DIA . 10 3.00 10 (c) Determine the design factor of safety for bearing at the most highly loaded bolt hole in the fitting

Answer 1

Find the reaction force on each bolt in x - direction. 2750 Px = P2x = P3x = P x = Psx = P = P = = = 458.333 lbf 6 Find the reaction force on each bolt in y- direction. 1150 Py = P2, = P3, = P4, = Ps, = P, = P == = 191.666 lbf Find the resultant of above forces on each bolt. Piz = P,2 = Pse = P4z = Ps2 = Poz = P = VP?? +P, P. = V458.333 +191.666² = 496.794 lbf Find the eccentric load on each bolt. Pieli = Pza12 = Ps3 = Pieta = Psels = Pelo = R & +R, kry? + kr + kr; + kr + kr + kr = 2750x3.08 +1150x1.17 2750 x 3.08 +1150x1.17 4x1.6192 + 2 x 0.612 Apply eccentric loading of bolt. P, cr P. = kr 2750 x 3.08 +1150x1.171619 Pie = kr = 4x1.6192 + 2 x 0.61 Pe = 1415.22 lbf = Pse = Pue = Poe k= 2750 x 3.08 +1150x1.17 70 61 P2 = kr AX 1.6192 +2 x 0.61% Pne = 533.221 lbf = Pse (a) Find the resultant force on bolt 1. As P, and Pe are perpendicular to each other. Therefore, Pri = VP? + P. ? = _496.7942 +1415.222 Px = 1496.794 +1415.222 Pri = 1499.88 lbf Find the resultant force on bolt 2.

As Pr2 = Px + Pne = 458.333 +533.221 = 991.551 lbf Pvz = P, = 191.666 lbf Paz = P2? + PZ = 1991.551 +191.6662 PR2 = 1009.905 lbf Find the resultant force on bolt 3. Angle between P, and Pze is 45º. Therefore, Pa3 = VP,? + Pg2 +2(2)(Ps.)cos 45° Pg = V496.794? +1415.22? + 2 x 496.794 x 1415.22 x cos 45 PR3 = 1801.049 lbf Find the resultant force on bolt 4. Like the case of bolt 1. Therefore, Pax = VP? + P2 = 1496.7942 +1415.222 Pre = V496.7942 +1415.222 P24 = 1499.88 lbf Find the resultant force on bolt 5. As Prs = Pzx + Pse = 533.221-458.333 = 74.888 lbf Pys = Ps, = 191.666 lbf ? +P. ? = V74.888? +191.6662 Pas = 205.788 lbf Find the resultant force on bolt 6. Like the case of bolt 3. Pro = VP? + Pre +2(P)(P..)cos 45º PRO = 1496.794 +1415.222 + 2 x 496.794x1415.22 x cos 45 Peo = 1801.049 lbf (b) Bolt 3 and bolt 6 are critical as they are most highly loaded bolt in connection. Therefore,

Failure load Factor of safety Applied load Factor of safety = 1.0549 1900 1801.049 (c) The lug will be experiencing tensile load. Maximum allowable tensile load = 1.15 x Failure load =1.15x1900 Maximum allowable tensile load = 2185 lbf Resultant load on the slug= V275002 +11502 = 2980.77 lbf Resultant load on each slug=0.6x 2980.77 = 1788.462 lbf Therefore, 2185 - = Maximum allowable tensile load Factor of safety = Max Applied load Factor of safety = 1.221 1788.462