Find: Ae [in2] dbolt=1/2 [in] tdamaged=1/32 [in] 2 [in] A) 0.3 plate C) 0.5 D) 0.6 plate bolts P Find the effective area of the L beam, in inches squared. [pause] In this probelm, --- L 2 x 2 x 1/8
Find: Ae [in2] dbolt=1/2 [in] tdamaged=1/32 [in] 2 [in] A) 0.3 plate C) 0.5 D) 0.6 plate bolts P a 2 by 2 by 1/8 L beam is bolted to a plate and subjected to a force P. The bolts are spaced 2 inches apart --- L 2 x 2 x 1/8
Find: Ae [in2] dbolt=1/2 [in] tdamaged=1/32 [in] 2 [in] A) 0.3 plate C) 0.5 D) 0.6 plate bolts P with a given bolt diamter, d bolt. The effective area of a steel member, A e, --- L 2 x 2 x 1/8
Find: Ae [in2] dbolt=1/2 [in] Ae = Anet * U tdamaged=1/32 [in] 2 [in] plate equals, the net area of that member, A net, times, --- L 2 x 2 x 1/8
Find: Ae [in2] effective area dbolt=1/2 [in] Ae = Anet * U tdamaged=1/32 [in] net reduction 2 [in] area plate coefficient the reduction coefficient, U. In this equation, the net area, equals, --- L 2 x 2 x 1/8
Find: Ae [in2] net area dbolt=1/2 [in] Ae = Anet * U tdamaged=1/32 [in] Anet = Agross - Aholes 2 [in] plate the gross area of the member, minus, the area from the holes. From a standard steel reference manual, --- L 2 x 2 x 1/8
Find: Ae [in2] net area dbolt=1/2 [in] Ae = Anet * U tdamaged=1/32 [in] Anet = Agross - Aholes 2 [in] plate we can lookup that a 2 by 2 by 1/8 L beam has gross area of ---- L 2 x 2 x 1/8
Find: Ae [in2] net area dbolt=1/2 [in] Ae = Anet * U tdamaged=1/32 [in] Anet = Agross - Aholes 2 [in] plate Agross=0.484 [in2] 0.484 inches squared. [pause] The area from the holes, equals, the quantity of holes --- L 2 x 2 x 1/8
Find: Ae [in2] net area dbolt=1/2 [in] Ae = Anet * U tdamaged=1/32 [in] Anet = Agross - Aholes 2 [in] plate 0.484 [in2] Aholes = (# of holes) (dhole+2*tdamaged) for a given facture path, times the sum of, the diamter of a single hole plus 2 times the damaged thickness around the hole, times the thickness of beam. * beam thickness * L 2 x 2 x 1/8
Find: Ae [in2] 1 net area dbolt=1/2 [in] Ae = Anet * U tdamaged=1/32 [in] Anet = Agross - Aholes 2 [in] plate 0.484 [in2] 1 Aholes = (# of holes) (dhole+2*tdamaged) The number of holes, equals, 1, because the beam will fracture through ---- * beam thickness * L 2 x 2 x 1/8
Find: Ae [in2] 1 net area dbolt=1/2 [in] Ae = Anet * U tdamaged=1/32 [in] Anet = Agross - Aholes 2 [in] plate 0.484 [in2] 1 Aholes = (# of holes) (dhole+2*tdamaged) just one of the bolt holes. [pause] As a typical value, we’ll use, 1/32 of an inch --- * beam thickness * fracture path L 2 x 2 x 1/8
Find: Ae [in2] 1 net area dbolt=1/2 [in] Ae = Anet * U tdamaged=1/32 [in] Anet = Agross - Aholes 2 [in] plate 0.484 [in2] 1 Aholes = (# of holes) (dhole+2*tdamaged) as the damagaed thickness around the bolt hole, and the diamter of the hole itself, equals, --- * beam thickness * fracture path L 2 x 2 x 1/8
Find: Ae [in2] 1 net area dbolt=1/2 [in] Ae = Anet * U tdamaged=1/32 [in] Anet = Agross - Aholes 2 [in] plate 0.484 [in2] 1 Aholes = (# of holes) (dhole+2*tdamaged) the diamter of the bolt, plus, 1/16 of an inch. Knowing the diamter of the bolt, --- * beam thickness * dhole= dbolt + 1/16 [in] L 2 x 2 x 1/8
Find: Ae [in2] 1 net area dbolt=1/2 [in] Ae = Anet * U tdamaged=1/32 [in] Anet = Agross - Aholes 2 [in] plate 0.484 [in2] 1 Aholes = (# of holes) (dhole+2*tdamaged) equals, half an inch, the diamter of the hole, equals, --- * beam thickness * dhole= dbolt + 1/16 [in] L 2 x 2 x 1/8
Find: Ae [in2] 1 net area dbolt=1/2 [in] Ae = Anet * U tdamaged=1/32 [in] Anet = Agross - Aholes 2 [in] plate 0.484 [in2] 1 Aholes = (# of holes) (dhole+2*tdamaged) 9/16 of an inch. [pause] The beam thickness for this particular sized beam, --- * beam thickness * dhole= dbolt + 1/16 [in] dhole= 9/16 [in] L 2 x 2 x 1/8
Find: Ae [in2] 1 net area dbolt=1/2 [in] Ae = Anet * U tdamaged=1/32 [in] Anet = Agross - Aholes 2 [in] plate 0.484 [in2] 1 Aholes = (# of holes) (dhole+2*tdamaged) can also be looked up in a standard steel reference manual, and equals, --- * beam thickness * dhole= 9/16 [in] L 2 x 2 x 1/8
Find: Ae [in2] 1 net area dbolt=1/2 [in] Ae = Anet * U tdamaged=1/32 [in] Anet = Agross - Aholes 2 [in] plate 0.484 [in2] 1 Aholes = (# of holes) (dhole+2*tdamaged) 1/8 of an inch. [pause] After plugging in, --- * beam thickness * dhole= 9/16 [in] t = 1/8 [in] L 2 x 2 x 1/8
Find: Ae [in2] 1 net area dbolt=1/2 [in] Ae = Anet * U tdamaged=1/32 [in] Anet = Agross - Aholes 2 [in] plate 0.484 [in2] 1 Aholes = (# of holes) (dhole+2*tdamaged) all the variables, the area from holes, equals, --- * beam thickness * dhole= 9/16 [in] t = 1/8 [in] L 2 x 2 x 1/8
Find: Ae [in2] 1 net area dbolt=1/2 [in] Ae = Anet * U tdamaged=1/32 [in] Anet = Agross - Aholes 2 [in] plate 0.484 [in2] 1 Aholes = (# of holes) (dhole+2*tdamaged) 0.078 inches squared. [pause] Therefore, the net area of the beam, equals, --- * beam thickness * dhole= 9/16 [in] 1/8 [in] Aholes = 0.078 [in2] L 2 x 2 x 1/8
Find: Ae [in2] 1 net area dbolt=1/2 [in] Ae = Anet * U tdamaged=1/32 [in] 2 [in] plate Anet = Agross - Aholes 1 0.484 [in2] Aholes = (# of holes) (dhole+2*tdamaged) 0.484 inches squared minus 0.078 inches squared, which is, ---- * beam thickness * dhole= 9/16 [in] 1/8 [in] Aholes = 0.078 [in2] L 2 x 2 x 1/8
Find: Ae [in2] 1 net area dbolt=1/2 [in] Ae = Anet * U tdamaged=1/32 [in] Anet = 0.406 [in2] 2 [in] plate Anet = Agross - Aholes 1 0.484 [in2] Aholes = (# of holes) (dhole+2*tdamaged) 0.406 inches squared. [pause] Next we need to determine, --- * beam thickness * dhole= 9/16 [in] 1/8 [in] Aholes = 0.078 [in2] L 2 x 2 x 1/8
Find: Ae [in2] Anet = 0.406 [in2] dbolt=1/2 [in] Ae = Anet * U tdamaged=1/32 [in] reduction 2 [in] plate coefficient 0.484 [in2] Aholes = (# of holes) (dhole+2*tdamaged) the reduction coefficient, U. [pause] The reduction coefficient, equals, --- * beam thickness * dhole= 9/16 [in] 1/8 [in] Aholes = 0.078 [in2] L 2 x 2 x 1/8
Find: Ae [in2] Anet = 0.406 [in2] dbolt=1/2 [in] Ae = Anet * U tdamaged=1/32 [in] reduction 2 [in] plate coefficient x U=min 0.9, 1- L the smaller of 0.9, and the quantity 1 minus x bar over L. In this equation, L refers to --- L 2 x 2 x 1/8
Find: Ae [in2] Anet = 0.406 [in2] dbolt=1/2 [in] Ae = Anet * U tdamaged=1/32 [in] reduction 2 [in] plate coefficient x U=min 0.9, 1- L L the distance between the centers of the first and last bolt holes, which equals, --- L 2 x 2 x 1/8
Find: Ae [in2] Anet = 0.406 [in2] dbolt=1/2 [in] Ae = Anet * U tdamaged=1/32 [in] reduction 2 [in] plate coefficient x U=min 0.9, 1- L L 2 inches. [pause] Next, if we take a cross section, --- L 2 x 2 x 1/8
Find: Ae [in2] A A’ Anet = 0.406 [in2] dbolt=1/2 [in] Ae = Anet * U tdamaged=1/32 [in] reduction A plate coefficient x U=min 0.9, 1- 2 [in] L of the beam and plate, through A A prime, --- A’
Find: Ae [in2] A A’ L 2 x 2 x 1/8 dbolt=1/2 [in] tdamaged=1/32 [in] plate 2 [in] we can identify the value of x bar as --- A’ section A-A’
Find: Ae [in2] A A’ x L 2 x 2 x 1/8 dbolt=1/2 [in] tdamaged=1/32 [in] plate 2 [in] the distance between the centroid of the beam, and the plane of connection, --- A’ section A-A’
Find: Ae [in2] A A’ x L 2 x 2 x 1/8 dbolt=1/2 [in] tdamaged=1/32 [in] plate 2 [in] which for a 2 by 2 by 1/8 L beam, equals, ---- A’ section A-A’
Find: Ae [in2] A A’ x = x L 2 x 2 x 1/8 dbolt=1/2 [in] 0.546 [in] tdamaged=1/32 [in] x A plate 2 [in] 0.546 inches. Therefore, after plugging in the values for --- A’ section A-A’
Find: Ae [in2] x x = Anet = 0.406 [in2] dbolt=1/2 [in] Ae = Anet * U tdamaged=1/32 [in] reduction coefficient x x U=min 0.9, 1- L L = 2 [in] x bar, and, the length L, the reduction coefficient is the smaller of --- x = 0.546 [in] section view
Find: Ae [in2] x x = Anet = 0.406 [in2] dbolt=1/2 [in] Ae = Anet * U tdamaged=1/32 [in] reduction coefficient x x U=min 0.9, 1- L L = 2 [in] 0.9 and 0.727, which equals, 0.727. ---- x = 0.546 [in] U = min (0.9, 0.727) section view
Find: Ae [in2] x x = Anet = 0.406 [in2] dbolt=1/2 [in] Ae = Anet * U tdamaged=1/32 [in] reduction coefficient x x U=min 0.9, 1- L L = 2 [in] [pause] After plugging in the net area and reduction coefficient, --- x = 0.546 [in] U = min (0.9, 0.727) U = 0.727 section view
Find: Ae [in2] x x = Anet = 0.406 [in2] dbolt=1/2 [in] Ae = Anet * U tdamaged=1/32 [in] reduction coefficient x x U=min 0.9, 1- L L = 2 [in] the effective area of the L beam, equals, --- x = 0.546 [in] U = min (0.9, 0.727) U = 0.727 section view
Find: Ae [in2] x x = Anet = 0.406 [in2] dbolt=1/2 [in] Ae = Anet * U tdamaged=1/32 [in] reduction Ae = 0.295 [in2] coefficient x x U=min 0.9, 1- L L = 2 [in] 0.295 inches squared. [pause] x = 0.546 [in] U = min (0.9, 0.727) U = 0.727 section view
Find: Ae [in2] x = Anet = 0.406 [in2] dbolt=1/2 [in] Ae = Anet * U tdamaged=1/32 [in] reduction Ae = 0.295 [in2] coefficient x U=min 0.9, 1- L A) 0.3 B) 0.4 C) 0.5 D) 0.6 L = 2 [in] When reviewing the possible solutions, --- x = 0.546 [in] U = min (0.9, 0.727) U = 0.727
Find: Ae [in2] x = Anet = 0.406 [in2] dbolt=1/2 [in] Ae = Anet * U tdamaged=1/32 [in] reduction Ae = 0.295 [in2] coefficient answerA x U=min 0.9, 1- L A) 0.3 B) 0.4 C) 0.5 D) 0.6 L = 2 [in] the answer is A. x = 0.546 [in] U = min (0.9, 0.727) U = 0.727