Question

2. (30 pts) Determine the block shear strength of the tension member shown below. Bolts diameter = 7/8 in, A36 steel is used (Table 2-4 XN, A16 s 58 kse DO 3232 -in. bolts 19 (3) /8" bolts 3" bolt spacing (fy = 36 ksi P. 2-487 (Table 1-7) Ag = 2.50 in ² p. 1-48 - 1.00 in t. 3/8" And = [3.5"-(7/6" + /6")] (3/8") = 10.94 in? Ag. = (7.5") (2/6") - 2.81 in? Anu = [7.5 - (2.5)(1/" />")](3/8") - 1.875 mm shear block: ØR n = 0 (0.6 Fu Anu + Ups Fu Ant) = $10.6 Fy Agu + Ubs fu Ant) -0.75/0.6 & 58ksi x 1.875 in² + 1 + 58ksi x 0.94 in) - 89.8 xips 0.75(0.6+ 36ksi x 2.81in" & 1 58ksi & 0.94.n") = 106.4 kis Block Shear strength: 864 kips)

Answer 1

The shear areas are Gross shear area = Agv = 3/8 (7.5) = 2.813 in? There are 2.5 bolt holes Net shear area = Any = 3/8(7.5 – 2.5(7/8 + 1/8)) = 1.875 in? The tension area is given by considering one half of bolt hole Ant = 3/8(1.5 – 0.5(778 + 1/8)) = 0.375 in? Since the block shear failure will occur in angle , Ubs = 1 Nominal block shear strength is given by Rn = 0.6FųAny + UbsFųAnt = 0.6x58x1.875 + 1x58x0.375 = 87 kips Check for upper limit Rp = 0.6FyAgv + UbsFųAnt = 0.6x36x2.813 + 1x58x0.375 = 82.51 kips Therefore nominal block shear strength is 82.51 kips Design block shear strength for LRFD =ØR, = 0.75x82.51 = 61.9 kips Allowable block shear strength For ASD=R2/22 = 82.51/2 = 41.3 kips

Answer 2

I believe the solution is wrong. Along the shear plane (Anv), there is 1/2 + 1/2 + 1/4 hole = 1.25 holes, not 2.5 holes. The shear plane runs through the center of the holes, not the entire hole. Same with the tension plane (Ant).