Example 2.3 Determine the effective net area and the corresponding design strength for the single angle tension member of The tension member is an 4 x 4 x 3/8 in. made Example 2.2. L from A36 steel. It is connected to a gusset plate with 5/8 in. diameter bolts, as shown in Figure below. The spacing between the bolts is 3 in. center-to-center. Compare your results with those obtained for Example .2. Gross area of angle = Ag = 2.86 in2 (from Table 1-7 on page 1-4 of AISC) Net section area = An Bolt diameter = 5/8 in. Hole diameter for calculating net area = 11/16 + 1/16 = 3/4 in. Net section area = Ag (3/4) x 3/8 = 2.86 3/4 x 3/8 = 2.579 in2 – – is the distance from the centroid of the area connected to the plane of connection For this case is equal to the distance of centroid of the angle from the edge. This value is given in the the manual. Table 1-7 on page 1- of AISC = 1.13 in. L is the length of the connection, which for this case will be equal to 2 x 3.0 in. L = 6.0 in. = 1- = 1- = 0.8116 in. U Alternately, as per Case 8 of Table D3.1. Larger value of 0.8116 and 0.60 is used. U = 0.60 Effective net area = Ae = 0.8116 x 2.579 in2 = 2.093 in2 Gross yielding design strength = Ag Fy = 0.9 x 2.86 in2 x 36 ksi = 92.664 kips t Net section fracture = Ae Fu = 0.75 x 2.093 in2 x 58 ksi = 91.045 kips t Design strength = 91.045 kips (net section fracture governs)
Example 2.3 Determine the effective net area and the corresponding design strength for the single angle tension member of Example 2.2. The tension member is an L 4 x 4 x 3/8 in. made from A36 steel. It is connected to a gusset plate with 5/8 in. diameter bolts, as shown in Figure below. The spacing between the bolts is 3 in. center-to-center. Compare your results with those obtained for Example .2. Gross area of angle = Ag = 2.86 in2 (from Table 1-7 on page 1-4 of AISC) Net section area = An Bolt diameter = 5/8 in. Hole diameter for calculating net area = 11/16 + 1/16 = 3/4 in. Net section area = Ag – (3/4) x 3/8 = 2.86 – 3/4 x 3/8 = 2.579 in2 is the distance from the centroid of the area connected to the plane of connection For this case is equal to the distance of centroid of the angle from the edge. This value is given in the Table 1-7 on page 1- of the AISC manual. = 1.13 in. L is the length of the connection, which for this case will be equal to 2 x 3.0 in. L = 6.0 in. U = 1- = 1- = 0.8116 in. Alternately, U = 0.60 as per Case 8 of Table D3.1. Larger value of 0.8116 and 0.60 is used. Effective net area = Ae = 0.8116 x 2.579 in2 = 2.093 in2 Gross yielding design strength = t Ag Fy = 0.9 x 2.86 in2 x 36 ksi = 92.664 kips Net section fracture = t Ae Fu = 0.75 x 2.093 in2 x 58 ksi = 91.045 kips Design strength = 91.045 kips (net section fracture governs)
CE470-Ch-2-Tension%20members.docx