A Beginner's Guide to the Steel Construction Manual, 13th ed. (old) Chapter 4 - Bolted Connections |
|
Section 4.9 Example Problem 4.1 The example problems presented in this section refer to a spreadsheet solution. Click on the link below to get the file: Chapter 4: Excel Spreadsheet Solutions Given: A lap splice is connected with (11) A325-X bolts as shown in Figure 4.9.1.1. Assume that the applied load consists of 40% dead load and 60% live load. The steel plates are ASTM A36 stel, the bolts are fully pretensioned 3/4" diameter ASTM A325-X bolts in standard holes. The faying surface is class A. Figure 4.9.1.1 Wanted: Determine the capacity of the connection based on bolt strength as specified below. Consider both LRFD and ASD. Express your results in terms of service load levels.
Solution: Since the load combination only includes live and dead load, ASCE 7 load combination 2 (for both LRFD and ASD) will be the controlling case. The relationship between factored and unfactored load is: Pu = 1.2D +1.6L = 1.2(0.40Ps,eq) + 1.6(0.60Ps,eq) = 1.44 Ps,eq The ASD case is simply the D + L: Pa = D +L = (0.40Ps,eq) + (0.60Ps,eq) = 1.0 Ps,eq a) Slip Capacity - (See specification section J3.8) rn = mDuhscTbNs = 11.1 kips/per bolt where: m = 0.35, Du = 1.13, hsc = 1.00, Tb = 28 k (from Table J3.1), and Ns = 1 (each bolt has only 1 shear plane). The connection has 11 bolts, so, for the connection: Rn = (11.1 k/bolt)(11 bolts) = 122 kips/connection For LRFD, the capacity is determined by: Pu = 1.44 Ps,eq < fRn = 1.0(122 kips/connection) = 122 kips/connection Ps,eq < 84.6 k/connection (LRFD) For ASD, the capacity is determined by: Pa = 1.0 Ps,eq < Rn/W = (122 kips/connection)/1.5 Ps,eq < 81.2 k/connection (ASD) Observations:
b) Bearing Capacity - (See specification J3.6) The nominal capacity of a single shear plane is computed by: rn = FnAb = 26.5 kips/shear plane Where Fn = 60 ksi (from Table J3.2) and Ab = .4418 in2 (area of a 3/4" diameter circle). The connection has 11 bolts, all in single shear, so, for the connection: Rn = (26.5 k/shear plane)(11 shear planes) = 292 kips/connection For LRFD, the capacity is determined by: Pu = 1.44 Ps,eq < fRn = 0.75 (292 kips/connection) = 219 kips/connection Ps,eq < 152 k/connection (LRFD) For ASD, the capacity is determined by: Pa = 1.0 Ps,eq < Rn/W = (292 kips/connection)/2.0 Ps,eq < 146 k/connection (ASD) Observations:
Solution Summary The following values are the capacities of the CONNECTION based on two bolt limit states. The total capacity of the connection is also a function of the capacity of members in the connection.
Observations:
|