A Beginner's Guide to the Steel Construction Manual, 14^th ed.

Chapter 4 - Bolted Connections

© 2006, 2007, 2008, 2011 T. Bartlett Quimby

Section 4.9

Example Problem 4.5

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 bracket is connected to a column flange as shown in Figure 4.9.5.1.  Assume that the applied load is 50% D and 50% L.  The bolts are pretensioned 7/8" diameter A325-N bolts.  The faying surface is class A.

Figure 4.9.5.1
Out-of-Plane Eccentricity
Click on image for larger view

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 comparable service level loads.

1. Determine the bearing capacity of the connection using the SCM Case I method.
2. Determine the bearing capacity of the connection using the SCM Case II method.

Solution:

The problem has been solved in the spreadsheet provided for this problem.  The following explanation is intended to help you understand the spreadsheet computation.

The composite load factors will be needed later in the problem to convert values for P_u and P_a to comparable P_s,eq values so they have been computed first.

Shear Capacity

For both of these problems, the shear limit states are identical so let's look at that first:

The design inequality for the shear case is:

LRFD:    r_uv < fr_nv         ASD:    r_av < r_nv / W

Where:

r_nv = F_nvA_b
r_uv = P_u / (n_bA_b)
r_av = P_a / (n_bA_b)

Substituting these values into the limit state equations above, it is fairly simple task to solve for P_a and P_u.  The resulting values are converted to P_s,eq values for comparison.  Note that the computation for both Case I and Case II are identical for the shear limit state in this case.  This is because the shear capacity is not modified for the presence of tension for bearing type (-N and -X) bolted connections.

a)  Case I Tension Capacity

Again, for this case we need to start with the design inequalities:

LRFD:    r_ut < fr_nt         ASD:    r_at < r_nt / W

Where:

r_nt = F'_ntA_b
r_ut = P_u e c A_b/ I_x
r_at = P_a e c A_b/ I_x
c = maximum distance from the N.A. to the center of a bolt.

Determining the section properties I_x and c are explained in the SCM on pages 7-10 and 7-11.  The approach is to locate the neutral axis by finding where the areas above and below the axis are equal.  In this case, the bolts in the compression zone are ignored, which makes the computation a bit iterative.  Once the N.A. is located then the moment of inertia, I_x, can be computed as long as the maximum value of c.

The equations for F'_nt can be found in the SCM as equations J3-3.

After making substitutions, the resulting design inequalities become:

LRFD:  P_u e c A_b/ I_x < f min[1.3F_nt - (P_u/(n_bA_b))(F_nt/(fF_nv)) , F_nt] A_b

ASD:  P_a e c A_b/ I_x < min[1.3F_nt - (P_u/(n_bA_b))(WF_nt/F_nv) , F_nt] A_b / W

These equations can then be solved for P_u and P_a, which in turn can be converted to comparable P_s,eq.  The results are summarized in the results table.

Case II Tension Capacity

The approach is the same except the values for r_ut and r_at are computed differently.

This case is simpler in that there is no need to compute I_x or c.  The formulas for this method are presented in the SCM on page 7-12.

r_ut = P_u e / (n' d_m)
r_at = P_a e / (n' d_m)

After making substitutions, the resulting design inequalities become:

LRFD:  P_u e  / (n' d_m) < f min[1.3F_nt - (P_u/(n_bA_b))(F_nt/(fF_nv)) , F_nt] A_b

ASD:  P_a e  / (n' d_m) < min[1.3F_nt - (P_u/(n_bA_b))(WF_nt/F_nv) , F_nt] A_b / W

Again, these equations can then be solved for P_u and P_a, which in turn can be converted to comparable P_s,eq.  The results are summarized below.

Solution Summary:

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