A Beginner's Guide to the Steel Construction Manual, 16^th ed.

Chapter 8 - Bending Members

© 2006, 2007, 2008, 2011, 2017, 2023 T. Bartlett Quimby

Section 8.8.1

Example Problem 8.1

The example problems presented in this section have a spreadsheet solution. You will need this file to follow along with the presented solutions. You can click on the following link to get the file:

Chapter 8: Excel Spreadsheet Solutions

Given: A 40 ft long, simply support W 18x35 beam supports a point load, P, at mid-span. The load consists of equal parts of dead, live, and snow loads.

Wanted: Ignoring the self-weight of the beam, determine the maximum load, P_s,eq, that the beam can support for the given conditions:

a. The beam has full lateral support of the compression flange.
b. The beam has lateral support of the compression flange at ends as quarter points only.
c. The beam has lateral support of the compression flange at ends and mid-point only.
d. The beam has lateral support of the compression flange at ends only.

As the solution is wanted in both LRFD and ASD, we are reporting the results in terms of the comparable load P_s,eq. So, the first step is to determine the controlling ASCE 7 load cases and solve for the controlling composite load factors.

The approach taken in the spreadsheet solution is to solve for capacity for each of the limit states then take the lowest capacity of the applicable limit states as the capacity of the beam.

As the shear and deflection limit states are not affected by lateral support of the compression flange, their results are the same for all four parts of the problem. These capacities are computed first.

For the shear limit state, h/t_w is low enough that C_v equal 1, so shear yielding is the controlling equation. The value of the nominal shear capacity, V_n, is computed then altered according to the demands of the two design philosophies. Notice that the maximum shear in the beam equal P/2, so the maximum applied load, as limited by shear, is twice the shear capacity of the beam. You will note that the shear capacity is much larger than the capacity of the other limit states and never controls in this problem. This situation is not unusual for many steel beams.

The beam supports both snow and live load. Determining the load case to use for deflection becomes a case of engineer's judgment. It is unlikely that beam will ever see full snow load and full live load at the same time, so, in this engineer's judgment, it was decided to use 75% of each. For the total load case this is the same as ASD load case 4. For the live load only case, this is the same as ASD load case 4 with D equal to zero. With this in mind, composite load factors can be computed to convert the resulting load into terms of P_s,eq.

For a beam subject to a point load the deflection is computed by: _max = PL³/(48EI). This equation was used to compute the point load applied to the beam.

The flexural limit states for yielding and flange local buckling are also unaffected by the variation of laterally unbraced length, L_b, given in the problem. For these two limit states, we only solved for the nominal flexural capacity, M_n, instead of continuing the problem by solving the moment equation to get P_s,eq. This was done because the flexural capacity is the minimum over the flexural limit states so there is no need to go beyond M_n until the value of M_n due to lateral torsional buckling is determined.

To set up the lateral torsional buckling problem, we need to identify the laterally unbraced lengths for each of the four problems. Once these are found, then the value of C_b can be computed for each laterally unbraced length in each of the problems. To compute C_b, we need to find the relative magnitude of the moment at each quarter point of each laterally unbraced length. This is done in the spreadsheet and a graph of the moment diagram is shown.

For part (a), the beam has full lateral support (i.e., L_b = 0) so M_n equals M_p. A graph of the moment demand (M_u) vs. the moment capacity (fM_n) for the LRFD case is provided. A similar graph can be made for the ASD case.

For part (b), the beam has lateral support at the quarter points only. You will notice that, due to symmetry, the values of C_b are the same for the two end laterally unsupported lengths and the same for the two middle laterally unsupported lengths. This results in a different moment capacity for each of these regions as seen on the graph of the moment demand (M_u) vs. the moment capacity (fM_n) that is provided.

For parts (c) and (d), there is only one value of C_b in each case. In part (c) this is a result of symmetry. In part (d) the single laterally unbraced length results in a single C_b. In both cases the moment capacity is the same over the length of the beams, so a graph has not been provided.

Table 8.8.1.1 summarizes the results of the computations.

Table 8.8.1.1
Example 8.1 Results Summary
 P_s,eq (kips)

Observations:

• Shear never controls the overall capacity in this problem.
• Deflection is the controlling limit state for this situation when the beam has significant lateral support.
• Flexural strength controls the overall capacity as the length of the laterally unbraced segments increases.
• The flexural strength of the beam decreases substantially as the laterally unbraced segments increase in length. The beam has only 3% of it's maximum capacity in part (d)!

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