A Beginner's Guide to the Steel Construction Manual, 13th ed. (old)

Chapter 8 - Bending Members

© 2006, 2007, 2008 T. Bartlett Quimby

Introduction


Flexure


Shear


Deflection


Misc. Limit States


Beam Design

Chapter Summary

Example Problems

Homework Problems

References


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Section 8.8

Example Problem 8.1

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, Ps,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.

Solution:

As the solution is wanted in both LRFD and ASD, we are reporting the results in terms of the comparable load Ps,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 of the 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/tw is low enough that Cv equal 1, so shear yielding is the controlling equation.  The value of the nominal shear capacity, Vn, 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 Ps,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, Lb, given in the problem.  For these two limit states, we only solved for the nominal flexural capacity, Mn, instead of continuing the problem by solving the moment equation to get Ps,eq.  This was done because the flexural capacity is the minimum over the flexural limit states so there is no need to go beyond Mn until the value of Mn 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 Cb can be computed for each laterally unbraced length in each of the problems.  To compute Cb, 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. Lb = 0) so Mn equals Mp.  A graph of the moment demand (Mu) vs. the moment capacity (fMn) 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 Cb 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 (Mu) vs. the moment capacity (fMn) that is provided.

For parts (c) and (d), there is only one value of Cb in each case.  In part (c) this is a result of symmetry.  In part (d) the single laterally unbraced length results in a single Cb.  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

  Flexure   Shear   Deflection   Controlling Ps,eq

Controlling Limit State

part LRFD ASD LRFD ASD TL LL Only LRFD ASD LRFD ASD
a 19.69 19.91 251.53 254.88 15.41 17.12 15.41 15.41 TL Defl TL Defl
b 17.78 17.98 251.53 254.88 15.41 17.12 15.41 15.41 TL Defl TL Defl
c 1.91 1.93 251.53 254.88 15.41 17.12 1.91 1.93 Flexure Flexure
d 0.59 0.60 251.53 254.88 15.41 17.12 0.59 0.60 Flexure Flexure

Observations:

  • Shear never controls the overall capacity
  • 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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