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Contents | 1. Introduction to structural design | 2. Loads | 3. Wood |

Introduction to steel | Material properties | Sectional properties | Design approaches | Construction systems | Tension elements | Columns |

The design of steel wide-flange beams using the "allowable strength design" method is quite similar to the procedures used to design timber beams (see Chapter 3). Cross sections are selected based on their strength in bending, and then checked for shear and deflection.

Unlike wood beams, however, steel beams are designed based on their "available" strength, rather than on the more convention notion of an "allowable" stress. Whereas the strength of a wood beam corresponds to its outer fibers reaching a failure stress, steel beams do not fail when their outer fibers first begin to yield at the stress, *F _{y}*. A steel cross section is able to carry increased loads beyond the so-called elastic moment, shown in Figure 4.21

Figure 4.21: Bending stresses acting on steel wide-flange (-shaped) cross section corresponding to the (*a*) elastic moment and (*b*) plastic moment, with three examples: (*c*) and (*d*) illustrate elastic and plastic moments for a hypothetical section with all its area at the extreme fibers; (*e*) and (*f*) illustrate elastic and plastic moments for a typical W30 × 90 section; while (*g*) and (*h*) illustrate elastic and plastic moments for a rectangular section.

Previously, steel used an "allowable stress" method based on a limit state corresponding to the
elastic moment; and if the plastic moment were always stronger than the elastic moment to the
same extent for all cross sections, one could simply adjust the factor of safety for allowable strength
(plastic moment) design so that the method corresponded precisely to allowable stress (elastic moment)
design. However, it can be shown that the extra margin of safety gained by moving beyond
the elastic, to the plastic moment (i.e., from the condition of Figure 4.21*a* to Figure 4.21*b*), is not the same for all cross sections, so that allowable stress design for steel does not provide a consistent margin of safety against the limit state of complete yielding.

For wide-flange (-shaped) sections, the extremes can be represented by a hypothetical section with no web (i.e., consisting entirely of flanges of infinite density, or no thickness, as shown in Figures 4.21*c* and 4.21*d*); and, at the other extreme, a section whose flanges merge together at the neutral axis (i.e., a rectangular section, as shown in Figures 4.21*g* and 4.21*h*). In the first case, it is clear that the elastic moment and plastic moment coincide, and the so-called "shape factor" defining the ratio of plastic to elastic section modulus, *Z _{x}*/

Clearly, all -shaped sections must have a shape factor between these two extremes, i.e., between 1.0 and 1.5. The shape factor for a typical W-shape (W30 × 90), shown in Figures 4.21*e* and 4.21*f*, can be determined in the same manner, abstracting from the complexities of the actual shape by considering only perfectly rectangular flange and web areas. Using the dimensions shown, and performing the same equilibrium calculations as in the cases above, we get *S _{x}* = 240 in

The equation for plastic section modulus, *Z _{x}* = M/

(4.12)

where *M _{max}* = the maximum bending moment (in-kips),

We found earlier that the shape factor for a W30 × 90 section equals 1.16. By looking at the ratio of plastic to elastic section modulus for all wide-flange shapes, it can be seen that these shape factors fall between 1.098 (for a W14 × 90), and 1.297 (for a W14 × 730). One could therefore conservatively create a safety factor for elastic allowable stress design by assuming a shape factor of 1.10, and by multiplying this value by the safety factor for allowable strength design, 1/Ω = 1/1.67 = 0.60 (inverted to be consistent with the conventions for allowable stress safety factors). Equation 4.12 would then become *S _{req}* =

When the compression flange of a beam is not continuously braced, lateral-torsional buckling can
reduce the available bending moment below the value of *M _{p}*/Ω assumed earlier for laterally braced beams. How much this stress is reduced depends on whether the beam buckles before or after
the cross section begins to yield, and how bending stresses vary over the beam's effective length.
Figure 4.22 shows several possible stress stages for a cross-sectional shape as the bending moment
increases. At Figure 4.22

Figure 4.22: Elongation, strain, and stress diagrams for an elastic-plastic material such as steel showing (*a*) elongation and shortening of the actual material; (*b*) strain diagrams; (*c*) stress diagram at the point where the outer fiber has just yielded; (*d*) stress diagram corresponding to strain just beyond the elastic limit; (*e*) stress diagram corresponding to continued strain beyond the elastic limit; (*f*) stress diagram corresponding to the plastic moment (where the entire cross section has yielded); and (*g*) stress-strain diagram. [Diagram reproduced from fig. 1.64]

Being able to resist the full plastic moment represents an extra margin of safety: if a beam can develop this plastic moment without buckling, the maximum available bending moment of *M _{p}*/Ω is used, as shown in Equation 4.12. In addition to lateral-torsional buckling (Figure 4.23

Figure 4.23: Two modes of buckling limiting the strength of a wide-flange (-shaped) beam: (*a*) lateral-torsional buckling; and (*b*) local flange buckling

For sections that are compact and laterally braced, Equation 4.12 applies, and the full strength of the beam is utilized. However, as shown in Figure 4.24, this available strength must be reduced if either local flange buckling (where the section is not compact) or lateral-torsional buckling (where the section is not adequately braced) occurs before the plastic moment is reached.

Figure 4.24: Influence of lateral-torsional buckling and flange slenderness on available moment: three zones are defined for (*a*) lateral torsional buckling, with the boundaries established by the laterally unbraced length, *L*_{p} (the greatest unbraced length where the section can reach a plastic moment without lateral torsional buckling) and *L*_{r} (the greatest unbraced length where the section will buckle inelastically before reaching the plastic moment); and for (*b*) flange slenderness, with the boundaries established by the ratio of half the flange width to flange thickness, λ = *b*_{f} /(2*t*_{f}), set equal to λ_{p} (the greatest flange slenderness where the section can reach a plastic moment without local flange buckling) and λ_{r} (the greatest flange slenderness where the flange will buckle locally in an inelastic manner before reaching the plastic moment)

For beams with an unbraced length, *L _{b}*, that falls within the "middle zone" illustrated in Figure 4.24

However, all these equations are based on the assumption that the beam is subject to a uniform bending moment along its entire length; where the moment varies, as is almost always the case, this assumption is overly conservative, since lateral-torsional buckling is less likely to be triggered where bending stresses are not entirely at their maximum value along the whole length of an unbraced segment. For this reason, a coefficient, *C _{b}*, should be applied to the available strength of each unbraced segment of the beam, based on the distribution and magnitude of bending moments along that segment's length. This "lateral-torsional buckling modifier" is defined as follows for doubly-symmetric bending elements such as wide-flange beams:

(4.13)

where *M _{max}* is the greatest moment within the unbraced segment; and

Compact sections are proportioned so that neither the flange nor the web will buckle locally before
the onset of a plastic moment. Since all wide-flange webs meet the standards for compact sections,
only the flange slenderness, defined as λ = *b _{f}*/(2

Where a beam is both noncompact and laterally unbraced, both criteria illustrated in Figure 4.24
are tested, and the smaller capacity governs. For beams that are both compact and laterally braced, Appendix Table A-4.15 can be used to select the lightest W-shape for bending. For A992 wide-flange beams that are not adequately braced laterally (i.e, where *L _{b}* >

Once a selection is made based on bending stress, the section is then checked for shear and deflection.
The nominal shear strength, *V _{n}* , equals 0.6

required *A*_{w} = *V*/*F*_{v}

(4.12)

For a small group of wide-flange beams with slender webs, the safety factor for shear is increased from 1.5 to 1.67, and so the allowable shear stress becomes *F _{v}* = (0.6/1.67)

Where the top flange of a steel beam is coped (so that it may be fastened to the web of a girder while keeping the top surfaces of girder and beam flanges aligned), a mode of failure combining both shear and tension stresses in the beam web must be checked, with the shear and tension failure planes assumed to occur at the surface defined by the bolt centerline, as shown in Figure 4.25.

Figure 4.25: Block shear at coped beam with (*a*) coefficient *U*_{bs} = 1.0 where tension stress is uniform (single line of bolts); and (*b*) *U*_{bs} = 0.5 where tension stress has a triangular distribution (double line of bolts)

The nominal capacity of such a connection is found by adding the capacity of the net shear area subject to rupture (or the gross shear area subject to yielding) to the capacity of the net tension area subject to rupture. Where both net areas are subject to rupture, the capacity is defined as: *R _{n}* = 0.6

It is also possible that a mode of shear failure alone, with no tension component, could govern
the connection design. In such a case, both yielding on the gross area of the cross section and rupture
on the net area need to be checked. For yielding, the nominal capacity, *R _{n}*, equals 0.60

** Problem definition.** Find the capacity of a bolted double-angle connection to the web of a coped W18 × 86 wide-flange beam, considering only block shear in the web. Assume A992 steel (

Figure 4.26: Block shear in a coped beam, for Example 4.5

** Solution overview.** Find the smaller of the capacities based on rupture and yielding of shear area, rupture of tension area, and bolt bearing on the web.

*Problem solution*

**1.** Find capacity based on net areas subject to rupture. Lengths along net areas are found by subtracting the lengths of bolt hole diameters from the total (gross) dimension. The net area for shear, *A _{nv}* =

The capacity based on rupture of these net areas is defined as: *R _{n}* = 0.6

**2.** Find capacity based on gross area (yielding) for shear and net area (rupture) for tension. The gross area for shear, *A _{gv}* =

The capacity based on yielding of the shear area and rupture of the tension area is defined as: *R _{n}* = 0.6

**3.** The governing capacity is the smaller value from steps 1 and 2: *R _{n}* = 212.2 kips based on rupture of the net areas.

Deflection is based on the same criteria discussed in Chapter 3 for wood beams and involves a comparison of an allowable deflection, typically set at span/240 for total loads and span/360 for live loads on floor beams, to the actual computed deflection. Actual deflections can be computed based on the coefficients in Appendix Table A-4.17 (which are the same coefficients used for wood in Appendix Table A-3.15). Allowable deflection guidelines can be found in Appendix Table A-1.3 (also summarized in Appendix Table A-4.17).

** Problem definition.** Using A992 steel, design the typical beam and girder for the library stack area shown in Figure 4.27. Use the generic dead load for steel floor systems. Assume that the beams are continuously braced by the floor deck, and that the girders are braced only by the beams framing into them.

Figure 4.27: Framing plan for Example 4.6

** Solution overview.** Find loads; compute maximum bending moment and shear force; use appropriate tables to select beams for bending; then check for shear and deflection.

*Problem solution*

Find loads: From Appendix Table A-2.1, the dead load, *D* = 47 psf.

From Appendix Table A-2.2, the live load, *L* = 150 psf.

*Beam design*

**1.** Create load, shear and moment diagrams as shown in Figure 4.28 to determine critical (i.e., maximum) shear force and bending moment. The total distributed load, *w* = (dead + live)(tributary area for 1 linear foot) = (47 + 150)(6) = 1182 lb/ft = 1.18 kips/ft. Live load reduction would not apply even if the "influence" area was not less than 400 ft^{2}, because of the library stack occupancy (i.e., the probability of full loading makes live load reduction a dangerous assumption).

Figure 4.28: Load, shear, and moment diagrams for beam in Example 4.6

**2.** Find allowable bending stress: since the beam is laterally braced by the floor deck and the cross section is assumed to be compact, use Equation 4.12 to find *Z _{req}* = Ω

**3.** From Appendix Table A-4.15, select a W12 × 14 with actual *Z _{x}* = 17.4 in

**4.** Check section for shear: from Appendix Table A-4.3, the actual web area, *A _{w}* = d ×

**5.** From Equation 4.14, the required *A _{w}* = V/

**6.** From Appendix Table A-1.3 (also summarized in Appendix Table A-4.17), find the allowable total-load deflection for a floor beam: Δ* ^{T}_{allow}* = span/240 = 12(15)/240 = 0.75 in. and the allowable live-load deflection for a floor joist: Δ

**7.** From Appendix Table A-4.17, the actual total load deflection, Δ* ^{T}_{act}* =

*C* = 22.46.

*L* = 15 × 12 = 180 in. (the term, *L*, is used for both span and live load).

*P* = *w*(*L*/12) = (150 + 47)(6)(180/12) = 17,730 lb = 17.73 kips.

*E* = 29,000 ksi (Appendix Table A-4.1, Note 1).

*I* = 88.6 in^{4} (Appendix Table A-4.3).

Δ* ^{T}_{act}* = 22.46(17.73)(180/12)

Since Δ* ^{T}_{act}* = 0.523 in. ≤ Δ

**8.** From Appendix Table A-4.17, the actual live load deflection, Δ* ^{L}_{act}* =

*C* = 22.46.

*L* = 15 × 12 = 180 in.

*P* = *w*(*L*/12) = (150 × 6)(180/12) = 13500 lb = 13.5 kips (Use live load only!).

*E* = 29,000 ksi (Appendix Table A-4.1, Note 1).

*I* = 88.6 in^{4} (Appendix Table A-4.3).

Δ* ^{L}_{act}* = 22.46(13.5)(180/12)

Since Δ* ^{L}_{act}* = 0.398 in. ≤ Δ

**9.** Conclusion: The W12 × 14 section is OK for bending, shear and deflection. Therefore it is acceptable.

*Girder design*

**1.** Create load, shear and moment diagrams as shown in Figure 4.29 to determine the critical (i.e., maximum) shear force and bending moment. Each concentrated load is twice the typical beam reaction, or 17.73 kips. Alternatively, compute using tributary areas; that is, *P* = (47 + 150)(15 × 6) = 17,730 lb = 17.73 kips. Live load reduction does not apply even though the "influence" area is greater than 400 ft^{2}, because of the library stack occupancy (i.e., the probability of full loading makes live load reduction a dangerous assumption).

Figure 4.29: Load, shear, and moment diagrams for girder in Example 4.6

**2.** Find allowable bending stress: The girder is not continuously braced by the floor deck; rather, it is braced every 6 ft by the beams framing into it, so the unbraced length, *L _{b}* = 6 ft. Use Appendix Table A-4.16 to directly find the lightest cross section for bending, based on

Figure 4.30: Selection of W21×x44 beam based on available moment graphs (Appendix Table A-4.16) for Example 4.6

**3.** Check section for shear: from Appendix Table A-4.3, the actual web area, *A _{w}* = d ×

**4.** From Equation 4.14, the required *A _{w}* = V/

**5.** From Appendix Table A-1.3 (also summarized in Appendix Table A-4.17), find the allowable total-load deflection for a floor beam, Δ* ^{T}_{allow}* = span/240 = 12(24)/240 = 1.2 in., and the allowable live-load deflection for a floor joist, Δ

**6.** From Appendix Table A-4.17, the actual total-load deflection, Δ* ^{T}_{act}* =

*C* = 85.54.

*L* = 24 × 12 = 288 in.

*P* = (47 + 150)(15 × 6) = 17,730 lb = 17.73 kips.

*E* = 29,000 ksi (Appendix Table A-4.1, Note 1).

*I* = 843 in^{4} (Appendix Table A-4.3).

Δ* ^{T}_{act}* = 85.54(17.73)(288/12)

Since Δ* ^{T}_{act}* = 0.86 in. ≤ Δ

**7.** From Appendix Table A-4.17, the actual live-load deflection, Δ* ^{L}_{act}* =

*C* = 85.54.

*L* = 24 × 12 = 288 in.

*P* = 150(15 × 6) = 13,500 lb = 13.5 kips (Use live load only!).

*E* = 29,000 ksi (Appendix Table A-4.1, Note 1).

*I* = 843 in^{4} (Appendix Table A-4.3).

Δ* ^{L}_{act}* = 85.54(13.5)(288/12)

Since Δ* ^{L}_{act}* = 0.65 in. ≤ Δ

**8.** Conclusion: The W21 × 44 section is OK for bending, shear and deflection. Therefore it is acceptable.

** Problem definition.** Determine whether an HSS12 × 4 × 1/4 can be used as a typical beam for the library stack area shown in Example 4.6.

** Solution overview.** Find loads; compute maximum bending moment and shear force; check beam for bending, shear, and deflection.

*Problem solution*

**1.** Find loads and moment (same as Example 4.6):

The dead load, *D* = 47 psf.

The live load, *L* = 150 psf

Maximum moment, *M _{max}* = 399 in-kips

**2.** Find allowable bending stress: Since the beam is laterally braced by the floor deck and the cross section is assumed to be compact, use Equation 4.12 to find *Z _{req}* = Ω

**3.** From Appendix Table A-4.6, the actual plastic section modulus for an HSS12 × 4 × 1/4, *Z _{x}* = 25.6 in

**4.** Check section for shear: from Appendix Table A-4.2 (Note 3), the web area, *A _{w}* is taken as 2

**5.** From Equation 4.14, the required *A _{w}* = V/

**6.** From Appendix Table A-1.3 (also summarized in Appendix Table A-4.17), find the allowable total-load deflection for a floor beam: Δ* ^{T}_{allow}* = span/240 = 12(15)/240 = 0.75 in.; and the allowable live-load deflection for a floor joist: Δ

**7.** From Appendix Table A-4.17, the actual total-load deflection, Δ* ^{T}_{act}* =

*C* = 22.46.

*L* = 15 × 12 = 180 in.

*P* = *w*(*L*/12) = (47 + 150)(6)(180/12) = 17,730 lb = 17.73 kips.

*E* = 29,000 ksi (Appendix Table A-4.1, Note 1).

*I* = 119 in^{4} (Appendix Table A-4.6).

Δ* ^{T}_{act}* = 22.46(17.73)(180/12)

Since Δ* ^{T}_{act}* = 0.389 in. ≤ Δ

**8.** From Appendix Table A-4.17, the actual live-load deflection, Δ* ^{L}_{act}* =

*C* = 22.46.

*L* = 15 × 12 = 180 in.

*P* = *w*(*L*/12) = (150 × 6)(180/12) = 13500 lb = 13.5 kips (Use live load only!).

*E* = 29,000 ksi (Appendix Table A-4.1, Note 1).

*I* = 119 in^{4} (Appendix Table A-4.6).

Δ* ^{T}_{act}* = 22.46(13.5)(180/12)

Since Δ* ^{L}_{act}* = 0.300 in. ≤ Δ

**9.** Conclusion: The HSS12 × 4 × 1/4 section is OK for bending, shear and deflection. Therefore it is acceptable.

© 2020 Jonathan Ochshorn; all rights reserved. This section first posted November 15, 2020; last updated November 15, 2020.