## Example

Using LRFD, select the lightest W-section for the three-span continuous beam shown in Figure 4.10a to support a uniformly distributed dead load of 1.5 k/ft (22 kN/m) and a uniformly distributed live load of 3 k/ft (44 kN/m). The beam is laterally braced at the supports A, B, C, and D. Use A36 steel.

Load combinations: The beam is to be designed based on the worst load combination of Table 4.3. By inspection, the load combination 1.2D + 1.6L will control the design. Thus, the beam will be designed to support a factored uniformly distributed dead load of 1.2 x 1.5 = 1.8 k/ft and a factored uniformly distributed live load of 1.6 x 3 = 4.8 k/ft.

Placement of loads: The uniform dead load is to be applied over the entire length of the beam as shown in Figure 4.10b. The uniform live load is to be applied to spans AB and CD as shown in Figure 4.10c to obtain the maximum positive moment and it is to be applied to spans AB and BC as shown in Figure 4.10d to obtain the maximum negative moment.

Reduction of negative moment at supports: Assuming the beam is compact and Lb < Lpd (we shall check these assumptions later), a 10% reduction in support moment due to gravity load is allowed provided that the maximum moment is increased by 10 the average of the negative support moments. This reduction is shown in the moment diagrams as solid lines in Figure 4.10b and d. (The dotted lines in these figures represent the unadjusted moment diagrams.) This provision for support moment reduction takes into consideration the beneficial effect of moment redistribution in continuous beams, and it allows for the selection of a lighter section if the design is governed by negative moments. Note that no reduction in negative moments is made to the case when only spans AB and CD are loaded. This is because for this load case, the negative support moments are less than the positive in-span moments.

Determination of the required flexural strength, Mu: Combining load case 1 and load case 2, the maximum positive moment is found to be 256 kip ft. Combining load case 1 and load case 3, the maximum negative moment is found to be 266 kip ft. Thus, the design will be controlled by the negative moment and so Mu = 266 kip ft.

Beam selection: A beam section is to be selected based on Equation 4.38. The critical segment of the beam is span AB. For this span, the lateral unsupported length, Lb = 20 ft. For simplicity, the bending coefficient, Cb, is conservatively taken as 1. The selection of a beam section is facilitated by the use of a series of beam charts contained in the AISC-LRFD Manual (AISC 2001). Beam charts are plots of flexural design strength f bMn of beams as a function of the lateral unsupported length Lb based on Equation 4.39 to 4.41. A beam is considered satisfactory for the limit state of flexure if the beam strength curve envelopes the required flexural strength for a given Lb.

For the present example, Lb = 20 ft and Mu = 266 kip ft, the lightest section (the first solid curve that envelopes Mu = 266 kip ft for Lb = 20 ft) obtained from the chart is a W16 x 67 section. Upon adding the factored dead weight of this W16 x 67 section to the specified loads, the required flexural strength increases from 266 to 269 kip ft. Nevertheless, the beam strength curve still envelopes this required strength for Lb = 20 ft; therefore, the section is adequate.

Check for compactness: For the W16 x 67 section

Service dead load = 1.5 k/ft Service live load = 3 k/ft

Service dead load = 1.5 k/ft Service live load = 3 k/ft

FIGURE 4.10 Design of a three-span continuous beam (1 k = 4.45 kN, 1 ft = 0.305 m).

Therefore, the section is compact.

Check whether Lb < Lpd: Using Equation 4.62, with M1/M2 = 0, ry = 2.46 in., and Fy = 36 ksi, we have Lpd = 246 in. (or 20.5 ft). Since Lb = 20 ft is less than Lpd = 20.5 ft, the assumption made earlier is validated.

Check for the limit state of shear: The selected section must satisfy the shear strength criterion of Equation 4.48. From structural analysis, it can be shown that maximum shear occurs just to the left of support B under load case 1 (for dead load) and load case 3 (for live load). It has a magnitude of 81.8 kip. For the W16 x 67 section, h/tâ = 35.9, which is less than 2.4^(E/Fyw) = 69.5, so the design shear strength is given by Equation 4.47. We have, for Fyw = 36 ksi and Aw = dtw = (16.33)(0.395)

[fvVn = 0.90(0.60FywAw) = 125 kip] > [Vu = 81.8 kip]

Therefore, shear is not a concern. Normally, the limit state of shear will not control unless for short beams subjected to heavy loads.

Check for limit state of deflection: Deflection is a serviceability limit state. As a result, a designer should use service (not factored) load for deflection calculations. In addition, most beams are cambered to offset deflection caused by dead loads, so only live loads are considered in deflection calculations. From structural analysis, it can be shown that maximum deflection occurs in spans AB and CD when (service) live loads are placed on those two spans. The magnitude of the deflection is 0.297 in. Assuming the maximum allowable deflection is L/360 where L is the span length between supports, we have an allowable deflection of 20 x 12/360 = 0.667 in. Since the calculated deflection is less than the allowable deflection, deflection is not a problem.

Check for the limit state of web yielding and web crippling at points of concentrated loads: From structural analysis, it can be shown that maximum support reaction occurs at support B when the beam is subjected to loads shown as load case 1 (for dead load) and load case 3 (for live load). The magnitude of the reaction Ru is 157 kip. Assuming point bearing, that is, N = 0, we have for d = 16.33 in., k = 1.375 in., tf = 0.665 in., and tw = 0.395 in.

Web yielding: [fRn = Equation 4.54 = 97.8 kip] < [Ru = 157 kip] Web crippling: [fRn = Equation 4.56 = 123 kip] < [Ru = 157 kip]

Thus, both the web yielding and web crippling criteria are violated. As a result, we need to provide web stiffeners or bearing plate at support B. Suppose we choose the latter, the size of the bearing plate can be determined by solving Equations 4.54 and 4.56 for N using Ru = 157 kip. The result is N = 4.2 in. and 3.3 in., respectively. So, use N = 4.25 in. The width of the plate, B, should conform with the flange width, bf, of the W-section. The W16 x 67 section has a flange width of 10.235 in., so use B=10.5 in. The thickness of the bearing plate is to be calculated from the following equation (AISC 2001):

t v wy where Ru is the factored concentrated load at the support (kip), B is the width of the bearing plate (in.), k is the distance from the web toe of the fillet to the outer surface of the flange (in.), A is the area of the bearing plate (in.2), and Fy is the yield strength of the bearing plate.

Substituting Ru = 157 kip, B = 10.5 in., k = 1.375 in., A = 42 in.2, and Fy = 36 ksi into the above equation, we obtain t = 1.86 in. Therefore, use a 1| in. plate.

For uniformity, use the same size plate at all the supports. The bearing plates are to be welded to the supporting flange of the W-section. Use a W16 x 67 section. Provide bearing plates of size 1| in. x 4 in. x 10| in. at the supports.

### 4.5.3 Beam Bracing

The design strength of beams that are bent about their major axes depends on their lateral unsupported length Lb. The manner a beam is braced against out-of-plane deformation affects its design. Bracing can be provided by various means such as cross-frames, cross-beams or diaphragms, or encasement of the beam flange in the floor slab (Yura 2001). Two types of bracing systems are identified in the AISC-LRFD Specification â relative and nodal. A relative brace controls the movement of a braced point with respect to adjacent braced points along the span of the beam. A nodal (or discrete) brace controls the movement of a braced point without regard to the movement of adjacent braced points. Regardless of the type of bracing system used, braces must be designed with sufficient strength and stiffness to prevent out-of-plane movement of the beam at the braced points. Out-of-plane movement consists of lateral deformation of the beam and twisting of cross-sections. Lateral stability of beams can be achieved by lateral bracing, torsional bracing, or a combination of the two. For lateral bracing, bracing shall be attached near the compression flange for members bend in single curvature (except cantilevers). For cantilevers, bracing shall be attached to the tension flange at the free end. For members bend in double curvature, bracing shall be attached to both flanges near the inflection point. For torsional bracing, bracing can be attached at any cross-sectional location.

### 4.5.3.1 Stiffness Requirement for Lateral Bracing

The required brace stiffness of the bracing assembly in a direction perpendicular to the longitudinal axis of the braced member, in the plane of buckling, is given by

5 ,33Mu Cd

Lbrho

13. 3Mu Cd

Lbrho for relative bracing for nodal bracing

where Mu is the required flexural strength, Cd = 1.0 for single curvature bending; 2.0 for double curvature bending near the inflection point, Lbr is the distance between braces, and ho is the distance between flange centroids.

Lbr can be replaced by Lq (the maximum unbraced length for Mu) if Lbr < Lq. (For ASD, replace Mu in Equation 4.63 by 1.5Ma, where Ma is the required flexural strength based on ASD load combinations.)

4.5.3.2 Strength Requirement for Lateral Bracing

In addition to the stiffness requirement as stipulated above, braces must be designed for a required brace strength given by

MuCd

0 0