Chapter 6: Bending Members

Transcription

1 Chapter 6: Bending Members The following information is taken from Unified Design of Steel Structures, Second Edition, Louis F. Geschwindner, 2012, Chapter Bending Members in Structures A bending member carries loads applied normal to its longitudinal axis. In building construction, bending members (known as beams) provide support for floors and roofs. Beams may be simple span or continuous span. Beams transfer their load to other structural members such as columns, girders, and walls. Various shapes are used as bending members. Wide-flange sections are generally used as bending members. Angles and tees are commonly used as lintels over openings. Built-up shapes made by combining shapes and plates may be used; however, built-up shapes may not be economical because of the labor costs associated with fabrication. The most common and economical bending members (known as compact sections) are those that can reach the full material yield strength without being limited by buckling of any cross-sectional elements. 6.2 Strength of Beams As a load is applied to a bending member, stresses due to the bending moment are developed in the cross section. The stress at any point may be computed from the familiar flexural formula when the maximum computed stress in the beam is below the elastic limit. fy = My/I where M = the applied moment that stresses the section in the elastic range y = the distance from the neutral axis to the point where the stress is to be determined I = moment of inertia fy = the resulting bending stress at location y The stress at the extreme fiber is usually of greatest interest and the flexural formula becomes 6.1

2 fb = Mc/I = M/S where S = section modulus fb = the extreme fiber bending stress The moment that causes the extreme fiber to reach the yield stress, Fy, is called the yield moment, My (ref. Figure 6.2a, p. 166 of the textbook). If the moment in a ductile steel beam is increased beyond the yield moment, the strain in the extreme fibers increases but the stress in the outermost fibers stays the same (ref. Figure 6.2c, p. 166 of the textbook). The additional resisting moment is carried by the fibers nearer to the neutral axis As the moment increases this process continues with more parts of the beam cross section stressed to the yield stress. If the moment continues to increase, the portion of the cross section experiencing the yield stress continues to increase until the entire section experiences the yield stress (ref. Figure 6.2d, p. 166 of the textbook). When the stress distribution has reached this stage, a full plastic stress distribution is approached and a plastic hinge is said to have formed, and no additional moment can be resisted by the section. The moment that causes this full plastic stress distribution is called the plastic moment Mp. 6.2

3 A cross section that is capable of attaining a full plastic stress distribution and the corresponding moment is referred to as a compact section. - A compact section is one that has a sufficiently stocky cross section so that it is capable of developing a fully plastic stress distribution before buckling. At every stage of loading, equilibrium of the cross section requires that the total internal tension force be equal to the total internal compression force. For a doubly symmetric wide flange shape, equilibrium for the plastic moment occurs when the portion of the shape above the elastic neutral axis is stressed to the yield stress in compression while the portion below the elastic neutral axis is stressed to the yield stress in tension. For a non-symmetric shape, because the area above the elastic neutral axis is not equal to the area below the elastic neutral axis, a new axis (called the plastic neutral axis, PNA) must be defined that gives equal areas in tension and compression. - For symmetric shapes the elastic and plastic neutral axes coincide. Based on equilibrium requirements, the plastic moment (Mp) that corresponds to the fully yielded stress distribution is determined as follows. Mp = Fy (Ac yc) + Fy (At yt) If Ac and At are the equal compression and tension areas, respectively, and yc and yt are the distances from the centroids of these areas to the PNA, then the equation for the plastic moment may be simplified as follows. Mp = Fy (A/2) (yc + yt) The terms (A/2) (yc + yt) are the first moment of area of the cross section and are normally combined and called the plastic section modulus, Z and the plastic moment is given as Mp = Fy Z The plastic section modulus is listed for all available shapes in Part 1 of the Manual. For a given beam to attain its full plastic moment strength, it must be compact and satisfy the lateral support criteria established in Specification Section F2. If these criteria are not met, the strength is less than the plastic moment Mp. The criteria to be satisfied are defined by two limit states in addition to yielding: local buckling (noncompact shape) and lateral torsional buckling. 6.3

4 In this chapter the buckling moments of a series of compact ductile steel beams with differing lateral or torsional bracing situations are considered. The differing lateral or torsional bracing situations include the following. 1. First, the beams will be assumed to have continuous lateral bracing for the compression flange. 2. Next, the beams will be assumed to be braced laterally at short intervals. 3. Finally, the beams will be assumed to be braced laterally at larger intervals. In the figure at the right a typical curve showing the nominal resisting or buckling moments of a beam with varying unbraced lengths is presented. There are three distinct ranges, or zones, of behavior, depending on the lateral bracing situation. Zone 1: There is continuous or closely spaced lateral bracing; the beam will experience yielding of the entire cross section. Zone 2: As the distance between lateral bracing is increased, the beam begins to fail elastically at smaller moments. Zone 3: With even larger unbraced lengths, the beam will fail elastically at even smaller moments. Plastic Behavior (Zone 1) Consider a compact beam with continuous lateral bracing of the compression flange. The beam could be loaded until its full plastic moment Mp is reached at some point or at multiple points along the beam. Further loading produces a redistribution of moments (or stresses) in the beam. Consider a compact beam with closely spaced intermittent lateral bracing of the compression flange. The beam can be loaded until its full plastic moment Mp is reached if the spacing between braced points does not exceed a certain value, called Lp. - Lp is the limiting laterally unbraced length for the limit state of yielding. - The value of Lp is dependent on the dimensions of the beam cross section and on its yield stress. 6.4

5 Inelastic Buckling (Zone 2) If the spacing between the points of lateral or torsional bracing is increased, the section may be loaded until some, but not all, of the compression fibers are stressed to the yield stress Fy. The section will not have sufficient rotation capacity to permit full moment redistribution. Buckling occurs before the yield stress is reached in some of the compression elements (referred to as inelastic buckling). As the unbraced length is increased, the moment that the section can resist decreases, until finally the beam will buckle before the yield stress is reached anywhere in the cross section. The maximum unbraced length Lr at which the yield stress Fy is reached at one point marks the end of the inelastic range. - Lr is the limiting laterally unbraced length for the limit state of inelastic lateral-torsional buckling. - The value of Lr is dependent on the dimensions of the beam cross section, the yield stress, and residual stresses in the beam. - At this point, as soon as the moment causes the yield stress to be reached at some point in the cross section, the section will buckle. Elastic Buckling (Zone 3) If the unbraced length is greater than Lr, the section will buckle elastically before the yield stress is reached anywhere. As the unbraced length is further increased, the buckling moment becomes even smaller. As the moment is increased in such a beam, the beam will deflect transversely until a critical moment value Mcr is reached. - At this time, the beam cross section will twist and the compression flange will move laterally. 6.3 Design of Compact Laterally Supported Wide-Flange Beams Chapter F of the AISC Specification outlines the provisions for the design of members for flexure. Section F1(2) of the Specification requires that all beams be restrained against twist about their longitudinal axis at support points. 6.5

6 For a compact beam to attain its full plastic moment strength, it must also be laterally supported at certain intervals along its compression flange. When this is the case, such a beam is said to have full lateral support. The nominal strength of a compact member with full lateral support is determined only by the limit state of yielding. - The limit state of lateral torsional buckling does not apply. According to Section F2, if the unbraced length Lb of the compression flange of a compact I-shaped or C-shaped section does not exceed Lp (for elastic analysis), then the member s bending strength about its major axis may be determined using the following equations. Mn = Mp = Fy Zx AISC Equation F2-1 LRFD design strength: φb Mn = φb Fy Zx where φb = 0.90 ASD allowable strength: Mn/Ωb = Fy Z/Ωb where Ωb = 1.67 When a conventional elastic analysis is used to establish member forces, the unbraced length Lb may not exceed the value Lp if the nominal moment Mn is to equal Fy Zx. - Lb is the length between points that are either braced against lateral displacement of the compression flange or braced against twist of the cross section. - The value of Lp is determined by the AISC Specification using the following equation. Lp = 1.76 ry (E/Fy) 1/2 AISC Equation F2-5 Values of Lp are listed in Table 3-2 of the AISC Manual for W-shapes. There is no limit of the unbraced length for circular or square cross sections or for I-shaped beams bent about their minor axis. If I-shaped sections are bent about their minor (or y-axes), they will not buckle before the full plastic moment Mp about the y-axis is developed, as long as the flange element is compact. In beam design the following items need to be considered: moments, shears, deflections, crippling, lateral bracing for the compression flange, and fatigue. Beams are generally selected to provide sufficient design moment capacities (φbmn or Mn/Ωb) and then checked to determine if any of the other items are critical. 6.6

7 Rather than work with formulas, Part 3 of the AISC Manual simplifies beam design. First, the moment Mu or Ma is computed. Then, a section having the required moment capacity is selected from Table 3-2 of the AISC Manual, entitled W shapes Selection by Zx. - In Table 3-2 W-shapes are sorted in descending order by strong-axis flexural strength and then grouped in ascending order by weight with the lightest W-shape in each range in bold. Two important items should be remembered in selecting shapes. 1. Cost: the cost of steel is based on weight. It is desirable to select the lightest possible shape having the required plastic modulus. - Normally the deeper sections will have the lightest weights for the required plastic modulus. - Depth limitations may be considered if the deeper section causes a problem with headroom or significantly adds to the overall building height. 2. The beam orientation. The plastic modulus values Zx in Table 3-2 are given with respect to the horizontal axis for beams in the upright position. - If a beam is turned on its side, the proper plastic modulus Zy about the y-axis can be found in Table 3-4 of the AISC Manual or in tables giving the dimensions and shape properties in Part 1 of the AISC Manual. Beam Weight Estimates In design, the weight of the beam is usually included in the calculation for the bending moment because the beam must support itself as well as the external loads. Following is a simple method that allows the designer to estimate the beam weight quickly and accurately. 1. First, calculate the maximum bending moment, not including the weight of the beam. 2. Then, pick a beam section from AISC Table 3-2 based on the calculated moment. 3. Recalculate the bending maximum moment using the weight of the selected beam section as the estimated beam weight. - The beam weight will likely be very close to the weight of the member that is selected for the final design. 6.7

8 Example Problems Design of Compact, Laterally Supported Wide-Flange Beams Example Given: Beam loaded as shown. Steel: Fy = 50 ksi The beam is compact. The compression flange is laterally braced. Find: Determine if the beam is adequate to support the loads shown. Check both LRFD and ASD methods. Solution W21 x 44 (Zx = 95.4 in 4 ) LRFD (φb = 0.90) wu = 1.2 D L = 1.2 ( ) (3.0) = 6.05 kips/ft Mu = wu L 2 /8 = 6.05(21) 2 /8 = kip-ft Mn = Fy Zx = 50 (95.4) = 4770 kip-inch (397.5 kip-ft) φb Mn = 0.90 (397.5) = kip-ft > Mu = kip-ft The beam is adequate to support the loads shown. OK ASD (Ωb = 1.67) wa = D + L = ( ) = Ma = wa L 2 /8 = 4.044(21) 2 /8 = kip-ft Mn = Fy Zx = 50 (95.4) = 4770 kip-inch (397.5 kip-ft) Mn/Ωb = 397.5/1.67 = kip-ft > Ma = kip-ft The beam is adequate to support the loads shown. (Same as before.) OK Using AISC Table 3-2, for the W21 x 44: LRFD: φb Mpx = 358 kip-ft > Mu = kip-ft ASD: Mpx/Ωb = 238 kip-ft > Ma = kip-ft (Note: Mpx represents the plastic moment of a section about its x-axis.) These values agree with the preceding calculations. 6.8

9 Example Given: Beam loaded as shown. P L = 36 k P L = 36 k Steel: Fy = 50 ksi The beam is compact. The compression flange is laterally braced. Find: Select the most economical section. Use both LRFD and ASD methods. Solution LRFD Calculate the factored loads (not including the weight of the beam). wu = 1.2 D L = 1.2 (2.0) (0) = 2.40 kips/ft Pu = 1.2 D L = 1.2 (0) (36) = 57.6 kips Calculate the moment (not including the weight of the beam): use AISC Table 3-23 (Cases 1 and 9). Mu = wu L 2 /8 + Pu (L/3) = 2.40(30) 2 / (30/3) = 846 kip-ft Select a beam section from AISC Table 3-2 to use to estimate the beam weight. Select W27 x 84 φbmpx = 915 kip-ft > Mu = 846 kip-ft OK Recalculate the factored uniformly distributed load (including the weight of the beam). wu = 1.2 D L = 1.2 ( ) (0) = 2.50 kips/ft Recalculate the moment (including the weight of the beam): use AISC Table 3-23 (Cases 1 and 9). Mu = wu L 2 /8 + P (L/3) = 2.50(30) 2 / (30/3) = kip-ft Select a beam section from AISC Table 3-2 for final design. Select W27 x 84 φbmpx = 915 kip-ft > Mu = kip-ft OK 6.9

10 ASD Calculate the load combinations (not including the weight of the beam). wa = D + L = = 2.0 kips/ft Pa = D + L = = 36 kips Calculate the moment (not including the weight of the beam): use AISC Table 3-23 (Cases 1 and 9). Ma = wa L 2 /8 + Pa (L/3) = 2.0(30) 2 / (30/3) = 585 kip-ft Select a beam section from AISC Table 3-2 to use to estimate the beam weight. Select W27 x 84 Mpx/Ωb = 609 kip-ft > Ma = 585 kip-ft OK Recalculate the uniformly distributed load combination (including the weight of the beam). wa = D + L = ( ) + 0 = kips/ft Recalculate the moment (including the weight of the beam) using AISC Table 3-23 (Cases 1 and 9). Ma = wa L 2 /8 + P (L/3) = 2.084(30) 2 / (30/3) = kip-ft Select a beam section from AISC Table 3-2 for final design. Select W27 x 84 Mpx/Ωb = 609 kip-ft > Ma = kip-ft OK 6.10

11 6.4 Design of Compact Laterally Unsupported Wide-Flange Beams Most steel beams are used in such a manner that their compression flanges are fully restrained against lateral buckling (Zone 1). Concrete floors are poured in such a manner that the concrete provides lateral support for the compression flange of the beam. If lateral support of the compression flange is not provided by a floor slab, it is possible that such support may be provided with connecting beams or with special members used for the purpose of bracing. Beams that frame into the sides of a beam or girder and are connected to the compression flange can usually be counted on to provide full lateral support at the connection. - If the connection is made primarily to the tension flange, little lateral support is provided to the compression flange. The intermittent welding of metal roof or floor decks to the compression flanges of beams will generally provide sufficient lateral bracing. - The corrugated sheet-metal roofs that are usually connected to the purlins with metal straps probably furnish only partial lateral support. When wood flooring is bolted to supporting steel beams only partial lateral support is provided. If there is doubt in the designer s mind as to the degree of lateral support provided, it is advisable to assume that there is no lateral support. Lateral Torsional Buckling The compression region of a bending member cross section has a tendency to buckle similarly to how a pure compression member buckles. The upper half of the wide flange member in bending acts as a T in pure compression. - The upper half of the wide flange is fully braced about its horizontal axis by the web so it will not buckle vertically. - The upper half of the wide flange may be unbraced about its vertical axis and thus will have a tendency to buckle laterally. The tension region tends to restrain the lateral buckling and, as a result, the beam deflects downward and buckles laterally, causing it to twist. 6.11

12 In order to resist this tendency to buckle, Specification Sections B3.6 and F1(2) require that all bending members be restrained against rotation about their longitudinal axis at their support points. If the beam has sufficient lateral and torsional support along its length, a compact section can develop the yield stress before buckling occurs. If the beam tends to buckle before the yield stress is reached, the nominal moment strength is less than the plastic moment. To ensure that a beam cross section can develop its full plastic moment strength without lateral torsional buckling, Specification Section F2.2 limits the slenderness as follows. where Lb/ry 1.76 (E/Fy) 1/2 Lb = unbraced length of the compression flange ry = radius of gyration for the shape about the y-axis The requirement for attaining the full plastic moment strength is given in Specification F2. where Lb Lp = 1.76 ry (E/Fy) 1/2 AISC Equation F2-5 Lp = the limiting laterally unbraced length for the limit state of yielding Lp is the maximum unbraced length that permits the shape to reach its plastic moment strength. When the unbraced length of a beam exceeds Lp, its strength is reduced because the member will have a tendency to buckle before the plastic moment is reached. Zone 2 When Lp < Lb Lr, the section may be loaded until some, but not all, of the compression fibers are stressed to the yield stress Fy. The nominal moment strength is calculated using the following equation. where Mn = Cb {Mp (Mp 0.7 Fy Sx)[(Lb Lp)/(Lr Lp)]} Mp AISC Equation F2-2 Lr = limiting laterally unbraced length for the limit state of inelastic lateraltorsional buckling 6.12

13 The Specification sets the level of residual stress at 0.3Fy so that only 0.7Fy is available to resist a bending moment elastically. - The definition of plastic moment Mp = Fy Zx for beams in Zone 1 is not affected by residual stresses. The sum of the compressive stresses equals the sum of the tensile stresses in the section and the net effect is, theoretically, zero. Lr is a function of several properties of the section, such as its cross sectional area, modulus of elasticity, yield stress, and warping and torsional properties. - Complex formulas to determine the value of Lr are given in Section F1 of the AISC Specification. - Fortunately, numerical values of Lr (as well as values of Lp) have been determined for sections that are normally used as beams and are given in Table 3-2 of the AISC Manual, entitled W Shapes Selected by Zx. Simplified expressions of AISC Equation F2-2 that follow are presented in the AISC Manual (p. 3-9). LRFD: φbmn = Cb [φb Mpx φb BF (Lb Lp)] φb Mpx Equation 3-4a ASD: Mn/Ωb = Cb [Mpx/Ωb (BF/Ωb) (Lb Lp)] Mpx/Ωb Equation 3-4b where BF = (Mpx 0.7 Fy Sx)/(Lr Lp) The bending factors (BF) represent certain factors of AISC Equation F2-2 as can be seen by comparing the equations. Numerical values (in kips) of the bending factors (BF) for W-shapes are listed in Table 3-2 of the AISC Manual. 6.13

14 Example Problems Moment Capacities, Zone 2 Example Given: Beam section W24 x 62 Steel: Fy = 50 ksi Lb = 8.0 Cb = 1.0 Find: LRFD design moment capacity φbmnx ASD allowable moment capacity Mnx/Ωb Solution From AISC Table 3-2 for the W24 x 62: φbmpx = 574 kip-ft Mpx/Ωb = 382 kip-ft Lp = 4.87 Lr = 14.4 φbbf (LRFD) = 24.1 kips BF/Ωb (ASD) = 16.1 kips Lp < Lb < Lr, thus the beam section falls into Zone 2. LRFD Calculate φbmnx using the simplified expression of AISC Equation F2-2 that is presented in the AISC Manual (p. 3-9). φbmnx = Cb [φbmpx φbbf (Lb Lp)] φbmpx φbmnx = 1.0 [ ( )] = kip-ft < φbmpx = 574 kip-ft OK ASD Calculate Mnx/Ωb using the simplified expression of AISC Equation F2-2 that is presented in the AISC Manual (p. 3-9). Mnx/Ωb = Cb [Mpx/Ωb (BF/Ωb) (Lb Lp)] Mpx/Ωb Mnx/Ωb = 1.0 [ ( )] = kip-ft < Mpx/Ωb = 382 kip-ft OK 6.14

15 Example Given: Beam section W21 x 73 Steel: Fy = 50 ksi Cb = 1.0 Find: LRFD design moment capacity (φbmnx) and ASD allowable moment capacity (Mnx/Ωb) for simple spans of 6 and 12 feet with lateral bracing for the compression flange provided at the end only. Solution From AISC Table 3-2 for the W21 x 73: φb Mpx = 645 kip-ft Mpx/Ωb = 429 kip-ft Lp = 6.39 Lr = 19.2 φbbf (LRFD) = 19.4 kips BF/Ωb (ASD) = 12.9 kips 6 Span Lb < Lp, thus the beam section falls into Zone 1. Using AISC Table 3-2 for the W21 x 73: LRFD: φbmnx = φbmpx = 645 kip-ft ASD: Mnx/Ωb = Mpx/Ωb = 429 kip-ft 12 Span Lp < Lb < Lr, thus the beam section falls into Zone 2. Calculate φbmnx using the simplified expression of AISC Equation F2-2 that is presented in the AISC Manual (p. 3-9). φbmnx = Cb [φbmpx φbbf (Lb Lp)] φbmpx φbmnx = 1.0 [ ( )] = kip-ft < φbmpx = 645 kip-ft OK Calculate Mnx/Ωb using the simplified expression of AISC Equation F2-2 that is presented in the AISC Manual (p. 3-9). Mnx/Ωb = Cb [Mpx/Ωb (BF/Ωb) (Lb Lp)] Mpx/Ωb Mnx/Ωb = 1.0 [ ( )] = kip-ft < Mpx/Ωb = 429 kip-ft OK 6.15

16 Zone 3 When Lb > Lr, the section will buckle elastically before the yield stress is reached anywhere in the section. The nominal moment strength is calculated using the following equation. Mn = Fcr Sx Mp AISC Equation F2-3 Fcr = (Cbπ 2 E)/(Lb/rts) 2 { [Jc/(Sx ho)](lb/rts) 2 } 1/2 AISC Equation F2-4 where rts = effective radius of gyration (listed in AISC Table 1-1) J = torsional constant (listed in AISC Table 1-1) c = 1.0 for doubly symmetrical I-shape ho = distance between flange centroids (listed in AISC Table 1-1) Lateral-torsional buckling will not occur if the moment of inertia of the section about the bending axis is less than or equal to the moment of inertia out of plane. For this reason the limit state of lateral torsional buckling is not applicable for shapes bent about their minor axes, for shapes with Ix Iy, or for circular or square shapes. Furthermore, yielding controls if the section is noncompact. 6.16

17 Example Problem Elastic Buckling, Zone 3 Example Given: Beam section W18 x 97 Steel: Fy = 50 ksi Cb = 1.0 Lb = 38 Find: LRFD design moment capacity (φbmnx) ASD allowable moment capacity (Mnx/Ωb) Solution W18 x 97 (Lr = 30.4, rts = 3.08, ho = 17.7, J = 5.86 in 4, Sx = 188 in 4, Zx = 211 in 3 ) Lb = 38 > Lr = 30.4, thus the beam section falls into Zone 3. Calculate Fcr using AISC Equation F2-4. Fcr = (Cbπ 2 E)/(Lb/rts) 2 { [Jc/(Sx ho)](lb/rts) 2 } 1/2 = 1.0π 2 (29,000)/(38x12/3.08) 2 { [5.86(1.0)/(188)(17.7)](38x12/3.08) 2 } 1/2 = (2.003) = ksi Calculate Mnx using AISC Equation F2-3. Mnx = Fcr Sx = 26.15(188) = kip-inch (409.7 kip-ft) < Mp = Fy Zx = 50 (211) = 10,550 kip-inch (879.2 kip-ft) Compute the LRFD design moment capacity (φbmnx) and the ASD allowable moment capacity (Mnx/Ωb). LRFD (φb = 0.90): φbmnx = 0.90 (409.7) = kip-ft ASD (Ωb = 1.67): Mnx/Ωb = 409.7/1.67 = kip-ft 6.17

18 Moment Gradient A bending moment coefficient Cb is included in formulas to account for the fact that moment is not usually uniform across the entire length of the beam. Cb is called the lateral-torsional buckling modification factor for nonuniform moment diagrams when both ends of the unsupported segment are braced. Lateral buckling may be affected by the end restraint and by the loading conditions of the member. The moment in the unbraced beam in part (a) of the figure below causes a more severe compression situation than does the moment in the unbraced beam in part (b) of the figure. The upper flange of the beam in the figure at the left is in compression for its entire length. In the figure at the right, the length of the column (i.e. the length of the upper flange that is in compression) is much less (in effect, a much shorter column ). The basic moment capacity equations for Zones 2 and 3 were developed for laterally unbraced beams subject to single curvature with Cb = 1.0. Frequently beams are not bent in single curvature (e.g. the fixed beam in part b of the figure above), with the result that they can resist more moment. - To handle this situation, the AISC Specification provides Cb coefficients larger than 1.0 that are to be multiplied by the computed Mn values. A Cb value equal to 1.0 may be used conservatively, but the designer is missing out on the possibility of significant savings in steel weight for some situations. 6.18

19 It is important to note, when using Cb values, the moment capacity obtained by multiplying Mn by Cb may not be larger than the plastic moment Mn of Zone 1, where Mn = Fy Zx. The value of Cb is determined from the following expression. Cb = 12.5 Mmax/(2.5 Mmax + 3 MA + 4 MB + 3 MC) AISC Equation F1-1 where Mmax = the largest moment in an unbraced segment of a beam MA, MB, MC = the moments at the ¼ point, ½ point, and ¾ point of the beam segment, respectively Some typical values for Cb include the following. Cb = 1.14 for a simply supported beam with a uniformly distributed load with bracing only at the supports. Cb = 2.38 for a fixed-end beam with a uniformly distributed load with bracing only at the supports, or with bracing at the supports and mid-span. Cb = 1.92 for a fixed-end beam with a concentrated load at mid-span with bracing only at the supports. Cb = 2.27 for a fixed-end beam with a concentrated load with bracing only at the supports and mid-span. Cb = 1.0 for cantilevers or overhangs with a concentrated load at the free end with bracing only at the fixed support (i.e. the free end is unbraced). Other typical values of Cb are shown in AISC Table 3-1 (p. 3-18) for various loading situations for simply supported beams. Design Charts Fortunately the values of the LRFD design moment capacity (φbmn) and the ASD allowable moment capacity (Mn/Ωb) for sections normally used as beams have been computed by the AISC. Values of φbmn and Mn/Ωb have been plotted for a wide range of unbraced lengths. - These plots are shown as Table 3-10 in the AISC Manual. These diagrams enable a designer to solve any of the problems previously considered in this chapter in a short period of time. The values provided include unbraced lengths in the plastic range (Zone 1), in the inelastic buckling range (Zone 2), and in the elastic buckling range (Zone 3). 6.19

20 The values are plotted for Fy = 50 ksi and Cb = Lp is indicated with a solid circle ( ). - Lr is indicated with a hollow circle ( ). The dashed lines indicate that the sections will provide the necessary moment capacities, but are not economical. When the plotted line is solid, the W-shape for that curve is the lightest cross section for a given combination of available flexural strength and unbraced length. To select a member using AISC Table 3-10, enter the chart with the unbraced length Lb and the LRFD factored design moment Mu or the ASD allowable moment Ma. Example (LRFD solution) Assume that Cb = 1.0, Fy = 50 ksi; using AISC Table 3-10, select a beam with Lb = 18, and Mu = 544 kip-ft. First, proceed up from the bottom of the chart for an unbraced length of 18 until you reach a horizontal line corresponding with φbmn = 544 kip-ft. - Any section to the right and above this intersection point will have a greater design moment capacity. The first sections that are encountered are the W16 x 89 and W14 x 90, shown as dashed lines in this section of the chart. Proceeding further upward and to the right, the first solid line encountered is for a W24 x 84 and represents the lightest satisfactory section. Example (ASD solution) Assume that Cb = 1.0, Fy = 50 ksi; using AISC Table 3-10, select a beam with Lb = 18, and Ma = 370 kip-ft. First, proceed up from the bottom of the chart for an unbraced length of 18 until we reach a horizontal line corresponding with Mn/Ωb = 370 kip-ft. - Any section to the right and above this intersection point will have a greater design moment capacity. In this case, the first section that is encountered is the W24 x 84, shown as a solid line in this section of the chart. To use AISC Table 3-10 when Cb > 1.0, use the following procedure. 1. Compute an effective moment value. For LRFD, (Mu)effective = Mu/Cb For ASD, (Ma)effective = Ma/Cb 6.20

21 2. Use the effective moment value and the unbraced length with AISC Table 3-10 to select a trial section. 3. Determine whether the beam is in Zone 1, 2 or 3 by comparing the unbraced length Lb with Lp and Lr of the selected trial section, then proceed with the necessary checks, as outlined below. For beams in Zone 1, use AISC Table 3-2 to check the moment strength of the selected section: that is, φbmn for LRFD or Mn/Ωb for ASD. The calculated moment Mu for LRFD design, or Ma for ASD design, must not exceed the moment strength of the selected section. For LRFD, Mu φbmn = φbmpx φbmpx = φb Fy Zx For ASD, Ma Mn/Ωb = Mpx/Ωb Mpx/Ωb = Fy Zx/Ωb If the calculated moment (Mu or Ma) is greater than the moment strength of the selected trial section (i.e. φbmn for LRFD, or Mn/Ωb for ASD), then select a larger section from AISC Table 3-2 that provides the required moment strength, so that For LRFD, For ASD, φbmn = φbmpx Mu Mn/Ωb = Mpx/Ωb Ma For beams in Zone 2, use AISC Table 3-2 to check the moment strength of the selected section: that is, φbmn for LRFD or Mn/Ωb for ASD. As an initial check, the calculated moment Mu for LRFD design, or Ma for ASD design, must not exceed the moment strength of the selected section. For LRFD, For ASD, Mu φbmn = φbmpx Ma Mn/Ωb = Mpx/Ωb - Note that φbmpx for LRFD and Mpx/Ωb for ASD are maximum values of the moment strength for the selected section; however, the actual values for the selected beam section may be less than the maximum values listed in AISC Table 3-2 because of the effects of the unbraced length. If the calculated moment (Mu or Ma) is greater than the moment strength of the selected trial section (i.e. φbmn for LRFD, or Mn/Ωb for ASD), then select a larger section from AISC Table 3-2 that may provide the required moment strength, so that For LRFD, For ASD, φbmn = φbmpx Mu Mn/Ωb = Mpx/Ωb Ma Use the simplified expression of AISC Equation F2-2 that is presented in the AISC Manual (p. 3-9) to compute φbmn for LRFD and Mn/Ωb for ASD to verify 6.21

22 the actual moment strength of the selected section and compare with the required moment strength. For LRFD, For ASD, φbmn Mu Mn/Ωb Ma - If the calculated moment (Mu or Ma) is greater than the moment strength of the selected trial section (i.e. φbmn for LRFD, or Mn/Ωb for ASD), then select and check a larger section from AISC Table 3-2 until a section that provides the required moment strength is found. For beams in Zone 3, use AISC Table 3-2 to check the moment strength of the selected section: that is, φbmn for LRFD or Mn/Ωb for ASD. As an initial check, the calculated moment Mu for LRFD design, or Ma for ASD design, must not exceed the moment strength of the selected section. For LRFD, For ASD, Mu φbmn = φbmpx Ma Mn/Ωb = Mpx/Ωb - Note that φbmpx for LRFD and Mpx/Ωb for ASD are maximum values of the moment strength for the selected section; however, the actual values for the selected beam section may be less than the maximum values listed in AISC Table 3-2 because of the effects of the unbraced length. If the calculated moment (Mu or Ma) is greater than the moment strength of the selected trial section (i.e. φbmn for LRFD, or Mn/Ωb for ASD), then select a larger section from AISC Table 3-2 that may provide the required moment strength, so that For LRFD, For ASD, φbmn = φbmpx Mu Mn/Ωb = Mpx/Ωb Ma Use AISC Equations F2-3 and F2-4 to compute the nominal moment strength Mn for the trial section. Compute φbmn for LRFD and Mn/Ωb for ASD to verify the actual moment strength of the selected section and compare with the required moment strength. For LRFD, For ASD, φbmn = φbmpx Mu Mn/Ωb = Mpx/Ωb Ma - If the calculated moment (Mu or Ma) is greater than the moment strength of the selected trial section (i.e. φbmn for LRFD, or Mn/Ωb for ASD), then select and check a larger section from AISC Table 3-2 until a section that provides the required moment strength is found. 6.22

23 Example Problem Design Charts with Cb > 1.0 Example Given: Beam loaded as shown. Steel: Fy = 50 ksi Bracing is provided only at the ends and center line of the member: Cb = 1.67 Find: Select the lightest member using both the LRFD and ASD methods. Solution LRFD Calculate the factored loads (assume a beam weight of 84 lb/ft). wu = 1.2 D L = 1.2 ( ) (0) = kip/ft Pu = 1.2 D L = 1.2 (30) (40) = 100 kips Calculate the moment using AISC Table 3-23 (Cases 1 and 7). Mu = wul 2 /8 + PuL/4 = (34) 2 / (34)/4 = kip-ft Since Cb > 1.0, calculate the effective moment. (Mu)effective = Mu/Cb = 864.6/1.67 = 518 kip-ft Enter AISC Table 3-10 (Lb = 17, φbmn = 518 kip-ft) and select a trial beam section. Select W24 x 76 Check the selected trial section. From Table 3-2: W24 x 76 φbmpx = 750 kip-ft < Mu = kip-ft Check the next larger section. From Table 3-2: for W24 x 84 φbmpx = 840 kip-ft < Mu = kip-ft Check the next larger section. From Table 3-2: for W27 x 84 φbmpx = 915 kip-ft > Mu = kip-ft NG NG May be OK Verify the actual moment strength of the selected section. From Table 3-2: W27 x 84 (φbmpx = 915 kip-ft, φbbf = 26.4, Lp = 7.31, Lr = 20.8 ) Lp = 7.31 < Lb = 17 < Lr = 20.8, thus the beam section falls into Zone

24 Calculate φbmnx using the simplified expression of AISC Equation F2-2 that is presented in the AISC Manual (p. 3-9). φbmnx = Cb [φbmpx φbbf (Lb Lp)] φbmpx φbmnx = 1.67 [ ( )] = kip-ft > φbmpx = 915 kip-ft (Use φbmnx = 915 kip-ft) φbmnx = 915 kip-ft > Mu = kip-ft Select W27 x 84 OK ASD Calculate the load combinations (assume a beam weight of 84 lb/ft). wa = D + L = ( ) + 0 = kip/ft Pa = D + L = = 70 kips Calculate the moment using AISC Table 3-23 (Cases 1 and 7). Ma = wal 2 /8 + PaL/4 = (34) 2 /8 + 70(34)/4 = kip-ft Since Cb > 1.0, calculate the effective moment. (Ma)effective = Ma/Cb = 607.1/1.67 = 363 kip-ft Enter AISC Table 3-10 (Lb = 17, Mn/Ωb = 363 kip-ft) and select a trial beam section. Select W24 x 84 Check the selected trial section. From Table 3-2: W24 x 84 Mpx/Ωb = 559 kip-ft < Ma = kip-ft NG Check the next larger section. From Table 3-2: W27 x 84 Mpx/Ωb = 609 kip-ft > Ma = kip-ft May be OK Verify the actual moment strength of the selected section. From Table 3-2: W27 x 84 (Mpx/Ωb = 609 kip-ft, BF/Ωb = 17.6, Lp = 7.31, Lr = 20.8 ) Lp = 7.31 < Lb = 17 < Lr = 20.8, thus the beam section falls into Zone 2. Calculate Mnx/Ωb using the simplified expression of AISC Equation F2-2 that is presented in the AISC Manual (p. 3-9). Mnx/Ωb = Cb [Mpx/Ωb (BF/Ωb) (Lb Lp)] Mpx/Ωb Mnx/Ωb = 1.67 [ ( )] = kip-ft > Mpx/Ωb = 609 kip-ft (Use Mnx/Ωb = 609 kip-ft) Mnx/Ωb = 609 kip-ft > Ma = kip-ft Select W27 x 84 OK 6.24

25 6.5 Design of Noncompact Beams Local Buckling Local buckling occurs when a compression element of a cross section buckles under the load before that element reaches the yield stress. The section is not compact and the strength of the member is less than the plastic moment Mp. Local buckling failures occur when the flange or web are slender. The projecting flange of a wide-flange member is considered an unstiffened element because the web supports only one edge, while the other edge is unsupported and free to rotate. The wide-flange web is connected at both its edges to the flanges, so it is considered a stiffened element. Table B4.1b of the Specification provides the limiting slenderness values λp for the flange and web to ensure that the full plastic moment strength can be reached. When both the flange and web have slenderness ratios (b/tf and h/tw, respectively) less than or equal to the λp values given in the table, the shapes are called compact shapes. - A compact section is one that has a sufficiently stocky cross section so that it is capable of developing a fully plastic stress distribution (assuming its compression flange has sufficient lateral bracing) before something buckles (web or flange). If either element exceeds this value, the shape cannot be called compact and the nominal strength must be reduced. - A noncompact section is one for which the yield stress can be reached in some, but not all, of its compression elements before buckling occurs. - A noncompact section is not capable of reaching a fully plastic stress distribution. For the flange of a W-shape to be compact, its width-to-thickness ratio must satisfy the following limit. λpf = bf/2tf λp = 0.38 (E/Fy) 1/2 Case 10, Table B4.1b λp = limiting slenderness parameter for a compact element For the web of a W-shape to be compact, its width-to-thickness ratio must satisfy the following limit. λpw = h/tw λp = 3.76 (E/Fy) 1/2 Case 15, Table B4.1b 6.25

26 Using common A992 steel with Fy = 50 ksi, these limits are as follows. For a compact flange: λpf = bf/2tf λp = 9.15 For a compact web: λpw = h/tw λp = 90.6 Comparing these limiting values with data given in AISC Manual Table 1-1, the majority of W-shapes have compact flanges and all have compact webs for Fy = 50 ksi. Flange Local Buckling The full range of nominal moment strength Mn of a cross section can be expressed as a function of flange slenderness λf. The three regions shown in Figure 6.18 (p. 190 of the textbook) identify three types of behavior. The first region represents plastic behavior and the shape is capable of attaining its full plastic moment strength (λf λp). - Shapes that fall into this region are called compact. The behavior exhibited in the middle region is inelastic and the shape is not capable of attaining its full plastic moment strength (i.e. λp < λf λr). - Shapes that fall into this category are called noncompact. Shapes that fall into the last region exhibit elastic buckling (λf > λr). - These shapes are called slender shapes. If a section has noncompact flanges (i.e. λp < λ λr), the value of Mn is given by the following equation. Mn = {Mp (Mp 0.7 Fy Sx) [(λ λpf)/(λrf λpf)]} AISC Equation F3-1 Values of λp and λr for different shapes are listed in Table B4.1b of the AISC Specification. If a section has slender flanges (i.e. λ > λr), the value of Mn is given by the following equation. Mn = 0.9 E kc Sx/λ 2 AISC Equation F

27 where λ = bf/2tf kc = 4/(h/tw) 1/2 (where 0.35 kc 0.76) Using common A992 steel with Fy = 50 ksi, the upper limit for the noncompact flange is as follows. λrf = 1.0 (E/Fy) 1/2 = Comparing this limiting value with values of bf/2tf given in AISC Manual Table 1-1, there are no W-shapes with flanges that exceed this limit. All wide-flange shapes have either compact or noncompact flanges (i.e. none have flanges with slender elements). Only a few W10 shapes have noncompact flanges. More generally, almost all of the standard hot-rolled W, M, S, and C shapes listed in the AISC Manual are compact, and none of them fall into the slender classification. All standard hot-rolled shapes have compact webs, but a few have noncompact flanges. - If a standard shape has a noncompact flange, it is so indicated in the various tables of the AISC Manual with a f footnote (e.g. W10 x 12, p. 3-27). - Where applicable, the numerical values shown in the tables are based on the reduced stresses caused by the non-compactness. The equations mentioned here were used, where applicable, to obtain the values used for the charts plotted in Table 3-10 of the AISC Manual. The designer should have little trouble with noncompact sections when Fy is no more than 50 ksi. However, the designer will have to use the formulas presented in Section F3 of the AISC Manual for shapes with Fy values larger than 50 ksi. Web Local Buckling Comparing the slenderness criteria for local web buckling (Case 15, Table B4.1b) with values of h/tw give in AISC Manual Table 1-1, all wide-flange shapes have compact webs. The consideration of noncompact W-shapes is a consideration of only flange local buckling and W-shapes contain no slender elements. Slender webs are a consideration for built-up members, such as plate girders. 6.27

28 Example Problem Noncompact Sections Example Given: Beam section consisting of a W12 x 65 with full lateral bracing. Steel: Fy = 50 ksi Find: The LRFD flexural design stress and the ASD allowable flexural stress. Solution W12 x 65 (bf = 12.0, tf = 0.605, bf/2tf = 9.92, Sx = 87.9 in 3, Zx = 96.8 in 3 ) Determine whether the flange is compact. Reminder: All standard hot-rolled shapes have compact webs, but a few have noncompact flanges. From AISC Table B4.1b (Case 10): λp = 0.38 (E/Fy) 1/2 = 0.38 (29,000/50) 1/2 = 9.15 λr = 1.0 (E/Fy) 1/2 = 1.0(29,000/50) 1/2 = From AISC Table 1-1: λ = bf/2tf = 9.92 λp = 9.15 < λ = 9.92 < λr = 24.08, so the flange is noncompact. Calculate the nominal flexural stress. Mp = Fy Zx = 50 (96.8) = 4840 kip-inch Mn = {Mp (Mp 0.7 Fy Sx) [(λ λpf)/(λrf λpf)]} AISC Equation F3-1 = {4840 [ (50) (87.9)] [( )/( )]} = kip-inch (395.7 kip-ft) Determine φbmn and Mn/Ωb. LRFD (φb = 0.90): φbmn = 0.90 (395.7) = 356 kip-ft ASD (Ωb = 1.67): Mn/Ωb = 395.7/1.67 = 237 kip-ft These values correspond with the values listed in Table 3-2 of the AISC Manual. 6.28

29 6.6 Design of Beams for Weak Axis Bending Beams are not normally oriented for bending about the weak axis. However, beams occasionally must resist bending about the weak axis due to lateral loads. The design of I-shaped beams for weak axis bending is relatively easy. Section F6 of the Specification addresses I-shaped members and channels bent about their weak axis. Two limit states are identified by the Specification: yielding and flange local buckling. - For the limit state of yielding, the following equation is used. Mn = Mp = Fy Zy 1.6 Fy Sy Equation F6-1 - For sections with compact flanges the limit state of flange local buckling does not apply. - For the few W-shapes with noncompact flanges, Equation F6-2 is used to determine the nominal flexural strength Mn. Equation F6-2 is similar to Equation F3-1, except Sy is used in lieu of Sx. - For sections with slender flanges (e.g. built-up members), Equation F6-3 is used to determine the nominal flexural strength Mn. 6.7 Design of Beams for Shear As a member bends, shear stresses occur because of the changes in the length of the longitudinal fibers. Generally shear is not a problem in steel beams. The webs of rolled shapes are capable of resisting large shear forces, except in the following cases. 1. If large concentrated loads are placed near beam supports, these loads will cause large internal forces without corresponding increases in bending moment. Example: Upper columns may be offset with respect to the columns below. 2. Shear may be a problem if two members (such as a beam and a column) are rigidly connected so that their webs are in the same plane. Example: The junction of columns and beams (or rafters) in rigid frame structures. 3. Shear may be a problem if beams are notched or coped or where holes are cut in beam webs for ductwork. 6.29

30 4. Very heavily loaded beams can have excessive shears. 5. Shear may be a problem for ordinary loadings when very thin webs are used, as in plate girders. The LRFD design shear strength, φvvn, and the ASD allowable shear strength, Vn/Ωv, are determined according to the provisions of Chapter G of the AISC Specification. LRFD: ASD: Vu φvvn Va Vn/Ωv The nominal shear strength Vn of unstiffened or stiffened webs is specified as Vn = 0.6 Fy Aw Cv AISC Equation G2-1 where Aw = area of the web = d tw Cv = web shear coefficient (defined below) d = overall depth of the member tw = web thickness For the webs of rolled I-shaped members, when h/tw 2.24 (E/Fy) 1/2 Cv = 1.0 φv = 1.00 and Ωv = 1.50 Almost all current W, S, and HP shapes for Fy = 50 ksi meet this criteria. Exceptions are listed by User Note in Section G2 of the AISC Specification. For webs of all other doubly symmetric shapes, singly symmetric shapes, and channels (not including round HSS), Cv is determined from the following equations. a. When h/tw 1.10 (kv E/Fy) 1/2 then Cv = 1.0 AISC Equation G2-3 kv = web plate buckling coefficient (defined by AISC Section G2.1) For webs without transverse stiffeners and with h/tw < 260, kv = 5 (except for the stem of tees) kv = 1.2 for the stem of tees For webs with transverse stiffeners, kv = 5 + 5/(a/h) 2 kv = 5 when a/h > 3.0 or a/h > [260/(h/tw)]

31 where a = clear distance between transverse stiffeners h = for rolled shapes, the clear distance between flanges less the fillet or corner radii b. When 1.10 (kv E/Fy) 1/2 < h/tw 1.37 (kv E/Fy) 1/2 then Cv = 1.10 (kv E/Fy) 1/2 /(h/tw) AISC Equation G2-4 c. When h/tw > 1.37 (kv E/Fy) 1/2 then Cv = 1.51 kv E/[(h/tw) 2 Fy] AISC Equation G2-5 For all other shapes other than rolled I-shaped members, φv = 0.90 (LRFD) and Ωv = 1.67 (ASD). Notes 1. The values of φvvnx and Vnx/Ωv with Fy = 50 ksi are given for W-shapes in Table 3-2 of the AISC Manual. 2. AISC Table 3-6 is provided for determining the maximum uniform load each W shape can support for various spans. The values are based on Fy = 50 ksi. The values are controlled by maximum moments or shears, as specified by LRFD or ASD. If Vu or Va for a particular beam exceeds the AISC specified shear strength of the member, the usual procedure is to select a slightly heavier section. If a much heavier section is required than what is required for moment, doubler plates may be welded to the beam web, or stiffeners may be connected to the webs in zones of high shear. - Doubler plates must meet the width-thickness requirements for compact stiffened elements (ref. Section B4 of the AISC Specification). - Doubler plates must be welded sufficiently to the member webs to develop their proportionate share of the load. The AISC specified shear strengths of a beam or girder are based on the entire area of the web. Sometimes, however, a connection is made to only a portion of the web. - For such a case, the designer may decide to assume that the shear is spread over only part of the web depth for purposes of computing shear strength. 6.31

32 - Thus, the designer may compute Aw as being equal to tw times the smaller depth for use in the shear strength expression. When beams that have their top flanges at the same elevations (the usual situation) are connected to each other, it is frequently necessary to cope one of them as shown in the figure at the right. For such cases, there is a distinct possibility of a block shear failure along the broken lines shown. 6.32

33 Example Problem Shear Example Given: Beam (W21 x 55) loaded as shown. Steel: Fy = 50 ksi Find: Check the adequacy of the beam in shear. Solution W21 x 55 (d = 20.8, tw = 0.375, h/tw = 50.0) h/tw = 50.0 < 2.24 (E/Fy) 1/2 = 2.24 (29,000/50) 1/2 = and Cv = 1.0 Vn = 0.6 Fy Aw Cv = 0.6 (50)(20.8 x 0.375)(1.0) = kips Equation G2-1 LRFD (φv = 1.00) wu = 1.2 D L = 1.2 (2.0) (4.0) = 8.80 kips/ft Vu = wu L/2 = 8.80(20)/2 = 88.0 kips φvvn = 1.00 (234.0) = kips > Vu = 88.0 kips OK The beam is adequate for shear. ASD (Ωv = 1.50) wa = D + L = = 6.0 kips/ft Va = wa L/2 = 6.0(20)/2 = 60.0 kips Vn/Ωv = 234.0/1.50 = kips > Va = 60.0 kips The beam is adequate for shear. OK The values computed above for φvvn and Vn/Ωv correspond with the values listed Table 3-2 of the AISC Manual. LRFD: φvvnx = 234 kips ASD: Vnx/Ωv = 156 kips 6.33

34 6.8 Continuous Beams Beams that span over more than two supports are called continuous beams. Continuous beams are statically indeterminate and must be analyzed using more than just the equations of equilibrium. The AISC Manual includes shears, moments, and deflections for several continuous beams with various load patterns in AISC Table The ductile nature of steel permits steel members to redistribute load. When one section of a member becomes overloaded, the member can redistribute a portion of its load to a less loaded section. Advanced analysis methods (e.g. plastic analysis) may be used in design through the provisions of Appendix 1 of the Specification. - If plastic analysis is used, this advantage is automatically included in the analysis. Appendix Section 1.3 also permits use of the simplified plastic analysis approach for continuous beams. To allow the designer to take advantage of some of the redistribution that is accounted for in plastic analysis, Section B3.7 gives provisions for moment redistribution in beams by a rule of thumb approach that approximates the real plastic behavior if an elastic analysis is used. Design of beams and girders that are compact and have sufficiently braced compression flanges may take advantage of this simplified redistribution approach. To use the simplified redistribution, for doubly symmetric I-shaped beams the unbraced length of the compression flange Lb must be less than that given in Specification Section F13.5, as determined by the following equation. where Lm = [ (M1/M2)] (E/Fy) ry Equation F13-8 M1 = the smaller moment at the end of the unbraced length M2 = the larger moment at the end of the unbraced length M1/M2 is positive when the moments cause reverse curvature and negative for single curvature For continuous compact section, Section B3.7 of the AISC Specification states: The required flexural strength of beams composed of compact sections, as defined in Section B4.1, and satisfying the unbraced length requirements of 6.34