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Beams Stephen Krone, DSc, PE University of Toledo

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Shear and Moment A beam is a structural member subjected to transverse loads and negligible axial loads.

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Internal Shear Causes Moment These internal shear forces and bending moments cause longitudinal axial stresses and shear stresses in the cross-section.

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Increasing Loads Maximum bending moment is reached at extreme fibers first, where the stress reaches the yield stress limit.

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The Moment - Curvature (M- f ) response With Increasing Moment

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M y is the yield and M p is the plastic moment capacity of the cross-section. The ratio M p to M y is the shape factor f. For a rectangular section, f is equal to 1.5. For a wide-flange section, f is equal to 1.1.

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Developing the Plastic Moment

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LRFD Raised the ASD Limit State

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Columns’ Slenderness Ratio Controls the Critical Buckling Stress

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Lateral Torsional Buckling Behavior Compression flange begins to buckle out of plane. Bracing on the compression flange makes it more difficult to fail.

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Methods of Bracing Compression Flange

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Beam and Column Similarities The design of the beam is usually based on the compression flange. The braced length of support of the compression flange, L b, usually determines the type of failure. There are three zones of failure: plastic, inelastic buckling, and elastic buckling. Each failure zone has its own set of equations.

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Lateral Torsional Buckling: L b L p AISC F2-1 AISC F2-2 M n = F cr S x <= M p AISC F2-3 AISC F2-5 L p = 1.76 r y L r AISC F2-6 AISC 16.1-48 Plastic Inelastic Buckling Elastic Buckling

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Plastic Buckling Most beams fall in the plastic zone. AISC Table 3-10 Fully braced reaches full M p. This is closely spaced lateral bracing, where L b < L p AISC F2-1 AISC F2-2 M n = F cr S x <= M p AISC F2-3

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Plastic Section Modulus

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M p, Plastic Moment and Plastic Section Modulus

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W Shapes – Braced => Mp

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M p, Plastic Moment Concrete floor slab provides lateral bracing. L b = 0.

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Beam Example: Compact & Braced Design a simply supported beam subjected to uniformly distributed dead load of 450 lbs/ft. and a uniformly distributed live load of 550 lbs/ft. The dead load does not include the self- weight of the beam. Step I. Calculate the factored design loads (without self-weight). w U = 1.2 w D + 1.6 w L = 1.42 kips / ft. M U = w u L 2 /8 = 1.42 x 30 2 /8 = 159.75 kip-ft.

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Beam Example (Cont.) Step II. Select the lightest section from the AISC Manual design tables. From the AISC manual, select W16 x 26 made from 50 ksi steel with f b M p = 166.0 kip-ft. Step III. Add self-weight of designed section and check design w sw = 26 lbs/ft Therefore, w D = 476 lbs/ft = 0.476 lbs/ft. w u = 1.2 x 0.476 + 1.6 x 0.55 = 1.4512 kips/ft. Therefore, Mu = 1.4512 x 302 / 8 = 163.26 kip-ft. < f b M p of W16 x 26. OK!

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Step IV. Check deflection at service loads. w = 0.45 + 0.026 + 0.55 kips/ft. = 1.026 kips/ft. D = 5 w L 4 / (384 E Ix) = 5 x(1.026/12) x (30 x 12) 4 /(384 x 29000 x 301) D = 2.142 in. > L/360 Step V. Redesign with service-load deflection as design criteria L /360 = 1.0 in. > 5 w L 4 /(384 E I x ) Therefore, I x > 644.8 in 4 Beam Example (Cont.)

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Example (Cont.) Select the section from the moment of inertia tables in the AISC Table 3-3. (p. 3-21) Ix > 644.8 in 4 Select W21 x 44. W21 x 44 with Ix = 843 in 4 and f b M p = 358 kip-ft. (50 ksi steel). Deflection at service load D = 0.765 in. < L/360 - OK!

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Step VI. Check shear for W16 x 26. w u = 1.2 w D + 1.6 w L = 1.42 kips / ft. V = wl/2 = 1.42(30)/2 = 21.3 kips V n = 0.6F y A w C v = 0.6 (50ksi)(0.25x15.7)(1.0) = 118 kips > 21.3 kips V n can be compared with Z Table (3-2) of f v V nx = 106.0 kips > 21.3 kips Beam Example (Cont.)

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Non-plastic Failure The development of a plastic stress distribution over the cross-section can be hindered by two different length effects: (1)Local buckling of the individual plates (flanges and webs) of the cross-section fail before they develop the compressive yield stress s y. (2) Lateral-torsional buckling of the unsupported length of the beam / member fail before the cross-section develops the plastic moment Mp.

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Lateral-Torsional Buckling L b of a beam-member can undergo lateral-torsional buckling due to the applied flexural loading (bending moment). Lateral-torsional buckling is similar to the flexural buckling or flexural-torsional buckling of a column subjected to axial loading. There is one very important difference. For a column, the axial load causing buckling remains constant along the length. But, for a beam, usually the lateral-torsional buckling causing bending moment varies along the unbraced length.

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LTB: Lateral-Torsional Buckling b Mn is a function of: –Beam section properties –(Zx, ry, x1, x2, Sx, G, J, A, Cw, Iy) –Unbraced length, Lb –L b = distance between points which are either braced against lateral displacement of compression flange or braced against twist of the cross section –Lp : Limiting L b for full plastic bending capacity –Lr : Limiting L b for inelastic LTB E : Modulus of elasticity, ksi C b : Bending coefficient based on moment gradient.

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Lateral Torsional Buckling: L p L b L r If the laterally unbraced length L b is less than or equal to a plastic length L p then lateral torsional buckling is not a problem and the beam will develop its plastic strength Mp. - for I members & Channels (See Pg. 16.1-48, F2-5) If L b is greater than L p then lateral torsional buckling will occur and the moment capacity of the beam is reduced below the plastic strength Mp. L p = 1.76 r y

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Lateral Torsional Buckling: L p L b L r AISC F2-1 AISC F2-2 M n = F cr S x <= M p AISC F2-3 AISC F2-5 L p = 1.76 r y L r AISC F2-6 AISC 16.1-48

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Lateral Torsional Buckling: L p L b L r When L p L b L r, then the lateral-torsional buckling Mn is given by: M n = C b [M p – (M p -0.7F y S x )(L b - L p / L r –L p )] Linear interpolation between (L p, M p ) and (L r, M r ) AISC Table 3-2 - L p, L r, M p, M r MpMp MrMr LpLp LbLb LrLr

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As the spacing of lateral bracing increases but not all compression fibers will reach F y. This is referred to as inelastic buckling. Easier to use BF: –M n = M p – (BF )(L b - L p ) L r is the point where the transition is to elastic buckling. Lateral Torsional Buckling: L p L b L r

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Lateral Torsional Buckling: L b > L r AISC F2-1 AISC F2-2 AISC F2-3 Elastic Buckling The buckling moment becomes smaller and smaller as the unbraced length increases. AISC p. 16.1-47

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Lateral Torsional Buckling: L b > L r AISC F2-1 AISC F2-2 M n = F cr S x <= M p AISC F2-3 AISC F2-5 L p = 1.76 r y L r AISC F2-6 AISC 16.1-48 Plastic Inelastic Buckling Elastic Buckling

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Compact, Non-Compact, and Slender Slender sections cannot develop Mp due to elastic local buckling. Non-compact sections can develop My but not Mp before local buckling occurs. Only compact sections can develop the plastic moment Mp. Applies to major and minor axis bending

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Local Buckling of Flange Due to Compressive Stress Equations for local buckling of steel plates have limiting slenderness ratios for the individual plate elements of the cross-sections. AISC B4 (page 16.14), Table B4.1 (16.1-13) and Page 16.1-223 Steel sections are classified as compact, non-compact, or slender depending upon the slenderness (l) ratio of the individual plates of the cross-section.

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Local Buckling

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Limit States for Local Buckling Table B4.1 (p 16.1-16)

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Strength Limit State for Local Buckling = Slenderness parameter; must be calculated for flange and web buckling (actually no web slender for W, M, C) p = limiting slenderness parameter for compact element r = limiting slenderness parameter for non-compact element p : Section capable of developing fully plastic stress distribution

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Strength Limit State for Local Buckling p r : Section capable of developing yield stress before local buckling occurs; will buckle before fully plastic stress distribution can be achieved. r : Slender compression elements; will buckle elastically before yield stress is achieved.

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Compact, Non Compact, or Slender For inelastic buckling because of unbraced length: M n = M p – (M p -0.7F y S x )(L b - L p / L r –L p )] For Flange Local Buckling because flange in not compact: p r - p M n = M p – (M p -0.7F y S x )( - p / r - p ) AISC F2-2 AISC F3-1

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The Bending Coefficient, C b Worse case scenario is for compression flange under constant uniform moment Moments usually vary over unbraced length, so AISC uses C b to modify for only segment that undergoes LTB under the highest moment. This factor is similar to the effective length K that we modified in column buckling. –(Old formula like G/G) –C b =1.75 + 1.05(M 1 /M 2 )+0.3(M 1 /M 2 )

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C b can always be taken as 1. C b = 12.5 M max /(2.5M max +3M a +4M b + 3M c ) M max = maximum value in braced section M a = value at quarter point M b = value at center point M c = value at three quarter point The Bending Coefficient, C b

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The AISC Specification says that: C b is a multiplier. C b is always greater than 1.0 for non- uniform bending moment. –C b is equal to 1.0 for uniform bending moment. –If you cannot calculate or figure out C b, then it can be conservatively assumed as 1.0.

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Moment Capacity versus L b (for non-uniform moment case)

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Bending Coefficient, C b AISC Table 3-1

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Example Design the beam shown below. The unfactored uniformly distributed live load is equal to 3 kips/ft. There is no dead load. Lateral support is provided at the end reactions.

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Step I. Calculate the factored loads assuming a reasonable self-weight. Assume self-weight = w sw = 100 lbs/ft. Dead load = w D = 0 + 0.1 = 0.1 kips/ft. Live load = w L = 3.0 kips/ft. Ultimate load = –w u = 1.2w D + 1.6w L = 4.92 kips/ft. Factored ultimate moment = –Mu = w u L 2 /8 = 354.24 kip-ft.

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Step II. Determine unsupported length L b & C b One unsupported span with L b = 24 ft. C b = 1.14 for the parabolic bending moment diagram, See values of C b shown in Table 3-1. Step III. Select a wide-flange shape The moment capacity of the selected section f b Mn > Mu (Note f b = 0.9) f b Mn = moment capacity = C b x ( f b Mn for the case with uniform moment) f b Mn Table 3-10 in the AISC-LRFD manual shows plots of f b Mn - L b for the case of uniform bending moment (C b =1.0)

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Therefore, in order to select a section, calculate Mu/ C b and use it with L b to find the first section with a solid line. Mu/ C b = 354.24/1.14 = 310.74 kip-ft. Select W16 x 67 (50 ksi steel) f b Mn =357 kip-ft. for L b = 24 ft. and C b =1.0 For the case with C b = 1.14, f b Mn = 1.14 x 357 = 406.7 kip-ft., which must be f b Mn = 491 kip-ft. OK! Thus, W16 x 67 made from 50 ksi steel with moment capacity equal to 406.7 kip-ft. for an unsupported length of 24 ft. is the designed section.

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Step IV. Check for local buckling. Also found in footnotes of sections. l = b f / 2t f = 7.7; l p = 0.38 (E/Fy)0.5 = 9.192 Therefore, l < l p - compact flange l = h/t w = 34.4; l p = 3.76 (E/Fy)0.5 = 90.5 Therefore, l < l p- compact web

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Step V. Check shear. (W16 x 67 made from 50 ksi steel) w u = 1.2w D + 1.6w L = 4.92 kips/ft. V = wl/2 = 4.92(24)/2 = 59 kips V n = 0.6F y A w C v = 0.6 (50ksi)(0.395x16.3)(1.0) = 193 kips OK

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Step VI. Check deflection. (W16 x 67 made from 50 ksi steel) D = Wl 3 /384 EI D = (4.92 klf x 24ft)(24 x 12) 3 /384(29,000 ksi)(954 in 4 ) D = 0.27in D allow = L /360 = 24 x 12/ 360 = 0.8 in. OK

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Summary LRFD Beam Design 1.Determine the location and magnitude of the loads and draw a load diagram. 2.Determine if the shape is compact by calculating slenderness parameter p, p Full plastic moment, M p p r p r - p p r M n = M p – (M p 0.7F y S x )( - p / r - p ) > r > r M n = F cr S x

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Summary LRFD Beam Design Check Shape: If the shape is non compact because of the flange, (calculate slenderness parameter) p, p Full plastic moment, M p p r p r - p p r M n = M p – (M p 0.7F y S x )( - p / r - p ) > r > r M n = F cr S x

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Summary LRFD Beam Design 3. Find Moment - Lateral Torsional Buckling AISC F2-1 AISC F2-2 M n = F cr S x <= M p AISC F2-3 AISC F2-5 L p = 1.76 r y L r AISC F2-6 AISC 16.1-48

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Summary LRFD Beam Design 4. Check shear in the web AISC G2-3 AISC G2-1 AISC G2-3 and AISC G2-3 h/t w = 2.24 Most cases with W shapes f v = 1 and C v = 1, f v = 0.9

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Summary LRFD Beam Design 5. Check deflection.

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Summary Local buckling is not an issue for most beams for A992 beams Effective bracing is provided at ends to restrain rotation about the longitudinal axis. Effective bracing reduces L b and prevents twist of the cross section and/or lateral movement of the compression flange. C b is important in economical design when plastic moment is not developed

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Works Cited Segui, William T. 2007. Steel Design. 4th ed. Thompson Engineering. Steel Construction Manual, 13th ed. AISC, 2005. CE 405: Design of Steel Structures –Dr. A. Varma

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