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Beams Stephen Krone, DSc, PE University of Toledo
Shear and Moment A beam is a structural member subjected to transverse loads and negligible axial loads.
Internal Shear Causes Moment These internal shear forces and bending moments cause longitudinal axial stresses and shear stresses in the cross-section.
Increasing Loads Maximum bending moment is reached at extreme fibers first, where the stress reaches the yield stress limit.
The Moment - Curvature (M- f ) response With Increasing Moment
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.
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.
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
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
M p, Plastic Moment Concrete floor slab provides lateral bracing. L b = 0.
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.
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!
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.)
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!
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.)
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.
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.
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.
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
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
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
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
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
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
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
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.
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
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.
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
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 )
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
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.
Moment Capacity versus L b (for non-uniform moment case)
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.
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.
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)
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.
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
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
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
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
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
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
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
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
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