2 Shear and MomentA beam is a structural member subjected to transverse loads and negligible axial loads.
3 Internal Shear Causes Moment These internal shear forces and bending moments cause longitudinal axial stresses and shear stresses in the cross-section.
4 Increasing LoadsMaximum bending moment is reached at extreme fibers first, where the stress reaches the yield stress limit.
5 The Moment - Curvature (M-f) response With Increasing Moment
6 My is the yield and Mp is the plastic moment capacity of the cross-section. The ratio Mp to My is the shape factor f. For a rectangular section, f is equal to For a wide-flange section, f is equal to 1.1.
12 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, Lb, 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.
18 Mp, Plastic Moment Concrete floor slab provides lateral bracing.
19 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).wU = 1.2 wD wL = 1.42 kips / ft.MU = wuL2/8 = 1.42 x 302/8 = kip-ft.
20 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 fbMp = kip-ft.Step III. Add self-weight of designed section and check designwsw = 26 lbs/ftTherefore, wD = 476 lbs/ft = lbs/ft.wu = 1.2 x x 0.55 = kips/ft.Therefore, Mu = x 302 / 8 = kip-ft. < fbMp of W16 x 26.OK!
21 Beam Example (Cont.) Step IV. Check deflection at service loads. w = kips/ft. = kips/ft.D = 5 w L4 / (384 E Ix)= 5 x(1.026/12) x (30 x 12)4/(384 x x 301)D = in. > L/360Step V. Redesign with service-load deflection as design criteriaL /360 = 1.0 in. > 5 w L4/(384 E Ix)Therefore, Ix > in4
22 Example (Cont.)Select the section from the moment of inertia tables in the AISC Table (p. 3-21)Ix > in4Select W21 x 44.W21 x 44 with Ix = 843 in4 and fbMp = 358 kip-ft. (50 ksi steel).Deflection at service loadD = in. < L/ OK!
23 Beam Example (Cont.) Step VI. Check shear for W16 x 26 . wu = 1.2 wD wL = 1.42 kips / ft.V = wl/2 = 1.42(30)/2 = 21.3 kipsVn = 0.6FyAwCv= 0.6 (50ksi)(0.25x15.7)(1.0)= 118 kips > 21.3 kipsVn can be compared with Z Table (3-2) offvVnx = kips > 21.3 kips
24 Non-plastic FailureThe development of a plastic stress distribution over the cross-section can be hindered by two different length effects:Local buckling of the individual plates (flanges and webs) of the cross-section fail before they develop the compressive yield stress sy.(2) Lateral-torsional buckling of the unsupported length of the beam / member fail before the cross-section develops the plastic moment Mp.
25 Lateral-Torsional Buckling Lb 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.
26 LTB: Lateral-Torsional Buckling bMn is a function of:Beam section properties(Zx, ry, x1, x2, Sx, G, J, A, Cw, Iy)Unbraced length, LbLb = distance between points which are either braced against lateral displacement of compression flange or braced against twist of the cross sectionLp : Limiting Lb for full plastic bending capacityLr : Limiting Lb for inelastic LTBE : Modulus of elasticity, ksiCb: Bending coefficient based on moment gradient.
27 Lateral Torsional Buckling: Lp Lb Lr If the laterally unbraced length Lb is less than or equal to a plastic length Lp then lateral torsional buckling is not a problem and the beam will develop its plastic strength Mp.- for I members & Channels (See Pg , F2-5)If Lb is greater than Lp then lateral torsional buckling will occur and the moment capacity of the beam is reduced below the plastic strength Mp.Lp = 1.76 ry
29 Lateral Torsional Buckling: Lp Lb Lr When Lp Lb Lr, then the lateral-torsional buckling Mn is given by:Mn= Cb[Mp – (Mp -0.7Fy Sx )(Lb- Lp / Lr–Lp)]Linear interpolation between (Lp, Mp) and (Lr, Mr)AISC Table Lp, Lr, Mp, MrMpMrLpLbLr
30 Lateral Torsional Buckling: Lp Lb Lr As the spacing of lateral bracing increases but not all compression fibers will reach Fy.This is referred to as inelastic buckling.Easier to use BF:Mn= Mp – (BF )(Lb- Lp)Lr is the point where the transition is to elastic buckling.
31 Lateral Torsional Buckling: Lb > Lr Elastic BucklingThe buckling moment becomes smaller and smaller as the unbraced length increases.AISC pAISC F2-1AISC F2-2AISC F2-3
33 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
34 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 ( ) and PageSteel sections are classified as compact, non-compact, or slender depending upon the slenderness (l) ratio of the individual plates of the cross-section.
36 Limit States for Local Buckling Table B4.1 (p 16.1-16)
37 Limit States for Local Buckling Table B4.1 (p 16.1-16)
38 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 elementr = limiting slenderness parameter for non-compact element p : Section capable of developing fully plastic stress distribution
39 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.
40 Compact, Non Compact, or Slender For inelastic buckling because of unbraced length:Mn= Mp – (Mp -0.7Fy Sx )(Lb- Lp / Lr–Lp)]For Flange Local Buckling because flange in not compact:Mn= Mp – (Mp -0.7Fy Sx )( - p / r - p)AISC F2-2AISC F3-1
41 The Bending Coefficient, Cb Worse case scenario is for compression flange under constant uniform momentMoments usually vary over unbraced length, so AISC uses Cb 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)Cb = (M1/M2)+0.3(M1/M2 )
42 The Bending Coefficient, Cb Cb can always be taken as 1.Cb =12.5 Mmax/(2.5Mmax+3Ma +4Mb + 3Mc)Mmax = maximum value in braced sectionMa = value at quarter pointMb = value at center pointMc = value at three quarter point
43 The AISC Specification says that: Cb is a multiplier.Cb is always greater than 1.0 for non-uniform bending moment.Cb is equal to 1.0 for uniform bending moment.If you cannot calculate or figure out Cb, then it can be conservatively assumed as 1.0.
44 Moment Capacity versus Lb (for non-uniform moment case)
48 Step II. Determine unsupported length Lb & Cb One unsupported span with Lb = 24 ft.Cb = 1.14 for the parabolic bending moment diagram, See values of Cb shown in Table 3-1.Step III. Select a wide-flange shapeThe moment capacity of the selected sectionfb Mn > Mu (Note fb = 0.9)fbMn = moment capacity = Cb x (fbMn for the case with uniform moment) fbMnTable 3-10 in the AISC-LRFD manual shows plots of fbMn - Lb for the case of uniform bending moment (Cb =1.0)
49 Therefore, in order to select a section, calculate Mu/ Cb and use it with Lb to find the first section with a solid line.Mu/ Cb = /1.14 = kip-ft.Select W16 x 67 (50 ksi steel) fbMn =357 kip-ft. for Lb = 24 ft. and Cb =1.0For the case with Cb = 1.14,fbMn = 1.14 x 357 = kip-ft., which must be fbMn = 491 kip-ft. OK!Thus, W16 x 67 made from 50 ksi steel with moment capacity equal to kip-ft. for an unsupported length of 24 ft. is the designed section.
50 Therefore, l < lp - compact flange Step IV. Check for local buckling. Also found in footnotes of sections.l = bf / 2tf = 7.7; lp = 0.38 (E/Fy)0.5 = 9.192Therefore, l < lp - compact flangel = h/tw = 34.4; lp = 3.76 (E/Fy)0.5 = 90.5Therefore, l < lp - compact web
51 Step V. Check shear. (W16 x 67 made from 50 ksi steel) wu = 1.2wD + 1.6wL = 4.92 kips/ft.V = wl/2 = 4.92(24)/2 = 59 kipsVn = 0.6FyAwCv= 0.6 (50ksi)(0.395x16.3)(1.0)= 193 kipsOK
52 Step VI. Check deflection. (W16 x 67 made from 50 ksi steel) D = Wl3/384 EID = (4.92 klf x 24ft)(24 x 12) 3/384(29,000 ksi)(954 in4)D = 0.27inDallow = L /360 = 24 x 12/ 360 = 0.8 in.OK
53 Summary LRFD Beam Design Determine the location and magnitude of the loads and draw a load diagram.Determine if the shape is compact by calculating slenderness parameter p Full plastic moment, Mpp r Mn= Mp – (Mp 0.7Fy Sx )( - p / r - p) > r Mn= Fcr Sx
54 Summary LRFD Beam Design Check Shape: If the shape is non compact because of the flange, (calculate slenderness parameter) p Full plastic moment, Mpp r Mn= Mp – (Mp 0.7Fy Sx )( - p / r - p) > r Mn= Fcr Sx
58 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 Lb and prevents twist of the cross section and/or lateral movement of the compression flange.Cb is important in economical design when plastic moment is not developed
59 Works CitedSegui, William T Steel Design. 4th ed. Thompson Engineering.Steel Construction Manual, 13th ed. AISC, 2005.CE 405: Design of Steel Structures –Dr. A. Varma