Report

62323: Architectural Structures II Design of Beam-Columns Monther Dwaikat Assistant Professor Department of Building Engineering An-Najah National University 68402 Slide # 1 Beam-Column - Outline Beam-Columns Moment Amplification Analysis Braced and Unbraced Frames Analysis/Design of Braced Frames Design of Base Plates 68402 Slide # 2 Design for Flexure – LRFD Spec. Commonly Used Sections: • I – shaped members (singly- and doubly-symmetric) • Square and Rectangular or round HSS 68402 Slide # 3 Beam-Columns Likely failure modes due to combined bending and axial forces: • • • • • • Bending and Tension: usually fail by yielding Bending (uniaxial) and compression: Failure by buckling in the plane of bending, without torsion Bending (strong axis) and compression: Failure by LTB Bending (biaxial) and compression (torsionally stiff section): Failure by buckling in one of the principal directions. Bending (biaxial) and compression (thin-walled section): failure by combined twisting and bending Bending (biaxial) + torsion + compression: failure by combined twisting and bending 68402 Slide # 4 Beam-Columns Structural elements subjected to combined flexural moments and axial loads are called beam-columns The case of beam-columns usually appears in structural frames The code requires that the sum of the load effects be smaller than the resistance of the elements Q i i Rn 1.0 Thus: a column beam interaction can be written as M ux M uy Pu 1.0 c Pn b M nx b M ny This means that a column subjected to axial load and moment will be able to carry less axial load than if no moment would exist. 68402 Slide # 5 Beam-Columns AISC code makes a distinct difference between lightly and heavily axial loaded columns P for u 0.2 c Pn M uy Pu 8 M ux 1.0 c Pn 9 b M nx b M ny AISC Equation P for u 0.2 c Pn M ux M uy Pu 1.0 2c Pn b M nx b M ny AISC Equation 68402 Slide # 6 Beam-Columns Definitions Pu = factored axial compression load Pn = nominal compressive strength Mux = factored bending moment in the x-axis, including second-order effects Mnx = nominal moment strength in the x-axis Muy = same as Mux except for the y-axis Mny = same as Mnx except for the y-axis c = Strength reduction factor for compression members = 0.90 b = Strength reduction factor for flexural members = 0.90 68402 Slide # 7 Beam-Columns The increase in slope for lightly axial-loaded columns represents the less effect of axial load compared to the heavily axial-loaded columns Pu/cPn Unsafe Element Safe Element 0.2 Mu/bMn These are design charts that are a bit conservative than behaviour envelopes 68402 Slide # 8 Moment Amplification When a large axial load exists, the axial load produces moments due to any element deformation. x d P P d M The final moment “M” is the sum of the original moment and the moment due to the axial load. The moment is therefore said to be amplified. As the moment depends on the load and the original moment, the problem is nonlinear and thus it is called second-order problem. 68402 Slide # 9 Braced and Unbraced Frames Two components of amplification moments can be observed in unbraced frames: Moment due to member deflection (similar to braced frames) Moment due to sidesway of the structure Unbraced Frames Member deflection Member sidesway 68402 Slide # 10 Unbraced and Braced Frames In braced frames amplification moments can only happens due to member deflection Braced Frames Sidesway bracing system Member deflection 68402 Slide # 11 Unbraced and Braced Frames Braced frames are those frames prevented from sidesway. In this case the moment amplification equation can be simplified to: M ux B1x M ntx EAg 2 Pe Cm B1 Pu 1 Pe KL / r 1 M uy B1 y M nty AISC Equation 2 KL/r for the axis of bending considered K ≤ 1.0 68402 Slide # 12 Unbraced and Braced Frames The coefficient Cm is used to represent the effect of end moments on the maximum deflection along the element (only for braced frames) Cm M1 0.6 0.4 M 2 M1 ve M2 M1 ve M2 When there is transverse loading on the beam either of the following case applies Conservatively Cm 1.00 68402 Slide # 13 Ex. 5.1- Beam-Columns in Braced Frames A 3.6-m W12x96 is subjected to bending and compressive loads in a braced frame. It is bent in single curvature with equal and opposite end moments and is not loaded transversely. Use Grade 50 steel. Is the section satisfactory if Pu = 3200 kN and first-order moment Mntx = 240 kN.m Step I: From Section Property Table W12x96 (A = 18190 mm2, Ix = 347x106 mm4, Lp = 3.33 m, Lr = 14.25 m, Zx = 2409 mm3, Sx = 2147 mm3) 68402 Slide # 14 Ex. 5.1- Beam-Columns in Braced Frames Step II: Compute amplified moment - For a braced frame let K = 1.0 KxLx = KyLy = (1.0)(3.6) = 3.6 m - From Column Chapter: cPn = 4831 kN Pu/cPn = 3200/4831 = 0.662 > 0.2 Use eqn. - There is no lateral translation of the frame: Mlt = 0 Mux = B1Mntx Cm = 0.6 – 0.4(M1/M2) = 0.6 – 0.4(-240/240) = 1.0 Pe1 = 2EIx/(KxLx)2 = 2(200)(347x106)/(3600)2 = 52851 kN 68402 Slide # 15 Ex. 5.1- Beam-Columns in Braced Frames B1 Cm 1.0 1.073 1.0 P 3200 1 u 1 52851 Pe1 (OK ) Mux = (1.073)(240) = 257.5 kN.m Step III: Compute moment capacity Since Lb = 3.6 m Lp < Lb< Lr b M n 739 kN.m 68402 Slide # 16 Ex. 5.1- Beam-Columns in Braced Frames Step IV: Check combined effect M uy Pu 8 M ux c Pn 9 b M nx b M ny 3200 8 257.5 0 0.972 1.0 4831 9 739 Section is satisfactory 68402 Slide # 17 Ex. 5.2- Analysis of Beam-Column Check the adequacy of an ASTM A992 W14x90 column subjected to an axial force of 2200 kN and a second order bending moment of 400 kN.m. The column is 4.2 m long, is bending about the strong axis. Assume: • • ky = 1.0 Lateral unbraced length of the compression flange is 4.2 m. 68402 Slide # 18 Ex. 5.2- Analysis of Beam-Column Step I: Compute the capacities of the beam-column cPn = 4577 kN Mny = 380 kN.m Mnx = 790 kN.m Step II: Check combined effect Pu 2200 0.481 0.2 c Pn 4577 M uy Pu 8 M ux c Pn 9 b M nx b M ny 2200 8 400 0 0.931 1.0 4577 9 790 OK 68402 Slide # 19 Design of Beam-Columns Trial-and-error procedure • Select trial section • Check appropriate interaction formula. • Repeat until section is satisfactory 68402 Slide # 20 Ex. 5.3 – Design-Beam Column Select a W shape of A992 steel for the beam-column of the following figure. This member is part of a braced frame and is subjected to the service-load axial force and bending moments shown (the end shears are not shown). Bending is about the strong axis, and Kx = Ky = 1.0. Lateral support is provided only at the ends. Assume that B1 = 1.0. 68402 PD = 240 kN PL = 650 kN MD = 24.4 kN.m ML = 66.4 kN.m 4.8 m MD = 24.4 kN.m ML = 66.4 kN.m Slide # 21 Ex. 5.3 – Design-Beam Column Step I: Compute the factored axial load and bending moments Pu = 1.2PD + 1.6PL = 1.2(240)+ 1.6(650) = 1328 kN. Mntx = 1.2MD + 1.6ML = 1.2(24.4)+ 1.6(66.4) = 135.5 kN.m. B1 = 1.0 Mux = B1Mntx = 1.0(135.5) = 135.5 kN.m Step II: compute Mnx, Pn • • • The effective length for compression and the unbraced length for bending are the same = KL = Lb = 4.8 m. The bending is uniform over the unbraced length , so Cb=1.0 Try a W10X60 with Pn = 2369 kN and Mnx = 344 kN.m 68402 Slide # 22 Ex. 5.3 – Design-Beam Column Step III: Check interaction equation Pu 1328 0.56 0.2 c Pn 2369 M uy Pu 8 M ux c Pn 9 b M nx b M ny 1328 8 135.5 0 0.91 1.0 2369 9 344 OK Step IV: Make sure that this is the lightest possible section. Try W12x58 with Pn = 2247 kN and Mnx = 386 kN.m Pu 1328 0.59 0.2 c Pn 2247 M uy Pu 8 M ux c Pn 9 b M nx b M ny 1328 8 135.5 0 0.90 1.0 2247 9 386 Use a W12 x 58 section 68402 Slide # 23 Design of Base Plates We are looking for design of concentrically loaded columns. These base plates are connected using anchor bolts to concrete or masonry footings The column load shall spread over a large area of the bearing surface underneath the base plate AISC Manual Part 16, J8 68402 Slide # 24 Design of Base Plates The design approach presented here combines three design approaches for light, heavy loaded, small and large concentrically loaded base plates Area of Plate is computed such that n m 0.8 bf 0.95d N The dimensions of the plate are computed such that m and n are approximately equal. B Pp Pu where: 0.6 If plate covers the area of the footing PP 0.85 f cA1 If plate covers part of the area of the footing PP 0.85 f cA1 A2 1.7 f cA1 A1 A1 = area of base plate A2 = area of footing f’c = compressive strength of concrete used for footing 68402 Slide # 25 Design of Base Plates Thickness of plate t pl l m l maxn n ' 2 Pu Pu 1.5 l 0.9 B N Fy B NFy N 0.95 d m 2 n B 0.8 b f 1 dbf 4 2 X 1 1 X n ' 2 4dbf Pu X 2 ( d b ) c Pp f However may be conservatively taken as 1 c 0.6 Pp Nominal bearing strength 68402 Slide # 26 Ex. 5.4 – Design of Base Plate • For the column base shown in the figure, design a base plate if the factored load on the column is 10000 kN. Assume 3 m x 3 m concrete footing with concrete strength of 20 MPa. 0.95d W14x211 0.8bf B 68402 Slide # 27 N Ex. 5.4 - Design of Base Plate Step I: Plate dimensions • Assume A2 2 A1 thus: Pp 1.7 f cA1 Pu 0.6 1.7 20 A1 10000103 A1 490.2 103 m m2 • A2 4.28 2 A1 Assume m = n N 0.95d 2m 0.95 399 2m 379 2m B 0.8b f 2m 0.8 401 2m 321 2m • A1 NB 379 2m 321 2m 490.2 103 m 175.4 m m N = 729.8 mm say N = 730 mm B = 671.8 mm say B = 680 mm 68402 Slide # 28 Ex. 5.4 - Design of Base Plate Step II: Plate thickness t p 1.5( m ,n ,or n' ) fp Fy m ( N 0.95d ) / 2 175.5 m m n ( B 0.8b f ) / 2 179.5 m m 1 n' dbf 100m m 4 68402 Slide # 29 Ex. 5.4 - Design of Base Plate Selecting the largest cantilever length 10000103 fp 20.14 MPa 680 730 20.14 t req 1.5(179.5) 76.7 m m 248 use 730 mm x 670 mm x 80 mm Plate 68402 Slide # 30