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Monther Dwaikat Assistant Professor Department of Building Engineering

Monther Dwaikat Assistant Professor Department of Building Engineering An-Najah National University. 68402: Structural Design of Buildings II 61420: Design of Steel Structures 62323: Architectural Structures II. Design of Beam-Columns. Beam-Column - Outline. Beam-Columns

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Monther Dwaikat Assistant Professor Department of Building Engineering

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  1. Monther Dwaikat Assistant Professor Department of Building Engineering An-Najah National University 68402: Structural Design of Buildings II61420: Design of Steel Structures62323: Architectural Structures II Design ofBeam-Columns 68402

  2. Beam-Column - Outline • Beam-Columns • Moment Amplification Analysis • Second Order Analysis • Compact Sections for Beam-Columns • Braced and Unbraced Frames • Analysis/Design of Braced Frames • Analysis/Design of Unbraced Frames • Design of Bracing Elements 68402

  3. Design for Flexure – LRFD Spec. • Commonly Used Sections: • I – shaped members (singly- and doubly-symmetric) • Square and Rectangular or round HSS 68402

  4. Beam-Columns 68402

  5. Beam-Columns 68402

  6. Beam-Columns 68402

  7. Beam-Columns 68402

  8. Beam-Columns 68402

  9. Beam-Columns 68402

  10. 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

  11. 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 • Thus: a column beam interaction can be written as • 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

  12. Beam-Columns • AISC code makes a distinct difference between lightly and heavily axial loaded columns AISC Equation AISC Equation 68402

  13. 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

  14. 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 Unsafe Element Pu/fcPn Safe Element 0.2 Mu/fbMn These are design charts that are a bit conservative than behaviour envelopes 68402

  15. P P d M d x Moment Amplification • When a large axial load exists, the axial load produces moments due to any element deformation. • 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

  16. Moment Amplification Second-order Moments, Puδ and Pu Moment amplification in column braced against sidesway Mu = Mnt + Puδ Moment amplification in unbraced column Mu = Mlt + Pu 68402

  17. Moment Amplification • Using first principles we can prove that the final moment Mmax is amplified from M0 as • The amplification factor B can be Where 68402

  18. Second Order Analysis 68402

  19. Second Order Analysis 68402

  20. Second Order Analysis 68402

  21. Second Order Analysis 68402

  22. Second Order Analysis 68402

  23. Second Order Analysis 68402

  24. Second Order Analysis 68402

  25. Compact Sections for Beam-Columns • The axial load affects the ratio for compactness. When the check for compactness for the web is performed while the web is subjected to axial load the following ratios shall be Flange limit is similar to beams 68402

  26. 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

  27. Unbraced and Braced Frames • In braced frames amplification moments can only happens due to member deflection Braced Frames Sidesway bracing system Member deflection 68402

  28. Unbraced and Braced Frames • The AISC code approximate the effect by using two amplification factors B1 and B2 AISC Equation AISC Equation • Where • B1 amplification factor for the moment occurring in braced member • B2 amplification factor for the moment occurring from sidesway • Mnt and Pnt is the maximum moment and axial force assuming no sidesway • Mlt and Plt is the maximum moment and axial force due to sidesway • Pr is the required axial strength 68402

  29. Unbraced and Braced Frames • Braced frames are those frames prevented from sidesway. • In this case the moment amplification equation can be simplified to: AISC Equation • KL/r for the axis of bending considered • K ≤ 1.0 68402

  30. 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) • When there is transverse loading on the beam either of the following case applies 68402

  31. Unbraced and Braced Frames • AISC requires stability bracing to have • Specific strength to resist the lateral load • Specific axial stiffness to limit the lateral deformation. Braced Frames Unbraced Frames • Where Pu is the sum of factored axial load in the braced story • Pbr is bracing strength and bbr is braced or unbraced frame stiffness (f = 0.75) 68402

  32. Unbraced and Braced Frames • Unbraced frames can observe loading + sidesway • In this case the moment amplification equation can be simplified to: BMD AISC Equation 68402

  33. Unbraced and Braced Frames • A minimum lateral load in each combination shall be added so that the shear in each story is given by: 68402

  34. Analysis of Unbraced Frames is the sum of factored axial loads on all columns in floor is the drift due to the unfactored horizontal forces is the story height story shear produced by unfactored horizontal forces is the drift index (is generally between 1/500 to 1/200) is the sum of Euler buckling loads of all columns in floor is the factored axial load in the column RM can be conservatively taken as 0.85 68402

  35. 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

  36. 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

  37. Ex. 5.1- Beam-Columns in Braced Frames Mux= (1.073)(240) = 257.5 kN.m Step III: Compute moment capacity Since Lb = 3.6 m Lp< Lb< Lr 68402

  38. Ex. 5.1- Beam-Columns in Braced Frames Step IV: Check combined effect  Section is satisfactory 68402

  39. 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

  40. Ex. 5.2- Analysis of Beam-Column • Step I: Compute the capacities of the beam-column cPn = 4577 kN Mnx = 790 kN.m Mny = 380 kN.m • Step II: Check combined effect OK 68402

  41. Design of Beam-Columns • Trial-and-error procedure • Select trial section • Check appropriate interaction formula. • Repeat until section is satisfactory 68402

  42. Design of Unbraced Frames • Design can be performed using the following procedure: • Use a procedure similar to that of braced frames • To start the design assume B1 = 1.0 and compute B2 by assuming the ratio • Compute Mu and perform same procedure used for braced frames 68402

  43. Ex. 5.3- Analysis-External Column • Check the exterior column of an unbraced frame shown in the figure for the following load combination. All columns are 3.8 m long and all beams are 9 m long. Assume A992 steel. W24x76 W14x90 For this frame W24x76 68402

  44. Ex. 5.3- Analysis-External Column Step I: Calculate Kx andKy Effective length, Ky , assumed braced frame W24x76 W14x90 W24x76 68402

  45. Ex. 5.3- Analysis-External Column Step II: Calculate Pnandp 68402

  46. Ex. 5.3- Analysis-External Column Step III: Determine second-order moments-No translation, Mnt Due to lack of information, assume Cm = 1.0 68402

  47. Ex. 5.3- Analysis-External Column Step IV: Determine second-order moments - Translation, Mlt. Don’t know all columns in story, thus assume the frame will have a deflection limit For this frame Thus, 68402

  48. Ex. 5.3- Analysis-External Column • Step V: Second-order moment • Step VI: Check combined effect OK • Thus, the W14x90, Fy = 344 MPa will work for this loading case. • Now it should be checked for any other load case, such as 1.2D+1.6L 68402

  49. Ex. 5.4 – 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. 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 68402

  50. Ex. 5.4 – 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

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