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68402: Structural Design of Buildings II 61420: Design of Steel Structures 62323: Architectural Structures II

68402: Structural Design of Buildings II 61420: Design of Steel Structures 62323: Architectural Structures II

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## 68402: Structural Design of Buildings II 61420: Design of Steel Structures 62323: Architectural Structures II

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**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 of Beams for Flexure 68402**Design of Beams for Flexure**• Introduction • Moment Curvature Response • Sectional Properties • Serviceability Requirements (Deflections) • Compact, Non-compact and Slender Sections • Lateral Torsional Buckling • Design of Beams 68402**Beams under Flexure**• Members subjected principally to transverse gravity loading • Girders (important floor beams, wide spacing) • Joists (less important beams, closely spaced) • Purlins (roof beams, spanning between trusses) • Stringers (longitudinal bridge beams) • Lintels (short beams above window/door openings) 68402**Design for Flexure**• Limit states considered • Yielding • Lateral-Torsional Buckling • Local Buckling • Compact • Non-compact • Slender 68402**Design for Flexure – LRFD Spec.**• Commonly Used Sections: • I – shaped members (singly- and doubly-symmetric) • Square and Rectangular or round HSS • Tees and Double Angles • Rounds and Rectangular Bars • Single Angles • Unsymmetrical Shapes Will not be covered in this course 68402**Section Force-Deformation Response & Plastic Moment (MP)**• A beam is a structural member that is subjected primarily to transverse loads and negligible axial loads. • The transverse loads cause internal SF and BM in the beams as shown in Fig. 1 Fig. 1- SF & BM in a SS Beam 68402**Section Force-Deformation Response & Plastic Moment (MP)**• These internal SF & BM cause longitudinal axial stresses and shear stresses in the cross-section as shown in the Fig. 2 • Curvature = = 2/d (Planes remain plane) Fig. 2 - Longitudinal axial stresses caused by internal BM 68402**Section Force-Deformation Response & Plastic Moment (MP)**• Steel material follows a typical stress-strain behavior as shown in Fig 3 below. E = 200 GPa Fig 3 - Typical steel stress-strain behavior. 68402**Section Force-Deformation Response & Plastic Moment (MP)**• If the steel stress-strain curve is approximated as a bilinear elasto-plastic curve with yield stress equal to y, then the section Moment - Curvature (M-) response for monotonically increasing moment is given by Fig. 4. • In Fig. 4, My is the moment corresponding to first yield and Mp is the plastic moment capacity of the cross-section. • The ratio of Mp to My - the shape factor f for the section. • For a rectangular section, f = 1.5. For a wide-flange section, f≈ 1.1. 68402**Moment-Curvature (NEW)**• Beam curvatureis related to its strain and thus to the applied moment e y (1) (2) (3) (4) 68402**Where Z is the plastic section modulus 1.1 S**Moment-Curvature (NEW) • When the section is within elastic range Where S is the elastic section modulus • When the moment exceeds the yield moment My • Then 68402**Section Force-Deformation Response & Plastic Moment (MP)**Fig. 4 - M- response of a beam section 68402**Section Force-Deformation Response & Plastic Moment (MP)**• Calculation of MP: Cross-section subjected to either +y or - y at the plastic limit. See Figure 5 below. (a) General cross-section (c) Force distribution (b) Stress distribution Figure 5. Plastic centroid and MP for general cross-section. 68402**Moment-Curvature**• When the whole section is yielding a plastic hinge will be formed plastic hinge • Structural analysis by assuming collapsing mechanisms of a structure is known as “Plastic analysis” • The plastic moment Mp is therefore the moment needed at the section to form a plastic hinge 68402**Section Force-Deformation Response & Plastic Moment (MP)**• The plastic centroid for a general cross-section corresponds to the axis about which the total area is equally divided, i.e., A1 = A2 = A/2 • The plastic centroid is not the same as the elastic centroid or center of gravity (c.g.) of the cross-section. • As shown below, the c.g. is defined as the axis about which A1y1 = A2y2. 68402**Section Force-Deformation Response & Plastic Moment (MP)**• For a cross-section with at-least one axis of symmetry, the neutral axis corresponds to the centroidal axis in the elastic range. However, at Mp, the neutral axis will correspond to the plastic centroidal axis. • For a doubly symmetric cross-section, the elastic and the plastic centroid lie at the same point. • Mp = y x A/2 x (y1+y2) • As shown in Figure 5, y1 and y2 are the distance from the plastic centroid to the centroid of area A1 and A2, respectively. 68402**Section Force-Deformation Response & Plastic Moment (MP)**• A/2 x (y1+y2) is called Z, the plastic section modulus of the cross-section. Values for Z are tabulated for various cross-sections in the properties section of the LRFD manual. • bMp = 0.90 Z Fy b - strength reduction factor Mp - plastic moment, which must be 1.5 My for homogenous cross-sections My - moment corresponding to onset of yielding at the extreme fiber from an elastic stress distribution = Fy S for homogenous cross-sections and = Fyf S for hybrid sections. Z - plastic section modulus from the Properties section of the AISC manual. S - elastic section modulus, also from the Properties section of the AISC manual. 68402**Ex. 4.1 – Sectional Properties**• Determine the elastic section modulus, S, plastic section modulus, Z, yield moment, My, and the plastic moment MP, of the cross-section shown below. What is the design moment for the beam cross-section. Assume A992 steel. 300 mm 15 mm 400 mm 10 mm 25 mm 400 mm 68402**Ex. 4.1 – Sectional Properties**• Ag = 300 x 15 + (400 - 15 - 25) x 10 + 400 x 25 = 18100 mm2 Af1 = 300 x 15 = 4500 mm2 Af2 = 400 x 25 = 10000 mm2 Aw = 10 x (400 - 15 - 25) = 3600 mm2 • distance of elastic centroid from bottom = Ix = 400x253/12 +10000(12.5-145.3)2 + 10x3603/12 +3600(205-145.3)2 + 300x153/12 +4500(392.5-145.3)2 = 503.7x106 mm4 Sx = 503.7x106 / (400-145.3) = 1977.5x103 mm3 My-x = Fy Sx = 680.2 kN-m. Sx - elastic section modulus 68402**Ex. 4.1 – Sectional Properties**• distance of plastic centroid from bottom = • y1 = centroid of top half-area about plastic centroid = mm • y2 = centroid of bottom half-area about plas. cent. = mm • Zx = A/2 x (y1 + y2) = 9050 x (256.7 + 11.3) = 2425400 mm3 • Zx - plastic section modulus 68402**Ex. 4.1 – Sectional Properties**• Mp-x = Zx Fy = 2425400 x 344/106 = 834.3 kN.m • Design strength according to AISC Spec. F1.1= bMp= 0.9 x 834.3 = 750.9 kN.m • Check = Mp 1.5 My • Therefore, 834.3 kN.m < 1.5 x 680.2 = 1020.3 kN.m - OK! 68402**Flexural Deflection of Beams - Serviceability**• Steel beams are designed for the factored design loads. The moment capacity, i.e., the factored moment strength (bMn) should be greater than the moment (Mu) caused by the factored loads. • A serviceable structure is one that performs satisfactorily, not causing discomfort or perceptions of unsafety for the occupants or users of the structure. • For a steel beam, being serviceable usually means that the deformations, primarily the vertical slag, or deflection, must be limited. • The maximum deflection of the designed beam is checked at the service-level loads. The deflection due to service-level loads must be less than the specified values. 68402**Serviceability Requirements**• Steel beams need to satisfy SLS in addition to ULS • Serviceability limit states are usually checked using non-factored loads. • Deflection under live loads shall be limited to L/360 • Dead load deflections can be compensated by cambering beams. • SLS might also include limiting stresses in bottom or top flanges if fatigue is a concern in design (Will be further discussed with plate girders). • Standard equations to calculate deflection for different load cases: 68402**Flexural Deflection of Beams - Serviceability**• The AISC Specification gives little guidance other than a statement, “Serviceability Design Considerations,” that deflections should be checked. Appropriate limits for deflection can be found from the governing building code for the region. • The following values of deflection are typical max. allowable deflections. LL DL+LL • Plastered floor construction L/360 L/240 • Unplastered floor construction L/240 L/180 • Unplastered roof construction L/180 L/120 • DL deflection – normally not considered for steel beams 68402**Flexural Deflection of Beams - Serviceability**• In the following examples, we will assume that local buckling and lateral-torsional buckling are not controlling limit states, i.e, the beam section is compact and laterally supported along the length. 68402**Ex. 4.2 - Deflections**• Design a 9 m long simply supported beam subjected to UDL of 6 kN/m dead load and a UDL of 8 kN/m live load. 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 + 1.6 wL = 20 kN/m Mu = wu L2 / 8 = 20 x 92 / 8 = 202.5 kN.m (SS beam) • Step II. Select the lightest section from the AISC Manual design tables. Zx = Mu/(bFy) = 202.5x106/(0.9x344) = 654x103 select W16 x 26 made from A992 steel with bMp = 224 kN.m 68402**Ex. 4.2 - Deflections**• Step III. Add self-weight of designed section and check design wsw = 0.38 kN/m Therefore, wD = 6.38 kN/m wu = 1.2 x 6.38 + 1.6 x 6 = 20.46 kN/m Therefore, Mu = 20.46 x 92 / 8 = 207.2 kN.m < bMp of W16 x 26. OK! • Step IV. Check deflection at service loads. w = 8 kN/m D = 5 w L4 / (384 E Ix) = 5 x (8) x103 x (9)4 / (384 x 200x125) D = 27.3 mm > L/360 - for plastered floor construction • Step V. Redesign with service-load deflection as design criteria L /360 = 25 mm > 5 w L4/(384 E Ix) Therefore, Ix > 136.7x106 mm4 68402**Ex. 4.2 - Deflections**Select the section from the moment of inertia selection from Section Property Tables – select W16 x 31 with Ix = 156x106 mm4 • Note that the serviceability design criteria controlled the design and the section 68402**Ex. 4.3 – Beam Design**• Design the beam shown below. The unfactored dead and live loads are shown in Fig. 6 below. • Step I. Calculate the factored design loads (without self-weight). wu = 1.2 wD + 1.6 wL = 1.2 x 10 + 1.6 x 11 = 29.6 kN/m Pu = 1.2 PD + 1.6 PL = 1.2 x 0 + 1.6 x 40 = 64 kN Mu = wU L2 / 8 + PU L / 4 = 299.7 + 144 = 443.7 kN.m 40 kN 10 kN/m 11 kN/m 4.5 m 9 m 68402**Ex. 4.3 – Beam Design**• Step II. Select W21 x 44 Zx = 1563x103 mm3 bMp = 0.9x1563x103x344/1000000 = 483.9 kN.m Self-weight = wsw = 0.64 kN/m. • Step III. Add self-weight of designed section and check design wD = 10 + 0.64 = 10.64 kN/m wu = 1.2 x 10.64 + 1.6 x 11 = 30.4 kN/m Therefore, Mu = 30.4 x 92/8 + 144 = 451.8 < bMp of W21 x 44. OK! • Step IV. Check deflection at service live loads. Service loads • Distributed load = w = 11 kN/m • Concentrated load = P = L = 40 kN 68402**Ex. 4.3 – Beam Design**Deflection due to uniform distributed load = d = 5 w L4 / (384 EI) Deflection due to concentrated load = c = P L3 / (48 EI) Therefore, service-load deflection = = d + c = 5x11x94x109/(384x351x106x200) +40x93x109/(48x351x106x200) = 13.4 + 8.7 = 22.1 mm L = 9 m. Assuming unplastered floor construction, max = L/240 = 9000/240 = 37.5 mm Therefore, < max- OK! 68402**Local Buckling of Beam Section – Compact and Non-compact**• Mp, the plastic moment capacity for the steel shape, is calculated by assuming a plastic stress distribution (+ or - y) over the cross-section. • The 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 before they develop the compressive yield stress y. • Lateral-torsional buckling of the unsupported length of the beam / member before the cross-section develops the plastic moment Mp. • The analytical equations for local buckling of steel plates with various edge conditions and the results from experimental investigations have been used to develop limiting slenderness ratios for the individual plate elements of the cross-sections. 68402**Local Buckling of Beam Section – Compact and Non-compact**Figure 7. Local buckling of flange due to compressive stress (s) 68402**Local Buckling of Beam Section – Compact and Non-compact**• Steel sections are classified as compact, non-compact, or slender depending upon the slenderness (l) ratio of the individual plates of the cross-section. • Compact section if all elements of cross-section have p • Non-compact sections if any one element of the cross-section has pr • Slender section if any element of the cross-section has r • It is important to note that: • If p, then the individual plate element can develop and sustain y for large values of e before local buckling occurs. • If pr, then the individual plate element can develop y at some locations but not in the entire cross section before local buckling occurs. • If r, then elastic local buckling of the individual plate element occurs. 68402**Classification of Sections**• Classifications of bending elements are based on limits of local buckling • The dimensional ratio l represents • Two limits exist p and r • p represents the upper limit for compact sections • r represents the upper limit for non-compact sections 68402**Local Buckling of Beam Section – Compact and Non-compact**• Thus, 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. Figure 8. Stress-strain response of plates subjected to axial compression and local buckling. 68402**Local Buckling of Beam Section – Compact and Non-compact**• All rolled wide-flange shapes are compact with the following exceptions, which are non-compact. • W21x48, W40x174, W14x99, W14x90, W12x65, W10x12, W8x10, W6x15 (made from A992) 68402**Classification of Sections**• The limits are Flange Web 68402**Lateral-Torsional Buckling (LTB)**• The laterally unsupported length of a beam-member can undergo LTB due to the applied flexural loading (BM). Figure 9. Lateral-torsional buckling of a wide-flange beam subjected to constant moment. 68402**Lateral-Torsional Buckling (LTB)**• LTB is fundamentally similar to the flexural buckling or flexural-torsional buckling of a column subjected to axial loading. • The similarity is that it is also a bifurcation-buckling type phenomenon. • The differences are that lateral-torsional buckling is caused by flexural loading (M), and the buckling deformations are coupled in the lateral and torsional directions. • 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 LTB causing bending moment M(x) varies along the unbraced length. • The worst situation is for beams subjected to uniform BM along the unbraced length. Why? 68402**Lateral-Torsional Buckling (LTB) –Uniform BM**• Consider a beam that is simply-supported at the ends and subjected to four-point loading as shown below. The beam center-span is subjected to uniform BM (M). Assume that lateral supports are provided at the load points. • Laterally unsupported length = Lb. • 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. 68402**Lateral-Torsional Buckling (LTB) – Uniform BM**• Lp = 1.76 ry x - for I members & channels • If Lb is greater than LP then lateral torsional buckling will occur and the moment capacity of the beam will be reduced below the plastic strength MP as shown in Fig. 10. • As shown in Fig. 10, the lateral-torsional buckling moment (Mn = Mcr) is a function of the laterally unbraced length Lb and can be calculated using the equation: 68402**Lateral-Torsional Buckling (LTB) – Uniform BM**• Mn = Mcr = Mn - moment capacity Lb - laterally unsupported length. Mcr - critical lateral-torsional buckling moment. E – 200 GPa; G – 77 GPa Iy - moment of inertia about minor or y-axis (mm4) J - torsional constant (mm4) from the Section Property Tables. Cw - warping constant (mm6) from the Section Property Tables. • This Eq. is valid for ELASTIC LTB only (like the Euler equation). This means it will work only as long as the cross-section is elastic and no portion of the cross-section has yielded. 68402**(0.7Fy)**Lateral-Torsional Buckling (LTB) – Uniform BM Plastic No Instability Inelastic LTB Elastic LTB Fig. 10 Lateral Torsional Buckling (Uniform Bending) 68402**Lateral-Torsional Buckling (LTB) – Uniform BM**• As soon as any portion of the cross-section reaches the yield stress Fy, the elastic LTB equation cannot be used. • Lr is the unbraced length that corresponds to a LTB moment Mr = Sx (0.7Fy). • Mr will cause yielding of the cross-section due to residual stresses. • When the unbraced length is less than Lr, then the elastic LTB Eq. cannot be used. • When the unbraced length (Lb) is less than Lr but more than the plastic length Lp, then the LTB Mn is given by the Eq. below: 68402**Lateral-Torsional Buckling – Uniform BM**• If Lp Lb Lr, then • This is linear interpolation between (Lp, Mp) and (Lr, Mr) • See Fig. 10 again. • For a doubly symmetric I-shape: c = 1 • h0= distance between the flange centroids (mm) 68402**Moment Capacity of Beams Subjected to Non-uniform BM**• As mentioned previously, the case with uniform bending moment is worst for lateral torsional buckling. • For cases with non-uniform bending moment, the LTB moment is greater than that for the case with uniform moment. • The AISC specification says that: • The LTB moment for non-uniform bending moment case • Cb x lateral torsional buckling moment for uniform moment case. 68402**Moment Capacity of Beams Subjected to Non-uniform BM**• Cb is always greater than 1.0 for non-uniform bending moment. • Cb is equal to 1.0 for uniform bending moment. • Sometimes, if you cannot calculate or figure out Cb, then it can be conservatively assumed as 1.0. for doubly and singly symmetric sections Mmax - magnitude of maximum bending moment in Lb MA - magnitude of bending moment at quarter point of Lb MB - magnitude of bending moment at half point of Lb MC - magnitude of bending moment at three-quarter point of Lb • Use absolute values of M 68402**MA**@ quarter MC Mmax MB @ three-quarter @ mid Flexural Strength of Compact Sections Moments determined between bracing points 68402**Values of Cb**3-1 68402