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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 Compression Members 68402**Design of Compression Members**• Short and long columns • Buckling load and buckling failure modes • Elastic and Inelastic buckling • Local buckling • Design of Compression Members • Effective Length for Rigid Frames • Torsional and Flexural-Torsional Buckling • Design of Singly Symmetric Cross Sections 68402**Axially Loaded Compression Members**• Columns • Struts • Top chords of trusses • Diagonal members of trusses Column and Compression member are often used interchangeably 68402**Axially Loaded Compression Members**• Commonly Used Sections: • W/H shapes • Square and Rectangular or round HSS • Tees and Double Tees • Angles and double angles • Channel sections 68402**Columns**• Failure modes (limit states): • Crushing (for short column) • Flexural or Euler Buckling (unstable under bending) • Local Buckling (thin local cross section) 68402**Short Columns**• Compression Members: Structural elements that are subjected to axial compressive forces only are called columns. Columns are subjected to axial loads thru the centroid. • Stress: The stress in the column cross-section can be calculated as f - assumed to be uniform over the entire cross-section. Short columns - crushing 68402**Long Columns**• This ideal state is never reached. The stress-state will be non-uniform due to: • Accidental eccentricity of loading with respect to the centroid • Member out-of –straightness (crookedness), or • Residual stresses in the member cross-section due to fabrication processes. • Accidental eccentricity and member out-of-straightness can cause bending moments in the member. However, these are secondary and are usually ignored. • Bending moments cannot be neglected if they are acting on the member. Members with axial compression and bending moment are called beam-columns. • “Long” columns 68402**Long Columns**• The larger the slenderness ratio (L/r), the greater the tendency to buckle under smaller load • Factors affecting tendency to buckle: • end conditions • unknown eccentricity (concentric & eccentric loads) • imperfections in material • initial crookedness • out of plumbness • residual stress • buckling can be on one or both axes (major or minor axis) 68402**Column Buckling**• Consider a long slender compression member. If an axial load P is applied and increased slowly, it will ultimately reach a value Pcr that will cause buckling of the column. Pcr is called the critical buckling load of the column. Figure 1. Buckling of axially loaded compression members 68402**d**d x x Buckling Load • Now assume we have a pin connected column. If we apply a similar concept to that before here we find • The internal resisting moment M in the column is Pcr Pcr • We can write the relationship between the deflected shape and the Moment M d M - The load at which bucking starts to happen (Critical buckling load) P P 68402**Column Buckling**• What is buckling? Buckling occurs when a straight column subjected to axial compression suddenly undergoes bending as shown in the Figure 1(b). Buckling is identified as a failure limit-state for columns. • The critical buckling load Pcr for columns is theoretically given by Equation (3.1): • [3.1] I - moment of inertia about axis of buckling. K - effective length factor based on end boundary conditions. 68402**Effective Length**KL- Distance between inflection points in column. K- Effective length factor L- Column unsupported length KL=0.7L L KL=L KL=0.5L L K = 1.0 K = 0.5 K = 0.7 68402**Column Buckling**Table C-C2.2 Approximate Values of Effective Length Factor, K Boundary conditions 68402**Ex. 3.1- Buckling Loads**Determine the buckling strength of a W 12 x 50 column. Its length is 6 m. For minor axis buckling, it is pinned at both ends. For major buckling, is it pinned at one end and fixed at the other end. 68402**Ex. 3.1- Buckling Loads**Step I. Visualize the problem • For the W12 x 50 (or any wide flange section), x is the major axis and y is the minor axis. Major axis means axis about which it has greater moment of inertia (Ix >Iy). Step II. Determine the effective lengths • According to Table C-C2.2: • For pin-pin end conditions about the minor axis • Ky = 1.0 (theoretical value); and Ky = 1.0 (recommended design value) • For pin-fix end conditions about the major axis • Kx = 0.7 (theoretical value); and Kx = 0.8 (recommended design value). • According to the problem statement, the unsupported length for buckling about the major (x) axis = Lx = 6 m. 68402**Ex. 3.1- Buckling Loads**• The unsupported length for buckling about the minor (y) axis = Ly = 6 m. • Effective length for major (x) axis buckling = Kx Lx = 0.8 x 6 = 4.8 m. • Effective length for minor (y) axis buckling = Ky Ly = 1.0 x 6 = 6 m. Step III. Determine the relevant section properties • Elastic modulus of elasticity = E = 200 GPa (constant for all steels) • For W12 x 50: Ix = 163x106 mm4. Iy = 23x106 mm4 68402**Ex. 3.1- Buckling Loads**Step IV. Calculate the buckling strength • Critical load for buckling about x - axis = Pcr-x = Pcr-x = = 13965 kN. Critical load for buckling about y-axis = Pcr-y = Pcr-y = = 1261 kN. Buckling strength of the column = smaller (Pcr-x, Pcr-y) = Pcr = 1261 kN. Minor (y) axis buckling governs. 68402**Ex. 3.1- Buckling Loads**Notes: • Minor axis buckling usually governs for all doubly symmetric cross-sections. However, for some cases, major (x) axis buckling can govern. • Note that the steel yield stress was irrelevant for calculating this buckling strength. 68402**Inelastic Column Buckling**• Let us consider the previous example. According to our calculations Pcr = 1261 kN. This Pcr will cause a uniform stress f = Pcr/A in the cross-section. • For W12 x 50, A = 9420 mm2. Therefore, for Pcr = 1261 kN; f = 133.9 MPa. • The calculated value of f is within the elastic range for a 344 MPa yield stress material. • However, if the unsupported length was only 3 m, Pcr =would be calculated as 5044 kN, and f = 535.5 MPa. • This value of f is ridiculous because the material will yield at 344 MPa and never develop f = 535.5 kN. The member would yield before buckling. 68402**Inelastic Column Buckling**• Eq. (3.1) is valid only when the material everywhere in the cross-section is in the elastic region. If the material goes inelastic then Eq. (3.1) becomes useless and cannot be used. • What happens in the inelastic range? • Several other problems appear in the inelastic range. • The member out-of-straightness has a significant influence on the buckling strength in the inelastic region. It must be accounted for. • The residual stresses in the member due to the fabrication process causes yielding in the cross-section much before the uniform stress f reaches the yield stress Fy. • The shape of the cross-section (W, C, etc.) also influences the buckling strength. • In the inelastic range, the steel material can undergo strain hardening. • All of these are very advanced concepts and beyond the scope of this course. 68402**AISC Specifications for Column Strength**• The AISC specifications for column design are based on several years of research. • These specifications account for the elastic and inelastic buckling of columns including all issues (member crookedness, residual stresses, accidental eccentricity etc.) mentioned above. • The specification presented here will work for all doubly symmetric cross-sections. • The design strength of columns for the flexural buckling limit state is equal to cPn • Where, c = 0.9 (Resistance factor for compression members) 68402**Inelastic Buckling of Columns**• Elastic buckling assumes the material to follow Hooke’s law and thus assumes stresses below elastic (proportional) limit. • If the stress in the column reaches the proportional limit then Euler’s assumptions are violated. Stress “F” Elastic Buckling (Long Columns) Proportional limit Euler assumptions Inelastic Buckling (Short columns) L/r 68402**AISC Specifications for Column Strength**Pu Pn Pn = Ag Fcr[E3-1] [E3-4] Fe- Elastic critical Euler buckling load Ag - gross member area K - effective length factor L - unbraced length of the member r - governing radius of gyration The 0.877 factor in Eq (E3-3) tries to account for initial crookedness. Inelastic [E3-2] Elastic [E3-3] 68402**AISC Specifications For Column Strength**Elastic Buckling (Long columns) Inelastic Buckling (Short columns) 68402**AISC Specifications For Column Strength**• For a given column section: • Calculate I, Ag, r • Determine effective length K L based on end boundary conditions. • Calculate KL/r • If KL/r is greater than , elastic buckling occurs and use Equation (E3.4) • If KL/r is less than or equal to , inelastic buckling occurs and use Eq. (E3.3) • Note that the column can develop its yield strength Fy as KL/r approaches zero. 68402**Ex. 3.2 - Column Strength**• Calculate the design strength of W14 x 74 with length of 6 m and pinned ends. A36 steel is used. • Step I. Calculate the effective length and slenderness ratio for the problem Kx = Ky = 1.0 Lx = Ly = 6 m Major axis slenderness ratio = KxLx/rx = 6000/153.4 = 39.1 Minor axis slenderness ratio = KyLy/ry = 6000/63 = 95.2 • Step II. Calculate the buckling strength for governing slenderness ratio The governing slenderness ratio is the larger of (KxLx/rx, KyLy/ry) 68402**Ex. 3.2 - Column Strength**• KyLy/ry is larger and the governing slenderness ratio; • MPa • Therefore, MPa • Design column strength = cPn = 0.9 (Ag Fcr) = 0.9 (14060x154)/1000= 1948.7 kN. • Design strength of column = 1948.7 kN. 68402**Local Buckling Limit State**• The AISC specifications for column strength assume that column buckling is the governing limit state. However, if the column section is made of thin (slender) plate elements, then failure can occur due to localbuckling of the flanges or the webs. Figure 4. Local buckling of columns 68402**Local Buckling Limit State**• Local buckling is another limitation that represents the instability of the cross section itself. • If local buckling occurs, the full strength of the cross section can not be developed. 68402**Local Buckling Limit State**• If localbuckling of the individual plate elements occurs, then the column may not be able to develop its buckling strength. • Therefore, the local buckling limit state must be prevented from controlling the column strength. • Local buckling depends on the slenderness (width-to-thickness b/t ratio) of the plate element and the yield stress (Fy) of the material. • Each plate element must be stocky enough, i.e., have a b/t ratio that prevents local buckling from governing the column strength. • The AISC specification provides the slenderness (b/t) limits that the individual plate elements must satisfy so that local buckling does not control. 68402**Local Buckling Limit State**• Local buckling can be prevented by limiting the width to thickness ratio known as “” to an upper limit r 68402**Local Buckling Limit State**• The AISC specification provides two slenderness limits (p and r) for the local buckling of plate elements. 68402**Local Buckling Limit State**• If the slenderness ratio (b/t) of the plate element is greater than r then it isslender. It will locally buckle in the elastic range before reaching Fy • If the slenderness ratio (b/t) of the plate element is less than r but greater than p, then it is non-compact. It will locally buckle immediately after reaching Fy • If the slenderness ratio (b/t) of the plate element is less than p, then the element is compact. It will locally buckle much after reaching Fy • If all the plate elements of a cross-section are compact, then the section is compact. • If any one plate element is non-compact, then the cross-section is non-compact • If any one plate element is slender, then the cross-section is slender. 68402**Local Buckling Limit State**• Cross section can be classified as “compact”, “non compact” or “slender” sections based on their width to thickness ratios • If the cross-section does not satisfy local buckling requirements its critical buckling stress Fcr shall be reduced • If then the section is slender, a reduction factor for capacity shall be computed from • It is not recommended to use slender sections for columns. AISC Manual for Steel Design 68402**Local Buckling Limit State**• The slenderness limits p and r for various plate elements with different boundary conditions are given in the AISC Manual. • Note that the slenderness limits (p and r) and the definition of plate slenderness (b/t) ratio depend upon the boundary conditions for the plate. • If the plate is supported along two edges parallel to the direction of compression force, then it is a stiffened element. For example, the webs of W shapes • If the plate is supported along only one edge parallel to the direction of the compression force, then it is an unstiffened element. Ex., the flanges of W shapes. • The local buckling limit state can be prevented from controlling the column strength by using sections that are compact and non-compact. • Avoid slender sections 68402**Ex. 3.3 – Local Buckling**• Determine the local buckling slenderness limits and evaluate the W14 x 74 section used in Example 3.2. Does local buckling limit the column strength? • Step I. Calculate the slenderness limits See Tables in previous slide. For the flanges of I-shape sections in pure compression For the webs of I-shapes section in pure compression Use E = 200000 MPa 68402**Ex. 3.3 – Local Buckling**• Step II. Calculate the slenderness ratios for the flanges and webs of W14 x 74 • For the flanges of I-shape member, b = bf/2 = flange width / 2 Therefore, b/t = bf/2tf. For W 14 x 74, bf/2tf = 6.41 (See Section Property Table) • For the webs of I shaped member, b = h h is the clear distance between flanges less the fillet / corner radius of each flange For W14 x 74, h/tw = 25.4 (See Section Property Table). • Step III. Make the comparisons and comment For the flanges, b/t < r. Therefore, the flange is non-compact For the webs, h/tw < r. Therefore the web is non-compact Therefore, the section is non-compact Therefore, local buckling will not limit the column strength. 68402**Design of Compression Members**• Steps for design of compression members • Calculate the factored loads Pu • Assume (a cross section) or (KL/r ratio between 50 to 90) • Calculate the slenderness ratio KL/r and the ratio Fe • Calculate c Fcr based on value ofFe • Calculate the Area required Ag • Choose a cross section and get (KxL/rx )and (KyL/ry) (KL/r) max • Recalculate c Fcrand thus check • Check local buckling requirements 68402**Ex. 3.4 – Design Strength**• Determine the design strength of an ASTM A992 W14 x 132 that is part of a braced frame. Assume that the physical length L = 9 m, the ends are pinned and the column is braced at the ends only for the X-X axis and braced at the ends and mid-height for the Y-Y axis. • Step I. Calculate the effective lengths. From Section Property Table For W14 x 132: rx = 159.5 mm; ry = 95.5 mm; Ag =25030 mm2 Kx = 1.0 and Ky = 1.0 Lx = 9 m and Ly = 4.5 m KxLx = 9 m and KyLy = 4.5 m 68402**Ex. 3.4 – Design Strength**• Step II. Determine the governing slenderness ratio KxLx/rx = 9000/159.5 = 56.4 KyLy/ry = 4500/95.5 = 47.1 The larger slenderness ratio, therefore, buckling about the major axis will govern the column strength. • Step III. Calculate the column strength MPa MPa 68402**Ex. 3.4 – Design Strength**• Step IV. Check the local buckling limits For the flanges, bf/2tf = 7.15 < For the web, h/tw = 17.7< Therefore, the section is non-compact. OK. 68402**Ex. 3.5 – Column Design**• A compression member is subjected to service loads of 700 kN DL and 2400 kN of LL. The member is 7.8 m long & pinned at each end. Use A992 steel and select a W shape. • Step I. Calculate the factored design load Pu Pu = 1.2 PD + 1.6 PL = 1.2 x 700 + 1.6 x 2400 = 4680 kN. • Step II. Calculate Fcr by assuming KL/r = 80 MPa MPa 68402**Ex. 3.5 – Column Design**• Step III. Calculate the required area of steel A = 4680*1000/(0.9*215.3) = 24156 mm2 • Step IV. Select a W shape from the Section Property Tables Select W14 x 132. It has A = 25030 mm2 OR W12 x 136 A = 25740 mm2 Select W14 x 132 because it has lower weight. KyLy/ry =7800/95.5 = 81.7 Fe = 295.7 Fcr = 211.4 Pn = 4897.3 kN OK W14 x 145 is the lightest. • Section is non-compact but students have to check for that • Note that column sections are usually W12 or W14. Generally sections bigger than W14 are not used as columns. 68402**Effective Length**• Specific Values of K shall be known Recommended design values (not theoretical values) • Values for K for different end conditions range from 0.5 for theoretically fixed ends to 1.0 for pinned ends and are given by: Table C-C2.2 AISC Manual • For compression elements connected as rigid frames the effective length is a function of the relative stiffness of the element compared to the overall stiffness of the joint. This will be discussed later in this chapter 68402**K Factor for Rigid Frames**• If we assume all connections are pinned then: Kx L = 3 m and Ky L = 6 m • However the rigidity of the beams affect the rotation of the columns. Thus in rigid frames the K factor can be determined from the relative rigidity of the columns • Determine a G factor 3 m 6 m • Where “c” represents column and “g” represents girder • The G value is computed at each end of the member and K is computed factor from the monograms in AISC Manual – Figure C-C2.2 68402**Effective Length of Columns in Frames**• So far, we have looked at the buckling strength of individual columns. These columns had various boundary conditions at the ends, but they were not connected to other members with moment (fix) connections. • The effective length factor K for the buckling of an individual column can be obtained for the appropriate end conditions from Table C-C2.2 of the AISC Manual . • However, when these individual columns are part of a frame, their ends are connected to other members (beams etc.). • Their effective length factor K will depend on the restraint offered by the other members connected at the ends. • Therefore, the effective length factor K will depend on the relative rigidity (stiffness) of the members connected at the ends. 68402**Effective Length of Columns in Frames**• The effective length factor for columns in frames must be calculated as follows: • First, you have to determine whether the column is part of a braced frame or an unbraced (moment resisting) frame. • If the column is part of a braced frame then its effective length factor 0 < K ≤ 1 • If the column is part of an unbraced frame then 1 < K ≤ ∞ • Then, you have to determine the relative rigidity factor G for both ends of the column • G is defined as the ratio of the summation of the rigidity (EI/L) of all columns coming together at an end to the summation of the rigidity (EI/L) of all beams coming together at the same end. It must be calculated for both ends of the column c: for columns b: for beams 68402**Effective Length of Columns in Frames**• Then, you can determine the effective length factor K for the column using the calculated value of G at both ends, i.e., GA and GB and the appropriate alignment chart • There are two alignment charts provided by the AISC manual, • One is for columns in braced (sidesway inhibited) frames. 0 < K ≤ 1 • The second is for columns in unbraced (sidesway uninhibited) frames. 1 < K ≤ ∞ • The procedure for calculating G is the same for both cases. 68402**Effective Length**Monograph or Jackson and Moreland Alignment Chart for Unbraced Frame 68402