东南大学：《建筑力学 Architectural Mechanics》课程教学课件（英文讲稿）A19 Column Buckling

Column Bucklingmi@se.er.cn

Column Buckling mi@seu.edu.cn

Contents·StabilityandBuckling（稳定性与失稳）ExamplesofColumns（压杆应用示例）·Conventional DesignofColumns（压杆的常规设计方法）·EulersFormulaforPin-endedColumns（端部铰接压杆欧拉公式）·BucklingModes（失稳模态）·ExtensionofEuler'sFormula（欧拉公式的扩展）·BucklinginOrthogonalPlanes（相互垂直平面内的失稳）·WaystoImproveColumnStability（提高压杆稳定性的途径）·ApplicabilityofEuler'sFormula（欧拉公式的适用范围）·FailureDiagramofColumns（压杆的失效总图）·Critical StressofColumns（压杆临界应力的确定）·DesignofColumns（压杆的稳定性设计方法）·EccentricallyLoadedColumns（偏心压杆）2

• Stability and Buckling（稳定性与失稳） • Examples of Columns（压杆应用示例） • Conventional Design of Columns（压杆的常规设计方法） • Euler’s Formula for Pin-ended Columns（端部铰接压杆欧拉公式） • Buckling Modes（失稳模态） • Extension of Euler’s Formula（欧拉公式的扩展） • Buckling in Orthogonal Planes（相互垂直平面内的失稳） • Ways to Improve Column Stability（提高压杆稳定性的途径） • Applicability of Euler’s Formula（欧拉公式的适用范围） • Failure Diagram of Columns（压杆的失效总图） • Critical Stress of Columns（压杆临界应力的确定） • Design of Columns（压杆的稳定性设计方法） • Eccentrically Loaded Columns（偏心压杆） 2 Contents

Stability and Buckling? Stability is characterized as the ability of a structure to maintainits (stable) equilibrium under working conditions: Buckling is the behavior of a structure losing its equilibriumunder working conditions. This is another type of failure criterionin addition to strength (fracture/yielding), stiffness (deformation)andfatiguecriteria: Buckling occurs suddenly and results in catastrophic accident3

• Stability is characterized as the ability of a structure to maintain its (stable) equilibrium under working conditions. Stability and Buckling • Buckling is the behavior of a structure losing its equilibrium under working conditions. This is another type of failure criterion, in addition to strength (fracture/yielding), stiffness (deformation) and fatigue criteria. • Buckling occurs suddenly and results in catastrophic accident. 3

Examples of ColumnsV

4 Examples of Columns

Conventional Design of Columns: In the design of columns, cross-sectional area is selected such that- allowable stress is not exceededP[o]A- deformation falls withinspecificationspAL[s]5LAEBB: After these design calculations, manydiscover that the column is unstableunder loading and that it suddenlybecomes sharply curved or buckles5

• In the design of columns, cross￾sectional area is selected such that - allowable stress is not exceeded   P A     - deformation falls within specifications   L P L AE       • After these design calculations, many discover that the column is unstable under loading and that it suddenly becomes sharply curved or buckles. Conventional Design of Columns 5

Euler's Formula for Pin-ended ColumnsF. Consider an axially loaded beamWWAfter a small perturbation, thex=0,9=7WWsystem reaches a neutralxequilibrium configuration such thatQPMdw*C7Wdr?EIEId?wPW=0dx?EIPX=LV=OIB=w"+k2w=0. k?CTEI= w = Asin kx + Bcos kx0 = w(0)= w(L)B= 0;EIn元kL= nπ =L6

• Consider an axially loaded beam. After a small perturbation, the system reaches a neutral equilibrium configuration such that     2 2 2 2 2 2 2 2 2 0 " 0, sin cos 0 0 0; cr cr cr cr d w M P w dx EI EI d w P w dx EI P w k w k EI w A kx B kx w w L B EIn kL n P L                             Euler’s Formula for Pin-ended Columns 6 w w w w

Buckling ModesV9Pe9EI元EI元4EI元PDPcrL?crLC71

2 2 2 2 2 2 4 9 ; ; cr cr cr EI EI EI P P P L L L       Buckling Modes 7

Cantileyered Columns A column with one fixed and onePfree end, will behave as the upperhalf of a pin-connected column.. The critical loading is calculatedLfrom Euler's formula.BBL. =2L元?EI元?EIPL.4L2元?E元?EAO(L/i,) 4(L/i)PIL, = 2L = equivalent length8

• A column with one fixed and one free end, will behave as the upper￾half of a pin-connected column. • The critical loading is calculated from Euler’s formula,     2 equivalent length 4 4 2 2 2 2 2 2 2 2       L L L i E L i E L EI L EI P e e r r cr e cr      Cantilevered Columns 8 L P A B 2 L L e  P P A A B

Columns with Two Fixed Ends. The symmetry of the supports and ofthe loading requires that the shear at+C and the horizontal reactions at bothL/4+L/2Dends be zero.L/4C+C. The equation of the deflection curveinvolves sine and cosine functions.6: Point D must be a point of inflectionwhere the bending moment is zero.: It follows that the portion DE of thecolumn must behave as a pin endedcolumn1元?EI4元2EIDL。= 0.5LUL.L2B9

Columns with Two Fixed Ends • The symmetry of the supports and of the loading requires that the shear at C and the horizontal reactions at both ends be zero. • The equation of the deflection curve involves sine and cosine functions. • Point D must be a point of inflection, where the bending moment is zero. • It follows that the portion DE of the column must behave as a pin ended column. 2 2 2 2 4 0.5 e cr e EI EI L L P L L       9

Columns with One Fixed End and One Free End. The differential equation(Pw-Vx)d'w[x= 0,y =0]dr?EIWAAd'wpVxdr?EIEIPVxTI=w=Asinkx+Bcoskx+PEIVL0 = w(0)= w(L))= B=O:AsinkLpBBV0 = w'(L)= AkcoskL:[x= L,y = 0]P[x = L, dyldx =0]= k2 = 20.19/L3 tan kL = kL元?EI20.19EI= P.=EIk?L(0.699L)). Equivalent length: L。~ 0.7L10

w w Columns with One Fixed End and One Free End           2 2 2 2 2 2 2 2 2 2 2 sin cos , 0 0 0; sin 0 cos tan 20.19 20.19 0.699 cr d w Pw Vx dx EI d w P Vx w dx EI EI Vx P w A kx B kx k P EI VL w w L B A kL P V w L Ak kL P kL kL k L EI EI P EIk L L                                     0.7 • Equivalent length: L L e  • The differential equation 10 w