高等材料力學 關百宸 老師 海洋科學科學(工程)與水下科技人才培育方案 磨課師 示範教學 Buckling Analysis of Beam Column Review Derive the critical loading for column structural members with Euler’s Equation Practice the calculation and learn how to deal with beam column with different boundary conditions Beam Column Find the effect due to distributed loading Assumption of Ideal Column Ideal column: Perfectly straight before loading Made of homogeneous material Load is applied through the centroid of the cross section Linear-elastic material Buckle and bend the beam in a single plane Ideal Column with Pin Supports Theoretically, when the axial load P increases, the column fails due to : Material Yielding: Plastic deformation Fracture : Crack forming and propagating P P P fracture yielding Ideal Column with Pin Supports When the axial load P reaches the critical load Pcr, the column becomes unstable. Any small lateral force F will cause the column in the deflected position, even after the force F is removed. P Pcr F Pcr Ideal Column with Pin Supports Free body diagram of Column under an axial load P P P Momentum equilibrium at point A: M A F 0 where x A ν P M F d 2 M EI 2 dx Axial force equilibrium: F P d 2 EI 2 P 0 dx Ideal Column with Pin Supports Ordinary differential equation of Column d 2 EI 2 P 0 dx Let v e x EI 2e x Pe x 0 2 P 0 EI i P P , i EI EI Ideal Column with Pin Supports The general solution of buckling B1e i P x EI B2e i P x EI P P C1 sin x C2 cos x EI EI Euler's formula: e i x cos x i sin x 還記得吧… Ideal Column with Pin Supports Apply the boundary condition of pin-pin ends P v P P C1 sin x C2 cos x EI EI x L EI Boundary conditions: when x = 0, v = 0 C2 0 Ideal Column with Pin Supports Apply the boundary condition of pin-pin ends P when x = L, v = 0 P C1 sin L 0 EI v x L EI Two possible solution (a) C1 0 (b) P sin L 0 EI Ideal Column with Pin Supports when P sin L 0 EI P L n EI Where n = 1, 2, 3, … Function of sin(x) EI 2 n 2 P L2 When n = 1 the critical load EI 2 Pcr 2 L This critical load is sometimes referred to as Euler load Ideal Column with Pin Supports For different mode n we can get the corresponding buckled shape of pinpin support Beam-Column: 2 2 n v C1 sin L x 0 EI Pcr 2 L n=1 4 EI Pcr L2 n=2 9 EI 2 Pcr L2 n=3 16 EI 2 Pcr L2 n=4 P P 但是如果樑-柱上有均 佈力怎麼辦? q(x) Linear Elastic Beam-Column Free body diagram of Beam-Column under a axial load P and distribution load q(x) Momentum equilibrium at point A: P M V ∆x M V x P qdxx 0 q(x) ∆ν A V+ ∆ V M+∆M P where α is a temporary variable Dividing by ∆x V M P qdx x x Linear Elastic Beam-Column Take the limits as x 0 V P M V ∆x where dM d P dx dx d 2 M EI 2 dx q ∆ν V+ ∆ V M+∆M P d 3 d V EI 3 P dx dx Linear Elastic Beam-Column Equilibrium in the v-direction gives V P x q V V V qdx 0 M V ∆x ∆x v V qdx 0 V+ ∆ V q ∆ν x 0 V+ ∆ V M+∆M P q dV dx Linear Elastic Beam-Column The ODE of Beam-Column: d 3 d V EI 3 P dx dx P dV d 4 d 2 EI 4 P 2 q dx dx dx d 4 d 2 EI 4 P 2 q dx dx M V ∆x q ∆ν V+ ∆ V M+∆M P Linear Elastic Beam-Column The general solution of Beam-Column equation d 4 d 2 EI 4 P 2 q dx dx The general solution: v vh v p vh is the homogeneous solution vp is the particular solution Linear Elastic Beam-Column Solving the homogeneous solution vh The homogeneous equation: Let d 4 h d 2 h EI P 2 0 4 dx dx vh e x 4 e x P 2 x e 0 EI P 2 2 0 EI 0, 0, i P EI i P EI Linear Elastic Beam-Column the homogeneous solution vh vh A1e0 x A2 xe0 x A3e i P x EI A4e i P x EI vh C1 C2 x C3 cos x C4 sin x The general solution v: v C1 C2 x C3 cos x C4 sin x v p 同學有沒有甚麼問題? Example Find the equation for the elastic curve v(x) for the uniformly loaded beam-columns shown q0 (a) P x EI v q0 (b) P x EI v Example Beam-Column with Side Sway Buckling Derive the critical loads for the classical Euler buckling problems in the side sway problems (a) Fix-Fix ends (b) Pin-Fix ends q0 q0 P P x EI v l x EI v l