notes_27_buckling

# notes_27_buckling - Buckling general Up to this point, the...

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qx Vx yx Buckling general Up to this point, the stress and deflections have been proportional to an applied load: e.g. σ x = M y bending stress proportional to moment I maximum deflection of a simply supported beam subject to uniform load per unit length q => = 5 qL 4 deflection proportional to the uniform load. y max 384 EI This is not always the case, such as when compressive loads with/without lateral loads act on a column (beam). Moments, stresses and deflections will NOT be proportional to axial loads, but will be dependent (not proportional) to deflections, thus sensitive to slight initial deflections and/or eccentricities in the application of the load. Euler buckling (derived from general case of beam-columns including lateral load q(x)) consider: x dy y q P P M+dM/dx*dx q P P dx M V V+dV/dx*dx Σ F y_dir = 0 => V + d d x Vdx V + qdx = 0 => () := d d x () (1) and Σ M A = 0 => M + d M dx M V + d V dx dx dx dx P d ydx = 0 => dx dx 2 dx d M P d y = V (2) dx dx as in previous bending: Mx = d 4 := E I d 2 2 => d 2 2 4 dx dx dx using (1) and (-) the derivative of (2) wrt x; d d M P d y  = d V = q(x) => d d xx dx  dx d 4 d 2 4 + P 2 = q(x) or setting k := P => dx dx d 4 4 + k 2 d 2 2 = euler buckling uses q = 0 which we will do now dx dx 1 of 33 notes_27_buckling_notes.mcd

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yx y0 yL general solution is: yx ( ) := A sin(k x) + B cos(k + Cx + D check => d 4 4 () + k 2 d 2 2 0 dx dx 0 now apply to column with pinned ends: x P y P boundary conditions are: y(0) = y(L) = 0 and d 2 () = d 2 = 0 (no bending moment at the ends) 2 2 dx dx y(0) = 0 y(L) = 0 A sin k L ( B cos k L ( + CL + D + = 0 2 x d d 2 = 0 k 2 A sin k 0 ( k 2 B cos k 0 ( = - k 2 B = 0 2 x d d 2 = 0 k 2 A sin k L ( = 0 => A sin k L ( = 0 relation above this leaves A sin k L ( = 0 sin k L ( = 0 π recall that k^2 = P/(E*I) n π L 2 = P/(E*I) P cr when P cr n π ( 2 EI L 2 := the displacement is then: ( ) A sin k x ( := where A can be any value. i.e. with P < Pcr the trival solution applies y = 0, but at Pcr y(x) can be >0 and arbitrary. minimum P cr occurs when n = 1 => P cr π L 2 := sometimes labeled P E => B + D = 0 => ) ) => ) ) => B and D = 0 => ) ) C = 0 from the y(L) = 0 => ) which has a non trivial solution only when ) => k*L = n* = => k^2 solution defining that force P as or . ... ) ) 2 of 33 notes_27_buckling_notes.mcd
let's look at a few other sets of boundary conditions and determine the P cr by inspection: A. clamped - free x y P B. clamped - clamped x y P C. clamped - clamped, free to translate x y D. clamped - pinned, not free to translate x y P P cr P cr P cr P cr this one is not obvious (at least to me!!) let's apply boundary conditions to the general solution: yx ( ) := A sin(k x) + B cos(k + Cx + D 3 of 33 notes_27_buckling_notes.mcd

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y0 yL
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## This note was uploaded on 02/24/2012 for the course MECHANICAL 2.082 taught by Professor Davidburke during the Spring '03 term at MIT.

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notes_27_buckling - Buckling general Up to this point, the...

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