{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

notes_27_buckling

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

This preview shows pages 1–4. Sign up to view the full content.

q x V x q x y x y x y x y x y x y x q x 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 q L 4 deflection proportional to the uniform load. y max 384 E I 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 V dx V + q dx = 0 => ( ) := d d x ( ) (1) and Σ M A = 0 => M + d M dx M V + d V dx dx ( ) dx dx P d y dx = 0 => dx dx 2 dx d M P d y = V (2) dx dx as in previous bending: M x ( ) = E I d 4 ( ) ( ) := E I d 2 2 ( ) => d 2 2 M x 4 dx dx dx using (1) and (-) the derivative of (2) wrt x; d d M P d y  = d V = q(x) => d dx x dx  dx d 4 d 2 E I 4 ( ) + P 2 ( ) = q(x) or setting k := P E I => dx dx d 4 4 ( ) + k 2 d 2 2 ( ) = ( ) euler buckling uses q = 0 which we will do now dx dx E I 1 of 33 notes_27_buckling_notes.mcd

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
y x y x y 0 y L general solution is: y x ( ) := A sin(k x) + B cos(k x) + C x + 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 ( + C L + D + = 0 2 x y 0 ( ) d d 2 = 0 k 2 A sin k 0 ( k 2 B cos k 0 ( = - k 2 B = 0 2 x y L ( ) 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 E I L 2 := the displacement is then: y x ( ) 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 2 E I 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: y x

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 24

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

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online