MIT - 16.20 Fall, 2002 Unit 16 Bifurcation Buckling Readings : Rivello 14.1, 14.2, 14.4 Paul A. Lagace, Ph.D. Professor of Aeronautics & Astronautics and Engineering Systems Paul A. Lagace ? 2001 MIT - 16.20 Fall, 2002 V. Stability and Buckling Paul A. Lagace ? 2001 Unit 16 - 2 MIT - 16.20 Fall, 2002 Now consider the case of compressive loads and the instability they can cause. Consider only static instabilities (static loading as opposed to dynamic loading [e.g., flutter]) From Unified, defined instability via: “A system becomes unstable when a negative stiffness overcomes the natural stiffness of the structure. ” (Physically, the more you push, it gives more and builds on itself) Review some of the mathematical concepts. Limit initial discussions to columns. Generally, there are two types of buckling/instability ? Bifurcation buckling ? Snap-through buckling Paul A. Lagace ? 2001 Unit 16 - 3 MIT - 16.20 Fall, 2002 Bifurcation Buckling There are two (or more) equilibrium solutions (thus the solution path “bifurcates ” ) from Unified … Figure 16.1 Representation of initially straight column under compressive load Paul A. Lagace ? 2001 Unit 16 - 4 MIT - 16.20 Fall, 2002 Figure 16.2 Basic load-deflection behavior of initially straight column under compressive load Actual behavior Note : Bifurcation is a mathematical concept. The manifestations in an actual system are altered due to physical realities/imperfections. Sometimes these differences can be very important. (first continue with ideal case … ) Perfect ABC - Equilibrium position, but unstable behavior BD - Equilibrium position There are also other equilibrium positions Imperfections cause the actual behavior to only follow this as asymptotes (will see later) Paul A. Lagace ? 2001 Unit 16 - 5 MIT - 16.20 Fall, 2002 Snap-Though Buckling Figure 16.3 Representation of column with curvature (shallow arch) with load applied perpendicular to column Figure 16.4 Basic load-deflection behavior of shallow arch with transverse load arch “ snaps through ” to F when load reaches C Thus, there are nonlinear load-deflection curves in this behavior Paul A. Lagace ? 2001 Unit 16 - 6 MIT - 16.20 Fall, 2002 For “deeper ” arches, antisymetric behavior is possible Figure 16.5 Representation of antisymetric buckling of deeper arch under transverse load (flops over) before snapping through Figure 16.6 Load-deflection behavior of deeper arch under transverse load ABCDEF - symmetric snap- through ABF - antisymmetric behavior A D E ? ? ? Paul A. Lagace ? 2001 Unit 16 - 7 MIT - 16.20 Fall, 2002 Will deal mainly with … Bifurcation Buckling First consider the “perfect” case: uniform column under end load. First look at the simply-supported case …column is initially straight ? Load is applied along axis of beam ? “Perfect” column ? only axial shortening occurs (before instability), i.e., no bending Figure 16.7 Simply-supported column under end compressive load E I = constant Paul A. Lagace ? 2001 Unit 16 - 8 MIT - 16.20 Fall, 2002 Recall the governing equation : 4 2 EI dw + P dw = 0 dx 4 dx 2 --> Notice that P does not enter into the equation on the right hand side (making the differential equation homogenous), but enters as a coefficient of a linear differential term This is an eigenvalue problem. Let: λ x w = e this gives: λ 4 + P λ 2 = 0 EI ? λ =± P EI i 0, 0 repeated roots ? need to look for more solutions End up with the following general homogenous solution : w = A sin P EI x + B cos P EI x + C + D x Paul A. Lagace ? 2001 Unit 16 - 9 MIT - 16.20 Fall, 2002 where the constants A, B, C, D are determined by using the Boundary Conditions For the simply-supported case, boundary conditions are: @ x = 0 w = 0 2 M = E I dw = 0 dx 2 @ x = l w = 0 M = 0 From: w(x = 0) = 0 ? B + C = 0 B = 0 ? M(x = 0) = 0 ? ? EI P B = 0 C = 0 EI w(x = l ) = 0 ? A sin P EI l + D l = 0 ? D = 0 M(x = l ) = 0 ? ? EI P A sin P EI l = 0 EI Paul A. Lagace ? 2001 Unit 16 - 1 0 MIT - 16.20 Fall, 2002 and can see that: AP sin P EI l = 0 This occurs if: ? P = 0 (not interesting) ? A = 0 (trivial solution, w = 0) ? sin P EI l = 0 ? P EI l = n π Thus, the critical load is: 22 n π E I P = l 2 associated with each is a shape ( mode ) nx π w = A sin l Paul A. Lagace ? 2001 Unit 16 - 1 1 MIT - 16.20 Fall, 2002 A is still undefined (instability ? w --> ∞ ) So have buckling loads and associated mode shapes Figure 16.8 Representation of buckling loads and modes for simply- supported columns 2nd mode 1st mode P 2 = 4P 1 The lowest value is the one where buckling occurs: P = π 2 EI Euler buckling load cr l 2 Paul A. Lagace ? 2001 Unit 16 - 1 2 MIT - 16.20 Fall, 2002 The higher loads can be reached if the column is “ artificially restrained ” at lower bifurcation points. Other Boundary Conditions There are 3 (/4) allowable boundary conditions for w (need two on each end) which are homogeneous ( ? … = 0) w = 0 ? Simply-supported (pinned) 2 M = E I dw = 0 dx 2 w = 0 ? Fixed end (clamped) dw = 0 dx 2 M = E I dw = 0 dx 2 ? Free end 2 S = d ?? EI dw ? 2 ? = 0 dx ? dx ? S = 0 ? Sliding dw = 0 dx Paul A. Lagace ? 2001 Unit 16 - 1 3 MIT - 16.20 Fall, 2002 There are others of these that are homogeneous and inhomogeneous Boundary Conditions Examples: ? Free end with an axial load ? Vertical spring ? Torsional spring Solution Procedure for P cr : M = 0 S = ? P dw 0 dx M = 0 S = k f w w = 0 dw M = ? k T dx ? Use boundary conditions to get four equations in four unknowns (the constants A, B, C, D) ? Solve this set of equations to find non-trivial value of P Paul A. Lagace ? 2001 Unit 16 - 1 4 MIT - 16.20 Fall, 2002 ? xxxx ? ? A ? ? ? ? ? xxxx ? B ? ? ? ? ? = 0 homogeneous ? xxxx ? ? C ? equation ? ? ? ? ? xxxx ? ? D ? matrix ? Set determinant of matrix to zero ( ? = 0) and solve resulting equation. Will find, for example, that for a clamped-clamped column: 4 π 2 EI P cr = l 2 (need to do solution geometrically) with the associated eigenfunction ? 1 ? cos 2 π x ? ? l ? Figure 16.9 Representation of clamped-clamped column under end load Paul A. Lagace ? 2001 Unit 16 - 1 5 MIT - 16.20 Fall, 2002 Figure 16.10 Representation of buckling mode of clamped-clamped column Note terminology : buckling load = eigenvalue buckling mode = eigenfunction Notice that this critical load has the same form as that found for the simply-supported column except it is multiplied by a factor of 4 Can express the critical buckling load in the generic case as: cE I π 2 P = cr l 2 where: c = coefficient of edge fixity Paul A. Lagace ? 2001 Unit 16 - 1 6 MIT - 16.20 Fall, 2002 Figure 16.11 Representation of buckling of columns with different end conditions spring k T torsional c = 1 c= 4 c = 0.25 c is between 1 and 4 Generally use c ≈ 2 for aircraft work with “ fixed ends ” ? Cannot truly get perfectly clamped end ? Simply-supported is too conservative ? Empirically, c = 2 works well and remains conservative Paul A. Lagace ? 2001 Unit 16 - 1 7 MIT - 16.20 Fall, 2002 Other important parameters: radius of gyration = ρ = ( IA 12 ) slenderness ration = L ρ L effective length = L ′ = c See Rivello Considerations for Orthotropic or Composite Beams If maintain geometrical restrictions of a column ( l >> in-plane directions), only the longitudinal properties, E I , are important. Thus, use techniques developed earlier: ? E L for orthotropic ? E 1 I * for composite Note : Consider general cross-section Buckling could occur in y or z direction (or any direction transverse to x, for that matter). Paul A. Lagace ? 2001 Unit 16 - 1 8 MIT - 16.20 Fall, 2002 --> must evaluate I * for each direction and see which is less… buckling occurs for the case where I * is smaller --> anywhere in y-z plane --> use Mohr’ s circle Note : May need to be corrected for shearing effects See Timoshenko and Gere, Theory of Elastic Stability , pp. 132-135 Effects of Initial Imperfections Figure 16.12 Representation of column with initial imperfection initial deflection (imperfection due to OR manufacturing, etc.) Paul A. Lagace ? 2001 Unit 16 - 1 9 MIT - 16.20 Fall, 2002 Figure 16.13 Representation of column loaded eccentrically e = “ eccentricity ” load n o t applied along center line of column These two cases are basically handled the same -- a moment is applied in each case ? Case 1 -- due to initial imperfection ? Case 2 -- since load is not applied along axis of column (beam) Look closely at second case: Figure 16.14 Representation of full geometry of simply-supported column loaded eccentrically The governing equation is still the same: Paul A. Lagace ? 2001 Unit 16 - 2 0 MIT - 16.20 Fall, 2002 4 2 EI dw + P dw = 0 dx 4 dx 2 and thus the basic solution is the same: w = A sin P EI x + B cos P EI x + C + D x What changes are the Boundary Conditions For the specific case of Figure 16.14: @ x = 0 w = 0 --> B + C = 0 2 M = E I dw = ?null Pe ?null B = e dx 2 C = -e P ?null ? EI B = ?null P e EI Figure 16.15 Representation of end moment for column loaded eccentrically Paul A. Lagace ? 2001 Unit 16 - 2 1 MIT - 16.20 Fall, 2002 w = 0 ?null A sin P EI l + e cos P EI l ?null e + D l = 0 @ x = l 2 M = E I dw = ?null Pe ?null dx 2 ? EI P A sin P EI l ?null E I P e cos P EI l = ?null P e EI EI Find: D = 0 ?null P EI ?null e ? 1 ?null cos l ? ?null ? A = sin P EI l Putting this all together, find: w P EI l P EI l P EI x P EI x = ?null ?null? ?null ?null? ?null + ?null ?null?null ?null?null?null ?null?null ?null? ?null?null?null ?null?null 1 1 cos sin sin cos e Paul A. Lagace ? 2001 Unit 16 - 2 2 MIT - 16.20 Fall, 2002 Deflection is generally finite (this is not an eigenvalue problem). π 2 EI However, as P approaches P cr = l 2 , w again becomes unbounded (w --> ∞ ) Figure 16.16 Load-deflection response for various levels of eccentricity of end-loaded column π 2 EI = l 2 P cr increasing eccentricity (e/ llll ) --> Nondimensional problem via e/ l So, w approaches perfect case as P approaches P cr . But, as e/ l increases, behavior is less like perfect case. Paul A. Lagace ? 2001 Unit 16 - 2 3 MIT - 16.20 Fall, 2002 Bending Moment now: ? ?null P EI ?null ?null ?null ? 1 ?null cos l ?null P EI P EI ? 2 ? ?null ?null ? M = E I dw = ?null eP ?null P EI sin x + cos x ? dx 2 ?null sin l ?null ?null ? ?null ?null As P goes to zero, M --> - e P This is known as the primary bending moment (i.e., the bending moment due to axial loading) Also note that as P EI l --> π (P --> P cr ), M --> ∞null (This is due to the fact that there is an instability as w --> ∞ . This cannot happen in real life) Paul A. Lagace ? 2001 Unit 16 - 2 4 MIT - 16.20 Fall, 2002 Figure 16.17 Moment-load response for eccentrically loaded column eP (primary) actual behavior is bonded by two asymptotes Overall: Primary Bending Moment M = -Pe - Pw Secondary (due to deflection) Bending Moment Paul A. Lagace ? 2001 Unit 16 - 2 5 MIT - 16.20 Fall, 2002 Figure 16.18 Representation of moments due to eccentricity and deflection Note : All this is good for small deflections. As deflections get large, have post buckling considerations. (Will discuss later) Paul A. Lagace ? 2001 Unit 16 - 2 6