Elastic Buckling Behavior of Beams CE579 - Structural Stability and Design ELASTIC BUCKLING OF BEAMS Going back to the original three second-order differential equations: 3 Therefore, z z E I v P v P 0 M BY MTY M BY M MTX M L L z z E I y u P u P y0 M (MM TY M BY M BY MTX M L TX +M ) (M TY +M BY ) L z E Iw ( G KT K) u ( M ( M MTX ) P y0) L z v u v ( M BY ( M BY MTY ) P 0) ( MTY M BY ) ( MTX M ) 0 L L L
ELASTIC BUCKLING OF BEAMS Consider the case of a beam subjected to uniaial bending only: because most steel structures have beams in uniaial bending Beams under biaial bending do not undergo elastic buckling P=0; M TY =M BY =0 The three equations simplify to: 3 z E I v M MTX M L z E I y u M MTX M L z u E Iw ( G KT K) u M ( M MTX ) ( MTX M ) 0 L L Equation () is an uncoupled differential equation describing in-plane bending behavior caused by M TX and M ELASTIC BUCKLING OF BEAMS Equations () and (3) are coupled equations in u and that describe the lateral bending and torsional behavior of the beam. In fact they define the lateral torsional buckling of the beam. The beam must satisfy all three equations (,, and 3). Hence, beam in-plane bending will occur UNTIL the lateral torsional buckling moment is reached, when it will take over. Consider the case of uniform moment (M o ) causing compression in the top flange. This will mean that -M = M TX = M o
Uniform Moment Case For this case, the differential equations ( and 3) will become: E I u M 0 y o E I ( G K K) u M 0 w T o where : K Wagner ' s effect due to warping caused by torsion K A a da M o But, y neglecting higher order terms I M o K y ( o ) ( yo y) da I A M o K y o 0 yo y yy 0 da I A M o K o y da y y da 0 y da yo y da yo y da I A A A A A ELASTIC BUCKLING OF BEAMS M o K y y da yoi I A y y da A K M o yo I y y da A K M where, I y o o is a new sectional property The beam buckling differential equations become : () E I u M 0 y o (3) E I ( G K M ) u M 0 w T o o 3
ELASTIC BUCKLING OF BEAMS M o Equation () gives u EI Substituting u from Equation () in (3) gives : iv M o E Iw ( G KT M o ) 0 EI For doubly symmetric sec tion : 0 iv GK M 0 E I Let, T o w E I yiw GK T o and E Iw E I yiw iv 0 becomes the combined d.. e of LTB y M y ELASTIC BUCKLING OF BEAMS Assume solution is of the form z e 0 4 0 4 4 4, e z 4 4, i Let,, and i Above are the four roots for C e C e C e C e z z i z i z 3 4 collecting real and imaginary terms G cosh( z) G sinh( z) G sin( z) G cos( z) 3 4 4
ELASTIC BUCKLING OF BEAMS Assume simply supported boundary conditions for the beam: (0) (0) ( L) ( L) 0 Solution for must satisfy all four b.. c 0 0 G 0 0 G 0 cosh( L) sinh( L) sin( L) cos( L) G 3 cosh( L) sinh( L) sin( L) cos( L) G 4 For buckling coefficient matri must be sin gular : det er min ant of matri 0 sinh( L) sinh( L) 0 Of these : only Ln sinh( L) 0 ELASTIC BUCKLING OF BEAMS n L 4 L L 4 L L L 4 4 L L M o GK T E I yiw L E Iw L GK T M o E I yiw L E Iw L EIy EIw Mo G K L L T 5
Uniform Moment Case The critical moment for the uniform moment case is given by the simple equations shown below. M o cr EI y L M o cr P y P r o EI w GK L T The AISC code massages these equations into different forms, which just look different. Fundamentally the equations are the same. The critical moment for a span with distance L b between lateral - torsional braces. P y is the column buckling load about the minor ais. P is the column buckling load about the torsional z- ais. Non-uniform moment The only case for which the differential equations can be solved analytically is the uniform moment. For almost all other cases, we will have to resort to numerical methods to solve the differential equations. Of course, you can also solve the uniform moment case using numerical methods z E I v M MTX M L z E I y u M MTX M L z u E Iw ( G KT K) u M ( M MTX ) ( MTX M ) 0 L L 6
What numerical method to use What we have is a problem where the governing differential equations are known. The solution and some of its derivatives are known at the boundary. This is an ordinary differential equation and a boundary value problem. We will solve it using the finite difference method. The FDM converts the differential equation into algebraic equations. Develop an FDM mesh or grid (as it is more correctly called) in the structure. Write the algebraic form of the d.e. at each point within the grid. Write the algebraic form of the boundary conditions. Solve all the algebraic equations simultaneously. Finite Difference Method f Backward difference f () Forward difference Central difference f(-h) f() f(+h) f ( h) f () hf () h f () h 3! 3! f () f () h f ( h) f () h h! f ( h) f () O(h) h h f () h 3! f () h 4 4! f iv () f () h 3 4! f iv () Forward differenceequation 7
Finite Difference Method f ( h) f () hf () h f () h 3! 3! f () f () f () f ( h) h h! f () f ( h) O(h) h f () h 3! f () h 4 4! f iv () f () h 3 4! f iv () Backwarddifferenceequation f ( h) f () hf () h! f ( h) f () hf () h! f () h 3 3! f () h 3 3! f () h 4 4! f iv () f () h 4 4! f iv () f () f () f ( h) f ( h) h h 3! f ( h) f ( h) O(h ) h f () Centraldifferenceequation Finite Difference Method The central difference equations are better than the forward or backward difference because the error will be of the order of h-square rather than h. Similar equations can be derived for higher order derivatives of the function f(). If the domain is divided into several equal parts, each of length h. h 3 i- i- i i+ i+ n At each of the nodes or section points or domain points the differential equations are still valid. 8
Finite Difference Method Central difference approimations for higher order derivatives: y i h y y i i y i h y i y i y i y i y h 3 i y i y i y i y i iv h 4 y i 4y i 6y i 4y i y i Notation y f ( ) y i f ( i) y i y i f ( i) f ( i) and so on FDM - Beam on Elastic Foundation Consider an interesting problemn --> beam on elastic foundation w()=w EI y iv k y() w() Fied end L K=elastic fdn. Pin support Convert the problem into a finite difference problem. 3 4 5 6 h =0. l EI y i iv k y i w Discrete form of differential equation 9
FDM - Beam on Elastic Foundation EI y i iv k y i w EI h 4 y i 4y i 6y i 4y i y i ky i w write 4 equationsfor i, 3, 4, 5 0 3 4 5 6 7 h =0. l Need two imaginary nodes that lie within the boundary Hmm. These are needed to only solve the problem They don t mean anything. FDM - Beam on Elastic Foundation At i : 65 EI y L 4 0 4y 6y 4y 3 y 4 ky w At i 3: 65 EI y L 4 4y 6y 3 4y 4 y 5 ky 3 w At i 4 : 65 EI y L 4 4y 3 6y 4 4y 5 y 6 ky 4 w At i 5 : 65 EI y L 4 3 4y4 6y 5 4y 6 y 7 ky 5 w Lets consider the boundary conditions: y(0) 0 y 0 () y( L) 0 y 0 () 6 M( L) 0 (3) (0) 0 y(0) 0 (4) 0
FDM - Beam on Elastic Foundation y(0) 0 y 0 () 7 5 0 y( L) 0 y 0 () 6 M( L) 0 (3) EI y( L) 0 y 0 6 y 6 y5 y6 y7 0 h y y (3 ) (0) 0 y(0) 0 (4) y y y0 0 h y y (4 ) FDM - Beam on Elastic Foundation Substituting the boundary conditions: Let a = kl 4 /65EI At i : 7y 4y 3 y 4 kl4 65 EI y wl4 65 EI At i 3: 4y 6y 3 4y 4 y 5 kl4 65 EI y 3 wl4 65 EI At i 4 : y 4y 3 6y 4 4y 5 kl4 65EI y 4 wl4 65EI At i 5 : y 3 4y 5y 5 kl4 65 EI y 5 wl4 65 EI 7 a 4 0 4 6 a 4 4 6 a 4 0 4 5 a y y 3 y 4 y 5 wl 4 65 EI
FDM - Column Euler Buckling w L P Buckling problem: Find aial load P for which the nontrivial Solution eists. OrdinaryDifferential Equation y iv () P EI y () w EI Finite difference solution. Consider case Where w=0, and there are 5 stations 0 3 4 5 P 6 h=0.5l Finite difference method iv P yi y i 0 EI At stations i, 3, 4 y 0 y y0 0 h 5 0 EI y 0 5 ( y y y ) 0 6 5 4 FDM - Euler Column Buckling P y 4 i 4yi 6yi 4yi yi yi yi yi 0 h EI h Boundary conditions M y 0 () y 0 () 5 y y 0 y y 6 4 (3) (4)
FDM - Column Euler Buckling Final Equations 7y h 4 4y 3 y 4 P EI h (y y 3 ) 0 4y h 4 6y 3 4y 4 P EI h (y y 3 y 4 ) 0 h (y 4y 5y ) P 4 3 4 EI h (y y ) 0 3 4 Matri Form 7 4 y 0 y 0 4 6 4 y 3 PL 4 5 6EI y 3 0 0 0 y 4 y 4 FDM - Euler Buckling Problem [A]{y}+[B]{y}={0} How to find P? Solve the eigenvalue problem. Standard Eigenvalue Problem [A]{y}={y} Where, = eigenvalue and {y} = eigenvector Can be simplified to [A-I]{y}={0} Nontrivial solution for {y} eists if and only if A-I =0 One way to solve the problem is to obtain the characteristic polynomial from epanding A-I =0 Solving the polynomial will give the value of Substitute the value of to get the eigenvector {y} This is not the best way to solve the problem, and will not work for more than 4or 5th order polynomial 3
FDM - Euler Buckling Problem For solving Buckling Eigenvalue Problem [A]{y} + [B]{y}={0} [A+ B]{y}={0} Therefore, det A+ B =0 can be used to solve for 7 4 A 4 6 4 4 5 and PL 6EI 7 4 4 6 4 0 4 5.075 0 B 0 PL 6EI.075 P cr 7.77 EI L Eact solutionis0.4 EI L FDM - Euler Buckling Problem % error in solution from FDM {y}= {0.484.0 0.8896} T 0 3 4 5 P 6 4
FDM Euler Buckling Problem Inverse Power Method: Numerical Technique to Find Least Dominant Eigenvalue and its Eigenvector Based on an initial guess for eigenvector and iterations Algorithm ) Compute [E]=-[A] - [B] ) Assume initial eigenvector guess {y} 0 3) Set iteration counter i=0 4) Solve for new eigenvector {y} i+ =[E]{y} i 5) Normalize new eigenvector {y} i+ ={y} i+ /ma(y j i+ ) 6) Calculate eigenvalue = /ma(y j i+ ) 7) Evaluate convergence: i+ - i < tol 8) If no convergence, then go to step 4 9) If yes convergence, then = i+ and {y}= {y} i+ Inverse Iteration Method 5
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Different Boundary Conditions Beams with Non-Uniform Loading Let M o cr be the lateral-torsional buckling moment for the case of uniform moment. If the applied moments are non-uniform (but varying linearly, i.e., there are no loads along the length) Numerically solve the differential equation using FDM and the Inverse Iteration process for eigenvalues Alternately, researchers have already done these numerical solution for a variety of linear moment diagrams The results from the numerical analyses were used to develop a simple equation that was calibrated to give reasonable results. 8
Beams with Non-uniform Loading Salvadori in the 970s developed the equation below based on the regression analysis of numerical results with a simple equation M cr = C b M o cr Critical moment for non-uniform loading = C b critical moment for uniform moment. Beams with Non-uniform Loading 9
Beams with Non-uniform Loading Beams with Non-Uniform Loading In case that the moment diagram is not linear over the length of the beam, i.e., there are transverse loads producing a nonlinear moment diagram The value of C b is a little more involved 0
Beams with non-simple end conditions M o cr = (P y P r o ) 0.5 P Y with K b P with K t Beam Inelastic Buckling Behavior Uniform moment case
Beam Inelastic Buckling Behavior Non-uniform moment Beam In-plane Behavior Section capacity M p Section M- behavior
Beam Design Provisions CHAPTER F in AISC Specifications 3