# Elastic Buckling Behavior of Beams

Document Sample

```					Elastic Buckling Behavior of Beams

CE579 - Structural Stability and Design
ELASTIC BUCKLING OF BEAMS

• Going back to the original three second-order differential
equations:

Therefore,
                 z                        z
1   E I x v  P v    P x0  M BY   M TY  M BY    M BX   M TX  M BX 
                L                         L

 M TY  M BY  
z                            z
2    E I y u   P u     P y0  M BX  (MTX+MBX)    M BY   M TX +MBY)
(MTY  M BX
                   L                           L
z
3   E I w    (G KT  K )    u  ( M BX  ( M BX  M TX )  P y0 )
L
z                            v                 u
 v ( M BY  ( M BY  M TY )  P x0 )  ( M TY  M BY )  ( M TX  M BX )  0
L                             L                 L
ELASTIC BUCKLING OF BEAMS

• Consider the case of a beam subjected to uniaxial bending only:
 because most steel structures have beams in uniaxial bending
 Beams under biaxial bending do not undergo elastic buckling
• P=0; MTY=MBY=0
• The three equations simplify to:

z
1   E I x v  M BX      M TX  M BX 
L
z
2   E I y u   M BX   M TX  M BX  
L
         z                 u
3   E I w    (G KT  K )    u   M BX  ( M BX  M TX )   ( M TX  M BX )  0
         L                 L
• Equation (1) is an uncoupled differential equation describing in-plane
bending behavior caused by MTX and MBX
ELASTIC BUCKLING OF BEAMS

• Equations (2) 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 (1, 2, 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 (Mo) causing compression in
the top flange. This will mean that
 -MBX = MTX = Mo
Uniform Moment Case

• For this case, the differential equations (2 and 3) will become:
E I y u    M o  0
E I w    (G KT  K )    u   M o   0
where :
K  Wagner ' s effect due to warping caused by torsion
K    a 2 dA
A

Mo
But ,      y  neglecting higher order terms
Ix
Mo
K           y ( xo  x) 2  ( yo  y ) 2  dA
A
Ix                              
Mo
K             
y  xo  x 2  2 xx0  yo  y 2  2 yy0  dA
2                  2

Ix                                            
A

M           2                                                                    
K  o          xo  y dA   y  x  y  dA  x0  2xy dA  yo  y dA  2 yo  y dA 
2   2                      2                  2
        
Ix          A           A                     A             A             A      
ELASTIC BUCKLING OF BEAMS

Mo                              
K       y  x  y  dA  2 yo I x 
2    2

Ix  A            

 y  x 2  y 2  dA        
                        
K  Mo  A                   2 yo 
          Ix               
                           
 
y  x 2  y 2  dA

 K  M ox          where,  x     A
 2 yo
Ix
 x is a new sec tional property
The beam buckling differential equations become :
(2) E I y u    M o  0
(3) E I w    (G KT  M o  x )    u   M o   0
ELASTIC BUCKLING OF BEAMS

Mo
Equation (2) gives u           
E Iy
Substituting u  from Equation (2) in (3) gives :
M o2
E I w  iv  (G KT  M o  x )          0
E Iy
For doubly symmetric sec tion :  x  0
G KT        M o2
 
iv
   2        0
E Iw       E I y Iw
G KT                 M o2
Let , 1           and    2  2
E Iw                E I y Iw
 iv  1    2  0  becomes the combined d .e. of LTB
ELASTIC BUCKLING OF BEAMS

Assume solution is of the form   e z
                   
  4  1  2  2 e z  0
  4  1  2  2  0
1  12  42              12  42  1
   2
, 
2                          2
1  12  42                 1  12  42
                              , i
2                             2

 Let ,   1 , and            i 2

Above are the four roots for 
  C1e1z  C2e 1z  C3ei 2 z  C4e i 2 z
 collecting real and imaginary terms
  G1 cosh(1 z )  G2 sinh(1 z )  G3 sin( 2 z )  G4 cos( 2 z )
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.
       1              0              0                 1        G1 
      12             0              0                 2
2     G 
 
                                                               2  0
 cosh(1 L)       sinh(1 L)     sin( 2 L)        cos( 2 L)  G3
 
12 cosh(1 L) 12 sinh(1 L)  2 sin( 2 L)  2 cos( 2 L)  G4 

2                 2
  
For buckling coefficient matrix must be sin gular :
 det er min ant of matrix  0
          
 12   2  sinh(1 L)  sinh( 2 L)  0
2

Of these :
only sinh( 2 L)  0
 2 L  n
ELASTIC BUCKLING OF BEAMS
n
 2 
L
12  42  1           
                          
2               L
2 2
   42  1  2
1
2

L
2
 2 2               2 2    2 2 
 2  1   1  2  21  2 
2

 2                             L 
L                    L
4                   4
2        2 
 2   2  1  2                                   2E I y   2E Iw          
L        L                          Mo                         G KT 
L2        L2             
M o2      2 G KT    2 
 2  2       2         
E I y Iw  L   E I w   L2 

  2 G KT    2 
 M o   E I y Iw   2 
2
 
 L   E I w   L2 
Uniform Moment Case

• The critical moment for the uniform moment case is given by
the simple equations shown below.
 2 EIy   2 EIw         
M 
o
cr       2
 2          GKT 
L      L               
M cr  Py  P  ro2
o

• 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 Lb between lateral
- torsional braces.
 P is the column buckling load about the torsional z- axis.
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 x v  M BX      M TX  M BX 
L
z
E I y u   M BX   M TX  M BX  
L
         z                 u
E Iw     (G KT  K )    u   M BX  ( M BX  M TX )   ( M TX  M BX )  0
         L                 L
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
Forward difference
Backward difference
f’(x)
Central difference

f(x-h)            f(x)      f(x+h)

x
h      h
h2           h3           h 4 iv
f ( x  h)  f ( x) hf ( x)     f ( x)  f ( x)      f ( x)
2!           3!           4!
f ( x  h)  f ( x) h              h2           h 3 iv
 f ( x)                       f ( x)  f ( x)         f ( x)
h            2!           3!            4!
f ( x  h)  f ( x)
 f ( x)                       O(h)         Forward differenceequation
h
Finite Difference Method

h2            h3           h 4 iv
f ( x  h)  f ( x) hf ( x)      f ( x)  f ( x)       f ( x)
2!            3!           4!
f ( x)  f ( x  h) h               h2            h 3 iv
 f ( x)                       f ( x)  f ( x)           f ( x)
h             2!           3!            4!
f ( x)  f ( x  h)
 f ( x)                       O(h)          Backward differenceequation
h

h2           h3           h 4 iv
f ( x  h)  f ( x) hf ( x)     f ( x)  f ( x)      f ( x)
2!           3!           4!

h2           h3           h 4 iv
f ( x  h)  f ( x) hf ( x)     f ( x)  f ( x)     f ( x)
2!           3!           4!

f ( x  h)  f ( x  h) 2h 2
 f ( x)                               f ( x)
2h              3!
f ( x  h)  f ( x  h)
 f ( x)                           O(h 2 )        Central differenceequation
2h
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(x).
• If the domain x is divided into several equal parts, each of
length h.
h

1   2   3       i-2   i-1   i   i+1   i+2     n

• At each of the ‘nodes’ or ‘section points’ or ‘domain points’
the differential equations are still valid.
Finite Difference Method

• Central difference approximations for higher order
derivatives:                             Notation
y  f ( x)
yi   1 y  y                                      y i  f ( x  i)
i1    i1
2h                                            y i  f ( x  i)
yi   1 y  2y  y                                y i  f ( x  i) and so on
i1      i     i1
h2
yi   1 y  2y  2y  y 
i2     i1     i1    i2
2h 3                               
1
y i  4 y i2  4 y i1  6y i  4 y i1  y i2 
iv

h
FDM - Beam on Elastic Foundation

• Consider an interesting problemn --> beam on elastic
foundation
w(x)=w

EI y iv  k y( x)  w( x)
Fixed end                                                Pin support
K=elastic fdn.
x               L


• Convert the problem into a finite difference problem.
1          2   3        4           5        6                     iv
EI y i  k y i  w
h =0.2 l

Discrete form of differential equation
FDM - Beam on Elastic Foundation

iv
EI y i  k y i  w
EI
 4 y i2  4 y i1  6y i  4 y i1  y i2   kyi  w
h
write 4 equations for i  2, 3, 4, 5

0          1          2   3   4        5      6       7

h =0.2 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

625EI
At i  2 :       y 0  4 y1  6y 2  4 y 3  y 4   ky2  w
L4
625EI
At i  3 :    4  y1  4 y 2  6y 3  4 y 4  y 5   ky3  w
L
625EI
At i  4 :     4 y 2  4 y 3  6y 4  4 y 5  y 6   ky4  w
L
625EI
At i  5 :       y 3  4 y41  6y 5  4 y 6  y 7   ky5  w
L4
• Lets consider the boundary conditions:
          y (0)  0             y1  0                    (1)
y ( L)  0            y6  0                     (2)
M ( L)  0                                        (3)
 (0)  0          y(0)  0                    (4)
FDM - Beam on Elastic Foundation

y (0)  0             y1  0      (1)
y ( L)  0            y6  0      (2)
M ( L)  0                          (3)
 EI y( L)  0        
y6  0
1
2  5

 y6        y  2 y6  y7   0
h
 y7   y5                          (3)
 (0)  0       y(0)  0               (4)
1

 y1      y2  y0   0
2h
 y2  y0                                (4)
FDM - Beam on Elastic Foundation

• Substituting the boundary conditions:
kL4          wL4
At i  2 :   7y 2  4 y 3  y 4            y2 
625EI         625EI
kL4        wL4
At i  3 :   4 y 2  6y 3  4 y 4  y 5          y3 
625EI       625EI
kL4        wL4
At i  4 :   y 2  4 y 3  6y 4  4 y 5          y4 
625EI       625EI
kL4           wL4
At i  5 :   y 3  4 y1  5y 5            y5 
625EI         625EI
Let a = kl4/625EI


7  a 4   1    0  y 2  1
                     
 4 6  a 4    1  y 3  1 wL4
  
 1    4 6  a 4  y 4  1 625EI
                  
  0    1  4 5  a y 5  1
  
FDM - Column Euler Buckling

w
P   Buckling problem: Find axial load
x                                        P for which the nontrivial
L                           Solution exists.

Ordinary DifferentialEquation
P             w
y iv ( x)     y ( x) 
EI            EI

Finite difference solution. Consider case
Where w=0, and there are 5 stations

0      1     2          3   4      5   P 6

x

h=0.25L
FDM - Euler Column Buckling
Finite difference method
P 
yiiv        yi  0
EI
At stations i  2, 3, 4
1                                                   P 1
 yi 2  4 yi 1  6 yi  4 yi 1  yi  2    2  yi 1  2 yi  yi 1   0
h4                                                  EI h
Boundary conditions
y1  0                 (1)
y5  0                 (2)

y1  0
1
       y2  y0   0
2h
 y0  y2                      (3)
M5  0

 EI  y5  0
 ( y6  2 y5  y4 )  0
 y6   y4                     (4)
FDM - Column Euler Buckling

• Final Equations
1                            P 1
7y 2  4 y 3  y 4    2 (2y 2  y 3 )  0
h4                           EI h
1                              P 1
4 
4 y 2  6y 3  4 y 4      ( y 2  2y 3  y 4 )  0
h                              EI h 2
1                            P 1
4
( y 2  4 y 3  5y 4 )     ( y 3  2y 4 )  0
h                            EI h 2

 Matrix Form
 7 4 1  y 2     2 1 0  y 2  0
          PL2               
4 6 4 y 3      1 2 1  y 3   0
 y  16EI  0 1 2 y  0
 1 4 5  4 
                            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} exists if and only if
| A-I|=0
 One way to solve the problem is to obtain the characteristic
polynomial from expanding | 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
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 1                 2 1 0 
                                
A  4 6 4             B   1 2 1 
 1 4 5 
                         0 1 2
        
PL2                                  PL2
and                                               1.11075
16EI                                 16EI
7  2  4     1                                        EI
4   6  2  4    0                   Pcr  17.772 2
L
1     4     5  2                                              EI
Exact solution is 20.14
   1.11075                                                        L2
FDM - Euler Buckling Problem

• 11% error in solution from FDM
• {y}= {0.4184 1.0 0.8896}T

0     1   2   3    4     5   P 6

x
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
   1) Compute [E]=-[A]-1[B]
   2) Assume initial eigenvector guess {y}0
   3) Set iteration counter i=0
   4) Solve for new eigenvector {y}i+1=[E]{y}i
   5) Normalize new eigenvector {y}i+1={y}i+1/max(yji+1)
   6) Calculate eigenvalue = 1/max(yji+1)
   7) Evaluate convergence: i+1-i < tol
   8) If no convergence, then go to step 4
   9) If yes convergence, then = i+1 and {y}= {y}i+1
Inverse Iteration Method
Different Boundary Conditions

• Let Mocr 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.

• Salvadori in the 1970s developed the equation below based
on the regression analysis of numerical results with a simple
equation
 Mcr = Cb Mocr
for uniform moment.

• In case that the moment diagram is not linear over the length
of the beam, i.e., there are transverse loads producing a non-
linear moment diagram
 The value of Cb is a little more involved
Beams with non-simple end conditions

• Mocr = (Py P ro2)0.5
 PY with Kb
 P with Kt
Beam Inelastic Buckling Behavior

• Uniform moment case
Beam Inelastic Buckling Behavior

• Non-uniform moment
Beam In-plane Behavior

• Section capacity Mp
• Section M- behavior
Beam Design Provisions

CHAPTER F in AISC Specifications

```
DOCUMENT INFO
Shared By:
Categories:
Tags:
Stats:
 views: 18 posted: 2/8/2012 language: pages: 45
How are you planning on using Docstoc?