beam buckling

Elastic Buckling Behavior of Beams

CE579 - Structural Stability and Design

ELASTIC BUCKLING OF BEAMS

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

0

0

0

0

,

( ) ( ( ) )

( ( ) ) ( ) ( ) 0

x BY TY BY BX TX BX

y BX TY BY BY TX BX

w T BX BX TX

BY BY TY TY BY TX BX

Therefore

z zE I v P v P x M M M M M M

L L

z zE I u P u P y M M M M M M

L L

zE I G K K u M M M P y

Lz v u

v M M M P x M M M ML L L

1

2

3

(MTX+MBX) (MTY+MBY)


• 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:

• Equation (1) is an uncoupled differential equation describing in-plane bending behavior caused by MTX and MBX

( ) ( ) ( ) 0

x BX TX BX

y BX TX BX

w T BX BX TX TX BX

zE I v M M M

Lz

E I u M M ML

z uE I G K K u M M M M M

L L

1

2

3


• 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:

2

2 2

2 2 2 20 0

0

( ) 0

:

'

,

( ) ( )

2 2

y o

w T o

A

o

x

oo o

xA

oo o

x

E I u M

E I G K K u M

where

K Wagner s effect due to warping caused by torsion

K a dA

MBut y neglecting higher order terms

I

MK y x x y y dA

I

MK y x x xx y y yy dA

I

2 2 2 2 20 2 2

A

oo o o

x A A A A A

MK x y dA y x y dA x xy dA y y dA y y dA

I


2 2

2 2

2 2

2

2

, 2

sec

oo x

x A

Ao o

x

Ao x x o

x

x

MK y x y dA y I

I

y x y dA

K M yI

y x y dA

K M where yI

is a new tional property

:

(2) 0

(3) ( ) 0

y o

w T o x o

The beam buckling differential equations become

E I u M

E I G K M u M


2

2

2

2

1 2 2

1 2

(2)

(2) (3) :

( ) 0

sec : 0

0

,

0

o

y

iv ow T o x

y

x

iv oT

w y w

oT

w y w

iv

MEquation gives u

E I

Substituting u from Equation in gives

ME I G K M

E I

For doubly symmetric tion

MG K

E I E I I

MG KLet and

E I E I I

. .becomes the combined d e of LTB


1 1 2 2

4 21 2

4 21 2

2 21 1 2 1 2 12

2 21 1 2 1 1 2

1 2

1 2 3 4

0

0

4 4,

2 2

4 4,

2 2

, ,

z

z

z z i z i z

Assume solution is of the form e

e

i

Let and i

Above are the four roots for

C e C e C e C e

collect

1 1 2 1 3 2 4 2cosh( ) sinh( ) sin( ) cos( )

ing real and imaginary terms

G z G z G z G z


• Assume simply supported boundary conditions for the beam:

12 21 2 2

1 1 2 2 32 2 2 2

41 1 1 1 2 2 2 2

(0) (0) ( ) ( ) 0

. .

1 0 0 10 0

0cosh( ) sinh( ) sin( ) cos( )cosh( ) sinh( ) sin( ) cos( )

L L

Solution for must satisfy all four b c

GG

L L L L GGL L L L

For buckl

2 21 2 1 2

2

2

sin :

det min 0

sinh( ) sinh( ) 0

:

sinh( ) 0

ing coefficient matrix must be gular

er ant of matrix

L L

Of these

only L

L n


2

21 2 1

221 2 1 2

22 2 22

1 1 12 2 2

2

2 2

2 12 2

2 2 2

2 2 2 2

2 22

2 2

4

2

24

2 2 22

4 4

o T

y w w

To y w

w

n

L

L

L

L L L

L L

M G K

E I I L E I L

G KM E I I

L E I L

2 2

2 2

y wo T

E I E IM G K

L L

Uniform Moment Case

• The critical moment for the uniform moment case is given by the simple equations shown below.

• 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. Py is the column buckling load about the minor axis.

P is the column buckling load about the torsional z- axis.

Mcro

2EIy

L2

2EIw

L2GKT

Mcro Py P r o

2

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

( ) ( ) ( ) 0

x BX TX BX

y BX TX BX

w T BX BX TX TX BX

zE I v M M M

Lz

E I u M M ML

z uE I G K K u M M M M M

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 ( x h) f ( x)h f ( x) h2

2!f ( x)h3

3!f ( x) h4

4!f iv ( x)

f ( x)f ( x h) f ( x)

h

h

2!f ( x)

h2

3!f ( x)

h3

4!f iv (x)

f ( x) f ( x h) f ( x)h

O(h) Forward difference equation

h h

f(x)f(x-h)

f

x

f’(x)

Forward differenceBackward difference

Central difference

f(x+h)


f ( x h) f ( x)h f ( x) h2

2!f ( x)h3

3!f ( x) h4

4!f iv ( x)

f ( x h) f ( x) h f ( x)h2

2!f ( x)

h3

3!f ( x)

h4

4!f iv ( x)

f ( x) f ( x h) f ( x h)

2h2h2

3!f ( x)

f ( x)f ( x h) f ( x h)

2hO(h2 ) Central difference equation

f ( x h) f ( x) h f ( x) h2

2!f ( x) h3

3!f ( x) h4

4!f iv ( x)

f ( x)f ( x) f ( x h)

h

h

2!f ( x)

h2

3!f ( x)

h3

4!f iv (x)

f ( x) f ( x) f ( x h)

hO(h) Backward difference equation


• 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.

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

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

h


• Central difference approximations for higher order derivatives:

Notation

y f ( x)

y i f ( x i)

y i f ( x i)

y i f ( x i) and so on

y i 12h

y i1 y i 1

y i

1

h2y i1 2y i y i 1

y i 1

2h3y i2 2y i1 2y i 1 y i 2

y iiv

1

h4y i2 4y i1 6y i 4y i 1 y i 2

FDM - Beam on Elastic Foundation

• Consider an interesting problemn --> beam on elastic foundation

• Convert the problem into a finite difference problem.

Fixed end Pin support

x L

w(x)=w

K=elastic fdn.

EI y iv k y( x)w(x)

1 2 3 4 5 6

h =0.2 l

EI y iiv k y i w

Discrete form of differential equation


EI y iiv k y i w

EI

h4y i 2 4y i 1 6y i 4y i1 y i2 ky i w

write 4 equations for i2, 3, 4, 5

1 2 3 4 5 6

h =0.2 l

70

Need two imaginary nodes that lie within the boundary Hmm…. These are needed to only solve the problemThey don’t mean anything.


• Lets consider the boundary conditions:

At i2 :625EI

L4y0 4y1 6y2 4y3 y4 ky2 w

At i3 :625EI

L4y1 4y2 6y3 4y4 y5 ky3 w

At i4 :625EI

L4y2 4y3 6y4 4y5 y6 ky4 w

At i5 :625EI

L4y3 4y41 6y5 4y6 y7 ky5 w

1

6

(0) 0 0 (1)

( ) 0 0 (2)

( ) 0 (3)

(0) 0 (0) 0 (4)

y y

y L y

M L

y


1

6

6

6 5 6 72

7 5

1 2 0

2 0

(0) 0 0 (1)

( ) 0 0 (2)

( ) 0 (3)

( ) 0 0

12 0

(3 )

(0) 0 (0) 0 (4)

10

2(4 )

y y

y L y

M L

EI y L y

y y y yh

y y

y

y y yh

y y


• Substituting the boundary conditions:

Let a = kl4/625EI

At i2 : 7y2 4y3 y4 kL4

625EIy2

wL4

625EI

At i3 : 4y2 6y3 4y4 y5 kL4

625EIy3

wL4

625EI

At i4 : y2 4y3 6y4 4y5 kL4

625EIy4

wL4

625EI

At i5 : y3 4y1 5y5 kL4

625EIy5

wL4

625EI

7 a 4 1 0

4 6 a 4 1

1 4 6 a 40 1 4 5 a

y2y3y4y5

1

1

1

1

wL4

625EI

FDM - Column Euler Buckling

P

w

xL

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

Ordinary Differential Equation

y iv (x) P

EIy (x) w

EI

Finite difference solution. Consider caseWhere w=0, and there are 5 stations

P

x

0 1 2 3 4 5 6

h=0.25L

FDM - Euler Column Buckling

2 1 1 2 1 14 2

1

5

1

2 0

0 2

5

5

6 5 4

6 4

0

2, 3, 4

1 14 6 4 2 0

0 (1)

0 (2)

0

10

2(3)

0

0

( 2 ) 0

(4)

ivi i

i i i i i i i i

Finite difference method

Py y

EIAt stations i

Py y y y y y y y

h EI hBoundary conditions

y

y

y

y yh

y y

M

EI y

y y y

y y

FDM - Column Euler Buckling

• Final Equations

1

h47y2 4y3 y4 P

EI1

h2( 2y2 y3 )0

1h4 4y2 6y3 4y4 P

EI 1

h2( y2 2y3 y4 )0

1

h4(y2 4y3 5y4 )

P

EI1

h2( y3 2y4 )0

Matrix Form

7 4 1

4 6 41 4 5

y2y3y4

PL2

16EI

2 1 0

1 2 1

0 1 2

y2y3y4

0

0

0

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


• 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

A7 4 1

4 6 41 4 5

B 2 1 0

1 2 1

0 1 2

and PL2

16EI7 2 4 1

4 6 2 4 1 4 5 2

0

1.11075

PL2

16EI1.11075

Pcr 17.772EI

L2

Exact solution is 20.14EI

L2


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

P

x

0 1 2 3 4 5 6

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

Beams with Non-Uniform Loading

• 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.

Beams with Non-uniform Loading

• Salvadori in the 1970s developed the equation below based on the regression analysis of numerical results with a simple equation Mcr = Cb Mo

cr

Critical moment for non-uniform loading = Cb x critical moment for uniform moment.

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 non-linear moment diagram The value of Cb is a little more involved

Beams with non-simple end conditions

• Mocr = (Py P ro

2)0.5

PY with Kb

Pwith 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

beam buckling

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