Buckling of coped steel beams and steel beams with partial endplates
Elastic Buckling Behavior of Beams CE579 - Structural Stability and Design.
Transcript of Elastic Buckling Behavior of Beams CE579 - Structural Stability and Design.
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)
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:
• 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
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:
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
ELASTIC BUCKLING OF BEAMS
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
ELASTIC BUCKLING OF BEAMS
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
ELASTIC BUCKLING OF BEAMS
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
ELASTIC BUCKLING OF BEAMS
• 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
ELASTIC BUCKLING OF BEAMS
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)
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 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
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.
• 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
Finite Difference Method
• 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
FDM - Beam on Elastic Foundation
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.
FDM - Beam on Elastic Foundation
• 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
FDM - Beam on Elastic Foundation
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
FDM - Beam on Elastic Foundation
• 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
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
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
FDM - Euler Buckling Problem
• 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
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