Transcript Finite Strip Analysis - Department of Civil Engineering

Finite Strip Analysis
buckling multiplier (stress, load, or moment)
Finite Strip Analysis
and the Beginnings of the Direct Strength Method
Toronto, July 2000
AISI Committee on Specifications
local
distortional
later-torsional
length of a half sine wave
1
Finite Strip Analysis
Overview
•
•
•
•
•
•
•
•
Introduction
Background
A simple verification problem
Analyze a typical section
Interpreting results (k, fcr, Pcr, Mcr)
Direct Strength Prediction
Improve a typical section
Individual analysis - “do it yourself”
2
Finite Strip Analysis
Introduction
• Understanding elastic buckling (stress, modes, etc.) is
fundamental to understanding the behavior and design of
thin-walled structures.
• Thorough treatment of plate buckling separates design of
cold-formed steel structures from typical structures.
• Hand solutions for plate buckling have taken us a great
distance but more modern approaches may be utilized now.
• Finite strip analysis is one efficient method for calculating
elastic buckling behavior.
3
Finite Strip Analysis
Introduction
• No new theory: Finite strip analysis uses the same “thin
plate” theory employed in classical plate buckling
solutions (e.g., k = 4) already in current use.
• Organized: The nuts and bolts of the analysis is organized
in a manner similar to the stiffness method for frames and
thus familiar to a growing group of engineers.
• Efficient: Single solutions and parameter studies can be
performed on PCs
• Free: Source code and programs for the finite strip
analysis is free
4
Finite Strip Analysis
Introduction
finite element
finite st rip
5
Finite Strip Analysis
What has to be defined?
property
give E, G, and v
nodes
give node number
give coordinates
indicate if any additional support
exist along the longitudinal edge
give applied stress on node
elements
give element number
give nodes that form the strip
give thickness of the strip
lengths
give all the lengths that elastic buckling should be examined at
6
Finite Strip Analysis
Finite Strip Software
• CU-FSM
– Matlab based full graphical version
– DOS engine only (execufsm.exe)
• Helen Chen has written a Windows front end
“procefsm.exe” which uses the CU-FSM DOS
engine (Thanks Helen!)
• Other programs with finite strip capability
– THINWALL from University of Sydney
– CFS available from Bob Glauz
7
Finite Strip Analysis
Overview
•
•
•
•
•
•
•
•
Introduction
Background
A simple verification problem
Analyze a typical section
Interpreting results (k, fcr, Pcr, Mcr)
Direct Strength Prediction
Improve a typical section
Individual analysis - “do it yourself”
8
Finite Strip Analysis
Theoretical Background
• Elastic buckling in matrix form
• Initial elastic stiffness [K]
– specialized shape functions
• Geometric stiffness [Kg]
• Forming the solution
• Elastic buckling solution
9
Finite Strip Analysis
Elastic Buckling (Matrix form)
F  Kd
• standard initial elastic
stiffness form
F  K Kg f d • consider effect of
stress on stiffness
F  K Kg f1 d • consider linear
multiples of constant
stress (f1)
Kd  Kg d
• eigenvalue problem
gives solution
10
Finite Strip Analysis
What is [K]?
• [K] is the initial elastic stiffness
b
y
a
x
1
v1

u2
u1
w1
v2
w2
z
F  Kd
 Fu1  
F  
 v1   
 Fu 2   

 
F
 v2  

 0
F
 w1  
 M 1  0

 
 Fw 2  0
M   0
 2
K uv 
0 0 0
0 0 0
0 0 0
0 0 0
0 0 0 0  u 1 
 
0 0 0 0  v1 
0 0 0 0  u 2 
 
0 0 0 0  v 2 
 
  w1 


   1 

  
w


 w 2 
 
  2 
K
11
Finite Strip Analysis
Shape Functions
b
y
a
x
1
v1

z
u2
u1
w1
v2
w2
 3x 2 2x 3 
w  Ym 1  2  3 
b
b 

 my 
Ym  sin 

 a 
 x   x   u 
u  1      1 Ym
 b   b  u 2 
 x   x   v1  a '
v  1      
Ym
 b   b  v 2  m
 w1 
 2x x 2   3x 2 2x 3   x 2 x  
 1 

x1 
 2   2  3  x 2    
b b   b
b  b
b   w 2 


 2 

these shape functions are
also known as [N]
12
Finite Strip Analysis
Strain-Displacement and [K]
K   t  BT DBdA
Plane stress Kuv comes
from these straindisplacement relations
Bending Kw comes
from these straindisplacement relations
 u1 
v 
u 
 1
   N    N d
v
u 2 
 v 2 
 w1 
 
w  N 1   Nd
w 2 
  2 
 x  
u x

  




v

y
 y 
  Bd  N'd
  u y  v x 

 xy  
 2w 
 2 
 2x 
    w2   Bd  N'd
 2 y 
  w 
 xy 


13
Finite Strip Analysis
Plane Stress Initial Stiffness
  aE1 abk 2m G 


 
2
b
6

 
 ak m  x E 2 ak m G 
 abk 2m E 2 aG 






4
4
6
2
b




K uv   t  
2
abk m G   ak m  x E 2 ak m G 
 aE1 abk 2m G 
aE
1
 





  


2
b
12
4
4
2
b
6

 


 
2
 ak  E
 abk m E 2 aG 
ak G
 ak m  x E 2 ak m G 
 m x 2  m 






4
4 
2b 
4
4 


 12
where: km 

symmetric 







2
 abk m E 2 aG 



6
2
b


Ey
m
Ex
E1 
E2 
a
1   x y
1   x y
 Fu1  
F  
 v1   
 Fu 2   

 
 Fv 2  

 0
 Fw1  
 M 1  0

 
 Fw 2  0
M   0
 2
K uv 
0 0 0
0 0 0
0 0 0
0 0 0
0 0 0 0  u 1 
 
0 0 0 0  v1 
0 0 0 0  u 2 
 
0 0 0 0  v 2 
 
  w1 
 

  1

  
w


 w 2 
 
   2  14
K
Finite Strip Analysis
Bending Initial Stiffness
  13ab 4
12a 2

  70 k m D y  5b k m D xy 




6a 2
6a


k
D

D


m
1
x
3

5b
b

  3a
a

 
k 2m D1  k 2m D xy 
5
5
 

2
  3a

11ab 4
   2 Dx 
k mDy 
420

K w    9abb
12
a


k 4m D y 
k 2m D xy 

5b

  140


6
a
6
a

2
   5b k m D1  b 3 D x 

2
  13ab k 4 D  a k 2 D 
 840 m y 5 m xy 


  a k 2m D1  3a D x 
2
10
b


 ab3 4

4ab 2

k mDy 
k m D xy 
15
 210



2ab 2
2a
k m D1  D x 
 
15
b


 13ab2 4

a 2

k m D y  k m D xy 
5
 840



a 2
3a
  k m D1  2 D x 
10
b


 3ab3 4

ab 2

k m D y  k m D xy 
15
 840



ab 2
a
  k m D1  D x

30
b


12a 2
 13ab 4

k mDy 
k m D xy 

5b
 70



6a 2
6a
  k m D1  3 D x 
5b
b


 11ab2 4

a 2

k m D y  k m D xy 
5
 420



3a 2
3a
  k m D1  2 D x 
5
b





symmetric














 ab3 4

4ab 2

k mDy 
k m D xy 
15
 210



2ab 2
2a


k
D

D

m 1
x 
15
b


 Fu1  
F  
 v1   
 Fu 2   

 
 Fv 2  

 0
 Fw1  
 M 1  0

 
 Fw 2  0
M   0
 2
K uv 
0 0 0
0 0 0
0 0 0
0 0 0
0 0 0 0  u 1 
 
0 0 0 0  v1 
0 0 0 0  u 2 
 
0 0 0 0  v 2 
 
  w1 
 

  1

  
w


 w 2 
 
  2 
K
15
Finite Strip Analysis
What is [Kg]?
• [Kg] is the stress dependent geometric stiffness,
(compressive stresses erode stiffness) The terms may
be derived through
– consideration of the total potential energy due to in-plane
forces, or
– equivalently consider equilibrium in the deformed geometry,
(i.e., consider the moments that develop in the deformed
geometry due to forces which are in-plane in the undeformed
geometry), also
– one can consider Kg as a direct manifestation of higher order
strain terms.
16
Finite Strip Analysis
Developing [Kg]
The tractions correspond to linear edge stresses f1 and f2 via T1 = f1t and T2 = f2t.
 
x T

K g     T1  T1  T2  G  G dxdy
b
0 0
y
a
a b
x
z
T1
T2
 u

 y
v
y
T
w 
  G d
y 
where {d} is the nodal displacements
the same shape functions as before are used,
therefore [G] is determined through
partial differentiation of [N].
17
Finite Strip Analysis
Geometric Stiffness
0
70T1  T2 
0
0
0
0
0
703T1  T2 



703T1  T2 
0
70T1  T2 
0
0
0
0




70T1  3T2 
0
0
0
0
0


70T1  3T2 
0
0
0
0


Kg  C

830T1  9T2  2b15T1  7T2 
54T1  T2 
 2b7T1  6T2  


b 2 5T1  3T2  2b6T1  7T2 
 3b 2 T1  T2  


symmetric
243T1  10T2   2b7T1  15T2 


b 2 3T1  5T2  

 
C  bm 1680a
2
 Fu1  
F  
 v1   
 Fu 2   

 
 Fv 2  

 0
 Fw1  
 M 1  0

 
 Fw 2  0
M   0
 2
K uv 
0 0 0
0 0 0
0 0 0
0 0 0
0 0 0 0  u 1 
 
0 0 0 0  v1 
0 0 0 0  u 2 
 
0 0 0 0  v 2 
 
  w1 
 

  1

  
w


 w 2 
 
  2 
K
18
Finite Strip Analysis
Forming the Complete Solution
• The stiffness matrix for the member is formed by
summing the element stiffness matrices (this is
done in exactly the same manner as the stiffness
solution for frame analysis)
– generate stiffness matrices in local coordinates
– transform to global coordinates ([k]n=Tk’
– add contribution of each strip to global stiffness,
symbolically:
#strips
K   k n
n 1
K    k 
#strips
g
n 1
g n
19
Finite Strip Analysis
Eigenvalue Solution
K d  K g d
or
K  K d  d
1
g
• The solution yields
–  the multiplier which gives the buckling stress
– {d} the buckling mode shapes
• The solution is performed for all lengths of
interest to develop a complete picture of the elastic
buckling behavior
20
Finite Strip Analysis
Overview
•
•
•
•
•
•
•
•
Introduction
Background
A simple verification problem
Analyze a typical section
Interpreting results (k, fcr, Pcr, Mcr)
Direct Strength Prediction
Improve a typical section
Individual analysis - “do it yourself”
21
Finite Strip Analysis
A Simple Verification Problem
• Find the elastic buckling stress of a simply
supported plate using finite strip analysis.
SS
SS
SS
SS
width = 6 in. (152 mm)
thickness = 0.06 in. (1.52 mm)
E = 29500 ksi (203000 MPa)
v = 0.3
Hand Solution:
2 E  t 
f cr  k
 
12 1   2  b 

2

 2 _____
f cr  ___
12 1  ___ 2

2

 ___ 

 = 10.665 ksi
___


22
Finite Strip Analysis
Finite Strip Analysis Notes
(analysis of SS plate)
•
•
•
double click procefsm.exe
enter elastic properties into the box
enter the node number, x coordinate, z coordinate, and applied stress
–
–
•
enter the element number, starting node number, ending node number, number of strips
between the nodes (at least 2 typically 4 or more) and the thickness
–
•
•
•
•
1,0.0,0.0,1.0 which means node 1 at 0,0 with a stress of 1.0
2,6.0,0.0,1.0 which means node 2 at 6,0 with a stress of 1.0
1,1,2,2,0.06 which means element 1 goes from node 1 to 2, put 2 strips in there and t=0.06
select plot cross-section to see the plate
enter a member length (say 6) and number of different half wavelengths (say 10)
do File - Save As - plate.inp
now select view/revise raw data file
23
Finite Strip Analysis
Finite Strip Analysis Notes
(analysis of SS plate) continued
•
View/Revise Raw Data File shows the actual text file that is used by the finite strip
analysis program. All detailed modifications must be made here before completing the
analysis. The format of the file is summarized as:
#nodes #elements #lengths
Ex Ey x y G
node# xco ordinate zco ordinate xDOF zDOF yDOF DOF stress
[repeat until all nodes are defined].
element# nodei nodej
t
[repeat until all elements are defined].
len1 len2 . . .[enter a total of #lengths lengths]
•
•
The x, y, z,  degrees of freedom are shown in the strip
to the right.
Supported degrees of freedom are supported along the
entire length (edge) of the strip. The ends of the strip
are simply supported (due to the selected shape
1
functions). Set a DOF variable to 0 to
support that DOF along the edge
w1
b
y
a
x
v1

v2
z
u2
u1
w2
24
Finite Strip Analysis
Finite Strip Analysis Notes
(analysis of SS plate) continued
•
First modify degrees of freedom so the plate is simply supported along the long edges
(the loaded edges are always simply supported). Put a pin along the left edge and a roller
along the right edge.
–
–
–
•
•
•
•
•
•
•
1 0 0 1 1 1 1 1.0
2 3 0 1 1 1 1 1.0
3 6 0 1 1 1 1 1.0
becomes
stays the same
becomes
1 0 0 0 0 1 1 1.0
1 6 0 1 0 1 1 1.0
Now delete the last line and replace it with the specific lengths that you want to use, say
for instance “3 4 5 6 7 8”
Now change the #lengths listed in the thrid column of the first line of the file to match
the selected number, in this example we have 6 different lengths
Now select Save for Finite Strip Analysis and save under the name plate.txt
Select Analysis - Open
Then type ‘plate.txt’ for the input file and ‘plate.out’ for the output file
Select Output - then plate.out - and open
Select Plot curve and plot mode, the result of this example is 10.69 ksi (vs. 10.665 ksi
hand solution - repeat using 4 strips - then result is 10.666 ksi)
25
Finite Strip Analysis
Overview
•
•
•
•
•
•
•
•
Introduction
Background
A simple verification problem
Analyze a typical section
Interpreting results (k, fcr, Pcr, Mcr)
Direct Strength Prediction
Improve a typical section
Individual analysis - “do it yourself”
26
Finite Strip Analysis
Analyze a typical section
• Pure bending of a C
– Quickie hand analysis
– Finite strip analysis using procefsm.exe
– Discussion
• Pure compression of a C
– Quickie hand analysis
– Finite strip analysis
– Discussion
• Comparisons and Further Discussion
27
Finite Strip Analysis
Strong Axis Bending of a C
2.44
0.84
• Approximate the buckling stress for pure bending.
2 E  t 
f cr  k
2  
12 1    b 
8.44

0.059
2

E  29500ksi
  0.3
 _____
comp. lip f cr  ___
12 1  ___ 2
2

 _____
comp. flange f cr  ___
12 1  ___ 2

2

 _____
web f cr  ___
12 1  ___ 2

2


 ___ 


 ___ 
2
 ___ 


 ___ 
2
 ___ 


 ___ 
2
28
Finite Strip Analysis
Strong Axis Bending of a C
2.44
0.84
2 E  t 
f cr  k
2  
12 1    b 
8.44

0.059
E  29500ksi
  0.3
• Approximate the buckling stress for pure bending.
2

 29500  0.059
lip f cr  0.43
  56.6ksi
2 
12 1  0.3  0.84 
2
2


2
*  29500  0.059
flange f cr  4
  62.4ksi
2 
12 1  0.3  2.44 
2


 29500  0.059
web f cr  24
  31.3ksi
2 
12 1  0.3  8.44 
2
2

* this k value would be fine-tuned by AISI B4.2

29
Finite Strip Analysis
Finite Strip Analysis Steps
(strong axis bending of a C)
•
•
•
•
•
•
•
•
•
•
•
•
•
Double click on procefsm.exe
Select File - Open - C.inp
Plot Cross Section
View/Revise Raw Data File
Go to bottom of text file and change lengths to “1 2 3 4 5 6 7 8 9 10 20 30 40 50 60 70
80 90 100”
Go to top of file (1st line 3rd entry) change the number of lengths from 20 to 19
Save for finite strip analysis as C.txt
Select Analysis - Open - execufsm
Enter in DOS window ‘C.txt’ return then ‘C.out’
Select output - ‘C.out’ - open
Check 2D, Check undef, push plot mode button
Push plot curve, set half wave-length to 5 rehit plot mode, set to 30 and plot
local buckling at 40ksi (~5 in. 1/2wvlngth), dist buckling at 52ksi (~30 in. 1/2wvlngth)
30
Finite Strip Analysis
local buckling
distortional
local
half-wavelength
buckling multiplier
31
Finite Strip Analysis
distortional
32
Finite Strip Analysis
Discussion
• Finite strip analysis identifies three distinct modes: local,
distortional, lateral-torsional
• The lowest multiplier for “each mode” is of interest. The
mode will “repeat itself” at this half-wavelength in longer
members
• Higher multipliers of the same mode are not of interest.
• The meaning of the
“half-wavelength” can be
readily understood from
the 3D plot. For example:
33
Finite Strip Analysis
Discussion
• How do I tell different modes?
– wavelength: local buckling should occur at wavelengths near or
below the width of the elements, longer wavelengths indicate a
different mode of behavior
– mode shape: in local buckling, nodes at fold lines should rotate
only, if they are translating then the local mode is breaking down
• What if more minimums occur?
– as you add stiffeners and other details more minima may occur,
every fold line in the plate adds the possibility of new modes.
Definitions of local and distortional buckling are not as well
defined in these situations. Use wavelength of the mode to help
you decide.
34
Finite Strip Analysis
Compression of a C
2.44
0.84
2 E  t 
f cr  k
2  
12 1    b 
8.44

0.059
E  29500ksi
  0.3
• Approximate the buckling stress for pure compression.
2

 29500  0.059
lip f cr  0.43
  56.6ksi
2 
12 1  0.3  0.84 
2
2


2
*  29500  0.059
flange f cr  4
  62.4ksi
2 
12 1  0.3  2.44 
2


 29500  0.059
web f cr  4
  5.2ksi
2 
12 1  0.3  8.44 
2
2

* this k value would be fine-tuned by AISI B4.2

35
Finite Strip Analysis
Finite Strip Analysis Steps
(compression of a C)
•
•
•
•
•
•
•
•
•
•
•
•
•
•
Double click on procefsm.exe
Select File - Open - C.inp
Change all applied stress to compression +1.0
Plot Cross Section
View/Revise Raw Data File
Go to bottom of text file and change lengths to “1 2 3 4 5 6 7 8 9 10 20 30 40 50”
Go to top of file (1st line 3rd entry) change the number of lengths from 20 to 14
Save for finite strip analysis as C.txt
Select Analysis - Open - execufsm
Enter in DOS window ‘C.txt’ return then ‘C.out’
Select output - ‘C.out’ - open
Check 2D, Check undef, push plot mode button
Push plot curve, set half wave-length to 6 and rehit plot mode
local buckling at 7.5ksi (pure compression)
36
Finite Strip Analysis
Finite Strip Analysis
Compression of a C
• fcr local = 7.5 ksi
• fcr distortional ~ 20 ksi (this value may be fine tuned by
selecting more lengths and re-analyzing)
• fcr overall at 80 in. = 29 ksi
37
Finite Strip Analysis
Comparision of Elastic Results
• Hand Analysis
– compression lip=56.6 flange=62.4 web=5.2 ksi
– bending
lip=56.6 flange=62.4 web=31.3 ksi
• Finite strip analysis
– compression local=7.5 distortional~20ksi
– bending
local=40 distortional=52 ksi
38
Finite Strip Analysis
Comparision of Results for
Buckling Stress of a C
• Hand Analysis
– Compression
• Lip = 56.6 ksi
• Flange = 62.4
• Web = 5.2
– Bending
• Lip = 56.6
• Flange = 62.4
• Web = 31.3
• Finite strip analysis
– Compression
• Local = 7.5 ksi
• Distortional ~ 20
– Bending
• Local = 40
• Distortional = 52
39
Finite Strip Analysis
Overview
•
•
•
•
•
•
•
•
Introduction
Background
A simple verification problem
Analyze a typical section
Interpreting results (k, fcr, Pcr, Mcr)
Direct Strength Prediction
Improve a typical section
Individual analysis - “do it yourself”
40
Finite Strip Analysis
Converting the results
• If f1 is the applied stress in the finite strip analysis and 
the multiplier that results from the elastic buckling then
fcr = f1 is known. How do we get k? Pcr? Mcr?
2
2
2
2


12
f
1



E
t
b




cr
• k is found via fcr  k
k 
 
2  
2


12
1


b

E
 
t
where:
b = element width of interest (flange, web, lip etc.)
• Pcr = Agfcr
• Mcr = Sgfcr
(as long as f1 is the extreme fiber stress of interest)
41
Finite Strip Analysis
Converting the results - Example
• For the C in pure compression what does the finite strip
analysis yield for the local buckling k of the web?


12f cr 1   2  b 
k
 
2 E
t

2

127.5 1  0.32  8.44 
k


 2 29500  0.059 
k  5.75
• For the flange?
127.51  0.32  
k
2 29500
k  0.48
2.44 


0
.
059


2
2
• Solutions are different when you recognize the interaction!
42
Finite Strip Analysis
Converting the results - Example
• For the C in compression what is the elastic critical local
buckling load? distortional buckling load? overall?
– From CU-FSM or hand calculation get the section properties
– (Pcr)local = Agfcr = 0.885in2*7.5ksi = 6.6 kips
– (Pcr)distortional = Agfcr = 0.885in2*20 ksi = 17.7 kips
– (Pcr)overall at 80 in. = Agfcr = 0.885in2*29 ksi = 25.7 kips
43
Finite Strip Analysis
Converting the results - Example
• For the C in bending what is the elastic critical local
buckling moment? distortional buckling moment?
– From CU-FSM or hand calculation get the section properties
– (Mcr)local = Sgfcr = 2.256in3*40 ksi = 90 in-kips
– (Mcr)distortional = Sgfcr = 2.256in3*52 ksi = 117 in-kips
44
Finite Strip Analysis
How can I use this information?
• Known
– local buckling load (Pcr)local from finite strip analysis
– distortional buckling load (Pcr)distortional from finite strip analysis
– overall or Euler buckling load (Pcr)Euler may be flexural, torsional,
or flexural-torsional in the special case of Kx=Ky=Kt then we may
use finite strip analysis results, in other cases hand calculations for
overall buckling of a column are used
– yield load (Py) from hand calculation
• Unknown
– design capacity Pn
• Methodology: Direct Strength Prediction
45
Finite Strip Analysis
Overview
•
•
•
•
•
•
•
•
Introduction
Background
A simple verification problem
Analyze a typical section
Interpreting results (k, fcr, Pcr, Mcr)
Direct Strength Prediction
Improve a typical section
Individual analysis - “do it yourself”
46
Finite Strip Analysis
Direct Strength Prediction
• The idea behind Direct Strength prediction is that with
(Pcr)local, (Pcr)distortional and (Pcr)Euler known an engineer
should be able to calculate the capacity reliably and
directly without effective width.
• Current work suggests the following approach for columns
– Find the inelastic long column buckling load (Pne) using the AISC
column curves already in the AISI Specification
– Check for local buckling using new curve (less conservative than
Winter) on the entire member with the max load limited to Pne
– Check for distortional buckling using Hancock’s curve (more
conservative than Winter) with the max load limited to Pne
– Design strength is the minimum
47
Finite Strip Analysis
Direct Strength for Columns
(1) Find inelastic long column buckling load
Pne
= 0.658 P
2c
y
 0.877 


1
.
5
for c
and  2  Py for c > 1.5
 c 
c= Py Pcre
Py = AgFy
Pcre = Minimum of the elastic column buckling load in
flexural, torsional, or torsional-flexural buckling, see Chapter C
(2) Local buckling strength*
Pnl
0.4

 P  0.4


P
= Pne for l  0.776 and 1  0.15 crl   crl  Pne for l > 0.776

 Pne   Pne 

l= Pne Pcrl
Pcrl = Elastic local column buckling load
48
Finite Strip Analysis
Direct Strength for Columns (cont.)
(3) Distortional buckling strength*
Pnd
0 .6
0 .6






P
P
= Pne for  d  0.561 and 1  0.25 crd   crd  Pne for d > 0.561

 Pne   Pne 

d= Pne Pcrd
Pcrd = Elastic distortional column buckling load
(4) Design Capacity
Pn
= minimum of Pnl , Pnd
 = …(ASD)
 = …(LRFD)
*these calculations include long column interaction, to ignore this interaction replace P ne with Py
49
Finite Strip Analysis
Column Example
• Consider the lipped C we have been analyzing. Assume 50
ksi yield, L=80 in. and Kx=Ky=Kt=1.0
• From finite strip analysis we know:
–
–
–
–
Pcrl = 6.6 kips
Pcrd = 17.7 kips
Pcre = 25.7 kips
also Py = Agfy = 0.885*50 = 44.25 kips
50
Pcrl = 6.6 kips
Pcrd = 17.7 kips
Pcre = 25.7 kips
Py = 44.25 kips
Finite Strip Analysis
Column Example (cont.)
(1) Find inelastic long column buckling load
c=
Py Pcre =
Pne
= 0.658 P
Pne
=
2c
y
 0.877 


1
.
5
for c
and  2  Py for c > 1.5
 c 
(2) Local buckling strength*
l=
Pne Pcrl =
Pnl
0 .4

 P  0.4


P
crl
  crl  Pne for l > 0.776
= Pne for l  0.776 and 1  0.15
Pne   Pne 



Pnl
=
51
Pcrl = 6.6 kips
Pcrd = 17.7 kips
Pcre = 25.7 kips
Py = 44.25 kips
Finite Strip Analysis
Column Example (cont.)
(3) Distortional buckling strength*
d=
Pnd
Pnd
Pne Pcrd =
0.6
0.6






P
P
= Pne for  d  0.561 and 1  0.25 crd   crd  Pne for d > 0.561

 Pne   Pne 

=
(4) Design Capacity
Pn
= minimum of Pnl , Pnd =
 = …(ASD)
 = …(LRFD)
*these calculations include long column interaction, to ignore this interaction replace P ne with Py
52
Pcrl = 6.6 kips
Pcrd = 17.7 kips
Pcre = 25.7 kips
Py = 44.25 kips
Finite Strip Analysis
Column Example (cont.)
(1) Find inelastic long column buckling load
c=
Pne
Pne
Py Pcre = 1.31


 0.877 
2
= 0.658 c Py for  c  1.5 and  2  Py for c > 1.5
 c 
= 21.58 kips
(2) Local buckling strength*
l=
Pne Pcrl = 1.81
Pnl
0 .4

 P  0.4


P
= Pne for l  0.776 and 1  0.15 crl   crl  Pne for l > 0.776
 Pne   Pne 

Pnl
= 12.2 kips
53
Pcrl = 6.6 kips
Pcrd = 17.7 kips
Pcre = 25.7 kips
Py = 44.25 kips
Pne = 21.58 kips
Finite Strip Analysis
Column Example (cont.)
(3) Distortional buckling strength*
d=
Pnd
Pnd
Pne Pcrd = 1.10
0 .6

 P  0.6


P
crd
  crd  Pne for d > 0.561
= Pne for  d  0.561 and 1  0.25
 Pne   Pne 

= 14.9 kips
(4) Design Capacity
Pn
= minimum of Pnl , Pnd = 12.2, 14.9 = 12.2 kips (local buckling limits)
 = …(ASD)
 = …(LRFD)
*these calculations include long column interaction, to ignore this interaction replace P ne with Py
54
Finite Strip Analysis
Direct Strength for Beams
• Direct Strength prediction for beams follows the same
format as for columns.
– Find the inelastic lateral buckling load (Mne) using the strength
curves already in the AISI Specification
– Check for local buckling using new curve (less conservative than
Winter) on the entire member with the max moment limited to Mne
– Check for distortional buckling using Hancock’s curve (more
conservative than Winter) with the max moment limited to Mne
– Design strength is the minimum
•
Note, all the beams studied at this time have been laterally braced - therefore
the interaction between local and lateral buckling and distortional and lateral
buckling has not been examined. For now, it is conservatively assumed that
these modes can interact in the same manner as completed for columns. (This
is what we do now when we use the effective section modulus)
55
Finite Strip Analysis
Direct Strength for Beams
(1) Find inelastic lateral buckling load
Mne
= My if M cre  2.78 M y Mcre if M cre  0.56 M y and
=
10
9


M y 1  36 M crey if 2.78My > Mcre > 0.56My
10 M
My = SgFy
Mcre = Elastic lateral buckling load of the beam
(2) Local buckling strength*
Mnl
0. 4

 M  0.4


M
= Mne for l  0.776 and 1  0.15 crl   crl  M ne for l > 0.776

 M ne   M ne 

l= M ne M crl
Mcrl = Elastic local beam buckling moment
56
Finite Strip Analysis
Direct Strength for Beams (cont.)
(3) Distortional buckling strength*
0.6

 M 0.6


M
Mnd = Mne for  d  0.561 and 1  0.25 crd   crd  M ne for d > 0.561

 M ne   M ne 

d= M ne M crd
Mcrd = Elastic distortional beam buckling moment
(4) Design Capacity
Mn
= minimum of Mnl , Mnd
 = …(ASD)
 = …(LRFD)
*these calculations include long column interaction, to ignore this interaction replace P ne with Py
57
Finite Strip Analysis
Beam Example
• Consider the lipped C we have been analyzing. Assume 50
ksi yield, assume the member is laterally braced, but
distortional buckling is still free to form and thus a
concern. Find the nominal capacity.
• From finite strip analysis we know:
–
–
–
–
Mcrl = 90 in-kips
Mcrd = 117 in-kips
Mcre = braced
also My = Sgfy = 2.256*50 = 113 in-kips
58
Mcrl = 90 in-kips
Mcrd = 117 in-kips
Mcre = braced
My = 113 in-kips
Finite Strip Analysis
Beam Example (cont.)
(1) Find inelastic lateral buckling load
Mne
= My if M cre  2.78 M y Mcre if M cre  0.56 M y and
=
10
9


M y 1  36 M crey if 2.78My > Mcre > 0.56My
10 M
My = SgFy
Mcre = Elastic lateral buckling load of the beam
Mne = My since “braced” = 113 in-kips
(2) Local buckling strength*
Mnl
0. 4

 M  0.4


M
= Mne for l  0.776 and 1  0.15 crl   crl  M ne for l > 0.776

 M ne   M ne 

l= M ne M crl
Mcrl = Elastic local beam buckling moment
Mnl = 89 in-kips
59
Mcrl = 90 in-kips
Mcrd = 117 in-kips
Mcre = braced
My = 113 in-kips
Finite Strip Analysis
Beam Example (cont.)
(3) Distortional buckling strength*
0.6

 M 0.6


M
Mnd = Mne for  d  0.561 and 1  0.25 crd   crd  M ne for d > 0.561

 M ne   M ne 

d= M ne M crd
Mcrd = Elastic distortional beam buckling moment
Mnd = 86 in-kips
(4) Design Capacity
Mn
= minimum of Mnl , Mnd
 = …(ASD)
 = …(LRFD)
Mn = 86 in-kips, distortional controls even though elastic critical is 30% higher
*these calculations include long column interaction, to ignore this interaction replace P ne with Py
60
Finite Strip Analysis
Direct Strength Verification
• Existing experimental data on laterally braced beams and
centrally loaded columns (unbraced) have been collected
and evaluated
– Laterally Braced Beams: Experimental data includes lipped and
unlipped C’s, Z’s, rectangular and trapezoidal decks w/ and w/o
int. stiffener(s) in the web and flange for a total of 574 members
– Columns: Experimental data includes lipped C's, Z's, lipped C's
with int. web stiffeners, racks, racks with compound lips for a total
of 227 members
• Finite strip analysis was conducted on each member, then
combined with the experimental results to compare vs. the
strength curves suggested for Direct Strength prediction
61
Finite Strip Analysis
Laterally Braced Beams
1.5
Distortional
Local
Winter
Distortional Curve
1
Local Curve
M test
My
0.5
0
0
1
2
max 
3
M y M cr
4
5
62
Finite Strip Analysis
Columns
1.6
1.4
1.2
local buckling controlled
1
u
strength F /F
n
distortional buckling controlled
0.8
0.6
local strength curve
0.4
0.2
distortional strength curve
0
0
1
2
3
4
5
6
slenderness of controlling mode (F
/F
n
cr
)
7
8
.5
63
Finite Strip Analysis
Direct Strength Verification (cont.)
• As the data shows, trends are clear, but large scatter can
exist for a particular member.
• Using separate strength curves given herein for local and
distortional limits
– Laterally braced beams have test to predicted ratio of 1.14 for all
data, 1.05 for local limits and 1.16 for distortional limits
– Columns have test to predicted ratio of 1.01 for all data
• Preliminary calibration using the strength curves suggested
herein appear to support the use of traditional =0.9 for
beams and =0.85 for columns (presumably ASD factors
would also remain unchanged)
64
Finite Strip Analysis
Deflection Calculations
It is anticipated that degradation of gross properties (i.e, Ag Ig)
due to local/distortional/overall buckling may be
approximated in the same manner as the degradation in the
strength, e.g.,
• Example strength & deflection
calc. completed at fy
• Example strength & deflection
calc. completed at fa
Pa  Ag f a
Py  Ag f y
Pne  0.658 y
P Pcre
Py
0.4

 P 0.4


P
Pnl  1  0.15 nl   nl  Pne

 Pne   Pne 

P
Aeff (at f y )  nl
fy
Pne  0.658Pa
( fa  f y )
Pcre
Pa
0.4

 P 0.4


P
Pnl  1  0.15 nl   nl  Pne

 Pne   Pne 

P
Aeff (at f a )  nl
fa
65
Finite Strip Analysis
Limitations of Direct Strength
• Conservative solution when one element is extraordinarily
slender (fcr approaches zero and the strength with it)
• Interaction of local-distortional-overall buckling not
studied thoroughly for beams and deserves further study
for columns
• Calibration of methods incomplete
• Impact of proposed changes incomplete
• Integration into existing design methods incomplete
• ...
66
Finite Strip Analysis
Overview
•
•
•
•
•
•
•
•
Introduction
Background
A simple verification problem
Analyze a typical section
Interpreting results (k, fcr, Pcr, Mcr)
Direct Strength Prediction
Improve a typical section
Individual analysis - “do it yourself”
67
Finite Strip Analysis
Improve a Section
• Examine the results for the C section in
compression that we previously evaluated
• Suggest alternative design
• Implement the design change
• Evaluate the finite strip results
• Recalculate the strength
68
Finite Strip Analysis
Previous Results
69
Finite Strip Analysis
Alternative Design Suggestions
• Intermediate stiffener in the web
• Multiple intermediate stiffeners in the web
• Decrease width of one flange in order to
incorporate a small fold in the web so that
sections could be nested together
• others...
– ________________________
– ________________________
70
Finite Strip Analysis
Suggested Alternative
• Old
• New
(0.50,0.00)
(0.50,1.50)
(2.44,0.0)
(2.44,0.84)
(0.00,2.00)
(2.44,7.60)
(0.00,8.44)
(2.44,8.44)
71
Finite Strip Analysis
72
Finite Strip Analysis
Convert stress to load
• For the C in compression what is the elastic critical local
buckling load? distortional buckling load? overall?
– From CU-FSM or hand calculation get the section properties
–
–
–
–
(Pcr)local = Agfcr = 0.868in2*13.2ksi = 11.4 kips
(Pcr)distortional = Agfcr = 0.868in2*28 ksi = 24.3 kips
(Pcr)overall at 80 in. = Agfcr = 0.868in2*26 ksi = 22.6 kips
Py for 50 ksi yield = Agfy = 0.868in2*50 ksi = 43.4 kips
73
Pcrl = 11.4 kips
Pcrd = 24.3 kips
Pcre = 22.6 kips
Py = 43.4 kips
Finite Strip Analysis
Direct Strength
(1) Find inelastic long column buckling load
c=
Py Pcre = 1.38
Pne
= 0.658 P
Pne
= 19.4 kips
2c
y
 0.877 


1
.
5
c
for
and  2  Py for c > 1.5
 c 
(2) Local buckling strength*
l=
Pne Pcrl = 1.30
Pnl
0 .4

 P  0.4


P
crl
  crl  Pne for l > 0.776
= Pne for l  0.776 and 1  0.15
 Pne   Pne 

Pnl
= 13.7 kips
74
Pcrl = 11.4 kips
Pcrd = 24.3 kips
Pcre = 22.6 kips
Py = 43.4 kips
Pne = 19.4 kips
Finite Strip Analysis
Direct Strength (cont.)
(3) Distortional buckling strength*
d= Pne Pcrd = 0.89
Pnd
0 .6

 P  0.6


P
= Pne for  d  0.561 and 1  0.25 crd   crd  Pne for d > 0.561
 Pne   Pne 

Pnd
= 15.8 kips
(4) Design Capacity
Pn
= minimum of Pnl , Pnd = 13.7, 15.8 = 13.7 kips (local buckling limits)
 = …(ASD)
 = …(LRFD)
*these calculations include long column interaction, to ignore this interaction replace P ne with Py
75
Finite Strip Analysis
Comparison
• “typical” C
• “nestable” C
•
•
•
•
•
•
•
Pcrl
Pcrd
Pcre
Py
Pne
Pnl
Pnd
= 6.6 kips
= 17.7 kips
= 25.7 kips
= 44.2 kips
= 21.6 kips
= 12.2 kips
= 14.9 kips
•
•
•
•
•
•
•
Pcrl
Pcrd
Pcre
Py
Pne
Pnl
Pnd
= 11.4 kips
= 24.3 kips
= 22.6 kips
= 43.4 kips
= 19.4 kips
= 13.7 kips
= 15.8 kips
•
Pn
= 12.2 kips
•
Pn
= 13.7 kips
76
Finite Strip Analysis
Extras
• Changing boundary conditions
– eliminate a mode you are not interested in by temporarily
supporting a DOF
– use symmetry and anti-symmetry to examine different modes and
behavior
– bound solutions by looking at fix vs. free conditions
77
Finite Strip Analysis
Extras
• Adding an elastic support
– you want to add elastic springs to the cross-section to model
external (continuous) support. This can be done only indirectly by
adding an unloaded strip to your model
– The elastic stiffness kx, ky, kz and k of a single strip is given
previously. Note selection of strip width, t, and boundary
conditions will generate all 4 k’s currently no method is available
for adding springs for only one DOF
78
Finite Strip Analysis
Extras
b
y
a
x
1
v1

z
u2
u1
w1
v2
w2
 Fu1  
F  
 v1   
 Fu 2   

 
 Fv 2  

 0
 Fw1  
 M 1  0

 
F
 w 2  0
M   0
 2
0 0
K uv 
0 0
0 0
0 0
0 0 0
0 0 0
0 0 0
0 0 0



K
0 0  u 1 
 
0 0  v1 
0 0  u 2 
 
0 0  v 2 
 
  w1 

   1 
w   w 
 2
 
  2 
0
0
0
0
1
2
defining a single unloaded strip fixed at the 1 edge
and attached at the 2 edge will add the above elastic
stiffness to the solution wherever the 2 edge is attached
to the member. Remember, this stiffness is along the
length of the member (the length of the strip)
in order to make an unloaded strip you will have to create
a very short dummy element near the 2 edge because loading
is defined at the nodes not the elements
79
Finite Strip Analysis
Extras
• Assemblage
– 2 C’s linked
together
vs. a single
C section
results are
in bending.
CU-FSM
used for
analysis
80
Finite Strip Analysis
Extras
• Beam-columns
– load with the expected stress distribution
instead of doing separate beam and column
analysis
– Make your own elastic buckling interaction
curve for a particular shape
81
Finite Strip Analysis
Extras
• Matlab version (free, but you need Matlab)
– Robust graphical interface
– Calculates section properties
– Add P and M, or combinations thereof directly instead of
adding stress
– Compare two analyses directly
– Perform parametric studies
– Completely free and available source, modify as you wish
– Write your own programs and software that call any of the
finite strip routines
82
Finite Strip Analysis
Limitations of Finite Strip
• Elastic only: elastic buckling stress is useful, but it is
– not an allowable stress,
– not necessarily the stress at which buckling “ensues”,
– not a conservative design ultimate stress.
• Optimum design
– Do not design so that the elastic buckling stress of all modes is at
or near the same stress - this invariably leads to coupled
instabilities and should be avoided - note that different strength
curves are used for different modes.
• Identification of minima
– Certain members and certain loadings blur the distinctions between
modes. Exercise judgment and remain conservative when in doubt,
i.e., the distortional buckling strength curve is more conservative
than the local buckling curve.
83
Finite Strip Analysis
Limitations of Finite Strip
• Mixed wavelength modes
– Finite strip analysis can not identify situations when the
wavelength in the flange differs from that in the web. Finite
element analysis indicates this can happen in certain
circumstances. Analysis to date shows that the finite strip
formulation does not lead to undo errors, but further work needed.
• Point-wise bracing, punch-outs...
– Any item that discretely varies along the length must be “smeared”
in the current approach. The benefit of individual braces is difficult
to quantify. This is a practical limit not a theoretical limit of the
current approach. For now, engineering judgment is required when
evaluating bracing, punch-outs etc..
• Shear interaction
84
Finite Strip Analysis
Overview
•
•
•
•
•
•
•
•
Introduction
Background
A simple verification problem
Analyze a typical section
Interpreting results (k, fcr, Pcr, Mcr)
Direct Strength Prediction
Improve a typical section
Individual analysis - “do it yourself”
85
Finite Strip Analysis
Geometry of Examples
3.43
2.44
2.44
0.97
0.84
3.0
4
0.81
8.44
0.060
3.95
8.44
4
0.083
0.059
C_nolip
0.059
15.93
L
C_rack
4
C
0.060
5.0
1.0
4
0.068
2.06
L_lip
C_deep
0.25
0.90@50o
2.44
10.0
0.68@50o
11.93
8.44
Z_deep
0.070
0.075
0.059
Z
6
plate
H
86
drawings not to scale, all dim. in inches
Finite Strip Analysis
END
87
Finite Strip Analysis
Classical Derivation for k
• Classical methods tend to use an energy
approach (following Timoshenko)
–
–
–
–
–
assume a series for deflected shape (must match boundary conditions)
determine the internal strain energy (independent of loading)
determine the external work (dependent on loading)
form the total potential energy
take the variation of the total potential energy with respect to the
amplitude coefficients of the series and set to zero
– determine the desired number of terms to be used in the series
– calculate the buckling stress and k
88
Finite Strip Analysis
Classical derivation for k (cont.)
consider solution for a simply supported plate with a stress gradient
 mx   ny 
w   a mn sin
 sin

 a   b 
m n
2
2
2
t/2
b
a w
E
 w
2
U plate 
z dz    2  2  dxdy
2  t / 2
0
0
21   
y 
 x
y  w 

1
Wplate  2   f1t 1   
 dxdy
b  x 

0 0
b a
displaced shape
internal energy
2
external work
  Uplate  Wplate
total potential energy

0
a mn
variation and solution
89