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1
Finite Strip Analysis
local distortional later-torsional
length of a half sine wave
buck
ling
mul
tiplie
r (s
tres
s, lo
ad,
or m
omen
t)
Finite Strip Analysisand the Beginnings of the Direct Strength Method
Toronto, July 2000
AISI Committee on Specifications
2
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”
3
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.
4
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
6
Finite Strip Analysis
What has to be defined?
nodesgive node numbergive coordinatesindicate if any additional support exist along the longitudinal edgegive applied stress on node
elementsgive element numbergive nodes that form the stripgive thickness of the strip
propertygive E, G, and v
lengthsgive all the lengths that elastic buckling should be examined at
7
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
8
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”
9
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
10
Finite Strip Analysis
Elastic Buckling (Matrix form)• standard initial elastic
stiffness form
• consider effect of stress on stiffness
• consider linear multiples of constant stress (f1)
• eigenvalue problem gives solution
dKF
dfKKF g
dfKKF 1g
dKdK g
11
Finite Strip Analysis
What is [K]?
x
y
z
u1
v1
w1
1u2
v2
w2
a
b
2
2
1
1
2
2
1
1
w
w
v
u
v
u
w
w
v
u
v
u
w
0000
0000
0000
00000000
0000
0000
0000
uv
M
F
M
F
F
F
F
F
K
K
2
2
1
1
2
2
1
1
dKF
• [K] is the initial elastic stiffness
12
Finite Strip Analysis
Shape Functions
m2
1 Yu
u
b
x
b
x1u
'm
2
1 Ym
a
v
v
b
x
b
x1v
a
ymsinYm
2
2
1
1
2
2
3
3
2
2
2
2
3
3
2
2
m w
w
b
x
b
xx
b
x2
b
x3
b
x
b
x21x
b
x2
b
x31Yw
x
y
z
u1
v1
w1
1u2
v2
w2
a
b
these shape functions arealso known as [N]
13
Finite Strip Analysis
Strain-Displacement and [K] dABDBtK T
dN
v
u
v
u
Nv
u
2
2
1
1
d'NdB
xvyu
yv
xu
xy
y
x
d'NdB
yx
wy
wx
w
2
2
2
2
2
Plane stress Kuv comesfrom these strain-displacement relations
dNw
w
Nw
2
2
1
1
Bending Kw comesfrom these strain-displacement relations
14
Finite Strip Analysis
Plane Stress Initial Stiffness
b2
aG
6
Eabk
4
Gak
4
Eak
b2
aG
12
Eabk
4
Gak
4
Eak
6
Gabk
b2
aE
4
Gak
4
Eak
12
Gabk
b2
aE
b2
aG
6
Eabk
4
Gak
4
Eak
symmetric6
Gabk
b2
aE
tK
22mm2xm2
2mm2xm
2m1m2xm
2
1
22mm2xm
2m1
uvm
where: km
am EEx
xy11 EEy
xy21
2
2
1
1
2
2
1
1
w
w
v
u
v
u
w
w
v
u
v
u
w
0000
0000
0000
00000000
0000
0000
0000
uv
M
F
M
F
F
F
F
F
K
K
2
2
1
1
2
2
1
1
15
Finite Strip Analysis
Bending Initial Stiffness
x12m
xy2my
4m
3
x212m
xy2my
4m
2
x12m
xy2my
4m
3
x212m
xy2my
4m
2
x312m
xy2my
4m
x212m
xy2my
4m
2
x312m
xy2my
4m
x12m
xy2my
4m
3
y4m
2
x2
xy2m1
2m
x312m
xy2my
4m
w
Db
a2Dk
15
ab2
Dk15
ab4Dk
210
ab
Db
a3Dk
5
a3
Dk5
aDk
420
ab11
Db
aDk
30
ab
Dk15
abDk
840
ab3
Db
a3Dk
10
a
Dk5
aDk
840
ab13
Db
a6Dk
b5
a6
Dkb5
a12Dk
70
ab13
Db
a3Dk
10
a
Dk5
aDk
840
ab13
Db
a6Dk
b5
a6
Dkb5
a12Dk
140
ab9
Db
a2Dk
15
ab2
Dk15
ab4Dk
210
ab
Dk420
ab11D
b
a3
Dk5
aDk
5
a3
symmetric
Db
a6Dk
b5
a6
Dkb5
a12Dk
70
ab13
K
2
2
1
1
2
2
1
1
w
w
v
u
v
u
w
w
v
u
v
u
w
0000
0000
0000
00000000
0000
0000
0000
uv
M
F
M
F
F
F
F
F
K
K
2
2
1
1
2
2
1
1
16
Finite Strip Analysis
• [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.
What is [Kg]?
17
Finite Strip Analysis
Developing [Kg]The tractions correspond to linear edge stresses f1 and f2 via T1 = f1t and T2 = f2t.
x
y
z
a
T T1
2
a
0
b
0
T211g dxdyGG
b
xTTTK
dGy
w
y
v
y
uT
where {d} is the nodal displacementsthe same shape functions as before are used,therefore [G] is determined through partial differentiation of [N].
18
Finite Strip Analysis
Geometric Stiffness
212
2121
212
21212
21212121
21
21
2121
2121
g
T5T3b
T15T7b2T10T324symmetric
TTb3T7T6b2T3T5b
T6T7b2TT54T7T15b2T9T308
0000T3T70
00000T3T70
0000TT700TT370
00000TT700TT370
CK
a1680mbC 2
2
2
1
1
2
2
1
1
w
w
v
u
v
u
w
w
v
u
v
u
w
0000
0000
0000
00000000
0000
0000
0000
uv
M
F
M
F
F
F
F
F
K
K
2
2
1
1
2
2
1
1
19
Finite Strip Analysis
• 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:
Forming the Complete Solution
strips#
1nnkK
strips#
1nngg kK
20
Finite Strip Analysis
Eigenvalue Solution
ddKK
or
dKdK
1g
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
21
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”
22
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
2
2
2
cr b
t
112
Ekf
2
2
2
cr ___
___
___112
________f
Hand Solution:
= 10.665 ksi
23
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– 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
• 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
– 1,1,2,2,0.06 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
24
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:
• The x, y, z, degrees of freedom are shown in the stripto the right.
• Supported degrees of freedom are supported along theentire length (edge) of the strip. The ends of the stripare simply supported (due to the selected shapefunctions). Set a DOF variable to 0 to support that DOF along the edge
#nodes #elements #lengthsEx Ey x y Gnode# xcoordinate zcoordinate 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]
x
y
z
u1
v1
w1
1u2
v2
w2
a
b
25
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 becomes 1 0 0 0 0 1 1 1.0
– 2 3 0 1 1 1 1 1.0 stays the same
– 3 6 0 1 1 1 1 1.0 becomes 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)
26
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”
27
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
28
Finite Strip Analysis
Strong Axis Bending of a C• Approximate the buckling stress for pure bending.
8.44
2.44
0.84
0.059 2
2
2
cr b
t
112
Ekf
3.0
ksi29500E
2
2
2
cr ___
___
___112
________f
2
2
2
cr ___
___
___112
________f
comp. lip
comp. flange
web
2
2
2
cr ___
___
___112
________f
29
Finite Strip Analysis
Strong Axis Bending of a C• Approximate the buckling stress for pure bending.
8.44
2.44
0.84
0.059 2
2
2
cr b
t
112
Ekf
3.0
ksi29500E
ksi3.3144.8
059.0
3.0112
2950024f
2
2
2
cr
ksi4.6244.2
059.0
3.0112
295004f
2
2
2*
cr
lip
flange
web
ksi6.5684.0
059.0
3.0112
2950043.0f
2
2
2
cr
* this k value would be fine-tuned by AISI B4.2
30
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)
33
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 bereadily understood fromthe 3D plot. For example:
34
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.
35
Finite Strip Analysis
Compression of a C• Approximate the buckling stress for pure
compression.8.44
2.44
0.84
0.059 2
2
2
cr b
t
112
Ekf
3.0
ksi29500E
ksi2.544.8
059.0
3.0112
295004f
2
2
2
cr
ksi4.6244.2
059.0
3.0112
295004f
2
2
2*
cr
lip
flange
web
ksi6.5684.0
059.0
3.0112
2950043.0f
2
2
2
cr
* this k value would be fine-tuned by AISI B4.2
36
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)
37
Finite Strip Analysis
Finite Strip AnalysisCompression 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
38
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
39
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
40
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”
41
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 thenfcr = f1 is known. How do we get k? Pcr? Mcr?
• k is found viawhere:b = element width of interest (flange, web, lip etc.)
• Pcr = Agfcr
• Mcr = Sgfcr (as long as f1 is the extreme fiber stress of interest)
2
2
2cr
2
2
2
cr t
b
E
1f12k
b
t
112
Ekf
42
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?
• For the flange?
• Solutions are different when you recognize the interaction!
75.5k
059.0
44.8
29500
3.015.712k
t
b
E
1f12k
2
2
2
2
2
2cr
48.0k
059.0
44.2
29500
3.015.712k
2
2
2
43
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
44
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
45
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
46
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”
47
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
48
Finite Strip Analysis
Direct Strength for Columns(1) Find inelastic long column buckling load
Pne = yP658.02c for 5.1c and y2
c
P877.0
for c > 1.5
c= crey PP
Py = AgFy
Pcre = Minimum of the elastic column buckling load inflexural, torsional, or torsional-flexural buckling, see Chapter C
(2) Local buckling strength*
Pnl = Pne for 776.0l and ne
4.0
ne
crl
4.0
ne
crl PP
P
P
P15.01
for l > 0.776
l= crlnePP
Pcrl = Elastic local column buckling load
49
Finite Strip Analysis
Direct Strength for Columns (cont.)(3) Distortional buckling strength*
Pnd = Pne for 561.0d and ne
6.0
ne
crd
6.0
ne
crd PP
P
P
P25.01
for d > 0.561
d= crdnePP
Pcrd = Elastic distortional column buckling load
*these calculations include long column interaction, to ignore this interaction replace Pne with Py
(4) Design CapacityPn = minimum of Pnl , Pnd = …(ASD) = …(LRFD)
50
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
51
Finite Strip Analysis
Column Example (cont.)(1) Find inelastic long column buckling load
c= creyPP =
Pne = yP658.02c for 5.1c and y2
c
P877.0
for c > 1.5
Pne =
(2) Local buckling strength*
l= crlne PP =
Pnl = Pne for 776.0l and ne
4.0
ne
crl
4.0
ne
crl PP
P
P
P15.01
for l > 0.776
Pnl =
Pcrl = 6.6 kipsPcrd = 17.7 kipsPcre = 25.7 kipsPy = 44.25 kips
52
Finite Strip Analysis
Column Example (cont.)(3) Distortional buckling strength*
d= crdne PP =
Pnd = Pne for 561.0d and ne
6.0
ne
crd
6.0
ne
crd PP
P
P
P25.01
for d > 0.561
Pnd =
*these calculations include long column interaction, to ignore this interaction replace Pne with Py
(4) Design CapacityPn = minimum of Pnl , Pnd = = …(ASD) = …(LRFD)
Pcrl = 6.6 kipsPcrd = 17.7 kipsPcre = 25.7 kipsPy = 44.25 kips
53
Finite Strip Analysis
Column Example (cont.)(1) Find inelastic long column buckling load
c= creyPP = 1.31
Pne = yP658.02c for 5.1c and y2
c
P877.0
for c > 1.5
Pne = 21.58 kips
(2) Local buckling strength*
l= crlne PP = 1.81
Pnl = Pne for 776.0l and ne
4.0
ne
crl
4.0
ne
crl PP
P
P
P15.01
for l > 0.776
Pnl = 12.2 kips
Pcrl = 6.6 kipsPcrd = 17.7 kipsPcre = 25.7 kipsPy = 44.25 kips
54
Finite Strip Analysis
Column Example (cont.)(3) Distortional buckling strength*
d= crdne PP = 1.10
Pnd = Pne for 561.0d and ne
6.0
ne
crd
6.0
ne
crd PP
P
P
P25.01
for d > 0.561
Pnd = 14.9 kips
*these calculations include long column interaction, to ignore this interaction replace Pne with Py
(4) Design CapacityPn = minimum of Pnl , Pnd = 12.2, 14.9 = 12.2 kips (local buckling limits) = …(ASD) = …(LRFD)
Pcrl = 6.6 kipsPcrd = 17.7 kipsPcre = 25.7 kipsPy = 44.25 kipsPne = 21.58 kips
55
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)
56
Finite Strip Analysis
Direct Strength for Beams(1) Find inelastic lateral buckling load
Mne = My if ycre M78.2M Mcre if ycre M56.0M and
= cre
y
M36
M10
y910 1M if 2.78My > Mcre > 0.56My
My = SgFy
Mcre = Elastic lateral buckling load of the beam
(2) Local buckling strength*
Mnl = Mne for 776.0l and ne
4.0
ne
crl
4.0
ne
crl MM
M
M
M15.01
for l > 0.776
l= crlne MM
Mcrl = Elastic local beam buckling moment
57
Finite Strip Analysis
Direct Strength for Beams (cont.)(3) Distortional buckling strength*
Mnd = Mne for 561.0d and ne
6.0
ne
crd
6.0
ne
crd MM
M
M
M25.01
for d > 0.561
d= crdne MM
Mcrd = Elastic distortional beam buckling moment
*these calculations include long column interaction, to ignore this interaction replace Pne with Py
(4) Design CapacityMn = minimum of Mnl , Mnd = …(ASD) = …(LRFD)
58
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
59
Finite Strip Analysis
Beam Example (cont.)(1) Find inelastic lateral buckling load
Mne = My if ycre M78.2M Mcre if ycre M56.0M and
= cre
y
M36
M10
y910 1M if 2.78My > Mcre > 0.56My
My = SgFy
Mcre = Elastic lateral buckling load of the beam
(2) Local buckling strength*
Mnl = Mne for 776.0l and ne
4.0
ne
crl
4.0
ne
crl MM
M
M
M15.01
for l > 0.776
l= crlne MM
Mcrl = Elastic local beam buckling moment
Mcrl = 90 in-kipsMcrd = 117 in-kipsMcre = bracedMy = 113 in-kips
Mne = My since “braced” = 113 in-kips
Mnl = 89 in-kips
60
Finite Strip Analysis
Beam Example (cont.)(3) Distortional buckling strength*
Mnd = Mne for 561.0d and ne
6.0
ne
crd
6.0
ne
crd MM
M
M
M25.01
for d > 0.561
d= crdne MM
Mcrd = Elastic distortional beam buckling moment
*these calculations include long column interaction, to ignore this interaction replace Pne with Py
(4) Design CapacityMn = minimum of Mnl , Mnd = …(ASD) = …(LRFD)
Mcrl = 90 in-kipsMcrd = 117 in-kipsMcre = bracedMy = 113 in-kips
Mnd = 86 in-kips
Mn = 86 in-kips, distortional controls even though elastic critical is 30% higher
61
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
62
Finite Strip Analysis
Laterally Braced Beams
0
0.5
1
1.5
0 1 2 3 4 5
Distortional
Local
Winter
Distortional Curve
Local Curve
max M My cr
M
Mtest
y
63
Finite Strip Analysis
Columns
0 1 2 3 4 5 6 7 80
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
local buckling controlled
distortional buckling controlled
slenderness of controlling mode (Fn
/Fc r
).5
stre
ngth
F u/F
n
local strength curve
distortional strength curve
64
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)
65
Finite Strip Analysis
Deflection Calculations
• Example strength & deflection calc. completed at fy
• Example strength & deflection calc. completed at fa
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.,
ygy fAP
yPP
ne PP crey658.0
nene
nl
ne
nlnl P
P
P
P
PP
4.04.0
15.01
)( yaaga fffAP
aPP
ne PP crea658.0
nene
nl
ne
nlnl P
P
P
P
PP
4.04.0
15.01
y
nlyeff f
PfatA ) (
a
nlaeff f
PfatA ) (
66
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
• ...
67
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”
68
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
70
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...– ________________________– ________________________
71
Finite Strip Analysis
Suggested Alternative
• Old • New
(2.44,0.84)
(2.44,0.0)(0.50,0.00)
(0.50,1.50)
(0.00,2.00)
(0.00,8.44) (2.44,8.44)
(2.44,7.60)
73
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
74
Finite Strip Analysis
Direct Strength(1) Find inelastic long column buckling loadc= creyPP = 1.38
Pne = yP658.02c for 5.1c and y2
c
P877.0
for c > 1.5
Pne = 19.4 kips
(2) Local buckling strength*
l= crlne PP = 1.30
Pnl = Pne for 776.0l and ne
4.0
ne
crl
4.0
ne
crl PP
P
P
P15.01
for l > 0.776
Pnl = 13.7 kips
Pcrl = 11.4 kipsPcrd = 24.3 kipsPcre = 22.6 kipsPy = 43.4 kips
75
Finite Strip Analysis
Direct Strength (cont.)(3) Distortional buckling strength*
d= crdne PP = 0.89
Pnd = Pne for 561.0d and ne
6.0
ne
crd
6.0
ne
crd PP
P
P
P25.01
for d > 0.561
Pnd = 15.8 kips
*these calculations include long column interaction, to ignore this interaction replace Pne with Py
(4) Design CapacityPn = minimum of Pnl , Pnd = 13.7, 15.8 = 13.7 kips (local buckling limits) = …(ASD) = …(LRFD)
Pcrl = 11.4 kipsPcrd = 24.3 kipsPcre = 22.6 kipsPy = 43.4 kipsPne = 19.4 kips
76
Finite Strip Analysis
Comparison
• “typical” C• Pcrl = 6.6 kips
• Pcrd = 17.7 kips
• Pcre = 25.7 kips
• Py = 44.2 kips
• Pne = 21.6 kips
• Pnl = 12.2 kips
• Pnd = 14.9 kips
• Pn = 12.2 kips
• “nestable” C• Pcrl = 11.4 kips
• Pcrd = 24.3 kips
• Pcre = 22.6 kips
• Py = 43.4 kips
• Pne = 19.4 kips
• Pnl = 13.7 kips
• Pnd = 15.8 kips
• Pn = 13.7 kips
77
Finite Strip Analysis
• 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
Extras
78
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
79
Finite Strip Analysis
Extras
2
2
1
1
2
2
1
1
w
w
v
u
v
u
w
w
v
u
v
u
w
0000
0000
0000
00000000
0000
0000
0000
uv
M
F
M
F
F
F
F
F
K
K
2
2
1
1
2
2
1
1
x
y
z
u1
v1
w1
1u2
v2
w2
a
b
defining a single unloaded strip fixed at the 1 edgeand attached at the 2 edge will add the above elasticstiffness to the solution wherever the 2 edge is attachedto the member. Remember, this stiffness is along thelength of the member (the length of the strip)
in order to make an unloaded strip you will have to createa very short dummy element near the 2 edge because loadingis defined at the nodes not the elements
00
00
12
80
Finite Strip Analysis
Extras
• Assemblage– 2 C’s linked
togethervs. a singleC sectionresults arein bending.CU-FSMused foranalysis
81
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
82
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
83
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.
84
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
85
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”
86
Finite Strip Analysis
Geometry of Examples
8.44
2.44
0.059
C_nolip
8.44
2.44
0.84
0.059
C
15.93
3.43
0.97
0.068
C_deep
3.95
3.0
0.81
0.083
C_rack
4
4
0.060
L
4
4
0.25
0.060
L_lip
10.0
5.0 1.0
0.075
H
8.44
2.44
0.68@50o
0.059
Z
0.90@50o
11.93
2.06
0.070
Z_deepplate
6
drawings not to scale, all dim. in inches
88
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
89
Finite Strip Analysis
Classical derivation for k (cont.)
m n
mn b
ynsin
a
xmsinaw
a
0
2
2
2
2
2b
0
2/t
2/t
22plate dxdy
y
w
x
wdzz
12
EU
b
0
a
0
2
121
plate dxdyx
w
b
y1tfW
plateplate WU
0amn
consider solution for a simply supported plate with a stress gradient
displaced shape
internal energy
external work
total potential energy
variation and solution