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Finite Strip Analysis and the Beginnings of the Direct Strength Method

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### Finite Strip Analysisand the Beginnings of the Direct Strength Method

Overview

Overview

Overview

Overview

Overview

buckling multiplier (stress, load, or moment)

distortional

later-torsional

local

length of a half sine wave

Toronto, July 2000

AISI Committee on Specifications

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”

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.

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

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

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

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”

Theoretical Background

- Elastic buckling in matrix form
- Initial elastic stiffness [K]
- specialized shape functions

- Geometric stiffness [Kg]
- Forming the solution
- Elastic buckling solution

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

What is [K]?

- [K] is the initial elastic stiffness

Strain-Displacement and [K]

Plane stress Kuv comesfrom these strain-displacement relations

Bending Kwq comesfrom these strain-displacement relations

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.

Developing [Kg]

where {d} is the nodal displacementsthe same shape functions as before are used,therefore [G] is determined through

partial differentiation of [N].

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=gTk’g)
- add contribution of each strip to global stiffness, symbolically:

Eigenvalue Solution

- The solution yields
- l 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

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”

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:

= 10.665 ksi

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

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, q 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

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)

- 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”

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

0.84

8.44

0.059

Strong Axis Bending of a C- Approximate the buckling stress for pure bending.

comp. lip

comp. flange

web

0.84

8.44

0.059

Strong Axis Bending of a C- Approximate the buckling stress for pure bending.

lip

flange

web

* this k value would be fine-tuned by AISI B4.2

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)

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:

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.

0.84

8.44

0.059

Compression of a C- Approximate the buckling stress for pure compression.

lip

flange

web

* this k value would be fine-tuned by AISI B4.2

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)

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

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

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

Comparision of Results for Buckling Stress of a C- 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”

Converting the results

- If f1 is the applied stress in the finite strip analysis and l the multiplier that results from the elastic buckling thenfcr = lf1 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)

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!

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

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

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

- 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”

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

Direct Strength for Columns (cont.)

*these calculations include long column interaction, to ignore this interaction replace Pne with Py

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

- Pcrl = 6.6 kips
- Pcrd = 17.7 kips
- Pcre = 25.7 kips
- Py = 44.25 kips

- Pcrl = 6.6 kips
- Pcrd = 17.7 kips
- Pcre = 25.7 kips
- Py = 44.25 kips

*these calculations include long column interaction, to ignore this interaction replace Pne with Py

- Pcrl = 6.6 kips
- Pcrd = 17.7 kips
- Pcre = 25.7 kips
- Py = 44.25 kips

- Pcrl = 6.6 kips
- Pcrd = 17.7 kips
- Pcre = 25.7 kips
- Py = 44.25 kips
- Pne = 21.58 kips

*these calculations include long column interaction, to ignore this interaction replace Pne with Py

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)

Direct Strength for Beams (cont.)

*these calculations include long column interaction, to ignore this interaction replace Pne with Py

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

- Mcrl = 90 in-kips
- Mcrd = 117 in-kips
- Mcre = braced
- My = 113 in-kips

Mne = My since “braced” = 113 in-kips

Mnl = 89 in-kips

- Mcrl = 90 in-kips
- Mcrd = 117 in-kips
- Mcre = braced
- My = 113 in-kips

Mnd = 86 in-kips

Mn = 86 in-kips, distortional controls even though elastic critical is 30% higher

*these calculations include long column interaction, to ignore this interaction replace Pne with Py

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

1.4

1.2

local buckling controlled

distortional buckling controlled

1

n

/F

u

0.8

strength F

0.6

local strength curve

0.4

0.2

distortional strength curve

0

0

1

2

3

4

5

6

7

8

.5

slenderness of controlling mode (F

/F

)

n

c

r

ColumnsDirect 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 f=0.9 for beams and f=0.85 for columns (presumably ASD factors would also remain unchanged)

Example strength & deflection calc. completed at fy

Example strength & deflection calc. completed at fa

Deflection CalculationsIt 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.,

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
- ...

- 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”

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

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...
- ________________________
- ________________________

New

Suggested Alternative(0.50,0.00)

(2.44,0.0)

(2.44,0.84)

(0.50,1.50)

(0.00,2.00)

(2.44,7.60)

(0.00,8.44)

(2.44,8.44)

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

- Pcrl = 11.4 kips
- Pcrd = 24.3 kips
- Pcre = 22.6 kips
- Py = 43.4 kips

- Pcrl = 11.4 kips
- Pcrd = 24.3 kips
- Pcre = 22.6 kips
- Py = 43.4 kips
- Pne = 19.4 kips

*these calculations include long column interaction, to ignore this interaction replace Pne with Py

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

ComparisonExtras

- 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

- 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 kq 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

Extras

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 loadingis defined at the nodes not the elements

Extras

- Assemblage
- 2 C’s linkedtogethervs. a singleC sectionresults arein bending.CU-FSMused foranalysis

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

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

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.

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

- 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”

2.44

2.44

3.0

0.84

0.97

0.81

3.95

8.44

8.44

0.083

4

0.059

0.059

C_rack

0.060

15.93

C_nolip

C

4

L

4

0.068

5.0

1.0

0.060

C_deep

4

2.06

2.44

L_lip

0.25

0.90@50o

10.0

0.68@50o

8.44

0.075

0.059

11.93

6

H

Z

Z_deep

plate

0.070

Geometry of Examplesdrawings not to scale, all dim. in inches

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

Classical derivation for k (cont.)

consider solution for a simply supported plate with a stress gradient

displaced shape

internal energy

external work

total potential energy

variation and solution

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