mech-intro 13.0 appa buckling

AA-1ANSYS, Inc. Proprietary

© 2010 ANSYS, Inc. All rights reserved.Release 13.0

November 2010

Introduction to ANSYS

Mechanical

Customer Training Material

Appendix A

Linear Buckling Analysis

Introduction to ANSYS Mechanical



November 2010

Customer Training MaterialChapter Overview

• In this chapter, performing linear buckling analyses in Mechanical

will be covered.

• Contents:

A. Background On Buckling

B. Buckling Analysis Procedure

C. Workshop 7-1

• The capabilities described in this section are generally applicable to

ANSYS DesignSpace Entra licenses and above.

– Some options discussed in this chapter may require more advanced

licenses, but these are noted accordingly.




November 2010

Customer Training MaterialA. Background on Buckling

• Many structures require an evaluation of their structural stability.

Thin columns, compression members, and vacuum tanks are all

examples of structures where stability considerations are important.

• At the onset of instability (buckling) a structure will have a very large

change in displacement {x} under essentially no change in the load

(beyond a small load perturbation).

F F

Stable Unstable




November 2010

Customer Training Material… Background on Buckling

• Eigenvalue or linear buckling analysis predicts the theoretical

buckling strength of an ideal linear elastic structure.

• This method corresponds to the textbook approach of linear elastic

buckling analysis.

– The eigenvalue buckling solution of a Euler column will match the

classical Euler solution.

• Imperfections and nonlinear behaviors prevent most real world

structures from achieving their theoretical elastic buckling strength.

• Linear buckling generally yields unconservative results by not

accounting for these effects.

• Although unconservative, linear buckling has the advantage of being

computationally cheap compared to nonlinear buckling solutions.




November 2010

Customer Training Material… Basics of Linear Buckling

• For a linear buckling analysis, the eigenvalue problem below is

solved to get the buckling load multiplier li and buckling modes yi:

Assumptions:

– [K] and [S] are constant:

• Linear elastic material behavior is assumed

• Small deflection theory is used, and no nonlinearities included

• It is important to remember these assumptions related to performing

linear buckling analyses in Mechanical.

0 ii SK yl




November 2010

Customer Training MaterialB. Buckling Analysis Procedure

• A Static Structural analysis will need to be performed prior to (or in

conjunction with) a buckling analysis. The steps in italics are specific

to buckling analyses.

– Attach Geometry

– Assign Material Properties

– Define Contact Regions (if applicable)

– Define Mesh Controls (optional)

– Include Loads and Supports

– Solve Static Structural Analysis

– Link a Linear Buckling Analysis

– Set Initial Conditions

– Request Results

– Solve the Model

– Review Results




November 2010

Customer Training Material… Geometry and Material Properties

• Any type of geometry supported by Mechanical may be used in

buckling analyses:

– Solid bodies

– Surface bodies (with appropriate thickness defined)

– Line bodies (with appropriate cross-sections defined)

• Only buckling modes and displacement results are available for line bodies.

– Although Point Masses may be included in the model, only inertial loads

affect point masses, so the applicability of this feature may be limited in

buckling analyses

• For material properties, Young’s Modulus and Poisson’s Ratio are

required as a minimum




November 2010

Customer Training Material… Contact Regions

• Contact regions are available in free vibration analyses,

however, contact behavior will differ for the nonlinear

contact types exactly as with modal analyses.

• Discussed earlier (see chapter 5).

Initially Touching Inside Pinball Region Outside Pinball Region

Bonded Bonded Bonded Free

No Separation No Separation No Separation Free

Rough Bonded Free Free

Frictionless No Separation Free Free

Contact TypeLinear Buckling Analysis




November 2010

Customer Training Material… Loads and Supports

• At least one structural load, which causes buckling, should be

applied to the model:

– All structural loads will be multiplied by the load multiplier (l) to

determine the buckling load (see below).

– Compression-only supports are not recommended.

– The structure should be fully constrained to prevent rigid-body motion.

F x l = Buckling Load

In a buckling analysis all applied

loads (F) are scaled by a

multiplication factor (l) until the

critical (buckling) load is reached




November 2010

Customer Training Material… Loads and Supports

• Special considerations must be given if constant and proportional

loads are present.

– The user may iterate on the buckling solution, adjusting the variable

loads until the load multiplier becomes 1.0 or nearly 1.0.

– Consider the example of a column with self weight WO and an externally

applied force A.

– A solution can be reached by iterating while adjusting the value of A until

l = 1.0. This insures the self weight = actual weight or WO * l WO .




November 2010

Customer Training Material… Buckling Setup

Buckling analyses are always coupled to a structural analysis within

the project schematic.

– The “Pre-Stress” object in the tree contains the results from a structural

analysis.

– The Details view of the “Analysis Settings” under the Linear Buckling

branch allows the user to specify the number of buckling modes to find.




November 2010

Customer Training Material… Solving the Model

• After setting up the model the buckling analysis can be solved along

with the static structural analysis.

– A linear buckling analysis is more computationally expensive than a

static analysis on the same model.

– The “Solution Information” branch provides detailed solution output.




November 2010

Customer Training Material… Reviewing Results

• After the solution is complete, the buckling modes can be reviewed:

– The Load Multiplier for each buckling mode is shown in the Details view

as well as the graph and chart areas. The load multiplier times the

applied loads represent the predicted buckling load.

Fbuckle = (Fapplied x l)




November 2010


• Interpreting the Load Multiplier (l):

– The tower model below has been solved twice. In the first case a unit

load is applied. In the second an expected load applied (see next page)




November 2010


• Interpreting the Load Multiplier (l):

LoadUnitadBucklingLo _*l

l adBucklingLo

LoadActualadBucklingLo _*l

FactorSafetyLoadActual

adBucklingLo_

_ l




November 2010


• The buckling load multipliers can be reviewed in the “Timeline”

section of the results under the “Linear Buckling” analysis branch

– It is good practice to request more than one buckling mode to see if the

structure may be able to buckle in more than one way under a given

applied load.




November 2010

Customer Training MaterialC. Workshop AA.1 – Linear Buckling

• Workshop WSAA.1 – Linear Buckling

• Goal:

– Verify linear buckling results in Mechanical for the pipe model

shown below. Results will be compared to closed form

calculations from a handbook.

../ANSYS_Mechanical-Intro_12.1_1st-edition/Workshops/WB-Mech_120_WS_07.1.ppt







November 2010

Customer Training MaterialGoals

• The goal in this workshop is to verify linear buckling results in

ANSYS Mechanical. Results will be compared to closed form

calculations from a handbook.

• Next we will apply an expected load of 10,000 lbf to the model and

determine its factor of safety.

• Finally we will verify that the structure’s material will not fail before

buckling occurs.




November 2010

Customer Training MaterialAssumptions

• The model is a steel pipe that is assumed to be fixed at one

end and free at the other with a purely compressive load

applied to the free end. Dimensions and properties of the

pipe are:

• OD = 4.5 in ID = 3.5 in. E = 30e6 psi, I = 12.7 in^4, L = 120 in.

• In this case we assume the pipe conforms to the following

handbook formula where P’ is the critical load:

• For the case of a fixed / free beam the parameter K = 0.25.

2

2

'L

IEKP




November 2010

Customer Training MaterialAssumptions

• Using the formula and data from the previous page we can

predict the buckling load will be:

lbf

eP 3.65648

)120(

771.1263025.0'

2

2




November 2010

Customer Training MaterialProject Schematic

1. Double click Static Structural in the Toolbox to create a new system.

2. Drag/drop a “Linear Buckling” system onto the “Solution” cell of the static structural system.

2.

1.




November 2010


• When the schematic is correctly set up it should appear as shown

here.

• The “drop target” from the previous page indicates the outcome of

the drag and drop operation. Cells A2 thru A4 from system (A) are

shared by system (B). Similarly the solution cell A6 is transferred to

the system B setup. In fact, the structural solution drives the

buckling analysis.

Project Schematic

“Drop Target”




November 2010

Customer Training MaterialProject Schematic

• Verify that the Project units are set to “US Customary (lbm, in, s, F, A,

lbf, V).

• Verify units are set to “Display Values in Project Units”.




November 2010

Customer Training Material. . . Project Schematic

3. From the static structural system (A), double click the Engineering Data cell.

4. To match the hand calculations referenced earlier, change the Young’s modulus of the structural steel.a. Highlight Structural Steel.

b. Expand “Isotropic Elasticity” and modify Young’s Modulus to 3.0E7 psi.

• Note : changing this property here does not affect the stored value for Structural Steel in the General Material library. To save a material for future use we would “Export” the properties as a new material to the material library.

3.

b.

a.




November 2010

Customer Training Material. . . Project Schematic

5. From the static structural

system (A), RMB the Geometry

cell and “Import Geometry”.

Browse to the file “Pipe.stp”.

6. Double click the Model cell to

start Mechanical.

• When the Mechanical application opens the tree

will reflect the setup from the project schematic.

5.

6.




November 2010

Customer Training MaterialPreprocessing

7. Set the working unit system to the U.S.

customary system:

a. U.S. Customary (in, lbm, psi, °F, s, V, A).

8. Apply constraints to the pipe:

a. Highlight the Static Structural branch (A5).

b. Select the surface on one end of the pipe.

c. “RMB > Insert > Fixed Support”.

a.

b.

a.

c.




November 2010

Customer Training MaterialEnvironment

9. Add buckling loads:

a. Select the surface on the opposite end of the pipe

from the fixed support.

b. “RMB > Insert > Force”.

c. In the force detail change the “Define by” field to

“Components”.

d. In the force detail enter “1” in the “Magnitude” field

for the “Z Component”.

c.

d.

a.

b.




November 2010

Customer Training Material. . . Environment

10.Solve the model:

a. Highlight the Solution branch for the Linear

Buckling analysis (B6) and Solve.

• Note, this will automatically trigger a solve for the

static structural analysis above it.

11.When the solution completes:

a. Highlight the buckling “Solution” branch (B6).

– The Timeline graph and the Tabular Data will

display the 1st buckling mode (more modes can be

requested).

b. RMB in the Timeline and choose “Select All”.

c. RMB > “Create Mode Shape Results” (this will add

a “Total Deformation” branch to the tree).

c.

b.

a.

a.




November 2010


– Click “Solve” to view the first mode

• Recall that we applied a unit (1) force thus the result compares well with our closed form calculation of 65648 lbf.

Results




November 2010

Customer Training Material. . . Results

12. Change the force value to the expected load

(10000 lbf):

a. Highlight the “Force” under the “Static

Structural (A5)” branch

b. In the details, change the “Z Component” of

the force to 10000.

13. Solve:

a. Highlight the Linear Buckling Solution branch

(B6), RMB and “Solve”.

11b.

11a.

12a.




November 2010

Customer Training Material. . . Results

• When the solution completes note the “Load Multiplier” field now

shows a value of 6.56. Since we now have a “real world” load

applied, the load multiplier is interpreted as the buckling factor of

safety for the applied load.

• Given that we have already calculated a buckling load of 65600 lbf,

the result is obviously trivial (65600 / 10000). It is shown here only for

completeness.




November 2010

Customer Training MaterialVerification

• A final step in the buckling analysis is added here as a “best

practices” exercise.

• We have already predicted the expected buckling load and calculated

the factor of safety for our expected load. The results so far ONLY

indicate results as they relate to buckling failure. To this point we

can say nothing about how our expected load will affect the stresses

and deflections in the structure.

• As a final check we will verify that the expected load (10000 lbf) will

not cause excessive stresses or deflections before it is reached.




November 2010

Customer Training Material. . . Verification

14. Review Stresses for 10,000lbf load:

a. Highlight the “Solution” branch under the

“Static Structural” environment (A6).

b. RMB > Insert > Stress > Equivalent Von Mises

Stress.

c. RMB > Insert > Deformation > Total.

d. Solve. a.

b.

c.




November 2010

Customer Training Material. . . Verification

• A quick check of the stress results shows the model as loaded is well

within the mechanical limits of the material being used (Engineering

Data shows compressive yield = 36,259 psi).

• As stated, this is not a required step in a buckling analysis but

should be regarded as good engineering practice.

mech-intro 13.0 appa buckling

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