Buckling Fatigue Analysis

BITS Pilani, Deemed to be University under Section 3 of UGC Act, 1956

Failure Analysis


• STATIC (LINEAR & NON-LINEAR)

• BUCKLING FAILURE

• FATIGUE FAILURE

• IMPACT FAILURE

• CREEP FAILURE

• FRACTURE FAILURE

Analysis Types Covered


BUCKLING ANALYSIS(STRUCTURAL ANALYSIS)


For an intermediate length compression member,kneeling occurs when some areas yield beforebuckling.

However, when a compression member becomes longer, the role ofthe geometry and stiffness (Young's modulus) becomes more andmore important. For a long (slender) column, buckling occurs waybefore the normal stress reaches the strength of the column material

The failure of a short compression member resulting fromthe compression axial force (YS or UTS Limit):

Compression Member

http://www.efunda.com/formulae/solid_mechanics/mat_mechanics/elastic_constants_E_nu.cfm

http://www.efunda.com/formulae/solid_mechanics/columns/theory.cfm


In practice, for a given material, the allowable stress in a compression memberdepends on the slenderness ratio Leff / r and can be divided into three regions:short, intermediate, and long.Short columns are dominated by the strength limit of the material. Intermediatecolumns are bounded by the inelastic limit of the member. Finally, long columnsare bounded by the elastic limit (i.e. Euler's formula).

Yield Strength

http://www.efunda.com/formulae/solid_mechanics/columns/inelastic.cfm

http://www.efunda.com/formulae/solid_mechanics/columns/columns.cfm


Buckling

Structures subject to compression load that

haven’t achieved material strength can show

failure mode called buckling.


Buckling Failure | Model Examples


Buckling Loads for Various BCs

CLASSICAL EULER SOLUTIONS



The demand for automobiles with less consumption of

fuel and less emission is increasing continuously.

Besides the light weight design, the increasing

usage of optimization software leads to thin walled and

slender components which tend to buckle under

compressive/lateral loading.



An interesting variation arises in the case of

automotive applications. In the case of front end

collision, the hood is expected to crumple (buckle) in

order to absorb the energy of collision, as well as to

save the passenger compartment.


How FE Analysis will help us?

If a structure has one or more dimensionsthat are small relative to the others (slender orthin-walled), and is subjected to compressiveloads, then a buckling analysis may benecessary.

From an FE analysis point of view, a bucklinganalysis is used to find the lowest buckling loadand to find the shape of the buckled structure.


FE Domain | Pre-Processing

• Geometry: 1-D, 2-D, 3-D

• Element: Structural: 1-D, 2-D, 3-D

• Material Prop: Structural (E, , )

• Geometrical Prop: Area, MI (I)

ν ρ


FE Domain | Solution

• Boundary Conditions:

Structural nodal constraints and Load Application

(depends upon Analysis Types)

• Solution Parameters:

Depends upon Analysis Type: Linear or Non-

Linear


How to Carry out Buckling FE Analysis Buckling loads are critical loads where certain typesof structures become unstable. Each load has an associated buckled mode shape.This is the shape that the structure assumes in abuckled condition. There are two primary means to perform a bucklinganalysis:

1. Eigen-value (Linear Analysis)Buckling is an Eigenvalue problem that is a functionof the material & geometric stiffness matrices.

2. Non-Linear (Non-Linear Analysis)(Geometrical Non-Linearity)


Eigen-value (Linear Analysis) Eigenvalue buckling analysis predicts the theoretical buckling

strength of an ideal elastic structure. This is known as classicalEuler buckling analysis.

This method is not recommended for accurate, real-worldbuckling prediction analysis.


Eigenvalue buckling analysis predicts the theoretical buckling strength (thebifurcation point) of an ideal linear elastic structure. (See Figure (b).) This methodcorresponds to the textbook approach to elastic buckling analysis: for instance,an eigenvalue buckling analysis of a column will match the classical Eulersolution. However, imperfections and nonlinearities prevent most real-worldstructures from achieving their theoretical elastic buckling strength. Thus,eigenvalue buckling analysis often yields unconservative results, and shouldgenerally not be used in actual day-to-day engineering analyses.


T1 | Buckling Analysis | Column

Theoretical Solution (Euler’s Formula)

Eigenvalue Solver (ANSYS) to get theoretical value

Non-Linear Analysis (ANSYS) to get more accurate value


Steel column (10X10 mm cross-section) is constrained atthe bottom. Objective is to calculate the required load tocause buckling.


E = 2E5

L = 100 and C/S: 10X10

I = 833.333 = [(10X103)/12]

P= (π2E*I)/4L2 = 41,081.65 Unit



BEAM3 ELEMENT

3 DoF = Ux, UYand RotZ




/TITLE,Eigenvalue Buckling Analysis

/PREP7 ! Enter the preprocessor

ET,1,BEAM3 ! Define the elt of the beam to be buckled

R,1,100,833.333,10 ! Real Consts: type 1, area (mm^2), I (mm^4), height (mm)

MP,EX,1,200000 ! Young's modulus (in MPa)MP,PRXY,1,0.3 ! Poisson's ratio

K,1,0,0 !Define the geometry of beam K,2,0,100L,1,2 ! Draw the lineESIZE,10 ! Set element size to 1 mmLMESH,ALL,ALL ! Mesh the lineFINISH



/SOLU ! Enter the solution mode

ANTYPE,STATIC ! Before you can do a buckling analysis, ANSYS! needs the info from a static analysis

PSTRES,ON ! Prestress can be accounted for - required! Eigenvalue buckling analysis

DK,1,ALL ! Constrain the bottom of beam

FK,2,FY,-1 ! Load the top vertically with a unit load.! The eignenvalue solver uses a unit force to

! determine the necessary buckling load.

SOLVEFINISH



/SOLU !Enter the solution mode again to solve bucklingANTYPE,BUCKLE !Buckling analysisBUCOPT,LANB,1 !Extraction Method (Block Lanczos) for mode-ISOLVEFINISH

!Check with General Postprocessor, List Results, Detailed Summary



Buckling Load = ~41,123 N


41,081.65 Unit


Buckling Load = 41,123 N

The eignenvalue solveruses a unit force todetermine the necessarybuckling load.

LINEAR ANALYSIS



T2NON-LINEAR ANALYSIS

Ensure that you have completed the Linear Analysis or Eigenvalue

Buckling Analysis before going for a Non-Linear Analysis


Non-Linear Analysis) Buckling loads for several configurations are readily available

from tabulated solutions. However, in real-life, structuralimperfections and nonlinearities prevent most real-worldstructures from reaching their eigenvalue predicted bucklingstrength; ie. it over-predicts the expected buckling loads

This method is not recommended for accurate, real-worldbuckling prediction analysis.


A more practical approach is to carry out a large displacement analysis,where buckling can be detected by the change of displacement in themodel.

A large displacement problem is non-linear in nature. Geometric non-linearity arises when deformations are large enough to significantly alterthe way load is applied, or load is resisted by the structure.

The approach to a non-linear buckling solution is achieved by applyingthe load slowly (dividing it into a number of small loads increments).

The model is assumed to behave linearly for each load increment, andthe change in model shape is calculated at each increment.

Stresses are updated from increment to increment, until the full appliedload is reached.

T2 | Buckling Non-Linear Analysis | Column


/TITLE,Non-Linear Buckling Analysis

/PREP7 ! Enter the preprocessor

ET,1,BEAM3 ! Define the element of the beam to !be buckled

R,1,100,833.333,10 ! Real Consts: type 1, area (mm^2), I (mm^4), height (mm)

MP,EX,1,200000 ! Young's modulus (in MPa)MP,PRXY,1,0.3 ! Poisson's ratio

K,1,0,0 ! Define the geometry of beam (100 !mm high)

K,2,0,100L,1,2 ! Draw the lineESIZE,10 ! Set element size to 1 mmLMESH,ALL,ALL ! Mesh the lineFINISH



/SOLUANTYPE,STATIC !Static analysis (not buckling)NLGEOM,ON ! Non-linear geometry solution OUTRES,ALL,ALL ! Stores bunches of outputNSUBST,20 ! Load broken into 20 load stepsNEQIT,1000 ! Use 20 sub-steps to find solutionAUTOTS,ON ! Auto time stepping

/ESHAPE,1 ! Plots the beam as a volume rather than !line

DK,1,ALL,0 ! Constrain bottomFK,2,FY,-50000 ! Apply load slightly greater than predicted

! buckling load to upper nodeFK,2,FX,-250 ! Add a horizontal load (~0.5% FY) to initiate

! Buckling (Thumb Rule)SOLVEFINISH



SOLUTION IS ON



View the deformed shapeT2 | Buckling Non-Linear Analysis | Column


View the DoF solution UYT2 | Buckling Non-Linear Analysis | Column`


In Time History: ADD UY at Node-2 (Top End) and FY (Reaction force)at Node-1 (Bottom-Node)



Force on X- axis and Deflection on Y Axis



Beam became UNSTABLE and BUCKLED with anapproximate load of 38,000 N.


Theoretical P = 41081 N

ANSYS P = 41,123 N

Non-Linear = 38,000 N


Buckling Features Buckling is a critical state of stress and deformation, atwhich a slight disturbance causes a gross additionaldeformation, or perhaps a total structural failure of the part. Structural behaviour of the part near or beyond'buckling' is not evident from the normal arguments ofstatics. Buckling failures do not depend on the strength of thematerial, but are a function of the component dimensions &modulus of elasticity. Therefore, materials with a high strength will buckle justas quickly as low strength ones.


FATIGUE ANALYSIS(EXTENSION OF TRANSIENT

DYNAMIC ANALYSIS)


S-N curve – A material property

Where, Su is the ultimatestrength and Se is theendurance limit (fatiguelimit). Assume the ratioSe/Su is equal to 0.5.


Miner’s rule


Miner’s rule• It states that if there are k different stress levels and the

average number of cycles to failure at the ith stress, Si, is Ni

(from S-N curve), then the damage fraction, C, is:

ni is the number of cycles accumulated at stress Si.

C is the fraction of life consumed by exposure to the

cycles at the different stress levels.

• When the damage fraction (C) reaches 1, failure occurs.


FE Domain | Pre-Processing

• Geometry: 1-D, 2-D, 3-D

• Element: Structural: 1-D, 2-D, 3-D

• Material Prop: Structural (E, ) and S-N curve

• Geometrical Prop: Area, MI (I)

ν


FE Domain | Solution• Boundary Conditions:

Structural nodal constraints and Load steps

• Solution Parameters:

General structural solution

• Post processing:

Miner’s rule the simplest and the most widely usedcumulative damage models for fatigue failure.


Problem definition | Geometry and applicationA flat leaf spring has been machined from AISI 1050 steel cold-drawn steel (Young’s Modulus is 210 GPa, Poisson’s ratio is 0.3,Yield Strength is 580 MPa and the Ultimate Strength is 690 MPa).

A cyclic load of ±100 N acts for 5,00,000 cycles and another cyclicload of ± 150 N acts for 5000 cycles at the free end of the leafspring. Assume one cycle of 20 seconds.


Fatigue Failure | Model Example

5 x 105 cycles

-100 N force in

Y direction

5 x 105 cycles

+100 N force in

Y direction


Fatigue Failure | Model Example

5 x 103 cycles

-150 N force in

Y direction

5 x 103 cycles

+150 N force in

Y direction


Problem definition | Type of loading

5 x 105 cycles

+100 N force in

Y direction

5 x 105 cycles

-100 N force in

Y direction

Load case 1

Load case 2


Problem definition | Type of loading

5 x 103 cycles

+150 N force in

Y direction

5 x 103 cycles

-150 N force in

Y direction

Load case 3

Load case 4


S-N table defined in ANSYS (POST-PROCESSOR)

ASSUMPTION: S-N DATA IS AVAILALE TO US


/PREP7ET,1,PLANE182

MP,EX,1,2.1E5MP,PRXY,1,.3

RECTNG,0,60,0,1

ESIZE,0.5

AMESH,1SAVEFINI

Pre-processing


/SOLUANTYPE,4 TRANSIENT DYNAMICDL,4,1,ALL!Define Time Intervals (10,20,30,40) for loading

TIME,10 !DEFINE LOAD STEP 1 AT TIME=10FK,2,FY,100LSWRITE,1,

FKDELE,ALL,ALL

TIME,20 !DEFINE LOAD STEP 2 AT TIME=20FK,2,FY,-100LSWRITE,2,

Solution


FKDELE,ALL,ALL

TIME,30 !DEFINE LOAD STEP 3 AT TIME=30FK,2,FY,150LSWRITE,3,

FKDELE,ALL,ALL

TIME,40 !DEFINE LOAD STEP 4 AT TIME=40FK,2,FY,-150LSWRITE,4,

LSSOLVE,1,4,1,

FINI

SAVE

Solution


/POST1 !DEFINE S-N TABLE (5 S-N VALUES)FP,1,10,1000,10000,100000,1000000 !5 KEY CYCLESFP,21,62100,62000,51200,42200,34800 ! Corresponding S values

FL,1,65 !STORE RESULTS AT A NODE IN FIXED END

Fatigue Analysis | Post-processingEvent

(Type of Cycle)

Load No.

Load Step No. Load Value No. of Repititions

(Cycles)

1 1 1 100 N 500,000

1 2 2 -100 N 500,000

2 1 3 150 N 5,000

2 2 4 -150 N 5,000


SET,,, ,,, ,1 !CALLS LOAD STEP 1 (Event-1, Load-1)!Defines the data set to be read from the results file!SET, Lstep, Sbstep, Fact, KIMG, TIME, ANGLE, NSET, ORDER!Data set number of the data set to be read. If a positive value!for NSET is entered,Lstep, Sbstep, KIMG, and TIME are ignored.

FSNODE,65,1,1 !Calculates and stores the stress components at a node for fatigue.

SET,,, ,,, ,2 !CALLS LOAD STEP 2 (Event-1, Load-2)FSNODE,65,1,2 !Calculates and stores the stress components at a node for fatigue



FE,1,500000 !ASSIGNS 500000 CYCLES TO EVENT NO.1FE,2,5000 !ASSIGNS 5000 CYCLES TO EVENT NO.2

FTCALC,1 !CALCULATES FATIGUE AT LOCATION 1 (NODE NO.65)FINISAVE

Fatigue Analysis | Post-processing


By Miner’s rule, Damage criteria (C = 0.73 < 1So, Fatigue failure will not occur in the model

Result for 2 Events [±100 & ±150]

Documents

Buckling Fatigue Analysis