buckling euler & column/load imperfections aersp 301 buckling euler & column/load...

AERSP 301BUCKLINGBUCKLING

EULER & COLUMN/LOAD IMPERFECTIONSEULER & COLUMN/LOAD IMPERFECTIONS

Jose Palacios

August 2008

TodayToday

• BUCKLING– EULER BUCKLING– COLUMN IMPERFECTIONS AND LOAD MISALIGNMENT– ENERGY METHODS AND APPROXIMATE SOLUTIONS

FINAL:

Thursday, August 14 from 10:00 am – 12 noon @ RCOE

Tentative Schedule:

M – Beam Buckling

T – Plate Theory

W – Hw # 7 Review

R – Intro to Vibration

F – Final Exam Review

STRUCTURAL INSTABILITY

• STRUCTURAL MEMBERS IN COMPRESSION ARE SUSCEPTIBLE TO FAILURE BY BUCKLING WHEN THE COMPRESSIVE LOAD EXCEEDS A CRITICAL LOAD (BUCKLING LOAD)

– THERE ARE MULTIPLE TYPES OF BUCKING

EULER BUCKLING OF COLUMNS

• FOR SMALL, ELASTIC DEFLECTIONS OF PERFECT, SLENDER COLUMNS

• VARIETY OF BOUNDARY CONDITIONS

• PHYSICALLY – IF YOU APPLY A COMPRESSIVE LOAD TO A COLUMN, AT SOME VALUE OF LOAD IT WILL SUDDENLY BOW (OR BUCKLE)

STRUCTURAL INSTABILITY (EULER)

• IN THEORY – FOR A PERFECT COLUMN LOADED PERFECTLY ALONG THE CENTROIDAL AXIS:

– THERE WILL ONLY BE A SHORTENING, NO BOWING (BUCKLING).– BUT WHAT HAPPENS IF A SMALL LATERAL LOAD IS APPLIED?

– DEPENDS ON THE LEVEL OF THE COMPRESSIVE LOAD…

– FOR:

• ADDITION OF LATERAL LOAD RESULTS IN DIFFERENT BEHAVOIR

• EULER BUCKLING – BEFORE AND AT CRITICAL LOAD, COLUMN IS RELATIVELY UNDEFORMED

• WHEN BUCKLING LOAD IS SURPASSED, SUDDEN, LARGE, DEFORMATION OCCURS

crcrcr PPPPPP

STRUCTURAL INSTABILTY (EULER)

• DETERMINATION OF BUCKLING LOAD FOR A PINNED-PINNED COLUMN:

• AT THE CRITICAL LOAD, Pcr, ANY ADDITIONAL LOAD WILL BUCKLE THE COLUMN AS SHOWN

zw

xPcr


• FROM BUCKLED SHAPE BENDING MOMENT AT ANY X LOCATION (show this)

wPx

wEIM cr

2

2

0

0

2

2

2

2

wEI

P

x

wor

wPx

wEI

cr

cr


• SET

• SOLUTION TO THIS HOMOGENEOUS ODE IS OF THE FORM:

– w – LATERAL DISPLACEMENT– A, B – CONSTANTS

EI

Pcr2Eigenvalue Problem

02 ww

xBxAw cossin


• USE BOUNDARY CONDITIONS TO DETERMINE CONSTANTS A & B:

xBxAw cossin

00 ,0@ Bwx

xAw sin

0sin0 ,@ LAwLx


• POSSIBLE SOLUTIONS:– A = 0 TRIVIAL SOLUTION

– OR SIN(λL) = 0: Non-Trivial Solution

– λ: EIGENVALUES (ALL POSSIBLE SOLUTIONS TO ODE)

0sin LA

3... 2, 1,n

0sin

L

n

nLL


• THEN:

3... 2, 1,n

0sin

L

n

nLL

EI

Pcr2

EIL

nP

EI

P

L

n

cr

cr

2

22

2


• NOW:

x

L

nAw

sin sin

xL

nIs called the buckling mode shape

9

3

4

2

1

2

2

2

2

2

2

L

EIPn

L

EIPn

L

EIPn

cr

cr

cr


• IN REALITY, BUCKLING OCCURS AT THE LOWEST VALUE

• HIGHER MODES WILL BE OBSERVED ONLY IF THERE ARE RESTRAINTS AT NODES OF THOSE MODES

12

2

L

EIPn cr

4

22

2

L

EIPn cr


• LATERAL RESTRAINT AT MID-POINT SUPPRESSES THE 1ST MODE AND CRITICAL BUCKLING LOAD

• LATERAL RESTRAINTS AT L/3 AND 2L/3 SUPPRESSES THE 1ST AND 2ND MODES AND CRITICAL BUCKLING LOAD IS INCREASED TO

9

32

2

L

EIPn cr


• DETERMINATION OF BUCKLING LOAD FOR A CLAMPED-FREE COLUMN

What will the moment be?


• BENDING MOMENT AT X (show this):

• EQUILIBRIUM EQUATION:

ntdisplaceme lateral tip theis where

wPM cr

crcr

cr

PwPx

wEIor

wPMx

wEI

2

2

2

2


• NON-HOMOGENOUS ODE SOLUTION (2 PARTS):– COMPLIMENTARY SOLUTION (SOLUTION TO HOMOGENOUS PART):

22

2

2

ww

EI

Pw

EI

P

dx

wd crcr

EI

Pcr2

02 ww

xBxAwh cossin


– PARTICULAR SOLUTION:

– FULL SOLUTION:

• APPLY BOUNDARY CONDITIONS:

pw

xBxAwww ph cossin

xBxAw sincos

0 0 0@ wwx


• ALSO, w(L) = δ

– THIS IMPLIES:

0 0 0@ wwx

00

0

AAw

BBw

L cos1

0cos L


• SO, 0cos L

2

12

0cos

0

nL

L

22

22

2

12

2

12

nL

EI

PL

nL

crn

n

For n = 1, 2, 3, 4,…


• BUCKLING LOAD – LOWEST VALUE FOR CLAMPED-FREE BEAM:

22

2

2

12

L

EInPcr

For n = 1, 2, 3, 4,…

22

4L

EIPcr


• SIMILARLY, IT CAN BE SHOWN THAT FOR A

• FROM THE ABOVE RESULTS, WE CAN WRITE:

• FOR ANY COLUMN, WHERE THE EQUIVALENT LENGTH, Le, DEPENDS ON THE BOUNDARY CONDITIONS

Clamped-Clamped Beam:2

24L

EIPcr

Clamped-Pinned Beam: 22046.2

L

EIPcr

22

ecr L

EIP


• Le DEPENDS ON BOUNDARY CONDITIONS:

22

ecr L

EIP

For a pinned-pinned: Le = LFor a clamped-clamped: Le = L/2For a clamped-free: Le = 2LFor a clamped-pinned: Le = 0.7L


• WE COULD ALSO WRITE:

C: COEFFICIENT OF CONSTRAINT OR END FIXITY FACTOR2

2

L

EICPcr

For a pinned-pinned: C = 1For a clamped-clamped: C = 4For a clamped-free: C = 0.25For a clamped-pinned: C = 2.046

STRUCTURAL INSTABILITY (IMPERFECTIONS)

COLUMN IMPERFECTIONS & LOAD MISALIGNMENT

• FORCE IS P, NOT Pcr

• UNLIKE PERFECTLY STRAIGHT COLUMN (WHERE BENDING OCCURS ONLY AFTER Pcr), WITH IMPERFECTIONS BENDING OCCURS IMMEDIATLEY ON APPLICATION OF COMPRESSIVE FORCE (DUE TO ITS OFFSET FROM THE SLIGHTLY CURVED CENTER LINE).


• BENDING MOMENT ALONG COLUMN:

totPwdx

wdEIM

2

2

002

2

0 wwEI

P

dx

wdwwwtot

EI

Psetting 2

022

2

2

wwdx

wd


• INITIAL SHAPE OF THE COLUMN IS A SINE FUNCTION:

• SOLUTION TO THIS NON-HOMOGENEOUS ODE:

(aoL IS THE AMPLITUDE. a0 IS THE DIMENSIONLESS IMPERFECTION AMPLITUDE –VERY SMALL NUMBER)

L

xLaw

sin00

L

xLaww

sin022

L

xLa

L

LxBxAw

sincossin 0222

22

Homogenous SolutionParticular Solution


• APPLY BOUNDARY CONDITIONS TO DETERMINE A & B:

0 0 0@ Bwx

L

xLa

L

LxAw

sinsin 0222

22

00sin 0 @ ALAwLx

L

xLa

L

Lw

sin0222

22


• SINCE:

L

xLa

L

Lw

L

xLaw

www

tot

o

tot

sin1

sin

0222

22

0

0


• @ X = L/2, LATERAL DEFLECTION TAKES ITS MAX. VALUE (CALL IT )

• USING

LaL 0222

2

La

EIPL 02

2

2

EI

Pcr2

LaPLEI

EI022

2


• RECALL, FOR PERFECT COLUMN, EULER’S CRITICAL BUCKLING LOAD WAS:

LaPLLP

LP

cr

cr022

2

222

2

LPEIL

EIP crcr

0/1

1 ; :Defining a

PPa

L

δa

cr


• a NON-DIMENSIONAL MID-PT. DISP

MAGNITUDE OF INITIAL IMPERFECTION, ao, AFFECTS THE AMPLITUDE OF DEFLECTION, BUT NOT THE LIMITING (BUCKLING) LOAD

IF ao = 0, BUCKLES LIKE EULER COLUMN (NO BENDING UNTIL LOAD PASSES Pcr)

0/1

1 a

PPa

cr

00 1

1a

P/Pa

a

cr

0

1a

a

P

P

cr

or


• PREVIOUS COLUMN, LOADED PERFECTLY BUT GEOMETRICALLY IMPERFECT

• NOW COLUMN IS GEOMETRICALLY PERFECT, BUT COMPRESSIVE LOAD P IS NOT ALIGNED WITH CENTROIDAL AXIS (LOAD IMPERFECTION, OFFSET BY ECCENTRICITY, e)


• BENDING MOMENT ON COLUMN:

• DETERMINE SOLUTION TO NON-HOMOGENEOUS ODE


• ODE SOLUTION:

• @ X = 0,


• @ X = L,


• MAX LATERAL DEFLECTION, , AT THE MID-PT. (X=L/2)


• a, ae, DIMENSIONLESS MID-SPAN DEFLECTION AND ECCENTRICITY


• SOLVING FOR


• FIGURE SHOWS THAT EVEN IF THERE IS A SMALL LOAD ECCENTRICITY, THE LOAD CAPACITY OF THE COLUMN IS DECREASED.

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