mom12 column buckling

8/9/2019 MOM12 Column Buckling

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Column Buckling

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Stability and Buckling

Examples of Columns Conventional Design of Columns

Eulers Formula for Pin-ended Columns

Buckling Modes

Eulers Formula for Cantilevered Columns

Extension of Eulers Formula

Buckling in Orthogonal Planes

Applicability of Eulers Formula Failure Diagram

Critical Stress for Intermediate Columns

Design of Columns

Ways to Improve Column Stability2

Contents


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Stability is characterized as the ability of a structure to maintain

its (stable) equilibrium under working conditions.


Buckling is the behavior of a structure losing its equilibriumunder working conditions. This is another type of failure criterion,

in addition to strength (fracture/yielding), stiffness (deformation)

and fatigue criteria.

Buckling occurs suddenly and results in catastrophic accident. 3


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4

Examples of Columns


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In the design of columns, cross-

sectional area is selected such that

- allowable stress is not exceeded

P

A

- deformation falls within

specifications

L P

L AE

After these design calculations, many

discover that the column is unstable

under loading and that it suddenly

becomes sharply curved or buckles.

Conventional Design of Columns

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Consider an axially loaded beam.

After a small perturbation, thesystem reaches a neutral

equilibrium configuration such that

2

2

2

2

2 2

2 2

2

0

" 0,

sin cos

0 0

0;

cr

cr

cr

cr

Pd w Mw

dx EI EI Pd w

wdx EI

Pw k w k

EI

w A kx B kx

w w L

B

EInkL n P

L


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2 2 2

2 2 2

4 9; ;cr cr cr

EI EI EI

P P PL L L

Buckling Modes

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A column with one fixed and one

free end, will behave as the upper-half of a pin-connected column.

The critical loading is calculated

from Eulers formula,

lengthequivalent2

4

4

2

2

2

2

2

2

2

2

LL

iLE

iLE

L

EI

L

EIP

e

rre

cr

e

cr


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tcoefficienlengthlength;equivalent

;2

2

2

2

2

2

2

2

e

rre

cr

e

cr

L

iL

E

iL

E

L

EI

L

EIP


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Assuming the same material and cross-sectional dimension, which

of the following four columns is most susceptible to buckling?

2l

5l 7l

9l

2

2 Euler Formulacr

EIP

l

(a) (b) (c) (d)

Sample Problem

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x

yz y

Buckling inx-z:

y = 12

2

y

cr y

EIP

L

Buckling inx-y:

z= 0.5

2

20.5

zcr z

EIP

L


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For the column shown, find the relationship between the critical

loads corresponding to buckling inx-y andx-zplane respectively.

l2b

b

z

y

Pcr

2

2

2

2

3

3

2 124

2 12

zcr z

z

y

cr y

y

cr z z

cr y y

EIP

l

EIP

l

b bP I

P I bb

Solution:

What if the cross-section is increased to 2b2b?

Sample Problem

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l

2r r

r

r

(a) (b) (c)

Pcr

For the column shown, find the relationship between the critical

loads corresponding to cross-sections described by (a), (b) and (c).

42 2 2 4

2 2 2

2 2 4

2 2

4

2 2 2 2 4

2 2 2

2

64 4: :

1 1: :

64 4 64 12

11: :16 3

12 12

acr a

cr a cr b cr c

bcr b

c

cr c

rEI E E rP

l l l P P P

EI E rP

l l

rEI E E rP

l l l

Solution:

Sample Problem

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Which of the following best describes the relationship between

criticalP1 andP2? (a)P1 =P2; (b)P1P2; (d) not sure.

a

a

P1

A B

C D

a

a

P2A B

C D1P

1P

0

012P 0

0

2P

2P

22P

Solution:

2 2

1 12 2

2 2

2 22 2

12

2 22N

N

:

:

AD

AB

E I E I a F P P

aa

E I E I b F P P

a a

Sample Problem

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Critical buckling stress

P: proportion limit;

p: critical slenderness ratio.

2

2 2 2 22

2 2 2

slenderness ratio

rcr

crcr

r

r

EIP

E AiP E EL

A L A L iP

A

L i

Applicability of Eulers formula

2

2cr P PP

E E

206 GPaQ235: 100

200 MPaP

P

E

Applicability of Eulers Formula

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Previous analyses assumed

stresses below the proportionallimit and initially straight,

homogeneous columns

Experimental data demonstrate

- for largeLe/ir, crfollowsEulers formula and

depends uponEbut not Y.

- for intermediateLe/ir, cr

depends on both YandE.

- for smallLe/ir,cr is

determined by the yield

strength Yand notE.

2 2

cr E Y

rL i

Failure Diagram

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O

Strength

analysis

Y

P

Y bacr

2 2

cr E

Y

P

Stability

analysis

Y cr Y 3. Short columns:

1. Long columns:2 2

2 2P cr p

P

E E

2. Intermediate columns:

YY P p cr Y a

a bb


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For the truss shown,F, , andLACare given. Find 0 < < /2, under

which columnAB andBCreach critical stability simultaneously.

2 2

2 2 2cos

cosAB

AB

EI EIF F

l l

2

2 22

2

2 2

coscos

cot tan

sin sin

EIF

l

EI

F l

A C

B

F

Solution:

1. Eulers equation for columnAB:

2 2

2 2 2sin

sinBC

CB

EI EIF F

l l

2. Eulers equation for columnBC:

3. If columnAB &BCreach critical stability simultaneously:

Sample Problem

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1. Method of safety factor

cr

st

st

cr

st

st

FF F

n

F

A n

2. Method of discount factor

st

F

A

: discount/stability factor

nst: safety factor

[Fst]: allowable load

[st]: allowable stress

Stability analysis of columns

- Stability check

- Cross-section design

- Allowable load/stress

Design of Columns

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2 2

22

2 2

2 2

cr

e

cr

r

EI EIP

L L

E E

L i

Increase moment of inertia for a given

cross-sectional area.

Ways to Improve Column Stability

Selection of materials.

Decrease effective column length (L).

For a group of columns, make each one

equally stable (avoid the last straw).

z r z y r yi i

Coordination of end conditions and

moment of inertia in orthogonal planes:

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A Q275 structural steel column is cylindrically pinned at both ends,

as shown in the following figure. IfFmax = 60 kN and nst= 3.5,check the column stability.

Solution:

lz=800m

m

h =45mm

x

y

ly=77m

m

b=20mm

x

z

F

lz

F

ly

- Inx-yplane, both ends are pinned.

1 80012.99 , 61.9

12.992 3

z z zz z

z

I lhi mm

A i

Sample Problem

Determine the principal buckling plane.

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- Inx-zplane, both ends are constrained.

0.5 7705.77 , 66.7

5.772 3

y y y

y y

y

I lbi mm

A i

y > z, hence column buckling first happens inx-zplane.

For Q-275 structural steel p = 96, and therefore y = 66.7 < p = 96

2

6 6

280 0.00872 241 MPa

(241 10 )(20 45 10 ) 217 kN

2173.62

60

cr y

cr cr

crst

F A

Fn n

F

The column is stable.

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For the composite beam and column structure shown,EAB =EBD =E,

dAB = dBD = d,L:d= 30, P.BD = 100. FindP, under whichBD reachesthe critical condition.

Solution:P

A C B

D

EI

EA L

LL: 120

4 P

L LBD

i d

2

2cr BD

EIF

L

Deformation compatibility:

33 85

6 3

BD BDF L F LPL

EI EI EA

3 2

18000

EdP

Sample Problem

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Examples of Columns Conventional Design of Columns


Buckling Modes




Applicability of Eulers Formula Failure Diagram


Design of Columns

Ways to Improve Column Stability 25

Contents

mom12 column buckling

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