Introduction to buckling

Thursday 28 January 2016

Introduction to buckling

Luca Sgambi

Images of an exercise in which Indian fakirs or Shaolin monks bend a steel bar with the upper part of their

chest. It often astonishes the crowd, however this can be explained with buckling.

https://www.youtube.com/watch?v=gLBHnGQgbmI

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In architecture and engineering, the phenomenon of buckling affects elements subject to compression,

typically columns (but also surfaces and beams).

H 50B

Art Institute of Chicago, Renzo Piano, 2009

Ancient Architecture Modern Architecture

H 10B

However, in common building we can hardly

find a H/B ratio higher than 20.

For columns subjected to compression, having

an elevated H/B ratio (slender element) the

theory of the beam, mentioned in the previous

lessons, turns out to be too much approximate.

(Slender columns)(Stocky columns)

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Uder these assumptions, the deflections are proportional to the acting forces and the principle of

superimposition is valid.

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The beam theory, studied in previous lessons, is based on the assumptions of small displacements. Under

these assumptions, we have derived the following equation:

2

2

d w x M xE Idx

The hypothesis of small displacements included two assumptions:

• It has assumed that displacements are very small, so as to be confused with their first-order

approximations.

• It is assumed that the balance of the structure may be imposed (for the calculation of displacements,

stresses, moments, etc.) on un-deformed condition.


In this theory, the collapse of a compressed

element occurs when the material has reached

its ultimate strength (su).

uNA

s s

When a column buckles, the collapse occurs even if the level of stresses, provided by a elastic linear

calculation, is maintained everywhere below the ultimate strength of the material.

This type of bending cannot be analysed with beam-theory in small displacements. In order to analyse it,

it’s necessary to review the beam-theory without the aid of the second assumption:

• It has assumed that displacements are very small, so as to be confused with their first-order

approximations.

• It is assumed that the balance of the structure may be imposed (for the calculation of displacements,

stresses, moments, etc.) on un-deformed condition.

In fact, approaching a given load value (a value which will be called critical load) significant inflections

may occur even in elements which are only compressed. These inflections produce stresses which are

much more elevated than what is provided in elastic linear calculation.

… and impose equilibrium in the deformed unknown position.

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Let’s consider a pinned-ends column subject only to an axial load P (the effect of the presence of a lateral

load will be evaluated in the next lessons).

Even if there’s no lateral load, we assume that there’s a small bending of the element and evaluate the

equilibrium in the deformed position w(x).

HF 0

VF 0

AM 0 AV P

AH 0

BH 0

Through the equilibrium we can obtain the external reactions.

P

A

B

M +

HB

HA

VA

w

x

w(x)

P

A

B

L

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2

2

d w xE I P w x

dx

2

2

d w x P w x 0E Idx

2PE I

2

22

d w xw x 0

dx

Even if there‘s no lateral load, in the deformed position a bending moment occurs due to the presence of axial load.

If we assume: We get the differential equation:

Let’s consider now only one part of the element and impose equilibrium to estimate the internal actions.

2

2

d w x M xE Idx

P

A

B

w

x

w(x)M +

P

M +

M(x)

M(x)

T(x)

N(x)

N(x)

T(x)

HF 0

VF 0

AM 0 AV P

AH 0

BH 0

w(x)

o

A

P

M(x)T(x)

N(x)

x

2

2

d w xx

dx

M x P w x M x P w x 0

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2

2

d w xE I P w x

dx

We can notice that the equation obtained doesn’t depend on conventions adopted.

2

2

d w xE I M x

dx

2

2

d w xE I M x

dx

M x P w x M x P w x

2

2

d w xE I M x

dx

M x P w x

2

22

d w xw x 0

dx

P

A

B

w

x

w(x)M +

w

w(x)M +

P

A

B

x

w

w(x)M +

P

A

B

x

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sin cosw x A x B x

cos sindw x

A x B xdx

sin cos

22 2

2

d w xA x B x

dx

sin cos sin cos2 2 2A x B x x B x 0 0 0

A and B impose boundary conditions, i.e. examining the existing constraints at the end of the element

and imposing known translations and rotations.

2

22

d w xw x 0

dx

The solution of this differential equation is given by this function:

Where A and B are two constants to be evaluated with boundary conditions. In fact, if we evaluate the

derivatives:

and we replace them inside of the differential equation, we obtain the identity:

2 PE I

with

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x 0 w x 0 A 0 B 1 0

sinx L w L 0 A L 0 sinA 0

L 0

In the present case, both in x = 0 and in x = L rotation

is free, while translation is blocked. Therefore:

B 0

The solution (A = 0 e B = 0) is the trivial solution and it represents the deflection curve w(x) that we would

find by assuming the validity of the hypothesis of small displacements.

We can notice that we have more than one solution:

A 0B 0

sinA 0

L 0

A 0B 0

P

A

B

2P

A

B

3P

A

B

A

B

P

A

B

w

x

w(x)M +

sin cosw x A x B x

w x 0

M x 0

NA

s

L

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2 PE I

22

2P

nE I L

22

2E I

P nL

sin L 0 L n nL

If we remember that we have assumed:

We can derive from the two previous equations: with n = 1, 2, 3, …

They are all critical loads for the column examined. In correspondence to them a deformation can occur:

sinw x A x where A remains indeterminate and, in an ideal case, can have any value

However, a structural element cannot deflect beyond a certain limit… therefore there may be a sudden

collapse.We can usually define “critical load” of the structural element the smallest of

the critical values found (n = 1).

2

cr 2E I

PL

32

sin(x)

x

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sinB 0

L 0

This solution is more interesting, because it admits infinite nontrivial solutions

which are the infinite points where the function sin(L) is equal to zero.

22

2E I

P nL

A

B

A

B

P < Pcr

A

B

P = Pcr

A

B

P = Pcr

2 PE I

22

2P

nE I L

22

2E I

P nL

If we remember that we have assumed:

We can derive from the two previous equations: with n = 1, 2, 3, …

32

sin(x)

x

sin L 0 L n nL

sinB 0

L 0

This solution is more interesting, because it admits infinite nontrivial solutions

which are the infinite points where the function sin(L) is equal to zero.

Critical deflection

sinw x A x

sinx

w x AL

if n = 1

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2 2

cr 2 20

E I E IP

Lk L

.k 0 5 .k 0 699 k 1 k 1 k 2 k 2

Similarly we can work for other boundary conditions. All these situations can be connected to a single

case (pinned-ends column) where, instead of length L of the element, we use the “effective length” L0.

We can evaluate L0 by multiplying L by a coefficient k which depends on the constraints present at the

edge of the element.

P P P P PP

L

00 sinw x A x

Flexural points

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L0L0

L0

L0

L0

L0

2 2

cr 2 20

E I 206000 491P 110 NL 3000

4 4

410I 491 mm64 64

s .u uP A 78 5 360 28260 N

,2 2

210A 78 5 mm4 4

Collapse for strength Collapse for buckling

With these results we can better understand what is the axial force which leads to the bar collapse, in

the experiment shown in the first slides. Since we do not know the real data, we will make the following

assumptions: = 10 mm; L = 3.0 m; S235 (E = 208000 MPa and su = 360 Mpa).

The load necessary to cause the bar collapse due to strength is equal to a SUV weight. The load

necessary to cause the bar collapse due to buckling is equal to the weight of 6 water bottles.

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E = Young’s Modulus depends on the material

Masorny = 2000 MPa

Legno = 11000 MPa (along grain)

Concrete = 28000 MPa

Steel = 206000 Mpa

I = Inertia depends on the section of the structural element

L0 = depends on constraints at the ends of the element

2

cr 20

E IPL

e

i

H

IPE, IPN, HE...B B

…but the real world is not…

4B12

3B H12 4

64 4 4

e i

64

(approximate)

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Pillar

Foundation

Confinement steel

Longitudinal bars

Pillar

Foundation

Compressible material

Confinement steel

Longitudinal bars

Welding joint

Steel plate

Anchor bolt

Column

Foundation

Column

Foundation

Bolted joint

Steel plate

Anchor bolt

The choice of the type of restraint (fixed end or pin) must be made carefully. Both restraints can be made either in steel or in reinforced concrete.

Let’s try to make a discussion about some real case.

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The Barcelona Pavilion, Ludwig Mies van der Rohe, 1929

3 3420 150 130 20I 5625000 86666 5711667 mm

12 12

2cr 2

206000 5711667 1P 1290 kN10003000

We can simplify the column composed by

angular elements designed by Mies to a

section with the following dimensions:

From geometry we can calculate the modulus of inertia

Moreover, let’s suppose that the column is a pinned-ends column and the inter floor height is 3 m.

And then the critical load

20 mm

150 mm

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The Barcelona Pavilion, Ludwig Mies van der Rohe, 1929

crPP 430 kN3

2i

430A 54 m8

Based on the dimensions obtainable from the picture it seems to be well-structured.

To stay away from buckling phenomenon

we assume a safety factor equal to 3 and

evaluate a “permissible load” as:

A normal roof level can have a load, per square meter, equal to 8 kN/m2 (5 kN/m2 of dead load and 3

kN/m2 of service load). Therefore, dividing the permissible load by the load per square meter we can

evaluate how many square meters the column is capable of supporting:

(In the next lessons the verification

according Eurocode 3 will be explained).

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New National Gallery, Ludwig Mies van der Rohe, 1968

. . . . . . . 4A 0 45 0 035 1 76 0 02 0 45 0 035 0 0667 mm

. .1p 0 0667 78 64 8 19 2 12804 kN

In Mies’s architecture, a cross-shaped column is often

present. However, it is not always possible to use the

pure cross of Barcelona Pavilion.

Let’s try to make some assessment both with a pure cross and with reinforced crossed shape which was

then actually used by Mies. Roof structure (64.8 x 64.8 m) is made up of 38 beams having the dimensions

reported in the figure. Let’s calculate their weight (p1).

Pure cross Reinforced cross

30 mm

960 mm

320 mm

320 mm

320 mm

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0.45 m

1.83 m

35 mm

( ) . .2p 5 3 64 8 64 8 33592 kN

totp 12804 33592 46396 kN

col46396P 5800 kN

8

To this weight we have add the roof slab wheight (we can estimate it at 5 kN/m2) and the service load (3

kN/m2).

The total load is supported by 8 columns. Therefore, each of them will have an axial load of :

New National Gallery, Ludwig Mies van der Rohe, 1968

In Mies’s architecture, a cross-shaped column is often

present. However, it is not always possible to use the

pure cross of Barcelona Pavilion.

Pure cross

30 mm

960 mm

320 mm

320 mm

320 mm

Reinforced cross

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Original sketch, Ludwig Mies van der Rohe, (MoMA – New York)

3 3430 960 930 30

I 2213932500 mm12 12

2

cr 2206000 2213932500 1

P 14866 kN10002 8700

crP 14866P 4955 kN

3 3

P

And the permissibile load for buckling safety

(as done in the previous example):

Section isn’t good, it could have buckling

problems!

The critical load is:1

2

1

2

We can assume for the column a static

cantilever scheme and thus evaluate the

critical load as:

(5800 kN)

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8.7 m

< !

3 3 3 3230 900 870 30 30 320 320 30I 2 2 30 320 465

12 12 12 12

2

cr 2206000 6141254500 1P 41238 kN

10002 8700

crP 41238

P 13746 kN3

4I 1822500000 1957500 2 81920000 2 720000 2075760000 6141257500 mm

crP 41238 kNcrP 14866 kN

Let’s evaluate the moment of inertia of the reinforced section:

We can evaluate the critical load and the permissible load (always using a safety factor equal to 3):

Now the section is well designed!30 mm

P P (5800 kN)

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(5800 kN)>

OK

Until now we have assumed that the critical deflection curve were contained in the design plan. However,

it is clear that the buckling will occur in the direction where the element opposes less resistance to

bending. If constraints impose the same restrictions in all directions, it occurs on the main plane which

involves the smallest moment of inertia.

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Be careful, boundary conditions can change along different directions!

Until now we have assumed that the critical deflection curve were contained in the design plan. However,

it is clear that the buckling will occur in the direction where the element opposes less resistance to

bending. If constraints impose the same restrictions in all directions, it occurs on the main plane which

involves the smallest moment of inertia.

P

P P

0L L

.0L 0 699 L

0L 2 2 L !

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Sendai Mediatheque, Toyo Ito, 2001

Toyo Ito uses these structural elements also for circulation of air, water, electricity, light and people inside

of the building.

One of the designs of the most famous modern architecture is Sendai Mediateque. In it, Toyo Ito uses as

vertical structural system some tubes made of elements in steel. These elements, variously inclined, are

joined by several horizontal ring beams. All elements are very slender.

Let’s try to do an assessment on one of them.

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Sendai Mediatheque, Toyo Ito, 2001

floorQ 7 5 144 6 10368 kN

2s

10368000A 43200 mm

240

The load area of the pipe is about 210 m2 (12.5 x 16.7 m). Let’s suppose that there is a dead load of 7

kN/m2 and a service load of 5 kN/m2, for each floor. Considering 6 loaded floors, on the ground floor we

have a vertical load of:

If we assume an admissible stress for steel equal to 240

MPa (S355), we need a section with a steel area equal to:

16.7 m

12.5 m

P

e = 830 mm

t = 25 mm

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4 4 4 4e i 4

700 660I 2471606540 mm

64 64

2

cr 2 20

E I 206000 2471606540 1P 102547 kN

1000L 7000

cr

crP 102547

P 34182 kN3 3

P 10368 kN

2s

43200A 3600 mm12

We can evaluate, as previously done, the safety in terms of buckling of this section (we can assume a L = 7

m and a pin-ended structure ):

The section is well dimensioned. Now we subdivide it into

12 elements to create a tubular structure built by Toyo Ito.

To the above-mentioned structure correspond elements with an outer diameter of 22 centimetres and

8 millimetres thick. If we revaluate safety with respect to buckling (I = 22375825 mm4):

2

cr 2 20

E I 206000 22375825 1P 928 kN

1000L 7000

cr

crP 928

P 309 kN3 3

10368

P 864 kN12

The new columns are under-sized!

e = 830 mm

t = 25 mm

e = 220 mm

t = 8 mm

>

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<<

!

OK

However, if we subdivide the tube with 5 ring beams, the effective length becomes 1/6 (1170 mm). We

evaluate again the safety with regard to buckling:

2

cr 2 20

E I 206000 22375825 1P 10578 kN

1000L 1170

crcr

P 10578P 3526 kN

3 3 P 864 kN

... and now, the structural element is safe!

The effective length has a great influence on the buckling behaviour!

P PP

ARCHITECTURE

STRUCTURE

L0 = 1170 mm

L0 = 7000 mm Ring beams

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>>OK

Homework

The example on Sendai Mediatheque has been included after having observed the spatiality of the

phenomenon. Why? What would happen if the elements which compose the columns were perfectly

vertical instead of inclined? Beam rings would work the same?

Based on this lecture, please perform similar considerations about the columns of the New Bordeaux

Stadium.

Nouveau Stade de Bordeaux, Herzog & de Meuron, 2015

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Remarks

Buckling is a phenomenon which leads to the collapse of structural elements due to stresses which are

much lower than those assessed by linear elastic theory.

To study the phenomenon of buckling we must partially remove the hypothesis of small displacements

and impose equilibrium in the deformed position.

For columns the critical load is summarised in the following formula, where the coefficient k depends on

the boundary conditions.

The phenomenon of buckling must be viewed in 3 dimensions (if we are designing a plane frame,

buckling plan cannot be the design plan!).

The effective length has a great importance because it’s squared. We must pay attention to the

constraints and how the structural system is composed.

2

cr 20

E IP

L

(The formula is also called Euler’s formula, named after the famous mathematician

who, in 1744, formulated and solved the buckling problem of a rod).

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Introduction to buckling

Engineering

Transcript of Introduction to buckling