Stability of structures - Solved examples

(1)

Reijo Kouhia Pekka Marjamäki

(2)

Example 1.1 Determineallequilibriumpathsofthe stru ture onsistingoftworigidbarsand

a linearelasti rotationalspring. Investigate alsothe stabilityof all paths.

P = λ4k/L

.

b

@@

_b

@@

_b

P

k

L/2

Solution: Let's assumethat the barsdispla e by anangle

ϕ

,then inthe middlepin the angle willbe 2

ϕ

.

b

b````

```

`

_b

b

@

_{b b}

@

P

k

ϕ

2

ϕ

The total potentialenergy

Π

of the stru ture is thus:

Π =

1

2 k(2ϕ)

2 − P L(1 − cos ϕ)

(1)

∂Π

∂ϕ

= 4kϕ − P L sin ϕ

(2)

∂

2 _Π

∂ϕ

2 = 4k − P L cos ϕ.

(3)

The stru turewillbeinequilibriumwhen thetotal potentialenergy attainsitsminimum,thus

the rst variationof the TPE willvanish.

δΠ =

∂Π

∂ϕ

δϕ = 0 ∀ δϕ 6= 0 ⇒

∂Π

∂ϕ

= 0

(4)

⇒







ϕ =

0

primary path

P =

4k

L

ϕ

sin ϕ

sek ondary path

(5)

Let us rst investigate the primary path. A point on an equilibrium path is stable, if a small

hange(disturban e) inthe equilibriumpositionwillwillin rease the valueof

Π

.Sin etherst variation is zero on an equilibrium path, then the se ond variationwill determine the hange

in the TPE. Sin e now

ϕ = 0

,

δ

2 Π =

∂

2 _Π

∂ϕ

2 (δϕ)

2 _{= 0}

(3)

⇒

∂

2 _Π

∂ϕ

2 = 4k − P L cos ϕ = 4k − P L

⇒ P

kr

= 4

k

L

(6)

The primary equilibriumpath is thusstable up to the point

(ϕ = 0, P

kr

)

. Next, the stabilityproperties of the se ondary pathis investigated.

∂

2 _Π

∂ϕ

2 = 4k − P L cos ϕ

(7)

Insering the equation of the se ondary path

P = 4kϕ/L sin ϕ

to the equation above,gives

∂

2 _Π

∂ϕ

2 P

II

= 4k

1 −

ϕ

tan ϕ

> 0 ∀ϕ

(8)

The se ondary path is this stable for all values of

ϕ

, ex ept the bifur ation point where

ϕ =

0, P

kr

= 4k/L

, and the se ond variation of

Π

iszero.

The equilibriumpaths are shown inthe

λ − ϕ

- oordinatesystem in the gure below.

0

0.2

0.4

0.6

0.8

1

1.2 -1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1 λ

ϕ

(4)

a linear elasti rotational spring. Investigate also the stability of all paths. Are there riti al

points onthe paths?

b

b````

```

`

_b

b

@@

_b

@@

_b

P

k

ϕ

0 cos ϕ

0 L/2

cos ϕ

0 L/2

Solution: The total potentialenergy expression isnow

Π =

1

2 k[2(ϕ − ϕ

0 )]

2 − P L(cos ϕ

0 − cos ϕ)

∂Π

∂ϕ

= 4k(ϕ − ϕ

0 ) − P L sin ϕ

∂

2 _Π

∂ϕ

2 = 4k − P L cos ϕ

The stru turewillbeinequilibriumwhen thetotal potentialenergy attainsitsminimum,thus

the rst variationof the TPE willvanish.

δΠ =

∂Π

∂ϕ

δϕ = 0 ∀ δϕ

⇒

∂Π

_∂ϕ

= 0

⇒ P =

4k(ϕ − ϕ

_{L sin ϕ}

0 )

An equilibriumstate is stableif the se ondvariationof the TPE ispositive

δ

2 _{Π =}

∂

2 Π

∂ϕ

2 (δϕ)

2 _{> 0}

⇒

∂

2 _Π

∂ϕ

2 > 0

⇒ P <

_{L cos ϕ}

4k

.

Inserting the equilibrium equation

P = 4k(ϕ − ϕ

0 )/L sin ϕ

in the expression above, gives the ondition for stability

4k

1 −

ϕ − ϕ

_{tan ϕ}

0 > 0

⇒

ϕ − ϕ

_{tan ϕ}

0 < 1,

whi hisvalidforallnon-negativevaluesof

ϕ

.Thusthis equilibriumpathdoesnot have riti al points.

(5)

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2 -2

-1.5

-1

-0.5

0

0.5

1

1.5

2 λ

ϕ

Inthegureabove,dottedlineshowstheequilibriumpathoftheperfe tstru ture

ϕ

0 = 0

, and solid line indi atesthe stable path when

ϕ

0 > 0

. The path in the negative part of

ϕ

axis shown by asolid lineis a omplementary path. The load parameter

λ

isdened as

λ = P/P

kr

= P L/(4k).

(9)

Noti e, that the omplementary path is not stable everywhere. Determine the unstable

and stable parths of the omplementary path! Note too, that this means an existen e of a

(6)

onsisting of two rigid bars (length

L/2

)and a linear elasti rotationalspring. Investigate also thestabilityof allpaths.Theperturbationload

F = ǫ4k/L

,where

ǫ

isadimensionless(se ond) perturbation parameter.

b

b````

```

`

_b

b

@@

_b

@@

_b

P

k

ϕ

0 ?

F

cos ϕ

0 L/2

cos ϕ

0 L/2

Solution: The total potentialenergy of the stru ture

Π

is

Π(ϕ; ϕ

0 , ǫ) =

1

2 k[2(ϕ − ϕ

0 )]

2 − P L(cos ϕ

0 − cos ϕ) −

1

2 F L(sin ϕ

0 − sin ϕ)

(10) A ne essary ondition of an equilibrium state is the stationarity of the TPE, thus the rst

variation of the total potentialenergy must vanish

δΠ =

dΠ

dϕ

δϕ =

4k(ϕ − ϕ

0 ) − P L sin ϕ +

1

2 F L cos ϕ

δϕ = 0

_{∀ δϕ 6= 0}

(11)

An equilibriumpath is thusdened by

P = 4

k

L

ϕ − ϕ

0 +

1 ₂

ǫ cos ϕ

sin ϕ

(12)

This equationdetermines aunique path with respe t to

ϕ

if the perturbationparameters does not satisfy the ondition

ǫ = 2ϕ

0

. In su h a ase the stru ture is a straight bar of length

L

at the unloaded state

P = 0

.Let usexamine this spe ial ase rst.

Case

ǫ = 2ϕ

0

The equilibriumequation isnow

dΠ

dϕ

= (ϕ − ϕ

0 ) − P L sin ϕ + 4kϕ

0 cos ϕ

(13)

= 4kϕ − P L sin ϕ + 4kϕ

0 (cos ϕ − 1) = 0

(14)

and the two solutionsare

ϕ = 0

primary path

P

I

,

(15)

P = 4

k

L

ϕ + ϕ

0 (cos ϕ − 1)

sin ϕ

se ondary path

P

II

(16)

An equilibriumstate is stableif the se ondvariationof

Π

:

δ

2 Π =

d

2 _Π

dϕ

2 (δϕ)

2

(7)

is positive.Let usrst examinestability of the primary path, i.e.when

ϕ = 0

,thus

δ

2 _Π|

_P

=

d

2 _Π

dϕ

2 P

(δϕ)

2 = (4k − P L)(δϕ)

2

(18)

The primary path is thus stable when

P < 4k/L

and unstable when

P > 4k/L

, and the riti al load is thus

P

cr

= 4k/L

. Let us examine wheather the riti al point is a symmetri or asymmetri bifur ationpoint.The expression of the third variation of the TPE is

δ

3 _Π|

_P

=

d

3 _Π

dϕ

3 P

(δϕ)

3

(19) where

d

3 _Π

dϕ

3 = −P L sin ϕ − 4kϕ

0 cos ϕ

(20) Atthe riti alpoint the valueof the third derivatve of the TPE is on

d

3 Π

dϕ

3 kr

= −4kϕ

0 6= 0

(21)

thus the riti al point is an asymmetri bifur ation point. The equilibrium path is drawn in

gure 1.

Case

ǫ 6= 2ϕ

0

Let us examine stability of the equilibrium path, dened in (12). The se ond variation of the TPE

δ

2 Π =

d

2 _Π

dϕ

2 (δϕ)

2

(22)

is obtained from the expression of the rst variation (11). An equilibrium state is stable if

the se ond variationof the TPE ispositivefor allkinemati allyadmissiblevariations

δϕ

, thus in this single degree of freedom example it is su ient to investigate the sign of the se ond

derivativeof the TPE

d

2 _Π

dϕ

2 = 4k − P L cos ϕ − 2kǫ sin ϕ

(23) Let's insert the expression of the equilibriumpath (12) in the expression above,gives

d

2 _Π

dϕ

2 = 4k

sin ϕ − (ϕ − ϕ

0 ) cos ϕ −

1 ₂

ǫ

sin ϕ

(24)

Letus examine the ases

ǫ > 2ϕ

0

and

ǫ < 2ϕ

0

separately.

In the ase

ǫ > 2ϕ

0

, the stru ture is below the horizonal line dened by the supports before applying the ompressive load, thus the stru ture will ontinue to displa e below the

supportline, thus

ϕ < 0

.Let usdene

ǫ = 2ϕ

0 + ¯ǫ

, and the expression (24) gives

d

2 _Π

dϕ

2 = 4k

sin ϕ − ϕ cos ϕ − ϕ

0 (1 − cos ϕ) − ¯ǫ

(8)

Sin e now

ϕ < 0

and both the nominator and denominator are negative, thus

δ

Π

is always

positive, i.e.the path is stable when

ǫ > 2ϕ

0

.

The ase

ǫ < 2ϕ

0

is more interesting. Now

ϕ > 0

and the denominator of the expression (24) isalwayspositivebut thenominator an have zeropoints.These roots anbesolved from

the trans endental equation

sin ϕ − (ϕ − ϕ

0 ) cos ϕ −

1 ₂

ǫ = 0.

(26)

Sin eanalyti alsolutionisimpossible,let'strytheasymptoti analysisassumingthattheangles

ϕ

and

ϕ

0

are small, thus

sin ϕ ≈ ϕ −

1

6 ϕ

3 _,

cos ϕ ≈ 1 −

1

2 ϕ

2 _,

and the expression (26) willhas a form

1

3 ϕ

3 −

1

2 ϕ

0 ϕ

2 _{+ (ϕ}

0 −

1 ₂

ǫ) = 0

(27)

Thethirdorderpolynomialabove anhavebothnegativeandpositivevaluesforpositivevalues

of

ϕ

. To show that,let ussrt al ulate the minumum point

ϕ

2 _{− ϕ}

0 ϕ = 0

=⇒

ϕ = ϕ

0 .

(28)

The minimum value of the fun tion dened in (27) (kun

ϕ > 0

) and the ondition for the nagativitywe get an inequality (let's dene

ǫ = ηϕ

0

)

−

1

3 ϕ

2

0 + 1 −

1

2 η < 0

=⇒

η > 2 −

1

3 ϕ

2

0

Takingthe ondition

ǫ < 2ϕ

0

intoa ountwe'llget a onditionfor the perturbationparameter

ǫ = ηϕ

0

:

2 > η > 2 −

1

3 ϕ

2

0

i.e.

2ϕ

0 > ǫ > (2 −

1

3 ϕ

2

0 )ϕ

0

forthe existen e ofa limitpoint ontheequilibriumpath. Inthe followinggure,some

equilib-rium paths are shown for some values of the perturbation parameter

ǫ

To sum up, the equilibrium pathsof this stru ture an have

•

atrivialequilibriumpathand anasymmetri bifur ationpointif

ǫ = 2ϕ

0

.These ondary path isdened inequation (16).

•

A stable equilibriumpath without riti al pointsif

ǫ > 2ϕ

0

or if

ǫ

/ (2 −

1

3 ϕ

2

0 )ϕ

0

.

•

An equilibriumpath has a limitpointif

(2 −

1

3 ϕ

2

(9)

0

0.5

1

1.5

2 -2

-1.5

-1

-0.5

0

0.5

1

1.5

2 λ

ϕ

ǫ = 2ϕ

0 ǫ = 1.99ϕ

0 ǫ = 1.8ϕ

0 ǫ = 2.2ϕ

0

(10)

The stru ture in the problem1 isan idealized olumnhaving a onstant bending stiness

EI

. Determine the spring oe ient

k

and how the riti al load will dier from the exa t beam solution.

Solution: The spring onstant

k

an be determined either by

•

lettingthe bifur ationloads tobeequal forboth models,

•

tomake the displa ementsat the middleequal under pointload at the middle,

•

tomake the displa ementsat the middleequal under uniformload. Dee tionunder a point load is

δ

_P

p

=

1

48 F L

3 EI

and for a unform load

δ

p

_q

=

5

384 qL

4 EI

.

Forthe spring-bar system the orresponding dee tions are

δ

_P

j

=

1

8 F L

2 k

and δ

j

q

=

1

16 qL

3 k

.

Let

δ

_P

j

= δ

_P

p

⇒ k

P

=

6EI

L

δ

j

_q

= δ

_q

p

⇒ k

q

=

12EI

L

The riti al load of the spring-bar system is thus

P

cr

=

4k

L

⇒ P

P

kr

=

24EI

L

2 and P

q

cr

=

48EI

L

2 .

One additionalway to ompute

k

is to make the bending strain energies equal under a uniform load.

(11)

rigidbarsand elasti springs.Investigatealsothestabilityofthe equilibriumpaths.Investigate

espe ially ases

k

1 = k

2

ja

k

1 = 5k

2

. What kindof real stru tures these models imitate?

Solution: The total potentialenergy expression is

Π = U + V

U =

1

2 k

1 L

2 _sin

2 _{ϕ + k}

2 u

2 +

1

2 k

2 [u − 2L(1 − cos ϕ)]

2 V = −P u

(29)

The equilibriumpaths an beobtained from the stationarity ondition of the TPE:

δΠ =

∂Π

∂ϕ

δϕ +

∂Π

∂u

δu = 0

(30)

Sin e the variationsof the displa ement

u

and rotation

ϕ

are arbitrary, the equilibrium paths are obtained from equations

∂Π

∂ϕ

= k

1 L

2 _{sin ϕ cos ϕ + k}

2 [u − 2L(1 − cos ϕ)](−2L sin ϕ) = 0

∂Π

∂u

= 2k

2 u + k

2 [u − 2L(1 − cos ϕ)] − P = 0

(31)

After some manipulationswe get

sin ϕ[k

1 L

2 cos ϕ − 2Lk

2 u + 4k

2 L

2 (1 − cos ϕ)] = 0

(32)

u =

P

3k

2 +

2

(12)

sin ϕ = 0

tai

(k

1 − 4k

2 )L

2 _{cos ϕ + 4k}

2 L

2 − 2k

2 Lu = 0

(34) Ifequation (33) is put into equation(34) and dene

k

2 = k

ja

k

1 = αk

, we get

⇒

P = kL 4 + (

3

2 α − 4) cos ϕ

(35)

whi histhe proje tion ofthese ondary pathontothe

(ϕ, P )

-plane.A ordinglyfromequation (33) we get

cos ϕ = 1 +

P

2kL

−

3

2 u

L

,

whi his substituted into(35)

⇒

P =

kL

4 − α

h

2α + (8 − 3α)

_L

u

i

,

whi hdes ribes the proje tion of the se ondary path onto the

(u, P )

-plane. The primary paths are dened as







ϕ =

0 u =

P

3k

and the se ondary paths











P = [4 + (

3 ₂

_{α − 4) cos ϕ]kL}

P =

kL

4 − α

h

2α + (8 − 3α)

_L

u

i

Let's investigatethe ases

α = 1

ja

α = 5

.

α = 1 ⇒







P = (4 −

5

2 cos ϕ)kL

P =

1

3 kL

2 + 5

u

L

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2 -0.3

-0.2

-0.1

0

0.1

0.2

0.3 P

kL

ϕ

(13)

0

0.5

1

1.5

2

2.5

3

3.5

4

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2 P

kL

u/L

0.333(2+5x)

3*x

We noti e, that displa ements are in reasing more rapidly on the se ondary path than

in the primary path. However, the load an still be in reased over the riti al value at the

bifur ation point.

(P

kr

=

3

2 kL

, thus the se ondary path is stable. In ompressed thin plates su h kind of behaviour an be obtained. The strong stability of the se ondary paths an be

utilizedalso indesign for some ases.

α = 5 ⇒







P = (4 +

7

2 cos ϕ)kL

P = −kL

10 − 7

_L

u

0

1

2

3

4

5

6

7

8 -0.3

-0.2

-0.1

0

0.1

0.2

0.3 P

kL

ϕ

(14)

0

1

2

3

4

5

6

7

8

0

0.5

1

1.5

2

2.5

3 P

kL

u/L

-10+7*x

3*x

In this ase the bifur ation load is mu h higher than in the preious one. However, the

se ondary equilibrium path is now unstable. Shells, espe ially exhibit su h kind of unstable

behaviourafter bifur ation.Ifthe post-bu klingregimeis unstable, su hstru tures are

imper-fe tionsensitive,whi hmeansthatthe riti alloadofanimperfe tstru tureismu hlowerthan

the theoreti al bifur ationload. Imperfe tions are due to e entri ities, geometri aldeviations

et .

Example 1.5 Investigate the ee t of imperfe tions in the previous example. Draw the

im-perfe tion sensitivity diagram for the ase

k

1 = 5k

2

.

Solution: Let's determine the riti al load asa fun tion of

ϕ

the imperfe tion amplitude

ϕ

0

. Now

U =

1

2 k

1 L

2 (sin ϕ − sin ϕ

0 )

2 + k

2 u

2 +

1

2 [u − 2L(cos ϕ

0 − cos ϕ)]

2

and

∂Π

∂ϕ

= k

1 L

2 (sin ϕ − sin ϕ

0 ) cos ϕ + k

2 [u − 2L(cos ϕ

0 − cos ϕ)](−2L sin ϕ) = 0

∂Π

∂u

= 2k

2 u + k

2 [u − 2L(cos ϕ

0 − cos ϕ)] − P = 0

Solving

u

fromthe equation above and substitute it intothe equationbelow, gives

u =

k

1 2k

2 sin ϕ − sin ϕ

0 tan ϕ

L − 2L(cos ϕ − cos ϕ

0 )

P = 3k

2 u − 2k

2 L((cos ϕ − cos ϕ

0 ) =

3k

1

2 sin ϕ − sin ϕ

0 tan ϕ

L − 8k

2 L((cos ϕ − cos ϕ

0 )

(15)

Figure2: The maximum load

λ

max

asa fun tion of the imperfe tionamplitude

ϕ

0

1

2

3

4

5

6

7

8

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1 P

kL

ϕ

ǫ

0 = 0

0.003

0.03

0.1

(16)

ation pointson the paths? (

EI = EA = ∞

)

c

a

c

c c

P k k L L L

Solution: Let's determine the displa ementsby using the following gure

c!!

!!

a

_b

b

_b

b

_b

b

_b

b

bb c

ϕ

0 ϕ

1 !!

ϕ

12 ϕ

2 ϕ

3 v

1 v

2 sin ϕ

0 =

v

1 L

, sin ϕ

12 =

v

2 − v

1 L

, sin ϕ

3 =

v

2 L

In addition

ϕ

1 = ϕ

0 − ϕ

12 = arcsin

v

1 L

− arcsin

v

2 − v

1 L

ϕ

2 = ϕ

3 + ϕ

12 = arcsin

v

2 L

+ arcsin

v

2 − v

1 L

Assuming small rotationswe an approximate

arcsin x ≈ x + 1/6 x

3

, and the rotationsat the

springs are

ϕ

1 = δ

1 +

1

6 δ

3

1 −

δ

2 − δ

1 +

1

6 (δ

2 − δ

1 )

2 = 2δ

1 − δ

2 +

1

3 δ

3

1 −

1

6 δ

3

2 +

1

2 δ

1 δ

2

2 −

1

2 δ

2

1 δ

2 ϕ

2 = δ

2 +

1

6 δ

3

2 +

δ

2 − δ

1 +

1

6 (δ

2 − δ

1 )

2 = 2δ

2 − δ

1 +

1

3 δ

3

2 −

1

6 δ

3

1 −

1

2 δ

1 δ

2

2 +

1

2 δ

2

1 δ

2

Dee tion underthe load

∆ = L

3 −

q

1 − δ

2

1 −

p1 − (δ

1 − δ

2 )

2 −

q

1 − δ

2

2 ≈ L

δ

2

1 + δ

2

2 − δ

1 δ

2 +

1

4 δ

4

1 +

1

4 δ

4

2 −

1

2 δ

3

1 δ

2 +

3

4 δ

2

1 δ

2

2 −

1

2 δ

1 δ

3

2

(17)

Π = Π(δ

1 , δ

2 ) =

1

2 kϕ

2

1 (δ

1 , δ

2 ) +

1

2 kϕ

2

2 (δ

1 , δ

2 ) − P ∆(δ

1 , δ

2 )

Equilibriumpaths are determined from the stationarity ondition

δΠ = 0

,whi hgives

∂Π

∂δ

1 = kϕ

1 ∂ϕ

1 ∂δ

1 + kϕ

2 ∂ϕ

2 ∂δ

1 − P

∂∆

∂δ

1 = 0

(36)

∂Π

∂δ

2 = kϕ

1 ∂ϕ

1 ∂δ

2 + kϕ

2 ∂ϕ

2 ∂δ

2 − P

∂∆

∂δ

2 = 0

(37) in whi h

∂ϕ

1 ∂δ

1 = 2 + δ

2 ₁

+

1

2 δ

2

2 − δ

1 δ

2 ∂ϕ

2 ∂δ

1 = −1 −

1

2 δ

2

1 −

1

2 δ

2

2 + δ

1 δ

2 ∂∆

∂δ

1 = L

2δ

1 − δ

2 + δ

1

3 −

3

2 δ

2

1 δ

2 +

3

2 δ

1 δ

2

2 −

1

2 δ

3

2 ∂ϕ

1 ∂δ

2 = −1 −

1

2 δ

2

1 −

1

2 δ

2

2 + δ

1 δ

2 ∂ϕ

2 ∂δ

2 = 2 + δ

2 ₂

_{− δ}

1 δ

2 +

1

2 δ

2

1 ∂∆

∂δ

2 = L

2δ

2 − δ

1 + δ

2

3 −

1

2 δ

3

1 +

3

2 δ

2

1 δ

2 −

3

2 δ

1 δ

2

Equations (36) and (37) are satised, if

ϕ

1 = ϕ

2 = ∆ = 0

i.e.

δ

1 = δ

2 = 0

. Let's investigate stabilityof this primarypath. The se ond variation of the total potentialenergy is

δ

2 Π =

∂

2 _Π

∂δ

2

1 [δ(δ

1 )]

2 + 2

∂

2 _Π

∂δ

1 ∂δ

2 δ(δ

1 )δ(δ

2 ) +

∂

2 _Π

∂δ

2

2 [δ(δ

2 )]

2 = (δ(δ

1 ) δ(δ

2 ))







∂

2 Π

∂δ

2

1 ∂

2 Π

∂δ

1 ∂δ

2 ∂

2 Π

∂δ

1 ∂δ

2 ∂

2 Π

∂δ

2

1 





δ(δ

1 )

δ(δ

2 )

!

,

in whi h the matrix

K = [∂

2 _Π/∂δ

i

∂δ

j

]

is the stability matrix. The path is stable, if and only if

δ

2 _{Π > 0}

. This is true if the stability matrix

K

is positive denite, whi h means that all its eigenvalues are positive.

On the primary path

δ

1 = δ

2 = 0

the elements of the stabilitymatrix are

∂

2 _Π

∂δ

2

1 = kϕ

1 ∂

2 _ϕ

1 ∂δ

2

1 + k

∂ϕ

1 ∂δ

1

2 + kϕ

2 ∂

2 _ϕ

2 ∂δ

2

1 + k

∂ϕ

2 ∂δ

1

2 − P

∂

2 _∆

∂δ

2

1

(38)

∂

2 _Π

∂δ

2

1 = 5k − 2P L

∂

2 _Π

∂δ

1 ∂δ

2 = −4k + P L

(18)

∂

2 _Π

∂δ

2

2 = 5k − 2P L

Denoting

P = λk/L

and solving the eigenvalues of

K

. The path is stable, when

K

is positive denite, i.e. allitseigenvalues are positive

K ¯

x = ω ¯

_{x ⇒ (K − ωI)¯x = 0 ⇒ det(K − ωI) = 0}

det

5 − 2λ − ω

λ − 4

5 − 2λ − ω

!

= 0

ω

1 = 1 − λ ja ω

2 = 9 − 3λ

The zero points of the eigenvalues o ur when

λ

have values

λ = 1

and

λ = 3

. The primary pathis stablewhen

λ < 1

. Theeigenmodedes of

K

i.e.the bu klingmodesof the stru tureare obtained from

λ = 1 ⇒

3 −3

−3

3 ! (

δ

1 δ

2 )

=

(

0

0 )

⇒ δ

1 = δ

2

The bu klingmode orrespondingto the riti al load

P

kr

= k/L

is:

c

a

```

a

`````

_{`` c}

c c

λ = 3 ⇒

−1

1

1 ₋₁

! (

δ

1 δ

2 )

=

(

0

0 )

⇒ δ

1 = −δ

2

And the bu lingmode orrespondingto the riti alload

P

kr

= 3k/L

is:

c

aa

_aa

aa

_aa

aaa a

c

c c

Let's nally investigate the post-bifur ation paths after the bran hing point at

λ = 1

. Substituting displa ements

δ

1 = δ

2 = δ ⇒ ϕ

1 = ϕ

2

intothe equation of equilibrium(36).

∂Π

∂δ

1 = 0 ⇒ k δ +

1

6 δ

3 2 +

1

2 δ

2 _{− 1 − P L δ −}

1

2 δ

3 _{= 0}

⇒

δ

k 1 +

1

6 δ

2 1 +

1 ₂

δ

2 _{− P L 1 +}

1

2 δ

2 = 0

⇒

δ = 0 tai P

II

= 1 +

1 ₆

δ

2 _k

L

(39)

The same orresponding to the higher bifur ationload:

λ = 3

kohdalla(

δ

1 = −δ

2 = δ ⇒ ϕ

2 =

−ϕ

1

)

∂Π

∂δ

1 = 0 ⇒ k 3δ +

3

2 δ

3 _{2 +}

5

2 δ

2 _{− (−1 − 2δ}

2 ₎

_{− P L 3δ +}

9

2 δ

3 _{= 0}

⇒ δ

3k 1 +

1

2 δ

2 3 +

9 ₂

δ

2 _{− P L 3 +}

9

2 δ

2 = 0

⇒ δ = 0 tai P

III

= 1 +

1 ₂

δ

2 _3k

L

(40)

(19)

0

0.5

1

1.5

2

2.5

3

3.5

4 -0.4

-0.2

0

0.2

0.4 λ

δ

P

II

P

III

Ifwewant toinvestigatestability properties ofthe paths

P

II

and

P

III

wehaveto substi-tute the equationsof the paths(39) and (40) intothe expression of the se ondvariationofthe

TPE (38). Forpath

P

II

it isvalid(

δ

1 = δ

2

)

∂

2 _Π

∂δ

2

1 P

II

= k

δ +

1

6 δ

3 δ + k

2 +

1

2 δ

2

2 + k

δ +

1

6 δ

3 · 0 + k(−1)

2 − k

1 +

1

6 δ

2 2 +

3

2 δ

2 = k

3 +

7

6 δ

2 ₊

1

6 δ

4 > 0 ∀ δ

Forthe path

P

III

(

δ

1 = −δ

2 , ϕ

1 = −ϕ

2

)

∂

2 _Π

∂δ

2

1 P

III

= k

3δ +

3

2 δ

3 3δ + k

2 +

5

2 δ

2

2 − k

3δ +

3

2 δ

3 (−2δ) + k(−1 − 2δ

2 )

2 −3k

1 +

1

2 δ

2 2 +

15

2 δ

2 = k

−1 +

7 ₂

δ

2 +

3

2 δ

4 < 0,

when

δ

issu iently small.Therefore the path

P

II

isstable andthe path

P

III

isunstable near the bifur ation point.

(20)

translational springs. Investigate alsostability of the paths. It is assumed that the point C is

not moving horizontally.

A C B

?

_v

θ

-

a

D

a

-?

P

6 ?

L

Solution: The total potentialenergy expression is

Π(v, θ) =

1

2 k∆

2 A

+

1

2 k∆

2 B

− P ∆

D

= k(v

2 + a

2 sin

2 _{θ) − P [v + L(1 − cos θ)]}

= ka

2 (u

2 + sin

2 _{θ) − P}

u +

1 α

(1 − cos θ)

,

in whi h

v = au

and

a = αL

.By dening

P = λka

, weget the form

Π

ka

2 = ˜

Π = u

2 _{+ sin}

2 θ − λ

u +

1 α

(1 − cos θ)

The equilibriumequations are

∂ ˜

Π

∂u

= 2u − λ = 0 ⇒ u =

λ

2 ∂ ˜

Π

∂θ

= 2 sin θ cos θ − λ

1 α

sin θ = 0

= sin θ

2 cos θ −

_α

1 λ

= 0

⇒

(

sin θ = 0 ⇒ θ = 0

λ = 2α cos θ

Thusthe primarypath isdened as

u = λ/2

ja

θ = 0

.Stabilityof the primarypath

∂

2 _Π

_˜

∂u

2 = 2,

∂

2 _Π

_˜

∂u∂θ

= 0,

∂

2 _Π

_˜

∂θ

2 = 2 cos 2θ −

λ

α

cos θ

Let's substitute the expressions of the primary path

u = λ/2

and

θ = 0

, into the stability matrix:

K =

"

2

0 0 2 −

α

λ

#

(21)

The riti alvalue ofthe load parameter

λ

kr

anbeobtained fromthe onditiondet

(K) = 0 ⇒

λ

kr

= 2αka = 2α

2 kL

.

On the se ondary equilibriumpath

u = λ/2

ja

λ = 2α cos θ

:

K =

"

2

0 0 2 cos 2θ − 2 cos

2 _θ

#

= 2

"

1

0 0 − sin

2 θ

#

The se ondary path isunstable sin e

K

2 2 = ∂

2 _Π/∂θ

˜

2 _{= − sin}

2 θ ≤ 0 ∀ θ

.

0

0.5

1

1.5

2 -0.4

-0.2

0

0.2

0.4 λ/α

θ

(22)

Example 2.1 Derive the expression of urvature for a plane beam using (a) the Lagrangian

and (b) the Eulerianapproa h.

Solution: The dieren e between the Lagrangian and Eulerian approa hes is the meaning of

theindependentvariable

x

.IntheLagrangianapproa hthe oordinateisatta hedtoamaterial point.The dispa ement atpoint

x

is the displa ementof the pointa upyingthe position

x

at the initialundeformed onguration. In the Eulerianapproa h the oordinateis referringonly

toa spatial point

x

.

da = dx

-x, u

?

y, v

ϕ

da

dv

u

u + du

Lagrange: It is seen fromthe gure

sin ϕ =

dv

da

=

dv

dx

= v

′

⇒ ϕ = arcsin v

′

sin e the urvature is

κ = 1/R = ϕ

′

we get

1 R

= ϕ

′

₌

v

′′

p1 − (v

′

₎

2 ,

d arcsin x

dx

=

1 √

1 − x

2 da

-x, u

?

y, v

ϕ

da

dv

dx

Euler: From the gure

(23)

tan ϕ =

dv

dx

= v

′

⇒ ϕ = arctan v

′

we obtainfor the urvature

κ =

1 R

=

∂ϕ

∂a

not

∂ϕ

∂x

!!

=

1 1 + v

′2

∂

∂a

dv

dx

where

d arctan x

dx

=

1 1 + x

2 =

1 1 + v

′2

d

2 _v

dx

2

1 √

1 + v

′2

(where da = dx

√

1 + v

′2

₎

=

v

′′

(1 + v

′2

₎

√

_{1 + v}

′2

=

v

′′

(1 + v

′2

₎

3/2

Note! When the higherorder terms are negle ted we get the same result forboth approa hes:

(24)

Example 2.2 Determine

P

cr

startingfrom the dierentialequation.

e e

2 EI EI L/2 L/2

-x

P Solution: In part 1 1

v

(4)

1 + k

2 v

1 ′′

= 0

, where

k

2 _{= P/2EI}

.(see problem ??) 2

v

(4)

2 = 0

BC : v

1 −

L

₂

= v

1 ′

−

L

2 = v

2 L

2 = v

2 ′

L

2 = 0

v

1 (0) = v

2 (0)

v

′

1 (0) = v

′

2 (0)

M

1 (0) = M

2 (0)

Q

1 (0) = Q

2 (0) + P v

2 ′

(0)

P

XX

X

z

M

1 -Q

1 XXXX

_XXXX

XXXX

P

Q

2 M

2

Solutionsfor the homogenious equationsare

v

1 = C

1 sin kx + C

2 cos kx + C

3 x + C

4 v

′

1 = C

1 k cos kx − C

2 k sin kx + C

3 v

′′

1 = −C

1 k

2 sin kx − C

2 k

2 cos kx

v

′′′

1 = −C

1 k

3 cos kx + C

2 k

3 sin kx

v

2 = C

5 x

3 + C

6 x

2 + C

7 x + C

8 v

′

2 = 3C

5 x

2 + 2C

6 x + C

7 v

′′

₂

= 6C

5 x + 2C

6 v

′′′

2 = 6C

5

Takingthe boundary onditions intoa ount

Q

1 (0) = Q

2 (0) + P v

2 ′

(0)

−2EIv

′′′

1 (0) = −EIv

2 ′′′

(0) + P v

2 ′

(0)

2C

1 k

3 = −6C

5 + 2k

2 C

7 C

5 = −

1

3 k

3 _C

1 +

1

3 k

2 _C

7

(25)

M

1 (0) = M

2 (0)

−2EIv

′′

1 (0) = −EIv

2 ′′

(0)

2C

2 k

2 = −2C

6 ⇒ C

6 = −k

2 C

2 v

′

1 (0) = v

2 ′

(0)

C

1 k + C

3 = C

7 ⇒ C

5 = −

1

3 k

3 _C

1 +

1

3 k

2 _(C

1 k + C

3 ) =

1

3 k

2 _C

3 v

1 (0) = v

2 (0)

C

2 + C

4 = C

8 v

1 −

L

₂

= 0 ⇒ C

4 = C

1 sin

kL

2 − C

2 cos

kL

2 + C

3 L

2 v

′

1 −

L

2 = 0 ⇒ C

3 = −k(C

1 cos

kL

2 + C

2 sin

kL

2 )

v

2 L

2 = 0 ⇒

1 ₃

k

2 C

3 L

2

3 − k

2 C

2 L

2

2 + (C

1 k + C

3 )

L

2 + C

4 = 0

⇒

kL

2 − kL cos

kL

2 1 +

1

24 (kL)

2 + sin

kL

2 C

1 +

1 −

1 ₄

(kL)

2 _{− cos}

kL

2 − kL sin

kL

2 1 +

1

24 (kL)

2 C

2 = 0

v

′

2 L

2 = 0 ⇒ k

2 C

3 L

2

2 − 2k

2 C

2 L

2 + C

1 k + C

3 = 0

⇒

1 −

1 +

1

4 (kL)

2 cos

kL

2 kC

1 +

−kL −

1 +

1

4 (kL)

2 sin

kL

2 C

2 = 0

⇒

"

a(kL) b(kL)

c(kL) d(kL)

# (

C

1 C

2 )

=

(

0

0 )

det = 0 ⇒ kL ≈ 7.55 ⇒ P

kr

= 114

EI

L

2

(26)

Example 2.3 Determine

P

cr

startingfrom the dierentialequation.(

α = 2, β = 1

)

@@

α

EI

β

EA EI EA L/2 L/2

-x

P

Solution: Sin e

β = 1 ⇒ (EA)

1 = (EA)

2 ⇒ P

1 = P

2 = P/2

.

Part 1:

v

(4)

1 + k

1

2 v

1 ′′

= 0

k

1

2 =

P/2

2EI

=

P

4EI

2:

v

(4)

2 − k

2

2 v

2 ′′

= 0

k

2

2 =

P/2

EI

=

P

2EI

k

2

2 = 2k

2

1 ⇒ k

2 =

√

2k

1 BC : v

1 −

L

₂

= v

′

1 −

L

2 = v

2 L

2 = v

2 ′

L

2 = 0

v

1 (0) = v

2 (0)

v

′

1 (0) = v

2 ′

(0)

M

1 (0) = M

2 (0)

Q

1 (0) = Q

2 (0) + P v

2 ′

(0)

P

2 XX

X

z

M

1 -Q

1 XXXX

_XXXX

XXXX

P

Q

2 M

2 P

2 XX

X

z

Solutionsfor the homogenious dierentialequations are

v

1 = C

1 sin kx + C

2 cos kx + C

3 x + C

4 v

1 ′

= C

1 k cos kx − C

2 k sin kx + C

3 v

′′

1 = −C

1 k

2 sin kx − C

2 k

2 cos kx

v

′′′

1 = −C

1 k

3 cos kx + C

2 k

3 sin kx

v

2 = C

5 sinh kx + C

6 cosh kx + C

7 x + C

8 v

′

2 = C

5 k cosh kx − C

6 k sinh kx + C

7 v

′′

2 = −C

5 k

2 sinh kx − C

6 k

2 cosh kx

v

′′′

2 = −C

5 k

3 cosh kx + C

6 k

3 sinh kx

Takingthe boundary onditions intoa ount

Q

1 (0) = Q

2 (0) + P v

′

2 (0)

−2EIv

′′′

1 (0) = −EIv

′′′

2 (0) + P v

1 ′

(0)

2C

1 k

3

1 = −C

5 k

2

3 + 4k

1

2 (C

1 k

1 + C

3 )

C

5 = −

1 k

3

2 2k

3 ₁

C

1 + 4k

2

1 C

3 =

1 √

2 C

1 +

2 √

2k

1 C

3

(27)

M

1 (0) = M

2 (0)

−2EIv

′′

1 (0) = −EIv

′′

2 (0)

2C

2 k

2

1 = −C

6 k

2

2 ⇒ C

6 = −2

k

1 k

2

2 C

2 = −C

2 v

′

1 (0) = v

′

2 (0)

C

1 k

1 + C

3 = C

5 k

2 + C

7 ⇒ C

7 = C

1 k

1 + C

3 − C

5 k

2 = −C

3 v

1 (0) = v

2 (0)

C

2 + C

4 = C

6 + C

8 ⇒ C

8 = C

2 + C

4 − C

6 = 2C

2 + C

4 v

1 −

L

2 = 0 ⇒ C

4 = C

1 sin

k

1 L

2 − C

2 cos

k

1 L

2 + C

3 L

2 v

′

1 −

L

₂

= 0 ⇒ C

3 = −k

1 C

1 cos

k

1 L

2 + C

2 sin

k

2 L

2 ⇒ C

4 =

sin

k

1 L

2 −

k

1 L

2 cos

k

1 L

2 C

1 −

cos

k

1 L

2 +

k

1 L

2 sin

k

1 L

2 C

2 v

2 L

2 =

1 √

2 C

1 +

2 √

2k

1 C

3 sinh

k

2 L

2 − C

2 cosh

k

2 L

2 − C

3 L

2 + 2C

2 + C

4 = 0

⇒

1 √

2 −

2 √

2 cos

k

1 L

2 sinh

k

2 L

2 + sin

k

1 L

2 C

1 −

√

2

2 sin

k

1 L

2 sinh

k

2 L

2 − cosh

k

2 L

2 + 2 − cos

k

1 L

2 C

2 = 0

v

′

2 L

2 =

1 √

2 C

1 −

2 √

2 C

1 cos

k

1 L

2 −

2 √

2 C

2 sin

k

1 L

2 k

2 cosh

k

2 L

2 −C

2 k

2 sinh

k

2 L

2 + k

1 C

1 cos

k

1 L

2 + k

1 C

2 sin

k

1 L

2 = 0

⇒

k

√

2

2 cosh

k

2 L

2 −

2k

2 √

2 cos

k

1 L

2 cosh

k

2 L

2 + k

1 cos

k

1 L

2 C

1 +

k

1 sin

k

1 L

2 − k

2 sinh

k

2 L

2 −

2k

2 √

2 sin

k

1 L

2 cosh

k

2 L

2 C

2 = 0

⇒

"

a(kL) b(kL)

c(kL) d(kL)

# (

C

1 C

2 )

=

(

0

0 )

The riti al load isobtained from the equation det[℄

= ad − bc = 0

⇒ k

1 L ≈ 10, 637 ⇒ P

kr

= 4k

1

2 EI

L

2 = 452, 6

EI

L

2

Ifthe dire tionof the load is reversed, the result is

247, 5EI/L

2

(28)

tensible beam and small dee tions. Solve the equations and determine the eigenmodes and

show that the eigenmodes are orthogonal.

EI

L

P

Solution: The total potentialenergy fun tionalis

Π(v) =

1

2 L

Z

0 EI(v

′′

₎

2 − P (v

′

₎

2 _dx,

wherethehorizontaldee tion

∆

undertheload

P

anbedeterminedas

ϕ

dx + du

dv

The Euler equations are obtained from the stationarity ondition ofthe fun tional

δΠ = Π

,v

δv =

L

Z

0 (EIv

′′

_δv

′′

_{− P v}

′

_δv

′

_{)dx = 0,}

where

δv

isthevariationofthe dee tion,i.e.anarbitraryfun tionsatisfyingthehomogenious kinemati al boundary onditions

v(0) = v

′

_{(0) = 0}

. After integration by parts we get the term

δv

as a ommonfa tor inside the integral

δΠ =

L

0 EIv

′′

_δv

′

₋

L

Z

0 (EIv

′′

₎

′

_δv

′

_{dx −}

L

0 P v

′

_{δv +}

L

Z

0 δvdx

=

L

0 EIv

′′

_δv

′

₋

L

0 (EIv

′′

₎

′

_{δv −}

L

0 P v

′

_{δv +}

L

Z

0 [(EIv

′′

₎

′′

_{+ (P v}

′

₎

′

_{] δvdx}

At the lower limit

δv(0) = δv

′

_{(0) = 0}

(29)

and shear for e:

M = −EIv

′′

sekä

Q = −(EIv

′′

₎

′

,we get

δΠ = −M(L)δv

′

_{(L) + [Q(L) − P v}

′

_{(L)]δv(L) +}

L

Z

0 [(EIv

′′

₎

′′

_{+ (P v}

′

₎

′

_{] δvdx = 0,}

sin e

δv

isarbitraryfun tion satisfyingthe boundary onditions

v(0) = v

′

_{(0) = 0}

, thusthe

fol-lowingequationshavetobesatised

(EIv

′′

₎

′′

_{+ (P v}

′

₎

′

_{= 0}

_{x ∈ (0, L)}

(Euleri equation)

M(L)

= 0

Q(L) − P v

′

_{(L) = 0}

)

natural boundary onditions

v(0)

= 0

v

′

_{(0) = 0}

)

essential boundary onditions

Ifthe bendingstiness

EI

and the ompressive for e

P

are onstants inthe domain,we get a homogeneousdierentialequation with onstant oe ients

EIv

(4)

+ P v

′′

_{= 0}

⇒ EIv

′′

_{+ P v = Cx + D, (C, D constants)}

⇒ v = A sin kx + B cos kx + Cx + D, k =

r P

_EI

The derivativesare

v

′

_{= Ak cos kx − Bk sin kx + C}

v

′′

_{= −Ak}

2 _{sin kx − Bk}

2 cos kx

v

′′′

_{= −Ak}

3 _{cos kx + Bk}

3 _{sin kx}

v(0) = 0 ⇒ B + D = 0

v

′

_{(0) = 0 ⇒ Ak + C = 0}

v

′′

_{(L) = 0 ⇒ A sin kL + B cos kL = 0}

−EIv

′′′

_{(L)−P v}

′

_{(L) = 0 ⇒ −EI(−Ak}

3 _{cos kL+Bk}

3 sin kL)−P (Ak cos kL−Bk sin kL+C) = 0

(41)

P = λ

EI

L

2 ⇒ k =

√

λ

L

It follows from equation (41)that

A = 0 ⇒ C = 0 ⇒ B cos kL = 0 ⇒ B = 0

or

cos kL = 0

. If

B = 0 ⇒ v ≡ 0

ityields atrivialsolution, hen e weshould have

(30)

⇒ λ

n

=

π

2 + nπ

2 ,

and the lowest bu klingload is

λ

0 =

π

2

2 ⇒ P

cr

=

π

2

4 EI

L

2

The eigenmode orresponding to the eigenvalue

λ

n

is

v

n

= B(cos k

n

x − 1), k

n

=

1 L

π

2 + nπ

It was asked to give the normalized eigenmodes. For that we should dene how this

normalizationshould be done. Itis usual to use the energy norm

||v

n

||

2 E

=

L

Z

0 EI(v

′′

_n

)

2 dx.

The energy orthogonalitythus means

L

Z

0 EIv

′′

n

v

m

′′

dx = 0, kun n 6= m.

Lets normalize the eigenmodes

v

n

su h, that

||v

n

||

E

= E

1

, where

E

1

is the energy unit and

[E

1 ] =

√

Nm

.

v

′′

n

= −Bk

n

2 cos k

n

x

⇒ E

1

2 = EIB

2 k

4 n

L

Z

0 cos

2 k

n

xdx

Lets hangevariablessu h, that

y = k

n

x, dx =

1 k

n

dy rajat

(

x = 0

_{⇒ y = 0}

x = L ⇒ y =

π

2 + nπ

⇒ E

1

2 = EIB

2 k

n

3 π

2 +nπ

Z

0 cos

2 ydy = EIB

2 k

3 _n

1

2 π

2 + nπ

⇒ B

2 ₌

2E

1

2 EIk

3 n

π

2 + nπ

=

2E

2

1 EI

π

₁₆

4 (1 + 2n)

4 ⇒ B =

4 √

2L

3/2

_E

1 π

2 _{(1 + 2n)}

2 √

_EI

The energy orthonormal eigenfun tionsare thus

v

n

(x) = b

n

(cos k

n

x − 1), B

n

=

4 √

2 π

2 _{(1 + 2n)}

2 E

1 L

3/2

√

EI

(31)

L

Z

0 EIv

n

′′

v

m

′′

dx = 0, kun n 6= m.

L

Z

0 v

′′

n

v

m

′′

dx =

L

Z

0 cos

h

π

2 (1 + 2n)

x

L

i

cos

h

π

2 (1 + 2m)

x

L

i

merk. y =

π

2 x

L

, dx =

2L

π

dy

=

2L

π

L

Z

0

(32)

span as afun tion of the ompressive for e

P

for the beam shown below.

e e

EI L

P

? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?

q

Solution: The dierential equationfor the beam- olumnis

v

(4)

+ k

2 v

′′

₌

q

EI

, where k

2 ₌

P

EI

The solutionis

v = C

1 sin kx + C

2 cos kx + C

3 x + C

4 + Ax

2 , where A =

q

2EIk

2 =

q

2P

.

Let's hoose the zero o-ordinate atthe midspan. From the boundary onditions we get

v

′

_{(0) = 0 ⇒ C}

1 k + C

3 = 0

v

′′′

_{(0) = 0 ⇒ C}

1 = 0 ⇒ C

3 = 0

v

′

L

2 = 0 ⇒ −C

2 k sin

kL

2 +

qL

2P

= 0

⇒ C

2 =

qL

2kP sin

kL

2 v

±

L

2 = 0 ⇒ C

2 cos

kL

2 + C

4 + A

L

2

4 = 0

⇒ C

4 = −

qL

2kP tan

kL

2 −

qL

2 8P

⇒ v(x) =

qL

2kP sin

kL

₂

cos kx −

qL

2kP sin

kL

₂

cos

kL

2 +

kL

4 sin

kL

2 +

q

2P

x

2

Denoting

P = λ EI/L

2

, thus

kL =

√

λ

.

⇒ v(0) =

qL

2kP sin

√

₂

λ

1 − cos

√

λ

2 !

−

qL

2 8P

⇒ M(x) = −EIv

′′

=

C

2 k

2 cos kx −

q

P

EI =

qL

2 λ

kL

2 cos kx

sin

kL

2 − 1

!

(33)

x = ±

L

₂

M

t

=

qL

2 λ

√

λ

2 tan

√

λ

2 − 1

!

The bending moment inthe midspan

(x = 0) M

k

=

qL

2 λ

√

λ

2 sin

√

₂

λ

− 1

!

P/P

kr

λ

M

t

/qL

2 M

k

/qL

2 v(0)/

qL

4 EI

0 0 -0.0833 0.0417 0.0026 0.5

2π

2

-0.1363 0.0908 0.0052 0.75

3π

2

-0.2390 0.1911 0.0103 0.9

3.6π

2

-0.5439 0.4944 0.0257

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9 M

qL

2 P/P

cr

M

t

M

k

Stability of structures - Solved examples

P = λ4k/L

b



 

b

b

@@

b

@@

b



P

k

L/2

L/2

ϕ

ϕ

b

b````

```

`

b



 

b

@

@

b b

@

@



P

k

ϕ

ϕ

Π

Π =

1

2

k(2ϕ)

2

− P L(1 − cos ϕ)

∂Π

∂ϕ

= 4kϕ − P L sin ϕ

∂

2

Π

∂ϕ

2

= 4k − P L cos ϕ.

δΠ =

∂Π

∂ϕ

δϕ = 0 ∀ δϕ 6= 0 ⇒

∂Π

∂ϕ

= 0

⇒







ϕ =

0

P =

4k

L

ϕ

sin ϕ

Π

ϕ = 0

δ

2

Π =

∂

2

Π

∂ϕ

2

_b

_b

_b

_{b b}

_Π

_Π

_{= 0}

_Π

_Π

_Π