FEM for buckling analysis

(1)

Jerzy Pamin

e-mail: jpamin@L5.pk.edu.pl

(2)

1 Introduction

2 Derivation of FEM equations Frame finite element

3 Example

(3)

forces and decreased by compressive forces.

A sufficiently large compressive force can reduce the bending stiffness to zero and structural buckling (instability mode) occurs.

Assumptions

linear elasticity: σ x = Eε x

static one-parameter load ideal system (no imperfections) equilibrium of buckled configuration

ε _x (x) = u ⁰ (x) − zw ⁰⁰ (x) + 1

2 (w ⁰ (x)) ²

w w

kr

λP 1 λP 2

λP ₃

x, u z, w

y

(4)

Derivation of FEM equations

Energetic criterion of equilibrium

Φ = U − W Φ – total energy

U – elastic energy: U = ¹ ₂ R

V

ε x σ x dV W – work of external forces: W = Q ^T P

Q – dof vector (nodal displacement vector)

P – external force vector

(5)

Φ – total energy

U – elastic energy: U = ¹ ₂ R

V

ε x σ x dV W – work of external forces: W = Q ^T P

Q – dof vector (nodal displacement vector) P – external force vector

Value of (compressive) normal force before buckling

∆l

l = ε _x , σ _x = N

A , σ _x = Eε _x , ε _x = u ⁰ =⇒ N (x) = EAu ⁰ (x)

(6)

Energetic criterion

Elastic energy

U = 1 2 Z

V

ε x σ x dV

U = 1 2

E

X

e=1

( Z

V ^e

E

u ê0 − zw ê00 + 1 2 w ê02

2 dV ^e

)

(7)

Energetic criterion

Elastic energy

U = 1 2 Z

V

ε x σ x

Eε x

dV

U = 1 2

E

X

e=1

( Z

V ^e

E

u ê0 − zw ê00 + 1 2 w ê02

2 dV ^e

)

(8)

Energetic criterion

Elastic energy

U = 1 2 Z

V

Eε ² _x dV

U = 1 2

E

X

e=1

( Z

V ^e

E

u ê0 − zw ê00 + 1 2 w ê02

2 dV ^e

)

(9)

Energetic criterion

Elastic energy

U = 1 2 Z

V

Eε ² _x dV

U = 1 2 Z

V

E

u ⁰ − zw ⁰⁰ + 1 2 w ⁰²

2 dV

(10)

U = 1 2 Z

V

Eε ² _x dV

U = 1 2 Z

V

E

u ⁰ − zw ⁰⁰ + 1 2 w ⁰²

2 dV

Elastic energy of discretized system

U = 1 2

E

X

e=1

( Z

V ^e

E

u ê0 − zw ê00 + 1 2 w ê02

2 dV ^e

)

(11)

U = 2 _V Eε _x dV

U = 1 2 Z

V

E

u ⁰ − zw ⁰⁰ + 1 2 w ⁰²

² dV

Elastic energy of discretized system

U = 1 2

E

X

e=1

( Z

V ^e

E

u ê0 − zw ê00 + 1 2 w ê02

² dV ^e

)

U= 1 2

E

X

e=1

( Z l ^e 0

Z

A ^e

E

u ^e02 +z ² w ^e002 + 1

4 w ê04 −2zu ê0 w ê00 +u ê0 w ê02 −zw ê00 w ê02

dA ^e

dx ^e

)

(12)

U = 2 _V Eε _x dV

U = 1 2 Z

V

E

u ⁰ − zw ⁰⁰ + 1 2 w ⁰²

² dV

Elastic energy of discretized system

U = 1 2

E

X

e=1

( Z

V ^e

E

u ê0 − zw ê00 + 1 2 w ê02

² dV ^e

)

U= 1 2

E

X

e=1

( Z l ^e 0

Z

A ^e

E

u ê02 +z ² w ê002 + ¹ ₄ w ê04

nonlinear term, upon linearization ∼ = 0

−2zu ê0 w ê00 +u ê0 w ê02 −zw ê00 w ê02 dA ê

dx ^e

)

(13)

U = 2 _V Eε _x dV

U = 1 2 Z

V

E

u ⁰ − zw ⁰⁰ + 1 2 w ⁰²

² dV

Elastic energy of discretized system

U = 1 2

E

X

e=1

( Z

V ^e

E

u ê0 − zw ê00 + 1 2 w ê02

² dV ^e

)

U= 1 2

E

X

e=1

( Z l ^e 0

Z

A ^e

E

u ê02 +z ² w ê002 −2zu ê0 w ê00 +u ê0 w ê02 −zw ê00 w ê02 dA ê

dx ^e

)

(14)

U = 2 _V Eε _x dV

U = 1 2 Z

V

E

u ⁰ − zw ⁰⁰ + 1 2 w ⁰²

² dV

Elastic energy of discretized system

U = 1 2

E

X

e=1

( Z

V ^e

E

u ê0 − zw ê00 + 1 2 w ê02

2 dV ^e )

U= 1 2

E

X

e=1

( Z l ^e 0

EA ê u ê02 +EI _y ê w ê002 +EA ê u ê0 w ê02 dx ê

)

(15)

U = 2 _V Eε _x dV

U = 1 2 Z

V

E

u ⁰ − zw ⁰⁰ + 1 2 w ⁰²

² dV

Elastic energy of discretized system

U = 1 2

E

X

e=1

( Z

V ^e

E

u ê0 − zw ê00 + 1 2 w ê02

2 dV ^e )

U= 1 2

E

X

e=1

( Z l ^e 0

EA ê u ê02 +EI _y ê w ê002 + EA ê u ê0 normal force N ê (x)

w ^e02 dx ^e

)

(16)

U = 2 _V Eε _x dV

U = 1 2 Z

V

E

u ⁰ − zw ⁰⁰ + 1 2 w ⁰²

² dV

Elastic energy of discretized system

U = 1 2

E

X

e=1

( Z

V ^e

E

u ê0 − zw ê00 + 1 2 w ê02

2 dV ^e )

U= 1 2

E

X

e=1

( Z l ^e 0

EA ê u ê02 +EI _y ê w ê002 + N ê (x)w ê02 dx ê

)

(17)

A, I

x ^e

0 l ^e

q ₂

q ₁ ^e q ^e ₃

q ₅

q ^e ₄ q ^e ₆

x ^e

i j

(18)

A, I

x ^e

0 l ^e

q ₂

q ₁ ^e q ^e ₃

q ₅

q ^e ₄ q ^e ₆

x ^e

i j

Displacement u(x) ê = N _u q ê , N _u = [L ê ₁ 0 0 L ê ₂ 0 0]

0 l ^e

1 x ê L ê ₁ (x ê ) = 1 − ^x _l e ê

0 l ^e

1 x ^e

L ê ₂ (x ê ) = ^x _l e ê

(19)

A, I

x ^e

0 l ^e

2 q ₁ ^e q ^e ₃

5 q ^e ₄ q ^e ₆

x ^e

i j

Deflection w(x) ê = N w q ê , N w = [0 H ₁ ê H ₂ ê 0 H ₃ ê H ₄ ê ]

0 l ^e

1 x ê H ₁ ê (x ê ) = 1 − 3

x ^e l ^e

2 + 2

x ^e l ^e

3 0 l ^e

1 x ê H ₃ ê (x ê ) = 3

x ^e l ^e

2 − 2

x ^e l ^e

3 0 l ^e x ^e

H ₂ ê (x ê ) = x ê h 1 −

x ^e l ^e

i 2

0 l ^e x ^e

H ₄ ê (x ê ) = x ê

x ^e l ^e

2 −

x ^e l ^e

(20)

Elastic energy for discretized structure

U = 1 2

E

X

e=1

( Z l ^e 0

EA ê u ê02 +EI _y ê w ê002 + N ê (x)w ê02 dx ê

)

u(x) ê = N u q ê = N u T I ê Q, w(x) ê = N w q ê = N w T I ê Q

(21)

U = 1 2

E

X

e=1

( Z l ^e 0

EA ê u ê02 +EI _y ê w ê002 + N ê (x)w ê02 dx ê

)

u(x) ê = N u q ê = N u T I ê Q, w(x) ê = N w q ê = N w T I ê Q

U= 1 2

E

X

e=1

( Z l ^e 0

EA ê Q ^T T I êT N _u ⁰ ^T N ⁰ _u T I ê Q+EI _y ê Q ^T T I êT N ⁰⁰ _w ^T N ⁰⁰ _w T I ê Q

+ N ê (x)Q ^T T I êT N _w ⁰ ^T N ⁰ _w T I ê Q

dx ^e o

(22)

U = 1 2

E

X

e=1

( Z l ^e 0

EA ê u ê02 +EI _y ê w ê002 + N ê (x)w ê02 dx ê

)

u(x) ê = N u q ê = N u T I ê Q, w(x) ê = N w q ê = N w T I ê Q

U= 1 2 Q ^T

E

X

e=1

( Z l ^e 0

EA ê T I êT N ⁰ _u ^T N ⁰ _u T I ê +EI _y ê T I êT N ⁰⁰ _w ^T N ⁰⁰ _w T I ê

+ N ê (x) I T êT N ⁰ _w ^T N ⁰ _w T I ê dx ê o

Q

(23)

U = 1 2

E

X

e=1

Z l 0

EA ê u ê02 +EI _y ê w ê002 + N ê (x)w ê02 dx ê

u(x) ê = N _u q ê = N _u T I ê Q, w(x) ê = N _w q ê = N _w T I ê Q

U = 1 2 Q ^T

( _E X

e=1

I T ^eT

Z l ^e 0

EA ê N ⁰ _u ^T N ⁰ _u + EI _y ê N ⁰⁰ _w ^T N ⁰⁰ _w dx ê T I ê

+

E

X

e=1

I T ^eT

Z l ^e 0

N ê (x)N ⁰ _w ^T N ⁰ _w dx ê T I ê )

Q

(24)

U = 1 2

X

e=1 0

EA ê u ê02 +EI _y ê w ê002 + N ê (x)w ê02 dx ê

u(x) ê = N _u q ê = N _u T I ê Q, w(x) ê = N _w q ê = N _w T I ê Q

U = 1 2 Q ^T







E

X

e=1

I T ^eT

Z l ^e 0

EA ê N ⁰ _u ^T N ⁰ _u + EI _y ê N ⁰⁰ _w ^T N ⁰⁰ _w dx ê k ê – linear stiffness matrix

I T ^e

+

E

X

e=1

I T ^eT

Z l ^e 0

N ê (x)N ⁰ _w ^T N ⁰ _w dx ê k ê _σ – initial stress matrix

I T ^e







Q

(25)

U = 1 2

E

X

e=1

( Z l ^e 0

EA ê u ê02 +EI _y ê w ê002 + N ê (x)w ê02 dx ê

)

u(x) ê = N u q ê = N u T I ê Q, w(x) ê = N w q ê = N w T I ê Q

U = 1 2 Q ^T

E

X

e=1

I

T êT k ê T I ê +

E

X

e=1

I T êT k ê _σ T I ê

!

Q

(26)

U = 1 2

E

X

e=1

( Z l ^e 0

EA ê u ê02 +EI _y ê w ê002 + N ê (x)w ê02 dx ê

)

u(x) ê = N u q ê = N u T I ê Q, w(x) ê = N w q ê = N w T I ê Q

U = 1 2 Q ^T





E

X

e=1

I T êT k ê T I ê

K

+

E

X

e=1

I T êT k ê _σ T I ê

K σ



 Q

(27)

U = 1 2

E

X

e=1

( Z l ^e 0

EA ê u ê02 +EI _y ê w ê002 + N ê (x)w ê02 dx ê

)

u(x) ê = N u q ê = N u T I ê Q, w(x) ê = N w q ê = N w T I ê Q

U = 1

2 Q ^T (K + K σ ) Q

(28)

FEM equations for stability

Total energy

Φ = 1

2 Q ^T (K + K _σ ) Q − Q ^T P

Minimization of energy

δΦ = 0 =⇒ (K + K _σ ) Q − P = 0

Equations for two adjacent equilibrium states – before and after buckling – eigenproblem

(K + K _σ )Q ¹ = P (K + K _σ )Q ² = P

− =⇒ (K + K σ )∆Q = 0

(29)

FEM equations for stability

Total energy

Φ = 1

2 Q ^T (K + K _σ ) Q − Q ^T P Minimization of energy

δΦ = 0 =⇒ (K + K _σ ) Q − P = 0

Equations for two adjacent equilibrium states – before and after buckling – eigenproblem

(K + K _σ )Q ¹ = P (K + K _σ )Q ² = P

− =⇒ (K + K σ )∆Q = 0

(30)

FEM equations for stability

Total energy

Φ = 1

2 Q ^T (K + K _σ ) Q − Q ^T P Minimization of energy

δΦ = 0 =⇒ (K + K _σ ) Q − P = 0

Equations for two adjacent equilibrium states – before and after buckling – eigenproblem

(K + K _σ )Q ¹ = P (K + K _σ )Q ² = P

− =⇒ (K + K σ )∆Q = 0

(31)

2 Minimization of energy

δΦ = 0 =⇒ (K + K _σ ) Q − P = 0

Equations for two adjacent equilibrium states – before and after buckling – eigenproblem

(K + K _σ )Q ¹ = P (K + K _σ )Q ² = P

− =⇒ (K + K σ )∆Q = 0 equation is satisfied when

det (K + K σ ) = 0 or ∆Q = 0

(32)

2 Minimization of energy

δΦ = 0 =⇒ (K + K _σ ) Q − P = 0

Equations for two adjacent equilibrium states – before and after buckling – eigenproblem

(K + K _σ )Q ¹ = P (K + K _σ )Q ² = P

− =⇒ (K + K σ )∆Q = 0 equation is satisfied when

det (K + K σ ) = 0

(33)

0 N ê = N ê _u N ê _w

, B ^e = LN ^e , L =





 d dx ^e

− d ² dx ^e2







, D ê = EA ê 0 0 EI ê

k ^e =





 EA

l 0 0 − EA

l 0 0

0 12EI l ³

6EI

l ² 0 − 12EI l ³

6EI l ²

0 6EI

l ² 4EI

l 0 − 6EI

l ² 2EI

l

− EA

l 0 0 EA

l 0 0

0 − 12EI l ³ − 6EI

l ² 0 12EI l ³ − 6EI

l ²

0 6EI

l ² 2EI

l 0 − 6EI

l ² 4EI

l







e

(34)

k ^e _σ = Z l ^e

0 N ^e (x)N ⁰ _w ^T N ⁰ _w dx ^e

k ê _σ = N ê (x) 30l ê







0 0 0 0 0 0

0 36 3l 0 −36 3l 0 3l 4l ² 0 −3l −l ²

0 0 0 0 0 0

0 −36 −3l 0 36 −3l 0 3l −l ² 0 −3l 4l ²







e

P = λP =⇒ k ^e _σ = λk ^e _σ

(35)

FEM algorithm

1 Statics – determination of normal forces KQ = P =⇒ N ^e =⇒ k ^e _σ

2 Buckling – eigenproblem

(K + λK σ )∆Q = 0 =⇒ λ kr =⇒ ∆Q – buckling mode

(36)

λ · 1 E, A, I, l

Q 2

Q 1

Q 3

Q 5

Q 4

Q 6

X Z

i j

N (x) = −1

(37)

Przykład

Cantilever

λ · 1 E, A, I, l

Q 2

Q 1

Q 3

Q 5

Q 4

Q 6

X Z

i j

Buckling

(K + λK σ )∆Q = 0





 EI

l ³







Al ²

I 0 0 - Âl _I ² 0 0 0 12 6l 0 -12 6l 0 6l 4l ² 0 -6l 2l ² - Âl _I ² 0 0 Âl _I ² 0 0

0 -12 -6l 0 12 -6l 0 6l 2l ² 0 -6l 4l ²





 +λ 1

30l







0 0 0 0 0 0 0 36 3l 0 -36 3l 0 3l 4l ² 0 -3l -l ² 0 0 0 0 0 0 0 -36 -3l 0 36 -3l 0 3l -l ² 0 -3l 4l ²



















∆Q 4

∆Q 5

∆Q 6







=





 0 0 0 0 0 0









 EI

l ³





Al ²

I 0 0

0 12 -6l 0 -6l 4l ²



 +λ 1 30l



 0 0 0 0 36 -3l 0 -3l 4l ²













∆Q 4

∆Q 5

∆Q 6



=



 0 0 0





EI l ³





Al ²

I 0 0

0 12 -6l 0 -6l 4l ²



 +λ 1 30l



 0 0 0 0 36 -3l 0 -3l 4l ²





= 0

⇒ λ 1 = 2.486 EI l ² λ 2 = 32.181 EI

l ²

(38)

Przykład

Cantilever

λ · 1 E, A, I, l

Q 2

Q 1

Q 3

Q 5

Q 4

Q 6

X Z

i j

Buckling

(K + λK σ )∆Q = 0





 EI

l ³







Al ²

I 0 0 - Âl _I ² 0 0 0 12 6l 0 -12 6l 0 6l 4l ² 0 -6l 2l ² - Âl _I ² 0 0 Âl _I ² 0 0

0 -12 -6l 0 12 -6l 0 6l 2l ² 0 -6l 4l ²





 +λ 1

30l







0 0 0 0 0 0 0 36 3l 0 -36 3l 0 3l 4l ² 0 -3l -l ² 0 0 0 0 0 0 0 -36 -3l 0 36 -3l 0 3l -l ² 0 -3l 4l ²



















∆Q 1

∆Q 2

∆Q 3

∆Q 4

∆Q 5

∆Q 6







=





 0 0 0 0 0 0









 EI

l ³





Al ²

I 0 0

0 12 -6l 0 -6l 4l ²



 +λ 1 30l



 0 0 0 0 36 -3l 0 -3l 4l ²













∆Q 4

∆Q 5

∆Q 6



=



 0 0 0





EI l ³





Al ²

I 0 0

0 12 -6l 0 -6l 4l ²



 +λ 1 30l



 0 0 0 0 36 -3l 0 -3l 4l ²





= 0

⇒ λ 1 = 2.486 EI l ² λ 2 = 32.181 EI

l ²

(39)

Przykład

Cantilever

λ · 1 E, A, I, l

Q 2

Q 1

Q 3

Q 5

Q 4

Q 6

X Z

i j

Buckling

(K + λK σ )∆Q = 0





 EI

l ³







Al ²

I 0 0 - Âl _I ² 0 0 0 12 6l 0 -12 6l 0 6l 4l ² 0 -6l 2l ² - Âl _I ² 0 0 Âl _I ² 0 0

0 -12 -6l 0 12 -6l 0 6l 2l ² 0 -6l 4l ²





 +λ 1

30l







0 0 0 0 0 0 0 36 3l 0 -36 3l 0 3l 4l ² 0 -3l -l ² 0 0 0 0 0 0 0 -36 -3l 0 36 -3l 0 3l -l ² 0 -3l 4l ²

















 0 0 0

∆Q 4

∆Q 5

∆Q 6







=





 0 0 0 0 0 0









 EI

l ³





Al ²

I 0 0

0 12 -6l 0 -6l 4l ²



 +λ 1 30l



 0 0 0 0 36 -3l 0 -3l 4l ²













∆Q 4

∆Q 5

∆Q 6



=



 0 0 0





EI l ³





Al ²

I 0 0

0 12 -6l 0 -6l 4l ²



 +λ 1 30l



 0 0 0 0 36 -3l 0 -3l 4l ²





= 0

⇒ λ 1 = 2.486 EI l ² λ 2 = 32.181 EI

l ²

(40)

Przykład

Cantilever

λ · 1 E, A, I, l

Q 2

Q 1

Q 3

Q 5

Q 4

Q 6

X Z

i j

Buckling

(K + λK σ )∆Q = 0





 EI

l ³







Al ²

I 0 0 - Âl _I ² 0 0 0 12 6l 0 -12 6l 0 6l 4l ² 0 -6l 2l ² - Âl _I ² 0 0 Âl _I ² 0 0

0 -12 -6l 0 12 -6l 0 6l 2l ² 0 -6l 4l ²





 +λ 1

30l







0 0 0 0 0 0 0 36 3l 0 -36 3l 0 3l 4l ² 0 -3l -l ² 0 0 0 0 0 0 0 -36 -3l 0 36 -3l 0 3l -l ² 0 -3l 4l ²

















 0 0 0

∆Q 4

∆Q 5

∆Q 6







=





 0 0 0 0 0 0









 EI

l ³





Al ²

I 0 0

0 12 -6l 0 -6l 4l ²



 +λ 1 30l



 0 0 0 0 36 -3l 0 -3l 4l ²













∆Q 4

∆Q 5

∆Q 6



=



 0 0 0





EI l ³





Al ²

I 0 0

0 12 -6l 0 -6l 4l ²



 +λ 1 30l



 0 0 0 0 36 -3l 0 -3l 4l ²





= 0

⇒ λ 1 = 2.486 EI l ² λ 2 = 32.181 EI

l ²

(41)

Przykład

Cantilever

λ · 1 E, A, I, l

Q 2

Q 1

Q 3

Q 5

Q 4

Q 6

X Z

i j

Buckling

(K + λK σ )∆Q = 0





 EI

l ³







Al ²

I 0 0 - Âl _I ² 0 0 0 12 6l 0 -12 6l 0 6l 4l ² 0 -6l 2l ² - Âl _I ² 0 0 Âl _I ² 0 0

0 -12 -6l 0 12 -6l 0 6l 2l ² 0 -6l 4l ²





 +λ 1

30l







0 0 0 0 0 0 0 36 3l 0 -36 3l 0 3l 4l ² 0 -3l -l ² 0 0 0 0 0 0 0 -36 -3l 0 36 -3l 0 3l -l ² 0 -3l 4l ²

















 0 0 0

∆Q 4

∆Q 5

∆Q 6







=





 0 0 0 0 0 0









 EI

l ³





Al ²

I 0 0

0 12 -6l 0 -6l 4l ²



 +λ 1 30l



 0 0 0 0 36 -3l 0 -3l 4l ²













∆Q 4

∆Q 5

∆Q 6



=



 0 0 0





EI l ³





Al ²

I 0 0

0 12 -6l 0 -6l 4l ²



 +λ 1 30l



 0 0 0 0 36 -3l 0 -3l 4l ²





= 0

⇒ λ 1 = 2.486 EI l ² λ 2 = 32.181 EI

l ²

(42)

Przykład

Cantilever

λ · 1 E, A, I, l

Q 2

Q 1

Q 3

Q 5

Q 4

Q 6

X Z

i j

Buckling

(K + λK σ )∆Q = 0





 EI

l ³







Al ²

I 0 0 - Âl _I ² 0 0 0 12 6l 0 -12 6l 0 6l 4l ² 0 -6l 2l ² - Âl _I ² 0 0 Âl _I ² 0 0

0 -12 -6l 0 12 -6l 0 6l 2l ² 0 -6l 4l ²





 +λ 1

30l







0 0 0 0 0 0 0 36 3l 0 -36 3l 0 3l 4l ² 0 -3l -l ² 0 0 0 0 0 0 0 -36 -3l 0 36 -3l 0 3l -l ² 0 -3l 4l ²

















 0 0 0

∆Q 4

∆Q 5

∆Q 6







=





 0 0 0 0 0 0









 EI

l ³





Al ²

I 0 0

0 12 -6l 0 -6l 4l ²



 +λ 1 30l



 0 0 0 0 36 -3l 0 -3l 4l ²













∆Q 4

∆Q 5

∆Q 6



=



 0 0 0





EI l ³





Al ²

I 0 0

0 12 -6l 0 -6l 4l ²



 +λ 1 30l



 0 0 0 0 36 -3l 0 -3l 4l ²





= 0 ⇒ λ 1 = 2.486 EI l ² λ 2 = 32.181 EI

l ²

(43)

λ · 1 E, A, I, l

Q 2

Q 1

Q 3

Q 5

Q 4

Q 6

X Z

i j





 EI

l ³







Al ²

I 0 0 - Âl _I ² 0 0 0 12 6l 0 -12 6l 0 6l 4l ² 0 -6l 2l ² - Âl _I ² 0 0 Âl _I ² 0 0

0 -12 -6l 0 12 -6l 0 6l 2l ² 0 -6l 4l ²





 +λ 1

30l







0 0 0 0 0 0 0 36 3l 0 -36 3l 0 3l 4l ² 0 -3l -l ² 0 0 0 0 0 0 0 -36 -3l 0 36 -3l 0 3l -l ² 0 -3l 4l ²

















 0 0 0

∆Q 4

∆Q 5

∆Q 6







=





 0 0 0 0 0 0









 EI

l ³





Al ²

I 0 0

0 12 -6l 0 -6l 4l ²



 +λ 1 30l



 0 0 0 0 36 -3l 0 -3l 4l ²













∆Q 4

∆Q 5

∆Q 6



=



 0 0 0





EI l ³





Al ²

I 0 0

0 12 -6l 0 -6l 4l ²



 +λ 1 30l



 0 0 0 0 36 -3l 0 -3l 4l ²





= 0 ⇒ λ 1 = 2.486 EI

l ² ⇒ P _cr ^anal = 2.467 EI l ² λ 2 = 32.181 EI

l ²

(44)

Buckling modes

Buckling modes are determined from one of two linearly independent equations upon substitution of respective eigenvalue

FEM for buckling analysis

Jerzy Pamin

e-mail: jpamin@L5.pk.edu.pl

1 Introduction

2 Derivation of FEM equations Frame finite element

3 Example

forces and decreased by compressive forces.

A sufficiently large compressive force can reduce the bending stiffness to zero and structural buckling (instability mode) occurs.

Assumptions

linear elasticity: σ x = Eε x

static one-parameter load ideal system (no imperfections) equilibrium of buckled configuration

ε x (x) = u 0 (x) − zw 00 (x) + 1

2 (w 0 (x)) 2

w w

kr

λP 1 λP 2

λP 3

x, u z, w

y

Derivation of FEM equations

Energetic criterion of equilibrium

Φ = U − W Φ – total energy

U – elastic energy: U = 1 2 R

V

ε x σ x dV W – work of external forces: W = Q T P

Q – dof vector (nodal displacement vector)

P – external force vector

Φ – total energy

U – elastic energy: U = 1 2 R

V

ε x σ x dV W – work of external forces: W = Q T P

Q – dof vector (nodal displacement vector) P – external force vector

Value of (compressive) normal force before buckling

∆l

l = ε x , σ x = N

A , σ x = Eε x , ε x = u 0 =⇒ N (x) = EAu 0 (x)

Energetic criterion

Elastic energy

U = 1 2 Z

V

ε x σ x dV

U = 1 2

E

X

e=1

( Z

V e

E



u e0 − zw e00 + 1 2 w e02

 2

dV e

)

Energetic criterion

Elastic energy

U = 1 2 Z

V

ε x σ x

Eε x

dV

U = 1 2

E

X

e=1

( Z

V e

E



u e0 − zw e00 + 1 2 w e02

 2

dV e

)

Energetic criterion

Elastic energy

U = 1 2 Z

V

Eε 2 x dV

U = 1 2

E

X

ε _x (x) = u ⁰ (x) − zw ⁰⁰ (x) + 1

2 (w ⁰ (x)) ²

λP ₃

U – elastic energy: U = ¹ ₂ R

ε x σ x dV W – work of external forces: W = Q ^T P

U – elastic energy: U = ¹ ₂ R

ε x σ x dV W – work of external forces: W = Q ^T P

l = ε _x , σ _x = N

A , σ _x = Eε _x , ε _x = u ⁰ =⇒ N (x) = EAu ⁰ (x)

V ^e

u ê0 − zw ê00 + 1 2 w ê02

2

dV ^e

V ^e

u ê0 − zw ê00 + 1 2 w ê02

2

dV ^e

Eε ² _x dV

V ^e

u ê0 − zw ê00 + 1 2 w ê02

2

dV ^e

Eε ² _x dV

u ⁰ − zw ⁰⁰ + 1 2 w ⁰²

2

Eε ² _x dV

u ⁰ − zw ⁰⁰ + 1 2 w ⁰²

2

V ^e

u ê0 − zw ê00 + 1 2 w ê02

2

dV ^e

U = 2 _V Eε _x dV

u ⁰ − zw ⁰⁰ + 1 2 w ⁰²

² dV

V ^e

u ê0 − zw ê00 + 1 2 w ê02

² dV ^e

( Z l ^e 0

Z

A ^e