Energetic criterion

Piotr Pluciński

e-mail: Piotr.Plucinski@pk.edu.pl

Jerzy Pamin

e-mail: Jerzy.Pamin@pk.edu.pl

Chair for Computational Engineering

Faculty of Civil Engineering, Cracow University of Technology URL: www.CCE.pk.edu.pl

1 Introduction

2 Derivation of FEM equations Frame finite element

3 Example

Observation

Bending stiffness is increased by tensile 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= Eεx

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

ε_x(x) = u⁰(x) − zw⁰⁰(x) + 1

2(w⁰(x))²

w w

P

P_cr

λ ¯P1 λ ¯P₂ λ ¯P3

x, u z, w

y

Derivation of FEM equations

Energetic criterion of equilibrium

Φ = U − W Φ – total energy

U – elastic energy: U = ¹₂R

V

εxσxdV W – work of external forces: W = d^Tf

d – dof vector (nodal displacement vector) f – external force vector

l = εx, σx=

A, σx= Eεx, εx= u⁰ =⇒ N (x) = EAu⁰(x)

Energetic criterion of equilibrium

Φ = U − W Φ – total energy

U – elastic energy: U = ¹₂R

V

εxσxdV W – work of external forces: W = d^Tf

d – dof vector (nodal displacement vector) f – external force vector

Value of (compressive) normal force before buckling

∆l

l = εx, σx=N

A, σx= Eεx, εx= u⁰ =⇒ N (x) = EAu⁰(x)

Energetic criterion

Elastic energy

U = 1 2 Z

V

εxσxdV

Elastic energy of discretized system

U = 1 2

E

X

e=1

(Z

V^e

E



u^e0− zw^e00+1 2w^e02

2

dV^e )

Energetic criterion

Elastic energy

U = 1 2 Z

V

εx σx

Eεx

dV

Elastic energy of discretized system

U = 1 2

E

X

e=1

(Z

V^e

E



u^e0− zw^e00+1 2w^e02

2

dV^e )

Energetic criterion

Elastic energy

U = 1 2 Z

V

Eε²_xdV

Elastic energy of discretized system

U = 1 2

E

X

e=1

(Z

V^e

E



u^e0− zw^e00+1 2w^e02

2

dV^e )

Energetic criterion

Elastic energy

U = 1 2 Z

V

Eε²_xdV

U =1 2 Z

V

E



u⁰− zw⁰⁰+1 2w⁰²

2

dV

U = 2

e=1 V^e

E u^e0− zw^e00+

2w^e02 dV^e

Elastic energy

U = 1 2 Z

V

Eε²_xdV

U =1 2 Z

V

E



u⁰− zw⁰⁰+1 2w⁰²

2

dV

Elastic energy of discretized system

U = 1 2

E

X

e=1

(Z

V^e

E



u^e0− zw^e00+1 2w^e02

² dV^e

)

Elastic energy

U = 1 2 Z

V

Eε²_xdV

U =1 2 Z

V

E



u⁰− zw⁰⁰+1 2w⁰²

2

dV

Elastic energy of discretized system

U = 1 2

E

X

e=1

(Z

V^e

E



u^e0− zw^e00+1 2w^e02

² dV^e

)

U=1 2

E

X

e=1

Z l^e 0

Z

A^e

E



u^e02+z²w^e002+1

4w^e04−2zu^e0w^e00+u^e0w^e02−zw^e00w^e02



dA^e



dx^e



Elastic energy

U = 1 2 Z

V

Eε²_xdV

U =1 2 Z

V

E



u⁰− zw⁰⁰+1 2w⁰²

2

dV

Elastic energy of discretized system

U = 1 2

E

X

e=1

(Z

V^e

E



u^e0− zw^e00+1 2w^e02

² dV^e

)

U=1 2

E

X

e=1

Z l^e 0

Z

A^e

E



u^e02+z²w^e002+¹₄w^e04

nonlinear term, upon linearization ∼= 0

−2zu^e0w^e00+u^e0w^e02−zw^e00w^e02



dA^e



dx^e



Elastic energy

U = 1 2 Z

V

Eε²_xdV

U =1 2 Z

V

E



u⁰− zw⁰⁰+1 2w⁰²

2

dV

Elastic energy of discretized system

U = 1 2

E

X

e=1

(Z

V^e

E



u^e0− zw^e00+1 2w^e02

² dV^e

)

U=1 2

E

X

e=1

Z l^e 0

Z

A^e

E u^e02+z²w^e002−2zu^e0w^e00+u^e0w^e02−zw^e00w^e02

dA^e



dx^e



Elastic energy

U = 1 2 Z

V

Eε²_xdV

U =1 2 Z

V

E



u⁰− zw⁰⁰+1 2w⁰²

2

dV

Elastic energy of discretized system

U = 1 2

E

X

e=1

(Z

V^e

E



u^e0− zw^e00+1 2w^e02

2

dV^e )

U=1 2

E

X

e=1

(Z l^e 0



EA^eu^e02+EI_y^ew^e002+EA^eu^e0w^e02 dx^e

)

Elastic energy

U = 1 2 Z

V

Eε²_xdV

U =1 2 Z

V

E



u⁰− zw⁰⁰+1 2w⁰²

2

dV

Elastic energy of discretized system

U = 1 2

E

X

e=1

(Z

V^e

E



u^e0− zw^e00+1 2w^e02

2

dV^e )

U=1 2

E

X

e=1

(Z l^e 0



EA^eu^e02+EI_y^ew^e002+ EA^eu^e0 normal force N^e(x)

w^e02 dx^e

)

Elastic energy

U = 1 2 Z

V

Eε²_xdV

U =1 2 Z

V

E



u⁰− zw⁰⁰+1 2w⁰²

2

dV

Elastic energy of discretized system

U = 1 2

E

X

e=1

(Z

V^e

E



u^e0− zw^e00+1 2w^e02

2

dV^e )

U=1 2

E

X

e=1

(Z l^e 0



EA^eu^e02+EI_y^ew^e002+ N^e(x)w^e02 dx^e

)

Frame finite element

A, I

x^e z^e

0 l^e

d^e₂

d^e₁ d^e₃

d^e₅

d^e₄ d^e₆

x^e z^e

i j

Frame finite element

A, I

x^e z^e

0 l^e

d^e₂

d^e₁ d^e₃

d^e₅

d^e₄ d^e₆

x^e z^e

i j

Displacement u(x)^e= Nud^e, Nu= [L^e₁ 0 0 L^e₂ 0 0]

0 l^e

1

x^e L^e₁(x^e) = 1 −^x_le^e

0 l^e

1

x^e L^e₂(x^e) = ^x_le^e

Frame finite element

A, I

x^e z^e

0 l^e

d^e₂

d^e₁ d^e₃

d^e₅

d^e₄ d^e₆

x^e z^e

i j

Deflection w(x)^e= N_wd^e, N_w= [0 H₁^e H₂^e 0 H₃^e H₄^e]

0 l^e

1

x^e H₁^e(x^e) = 1 − 3 ^x_le^e

²

+ 2 ^x_le^e

³

0 l^e

1

x^e H₃^e(x^e) = 3 ^x_le^e

²

− 2 ^x_le^e
3

0 l^e x^e

H₂^e(x^e) = x^e

1 − ^x_le^e

²

0 l^e x^e

H₄^e(x^e) = x^e

h x^e l^e

²

− ^x_le^e
i

Elastic energy for discretized structure

U =1 2

E

X

e=1

(Z l^e 0

EA^eu^e02+EI_y^ew^e002+ N^e(x)w^e02 dx^e

)

u(x)^e= N_ud^e= N_uTI^ed, w(x)^e= N_wd^e= N_wTI^ed

Elastic energy for discretized structure

U =1 2

E

X

e=1

(Z l^e 0



EA^eu^e02+EI_y^ew^e002+ N^e(x)w^e02 dx^e

)

u(x)^e= Nud^e= NuTI^ed, w(x)^e= Nwd^e= NwTI^ed

U=1 2

E

X

e=1

(Z l^e 0



EA^ed^TTI^eTN_u^0TN⁰_uTI^ed+EI_y^ed^TTI^eTN⁰⁰_w^TN⁰⁰_wTI^ed

+ N^e(x)d^TTI^eTN_w^{0 T}N⁰_wTI^ed dx^eo

Elastic energy for discretized structure

U =1 2

E

X

e=1

(Z l^e 0



EA^eu^e02+EI_y^ew^e002+ N^e(x)w^e02 dx^e

)

u(x)^e= Nud^e= NuTI^ed, w(x)^e= Nwd^e= NwTI^ed

U=1 2d^T

E

X

e=1

(Z l^e 0



EA^eTI^eTN⁰_u^TN⁰_uTI^e+EI_y^eTI^eTN⁰⁰_w^TN⁰⁰_wTI^e

+ N^e(x) IT^eTN⁰_w^TN⁰_wTI^e dx^eo

d

Elastic energy for discretized structure

U =1 2

E

X

e=1

(Z l^e 0



EA^eu^e02+EI_y^ew^e002+ N^e(x)w^e02 dx^e

)

u(x)^e= Nud^e= NuTI^ed, w(x)^e= Nwd^e= NwTI^ed

U = 1 2d^T

( _E X

e=1

I T^eT

Z l^e 0



EA^eN⁰_u^TN⁰_u+ EI_y^eN⁰⁰_w^TN⁰⁰_w dx^eTI^e

+

E

X

e=1

I T^eT

Z l^e 0

N^e(x)N⁰_w^TN⁰_wdx^eTI^e )

d

Elastic energy for discretized structure

U =1 2

E

X

e=1

(Z l^e 0



EA^eu^e02+EI_y^ew^e002+ N^e(x)w^e02 dx^e

)

u(x)^e= Nud^e= NuTI^ed, w(x)^e= Nwd^e= NwTI^ed

U = 1 2d^T







E

X

e=1

I T^eT

Z l^e 0



EA^eN⁰_u^TN⁰_u+ EI_y^eN⁰⁰_w^TN⁰⁰_w dx^e K^e– linear stiffness matrix

I T^e

+

E

X

e=1

I T^eT

Z l^e 0

N^e(x)N⁰_w^TN⁰_wdx^e K^e_σ – initial stress matrix

I T^e





 d

Elastic energy for discretized structure

U =1 2

E

X

e=1

(Z l^e 0



EA^eu^e02+EI_y^ew^e002+ N^e(x)w^e02 dx^e

)

u(x)^e= Nud^e= NuTI^ed, w(x)^e= Nwd^e= NwTI^ed

U = 1 2d^T

E

X

e=1

I

T^eTK^eTI^e+

E

X

e=1

I

T^eTK^e_σTI^e

! d

Elastic energy for discretized structure

U =1 2

E

X

e=1

(Z l^e 0



EA^eu^e02+EI_y^ew^e002+ N^e(x)w^e02 dx^e

)

u(x)^e= Nud^e= NuTI^ed, w(x)^e= Nwd^e= NwTI^ed

U =1 2d^T





E

X

e=1

I

T^eTK^eTI^e K

+

E

X

e=1

I

T^eTK^e_σTI^e Kσ



d

Elastic energy for discretized structure

U =1 2

E

X

e=1

(Z l^e 0

EA^eu^e02+EI_y^ew^e002+ N^e(x)w^e02 dx^e

)

u(x)^e= N_ud^e= N_uTI^ed, w(x)^e= N_wd^e= N_wTI^ed

U = 1

2d^T(K + Kσ) d

FEM equations for stability

Total energy

Φ = 1

2d^T(K + Kσ) d − d^Tf

Minimization of energy

δΦ = 0 =⇒ (K + Kσ) d − f = 0

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

(K + K_σ)d¹= f (K + K_σ)d²= f

− =⇒(K + K_σ)∆d = 0

FEM equations for stability

Total energy

Φ = 1

2d^T(K + Kσ) d − d^Tf

Minimization of energy

δΦ = 0 =⇒ (K + Kσ) d − f = 0

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

(K + K_σ)d¹= f (K + K_σ)d²= f

− =⇒(K + K_σ)∆d = 0

FEM equations for stability

Total energy

Φ = 1

2d^T(K + Kσ) d − d^Tf

Minimization of energy

δΦ = 0 =⇒ (K + Kσ) d − f = 0

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

(K + K_σ)d¹= f (K + K_σ)d²= f

− =⇒(K + K_σ)∆d = 0

Total energy

Φ = 1

2d^T(K + Kσ) d − d^Tf

Minimization of energy

δΦ = 0 =⇒ (K + Kσ) d − f = 0

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

(K + K_σ)d¹= f (K + K_σ)d²= f

− =⇒(K + K_σ)∆d = 0 equation is satisfied when

det (K + Kσ) = 0 or ∆d = 0

Total energy

Φ = 1

2d^T(K + Kσ) d − d^Tf

Minimization of energy

δΦ = 0 =⇒ (K + Kσ) d − f = 0

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

(K + K_σ)d¹= f (K + K_σ)d²= f

− =⇒(K + K_σ)∆d = 0 equation is satisfied when

det (K + Kσ) = 0

Linear stiffness matrix – frame element K^e=

Z l^e 0

B^eTD^eB^edx^e

N^e=

h_Ne u

N^e_w

i

, B^e= LN^e, L =





d dx^e

− d² dx^e2



, D^e=

h_EAe 0 0 EI^e

i

K^e=































EA

l 0 0 −EA

l 0 0

0 12EI

l³ 6EI

l² 0 −12EI l³

6EI l²

0 6EI

l² 4EI

l 0 −6EI

l² 2EI

l

−EA

l 0 0 EA

l 0 0

0 −12EI l³ −6EI

l² 0 12EI l³ −6EI

l²

0 6EI

l² 2EI

l 0 −6EI

l² 4EI

l































e

Initial stress matrix – frame element

K^e_σ= Z l^e

0

N^e(x)N⁰_w^TN⁰_wdx^e

k^e_σ= N^e(x) 30l^e

















0 0 0 0 0 0

0 36 3l 0 −36 3l 0 3l 4l² 0 −3l −l²

0 0 0 0 0 0

0 −36 −3l 0 36 −3l 0 3l −l² 0 −3l 4l²

















e

f = λ¯f =⇒ k^e_σ= λ¯k^e_σ

One-parameter loading f

λ ¯P1 λ ¯P2 λ ¯P3

Configurational loading ¯f

P¯₁ P¯2

P¯₃

FEM algorithm

1 Statics – determination of normal forces Kd = ¯f =⇒ N^e=⇒ ¯k^e_σ

2 Buckling – eigenproblem

(K + λ ¯K_σ)∆d = 0 =⇒ λ_kr=⇒ ∆d – buckling mode

λ · 1 E, A, I, l

d2

d1

d3

d5

d4

d6

X Z

i j

Statics

After computations of pre-buckling state:

N (x) = −1

Example

Cantilever

λ · 1 E, A, I, l

d2

d1

d3

d5

d4

d6

X Z

i j

Buckling

(K + λ ¯Kσ)∆d = 0













EI l³













Al²

I 0 0 -^Al_I² 0 0 0 12 6l 0 -12 6l 0 6l 4l² 0 -6l 2l² -^Al_I² 0 0 ^Al_I² 0 0

0 -12 -6l 0 12 -6l 0 6l 2l² 0 -6l 4l²













+λ 1 30l











0 0 0 0 0 0 0 36 3l 0 -36 3l 0 3l 4l²0 -3l -l² 0 0 0 0 0 0 0 -36 -3l 0 36 -3l 0 3l -l² 0 -3l 4l²

































∆d4

∆d5

∆d6











=











0 0 0 0 0 0











EI l³

"_Al2

I 0 0

0 12 -6l 0 -6l 4l²

#

+λ 1 30l

"

0 0 0 0 36 -3l 0 -3l 4l²

#!"

∆d4

∆d5

∆d6

#

=

"

0 0 0

#

EI l³

"_Al2

I 0 0

0 12 -6l 0 -6l 4l²

#

+λ 1 30l

"

0 0 0 0 36 -3l 0 -3l 4l²

#    

= 0

⇒ λ1= 2.486EI l² λ2= 32.181EI

l²

Example

Cantilever

λ · 1 E, A, I, l

d2

d1

d3

d5

d4

d6

X Z

i j

Buckling

(K + λ ¯Kσ)∆d = 0













EI l³













Al²

I 0 0 -^Al_I² 0 0 0 12 6l 0 -12 6l 0 6l 4l² 0 -6l 2l² -^Al_I² 0 0 ^Al_I² 0 0

0 -12 -6l 0 12 -6l 0 6l 2l² 0 -6l 4l²













+λ 1 30l











0 0 0 0 0 0 0 36 3l 0 -36 3l 0 3l 4l²0 -3l -l² 0 0 0 0 0 0 0 -36 -3l 0 36 -3l 0 3l -l² 0 -3l 4l²

































∆d1

∆d2

∆d3

∆d4

∆d5

∆d6











=











0 0 0 0 0 0











EI l³

"_Al2

I 0 0

0 12 -6l 0 -6l 4l²

#

+λ 1 30l

"

0 0 0 0 36 -3l 0 -3l 4l²

#!"

∆d4

∆d5

∆d6

#

=

"

0 0 0

#

EI l³

"_Al2

I 0 0

0 12 -6l 0 -6l 4l²

#

+λ 1 30l

"

0 0 0 0 36 -3l 0 -3l 4l²

#    

= 0

⇒ λ1= 2.486EI l² λ2= 32.181EI

l²

Example

Cantilever

λ · 1 E, A, I, l

d2

d1

d3

d5

d4

d6

X Z

i j

Buckling

(K + λ ¯Kσ)∆d = 0













EI l³













Al²

I 0 0 -^Al_I² 0 0 0 12 6l 0 -12 6l 0 6l 4l² 0 -6l 2l² -^Al_I² 0 0 ^Al_I² 0 0

0 -12 -6l 0 12 -6l 0 6l 2l² 0 -6l 4l²













+λ 1 30l











0 0 0 0 0 0 0 36 3l 0 -36 3l 0 3l 4l²0 -3l -l² 0 0 0 0 0 0 0 -36 -3l 0 36 -3l 0 3l -l² 0 -3l 4l²

































0 0 0

∆d4

∆d5

∆d6











=











0 0 0 0 0 0











EI l³

"_Al2

I 0 0

0 12 -6l 0 -6l 4l²

#

+λ 1 30l

"

0 0 0 0 36 -3l 0 -3l 4l²

#!"

∆d4

∆d5

∆d6

#

=

"

0 0 0

#

EI l³

"_Al2

I 0 0

0 12 -6l 0 -6l 4l²

#

+λ 1 30l

"

0 0 0 0 36 -3l 0 -3l 4l²

#    

= 0

⇒ λ1= 2.486EI l² λ2= 32.181EI

l²

Example

Cantilever

λ · 1 E, A, I, l

d2

d1

d3

d5

d4

d6

X Z

i j

Buckling

(K + λ ¯Kσ)∆d = 0













EI l³













Al²

I 0 0 -^Al_I² 0 0 0 12 6l 0 -12 6l 0 6l 4l² 0 -6l 2l² -^Al_I² 0 0 ^Al_I² 0 0

0 -12 -6l 0 12 -6l 0 6l 2l² 0 -6l 4l²













+λ 1 30l











0 0 0 0 0 0 0 36 3l 0 -36 3l 0 3l 4l²0 -3l -l² 0 0 0 0 0 0 0 -36 -3l 0 36 -3l 0 3l -l² 0 -3l 4l²

































0 0 0

∆d4

∆d5

∆d6











=











0 0 0 0 0 0











EI l³

"_Al2

I 0 0

0 12 -6l 0 -6l 4l²

#

+λ 1 30l

"

0 0 0 0 36 -3l 0 -3l 4l²

#!"

∆d4

∆d5

∆d6

#

=

"

0 0 0

#

EI l³

"_Al2

I 0 0

0 12 -6l 0 -6l 4l²

#

+λ 1 30l

"

0 0 0 0 36 -3l 0 -3l 4l²

#    

= 0

⇒ λ1= 2.486EI l² λ2= 32.181EI

l²