Non-Linear GeometryNon-Linear geometry, example - kinematics The lengths of the bar in undeformed and deformed configurations: (Truncated Taylor expansion) By insertion of the lengths,

Non-Linear Geometry

A brief introduction

Non-linear geometry, exampleP

P=0

A

PB

C

-PD

Non-Linear geometry, example- kinematics

The lengths of the bar in undeformed and deformed configurations: (Truncated Taylor expansion)

By insertion of the lengths, the strains may be written as:

The strains may be written as:

Non-Linear geometry, example- equilibrium

Choosing a linear elastic material: EAAN

Equilibrium of the central node:

since sinq(a+u)/L1

-P

N N

q

and

Non-Linear geometry, example

Tangential stiffness:

Kt

u

P

Derivation of the equilibrium equation:

Final form of tangential stiffness:

Ku= Ku(u)

K= K()

Non-Linear geometry , example

• First order theory: Kt=K0• Second order theory: Kt=K0+K• Third order theory: Kt=K0+K+Ku

Ku= Ku(u)

K= K()

Non-linear geometry

- General bar element

where

0000

0101

0000

0101

L

EA0K

1010

0000

1010

0000

L

NσK

uu

uu

σbb

bbK

3L

EA

First order:

Kt=K0 bar2e.m in Calfem

Second order:

Kt=K0 +K bar2g.m in Calfem

Third order:

Kt=K0 +K+Ku Not in Calfem

)u2(u)uuuu

)uuuu)u2(u

yyyxxy

yxxyxx

aba

baaub

)(u

)(u

)(

)(

24y

13x

12

12

uu

uu

yyb

xxa

and

Non-Linear geometry

- General solid element

The tangential element stiffness for solid elements may in most cases be written on the form:

• First order theory: Kt=K0• Second order theory: Kt=K0+K• Third order theory: Kt=K0+K+Ku

Solution of Non-linear Equations

Direct explicit method:

un = Kt-1 Pn

un+1= un+ un+1

R: residual, additive error

Kt1

u

P

P1

P2

P3

Kt2

Kt3

Kt4

u1 u2 u3

R

Divide into a number of load-steps

Out-of Balance Forces

• External forces: P

P = fb + fl

• Internal forces: element forces = I

• Equilibrium: P-I=0

• In the direct explicit method: P-I=R

• R: Force Residual (Out-of-balance forces)

AA

dAtdAt σBaDBBI TT

Newton-Raphson Method

u

P

P1=rn1

P2

un un+1

rn2

rn3 rn

4

dun1 dun

2 dun3

n=1Load steps n=1,2,…

Pn=Pn-1+Pnun

0=un-1

Iterations i=1,2,…calculate n from un

i

calculate residual Rni=Pn-1-In

i

calculate Kt ni-1

duni=(Kt n

i-1)-1Rni

uni=un

i-1+duni

stop iteration when reisdual is ok

end of load

Stability- Linear Buckling - example

P= −N

Bar with equilibrium in deformed configuration only:

Second order theory: Kt = K0+K

Note! N= −P and the second term becomes negative:

Tangent stiffness Kt= 0 when det(Kt) = 0

det(Kt) = 0 => P=kf L

0000

0101

0000

0101

L

EATK

1010

0000

1010

0000

L

N

f

T

kL

P

L

PL

EA

L

EAL

P

L

PL

EA

L

EA

00

00

00

00

K u1=u2=0

f

T

kL

PL

EA

0

0K

1

4

2

3

Stability- Linear Buckling- General problem

For a given the FE-equation becomes:

[K0 +K ()] a = F

For a certain critical load lP=Pcr the stiffness is zero and the stability limit is reached.

(l is a load multiplier)

[K0 +K ] a = 0

Homogeneous equation system, non-trivial solutions exist: (eigenvalue problem)

(K0 +liK )xi = 0

li = the eigenvalues (force multipliers)xi = the buckling mode shapes

P= −N

1

4

2

3

Linear Buckling in ABAQUS

• Apply loads, (for example 1 N)

li will then give the buckling loads.

• Choose ”Linear Perturbation” and then ”Buckle” as the step.

(First eigenvalue gives first buckling mode)

• Apply boundary conditions.

• Solve the eigenvalue problem.

• The solution gives the buckling modes and the force multipliers for the buckling loads.

Stability - Non-linear Buckling

• Element stiffness calculated with equilibrium in deformed configuration and updated displacement stiffness:

Third order theory: Kt = K0+K+Ku

• Includes all static effects in a physical problem. • Loading may be made until collapse is reached and post-

buckling analysis may be performed.

Non-linear Buckling in ABAQUS• Apply a load larger than the

anticipated buckling load

• Choose ”Static, General” problem as the step.

• Choose Nlgeom: on

• The time is fictive, dividing the load into load increments.

• Apply boundary conditions.

• A solution may not be found when a buckling load is reached.

• Preferably use displacement control.