Non-Linear GeometryNon-Linear geometry, example - kinematics The lengths of the bar in undeformed...
Transcript of Non-Linear GeometryNon-Linear geometry, example - kinematics The lengths of the bar in undeformed...
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Non-Linear Geometry
A brief introduction
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Non-linear geometry, exampleP
P=0
A
PB
C
-PD
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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:
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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
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Non-Linear geometry, example
Tangential stiffness:
Kt
u
P
Derivation of the equilibrium equation:
Final form of tangential stiffness:
Ku= Ku(u)
K= K()
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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()
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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
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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
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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
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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
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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
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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
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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
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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.
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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.
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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.