Stability of structures FE-based stability analysis
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Stability of structuresFE-based stability analysis
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()
General bar elementsee: S. Krenk, Non-Linear Modeling and Analysis of Solids and Structures
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
General solid elementsee: S. Krenk, Non-Linear Modeling and Analysis of Solids and Structures
The tangential element stiffness for solid elements may in many cases also be written on the form:
• First order theory: Kt=K0
• Second order theory: Kt=K0+K
• Third order theory: Kt=K0+K+Ku
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- Classical problem
• Look for displacements a when the tangent stiffness becomes zero:KC a = 0
where KC = K0+K is the tangent stiffness in the current state. This is a homogeneous equation system with non-trivial solutions a.
• In classical buckling analysis the current state is the unloaded base state.
• A homogeneous equation system may be formulated as an eigenvalue problem:
(K0 +liK )xi = 0
li = the eigenvalues (force multipliers)xi = the buckling mode shapes
• If the current state is the unloaded state, solve the second-order system for
loads f to get the stress distribution in the structure.
• The critical load fcr=lif (li becomes the load multiplier)
Example: classical buckling
• Simple frame
• Unloaded base state
• Differential load = -1N in y-dir at top of both pillars
• Fixed supports at base
Frame Base state
Example: classical buckling
1st eigenvalue=2.00 106
Critical load fcr=2.00 106 *(-1) Nfcr=-2.00 106 N
2nd eigenvalue=3.73 106
Critical load fcr=3.73 106 *(-1) Nfcr=-3.73 106 N
Classical Linear Buckling in ABAQUS
• Apply loads, (for example 1 N) and boundary conditions
• Choose ”Linear Perturbation” and then”Buckle” as the step.
(Give number of eigenvalues that you want, the first (lowest) eigenvalue gives the firstbuckling mode)
• Apply boundary conditions.
• Solve the eigenvalue problem.
• The solution gives the buckling modes and the force multipliers li for the buckling loads.
• fcr=lif will then give the buckling loads.
Stability - Linear Buckling- General problem
• If the current state is caused by pre-loads, fpre
KC = K0+K+Ku
is the tangent stiffness caused by the pre-loads.
• A homogeneous equation system may still be found as an eigenvalue problem:
(Kc +liK )xi = 0
li = the eigenvalues (force multipliers)xi = the buckling mode shapes
• K is now the differential stiffness at this state caused by the loads f.
• The critical load is now fcr= fpre + li f (where li is the load multiplier solved by the eigenvalue problem)
• If geometric nonlinearity is included, the base state geometry is the deformed geometry at the end of the last step.
Kc
u
P
Example: general linear buckling
• Same frame
• Preload with -1.9 106 N in y-dir at top of both pillars
• Differential load = -1N in y-dir at top of both pillars
• Fixed supports at base
Frame Base state
Example: general linear buckling
1st eigenvalue=0.11 106
Critical load fcr=(-1.9+0.11*(-1)) 106 Nfcr=-2.01 106 N
2nd eigenvalue=1.84 106
Critical load fcr=(-1.9+1.84*(-1)) 106 Nfcr=-3.74 106 N
Example2: general linear buckling
• Same frame
• Preload with -1.9 106 N in y-dir at top of both pillars and a load at the lefttop corner of 30 kN in negative x-dir
• Differential load = -1N in y-dir at top of both pillars
• Fixed support at base
Frame Base state
Example2: general linear buckling
1st eigenvalue=0.15 106
fcr=-2.05 106 N (+ stresses from pre-load)
Pre-load -1.9 106 N in y-dirand -20 kN in x-dir
1st eigenvalue=0.26 106
fcr=-2.16 106 N (+ stresses from pre-load)
Pre-load -1.9 106 N in y-dirand -40 kN in x-dir
General Linear Buckling in ABAQUS
• First, create a general static step (and non-linear geometry if desired)
• Create ”Linear Perturbation” and then”Buckle” as the second step.
• Apply pre-loads and boundary conditionsin the first step (general static step)
• Apply loads and boundary conditions in the second step (buckle), (for example 1 N)
• Solves first the pre-load step and then the eigenvalue problem with the base statefrom the pre-load.
• fcr= fpre + li f give 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.
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 σBaDBBITT
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+Pn
un0=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
Example 1: Non-linear buckling
• Same frame
• Load with -2.5 106 N in y-dir at top of both pillars
• Fixed support at base
• Solve with non-linear geometry
• No buckling!!!
• Why ?
Example2: Non-linear buckling , imperfections
No buckling!!
Load -2.5 106 N in y-dirand -1 kN in x-dir (top left corner)
Buckling at t=0.926fcr=-2.35 106 N
Load -2.5 106 N in y-dirand -10 kN in x-dir (top left corner)
Stability with imperfections
• General types of imperfections may be added to non-linear buckling analysis (2nd or 3rd order analysis)
1. Through adding eigenmode shapes on structure2. Through adding deformation from a previous static
analysis
Displacement -0.2m in y-dirand -1 kN in x-dir (top left corner)
Force at 0.2m displ = 2.09 106 N
force = 2.05 106 N
Example 3: Non-linear buckling- displacement control, imperfections
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.
ExamplesSnowloads on slender constructions
- and the finite element method
• Total last under en snörik vinter ca: 1.5 kN/m2• Ska klara ca 3.4 kN/m2• Spännvidd 48m
Ishall- underspänd tre-ledstakstol
UnstableCritical load ~ 1.2 kN/m2
Moment free joints
UnstableCritical load ~1.7 kN/m2
Spikplåtar
StableCritical load ~ 2.5 kN/m2
But too high stresses
Another similar ice-arena (span 52m)
Measured imperfections
Buckling shape analysed with imperfections
Buckling due to heat from fire
Brandförlopp temp.~ tid
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