Stability of structures FE-based stability analysis
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1 Stability of strctres FE-based stability analysis
2 Non-linear geometry, example P A B P C P= D -P
3 Non-Linear geometry, example - kinematics The strains may be written as: The lengths of the bar in ndeformed and deformed configrations: (Trncated Taylor expansion) By insertion of the lengths, the strains may be written as:
4 Non-Linear geometry, example - eqilibrim Choosing a linear elastic material: N A EA -P Eqilibrim of the central node: N q N since sinq(a+)/l and
5 Non-Linear geometry, example P K t Tangential stiffness: Derivation of the eqilibrim eqation: Final form of tangential stiffness: K = K () K = K ()
6 Non-Linear geometry, example First order theory: K t =K Second order theory: K t =K +K Third order theory: K t =K +K +K K = K () K = K ()
7 General bar element see: S. Krenk, Non-Linear Modeling and Analysis of Solids and Strctres where L EA K L N K σ σ b b b b K 3 L EA First order: K t =K bar2e.m in Calfem Second order: K t =K +K bar2g.m in Calfem Third order: K t =K +K +K Not in Calfem ) (2 ) ) ) (2 y y y x x y y x x y x x a b a b a a b ) ( ) ( ) ( ) ( 2 4 y 3 x 2 2 y y b x x a and
8 General solid element see: S. Krenk, Non-Linear Modeling and Analysis of Solids and Strctres The tangential element stiffness for solid elements may in many cases also be written on the form: First order theory: K t =K Second order theory: K t =K +K Third order theory: K t =K +K +K
9 Stability- Linear Bckling - example P= N Bar with eqilibrim in deformed configration only: Second order theory: K t = K +K Note! N= P and the second term becomes negative: Tangent stiffness K t = when det(k t ) = det(k t ) = => P=k f L L EA K T L N f T k L P L P L EA L EA L P L P L EA L EA K = 2 = f T k L P L EA K 4 2 3
10 Stability - Linear Bckling - Classical problem Look for displacements a when the tangent stiffness becomes zero: K C a = where K C = K +K is the tangent stiffness in the crrent state. This is a homogeneos eqation system with non-trivial soltions a. In classical bckling analysis the crrent state is the nloaded base state. A homogeneos eqation system may be formlated as an eigenvale problem: (K +l i K )x i = l i = the eigenvales (force mltipliers) x i = the bckling mode shapes If the crrent state is the nloaded state, solve the second-order system for loads f to get the stress distribtion in the strctre. The critical load f cr =l i f (l i becomes the load mltiplier)
11 Example: classical bckling Simple frame Unloaded base state Differential load = -N in y-dir at top of both pillars Fixed spports at base Frame Base state
12 Example: classical bckling st eigenvale=2. 6 Critical load fcr=2. 6 *(-) N fcr=-2. 6 N 2nd eigenvale= Critical load fcr= *(-) N fcr= N
13 Classical Linear Bckling in ABAQUS Apply loads, (for example N) and bondary conditions Choose Linear Pertrbation and then Bckle as the step. (Give nmber of eigenvales that yo want, the first (lowest) eigenvale gives the first bckling mode) Apply bondary conditions. Solve the eigenvale problem. The soltion gives the bckling modes and the force mltipliers l i for the bckling loads. f cr =l i f will then give the bckling loads.
14 Stability - Linear Bckling - General problem P K c If the crrent state is cased by pre-loads, f pre K C = K +K +K is the tangent stiffness cased by the pre-loads. A homogeneos eqation system may still be fond as an eigenvale problem: (K c +l i K )x i = l i = the eigenvales (force mltipliers) x i = the bckling mode shapes K is now the differential stiffness at this state cased by the loads f. The critical load is now f cr = f pre + l i f (where l i is the load mltiplier solved by the eigenvale problem) If geometric nonlinearity is inclded, the base state geometry is the deformed geometry at the end of the last step.
15 Example: general linear bckling Same frame Preload with N in y-dir at top of both pillars Differential load = -N in y-dir at top of both pillars Fixed spports at base Frame Base state
16 Example: general linear bckling st eigenvale=. 6 Critical load fcr=(-.9+.*(-)) 6 N fcr=-2. 6 N 2nd eigenvale=.84 6 Critical load fcr=( *(-)) 6 N fcr= N
17 Same frame Example2: general linear bckling Preload with N in y-dir at top of both pillars and a load at the left top corner of 3 kn in negative x-dir Differential load = -N in y-dir at top of both pillars Fixed spport at base Frame Base state
18 Example2: general linear bckling Pre-load N in y-dir and -2 kn in x-dir Pre-load N in y-dir and -4 kn in x-dir st eigenvale=.5 6 fcr= N (+ stresses from pre-load) st eigenvale=.26 6 fcr= N (+ stresses from pre-load)
19 General Linear Bckling in ABAQUS First, create a general static step (and nonlinear geometry if desired) Create Linear Pertrbation and then Bckle as the second step. Apply pre-loads and bondary conditions in the first step (general static step) Apply loads and bondary conditions in the second step (bckle), (for example N) Solves first the pre-load step and then the eigenvale problem with the base state from the pre-load. f cr = f pre + l i f give the bckling loads.
20 Stability - Non-linear Bckling Element stiffness calclated with eqilibrim in deformed configration and pdated displacement stiffness: Third order theory: K t = K +K +K Incldes all static effects in a physical problem. Loading may be made ntil collapse is reached and postbckling analysis may be performed.
21 Soltion of Non-linear Eqations Divide into a nmber of load-steps P K t4 P 3 K t3 R Direct explicit method: P 2 K t2 n = K t - P n P K t n+ = n + n+ R: residal, additive error 2 3
22 Ot-of Balance Forces External forces: P P = f b + f l Internal forces: element forces = I I Eqilibrim: P-I= A tb T DBdA a In the direct explicit method: P-I=R R: Force Residal (Ot-of-balance forces) A tb T σ da
23 Newton-Raphson Method P 2 P n= r n 3 r n 4 Load steps n=,2, P n =P n- +P n n = n- Iterations i=,2, calclate n from n i calclate residal R ni =P n- -I n i P =r n r n 2 d n d n 2 d n 3 n n+ calclate K t n i- d ni =(K i- t n ) - R i n ni = i- n +d i n stop iteration when reisdal is ok end of load
24 Example : Non-linear bckling Same frame Load with N in y-dir at top of both pillars Fixed spport at base Solve with non-linear geometry No bckling!!! Why?
25 Example2: Non-linear bckling, imperfections Load N in y-dir and - kn in x-dir (top left corner) Load N in y-dir and - kn in x-dir (top left corner) No bckling!! Bckling at t=.926 fcr= N
26 Stability with imperfections General types of imperfections may be added to non-linear bckling analysis (2 nd or 3 rd order analysis). Throgh adding eigenmode shapes on strctre 2. Throgh adding deformation from a previos static analysis
27 Example 3: Non-linear bckling - displacement control, imperfections Displacement -.2m in y-dir and - kn in x-dir (top left corner) force = N Force at.2m displ = N
28 Non-linear Bckling in ABAQUS Apply a load larger than the anticipated bckling load Choose Static, General problem as the step. Choose Nlgeom: on The time is fictive, dividing the load into load increments. Apply bondary conditions. A soltion may not be fond when a bckling load is reached. Preferably se displacement control.
29 Examples Snowloads on slender constrctions - and the finite element method
30 Ishall - nderspänd tre-ledstakstol Total last nder en snörik vinter ca:.5 kn/m2 Ska klara ca 3.4 kn/m2 Spännvidd 48m
31
32
33 Unstable Critical load ~.2 kn/m 2 Moment free joints
34 Unstable Critical load ~.7 kn/m 2 Spikplåtar
35 Stable Critical load ~ 2.5 kn/m 2 Bt too high stresses
36 Another similar ice-arena (span 52m)
37
38 Measred imperfections
39 Bckling shape analysed with imperfections
40 Bckling de to heat from fire Brandförlopp temp.~ tid
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