CA2b. NON-LINEAR FINITE ELEMENT ANALYSIS OF PLATE SUBJECTED TO SHEAR LOADING

Problem definition

In this assignment, you are going to implement the multiaxial elastic-plastic material response with linear isotropic hardeninginto a constitutive driver. In the spirit of the developments in computer assignment CA2a, we shall analyze the shear loadedcolumn in Fig. 1. In order to simplify the analysis, the columns may be considered as a "plate" in a state of plain strain. Asmay be seen in the Figure below, the plate is loaded by a shear load P[t], and the behavior of the plate during one loadingcycle will be analyzed whereby

P@tD = Pmax sinB 2 Π

TtF with T = 1

where Pmax is the maximum value of the load amplitude. It is determined by the bending moment at onset of plastic deforma-

tion according to beam theory My, cf. computer assignment 1, whereby

Pmax = 3My

awith My = MinBh

2 Ibl2 + 4 bl bu + bu2M Σy

12 H2 bl + buL ,h2 Ibl2 + 4 bl bu + bu

2M Σy

12 Hbl + 2 buL FThe height, width and thickness of the plate are a, h and t, respectively. Hence, the geometry data are chosen as follows:

a = 60 mm, h = 20 mmm, t = 10 mm

In the assignment, special focus will be placed on the assessment of the residual stress state of the column (obtained at theend of the time interval, i.e. t=1) as compared to the result obtained from the corresponding previous 1D analysis based onthe Euler-Bernoulli beam theory assumption.

Figure 1: "Plane strain" column subjected to prescribed shear loading

0.2 0.4 0.6 0.8 1.0t

-1.0

-0.5

0.5

1.0P@tD @´PmaxD

Cyclic prescribed shear load considered

Figure 2: Cyclic prescribed shear load variation versus time

The material behavior is described by the experimentally determined cyclic stress-strain curve shown in Figure 3. Like, inthe previous assignment, let us consider only the behavior in the first loading cycle, cf. Figure 3. In order to model thedescribed behavior, and at the same time simplifying our analysis, we shall consider the linear isotropic hardening rulecombined with the plasticity concept, i.e. r = 1 is used as a model parameter. Otherwise, the same model parameters as in

CA1 are used in the present assignment. The model response was fitted to the experimental curve using "manual" curvefitting and the resulting matching with r = 1 is shown in Fig. 4 below. It may be observed that the model response is far

from "perfect" when r = 1 is used. In particular,, the yield stress after one cycle is significantly overestimated by the linear

isotropic hardening model. The model parameters were obtained as: E = 200 GPa, H = E 10, Σy = 540 MPa and r =1,

corresponding to linear isotropic hardening.

Figure 3: Cyclic behavior of a typical railway steel in strain-controlled uniaxial test. Experimental results are due to Johan

Ahlström, Birger Karlsson, Wear 258, 1187–1193, (2005).

-0.010 -0.005 0.005 0.010Ε@tD

-500

500

Σ@tDIsotropic and kinematic hardening plasticity: cyclic loading

Figure 4: Linear isotropic hardening model response as compared to experimental values for the strain amplitudeDΕf 2 = 1 %, The model response has been obtained using the same parameters as in CA1 except for r=1.

Exercise problems

As alluded to in the problem definition, consider the assignment problems:

è Implement the elastic response of the model behavior. Check the response when Pmax =My

a. Almost no plastic

response should appear in the plate for this load magnitude? Discuss the result.

2 CA2.nb

è Implement the elastic-plastic response of the model using the discussed linear isotropic hardening law. Check the

response after one loading cycle when Pmax = 3My

a. In particular, consider the residual stress state at the base of

the column from the computed reaction forces. Discuss the obtained residual stress state in comparison with the corresponding result obtained from the 1D beam model behavior, already developed in the previous assignment.

Numerical procedures

Equilibrium condition

The equilibrium conditions for the considered column may be formulated as the balance of internal and external virtualwork:

àBΣ@uD : Ε@uD âB = à

Bf × u âB + à

G

t × u âB

where B is the 3D domain of the solid plate and ¶B is the external boundary thereof. In the present analysis, the plane strain

symmetry condition may be assumed, whereby the 3D problem may be reduced to a 2D problem. Hence, we may considerthe 3D domain restricted to the 2D region W so that B ® t ´ W, where t is the plate thickness. Likewise, the external

boundary may be reduced to ¶B ® t ´ G.

Next, finite element discretization is made with respect to the computational domain W into finite elements

We, e = 1, ... NEL. It is assumed that each element has the interpolation

u = âI=1

NODE

NI@xD uI Þ u = âI=1

NODE

NI@xD uI Þ l =¶u

¶x= â

I=1

NODE

uI Ä gI with gI =¶NI

¶x

where uI are the nodal displacement variables, and 8NI@xD<I=1,2,NODE are the element interpolation functions. In the

assignment, we restrict to CST elements corresponding linear displacement approximation within each triangle, i.e.NODE = 3. To this end, a local Ξ1, Ξ2-coordinate system is introduced, where the shape functions NI@Ξ1, Ξ2D are defined

(within a base triangle element shown in Fig. 5 below) as

N1 = 1 - HΞ1 + Ξ2L, N2 = Ξ1, N3 = Ξ2 Þ gI =¶NI

¶x= â

i=1

2 ¶NI

¶Ξi

Gi with Gi =¶Ξi

¶x

0.00.5

1.0Ξ1 0.0

0.5

1.0

Ξ20.00.51.0

,

0.00.5

1.0Ξ1 0.0

0.5

1.0

Ξ20.00.51.0

,0.0

0.51.0

Ξ1 0.0

0.51.0

Ξ20.00.51.0

Figure 5 Linear basis functions pertinent to the Constant Strain Triangle element

The corresponding discretized version of the equilibrium condition state above, cf. the lectures x-x, may be stated as balancebetween internal and external finite element nodal forces.

g@uD = b@uD - f@tD = 0 with b@u, tD = A

e=1

NEL àWe

Bet

Σ âW , f@tD = A

e=1

NELfeext

The problem is to find the total "displacement" u = 8uI<I=1,.. NNO so that the vector of out-of-balance forces g@uD = 0.

To this end, the non-linear solution strategy is used as discussed below.

Non-linear solution strategy - Newton's method

In order to solve the non-linear problem, we consider the balance of internal and external forces for each given time step as

CA2.nb 3

g@uD = b@uD - f@tD = 0

with

u = nu + v Dt and t = n+1t

For each time t = n+1t, consider the algorithm defined as: for each iterate i, evaluate the improved displacement incrementi+1

Ξ from the Newton step

iK i+1Ξ = -gAnu + iv DtE, i+1v = iv + i+1

Ξ

where K is the Jacobian (or tangent stiffness) matrix of the considered component. It is defined from the linearization

dg@uD = db@uD = K du

The tangent stiffness K is defined by

K =¶b

¶u= Dt

¶b

¶v=. ..

Preliminary results

Preliminary results in the form of a comparison between the residual stress states obtained from 1D-beam analysis modelingand the multiaxial finite element modeling are given in the Figure below.

-10 -5 5 10z

-400

-200

200

400Σ@zD

Fiber-stress response over cross section

From 1D beam-analysis

From FE analysis

Implementation: Pseudo code

Main routine: Input data and solution independent arrays

lh = 60; % horisontal lengthlv = 20; % vertical lengtht= 10;h=lh; bl=t; bu=t;nelv = 20 % # elements verticallynelh = 10 % # elememts horisontally

[nel, nno, edof, ex, ey, nol1, nol2, nol3, nol4] = plate_mesh_CST (nelv, nelh, lh, lv);

nva = 2*nno;bc = [2*nol1-1 2*nol1; zeros (1, length (nol3)) zeros (1, length (nol3))]';bcload = 2*nol3 - 1;

% loads and other constantsmtim = 400; % total number of timesteps

∆ = 10-6; % tolerance for equilibrium iterationsmite = 50; % max number of iterations

My=min(...

((bl^2 + 4*bl*bu + bu^2)*h^2*sigy)/(12.*(2*bl + bu)),... ((bl^2 + 4*bl*bu + bu^2)*h^2*sigy)/(12.*(bl + 2*bu)));

Pmax = 3 * Mel/lv;

% Material parameters for elastic plastic material

E = 200 103; Ν = 0.3; H = E/10; r = 1; Σy= 540; mp = [E Ν H r ΣyE;% Initiation of variablesu = zeros (nva, 1); %displacementv = zeros (nva, 1); %velocityf = zeros (nva, 1); %applied nodal loads

nΣ = zeros (4, nel); Σ = n

Σ;nΚ = zeros (1, nel); Κ = nΚ;

pdof = bc (: , 1); % prescribed degrees of freedomfdof = [1 : nva]'; fdof (pdof) = []; % "free" degrees of freedom

4 CA2.nb

lh = 60; % horisontal lengthlv = 20; % vertical lengtht= 10;h=lh; bl=t; bu=t;nelv = 20 % # elements verticallynelh = 10 % # elememts horisontally

[nel, nno, edof, ex, ey, nol1, nol2, nol3, nol4] = plate_mesh_CST (nelv, nelh, lh, lv);

nva = 2*nno;bc = [2*nol1-1 2*nol1; zeros (1, length (nol3)) zeros (1, length (nol3))]';bcload = 2*nol3 - 1;

% loads and other constantsmtim = 400; % total number of timesteps

∆ = 10-6; % tolerance for equilibrium iterationsmite = 50; % max number of iterations

My=min(...

((bl^2 + 4*bl*bu + bu^2)*h^2*sigy)/(12.*(2*bl + bu)),... ((bl^2 + 4*bl*bu + bu^2)*h^2*sigy)/(12.*(bl + 2*bu)));

Pmax = 3 * Mel/lv;

% Material parameters for elastic plastic material

E = 200 103; Ν = 0.3; H = E/10; r = 1; Σy= 540; mp = [E Ν H r ΣyE;% Initiation of variablesu = zeros (nva, 1); %displacementv = zeros (nva, 1); %velocityf = zeros (nva, 1); %applied nodal loads

nΣ = zeros (4, nel); Σ = n

Σ;nΚ = zeros (1, nel); Κ = nΚ;

pdof = bc (: , 1); % prescribed degrees of freedomfdof = [1 : nva]'; fdof (pdof) = []; % "free" degrees of freedom

Time stepping and non-linear Newton solution procedure

T = 1; Dt = T/mtim; time = 0;

for itim = 1 : mtim % for all time steps

time = time + Dt;

Ω = 2*Π/T; %one loading cyclev (bc (: , 1)) = bc (: , 2);f(nol3*2-1)=Pmax/length(nol3)*sin(omega*time); %distribute load in x-dir. on

line 3

err = 1; stop_eqiter = 0; iite = 0;while ((stop_eqiter == 0) & (iite < mite)) % equilibrium iterations

iite = iite + 1;K = sparse (nva, nva);g = zeros (nva, 1);for iel = 1 : nel % for all elements

leva = edof (iel, 2 : 7);[B, Area] = B_matrix (ex (iel, :), ey (iel, :));

due = v (leva)*Dt; Σtr = nΣ(: , iel) + Ee*B*due;% trial stress

[Σ (: , iel), Κ (: , iel), E_t] = IO_V _Mises(Σtr, nΚ (: , iel), mpL;

fext = zeros (6, 1);g (leva) = g(leva) + HBt Σ (: , iel)*Area*t - fext);K (leva, leva) = K (leva, leva) + Dt BtE_t B*Area*t;

end

g=g-f;

err = norm (g (fdof), 2); % Check convergenceif err < ∆

stop_eqiter = 1;else

Ξ = -K (fdof, fdof)\g (fdof);v (fdof) = v (fdof) + Ξ; % Update displacement increment (free nodes)

end

end

u = u + v*Dt; % update total displacementnΣ=Σ; n

Κ=Κ; % Save data from previous time stepend% post-processing...stop

CA2.nb 5

T = 1; Dt = T/mtim; time = 0;

for itim = 1 : mtim % for all time steps

time = time + Dt;

Ω = 2*Π/T; %one loading cyclev (bc (: , 1)) = bc (: , 2);f(nol3*2-1)=Pmax/length(nol3)*sin(omega*time); %distribute load in x-dir. on

line 3

err = 1; stop_eqiter = 0; iite = 0;while ((stop_eqiter == 0) & (iite < mite)) % equilibrium iterations

iite = iite + 1;K = sparse (nva, nva);g = zeros (nva, 1);for iel = 1 : nel % for all elements

leva = edof (iel, 2 : 7);[B, Area] = B_matrix (ex (iel, :), ey (iel, :));

due = v (leva)*Dt; Σtr = nΣ(: , iel) + Ee*B*due;% trial stress

[Σ (: , iel), Κ (: , iel), E_t] = IO_V _Mises(Σtr, nΚ (: , iel), mpL;

fext = zeros (6, 1);g (leva) = g(leva) + HBt Σ (: , iel)*Area*t - fext);K (leva, leva) = K (leva, leva) + Dt BtE_t B*Area*t;

end

g=g-f;

err = norm (g (fdof), 2); % Check convergenceif err < ∆

stop_eqiter = 1;else

Ξ = -K (fdof, fdof)\g (fdof);v (fdof) = v (fdof) + Ξ; % Update displacement increment (free nodes)

end

end

u = u + v*Dt; % update total displacementnΣ=Σ; n

Κ=Κ; % Save data from previous time stepend% post-processing...stop

Stress integration

function@Σ, Κ, EtD = IO_V _Mises HΣtr, nΚ, mpL

% Constitutive driver for von Mises model with linear isotropic hardening

% Note : Stress and strain by Voigt - representation Hfor input outputL!

% Σ = @Σ11, Σ22, Σ33, Σ12Dt

E = mp H1L; Ν = mp H2L; H = mp H3L; Σy = mp H5L;K = E 3 H1 - 2 * ΝL; G = E 2 H1 + ΝL;v1 = @1, 1 , 1, 0Dt; v2 = @0, 0, 0, 1Dt;

I = diag Hv1 + .5 * v2L;Idev = I -

1

3v1 * v1t;

Ee = 2 G Idev + K v1 * v1t;

Σdevtr =. ..;

Σetr =. ..;

Φtr = Σetr - Σy - nΚ;

if Φtr £ 0

% elastic loading

Σ = Σtr;

Κ = nΚ;

Et = Ee;

else

% plastic loading

Μ =. ..;

Σ =. ..;

Κ =. ..;

Et = Ee - ...;

end

Close up of FE element analysis

6 CA2.nb