Appunti completi corso Advanced computational mechanics

1D ELASTOPLASTIC CONSTITUTIVE MODEL, WITH LINEAR ISOTROPIC HARDENING

The equations for our rate problem are:

COMPLEMENTARITY CONDITION:

φ

If < 0 : at our time instance the material point is in the elastic range and so lambda dot =0 (mean that plastic strain cannot develop if we are in the elastic range);

φ

If = 0 (we are at the yield) : lambda dot can be different from 0, which mean >0 (so plasticity can be developed)

Where we have that:

λ represents accumulated plastic strain. ( so we accumulate plastic strains any time we develop it , even in tension or compression)

In the plastic flow rule we have that:

1D ELASTOPLASTICITY, LINEAR ISOTROPIC HARDENING (CONT'D):

We want to obtain the same previous relations but following a process which can be followed even in 2D or 3D wrt the previous one which was only in 1D.

The determination of the plastic multiplier (during plastic loading) is done using the consistency condition (if we are in a time instant in which we are at yield).

(φ=0) , to develop plasticity we must satisfy the following‘’phi dot’’ must be =0):relation (so(I)If we want to develop plasticity so wemust satisfy:But we have also to satisfy:So:So we get:Enforcing condition (I) (considering the case in which we are in plastic loading):The determination of elastoplastic tangent modulus is:(this link total strain to the stress)Increment of total stress = EH/ (E+H) * increment of total strain;LEZIONE 27/10/22INCREMENTAL ELASTOPLASTIC PROBLEM, FOR LINEAR ISOTROPIC HARDENINGWe have to proceed by Finite increment. Considering the pseudo timeaxis we will have a interval of interest. Then we consider a genericsubinterval going form tn to tn+1.Assuming that local state is known this mean that:Lambda is an internal variable which depend on the previous loading history of the material.So we are looking at the material element with a total strain given. Now we want to enforce the constitutiveequation at time tn+1, so we would like

assuming that the step is elastic are used also in the so called CORRECTIONPHASE. So even if the step is not elastic we can define quantities to make a check.

LINEAR ISOTROPIC HARDENING (CONT’D)

RETURNING MAPPING ALGHORITM, FOR

In this case we can rewrite this equation in this way ( since sign(x) make that any x can be expressed as x =|x| sign(x) we can express it as):

Here it strange cause we have 1 equation but 2 conclusions:

st- The 1 term is positive, which is a modulus. Since at any time this equation must hold, mean that if sign (sigma n+1) is positive also the trial correspondent one must be positive, the same if negative.

- Also this mean that:

At the end of the time step:

This cause going back to the governing equation if delta lambda > 0 phi must be =0

Here in the only UNKNOWN.

So we get:

By sub delta lambda, we get: –

RETURN-MAPPING ALGHORITM FOR 1D ELASTOPLASTICITY ISOTROPIC HARDENING

So at the end what the algorithm do is:

ELASTOPLASTIC BAR, WITH LINEAR ISOTROPIC HARDENING,

kinematic hardening. We have an initial elastic region and we know that when plasticity develop then the yield stress increase but simultaneously the whole elastic region translate upward. So we get a new yield stress in traction but the one in compression is different.

We have to introduce the BACK STRESS: Is the ordinate of the central point in the elastic region.

We can say that now: We how to define the evolution of alpha. This is done with the elastoplastic constitutive model (next slide).

Kinematic hardening predicts a lower yield stress in compression after yielding in tension (Bauschinger effect, for metals).

1D ELASTOPLASTIC CONSTITUTIVE MODEL, WITH LINEAR KINEMATIC HARDENING:

Where:

The difference between the actual stress and the back stress is the RELATIVE STRESS.

Enforcing CONSISTENCY CONDITION: (similar to the previous treatment done)

We have:

INCREMENTAL ELASTOPLASTIC PROBLEM FOR LINEAR KINEMATIC HARDENING:

Then we go to this step.

Also for this case we define:

INCREMENTAL ITERATIVE

SOLUTION PROCEDURE FOR ELASTOPLASTIC 2D TRUSSES

Now we think about a structural problem (we think at a truss system). We suppose a truss with m bars and a certain number of nodes. We define a vector of nodal displacement.

We assume that at time tn everything is known. Then we have an ext force vector known. At new time instant the values changes in Pn+1. We have to find the solution at tn+1 as illustrated up.

Recalling what is a stiffness matrix for a bar (NOT INTERESTING): s when we write INTERNAL FORCE = EXTERNAL FORCE on the node the force are positive if the are directed in the opposite way of the external one (directed as reference system)

We are in the middle of a NR iteration process so at a certain iteration k we know the value of the nodal displacement at the end of the step. This estimate can be written as:

The current estimate (with Un is known form the solution of the previous time step) can be written as on right.

Now for each element we can compute the current elongation at the end

of time step:

Since we know the elongation we can compute the strain by dividing by the length: (this strain is the one that we use in the return mapping algorithm):

So we get also an update of the axial force, from which we get the internal nodal forces:

Now we will modify this solution within the time step by modify the displacement.

Linearization is done in a neighborhood of the current estimate:

This elastoplastic stiffness matrix must be computed again at each iterations according to NR method.

INCREMENTAL ITERATIVE SOLUTION PROCEDURE FOR ELASTOPLASTIC SYSTEM OF 2BARS:

We consider 2 elastoplastic bars in parallel with a rigid link on right in order to have a common displacement, in order to have a scalar problem in terms of F and displacement, which is the response of the system.

The increment of external force is given so we know the external force at the end of the step.

We have to find at the end time step in terms of displacement, plastic strain lambda (govern the hardening) and

stress::Equilibrium at the end of the time step must be satisfied, and the equilibrium is given by(ext force is balanced by 2 axial forces):

Also the solution must be such that:

The external force must be balanced by the 2 external forces, and this is equilibrium. (cont’d):

INCREMENTAL-ITERATIVE SOLUTION; ELASTOPLASTIC SYSTEM OF 2 BARS

Considering a graph of the internal force in terms of displacement. From (Un, Fn) we want to increase the internal force, So we are incrementing the internal load in this range:

So that our solution is:

But the solution is unknown.

The curve represents the internal force (and we don’t know this). Also the curve is smooth but if we have linear hardening is not linear cause the response of an elastoplastic system like this will be PIECEWISE LINEAR (in reality).

If we suppose that during the time step we are at a certain iteration k we know the Un+1 . (is an estimate).

So we know the red point on x axis. Considering the tangent we will go from the red point

Computing deltaU we get Un+1 (k+1). Since we apply NR we have to evaluate for any estimate of the displ the value of the internal force and the derivative of it wrt the displ. To describe the iterations within the steps we suppose that we are in (we have an estimate of the displ):

So applying NR we linearize the internal forces in a neighborhood of displ. But to this value of displ correspond a stress. So we can define: (In this last formula in the bracket we apply the CHAIN RULE).

The problem is the evaluation of the algorithmic tangent modulus. This must be evaluated to apply NR.

ALGHORITMIC TANGENT MODULUS: if we want to compute an expression for this we have to differentiate the formula. So if we vary epsilon which is the variation of the output sigma? So in evaluating this quantity we have to take into account the formulas that compose the Algorithm.

On right we find the definition of the stresses.

LEZIONE 04/11/22 LINEAR ELASTICITY EQUATIONS (FOR ISOTROPIC MATERIALS) WRITTEN IN TERMS

We start by writing linear elasticity equation with lame’s constant useful to discuss von miseselastoplasticity. We know that the stress can be defined as:9 scalar equations; but only 6 are independent equations due to the symmetry of the stress and strain tensors

The 2 constants are:

So stresses are zero if the corresponding strain are zero. This is important to say that thus the principaldirections of stress and strain always coincide in an isotropic elastic material(Contracting = multiply the equation on both side by delta ij). For definition (delta ij)* (delta ij) =3

Also we have that:

So:

LINEAR ELASTICITY EQUATIONS WITH LAME’ CONSTANTS

Now we