Appunti Advanced Computational Mechanics

UNIAXIAL LOADING/UNLOADING OF AN ELASTOPLASTIC BAR, IN CASE OF LINEAR KINEMATIC HARDENING

We do the same thing with the kinematic hardening. We have:

We need to introduce the notion of back stress = the ordinate of the central point in the elastic region

Here we have the scheme of the hardening

We assume that:

Note that during elastoplastic loading both plastic and elastic strain are developed

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

Here we can see the evolution of the back stress (4 equations).

H>0 for the hardening.

Now similar to the previous treatment we have:

INCREMENTAL ELASTOPLASTIC PROBLEM, FOR LINEAR KINEMATIC HARDENING

RETURN MAPPING ALGORITHM, FOR LINEAR KINEMATIC HARDENING

INCREMENTAL-ITERATIVE SOLUTION PROCEDURE FOR ELASTOPLASTIC 2D TRUSSES

Suppose we have a truss with n bars with a certain number of nodes.

in each bar : stress, strains are homogeneous; isotropic linear hardening elastoplastic material.

We have a time discretization and the beginning we have the time t in which we know everything,

n

We have an external node vector P n+1

The solution must be a collection of: all the displacement, all the plastic strain of all bars,

NEWTON –RAPHSON SOLUTION PROCEDURE FOR AN ELASTOPLASTIC 2D TRUSS

This is the core of the lesson. We have an interaction index by k and a time instant and so the time discretization n+1

we have to imagine that we are inside the NR so:

We have Un the vector of displacement at the beginning at it’s known and we add a ∆U n+1

For each element we can compute the current elongation at each iteration k:

By dividing by the current length of the element we obtain the strain, with which we enter in the return mapping

algorithm

so we are working to the total strain to stress.

Here we have to change the vector of displacements U as we can see here we have different U

So it means writing the linear system: For NR the tangent elasto-plastic stiffness matrix must be always

updated

INCREMENTAL-ITERATIVE SOLUTION PROCEDURE FOR ELASTOPLASTIC SYSTEM OF TWO BARS

We have two elasto plastic bars in parallel, which is an isotropic linear hardening is assumed for both bars.

the two bars are fixed and the bars at the extreme are fixed in order to have the same displacement.

we have one force applied

The global curve can be described on the plain.

We discuss the incremental problem for this case so we need to find the solution in terms of displacement and

plastic strain and in terms of stresses.

The eq. must be satisfy by a scalar equation

We have a smooth curve of the internal force that we don’t know. In reality is peace wise.

We incrementing the external load and we obtain the solution of u the solution is obtained with F

n+1 n+1

= estimate of u by in iteration k, so just of an estimate

n+1

By using Nr we can evaluate the value of internal force required and the derivative of internal force in respect to the

displacement. To described the interaction between the step of loading we suppose at the first red on x axis which is

the estimate of the displacement, so we can apply NR. We need to linearize the internal force in the neighborhood

of that point. It correspond to a certain stress values. We need to perform so the derivative, which is multiplied by

an increment of the displacement

In correspond to this we have the subscript k quantities that are estimations at iteration k.

We need to do the linearization so we evaluate the tangent of the curve of the internal force by doing the derivative

in respect to u n+1

In particular: here the only unknown is

we solve it by the way we treat the elastic laws so we perform the tangent modulus.

By using the change rule we have:

The crucial problem sis the algorithmic tangent modulus, when we can evaluate this we can apply the NR.

ALGORITHMIC TANGENT MODULUS

To compute a closed-form expression of E one has to differentiate the update formulas of the return-mapping

n+1

algorithm.

In evaluating this quantity we are taking into account the formulas that composed the algorithm.

In other words: E represents the material tangent response whereas E reflects not only the material response but

t n+1

also all the artifacts of the numerical algorithm used for calculating starting from a give value of .

According to the return-mapping algorithm (isotropic hardening), for , one has the stress as :

4/11/22

LINEAR ELASTICITY EQUATIONS (FOR ISOTROPIC MATERIALS) WRITTEN IN TERMS OF LAMÈ CONSTANTS Λ , . µ

USEFUL IN DISCUSSING VON MISES ELASTOPLASTICITY

We start writing the linear Lamè equation, useful to discuss the von misses Elastoplasticity.

Elastoplasticity we can write as:

The stress components with different indices are equal to zero based on the Kronencker delta

Now we see why is useful to write the elastoplasticity with lame constant

Bulk modulus because the epsilon kk in which the relation holds it the dilation is link to the mean stress as:

K = Bulk modulus

LINEAR ELASTICITY EQUATIONS WITH LAMÉ CONSTANTS (CONTINUED)

We start again with the equation number 1 and we proceed in this way:

We are taking the deviatoric part of the equation, because it define the deviatoric stress

Deviatoric stress and strain tensors:

By taking the deviator part we obtain this equation that link the deviatoric stress and strain

This decoupling is useful for the elastic strain energy.

We rewrite this by expressing the new quantities:

We obtain 4 terms by computing the multiplications.

There are terms that goes to zero because the trace the deviatory stress is zero

It turn out that the von misis criterion put a limit to this J2 quantity and we will see it after. The von missis says that

when J2 reaches a value we are on the boundary on the elastic region.

3D ELASTOPLASTICITY - VON MISES YIELD CRITERION

Here we explain when we reach this values of J2

So this c coincide to that sigma zero. The corresponding value

at yielding

VON MISES YIELD SURFACE IN PRINCIPAL STRESS SPACE

Here h = unit vector along hydrostatic axis:

The sum of the director cosine is must be equal to 1, equal in respect to the axis.

Let’s consider a generic stress vector, whose components are the principal stresses.

So by the projection we obtain the modulus of OG:

We can say that sigma is almost equal

Now we analyze the vector GQ Multiplied by h in order to give him the vectorial nature

We are punting a threshold of the length to component generic stress vector OQ orthogonal to the hydrostatic axis.

So it is a way to say that In the principal stress space, von Mises yield surface is a right circular cylinder aligned with

the hydrostatic axis.

ELASTOPLASTIC CONSTITUTIVE MODEL – VON MISES ASSOCIATIVE MODEL WITH LINEAR ISOTROPIC HARDENING

Analogous to the 1D problem that we saw

We express the yield function phi, remember the expression obtained before:

We have the flow rule:

Von misis norm because the norm of the tensor is defined as like that. It is a scalar

Than we have these conditions:

Here we have to understand the lambda meaning and remember that we have expressed as:

Variable that causes the increase of yield stress.

We explain lambda dot by taking the flow rule and proceed in this way:

We are in a context in which we apply stress and in a certain time interval the stress is operating the development of

plastic strain is defined we have the definition of the component of the plastic strain along the various direction, but

they are condense into a scalar by this rules, condense to a quantity that contributes in integrating in time the

Harding that implies the hardening the elastic region.

In particular in terms of rated the plastic strain are equal to zero so don’t know imply any variation of volume.,

intrinsic to the law written up to now.

Let’s see the:

ELASTOPLASTIC CONSTITUTIVE MODEL – VON MISES ASSOCIATIVE MODEL WITH LINEAR ISOTROPIC HARDENING IN

VECTOR-MATRIX NOTATION

Set up in matrix notations. Psi here is the lambda before

Here we have the

stress matrix

multiplied by a

matrix that give us

the stress deviator

ELASTOPLASTIC CONSTITUTIVE MODEL – VON MISES ASSOCIATIVE MODEL WITH LINEAR ISOTROPIC HARDENING

By introducing the matrices we can compare the two notation. D = tangent matrix

In order to have the yielding we need to remain to the yield surface = meaning of the consistency condition

VON MISES, WITH LINEAR HARDENING (NOT AT EXAM)

ELASTO-PLASTIC EQUATIONS FOR INCREMENTAL PROBLEM

We have to do the same thing done in 1D so we do the increment, we are in the level of constitutive law so there is

no structure.

We will proceed in Finite quantities

We enforce the equations in t+1 time instant, so we obtain equations (1):

Pay attention to the indices

The trial stress is equal to the all the stress plus this elastic plastic coefficient multiplied to the total strain.

Where the sought stress is the expression (4).

The by taking the difference we have the expression (6), coincidence only if the step is elastic.

Then we can have this test:

Plastic corrector phase we need to have a lambda different to zero in order to have phi grater than zero.

Even is plastic we don’t need a correction for the hydrostatic stress and comes out to equation 10

Equation 12 compare to equation 10 have some term unchanged because from the stress we have the deviatory

stress They are proportional so

we can define Bheta

The hydrostatic stress that we find in term in the trial hydrostatic is the correct so we don’t need correction but just

in a radial way

We return to the yield surface by a radial correction

Since they are proportional we need just to find the value of the radial correction.

So we have this ration of the trial quantities, this ratio is important because it enter to the flow rule.

The conclusion is:

In terms of stress deviatory stress we need a correction which is the radial correction while for the mean stress this

correction is not needed

The problem is that the algorithm of tangent stiffness matrix is different to the continuum elastic tangent matrix. So

with the same reason of 1D we take into account the artifact with the backward algorithm that we are adopting.

The solution of the incremental problem through the plastic correction phase is given by the following relations:

EXAMPLE - ELASTOPLASTIC ANALYSIS OF A PERFORATED (PLANE STRESS) PLATE IN TENSION

By increasing the displacement

Here more finer because here we will have plasticity

If we make a plot of this quantity we can

decide to indicate in grey in which the Von

misses is greater than that values

At end of increment the von Misses has a

greater value so we have the initial develop

of plasticity

Most of the plasticity is concentrated to that

kind of area.

Here we are in the

elasticity

EXAMPLE OF ELASTOPLASTIC PROBLEM - FIXED END BEAM LOADED BY A VERTICAL F