Dimensional Elastic Modulus

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Alexander Hoover

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Oct 12, 2015, 1:39:03 PM10/12/15
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Hi Boyce,

From a previous discussion, we talked about

P = dW/dF ~ force / area   (PK1 Stress Tensor)

F = dchi/dX ~ dimensionless (Deformation Gradient) 

===> W ~ force/area  (Strain Dependent Elastic Energy Function)

For neo-Hookean,

   W = c1*(I1 - 3)

where

   I1 = tr(C) = tr(F^T F)

Here, c1 has units of force/area and is analogous to a (possibly rescaled) Young's modulus.

My issue is that I'm trying to match an experimental paper's flexural stiffness but wasn't sure how to rescale my elastic modulus used for a Neo-Hookean material. When rescaling the Young's modulus, should I use ds^2 for scaling my force/area? Qualitatively, I have things that look very similar to what I want I just need to find the right scaling parameter.


Best,
Alex

Boyce Griffith

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Oct 12, 2015, 1:54:27 PM10/12/15
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On Oct 12, 2015, at 1:39 PM, Alexander Hoover <hoover.a...@gmail.com> wrote:

Hi Boyce,

From a previous discussion, we talked about

P = dW/dF ~ force / area   (PK1 Stress Tensor)

F = dchi/dX ~ dimensionless (Deformation Gradient) 

===> W ~ force/area  (Strain Dependent Elastic Energy Function)

For neo-Hookean,

   W = c1*(I1 - 3)

where

   I1 = tr(C) = tr(F^T F)

Here, c1 has units of force/area and is analogous to a (possibly rescaled) Young's modulus.


My issue is that I'm trying to match an experimental paper's flexural stiffness but wasn't sure how to rescale my elastic modulus used for a Neo-Hookean material. When rescaling the Young's modulus, should I use ds^2 for scaling my force/area? Qualitatively, I have things that look very similar to what I want I just need to find the right scaling parameter.

The correct scaling will depend on how the data are reported in the paper (for instance, as force per unit undeformed area or force per unit deformed area).

-- Boyce

Alexander Hoover

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Oct 20, 2015, 3:05:37 PM10/20/15
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I guess the better question would be what are the units used when setting an elastic/Young's modulus in IBFE. For instance, in example 2 (below) you have "mu=10". Is the modulus' area dependent on element size or is it some other measure, e.g. Pascals?


namespace ModelData{ // Problem parameters.
static const double mu = 10.0;

// Stress tensor functions.
void
PK1_dev_stress_function(
    TensorValue<double>& PP,
    const TensorValue<double>& FF,
    const libMesh::Point& /*X*/,
    const libMesh::Point& /*s*/,
    Elem* const /*elem*/,
    const vector<NumericVector<double>*>& /*system_data*/,
    double /*time*/,
    void* /*ctx*/)
{
    PP = mu*FF;
    return;
}// PK1_dev_stress_function
void
PK1_dil_stress_function(
    TensorValue<double>& PP,
    const TensorValue<double>& FF,
    const libMesh::Point& /*X*/,
    const libMesh::Point& /*s*/,
    Elem* const /*elem*/,
    const vector<NumericVector<double>*>& /*system_data*/,
    double /*time*/,
    void* /*ctx*/)
{
    PP = -mu*tensor_inverse_transpose(FF,NDIM);
    return;
}// PK1_dil_stress_function
}

Boyce Griffith

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Oct 20, 2015, 3:10:23 PM10/20/15
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The code doesn't impose or require a specific system of units.

In the below case, the code is computing the 1st Piola-Kirchhoff elastic stress, which has units of force in the current configuration per area in the reference configuration. FF is dimensionless, so mu has units of force/area. The appropriate units depend on the units of force in the physical configuration and the units of length in the reference configuration. (Note that it is probably the case that length will have the same units in the current and reference configurations, but this does not have to be the case, e.g. if you use some initial coordinate mapping function that does not preserve length.)

-- Boyce

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Alexander Hoover

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Oct 20, 2015, 3:13:18 PM10/20/15
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That clears it up! Sorry to bother.
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