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.. only:: html

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.. rst-class:: sphx-glr-example-title

.. _sphx_glr_api_gallery_materials_elasticity_plot_03_lame_constants.py:


Lamé's elastic constants (1852)
===============================

Early elasticity theories built from molecular forces predicted a single
elastic constant for an isotropic solid, with Poisson's ratio fixed at 1/4.
Measurements disagreed. Gabriel Lamé's *Leçons sur la théorie mathématique de
l'élasticité* (1852) wrote the isotropic law with two constants,

.. math:: \sigma_{ij} = \lambda\,\varepsilon_{kk}\,\delta_{ij} + 2\mu\,\varepsilon_{ij},

and every other modulus follows from any two
(:func:`~engineeringkit.materials.lame_parameters`,
:func:`~engineeringkit.materials.shear_modulus`,
:func:`~engineeringkit.materials.bulk_modulus`). The example applies Lamé's
law to a tensile strain state and plots how the moduli depend on
Poisson's ratio.

.. GENERATED FROM PYTHON SOURCE LINES 21-26

.. code-block:: Python

    import matplotlib.pyplot as plt
    import numpy as np

    from engineeringkit.materials import bulk_modulus, lame_parameters, shear_modulus








.. GENERATED FROM PYTHON SOURCE LINES 27-33

Uniaxial tension from Lamé's law
--------------------------------

A bar stretched by :math:`\varepsilon` contracts sideways by
:math:`\nu\varepsilon`. Putting that strain into Lamé's law must give a
stress :math:`E\varepsilon` along the bar and none across it.

.. GENERATED FROM PYTHON SOURCE LINES 33-43

.. code-block:: Python


    E, nu, eps = 70e9, 0.33, 1e-3
    lam, mu = lame_parameters(E, nu)
    strain = np.diag([eps, -nu * eps, -nu * eps])
    stress = lam * np.trace(strain) * np.eye(3) + 2 * mu * strain
    print(f"aluminium: lambda = {lam / 1e9:.1f} GPa, mu = G = {mu / 1e9:.1f} GPa, K = {bulk_modulus(E, nu) / 1e9:.1f} GPa")
    print(f"stress from Lame's law: {np.round(np.diag(stress) / 1e6, 6)} MPa")
    assert np.allclose(np.diag(stress), [E * eps, 0, 0], atol=1e-3)
    assert np.isclose(lam + 2 * mu / 3, bulk_modulus(E, nu)) and mu == shear_modulus(E, nu)





.. rst-class:: sphx-glr-script-out

 .. code-block:: none

    aluminium: lambda = 51.1 GPa, mu = G = 26.3 GPa, K = 68.6 GPa
    stress from Lame's law: [70.  0.  0.] MPa




.. GENERATED FROM PYTHON SOURCE LINES 44-51

The moduli against Poisson's ratio
----------------------------------

As :math:`\nu \to 1/2` the material becomes incompressible:
:math:`\lambda` and :math:`K` diverge while :math:`G \to E/3`. A negative
:math:`\nu` (auxetic foams) is allowed down to -1, where :math:`G` diverges
and :math:`K \to 0`.

.. GENERATED FROM PYTHON SOURCE LINES 51-65

.. code-block:: Python


    nus = np.linspace(-0.9, 0.49, 300)
    lams, mus = lame_parameters(1.0, nus)
    fig, ax = plt.subplots(figsize=(6, 3.8))
    ax.semilogy(nus, mus, label="G / E")
    ax.semilogy(nus, bulk_modulus(1.0, nus), label="K / E")
    ax.semilogy(nus[lams > 0], lams[lams > 0], label="lambda / E")
    ax.axvline(0.25, color="0.6", ls=":", label="Poisson's 1/4")
    ax.set(xlabel="Poisson's ratio", ylabel="modulus / E", title="Isotropic elastic constants")
    ax.legend()
    fig.tight_layout()
    assert np.allclose(mus * (3 * lams + 2 * mus) / (lams + mus), 1)

    plt.show()



.. image-sg:: /api/gallery/materials/elasticity/images/sphx_glr_plot_03_lame_constants_001.png
   :alt: Isotropic elastic constants
   :srcset: /api/gallery/materials/elasticity/images/sphx_glr_plot_03_lame_constants_001.png
   :class: sphx-glr-single-img






.. rst-class:: sphx-glr-timing

   **Total running time of the script:** (0 minutes 0.052 seconds)


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