Interpretation of Lamé’s parameters in solid mechanics

Question

I think I have a pretty good understanding of the physical interpretation of Young's modulus $E$ and Poisson’s ratio $\nu$ in solid mechanics. However, I often find in mathematical papers that the equations are formulated in terms of Lamé's 1st and 2nd parameters $\mu$ and $\lambda$, respectively. I know there are formulas to relate the two in terms of Young's modulus and Poisson’s ratio, but I'm curious how can I best interpret their physical meaning.

Answer 1

$\mu$ is quite easy - it is the shear modulus. In engineering texts it is often called $G$.

I do not believe that $\lambda$ has a straightforward physical interpretation. However, the bulk modulus $\kappa=\lambda+2\mu/3$, so it is sometimes useful to think about $\lambda$ as something closely related to the bulk modulus. For example, the bulk modulus of a nearly incompressible material can become arbitrarily large; you can see from the formula above that in this case, asymptotically, $\kappa\approx\lambda$.

Answer 2

Physical meaning of Lame Parameters: A very elementary sequential explanation.

A vector is an entity that has a magnitude and a direction. A vector expressed in cartesian coordinates [Example: $3, 4$] or polar coordinates [$5, \arctan \left( 4/3 \right)$] still has the same length. Thus, for a vector, its magnitude (length) is the INVARIANT with change of Coordinate system.

A tensor, in our case, the Cauchy stress tensor, is a $3\times 3$ matrix showing $9$ stress components. Each component has a magnitude and two directions, whereas a vector has a magnitude and one direction. The two directions are given by the subscript indices ($\sigma_{ij}$),

$i \rightarrow$ Normal direction to the plane of action

$j \rightarrow$ Direction of the stress.

Like a vector, the tensor also has INVARIANTS. The trace (sum of diagonal elements) of the stress tensor in one such INVARIANT it will always be constant in any coordinate system. To verify this play around with $2 \times 2$ stress tensors and Mohr circles, you'll find the sum of diagonals are always constant.

Unlike the length of a vector, this invariant of the tensor doesn't have a perfectly intuitive physical picture, but it can be related to Volumetric stress. This invariant property, maintains the shape of the strained element. The volumetric stress only compresses the element while maintaining its geometric similarity to undeformed shape, a perfect cube becomes a smaller perfect cube.

The components off the diagonal in the stress tensor participate in shear deformations, they break the geometric similarity between undeformed and deformed shape. A cubic undeformed element becomes a rhomboid post deformation. Thus, the off-diagonal elements cause the deviatoric stress.

A question:

Volumetric or Deviatoric stresses, both cause deformations. Why do we need to split the two?

Answer: Volumetric compression will never cause an element to yield! Take a small cube of aluminum to the bottom of Mariana Trench and bring it back to surface, it’ll come back to its original size. On the other hand, Deviatoric stresses can cause yielding in the element. That’s why engineers/scientists want to split the stress (as well as strain) tensors into volumetric and deviatoric components and study them.

In one dimension, Hooke’s law states, force = stiffness $\times$ deformation. For an isotropic elastic solid in $3D$, Lame parameters help to split the Hooke’s law into Volumetric and Deviatoric components in a single equation.

stress = $LP_{1} \times$ Volumetric strain $+ LP_{2} \times$ Deviatoric strain.

$LP_{1} \leftarrow$ Lame's First parameter, usually $ \lambda $.

$LP_{2} \leftarrow$ Lame's Second parameter, usually $ \mu $.

Thus, $LP_{1}$ will always be strongly related to bulk modulus and $LP_{2}$ will always be related to shear modulus. Hope this explanation helps.

Answer 3

The Lame parameter $\lambda$ is equal to the off-diagonal term of the stiffness matrix, i.e. $C_{13}$. Where $\sigma_i=C_{ij} * \epsilon_j$. The stiffness matrix, $S$ is the inverse of the stiffness matrix, which has diagonal elements $1/E$ and off-diagonal elements $-v/E$ (for the upper left quadrant, the lower right is $1/G$ on the diagonal). Thus the top left corner of the $S$ matrix is:

$$S=\begin{bmatrix}1/E & -v/E & -v/E\\-v/E & 1/E & -v/E\\-v/E & -v/E & 1/E\end{bmatrix}$$

inverting gives $\frac{E(1-v)}{(1+v)(1-2v)}$ on the diagonal and $\frac{Ev}{(1+v)(1-2v)}$ (which is $\lambda$) on the off-diagonals. Some the lame parameter relates the transverse stress to uniaxial strain. Meanwhile, the diagonal terms are equivalent to the P-Wave modulus, or which relates axial stress to unaxial strain.