We can express all the equations in terms of the displacement field. Its primary advantage is that it combines the three sets of governing equations (equilibrium, kinematics, and constitutive) into a single vector equation, reducing the problem from 15 unknowns (stress, strain, displacement) to just 3 (the components of the displacement vector \(\mathbf{u}\)).

The derivation involves a systematic substitution, starting with the equilibrium equation and progressively replacing stress with strain, and then strain with displacement.

We begin with the three fundamental sets of equations in index notation.

1. Equilibrium Equation (Equation of Motion): This equation relates the divergence of the stress tensor to body forces and inertia. \[ \sigma_{ji,j} + \rho b_i = \rho \ddot{u}_i \tag{1} \] where \(\sigma_{ji}\) is the stress tensor, \(\rho\) is the density, \(b_i\) is the body force per unit mass, and \(u_i\) is the displacement vector.

Alternative form:

\[ \nabla \cdot \boldsymbol{\sigma} + \rho \mathbf{b} = \rho \ddot{\mathbf{u}}\tag{1'} \]

2. Constitutive Law (Generalized Hooke’s Law for Isotropic Materials): This law relates stress to strain using the two Lamé parameters, \(\lambda\) and \(\mu\) (the shear modulus, \(G\)). \[ \sigma_{ij} = \lambda \epsilon_{kk} \delta_{ij} + 2\mu \epsilon_{ij} \tag{2} \] where \(\epsilon_{ij}\) is the strain tensor, \(\epsilon_{kk} = \epsilon_{11}+\epsilon_{22}+\epsilon_{33}\) is the volumetric strain (trace of the strain tensor), and \(\delta_{ij}\) is the Kronecker delta.

Altenative form: \[ \boldsymbol{\sigma} = \lambda (\text{tr}(\boldsymbol{\epsilon})) \mathbf{I} + 2\mu \boldsymbol{\epsilon} \tag{2'} \]

3. Kinematic (Strain-Displacement) Relation: This equation defines strain in terms of displacement gradients for small deformations. \[ \epsilon_{ij} = \frac{1}{2} (u_{i,j} + u_{j,i}) \tag{3}\] or \[ \boldsymbol{\epsilon} = \frac{1}{2} \left[ \nabla \mathbf{u} + (\nabla \mathbf{u})^T \right]\tag{3'} \] The Substitution Process:

Step A: Express Stress in terms of Displacement First, substitute the kinematic relation (3) into the constitutive law (2). \[ \sigma_{ij} = \lambda \epsilon_{kk} \delta_{ij} + 2\mu \left[ \frac{1}{2} (u_{i,j} + u_{j,i}) \right] \] The volumetric strain term \(\epsilon_{kk}\) also needs to be expressed in terms of displacement: \[ \epsilon_{kk} = u_{k,k} = u_{1,1} + u_{2,2} + u_{3,3} \] Substituting this back gives the stress purely in terms of displacement: \[ \sigma_{ij} = \lambda u_{k,k} \delta_{ij} + \mu (u_{i,j} + u_{j,i}) \tag{4}\]

Alternative form: First, note that the trace of the strain tensor is the divergence of the displacement vector: \[ \text{tr}(\boldsymbol{\epsilon}) = \nabla \cdot \mathbf{u} \] Substitute this and the kinematic relation (3’) into the constitutive law (2’): \[ \boldsymbol{\sigma} = \lambda (\nabla \cdot \mathbf{u}) \mathbf{I} + 2\mu \left[ \frac{1}{2} (\nabla \mathbf{u} + (\nabla \mathbf{u})^T) \right] \] \[ \boldsymbol{\sigma} = \lambda (\nabla \cdot \mathbf{u}) \mathbf{I} + \mu (\nabla \mathbf{u} + (\nabla \mathbf{u})^T) \tag{4'}\]

Step B: Substitute Stress into the Equilibrium Equation Now, substitute this expression for stress (4) into the equilibrium equation (1). Since the stress tensor is symmetric (\(\sigma_{ij} = \sigma_{ji}\)), we can replace \(\sigma_{ji}\) with \(\sigma_{ij}\). \[ \left[ \lambda u_{k,k} \delta_{ij} + \mu (u_{i,j} + u_{j,i}) \right]_{,j} + \rho b_i = \rho \ddot{u}_i \] We now apply the derivative with respect to \(x_j\) to the terms in the bracket: \[ (\lambda u_{k,k} \delta_{ij})_{,j} + (\mu u_{i,j})_{,j} + (\mu u_{j,i})_{,j} + \rho b_i = \rho \ddot{u}_i \] Let’s analyze each term, assuming \(\lambda\) and \(\mu\) are constants: * Term 1: \((\lambda u_{k,k} \delta_{ij})_{,j} = \lambda (u_{k,k})_{,j} \delta_{ij}\). Due to the Kronecker delta \(\delta_{ij}\), this term is only non-zero when \(j=i\). So, the derivative becomes with respect to \(x_i\): \(\lambda (u_{k,k})_{,i} = \lambda u_{k,ki}\). * Term 2: \((\mu u_{i,j})_{,j} = \mu u_{i,jj}\). * Term 3: \((\mu u_{j,i})_{,j} = \mu u_{j,ij}\). Assuming the displacement field is sufficiently smooth, we can swap the order of differentiation: \(\mu u_{j,ji} = \mu (u_{j,j})_{,i}\).

Combining these terms gives: \[ \lambda u_{k,ki} + \mu u_{i,jj} + \mu u_{j,ji} + \rho b_i = \rho \ddot{u}_i \] Notice that \(u_{k,k}\) and \(u_{j,j}\) both represent the divergence of the displacement field. We can group the first and third terms: \[ \bbox[5px,border:1px #f2f2f2;background-color:#f2f2f2]{(\lambda + \mu) u_{j,ji} + \mu u_{i,jj} + \rho b_i = \rho \ddot{u}_i}\tag{5} \]

This is the Lamé-Navier equation in index notation.

Alternative form:

Now, take the divergence of the stress expression (4’) and substitute it into the equilibrium equation (1’): \[ \nabla \cdot \left[ \lambda (\nabla \cdot \mathbf{u}) \mathbf{I} + \mu (\nabla \mathbf{u} + (\nabla \mathbf{u})^T) \right] + \rho \mathbf{b} = \rho \ddot{\mathbf{u}} \] We use the following standard vector calculus identities: * \(\nabla \cdot (f \mathbf{I}) = \nabla f\) * \(\nabla \cdot (\nabla \mathbf{u}) = \nabla^2 \mathbf{u}\) (the vector Laplacian) * \(\nabla \cdot ((\nabla \mathbf{u})^T) = \nabla(\nabla \cdot \mathbf{u})\)

Applying these identities to our equation (assuming constant \(\lambda\) and \(\mu\)): \[ \lambda \nabla(\nabla \cdot \mathbf{u}) + \mu (\nabla^2 \mathbf{u}) + \mu \nabla(\nabla \cdot \mathbf{u}) + \rho \mathbf{b} = \rho \ddot{\mathbf{u}} \] Finally, group the terms with the gradient of the divergence: \[ \bbox[5px,border:1px #f2f2f2;background-color:#f2f2f2]{(\lambda + \mu) \nabla(\nabla \cdot \mathbf{u}) + \mu \nabla^2 \mathbf{u} + \rho \mathbf{b} = \rho \ddot{\mathbf{u}} }\tag{5'}\]

Boundary Conditions in Terms of Displacement

The boundary conditions that we discussed in the previous section \[ n_j\sigma_{ij}=t_i^\quad\text{on }\Gamma_\sigma \] \[ u_i=u_i^* \quad \text{on }\Gamma_u \] can be expressed in terms of the displacement filed as \[ \left[n_j \mu (u_{i,j}+u_{j,i})+n_i \lambda u_{k,k}\right]=t_i^*\quad\text{on }\Gamma_\sigma \] \[ u_i=u_i^*\quad\text{on }\Gamma_u. \]