Chapter 6
Strain and displacement
6.1 Displacement
The displacement field \(\v {u}(\v {r})\) describes how a point on our solid body moves during deformation. For a rigid body \(\v {u}\equiv 0\), but for a deformable body we obtain finite displacements during deformation. The point \(\v {r}\) then moves to the point \(\v {r}^\prime (\v {r}) = \v {r} + \v {u}(\v {r})\). The displacement field \(\v {u}(\v {r})=\v {r} - \v {r}^\prime (\v {r})\) is hence the difference of the deformed (“displaced”) point to its reference position \(\v {r}\). Note that the displacement field itself is defined as a function of this reference position \(\v {r}\).
6.2 Strain
The strain field in the small strain approximation is given by the symmetrized gradient of \(\v {u}(\v {r})\), \begin {equation} \label {eq:straingradient} \t {\varepsilon }(\v {r}) = \frac {1}{2} \left (\nabla \v {u} + (\nabla \v {u})^T\right ). \end {equation} The left hand side of Eq. \eqref{eq:straingradient} contains the gradient of a vector field, \(\nabla \v {u}\), which is a second rank tensor, \begin {equation} \nabla \v {u} = \begin {pmatrix} \frac {\partial u_x}{\partial x} & \frac {\partial u_x}{\partial y} & \frac {\partial u_x}{\partial z} \\ \frac {\partial u_y}{\partial x} & \frac {\partial u_y}{\partial y} & \frac {\partial u_y}{\partial z} \\ \frac {\partial u_z}{\partial x} & \frac {\partial u_z}{\partial y} & \frac {\partial u_z}{\partial z} \end {pmatrix}, \end {equation} whose components are given by \([\nabla \v {u}]_{ij}=\partial u_i/\partial r_j = u_{i,j}\). It is not the divergence, \(\nabla \cdot \v {u}\) which would give a scalar. This potential source of confusion can be avoided by writing the equation in the component-wise notation, \begin {equation} \label {eq:straingradientcomp} \varepsilon _{ij} = \frac {1}{2} (\partial _i u_j + \partial _j u_i) = \frac {1}{2} (u_{j,i} + u_{i,j}). \end {equation} (Note that these expression do not contain a sum since there is no repeated index.)
The geometric interpretation of the strain is that is converts a vector \(\v {r}\) that has a direction and length into the change this vector undergoes under deformation: \(\v {u}=\t {\varepsilon }\cdot \v {r}\) with the new (transformed) vector \(\v {r}'=\v {r}+\v {u}=(1+\t {\varepsilon })\cdot \v {r}\). In terms of thinking about tensors as a representation of a linear transformation, the strain tensor transforms a position into a displacement. The strain tensor is dimensionless.