Resources for Structural Engineers and Engineering Students

>>When you're done reading this section, check your understanding with the interactive quiz at the bottom of the page.

A constitutive model is the link between the stresses and strains of a structure. Of course, stresses are related to forces and strains are related to deformations. If we apply a certain set of forces to a structure, how will it elongate, compress, bend or move? It depends on the shape of the structure, but it also depends on what the structure is made of. Different materials deform differently depending on their material properties and the direction that they are loaded. A constitutive model describes the behaviour of an individual material under load.

The simplest type of constitutive relationship is for the linear axial deformation of a uniform bar:

\begin{align} \tag{1}\sigma &= E \epsilon \label{eq:Hooke} \end{align}

where $\sigma$ is stress, $E$ is the Young's Modulus, and $\epsilon$ is strain. In this relationship, $E$ is the constitutive model that links axial stress $\sigma$ (related to force) and the axial strain $\epsilon$. Of course, things get considerably more complex if we have to consider more than one direction of deformation, for example, if you have a beam (which can bend as well as deform axially), or if you have a plate or block that deforms. For these systems, it is typically necessary to represent the constitutive model as a matrix of values instead of using the single value $E$. For example, for a isotropic thin plate (i.e. the material behaves the same in all directions where there is no stress out of plane), the constitutive relationship may look like this:

\begin{equation}\tag{2} \begin{Bmatrix} \sigma_{11} \\ \sigma_{22} \\ \sigma_{12} \end{Bmatrix} = \frac{E}{1-\nu^2} \begin{bmatrix} 1 & \nu & 0 \\[6pt] \nu & 1 & 0 \\[6pt] 0 & 0 & \dfrac{1-\nu}{2} \end{bmatrix} \begin{Bmatrix} \epsilon_{11} \\ \epsilon_{22} \\ 2\epsilon_{12} \end{Bmatrix} \end{equation}

where $\sigma_{11}$, $\sigma_{22}$, $\epsilon_{11}$, and $\epsilon_{22}$ are axial stresses and strains in two dimensions, $\sigma_{12}$ and $\epsilon_{12}$ are shear stresses and strains, and $\nu$ is Poisson's ratio for the material. As you would imagine, the situation is considerably more complex in 3D.

If we stick with a uniform bar for now and assume a constant cross-sectional area $A$ and a length of $L$, then we can determine the relationship between force ($P$) and deformation ($\Delta$) by combining the constitutive relation from equation \eqref{eq:Hooke} with the definitions for stress and strain. Recall that stress is equal to force divided by area and strain is equal to change in elongation ($\delta$) divided by initial length ($L$):

\begin{equation} \tag{3}\sigma = \frac{P}{A} \quad \text{,} \quad \epsilon = \frac{\delta}{L} \end{equation}


\begin{align}\tag{4} \frac{P}{A} &= E \left( \frac{\delta}{L} \right) \\ \delta &= \frac{PL}{EA} \label{eq:PLEA} \tag{5}\end{align}

The relationship in equation \eqref{eq:PLEA} can be rewritten in terms of a Hooke's Law relationship:

\begin{equation} \tag{6} P = k \delta \end{equation}


\begin{equation} \tag{7} \label{eq:Stiffness} k = \frac{EA}{L} \end{equation}

In these equations, $k$ is the stiffness, which is the relationship between force and deformation (or force and displacement). The stiffness is the amount of force that must be applied in order to get a single unit of deformation. Therefore, the units for stiffness are force per unit distance. For example, if the stiffness is 20 kN\textbackslash m then this means that 20 kN must be applied to deform the structure by a unit of deformation, which in this case is 1 m. It also means that it would take 40 kN to deform by 2 m. A structure that is more stiff takes more force to move a given distance. It is also possible to have rotational stiffness for a structure that can bend, which is the relationship between moment and rotation.

The inverse of stiffness is called flexibility. Flexibility is given in units of distance per unit force. For the example in equation \eqref{eq:Stiffness}, the flexibility would be:

\begin{align} \tag{8} f &= \frac{1}{k} \\ \tag{9} f &= \frac{L}{EA} \end{align}

A structure that is more flexible takes less force to move a given distance.

Interactive Quiz