Latest revision as of 18:33, 4 December 2022

Numerical examples

The following pages describe several case studies in the field of elastostatics and elastodynamics. The subpages include the problem description (equations and boundary conditions), analytical solutions if obtainable and numerical results for given parameter values.

Basic equations of elasticity

To determine the distribution of static stresses and displacements in a solid body we must obtain a solution (either analytically or numerically) to the basic equations of the theory of elasticity, satisfying the boundary conditions on forces and/or displacements. The equations thus form a boundary value problem. For a general three dimensional solid object the equations governing its behavior are:

equations of equilibrium (3)
strain-displacement equations (6)
stress-strain equations (6)

where the number in brackets indicates the number of equations. The equations of equilibrium are three tensor partial differential equations for the balance of linear momentum, the strain-displacement equations are relations that stem from infinitesimal strain theory and the stress-strain equations are a set of linear algebraic constitutive relations (3D Hooke's law). In two dimensions these equations simplify to 8 equations (2 equilibrium, 3 strain-displacement, and 3 stress-strain).

A large amount of confusion regarding linear elasticity originates from the many different notations used in this field. For this reason we first provide an overview of the different notations that are used.

Direct tensor form

In direct tensor form (independent of coordinate system) the governing equations are:

Equation of motion (an expression of Newton's second law):

\[ \boldsymbol{\nabla} \cdot \boldsymbol{\sigma} + \boldsymbol{F} = \rho \ddot{\boldsymbol{u}} \]

Strain-displacement equations:

\[ \boldsymbol{\varepsilon} = \frac{1}{2}\left[ \boldsymbol{\nabla} \boldsymbol{u} + (\boldsymbol{\nabla} \boldsymbol{u})^T \right] \]

Stress-displacement equations (constitutive equations). For a linear elastic material this is Hooke's law:

\[ \boldsymbol{\sigma} = \boldsymbol{C} : \boldsymbol{\varepsilon} \]

where $\boldsymbol{\sigma}$ is the Cauchy stress tensor, $\boldsymbol{\varepsilon}$ is the infinitesimal strain tensor, $\boldsymbol{u}$ is the displacement vector, $\boldsymbol{C}$ is the fourth order stiffness tensor, $\boldsymbol{F}$ is the body force per unit volume (a vector quantity), $\rho$ is the mass density, $\boldsymbol{\nabla}(\bullet)$ is the gradient operator, $\boldsymbol{\nabla}\cdot(\bullet)$ is the divergence operator, $(\bullet)^T$ represents a transpose, $\ddot{(\bullet)}$ is the second derivative with respect to time, and $\boldsymbol{A}:\boldsymbol{B}$ is the inner product of two second order tensors (a tensor contraction).

Cartesian coordinate form

Using the Einstein summation convention (implied summations over repeated indexes) the equations are:

Equation of motion (an expression of Newton's second law):

\[ \sigma_{ji,j} + F_i = \rho \partial_{tt} u_i \]

Strain-displacement equations:

\[ \varepsilon_{ij} = \frac{1}{2}\left(u_{j,i}+u_{i,j}\right) \]

Stress-displacement equations (constitutive equations). For a linear elastic material this is Hooke's law:

\[ \sigma_{i,j} = C_{ijkl} \varepsilon_{kl}, \] where $i,j = 1,2,3$ represent, respectively, $x$, $y$ and $z$, the $(\bullet),j$ subscript is a shorthand for partial derivative $\partial(\bullet)/\partial x_j$ and $\partial_{tt}$ is shorthand notation for $\partial^2/\partial t^2$, $\sigma_{ij}=\sigma_{ji}$ is the Cauchy stress tensor (with 6 independent components), $F_i$ are the body forces, $\rho$ is the mass density, $u_i$ is the displacement, $\varepsilon_{ij} = \varepsilon_{ji}$ is the strain tensor (also with 6 independent components), and, finally $C_{ijkl}$ is the fourth-order stiffness tensor that due to symmetry requirements $C_{ijkl} = C_{klij} = C_{jikl} =C_{ijkl}$ can be reduced to 21 different elements.

Matrix-vector (FEM) notation

Equation of motion:

\[ \boldsymbol{L}^T\boldsymbol{\sigma} + \boldsymbol{F} = \rho\ddot{\boldsymbol{u}} \]

Strain-displacement equations:

\[ \boldsymbol{\varepsilon} = \boldsymbol{L}\boldsymbol{u} \]

Stress-displacement equations (constitutive equations). For a linear elastic material this is Hooke's law:

\[ \boldsymbol{\sigma} = \boldsymbol{C}\boldsymbol{\varepsilon} \] where $\boldsymbol{\sigma}$ is the Cauchy stress tensor represented in vector form (6 components), $\boldsymbol{L}$ is a differential operator matrix (size $3 \times 6$), $\bullet^T$ is transpose of a matrix $\bullet$, $\boldsymbol{F}$ is the body force vector, $\rho$ is the mass density, $\boldsymbol{u}$ is the displacement vector, $\boldsymbol{\varepsilon}$ is the strain tensor represented in vector form (6 components), and $\boldsymbol{C}$ is the symmetric stress-strain matrix (size $6 \times 6$ with 21 material constants $C_{ij} = C_{ji}$). Certain literature prefers using the symbol $\boldsymbol{D}$ instead of $\boldsymbol{C}$ for the stress-strain matrix. The symbol $C$ is then used as the "compliance tensor" that relates the strains to the stresses, e.g. $\boldsymbol{\varepsilon} = \boldsymbol{C}\boldsymbol{\sigma}$.

To solve the basic equations of elasticity two approaches exist according to the boundary conditions of the boundary value problem. In the displacement formulation the displacements are prescribed everywhere at the boundaries and the stresses and strains are eliminated from the formulation. The other possible option is that the surface tractions are prescribed everywhere on the surface boundary. The equations of elasticity are then manipulated to leave the stresses as the unknown to be solved for. This approach is known as the stress formulation and will not be considered here further.

Displacement formulation

The goal of the displacement formulation is to eliminate the strains and stresses from the basic equations of elasticity and leave only the displacements as the unknown to be solved for. The first step is to substitute the strain-displacement equations into Hooke's law, eliminating the strains as unknowns: \begin{equation}\label{eq:3d_hooke} \sigma_{ij} = \lambda\delta_{ij}\varepsilon_{kk} + 2\mu\varepsilon_{ij} \quad \Rightarrow \quad \begin{cases} \sigma_{xx} = \lambda (u_{x,x}+u_{y,y}+u_{z,z}) + 2\mu u_{x,x}\\ \sigma_{yy} = \lambda (u_{x,x}+u_{y,y}+u_{z,z}) + 2\mu u_{y,y}\\ \sigma_{zz} = \lambda (u_{x,x}+u_{y,y}+u_{z,z}) + 2\mu u_{z,z}\\ \sigma_{xy} = \mu (u_{x,y} + u_{y,x})\\ \sigma_{yz} = \mu (u_{y,z} + u_{z,y})\\ \sigma_{zx} = \mu (u_{z,x} + u_{x,z}) \end{cases} \end{equation} where $\lambda$ and $\mu$ are Lamé parameters. The formula above can also be written more concisely as \[ \sigma_{ij} = \lambda \delta_{ij} u_{k,k} + \mu(u_{i,j}+u_{j,i}). \] The next step is to substitute these equations into the equilibrium equation \[ \sigma_{ij,j} + F_i = \rho \partial_{tt} u_i \quad \Rightarrow \quad \begin{cases} \frac{\partial \sigma_{xx}}{\partial x} + \frac{\partial \sigma_{xy}}{\partial y} + \frac{\partial \sigma_{xz}}{\partial z} + F_x = \rho \frac{\partial^2 u_x}{\partial t^2} \\ \frac{\partial \sigma_{yx}}{\partial x} + \frac{\partial \sigma_{yy}}{\partial y} + \frac{\partial \sigma_{yz}}{\partial z} + F_y = \rho \frac{\partial^2 u_y}{\partial t^2} \\ \frac{\partial \sigma_{zx}}{\partial x} + \frac{\partial \sigma_{zy}}{\partial y} + \frac{\partial \sigma_{zz}}{\partial z} + F_z = \rho \frac{\partial^2 u_z}{\partial t^2} \end{cases} \] resulting in \begin{align*} \frac{\partial}{\partial x}\left( \lambda \left( \frac{\partial u_x}{\partial x} + \frac{\partial u_y}{\partial y} + \frac{\partial u_z}{\partial z} \right) + 2\mu \frac{\partial u_x}{\partial x} \right) + \frac{\partial}{\partial y}\left(\mu\left(\frac{\partial u_x}{\partial y}+\frac{\partial u_y}{\partial x}\right)\right) + \frac{\partial}{\partial z}\left(\mu\left(\frac{\partial u_x}{\partial z}+\frac{\partial u_z}{\partial x}\right)\right) + F_x &= \rho \frac{\partial^2 u_x}{\partial t^2} \\ \frac{\partial}{\partial x}\left(\mu\left(\frac{\partial u_y}{\partial x}+\frac{\partial u_x}{\partial y}\right)\right) + \frac{\partial}{\partial y}\left(\lambda \left( \frac{\partial u_x}{\partial x} + \frac{\partial u_y}{\partial y} + \frac{\partial u_z}{\partial z} \right) + 2\mu \frac{\partial u_y}{\partial y}\right) + \frac{\partial}{\partial z}\left(\mu\left(\frac{\partial u_y}{\partial z}+\frac{\partial u_z}{\partial y}\right)\right) + F_y &= \rho \frac{\partial^2 u_y}{\partial t^2}\\ \frac{\partial}{\partial x}\left(\mu\left(\frac{\partial u_z}{\partial x}+\frac{\partial u_x}{\partial z}\right)\right) + \frac{\partial}{\partial y}\left(\mu\left(\frac{\partial u_z}{\partial y}+\frac{\partial u_y}{\partial z}\right)\right) + \frac{\partial}{\partial z}\left(\lambda \left( \frac{\partial u_x}{\partial x} + \frac{\partial u_y}{\partial y} + \frac{\partial u_z}{\partial z} \right) + 2\mu \frac{\partial u_z}{\partial z}\right) + F_z &= \rho \frac{\partial^2 u_z}{\partial t^2} \end{align*} Using the assumption that Lamé parameters $\lambda$ and $\mu$ are constant we can rearrange to produce: \begin{align*} (\lambda + \mu) \frac{\partial}{\partial x}\left(\frac{\partial u_x}{\partial x}+\frac{\partial u_y}{\partial y}+\frac{\partial u_z}{\partial z}\right) + \mu\left(\frac{\partial^2 u_x}{\partial x^2}+\frac{\partial^2 u_x}{\partial y^2}+\frac{\partial^2 u_x}{\partial z^2}\right) + F_x &= \rho \frac{\partial^2 u_x}{\partial t^2} \\ (\lambda + \mu) \frac{\partial}{\partial y}\left(\frac{\partial u_x}{\partial x}+\frac{\partial u_y}{\partial y}+\frac{\partial u_z}{\partial z}\right) + \mu\left(\frac{\partial^2 u_y}{\partial x^2}+\frac{\partial^2 u_y}{\partial y^2}+\frac{\partial^2 u_y}{\partial z^2}\right) + F_y &= \rho \frac{\partial^2 u_y}{\partial t^2} \\ (\lambda + \mu) \frac{\partial}{\partial z}\left(\frac{\partial u_x}{\partial x}+\frac{\partial u_y}{\partial y}+\frac{\partial u_z}{\partial z}\right) + \mu\left(\frac{\partial^2 u_z}{\partial x^2}+\frac{\partial^2 u_z}{\partial y^2}+\frac{\partial^2 u_z}{\partial z^2}\right) + F_z &= \rho \frac{\partial^2 u_z}{\partial t^2} \\ \end{align*}

We can easily see that only the displacements are left in these equations. The equations obtaines in this manner are known as the Navier-Cauchy equations.

Navier-Cauchy equations

The Navier or Navier-Cauchy equations describe the dynamics of a solid through the displacement vector field $\b{u}$. The equations can also be expressed concisely in vector form as follows \begin{equation}\label{eq:navier_mu_lambda} \rho \frac{\partial^2 \b{u}}{\partial t^2} = (\lambda + \mu) \nabla(\nabla \cdot \b{u}) + \mu \nabla^2 \b{u} + \b{F} \end{equation} where $\mu$ and $\lambda$ are Lamé constants, $\rho$ is the object density and $\b{F}$ are the external forces. In certain cases we may prefer to use the Young modulus $E$ and the Poisson ratio $\nu$ instead of the Lamé constants. In this case the Navier equations are \begin{equation}\label{eq:navier_young_poisson} \rho \frac{\partial^2 \b{u}}{\partial t^2} = \frac{E}{2(1+\nu)}\left(\nabla^2 \b{u} + \frac{1}{1-2\nu}\nabla\left(\nabla \cdot \b{u}\right)\right) + \b{F} \end{equation}

Two-dimensional stress distributions

Many problems in elasticity can be simplified as two-dimensional problems described by plane theory of elasticity. In general there are two types of problems we may encounter in plane analysis: plane stress and plane strain. The first problem arises in analysis of thin plates loaded in the plane of the plate, while the second is used for elongated bodies of constant cross section subject to uniform loading.

Plane stress

Plane stress distributions build on the assumption that the normal stress and shear stresses directed perpendicular to the $x$-$y$ plane are assumed zero: \begin{equation}\label{eq:pstress_assump} \sigma_{zz} = \sigma_{zx} = \sigma_{zy} = 0. \end{equation} It is also assumed that the stress components do not vary through the thickness of the plate (the assumptions do violate some compatibility conditions, but are still sufficiently accurate for practical applications if the plate is thin).

Using (\ref{eq:pstress_assump}) the three-dimensional Hooke's law can be reduced to: \begin{equation}\label{eq:planestress} \boldsymbol{\sigma} = \boldsymbol{C}\boldsymbol{\varepsilon} \end{equation} in matrix form where for isotropic materials, we have \begin{equation}\label{eq:planestressmatrix} \boldsymbol{C} = \frac{E}{1-\nu^2} \begin{bmatrix} 1 & \nu & 0 \\ \nu & 1 & 0 \\ 0 & 0 & \frac{1-\nu}{2} \end{bmatrix} \qquad \text{(Plane stress)} \end{equation} and \begin{equation}\label{eq:2d_stress} \b{\sigma} = \begin{bmatrix} \sigma_{xx} \\ \sigma_{yy} \\ \sigma_{xy}\end{bmatrix}, \end{equation} \begin{equation}\label{eq:2d_strain} \b{\varepsilon} = \begin{bmatrix} \varepsilon_{xx} \\ \varepsilon_{yy} \\ \gamma_{xy} \end{bmatrix} = \begin{bmatrix} \varepsilon_{xx} \\ \varepsilon_{yy} \\ 2\varepsilon_{xy} \end{bmatrix} = \begin{bmatrix} \frac{\partial u_x}{\partial x} \\ \frac{\partial u_y}{\partial y} \\ \frac{\partial u_y}{\partial x} + \frac{\partial u_x}{\partial y} \end{bmatrix}. \end{equation} Note that the strain tensor uses the so-called engineering shear strain $\gamma_{ij}$. Under the plane stress assumption the Navier equations are given as: \begin{equation}\label{eq:navier_plane_stress} \rho \frac{\partial^2 \b{u}}{\partial t^2} = \frac{E}{2(1+\nu)} \left( \b{\nabla}^2 \b{u} + \frac{1+\nu}{1-\nu}\b{\nabla}\left( \b{\nabla}\cdot \b{u}\right) \right) + \b{F} \end{equation} or in component notation \begin{align} \rho \frac{\partial^2 u_x}{\partial t^2} &= \frac{E}{2(1+\nu)} \left( \frac{\partial^2 u_x}{\partial x^2} + \frac{\partial^2 u_x}{\partial y^2} \right) + \frac{E}{2(1-\nu)} \frac{\partial}{\partial x}\left( \frac{\partial u_x}{\partial x} + \frac{\partial u_y}{\partial y} \right) \\ \rho \frac{\partial^2 u_y}{\partial t^2} &= \frac{E}{2(1+\nu)} \left( \frac{\partial^2 u_y}{\partial x^2} + \frac{\partial^2 u_y}{\partial y^2} \right) + \frac{E}{2(1-\nu)} \frac{\partial}{\partial y}\left( \frac{\partial u_x}{\partial x} + \frac{\partial u_y}{\partial y} \right) \end{align}

Plane stress with Lamé constants

The plane stress stiffness tensor (\ref{eq:planestressmatrix}) may also be expressed in terms of Lamé constants $\lambda$ and $\mu$ by substituting $E$ and $\nu$ with \[E = \frac{\mu(3\lambda+2\mu)}{\lambda+\mu}\] \[\nu = \frac{\lambda}{2(\lambda+\mu)}\]

With these substitutions the Navier equations can be written in their component form as \begin{align} \rho \frac{\partial^2 u_x}{\partial t^2} &= \mu \left( \frac{\partial^2 u_x}{\partial x^2} + \frac{\partial^2 u_x}{\partial y^2} \right) + \left(\frac{2 \lambda \mu}{\lambda + 2 \mu}+\mu\right) \frac{\partial}{\partial x}\left( \frac{\partial u_x}{\partial x} + \frac{\partial u_y}{\partial y} \right) \\ \rho \frac{\partial^2 u_y}{\partial t^2} &= \mu \left( \frac{\partial^2 u_y}{\partial x^2} + \frac{\partial^2 u_y}{\partial y^2} \right) + \left(\frac{2 \lambda \mu}{\lambda + 2 \mu}+\mu\right) \frac{\partial}{\partial y}\left( \frac{\partial u_x}{\partial x} + \frac{\partial u_y}{\partial y} \right) \end{align}

Plane strain

The plane strain problem arises in analysis of walls, dams, tunnels where one dimension of the structure is very large in comparison to the other two dimensions ($x$- and $y$- coordinates). It is also appropriate for small-scale problems such as bars and rollers compressed by forces normal to their cross section. In all such problems the body may be imagined as a prismatic cylinder with one dimension much larger that the other two. The applied forces act in the $x$-$y$ plane and do not vary in the $z$ direction, leading to the assumption \begin{equation} \frac{\partial}{\partial z} = u_z = 0. \end{equation}

With the above assumption it follows immediately that \begin{equation} \varepsilon_{zz} = \varepsilon_{zx} = \varepsilon_{zy} = 0. \end{equation} The three dimensional Hooke's law can now be reduced to \begin{equation}\label{eq:planestrain} \boldsymbol{\sigma} = \boldsymbol{C}\boldsymbol{\varepsilon} \end{equation} where the matrix $C$ is given by \begin{equation}\label{eq:matrixplanestrain} \boldsymbol{C} = \frac{E}{(1+\nu)(1-2\nu)} \begin{bmatrix} 1-\nu & \nu & 0 \\ \nu & 1-\nu & 0 \\ 0 & 0 & \frac{1-2\nu}{2} \end{bmatrix} \qquad \text{(Plane strain)} \end{equation} The vectors $\boldsymbol{\sigma}$ and $\boldsymbol{\varepsilon}$ are the same as above for plane stress and are given in (\ref{eq:2d_stress}) and (\ref{eq:2d_strain}), respectively. In the case of plane strain we have additonal non-zero components of the stress tensor: \begin{equation}\label{eq:sigmazz} \sigma_{zz} = \nu(\sigma_{xx}+\sigma_{yy}) \end{equation} \begin{equation} \sigma_{yz} = \sigma_{zx} = 0 \end{equation} The reason $\sigma_{zz}$ is not included in the matrix stress-strain equation (\ref{eq:planestrain}) is because it is linearly dependent on the normal stresses $\sigma_{xx}$ and $\sigma_{zz}$.

The Navier-Cauchy equation that follows from the plane strain assumption can be found in (\ref{eq:navier_young_poisson}). For purposes of completeness we provide here the equations in component notation: \begin{align} \rho \frac{\partial^2 u_x}{\partial t^2} &= \frac{E}{2(1+\nu)} \left( \frac{\partial^2 u_x}{\partial x^2} + \frac{\partial^2 u_x}{\partial y^2} \right) + \frac{E}{2(1+\nu)(1-2\nu)} \frac{\partial}{\partial x}\left( \frac{\partial u_x}{\partial x} + \frac{\partial u_y}{\partial y} \right) \\ \rho \frac{\partial^2 u_y}{\partial t^2} &= \frac{E}{2(1+\nu)} \left( \frac{\partial^2 u_y}{\partial x^2} + \frac{\partial^2 u_y}{\partial y^2} \right) + \frac{E}{2(1+\nu)(1-2\nu)} \frac{\partial}{\partial y}\left( \frac{\partial u_x}{\partial x} + \frac{\partial u_y}{\partial y} \right) \end{align}

Plane strain with Lamé constants

Alternatively the stiffness tensor $\b{C}$ in the stress-strain equation $\b{\sigma} = \b{C}\b{\varepsilon}$ may be expressed in terms of $\mu$ and $\lambda$: \begin{equation} \boldsymbol{C} = \begin{bmatrix} 2\mu + \lambda & \lambda & 0 \\ \lambda & 2\mu + \lambda & 0 \\ 0 & 0 & \mu^* \end{bmatrix} \qquad \text{(Plane strain)} \end{equation} (*Note that this matrix requires the strain tensor to be defined using the engineering shear strain $\gamma_{xy} = 2\varepsilon_{xy}$. If we define the strain tensor as $\b{\varepsilon} = \{\varepsilon_{xx}\;\varepsilon_{yy}\;\varepsilon_{xy}\}^T$ then matrix component $C_{33}$ must be $2\mu$.)

The Navier equations in vector form can be found in (\ref{eq:navier_mu_lambda}). In component notation the equations are: \begin{align} \rho \frac{\partial^2 u_x}{\partial t^2} &= \mu \left( \frac{\partial^2 u_x}{\partial x^2} + \frac{\partial^2 u_x}{\partial y^2} \right) + (\lambda+\mu) \frac{\partial}{\partial x}\left( \frac{\partial u_x}{\partial x} + \frac{\partial u_y}{\partial y} \right) \\ \rho \frac{\partial^2 u_y}{\partial t^2} &= \mu \left( \frac{\partial^2 u_y}{\partial x^2} + \frac{\partial^2 u_y}{\partial y^2} \right) + (\lambda+\mu) \frac{\partial}{\partial y}\left( \frac{\partial u_x}{\partial x} + \frac{\partial u_y}{\partial y} \right) \end{align}

Connection between plane stress and plane strain

For isotropic materials with elastic modulus $E$ and Poisson's ratio $\nu$ it is possible to go from plane stress to plane strain, or vice-versa, by replacing $E$ and $\nu$ in the stress-strain matrix with a fictitious modulus $E^*$ and fictitious Poisson ratio $\nu^*$. This allows us to "reuse" a plane stress program to solve plane strain or again vice-versa (as long as the material is isotropic). A few exercises on this topic are given at this link (page 13).

To go from plane stress ($s$) to plane strain ($n$) insert the fictitious quantities \begin{equation}\label{eq:ston1} E_n^* = \frac{E_s}{1-\nu_s^2}, \end{equation} \begin{equation}\label{eq:ston2} \nu_n^* = \frac{\nu_s}{1-\nu_s}. \end{equation}

Substitution from plane stress to plane strain

Note: this derivation is shown just to confirm the above formulas. In a numerical program, we keep the stress-strain matrix and just use the above formulas (\ref{eq:ston1}) and (\ref{eq:ston2}) to update the values of $E$ and $\nu$. Also note we have omitted the indexes ($s$) and ($n$) in the following derivations.

We start with the plane stress matrix with the inserted fictitious values $E^*$ and $\nu^*$ \[ \frac{E^*}{1-{\nu^*}^2} \begin{bmatrix} 1 & \nu^* & 0 \\ \nu^* & 1 & 0 \\ 0 & 0 & \frac{1}{2}(1-\nu^*) \end{bmatrix} \] and make the above substitutions (\ref{eq:ston1}) and (\ref{eq:ston2}) leading to: \[ \frac{E}{\left(1-\nu^2\right)\left(1-\left(\frac{\nu}{1-\nu}\right)^2\right)} \begin{bmatrix} 1 & \frac{\nu}{1-\nu} & 0 \\ \frac{\nu}{1-\nu} & 1 & 0 \\ 0 & 0 & \frac{1}{2}(1-\frac{\nu}{1-\nu}) \end{bmatrix}. \] We can then use the rule to convert sums of squares into products as well as bring the factor $1/(1-v)$ out of the matrix: \[ \frac{E}{\left(1-\nu\right)\left(1+\nu\right)\left(1-\frac{\nu}{1-\nu}\right)\left(1+\frac{\nu}{1-\nu}\right)\left(1-\nu\right)} \begin{bmatrix} 1-\nu & \nu & 0 \\ \nu & 1-\nu & 0 \\ 0 & 0 & \frac{1}{2}(1-2\nu) \end{bmatrix}. \] By joining some of the factors to a common denominator and rearranging leads to \[ \frac{E\left(1-\nu\right)\left(1-\nu\right)}{\left(1-\nu\right)\left(1+\nu\right)\left(1-2\nu\right)1\left(1-\nu\right)} \begin{bmatrix} 1-\nu & \nu & 0 \\ \nu & 1-\nu & 0 \\ 0 & 0 & \frac{1}{2}(1-2\nu) \end{bmatrix}. \] The final step is canceling the factors that occur in both the numerator and denominator \[ \frac{E}{\left(1+\nu\right)\left(1-2\nu\right)} \begin{bmatrix} 1-\nu & \nu & 0 \\ \nu & 1-\nu & 0 \\ 0 & 0 & \frac{1}{2}(1-2\nu) \end{bmatrix} \] which leads exactly to the relationship for plane strain given in (\ref{eq:planestrain}).

In the opposite case we want to go from plane strain ($n$) to plane stress ($s$) we can use: \begin{equation}\label{eq:ntos1} E_s^* = \frac{E_n(1+2\nu_n)}{(1+\nu_n)^2}, \end{equation} \begin{equation}\label{eq:ntos2} \nu_s^* = \frac{\nu_n}{1+\nu_n}. \end{equation}

Substitution from plane strain to plane stress

Note that we have again omitted the indexes ($s$) and ($n$) since our wish is expressed by the bold title.

We start with the plane strain matrix with the inserted fictitious values $E^*$ and $\nu^*$ \[ \frac{E^*}{\left(1+\nu^*\right)\left(1-2\nu^*\right)} \begin{bmatrix} 1-\nu^* & \nu^* & 0 \\ \nu^* & 1-\nu^* & 0 \\ 0 & 0 & \frac{1}{2}(1-2\nu^*) \end{bmatrix} \] and make the above substitutions (\ref{eq:ntos1}) and (\ref{eq:ntos2}) leading to: \[ \frac{E(1+2\nu)}{\left(1+\nu\right)^2\left(1+\frac{\nu}{1+\nu}\right)\left(1-2\frac{\nu}{1+\nu}\right)} \begin{bmatrix} 1-\frac{\nu}{1+\nu} & \frac{\nu}{1+\nu} & 0 \\ \frac{\nu}{1+\nu} & 1-\frac{\nu}{1+\nu} & 0 \\ 0 & 0 & \frac{1}{2}(1-2\frac{\nu}{1+\nu}) \end{bmatrix}. \] Writing some of the sums with common denominators and rearranging leads to \[ \frac{E(1+2\nu)(1+\nu)(1+\nu)}{\left(1+\nu\right)^2\left(1+2\nu\right)\left(1-\nu\right)} \begin{bmatrix} \frac{1}{1+\nu} & \frac{\nu}{1+\nu} & 0 \\ \frac{\nu}{1+\nu} & \frac{1}{1+\nu} & 0 \\ 0 & 0 & \frac{1}{2}\left(\frac{1-\nu}{1+\nu}\right) \end{bmatrix}. \] We can also bring the factor $1/(1+\nu)$ out from the matrix components. After canceling all factors that occur in both denominator and numerator we are left with \[ \frac{E}{\left(1+\nu\right)\left(1-\nu\right)} \begin{bmatrix} 1 & \nu & 0 \\ \nu & 1 & 0 \\ 0 & 0 & \frac{1}{2}(1-\nu) \end{bmatrix}. \] Rewriting the product of sums as a sum of squares gives us the matrix for plane stress in (\ref{eq:planestress}).

Principal stresses

This are the stresses when we rotate our coordinate system so that there are no shear components. The principal stresses for plane stress conditions are given as eigenvalues of $\sigma$ with \[ \sigma_{1,2} = \frac12 (s_{xx} + s_{zz}) \pm \sqrt{\frac{1}{4}(s_{xx}-s_{zz})^2 + s_{xz}^2}. \] The principal shear stress or principal stress difference is the stress where only shear stresses are present. In case of plane stress it is calculated as \[ \tau_1 = \sqrt{\frac{1}{4}(s_{xx}-s_{zz})^2 + s_{xz}^2} = \frac{|\sigma_1 - \sigma_2|}{2}. \]

When the stress tensor is z 3x3 matrix, the formulas for eigenvalues are more complicated.

Von Mises stress

A more detailed explanation of the von Mises stress can be found on Wikipedia.

In materials science and engineering the von Mises yield criterion can be also formulated in terms of the von Mises stress or equivalent tensile stress, $\sigma_\mathrm{v}$. This is a scalar value of stress that can be computed from the Cauchy stress tensor $\boldsymbol{\sigma}$. In this case, a material is said to start yielding when the von Mises stress reaches a value known as yield strength, $\sigma_\mathrm{y}$.

Mathematically the von Mises yield criterion can be expressed as: \[ \sigma _\mathrm{v}=\sigma_\mathrm{y}={\sqrt {3J_{2}}}, \] where $J_2$ is the second deviatoric stress invariant.

In the case of general plane stress ($\sigma_{33}=0$,$\sigma_{23}=0$, $\sigma_{31}=0$) the von Mises stress can be calculated as: \[ \sigma_\mathrm{v} = \sqrt{\sigma_{11}^2 - \sigma_{11}\sigma_{22}+\sigma_{22}^2 + 3\sigma_{12}^2}. \]

computational solid mechanics

The goal in computational solid mechanics is to solve the equation of motion for a solid body \[\rho \frac{\partial^2 \b{u}}{\partial t^2} = \nabla\cdot\b{\sigma} + \b{F}\]

on the given domain $\Gamma = \Gamma_N \cup \Gamma_D$ with Dirichlet (essential) boundary conditions \[ \b{u} = \bar{\b{u}} \quad \text{ on } \Gamma_D,\] and traction (natural) boundary conditions \[ \b{\sigma}\cdot\b{n} = \bar{\b{t}},\quad \text{ on } \Gamma_N,\] where the quantities with the bar sign $\bar{}$ indicate prescribed displacements and surface tractions, respectively. Another type of boundary condition are mixed (Robyn) type boundary conditions. In this case the displacement might be given in one direction, and the traction will be given for the other direction. In case the right-hand side value is zero, the boundary conditions are called homogeneous. Homogeneous Dirichlet boundary conditions mean the the body is fully fixed (constrained), while zero traction boundary conditions allow the surface to move freely.

Using Hooke's law and the strain-displacement relation, the equation of motion can be transformed into the Navier-Cauchy equations \[\rho \frac{\partial^2 \b{u}}{\partial t^2} = \mu\nabla^2\b{u} + (\lambda+\mu)\nabla(\nabla\cdot\b{u}) + \b{F}.\] The solution of this equation with appropriate BC's will give us the desired displacement values across the domain.

To obtain the steady state solution we have two possible strategies:

Add a (linear) dampening term to the Navier equation, and simulate the dynamic behaviour until the motion ceases: \[\rho \frac{\partial^2 \b{u}}{\partial t^2} + \eta_C \frac{\partial \b{u}}{\partial t} = \mu\nabla^2\b{u} + (\lambda+\mu)\nabla(\nabla\cdot\b{u}) + \b{F}.\]
Directly solve for steady state: \[\mu\nabla^2\b{u} + (\lambda+\mu)\nabla(\nabla\cdot\b{u}) + \b{F} = 0.\]

Displacement boundary conditions are very simple to implement with both approaches. The desired values are simply prescribed for dynamic simulation with explicit methods, In case of implicit methods where we need to solve a system of equations the prescribed displacements are placed into the right-hand side of the system, and the value 1 is placed in the diagonal of the matrix (for a given node).

For the traction boundary condition

\[\begin{bmatrix} \sigma_{xx} & \sigma_{xy} \\ \sigma_{yx} & \sigma_{yy} \end{bmatrix} \cdot \begin{bmatrix} n_x \\ n_y \end{bmatrix} = \begin{bmatrix} \bar{t}_x \\ \bar{t}_y \end{bmatrix}\]

we first use Hooke's law and the strain-displacement relations to express the traction boundary in terms of displacements:

\[\begin{bmatrix} (2\mu+\lambda)\partial_x u_x + \lambda \partial_y u_y & \mu(\partial_x u_y + \partial_y u_x) \\ \mu(\partial_x u_y + \partial_y u_x) & \lambda \partial_x u_x + (2\mu + \lambda) \partial_y u_y \end{bmatrix} \cdot \begin{bmatrix} n_x \\ n_y \end{bmatrix} = \begin{bmatrix} \bar{t}_x \\ \bar{t}_y \end{bmatrix}\]

References

Slak J., Kosec G.; Refined meshless local strong form solution of Cauchy-Navier equation on an irregular domain. Engineering analysis with boundary elements. 2018;11 ; manuscript
Theory of matrix structural analysis

This list might be incomplete, check also reference on the Main Page.

Difference between revisions of "Solid Mechanics"

Latest revision as of 18:33, 4 December 2022

Contents

Numerical examples

Basic equations of elasticity

Direct tensor form

Cartesian coordinate form

Matrix-vector (FEM) notation

Displacement formulation

Navier-Cauchy equations

Two-dimensional stress distributions

Plane stress

Plane stress with Lamé constants

Plane strain

Plane strain with Lamé constants

Connection between plane stress and plane strain

Principal stresses

Von Mises stress

computational solid mechanics

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@@ Line 1: / Line 1: @@
+= Numerical examples =
+The following pages describe several case studies in the field of elastostatics and elastodynamics. The subpages include the problem description (equations and boundary conditions), analytical solutions if obtainable and numerical results for given parameter values.
+*[[Point contact]]
+*[[Cantilever beam]]
+*[[Hertzian contact]]
+*[[FWO case | Fretting fatigue contact]]
 =Basic equations of elasticity=
-To determine the distribution of static stresses and displacements in a solid body we must obtain a solution (either analytically or numerically) to the basic equations of the theory of elasticity, satisfying the boundary conditions on forces and/or displacements. For a general three dimensional solid object these equations are:
+To determine the distribution of static stresses and displacements in a solid body we must obtain a solution (either analytically or numerically) to the basic equations of the theory of elasticity, satisfying the boundary conditions on forces and/or displacements. The equations thus form a boundary value problem. For a general three dimensional solid object the equations governing its behavior are:
+* equations of equilibrium (3)
 * strain-displacement equations (6)
 * stress-strain equations (6)
-* equations of equilibrium (3)
+where the number in brackets indicates the number of equations. The equations of equilibrium are three tensor partial differential equations for the balance of linear momentum, the strain-displacement equations are relations that stem from infinitesimal strain theory and the stress-strain equations are a set of linear algebraic constitutive relations (3D Hooke's law). In two dimensions these equations simplify to 8 equations (2 equilibrium, 3 strain-displacement, and 3 stress-strain).
-where the number in brackets indicates the number of equations. We see there are 15 equations for 15 unknown variables (6 strains, 6 stresses and 3 displacements). In two dimensions this simplifies to 8 equations with 2 displacements, 3 stresses, and 3 strains.
+A large amount of confusion regarding linear elasticity originates from the many different notations used in this field. For this reason we first provide an overview of the different notations that are used.
+=== Direct tensor form ===
+In direct tensor form (independent of coordinate system) the governing equations are:
+*Equation of motion (an expression of Newton's second law):
+\[
+\boldsymbol{\nabla} \cdot \boldsymbol{\sigma} + \boldsymbol{F} = \rho \ddot{\boldsymbol{u}}
+\]
+*Strain-displacement equations:
+\[
+\boldsymbol{\varepsilon} = \frac{1}{2}\left[ \boldsymbol{\nabla} \boldsymbol{u} + (\boldsymbol{\nabla} \boldsymbol{u})^T  \right]
+\]
+*Stress-displacement equations (constitutive equations). For a linear elastic material this is Hooke's law:
+\[
+\boldsymbol{\sigma} = \boldsymbol{C} : \boldsymbol{\varepsilon}
+\]
+where $\boldsymbol{\sigma}$ is the Cauchy stress tensor, $\boldsymbol{\varepsilon}$ is the infinitesimal strain tensor, $\boldsymbol{u}$ is the displacement vector, $\boldsymbol{C}$ is the fourth order stiffness tensor, $\boldsymbol{F}$ is the body force per unit volume (a vector quantity), $\rho$ is the mass density, $\boldsymbol{\nabla}(\bullet)$ is the gradient operator, $\boldsymbol{\nabla}\cdot(\bullet)$ is the divergence operator, $(\bullet)^T$ represents a transpose, $\ddot{(\bullet)}$ is the second derivative with respect to time, and $\boldsymbol{A}:\boldsymbol{B}$ is the inner product of two second order tensors (a tensor contraction).
+=== Cartesian coordinate form ===
+Using the Einstein summation convention (implied summations over repeated indexes) the equations are:
+*Equation of motion (an expression of Newton's second law):
+\[
+\sigma_{ji,j} + F_i = \rho \partial_{tt} u_i
+\]
+*Strain-displacement equations:
+\[
+\varepsilon_{ij} = \frac{1}{2}\left(u_{j,i}+u_{i,j}\right)
+\]
+*Stress-displacement equations (constitutive equations). For a linear elastic material this is Hooke's law:
+\[
+\sigma_{i,j} = C_{ijkl} \varepsilon_{kl},
+\]
+where $i,j = 1,2,3$ represent, respectively, $x$, $y$ and $z$, the $(\bullet),j$ subscript is a shorthand for partial derivative $\partial(\bullet)/\partial x_j$ and $\partial_{tt}$ is shorthand notation for $\partial^2/\partial t^2$, $\sigma_{ij}=\sigma_{ji}$ is the Cauchy stress tensor (with 6 independent components), $F_i$ are the body forces, $\rho$ is the mass density, $u_i$ is the displacement, $\varepsilon_{ij} = \varepsilon_{ji}$ is the strain tensor (also with 6 independent components), and, finally $C_{ijkl}$ is the fourth-order stiffness tensor that due to symmetry requirements $C_{ijkl} = C_{klij} = C_{jikl} =C_{ijkl}$ can be reduced to 21 different elements.
+=== Matrix-vector (FEM) notation ===
+*Equation of motion:
+\[
+\boldsymbol{L}^T\boldsymbol{\sigma} + \boldsymbol{F} = \rho\ddot{\boldsymbol{u}}
+\]
+*Strain-displacement equations:
+\[
+\boldsymbol{\varepsilon} = \boldsymbol{L}\boldsymbol{u}
+\]
+*Stress-displacement equations (constitutive equations). For a linear elastic material this is Hooke's law:
+\[
+\boldsymbol{\sigma} = \boldsymbol{C}\boldsymbol{\varepsilon}
+\]
+where $\boldsymbol{\sigma}$ is the Cauchy stress tensor represented in vector form (6 components), $\boldsymbol{L}$ is a differential operator matrix (size $3 \times 6$), $\bullet^T$ is transpose of a matrix $\bullet$, $\boldsymbol{F}$ is the body force vector, $\rho$ is the mass density, $\boldsymbol{u}$ is the displacement vector, $\boldsymbol{\varepsilon}$ is the strain tensor represented in vector form (6 components), and $\boldsymbol{C}$ is the symmetric stress-strain matrix (size $6 \times 6$ with 21 material constants $C_{ij} = C_{ji}$). Certain literature prefers using the symbol $\boldsymbol{D}$ instead of $\boldsymbol{C}$ for the stress-strain matrix. The symbol $C$ is then used as the "compliance tensor" that relates the strains to the stresses, e.g. $\boldsymbol{\varepsilon} = \boldsymbol{C}\boldsymbol{\sigma}$.
+To solve the basic equations of elasticity two approaches exist according to the boundary conditions of the boundary value problem. In the '''displacement formulation''' the displacements are prescribed everywhere at the boundaries and the stresses and strains are eliminated from the formulation. The other possible option is that the surface tractions are prescribed everywhere on the surface boundary. The equations of elasticity are then manipulated to leave the stresses as the unknown to be solved for. This approach is known as the [https://en.wikipedia.org/wiki/Linear_elasticity#Stress_formulation '''stress formulation'''] and will not be considered here further.
+== Displacement formulation ==
+The goal of the displacement formulation is to eliminate the strains and stresses from the basic equations of elasticity and leave only the displacements as the unknown to be solved for. The first step is to substitute the strain-displacement equations into Hooke's law, eliminating the strains as unknowns:
+\begin{equation}\label{eq:3d_hooke}
+\sigma_{ij} = \lambda\delta_{ij}\varepsilon_{kk} + 2\mu\varepsilon_{ij} \quad \Rightarrow \quad
+\begin{cases}
+\sigma_{xx} = \lambda (u_{x,x}+u_{y,y}+u_{z,z}) + 2\mu u_{x,x}\\
+\sigma_{yy} = \lambda (u_{x,x}+u_{y,y}+u_{z,z}) + 2\mu u_{y,y}\\
+\sigma_{zz} = \lambda (u_{x,x}+u_{y,y}+u_{z,z}) + 2\mu u_{z,z}\\
+\sigma_{xy} = \mu (u_{x,y} + u_{y,x})\\
+\sigma_{yz} = \mu (u_{y,z} + u_{z,y})\\
+\sigma_{zx} = \mu (u_{z,x} + u_{x,z})
+\end{cases}
+\end{equation}
+where $\lambda$ and $\mu$ are [https://en.wikipedia.org/wiki/Lam%C3%A9_parameters Lamé parameters]. The formula above can also be written more concisely as
+\[
+\sigma_{ij} = \lambda \delta_{ij} u_{k,k} + \mu(u_{i,j}+u_{j,i}).
+\]
+The next step is to substitute these equations into the equilibrium equation
+\[
+\sigma_{ij,j} + F_i = \rho \partial_{tt} u_i \quad \Rightarrow \quad
+\begin{cases}
+\frac{\partial \sigma_{xx}}{\partial x} + \frac{\partial \sigma_{xy}}{\partial y} + \frac{\partial \sigma_{xz}}{\partial z} + F_x = \rho \frac{\partial^2 u_x}{\partial t^2} \\
+\frac{\partial \sigma_{yx}}{\partial x} + \frac{\partial \sigma_{yy}}{\partial y} + \frac{\partial \sigma_{yz}}{\partial z} + F_y = \rho \frac{\partial^2 u_y}{\partial t^2} \\
+\frac{\partial \sigma_{zx}}{\partial x} + \frac{\partial \sigma_{zy}}{\partial y} + \frac{\partial \sigma_{zz}}{\partial z} + F_z = \rho \frac{\partial^2 u_z}{\partial t^2}
+\end{cases}
+\]
+resulting in
+\begin{align*}
+\frac{\partial}{\partial x}\left( \lambda \left( \frac{\partial u_x}{\partial x} + \frac{\partial u_y}{\partial y} + \frac{\partial u_z}{\partial z} \right) + 2\mu \frac{\partial u_x}{\partial x} \right) + \frac{\partial}{\partial y}\left(\mu\left(\frac{\partial u_x}{\partial y}+\frac{\partial u_y}{\partial x}\right)\right) + \frac{\partial}{\partial z}\left(\mu\left(\frac{\partial u_x}{\partial z}+\frac{\partial u_z}{\partial x}\right)\right) + F_x &= \rho \frac{\partial^2 u_x}{\partial t^2} \\
+\frac{\partial}{\partial x}\left(\mu\left(\frac{\partial u_y}{\partial x}+\frac{\partial u_x}{\partial y}\right)\right) + \frac{\partial}{\partial y}\left(\lambda \left( \frac{\partial u_x}{\partial x} + \frac{\partial u_y}{\partial y} + \frac{\partial u_z}{\partial z} \right) + 2\mu \frac{\partial u_y}{\partial y}\right) + \frac{\partial}{\partial z}\left(\mu\left(\frac{\partial u_y}{\partial z}+\frac{\partial u_z}{\partial y}\right)\right) + F_y &= \rho \frac{\partial^2 u_y}{\partial t^2}\\
+\frac{\partial}{\partial x}\left(\mu\left(\frac{\partial u_z}{\partial x}+\frac{\partial u_x}{\partial z}\right)\right) + \frac{\partial}{\partial y}\left(\mu\left(\frac{\partial u_z}{\partial y}+\frac{\partial u_y}{\partial z}\right)\right) + \frac{\partial}{\partial z}\left(\lambda \left( \frac{\partial u_x}{\partial x} + \frac{\partial u_y}{\partial y} + \frac{\partial u_z}{\partial z} \right) + 2\mu \frac{\partial u_z}{\partial z}\right) + F_z &= \rho \frac{\partial^2 u_z}{\partial t^2}
+\end{align*}
+Using the assumption that Lamé parameters $\lambda$ and $\mu$ are constant we can rearrange to produce:
+\begin{align*}
+(\lambda + \mu) \frac{\partial}{\partial x}\left(\frac{\partial u_x}{\partial x}+\frac{\partial u_y}{\partial y}+\frac{\partial u_z}{\partial z}\right) + \mu\left(\frac{\partial^2 u_x}{\partial x^2}+\frac{\partial^2 u_x}{\partial y^2}+\frac{\partial^2 u_x}{\partial z^2}\right) + F_x &= \rho \frac{\partial^2 u_x}{\partial t^2} \\
+(\lambda + \mu) \frac{\partial}{\partial y}\left(\frac{\partial u_x}{\partial x}+\frac{\partial u_y}{\partial y}+\frac{\partial u_z}{\partial z}\right) + \mu\left(\frac{\partial^2 u_y}{\partial x^2}+\frac{\partial^2 u_y}{\partial y^2}+\frac{\partial^2 u_y}{\partial z^2}\right) + F_y &= \rho \frac{\partial^2 u_y}{\partial t^2} \\
+(\lambda + \mu) \frac{\partial}{\partial z}\left(\frac{\partial u_x}{\partial x}+\frac{\partial u_y}{\partial y}+\frac{\partial u_z}{\partial z}\right) + \mu\left(\frac{\partial^2 u_z}{\partial x^2}+\frac{\partial^2 u_z}{\partial y^2}+\frac{\partial^2 u_z}{\partial z^2}\right) + F_z &= \rho \frac{\partial^2 u_z}{\partial t^2} \\
+\end{align*}
+We can easily see that only the displacements are left in these equations. The equations obtaines in this manner are known as the Navier-Cauchy equations.
+=== Navier-Cauchy equations ===
+The Navier or Navier-Cauchy equations describe the dynamics of a solid through the displacement vector field $\b{u}$. The equations can also be expressed concisely in vector form as follows
+\begin{equation}\label{eq:navier_mu_lambda}
+\rho \frac{\partial^2 \b{u}}{\partial t^2} = (\lambda + \mu) \nabla(\nabla \cdot \b{u}) + \mu \nabla^2 \b{u} + \b{F}
+\end{equation}
+where $\mu$ and $\lambda$ are Lamé constants, $\rho$ is the object density and $\b{F}$ are the external forces. In certain cases we may prefer to use the Young modulus $E$ and the Poisson ratio $\nu$ instead of the Lamé constants. In this case the Navier equations are
+\begin{equation}\label{eq:navier_young_poisson}
+\rho \frac{\partial^2 \b{u}}{\partial t^2} = \frac{E}{2(1+\nu)}\left(\nabla^2 \b{u} + \frac{1}{1-2\nu}\nabla\left(\nabla \cdot \b{u}\right)\right) + \b{F}
+\end{equation}
-====Stress formulation====
+<!-- ==Strain-displacement equations==
-* [https://en.wikipedia.org/wiki/Linear_elasticity#Stress_formulation Stress formulation at Wikipedia]
-* [https://en.wikipedia.org/wiki/Stress_functions Stress functions at Wikipedia]
-==Strain-displacement equations==
-Under the action of applied forces, a point in the solid originally at $(x,y,z)$ moves to position $(X,Y,Z)$. This movement can be described completely by the displacement vector
+Under the action of applied forces, a point in the solid originally at $\boldsymbol{x} = (x,y,z)$ moves to position $\boldsymbol{X} = (X,Y,Z)$. This movement can be described completely by the displacement vector
+\begin{equation*}
+\boldsymbol{u}(\boldsymbol{x}) = \boldsymbol{X} - \boldsymbol{x},
+\end{equation*}
+consisting of three components
 \begin{equation}
-\boldsymbol{u(\boldsymbol{x})} = \{u_x(x,y,z) \quad u_y(x,y,z) \quad u_z(x,y,z)\},
+\boldsymbol{u} = \{u_x \quad u_y \quad u_z\}.
 \end{equation}
-the brackets $\{\}$ indicating that this is a column vector.
+The brackets $\{\}$ indicate that this is a column vector. The components of this vector are mutually orthogonal and their positive directions correspond to the positive directions of the coordinate axes. To prevent confusion with the shorthand notation for partial derivatives, the components of $\boldsymbol{u}$ are sometimes represented with the letter $u, v, w$, e.g. $\boldsymbol{u} = \{u \quad v \quad w\}$.
-==Stress-strain equations==
+The strains in the deformed structure can be expressed as partial derivatives of the displacements as follows: -->
-===Two-dimensional stress distributions===
+== Two-dimensional stress distributions ==
 Many problems in elasticity can be simplified as two-dimensional problems described by ''plane theory of elasticity''. In general there are two types of problems we may encounter in plane analysis: '''plane stress''' and '''plane strain'''. The first problem arises in analysis of thin plates loaded in the plane of the plate, while the second is used for elongated bodies of constant cross section subject to uniform loading.
-====Plane stress====
+=== Plane stress ===
 Plane stress distributions build on the assumption that the normal stress and shear stresses directed perpendicular to the $x$-$y$ plane are assumed zero:
@@ Line 32: / Line 151: @@
 Using (\ref{eq:pstress_assump}) the three-dimensional Hooke's law can be reduced to:
-\begin{equation}
+\begin{equation}\label{eq:planestress}
-\begin{bmatrix}
+\boldsymbol{\sigma} = \boldsymbol{C}\boldsymbol{\varepsilon}
-\sigma_{xx} \\
+\end{equation}
-\sigma_{yy} \\
+in matrix form where for isotropic materials, we have
-\sigma_{xy}
+\begin{equation}\label{eq:planestressmatrix}
-\end{bmatrix}
+\boldsymbol{C}
 = \frac{E}{1-\nu^2}
 \begin{bmatrix}
@@ Line 43: / Line 162: @@
 \nu & 1 & 0 \\
 & 0 & \frac{1-\nu}{2}
-\end{bmatrix}
+\end{bmatrix} \qquad \text{(Plane stress)}
-\begin{bmatrix}
+\end{equation}
-\varepsilon_{xx} \\
+and
-\varepsilon_{yy} \\
+\begin{equation}\label{eq:2d_stress}
-\varepsilon_{xy}
+\b{\sigma} = \begin{bmatrix} \sigma_{xx} \\ \sigma_{yy} \\ \sigma_{xy}\end{bmatrix},
-\end{bmatrix}
+\end{equation}
+\begin{equation}\label{eq:2d_strain}
+\b{\varepsilon} = \begin{bmatrix} \varepsilon_{xx} \\ \varepsilon_{yy} \\ \gamma_{xy} \end{bmatrix} =
+\begin{bmatrix} \varepsilon_{xx} \\ \varepsilon_{yy} \\ 2\varepsilon_{xy} \end{bmatrix} =
+\begin{bmatrix} \frac{\partial u_x}{\partial x} \\ \frac{\partial u_y}{\partial y} \\ \frac{\partial u_y}{\partial x} + \frac{\partial u_x}{\partial y} \end{bmatrix}.
+\end{equation}
+Note that the strain tensor uses the so-called engineering shear strain $\gamma_{ij}$. Under the plane stress assumption the Navier equations are given as:
+\begin{equation}\label{eq:navier_plane_stress}
+\rho \frac{\partial^2 \b{u}}{\partial t^2} = \frac{E}{2(1+\nu)} \left( \b{\nabla}^2 \b{u} + \frac{1+\nu}{1-\nu}\b{\nabla}\left( \b{\nabla}\cdot \b{u}\right)  \right) + \b{F}
 \end{equation}
+or in component notation
+\begin{align}
+\rho \frac{\partial^2 u_x}{\partial t^2} &= \frac{E}{2(1+\nu)} \left( \frac{\partial^2 u_x}{\partial x^2} + \frac{\partial^2 u_x}{\partial y^2} \right) + \frac{E}{2(1-\nu)} \frac{\partial}{\partial x}\left( \frac{\partial u_x}{\partial x} + \frac{\partial u_y}{\partial y} \right) \\
+\rho \frac{\partial^2 u_y}{\partial t^2} &= \frac{E}{2(1+\nu)} \left( \frac{\partial^2 u_y}{\partial x^2} + \frac{\partial^2 u_y}{\partial y^2} \right) + \frac{E}{2(1-\nu)} \frac{\partial}{\partial y}\left( \frac{\partial u_x}{\partial x} + \frac{\partial u_y}{\partial y} \right)
+\end{align}
+==== Plane stress with Lamé constants ====
+The plane stress stiffness tensor (\ref{eq:planestressmatrix}) may also be expressed in terms of Lamé constants $\lambda$ and $\mu$ by substituting $E$ and $\nu$ with
+\[E = \frac{\mu(3\lambda+2\mu)}{\lambda+\mu}\]
+\[\nu = \frac{\lambda}{2(\lambda+\mu)}\]
+With these substitutions the Navier equations can be written in their component form as
+\begin{align}
+\rho \frac{\partial^2 u_x}{\partial t^2} &= \mu \left( \frac{\partial^2 u_x}{\partial x^2} + \frac{\partial^2 u_x}{\partial y^2} \right) + \left(\frac{2 \lambda \mu}{\lambda + 2 \mu}+\mu\right) \frac{\partial}{\partial x}\left( \frac{\partial u_x}{\partial x} + \frac{\partial u_y}{\partial y} \right) \\
+\rho \frac{\partial^2 u_y}{\partial t^2} &= \mu \left( \frac{\partial^2 u_y}{\partial x^2} + \frac{\partial^2 u_y}{\partial y^2} \right) + \left(\frac{2 \lambda \mu}{\lambda + 2 \mu}+\mu\right) \frac{\partial}{\partial y}\left( \frac{\partial u_x}{\partial x} + \frac{\partial u_y}{\partial y} \right)
+\end{align}
-====Plane strain====
+=== Plane strain ===
 The plane strain problem arises in analysis of walls, dams, tunnels where one dimension of the structure is very large in comparison to the other two dimensions ($x$- and $y$- coordinates). It is also appropriate for small-scale problems such as bars and rollers compressed by forces normal to their cross section. In all such problems the body may be imagined as a prismatic cylinder with one dimension much larger that the other two. The applied forces act in the $x$-$y$ plane and do not vary in the $z$ direction, leading to the assumption
@@ Line 64: / Line 208: @@
 The three dimensional Hooke's law can now be reduced to
 \begin{equation}\label{eq:planestrain}
-\begin{bmatrix}
+\boldsymbol{\sigma} = \boldsymbol{C}\boldsymbol{\varepsilon}
-\sigma_{xx} \\
+\end{equation}
-\sigma_{yy} \\
+where the matrix $C$ is given by
-\sigma_{xy}
+\begin{equation}\label{eq:matrixplanestrain}
-\end{bmatrix}
+\boldsymbol{C}
 = \frac{E}{(1+\nu)(1-2\nu)}
 \begin{bmatrix}
@@ Line 74: / Line 218: @@
 \nu & 1-\nu & 0 \\
 & 0 & \frac{1-2\nu}{2}
-\end{bmatrix}
+\end{bmatrix} \qquad \text{(Plane strain)}
-\begin{bmatrix}
-\varepsilon_{xx} \\
-\varepsilon_{yy} \\
-\varepsilon_{xy}
-\end{bmatrix}
 \end{equation}
+The vectors $\boldsymbol{\sigma}$ and $\boldsymbol{\varepsilon}$ are the same as above for plane stress and are given in (\ref{eq:2d_stress}) and (\ref{eq:2d_strain}), respectively. In the case of plane strain we have additonal non-zero components of the stress tensor:
 \begin{equation}\label{eq:sigmazz}
 \sigma_{zz} = \nu(\sigma_{xx}+\sigma_{yy})
@@ Line 87: / Line 227: @@
 \sigma_{yz} = \sigma_{zx} = 0
 \end{equation}
-The reason $\sigma_{zz}$ is not included in the matrix stress-strain equation (\ref{eq:planestrain}) is because it is linearly dependent on the normal stresses $\sigma_{xx}$ and $\sigma_{zz}$ as shown in (\ref{eq:sigmazz}).
+The reason $\sigma_{zz}$ is not included in the matrix stress-strain equation (\ref{eq:planestrain}) is because it is linearly dependent on the normal stresses $\sigma_{xx}$ and $\sigma_{zz}$.
-====Connection between plane stress and plane strain====
+The Navier-Cauchy equation that follows from the plane strain assumption can be found in (\ref{eq:navier_young_poisson}). For purposes of completeness we provide here the equations in component notation:
+\begin{align}
+\rho \frac{\partial^2 u_x}{\partial t^2} &= \frac{E}{2(1+\nu)} \left( \frac{\partial^2 u_x}{\partial x^2} + \frac{\partial^2 u_x}{\partial y^2} \right) + \frac{E}{2(1+\nu)(1-2\nu)} \frac{\partial}{\partial x}\left( \frac{\partial u_x}{\partial x} + \frac{\partial u_y}{\partial y} \right) \\
+\rho \frac{\partial^2 u_y}{\partial t^2} &= \frac{E}{2(1+\nu)} \left( \frac{\partial^2 u_y}{\partial x^2} + \frac{\partial^2 u_y}{\partial y^2} \right) + \frac{E}{2(1+\nu)(1-2\nu)} \frac{\partial}{\partial y}\left( \frac{\partial u_x}{\partial x} + \frac{\partial u_y}{\partial y} \right)
+\end{align}
-==Equations of equilibrium==
+==== Plane strain with Lamé constants ====
-==Tonti diagram==
+Alternatively the stiffness tensor $\b{C}$ in the stress-strain equation $\b{\sigma} = \b{C}\b{\varepsilon}$ may be expressed in terms of $\mu$ and $\lambda$:
+\begin{equation}
+\boldsymbol{C}
+=
+\begin{bmatrix}
+\mu + \lambda & \lambda & 0 \\
+\lambda & 2\mu + \lambda & 0 \\
+& 0 & \mu^*
+\end{bmatrix} \qquad \text{(Plane strain)}
+\end{equation}
+(*Note that this matrix requires the strain tensor to be defined using the engineering shear strain $\gamma_{xy} = 2\varepsilon_{xy}$. If we define the strain tensor as $\b{\varepsilon} = \{\varepsilon_{xx}\;\varepsilon_{yy}\;\varepsilon_{xy}\}^T$ then matrix component $C_{33}$ must be $2\mu$.)
-=Point contact on a 2D half-plane=
+The Navier equations in vector form can be found in (\ref{eq:navier_mu_lambda}). In component notation the equations are:
+\begin{align}
+\rho \frac{\partial^2 u_x}{\partial t^2} &= \mu \left( \frac{\partial^2 u_x}{\partial x^2} + \frac{\partial^2 u_x}{\partial y^2} \right) + (\lambda+\mu) \frac{\partial}{\partial x}\left( \frac{\partial u_x}{\partial x} + \frac{\partial u_y}{\partial y} \right) \\
+\rho \frac{\partial^2 u_y}{\partial t^2} &= \mu \left( \frac{\partial^2 u_y}{\partial x^2} + \frac{\partial^2 u_y}{\partial y^2} \right) + (\lambda+\mu) \frac{\partial}{\partial y}\left( \frac{\partial u_x}{\partial x} + \frac{\partial u_y}{\partial y} \right)
+\end{align}
+=== Connection between plane stress and plane strain ===
-A starting point to solve problems in contact mechanics is to understand the effect of a point-load applied to a homogeneous, linear elastic, isotropic half-plane. This problem may be defined either as plane stress or plain strain (for solution with FreeFem++ we have choosen the latter). The traction boundary conditions for this problem are:
+For '''isotropic''' materials with elastic modulus $E$ and Poisson's ratio $\nu$ it is possible to go from plane stress to plane strain, or vice-versa, by replacing $E$ and $\nu$ in the stress-strain matrix with a fictitious modulus $E^*$ and fictitious Poisson ratio $\nu^*$. This allows us to "reuse" a plane stress program to solve plane strain or again vice-versa (as long as the material is isotropic). A few exercises on this topic are given [http://www.colorado.edu/engineering/cas/courses.d/IFEM.d/IFEM.Ch14.d/IFEM.Ch14.pdf at this link (page 13)].
-\begin{equation}\label{eq:bc}
-\sigma_{xy}(x,0) = 0, \quad \sigma_{yy}(x,y) = -P\delta(x,y)
-\end{equation}
-where $\delta(x,y)$ is the Dirac delta function. Together these boundary conditions state that there is a singular normal force $P$ applied at $(x,y) = (0,0)$ and there are no shear stresses on the surface of the elastic half-plane.
-The analytical relations for the stresses can be found from the [https://en.wikipedia.org/wiki/Flamant_solution Flamant solution] (stress distributions in a linear elastic wedge loaded by point forces a the tip. When the "wedge" is flat we get a half-plane. The derivation uses polar coordinates.) and are given as:
+To go from plane stres'''s''' ($s$) to plane strai'''n''' ($n$) insert the fictitious quantities
-\begin{equation}
+\begin{equation}\label{eq:ston1}
-\sigma_{xx} = -\frac{2P}{\pi} \frac{x^2y}{\left(x^2+y^2\right)^2},
+E_n^* = \frac{E_s}{1-\nu_s^2},
 \end{equation}
-\begin{equation}
+\begin{equation}\label{eq:ston2}
-\sigma_{xy} = -\frac{2P}{\pi} \frac{y^3}{\left(x^2+y^2\right)^2},
+\nu_n^* = \frac{\nu_s}{1-\nu_s}.
 \end{equation}
-\begin{equation}
-\sigma_{yy} = -\frac{2P}{\pi} \frac{xy^2}{\left(x^2+y^2\right)^2},
+<div class="toccolours mw-collapsible mw-collapsed">
-\end{equation}
+'''Substitution from plane stress to plane strain'''
-for some point $(x,y)$ in the half-plane. From this stress field the strain components and thus the displacements $(u,v)$ can be determined. The displacements are given by
+<div class="mw-collapsible-content">Note: this derivation is shown just to confirm the above formulas. In a numerical program, we keep the stress-strain matrix and just use the above formulas (\ref{eq:ston1}) and (\ref{eq:ston2}) to update the values of $E$ and $\nu$. Also note we have omitted the indexes ($s$) and ($n$) in the following derivations.
-\begin{align}
-u_x &= -\frac{P}{4\pi\mu}\left((\kappa-1)\theta - \frac{2xy}{r^2}\right), \label{eq:dispx}\\
+We start with the plane stress matrix with the inserted fictitious values $E^*$ and $\nu^*$
-u_y &= -\frac{P}{4\pi\mu}\left((\kappa+1)\log r - \frac{2y^2}{r^2}\right), \label{eq:dispy}
+\[
-\end{align}
+\frac{E^*}{1-{\nu^*}^2}
-where $$r = \sqrt{x^2+y^2}$$ and $$\tan \theta = \frac{x}{y}.$$ The symbol $\kappa$ is known as Dundars constant and is defined as
+\begin{bmatrix}
+& \nu^* & 0 \\
+\nu^* & 1 & 0 \\
+& 0 & \frac{1}{2}(1-\nu^*)
+\end{bmatrix}
+\]
+and make the above substitutions (\ref{eq:ston1}) and (\ref{eq:ston2}) leading to:
+\[
+\frac{E}{\left(1-\nu^2\right)\left(1-\left(\frac{\nu}{1-\nu}\right)^2\right)}
+\begin{bmatrix}
+& \frac{\nu}{1-\nu} & 0 \\
+\frac{\nu}{1-\nu} & 1 & 0 \\
+& 0 & \frac{1}{2}(1-\frac{\nu}{1-\nu})
+\end{bmatrix}.
+\]
+We can then use the rule to convert sums of squares into products as well as bring the factor $1/(1-v)$ out of the matrix:
+\[
+\frac{E}{\left(1-\nu\right)\left(1+\nu\right)\left(1-\frac{\nu}{1-\nu}\right)\left(1+\frac{\nu}{1-\nu}\right)\left(1-\nu\right)}
+\begin{bmatrix}
+-\nu & \nu & 0 \\
+\nu & 1-\nu & 0 \\
+& 0 & \frac{1}{2}(1-2\nu)
+\end{bmatrix}.
+\]
+By joining some of the factors to a common denominator and rearranging leads to
+\[
+\frac{E\left(1-\nu\right)\left(1-\nu\right)}{\left(1-\nu\right)\left(1+\nu\right)\left(1-2\nu\right)1\left(1-\nu\right)}
+\begin{bmatrix}
+-\nu & \nu & 0 \\
+\nu & 1-\nu & 0 \\
+& 0 & \frac{1}{2}(1-2\nu)
+\end{bmatrix}.
+\]
+The final step is canceling the factors that occur in both the numerator and denominator
 \[
-\kappa = \begin{cases} 3 - 4\nu & \qquad \text{plane strain}, \\
+\frac{E}{\left(1+\nu\right)\left(1-2\nu\right)}
-                       \cfrac{3 - \nu}{1+\nu} & \qquad \text{plane stress}. \end{cases}
+\begin{bmatrix}
+-\nu & \nu & 0 \\
+\nu & 1-\nu & 0 \\
+& 0 & \frac{1}{2}(1-2\nu)
+\end{bmatrix}
 \]
-The last remaining symbol is $\mu$ which represents the shear modulus (sometimes also denoted with $G$).
+which leads exactly to the relationship for plane strain given in (\ref{eq:planestrain}).
+</div>
+</div>
-==Numerical solution with [http://www.freefem.org/ FreeFem++]==
-Due to the known analytical solution the point-contact problem can be used for benchmarking numerical PDE solvers in terms of accuracy (as well as computational efficiency). The purpose of this section is to compare the numerical solution obtained by FreeFem++ with the analytical solution, as well as provide a reference numerical solution for the [http://www-e6.ijs.si/ParallelAndDistributedSystems/MeshlessMachine/wiki/index.php/Main_Page C++ library] developed in our laboratory.
-For purposes of simplicity we limit ourselves to the domain $(x,y) \in \Omega = [-1,1] \times[-1,-0.1]$ and prescribe Dirichlet displacement on the boundaries $\Gamma_D$ from the known analytical solution (\ref{eq:dispx}, \ref{eq:dispy}). This way we avoid having to deal with the Dirac delta traction boundary condition (\ref{eq:bc}). The problem can be describes as find $\boldsymbol{u(\boldsymbol{x})}$ that satisfies
+In the opposite case we want to go from plane strai'''n''' ($n$) to plane stres'''s''' ($s$) we can use:
-\begin{equation}
+\begin{equation}\label{eq:ntos1}
-\boldsymbol{\nabla}\cdot\boldsymbol{\sigma}= 0 \qquad \text{on }\Omega
+E_s^* = \frac{E_n(1+2\nu_n)}{(1+\nu_n)^2},
 \end{equation}
-and
+\begin{equation}\label{eq:ntos2}
-\begin{equation}
+\nu_s^* = \frac{\nu_n}{1+\nu_n}.
-\boldsymbol{u} = \boldsymbol{u}_{\text{analytical}} \qquad \text{on }\Gamma_D
 \end{equation}
-where $\boldsymbol{u}_\text{analytical}$ is given in equations (\ref{eq:dispx}) and (\ref{eq:dispy}).
-To solve the point-contact problem in FreeFem++ we must first provide the weak form of the balance equation:
-\begin{equation*}
-\boldsymbol{\nabla}\cdot\boldsymbol{\sigma} + \boldsymbol{b} = 0.
-\end{equation*}
-The corresponding weak formulation is
-\begin{equation}
-\int_\Omega \boldsymbol{\sigma} : \boldsymbol{\varepsilon}(\boldsymbol{v}) \, d\Omega - \int_\Omega \boldsymbol{b}\cdot\boldsymbol{v}\,d\Omega = 0
-\end{equation}
-where $:$ denotes the tensor scalar product (tensor contraction), i.e. $\boldsymbol{A}:\boldsymbol{B} =\sum_{i,j} A_{ij}B_{ij}$. The vector $\boldsymbol{v}$ is the test function or so-called "virtual displacement".
-===Stress and displacement fields===
+<div class="toccolours mw-collapsible mw-collapsed">
+'''Substitution from plane strain to plane stress'''
+<div class="mw-collapsible-content">
+Note that we have again omitted the indexes ($s$) and ($n$) since our wish is expressed by the bold title.
+We start with the plane strain matrix with the inserted fictitious values $E^*$ and $\nu^*$
+\[
+\frac{E^*}{\left(1+\nu^*\right)\left(1-2\nu^*\right)}
+\begin{bmatrix}
+-\nu^* & \nu^* & 0 \\
+\nu^* & 1-\nu^* & 0 \\
+& 0 & \frac{1}{2}(1-2\nu^*)
+\end{bmatrix}
+\]
+and make the above substitutions (\ref{eq:ntos1}) and (\ref{eq:ntos2}) leading to:
+\[
+\frac{E(1+2\nu)}{\left(1+\nu\right)^2\left(1+\frac{\nu}{1+\nu}\right)\left(1-2\frac{\nu}{1+\nu}\right)}
+\begin{bmatrix}
+-\frac{\nu}{1+\nu} & \frac{\nu}{1+\nu} & 0 \\
+\frac{\nu}{1+\nu} & 1-\frac{\nu}{1+\nu} & 0 \\
+& 0 & \frac{1}{2}(1-2\frac{\nu}{1+\nu})
+\end{bmatrix}.
+\]
+Writing some of the sums with common denominators and rearranging leads to
+\[
+\frac{E(1+2\nu)(1+\nu)(1+\nu)}{\left(1+\nu\right)^2\left(1+2\nu\right)\left(1-\nu\right)}
+\begin{bmatrix}
+\frac{1}{1+\nu} & \frac{\nu}{1+\nu} & 0 \\
+\frac{\nu}{1+\nu} & \frac{1}{1+\nu} & 0 \\
+& 0 & \frac{1}{2}\left(\frac{1-\nu}{1+\nu}\right)
+\end{bmatrix}.
+\]
+We can also bring the factor $1/(1+\nu)$ out from the matrix components. After canceling all factors that occur in both denominator and numerator we are left with
+\[
+\frac{E}{\left(1+\nu\right)\left(1-\nu\right)}
+\begin{bmatrix}
+& \nu & 0 \\
+\nu & 1 & 0 \\
+& 0 & \frac{1}{2}(1-\nu)
+\end{bmatrix}.
+\]
+Rewriting the product of sums as a sum of squares gives us the matrix for plane stress in (\ref{eq:planestress}).
+</div>
+</div>
+=== Principal stresses ===
+This are the stresses when we rotate our coordinate system so that there are no shear components.
+The principal stresses for plane stress conditions are given as eigenvalues of $\sigma$ with
+\[
+  \sigma_{1,2} = \frac12 (s_{xx} + s_{zz}) \pm \sqrt{\frac{1}{4}(s_{xx}-s_{zz})^2 + s_{xz}^2}.
+\]
+The principal shear stress or principal stress difference is the stress where only shear stresses are present. In case of plane stress it is calculated as
+\[
+  \tau_1 = \sqrt{\frac{1}{4}(s_{xx}-s_{zz})^2 + s_{xz}^2} = \frac{|\sigma_1 - \sigma_2|}{2}.
+\]
+When the stress tensor is z 3x3 matrix, the formulas for eigenvalues are more complicated.
+=== Von Mises stress ===
+A more detailed explanation of the von Mises stress can be found on [https://en.wikipedia.org/wiki/Von_Mises_yield_criterion Wikipedia].
+In materials science and engineering the von Mises yield criterion can be also formulated in terms of the von Mises stress or equivalent tensile stress, $\sigma_\mathrm{v}$. This is a scalar value of stress that can be computed from the Cauchy stress tensor $\boldsymbol{\sigma}$. In this case, a material is said to start yielding when the von Mises stress reaches a value known as yield strength, $\sigma_\mathrm{y}$.
+Mathematically the von Mises yield criterion can be expressed as:
+\[
+  \sigma _\mathrm{v}=\sigma_\mathrm{y}={\sqrt {3J_{2}}},
+\]
+where $J_2$ is the [https://en.wikipedia.org/wiki/Cauchy_stress_tensor#Stress_deviator_tensor second deviatoric stress invariant].
+In the case of general '''plane stress''' ($\sigma_{33}=0$,$\sigma_{23}=0$, $\sigma_{31}=0$) the von Mises stress can be calculated as:
+\[
+\sigma_\mathrm{v} = \sqrt{\sigma_{11}^2 - \sigma_{11}\sigma_{22}+\sigma_{22}^2 + 3\sigma_{12}^2}.
+\]
+= computational solid mechanics =
+The goal in computational solid mechanics is to solve the equation of motion for a solid body
+\[\rho \frac{\partial^2 \b{u}}{\partial t^2} = \nabla\cdot\b{\sigma} + \b{F}\]
+on the given domain $\Gamma = \Gamma_N \cup \Gamma_D$ with Dirichlet (essential) boundary conditions
+\[ \b{u} = \bar{\b{u}} \quad \text{ on } \Gamma_D,\]
+and traction (natural) boundary conditions
+\[ \b{\sigma}\cdot\b{n} = \bar{\b{t}},\quad \text{ on } \Gamma_N,\]
+where the quantities with the bar sign $\bar{}$ indicate prescribed displacements and surface tractions, respectively. Another type of boundary condition are mixed (Robyn) type boundary conditions. In this case the displacement might be given in one direction, and the traction will be given for the other direction.
+In case the right-hand side value is zero, the boundary conditions are called homogeneous. Homogeneous Dirichlet boundary conditions mean the the body is fully fixed (constrained), while zero traction boundary conditions allow the surface to move freely.
+Using Hooke's law and the strain-displacement relation, the equation of motion can be transformed into the Navier-Cauchy equations
+\[\rho \frac{\partial^2 \b{u}}{\partial t^2} = \mu\nabla^2\b{u} + (\lambda+\mu)\nabla(\nabla\cdot\b{u}) + \b{F}.\]
+The solution of this equation with appropriate BC's will give us the desired displacement values across the domain.
+To obtain the steady state solution we have two possible strategies:
+# Add a (linear) dampening term to the Navier equation, and simulate the dynamic behaviour until the motion ceases: \[\rho \frac{\partial^2 \b{u}}{\partial t^2} + \eta_C \frac{\partial \b{u}}{\partial t} = \mu\nabla^2\b{u} + (\lambda+\mu)\nabla(\nabla\cdot\b{u}) + \b{F}.\]
+# Directly solve for steady state: \[\mu\nabla^2\b{u} + (\lambda+\mu)\nabla(\nabla\cdot\b{u}) + \b{F} = 0.\]
+Displacement boundary conditions are very simple to implement with both approaches. The desired values are simply prescribed for dynamic simulation with explicit methods, In case of implicit methods where we need to solve a system of equations the prescribed displacements are placed into the right-hand side of the system, and the value 1 is placed in the diagonal of the matrix (for a given node).
-===Convergence studies===
+For the traction boundary condition
-<figure id="fig:point_contact_convergence">
+\[\begin{bmatrix}
-[[File:Convergence.png|thumb|upright=2|<caption>Convergence results for the point contact problem. The colours blue, red and green represent linear, quadratic and cubic finite elements, respectively.</caption>]]
+\sigma_{xx} & \sigma_{xy} \\
-</figure>
+\sigma_{yx} & \sigma_{yy}
+\end{bmatrix} \cdot
+\begin{bmatrix}
+n_x \\
+n_y
+\end{bmatrix} =
+\begin{bmatrix}
+\bar{t}_x \\
+\bar{t}_y
+\end{bmatrix}\]
-For the convergence between analytical and numerical solution we vary the number of nodes by increasing the grid size in both $x$- and $y$- directions simultaneously by powers of two from $2^2$ (16 nodes all together) to $2^7$ (16384 nodes all together).
+we first use Hooke's law and the strain-displacement relations to express the traction boundary in terms of displacements:
-The $L^2$ error norm is used to measure the "difference" between solutions. Since the displacements are the variable that obtain from FreeFem++, we use the displacement magnitude $|\boldsymbol{u}| = \sqrt{u_x^2+u_y^2}$ to define our $L^2$-error norm. The exact equation we have used is
-\begin{equation}
-L^2\text{-norm} = \sqrt{\frac{\int_\Omega (|\boldsymbol{u_{\text{numerical}}}|-|\boldsymbol{u_{\text{analytical}}}|)^2d\Omega}{\int_\Omega|\boldsymbol{u_{\text{analytical}}}|^2d\Omega}}.
-\end{equation}
-Results are shown in Figure \ref{fig:point_contact_convergence}
-=Contact between parallel cylinders=
+\[\begin{bmatrix}
+(2\mu+\lambda)\partial_x u_x + \lambda \partial_y u_y & \mu(\partial_x u_y + \partial_y u_x) \\
+\mu(\partial_x u_y + \partial_y u_x) & \lambda \partial_x u_x + (2\mu + \lambda) \partial_y u_y
+\end{bmatrix} \cdot
+\begin{bmatrix}
+n_x \\
+n_y
+\end{bmatrix} =
+\begin{bmatrix}
+\bar{t}_x \\
+\bar{t}_y
+\end{bmatrix}\]
 =References=
+* Slak J., Kosec G.; Refined meshless local strong form solution of Cauchy-Navier equation on an irregular domain. Engineering analysis with boundary elements. 2018;11 ; [http://comms.ijs.si/~gkosec/data/papers/31107623.pdf manuscript]
 * Theory of matrix structural analysis
+'''This list might be incomplete, check also reference on the Main Page.'''