Solid Mechanics
Contents
Case studies
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.
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 \[E = \frac{\mu(3\lambda+2\mu)}{\lambda+\mu}\] \[\nu = \frac{\lambda}{2(\lambda+\mu)}\]
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
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.
References
- Theory of matrix structural analysis
