Plane stress

Constitutive equations

Plane stress in curved surfaces

In certain cases, the plane stress model can be used in the analysis of gently curved surfaces. For example, consider a thin-walled cylinder subjected to an axial compressive load uniformly distributed along its rim, and filled with a pressurized fluid. The internal pressure will generate a reactive hoop stress on the wall, a normal tensile stress directed perpendicular to the cylinder axis and tangential to its surface. The cylinder can be conceptually unrolled and analyzed as a flat thin rectangular plate subjected to tensile load in one direction and compressive load in another other direction, both parallel to the plate.

Plane strain (strain matrix)

If one dimension is very large compared to the others, the principal strain in the direction of the longest dimension is constrained and can be assumed as constant, that means there will be effectively zero strain along it, hence yielding a plane strain condition (Figure 7.2). In this case, though all principal stresses are non-zero, the principal stress in the direction of the longest dimension can be disregarded for calculations. Thus, allowing a two dimensional analysis of stresses, e.g. a dam analyzed at a cross section loaded by the reservoir.

The corresponding strain tensor is:

${\displaystyle {\epsilon }_{ij}=\left[\begin{array}{ccc}{\epsilon }_{11}& {\epsilon }_{12}& 0\\ {\epsilon }_{21}& {\epsilon }_{22}& 0\\ 0& 0& 0\end{array}\right]}$

and the corresponding stress tensor is:

${\displaystyle {\sigma }_{ij}=\left[\begin{array}{ccc}{\sigma }_{11}& {\sigma }_{12}& 0\\ {\sigma }_{21}& {\sigma }_{22}& 0\\ 0& 0& {\sigma }_{33}\end{array}\right]}$

in which the non-zero ${\displaystyle {\sigma }_{33}}$ term arises from the Poisson's effect. However, this term can be temporarily removed from the stress analysis to leave only the in-plane terms, effectively reducing the analysis to two dimensions.

Stress transformation in plane stress and plane strain

Consider a point ${\displaystyle P}$ in a continuum under a state of plane stress, or plane strain, with stress components ${\displaystyle ({\sigma }_x,{\sigma }_y,{\tau }_{xy})}$ and all other stress components equal to zero (Figure 8.1). From static equilibrium of an infinitesimal material element at ${\displaystyle P}$ (Figure 8.2), the normal stress ${\displaystyle {\sigma }_{\mathrm{n}}}$ and the shear stress ${\displaystyle {\tau }_{\mathrm{n}}}$ on any plane perpendicular to the ${\displaystyle x}$-${\displaystyle y}$ plane passing through ${\displaystyle P}$ with a unit vector ${\displaystyle \mathbf{n}}$ making an angle of ${\displaystyle \theta }$ with the horizontal, i.e. ${\displaystyle \cos \theta }$ is the direction cosine in the ${\displaystyle x}$ direction, is given by:

${\displaystyle {\sigma }_{\mathrm{n}}=\frac{1}{2}({\sigma }_x+{\sigma }_y)+\frac{1}{2}({\sigma }_x-{\sigma }_y)\cos 2\theta +{\tau }_{xy}\sin 2\theta }$

${\displaystyle {\tau }_{\mathrm{n}}=-\frac{1}{2}({\sigma }_x-{\sigma }_y)\sin 2\theta +{\tau }_{xy}\cos 2\theta }$

These equations indicate that in a plane stress or plane strain condition, one can determine the stress components at a point on all directions, i.e. as a function of ${\displaystyle \theta }$, if one knows the stress components ${\displaystyle ({\sigma }_x,{\sigma }_y,{\tau }_{xy})}$ on any two perpendicular directions at that point. It is important to remember that we are considering a unit area of the infinitesimal element in the direction parallel to the ${\displaystyle y}$-${\displaystyle z}$ plane.

The principal directions (Figure 8.3), i.e., orientation of the planes where the shear stress components are zero, can be obtained by making the previous equation for the shear stress ${\displaystyle {\tau }_{\mathrm{n}}}$ equal to zero. Thus we have:

${\displaystyle {\tau }_{\mathrm{n}}=-\frac{1}{2}({\sigma }_x-{\sigma }_y)\sin 2\theta +{\tau }_{xy}\cos 2\theta =0}$

and we obtain

${\displaystyle \tan 2{\theta }_{\mathrm{p}}=\frac{2{\tau }_{xy}}{{\sigma }_x-{\sigma }_y}}$

This equation defines two values ${\displaystyle {\theta }_{\mathrm{p}}}$ which are ${\displaystyle {90}^{\circ }}$ apart (Figure 8.3). The same result can be obtained by finding the angle ${\displaystyle \theta }$ which makes the normal stress ${\displaystyle {\sigma }_{\mathrm{n}}}$ a maximum, i.e. ${\displaystyle \frac{d{\sigma }_{\mathrm{n}}}{d\theta }=0}$

The principal stresses ${\displaystyle {\sigma }_1}$ and ${\displaystyle {\sigma }_2}$, or minimum and maximum normal stresses ${\displaystyle {\sigma }_{\mathrm{m}\mathrm{a}\mathrm{x}}}$ and ${\displaystyle {\sigma }_{\mathrm{m}\mathrm{i}\mathrm{n}}}$, respectively, can then be obtained by replacing both values of ${\displaystyle {\theta }_{\mathrm{p}}}$ into the previous equation for ${\displaystyle {\sigma }_{\mathrm{n}}}$. This can be achieved by rearranging the equations for ${\displaystyle {\sigma }_{\mathrm{n}}}$ and ${\displaystyle {\tau }_{\mathrm{n}}}$, first transposing the first term in the first equation and squaring both sides of each of the equations then adding them. Thus we have

${\displaystyle {\left[{\sigma }_{\mathrm{n}}-\frac{1}{2}({\sigma }_x+{\sigma }_y)\right]}^2+{\tau }_{\mathrm{n}}^2={\left[\frac{1}{2}({\sigma }_x-{\sigma }_y)\right]}^2+{\tau }_{xy}^2}$

${\displaystyle ({\sigma }_{\mathrm{n}}-{\sigma }_{\mathrm{a}\mathrm{v}\mathrm{g}}{)}^2+{\tau }_{\mathrm{n}}^2=R^2}$

where

${\displaystyle R=\sqrt{{\left[\frac{1}{2}({\sigma }_x-{\sigma }_y)\right]}^2+{\tau }_{xy}^2}\text{and}{\sigma }_{\mathrm{a}\mathrm{v}\mathrm{g}}=\frac{1}{2}({\sigma }_x+{\sigma }_y)}$

which is the equation of a circle of radius ${\displaystyle R}$ centered at a point with coordinates ${\displaystyle [{\sigma }_{\mathrm{a}\mathrm{v}\mathrm{g}},0]}$, called Mohr's circle. But knowing that for the principal stresses the shear stress ${\displaystyle {\tau }_{\mathrm{n}}=0}$, then we obtain from this equation:

${\displaystyle {\sigma }_1={\sigma }_{\mathrm{m}\mathrm{a}\mathrm{x}}=\frac{1}{2}({\sigma }_x+{\sigma }_y)+\sqrt{{\left[\frac{1}{2}({\sigma }_x-{\sigma }_y)\right]}^2+{\tau }_{xy}^2}}$

${\displaystyle {\sigma }_2={\sigma }_{\mathrm{m}\mathrm{i}\mathrm{n}}=\frac{1}{2}({\sigma }_x+{\sigma }_y)-\sqrt{{\left[\frac{1}{2}({\sigma }_x-{\sigma }_y)\right]}^2+{\tau }_{xy}^2}}$

When ${\displaystyle {\tau }_{xy}=0}$ the infinitesimal element is oriented in the direction of the principal planes, thus the stresses acting on the rectangular element are principal stresses: ${\displaystyle {\sigma }_x={\sigma }_1}$ and ${\displaystyle {\sigma }_y={\sigma }_2}$. Then the normal stress ${\displaystyle {\sigma }_{\mathrm{n}}}$ and shear stress ${\displaystyle {\tau }_{\mathrm{n}}}$ as a function of the principal stresses can be determined by making ${\displaystyle {\tau }_{xy}=0}$. Thus we have

${\displaystyle {\sigma }_{\mathrm{n}}=\frac{1}{2}({\sigma }_1+{\sigma }_2)+\frac{1}{2}({\sigma }_1-{\sigma }_2)\cos 2\theta }$

${\displaystyle {\tau }_{\mathrm{n}}=-\frac{1}{2}({\sigma }_1-{\sigma }_2)\sin 2\theta }$

Then the maximum shear stress ${\displaystyle {\tau }_{\mathrm{m}\mathrm{a}\mathrm{x}}}$ occurs when ${\displaystyle \sin 2\theta =1}$, i.e. ${\displaystyle \theta ={45}^{\circ }}$ (Figure 8.3):

${\displaystyle {\tau }_{\mathrm{m}\mathrm{a}\mathrm{x}}=\frac{1}{2}({\sigma }_1-{\sigma }_2)}$

Then the minimum shear stress ${\displaystyle {\tau }_{\mathrm{m}\mathrm{i}\mathrm{n}}}$ occurs when ${\displaystyle \sin 2\theta =-1}$, i.e. ${\displaystyle \theta ={135}^{\circ }}$ (Figure 8.3):

${\displaystyle {\tau }_{\mathrm{m}\mathrm{i}\mathrm{n}}=-\frac{1}{2}({\sigma }_1-{\sigma }_2)}$