[[File:Plane stress.svg|400px|right|thumb|Figure 7.1 Plane stress state in a continuum.]] In [[continuum mechanics]], a material is said to be under '''plane stress''' if the [[stress (mechanics)|stress vector]] is zero across a particular surface. When that situation occurs over an entire element of a structure, the [[stress analysis]] is considerably simplified, as the stress state can be represented by a [[tensor]] of dimension 2 (a 2 × 2 matrix, instead of a 3 × 3 one). Plane stress typically occurs in thin flat plates that are acted upon only by load forces that are parallel to them. In certain situations, a gently curved thin plate may also be assumed to have plane stress for the purpose of stress analysis. This is the case, for example, of a thin-walled cylinder filled with a fluid under pressure. In such cases, stress components perpendicular to the plate are negligible compared to those parallel to it. In other situations, however, the bending stress of a thin plate cannot be neglected. One can still simplfy the analysis by using a two-dimensional domain, but the plane stress tensor each point must be complemented with bending terms. ==Mathematical definition== Mathematically, the stress at some point in the material is a plane stress if one of the three [[principal stress]]es (the [[eigenvalues and eigenvectors|eigenvalues]] of the [[Cauchy stress tensor]]) is zero. That is, there is [[Cartesian coordinate system]] in which the stress tensor has the form : \sigma = \begin{bmatrix} \sigma_{11} & 0 & 0 \\ 0 & \sigma_{22} & 0 \\ 0 & 0 & 0 \end{bmatrix} \equiv \begin{bmatrix} \sigma_{x} & 0 & 0 \\ 0 & \sigma_{y} & 0 \\ 0 & 0 & 0 \end{bmatrix} For example, consider a rectangular block of material measuring 10, 40 and 5 [[centimetre|cm]] along the x, y, and z, that is being stretched in the x direction and compressed in the y direction, by pairs of opposite forces with magnitudes 10 [[newton|N]] and 20 N, respectively, uniformly distributed over the corresponding faces. The stress tensor inside the block will be : \sigma = \begin{bmatrix} 500 [[pascal|Pa]] & 0 & 0 \\ 0 & -4000 Pa & 0 \\ 0 & 0 & 0 \end{bmatrix} More generally, if one chooses the first two coordinate axes arbitrarily but perpendicular to the direction of zero stress, the stress tensor will have the form : \sigma = \begin{bmatrix} \sigma_{11} & \sigma_{12} & 0 \\ \sigma_{21} & \sigma_{22} & 0 \\ 0 & 0 & 0 \end{bmatrix} \equiv \begin{bmatrix} \sigma_{x} & \tau_{xy} & 0 \\ \tau_{yx} & \sigma_{y} & 0 \\ 0 & 0 & 0 \end{bmatrix} and can therefore be represented by a 2 × 2 matrix, : \sigma_{ij} = \begin{bmatrix} \sigma_{11} & \sigma_{12} \\ \sigma_{21} & \sigma_{22} \end{bmatrix} \equiv \begin{bmatrix} \sigma_{x} & \tau_{xy} \\ \tau_{yx} & \sigma_{y} \end{bmatrix} ===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== [[File:Plane strain.svg|600px|right|thumb|Figure 7.2 Plane strain state in a continuum.]] {{main|Infinitesimal strain theory}} If one dimension is very large compared to the others, the [[Strain (materials science)|principal strain]] in the direction of the longest dimension is constrained and can be assumed as zero, 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. {{Clear}} The corresponding strain tensor is: : \varepsilon_{ij} = \begin{bmatrix} \varepsilon_{11} & \varepsilon_{12} & 0 \\ \varepsilon_{21} & \varepsilon_{22} & 0 \\ 0 & 0 & \varepsilon_{33}\end{bmatrix}\,\! in which the non-zero \varepsilon_{33}\,\! term arises from the [[Poisson's ratio|Poisson's effect]]. This strain term can be temporarily removed from the stress analysis to leave only the in-plane terms, effectively reducing the analysis to two dimensions. {{Clear}} ==Stress transformation in plane stress and plane strain== Consider a point P\,\! in a continuum under a state of plane stress, or plane strain, with stress components (\sigma_x, \sigma_y, \tau_{xy})\,\! and all other stress components equal to zero (Figure 7.1, Figure 8.1). From static equilibrium of an infinitesimal material element at P\,\! (Figure 8.2), the normal stress \sigma_\mathrm{n}\,\! and the shear stress \tau_\mathrm{n}\,\! on any plane perpendicular to the x\,\!-y\,\! plane passing through P\,\! with a unit vector \mathbf n\,\! making an angle of \theta\,\! with the horizontal, i.e. \cos \theta\,\! is the direction cosine in the x\,\! direction, is given by: :\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\,\! :\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 \theta\,\!, if one knows the stress components (\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 y\,\!-z\,\! plane. [[File:Stress transformation 2D.svg|460px|left|thumb|Figure 8.1 - Stress transformation at a point in a continuum under plane stress conditions.]] [[File:Stress at a plane 2D.svg|200px|none|thumb|Figure 8.2 - Stress components at a plane passing through a point in a continuum under plane stress conditions.]] 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 \tau_\mathrm{n}\,\! equal to zero. Thus we have: :\tau_\mathrm{n} = -\frac{1}{2}(\sigma_x - \sigma_y )\sin 2\theta + \tau_{xy}\cos 2\theta=0\,\! and we obtain :\tan 2 \theta_\mathrm{p} = \frac{2 \tau_{xy}}{\sigma_x - \sigma_y}\,\! This equation defines two values \theta_\mathrm{p}\,\! which are 90^\circ\,\! apart (Figure 8.3). The same result can be obtained by finding the angle \theta\,\! which makes the normal stress \sigma_\mathrm{n}\,\! a maximum, i.e. \frac{d\sigma_\mathrm{n}}{d\theta}=0\,\! The principal stresses \sigma_1\,\! and \sigma_2\,\!, or minimum and maximum normal stresses \sigma_\mathrm{max}\,\! and \sigma_\mathrm{min}\,\!, respectively, can then be obtained by replacing both values of \theta_\mathrm{p}\,\! into the previous equation for \sigma_\mathrm{n}\,\!. This can be achieved by rearranging the equations for \sigma_\mathrm{n}\,\! and \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 :\begin{align} \left[ \sigma_\mathrm{n} - \tfrac{1}{2} ( \sigma_x + \sigma_y )\right]^2 + \tau_\mathrm{n}^2 &= \left[\tfrac{1}{2}(\sigma_x - \sigma_y)\right]^2 + \tau_{xy}^2 \\ (\sigma_\mathrm{n} - \sigma_\mathrm{avg})^2 + \tau_\mathrm{n}^2 &= R^2 \end{align}\,\! where :R = \sqrt{\left[\tfrac{1}{2}(\sigma_x - \sigma_y)\right]^2 + \tau_{xy}^2} \quad \text{and} \quad \sigma_\mathrm{avg} = \tfrac{1}{2} ( \sigma_x + \sigma_y )\,\! which is the equation of a circle of radius R\,\! centered at a point with coordinates [\sigma_\mathrm{avg}, 0]\,\!, called [[Mohr's circle]]. But knowing that for the principal stresses the shear stress \tau_\mathrm{n} = 0\,\!, then we obtain from this equation: :\sigma_1 =\sigma_\mathrm{max} = \tfrac{1}{2}(\sigma_x + \sigma_y) + \sqrt{\left[\tfrac{1}{2}(\sigma_x - \sigma_y)\right]^2 + \tau_{xy}^2}\,\! :\sigma_2 =\sigma_\mathrm{min} = \tfrac{1}{2}(\sigma_x + \sigma_y) - \sqrt{\left[\tfrac{1}{2}(\sigma_x - \sigma_y)\right]^2 + \tau_{xy}^2}\,\! [[File:Principal stresses 2D.svg|400px|left|thumb|Figure 8.3 - Transformation of stresses in two dimensions, showing the planes of action of principal stresses, and maximum and minimum shear stresses.]] {{Clear}} When \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: \sigma_x = \sigma_1\,\! and \sigma_y = \sigma_2\,\!. Then the normal stress \sigma_\mathrm{n}\,\! and shear stress \tau_\mathrm{n}\,\! as a function of the principal stresses can be determined by making \tau_{xy}=0\,\!. Thus we have :\sigma_\mathrm{n} = \frac{1}{2} ( \sigma_1 + \sigma_2 ) + \frac{1}{2} ( \sigma_1 - \sigma_2 )\cos 2\theta\,\! :\tau_\mathrm{n} = -\frac{1}{2}(\sigma_1 - \sigma_2 )\sin 2\theta\,\! Then the maximum shear stress \tau_\mathrm{max}\,\! occurs when \sin 2\theta = 1\,\!, i.e. \theta = 45^\circ\,\! (Figure 8.3): :\tau_\mathrm{max} = \frac{1}{2}(\sigma_1 - \sigma_2 )\,\! Then the minimum shear stress \tau_\mathrm{min}\,\! occurs when \sin 2\theta = -1\,\!, i.e. \theta = 135^\circ\,\! (Figure 8.3): :\tau_\mathrm{min} = -\frac{1}{2}(\sigma_1 - \sigma_2 )\,\! == See also == * [[Plane strain]] == References == Meyers and Chawla (1999): "Mechanical Behavior of Materials," 66-75. [[Category:Metallurgy]] [[Category:Mechanical engineering]]