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| [[File:Plane stress.svg|400px|right|thumb|Figure 7.1 Plane stress state in a continuum.]]
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| 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, as is often the case for thin plates, the [[stress analysis]] is considerably simplified, as the stress state can be represented by a [[tensor]] of dimension 2 (representable as a 2 × 2 matrix rather than 3 × 3). <ref name=Meyers/> A related notion, [[plane strain]], is often applicable to very thick members.
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| 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.<ref name=Meyers/>
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| In other situations, however, the bending stress of a thin plate cannot be neglected. One can still simplify the analysis by using a two-dimensional domain, but the plane stress tensor each point must be complemented with bending terms.
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| ==Mathematical definition==
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| 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
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| : <math>\sigma =
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| \begin{bmatrix}
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| \sigma_{11} & 0 & 0 \\
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| 0 & \sigma_{22} & 0 \\
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| 0 & 0 & 0
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| \end{bmatrix}
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| \equiv
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| \begin{bmatrix}
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| \sigma_{x} & 0 & 0 \\
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| 0 & \sigma_{y} & 0 \\
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| 0 & 0 & 0
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| \end{bmatrix}</math>
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|
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| For example, consider a rectangular block of material measuring 10, 40 and 5 [[centimetre|cm]] along the <math>x</math>, <math>y</math>, and <math>z</math>, that is being stretched in the <math>x</math> direction and compressed in the <math>y</math> direction, by pairs of opposite forces with magnitudes 10 [[newton (unit)|N]] and 20 N, respectively, uniformly distributed over the corresponding faces. The stress tensor inside the block will be
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| : <math>\sigma =
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| \begin{bmatrix}
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| 500\mathrm{ Pa} & 0 & 0 \\
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| 0 & -4000\mathrm{ Pa} & 0 \\
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| 0 & 0 & 0
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| \end{bmatrix}
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| </math>
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| 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
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| : <math>\sigma =
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| \begin{bmatrix}
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| \sigma_{11} & \sigma_{12} & 0 \\
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| \sigma_{21} & \sigma_{22} & 0 \\
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| 0 & 0 & 0
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| \end{bmatrix}
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| \equiv
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| \begin{bmatrix}
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| \sigma_{x} & \tau_{xy} & 0 \\
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| \tau_{yx} & \sigma_{y} & 0 \\
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| 0 & 0 & 0
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| \end{bmatrix}</math>
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| and can therefore be represented by a 2 × 2 matrix,
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| : <math>\sigma_{ij} =
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| \begin{bmatrix}
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| \sigma_{11} & \sigma_{12} \\
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| \sigma_{21} & \sigma_{22}
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| \end{bmatrix}
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| \equiv
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| \begin{bmatrix}
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| \sigma_{x} & \tau_{xy} \\
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| \tau_{yx} & \sigma_{y}
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| \end{bmatrix}</math>
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| == Constitutive equations ==
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| :''See [[Hooke's law#Plane_stress]]''
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| ==Plane stress in curved surfaces ==
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| In certain cases, the plane stress model can be used in the analysis of gently curved surfaces. For example,
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| 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.
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| ==Plane stress ( strain matrix )==
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| [[File:Plane strain.svg|600px|right|thumb|Figure 7.2 Plane strain state in a continuum.]]
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| {{main|Infinitesimal strain theory}}
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| 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.
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| {{Clear}}
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| The corresponding strain tensor is:
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| : <math>\varepsilon_{ij} = \begin{bmatrix}
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| \varepsilon_{11} & \varepsilon_{12} & 0 \\
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| \varepsilon_{21} & \varepsilon_{22} & 0 \\
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| 0 & 0 & \varepsilon_{33}\end{bmatrix}\,\!</math>
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| in which the non-zero <math>\varepsilon_{33}\,\!</math> 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.<ref name=Meyers/>
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| {{Clear}}
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| ==Stress transformation in plane stress and plane strain==
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| Consider a point <math>P\,\!</math> in a continuum under a state of plane stress, or plane strain, with stress components <math>(\sigma_x, \sigma_y, \tau_{xy})\,\!</math> and all other stress components equal to zero (Figure 8.1). From static equilibrium of an infinitesimal material element at <math>P\,\!</math> (Figure 8.2), the normal stress <math>\sigma_\mathrm{n}\,\!</math> and the shear stress <math>\tau_\mathrm{n}\,\!</math> on any plane perpendicular to the <math>x\,\!</math>-<math>y\,\!</math> plane passing through <math>P\,\!</math> with a unit vector <math>\mathbf n\,\!</math> making an angle of <math>\theta\,\!</math> with the horizontal, i.e. <math>\cos \theta\,\!</math> is the direction cosine in the <math>x\,\!</math> direction, is given by:
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| :<math>\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\,\!</math>
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| :<math>\tau_\mathrm{n} = -\frac{1}{2}(\sigma_x - \sigma_y )\sin 2\theta + \tau_{xy}\cos 2\theta\,\!</math>
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| 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 <math>\theta\,\!</math>, if one knows the stress components <math>(\sigma_x, \sigma_y, \tau_{xy})\,\!</math> 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 <math>y\,\!</math>-<math>z\,\!</math> plane.
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| [[File:Stress transformation 2D.svg|460px|left|thumb|Figure 8.1 - Stress transformation at a point in a continuum under plane stress conditions.]]
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| [[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.]]
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| 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 <math>\tau_\mathrm{n}\,\!</math> equal to zero. Thus we have:
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| :<math>\tau_\mathrm{n} = -\frac{1}{2}(\sigma_x - \sigma_y )\sin 2\theta + \tau_{xy}\cos 2\theta=0\,\!</math>
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| and we obtain
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| :<math>\tan 2 \theta_\mathrm{p} = \frac{2 \tau_{xy}}{\sigma_x - \sigma_y}\,\!</math>
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| This equation defines two values <math>\theta_\mathrm{p}\,\!</math> which are <math>90^\circ\,\!</math> apart (Figure 8.3). The same result can be obtained by finding the angle <math>\theta\,\!</math> which makes the normal stress <math>\sigma_\mathrm{n}\,\!</math> a maximum, i.e. <math>\frac{d\sigma_\mathrm{n}}{d\theta}=0\,\!</math>
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| The principal stresses <math>\sigma_1\,\!</math> and <math>\sigma_2\,\!</math>, or minimum and maximum normal stresses <math>\sigma_\mathrm{max}\,\!</math> and <math>\sigma_\mathrm{min}\,\!</math>, respectively, can then be obtained by replacing both values of <math>\theta_\mathrm{p}\,\!</math> into the previous equation for <math>\sigma_\mathrm{n}\,\!</math>. This can be achieved by rearranging the equations for <math>\sigma_\mathrm{n}\,\!</math> and <math>\tau_\mathrm{n}\,\!</math>, first transposing the first term in the first equation and squaring both sides of each of the equations then adding them. Thus we have
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| :<math>\begin{align}
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| \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 \\
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| (\sigma_\mathrm{n} - \sigma_\mathrm{avg})^2 + \tau_\mathrm{n}^2 &= R^2 \end{align}\,\!</math>
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| where
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| :<math>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 )\,\!</math> | |
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| which is the equation of a circle of radius <math>R\,\!</math> centered at a point with coordinates <math>[\sigma_\mathrm{avg}, 0]\,\!</math>, called [[Mohr's circle]]. But knowing that for the principal stresses the shear stress <math>\tau_\mathrm{n} = 0\,\!</math>, then we obtain from this equation:
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| :<math>\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}\,\!</math>
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| :<math>\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}\,\!</math> | |
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| [[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.]]
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| {{Clear}}
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| When <math>\tau_{xy}=0\,\!</math> the infinitesimal element is oriented in the direction of the principal planes, thus the stresses acting on the rectangular element are principal stresses: <math>\sigma_x = \sigma_1\,\!</math> and <math>\sigma_y = \sigma_2\,\!</math>. Then the normal stress <math>\sigma_\mathrm{n}\,\!</math> and shear stress <math>\tau_\mathrm{n}\,\!</math> as a function of the principal stresses can be determined by making <math>\tau_{xy}=0\,\!</math>. Thus we have
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| :<math>\sigma_\mathrm{n} = \frac{1}{2} ( \sigma_1 + \sigma_2 ) + \frac{1}{2} ( \sigma_1 - \sigma_2 )\cos 2\theta\,\!</math>
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| :<math>\tau_\mathrm{n} = -\frac{1}{2}(\sigma_1 - \sigma_2 )\sin 2\theta\,\!</math>
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| Then the maximum shear stress <math>\tau_\mathrm{max}\,\!</math> occurs when <math>\sin 2\theta = 1\,\!</math>, i.e. <math>\theta = 45^\circ\,\!</math> (Figure 8.3):
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| :<math>\tau_\mathrm{max} = \frac{1}{2}(\sigma_1 - \sigma_2 )\,\!</math>
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| Then the minimum shear stress <math>\tau_\mathrm{min}\,\!</math> occurs when <math>\sin 2\theta = -1\,\!</math>, i.e. <math>\theta = 135^\circ\,\!</math> (Figure 8.3):
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| :<math>\tau_\mathrm{min} = -\frac{1}{2}(\sigma_1 - \sigma_2 )\,\!</math>
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|
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| == See also ==
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| * [[Plane strain]]
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| == References ==
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| <references>
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| <ref name=Meyers>
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| Meyers and Chawla (1999): "Mechanical Behavior of Materials," 66-75.
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| </ref>
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| </references>
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| [[Category:Metallurgy]]
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| [[Category:Mechanical engineering]]
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