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2.1: Stress Tensor

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    21476
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    We start with the presentation of simple concepts in one and two dimensions before introducing a general concept of the stress tensor. Consider a prismatic bar of a square cross-section subjected to a tensile force \(F\),

    2.1.1.png
    Figure \(\PageIndex{1}\): A long bar with three different cuts at \(\theta, \theta = 0\) and \(\pi/2 − \theta\).

    The force per unit area is called the surface traction \(T\):

    \[T = \sigma = \frac{\text{force}}{\text{area}} = \frac{F}{A_o} \left[ \frac{\mathrm{N}}{\mathrm{mm}^2} \right] \label{2.1.1}\]

    In the uniaxial case, the surface fraction is the only component of the stress tensor in the global coordinate system, commonly referred to as \(\sigma\).

    How can one apply a force to the end section of a bar? This can be done in a number of different ways (see Figure (\(\PageIndex{2}\))). A pin connection can be glued (or welded) to the end section, or a hole can be drilled through the bar to attach a pin.

    2.1.2.png
    Figure \(\PageIndex{2}\): How to apply tension to the end of a bar.

    Or an internal or external thread can be machined. Finally, axial force could be applied through frictional or mechanical grips. Except the welded or glued connector, a complex state of stress is created near the bar ends where the stress state is multi-axial. Such stress states is confined to a relatively short segment of the bar comparable with the height or diameter of the bar. Along this section a gradual transition takes place from the multi-axial state of stress to the uniaxial state, for which Equation \ref{2.1.1} holds.

    Note

    The above example an serve as a practical application of the Saint-Venant’s principle (1856). This principle named after the French elasticity theorist, Jean Claude Barre’ de Saint-Venant can be stated as: “the difference between the effects of two different but statically equivalent loads become very small at sufficiently large distances from load.”

    Think what are the “two” equivalent loads that are applied to the bar ends? We usually think of a cross-section being cut perpendicular to the axis of the bar. Consider now two cuts at the angles \(\theta\) and \(\left(\frac{\pi}{2} - \theta \right)\) to the normal direction. The planes are defined by the unit normal vector \(\boldsymbol{n}\).

    2.1.3.png
    Figure \(\PageIndex{3}\): Normal and tangential forces acting on the slant section of the bar.

    From the free body diagram, the components of the normal and tangential forces:

    \[F_{\mathrm{N}} = F \cos \theta \label{3.1.2}\]

    \[F_{\mathrm{n}} = F \cos (\frac{\pi}{2} - \theta)\]

    \[F_{\mathrm{T}} = F \sin \theta\]

    \[F_{\mathrm{t}} = F \sin (\frac{\pi}{2} - \theta) \label{3.1.5}\]

    The slant cross-section \(A\) is larger and is related to the reference cross-section by

    \[A_o = A_{\mathrm{A}} \cos \theta, \; A_o = A{\mathrm{B}} \cos (\frac{\pi}{2} - \theta)\]

    Consider now a unit volume cubic element located at the intersections of cuts A-A and B-B, Figure (\(\PageIndex{4}\)).

    2.1.4.png
    Figure \(\PageIndex{4}\): The volume element with surface traction acting on two adjacent facets.

    The surface traction (force per unit area) on the two perpendicular facets are

    \[\text{Facet parallel to A-A:} \quad T_{\mathrm{n}} = T \cos^2 \theta\]

    \[T_{\mathrm{t}} = T \sin \theta \cos \theta \label{3.1.8}\]

    \[\text{Facet parallel to B-B:} \quad T_{\mathrm{n}} = T \cos^2 (\frac{\pi}{2} - \theta)\]

    \[T_{\mathrm{t}} = T \sin (\frac{\pi}{2} - \theta) \cos (\frac{\pi}{2} - \theta)\]

    It can be observed that the tangential components of the surface traction vector on A-A and B-B cuts are identical. The normalized plots of the above quantities versus the orientation angle of the cross-section are shown in Figure (\(\PageIndex{5}\)).

    2.1.5.png
    Figure \(\PageIndex{5}\): Relative values of normal and shear components of the surface traction as a function of the orientation of the cut.

    It can be noted that the tangential component attains maximum at \(45^{\circ}\). This means that if the material fails due to shear loading, the fracture surface will always be oriented at \(45^{\circ}\). The above example teaches us that there are infinite combinations of normal and tangential components of surface tractions which are in equilibrium with the applied load. For each orientation of the cross-section there is a different pair of \(\{T_{\mathrm{n}}, T_{\mathrm{t}}\}\). The orientation of the surface element is uniquely defined by the unit normal vector \(\boldsymbol{n}\{n_1, n_2, n_3\}\). At the same time the components of the surface traction vector acting on the same element are \(\boldsymbol{T}\{T_1, T_2, T_3\}\).

    The components of the surface traction vector acting on this surface element are \(\boldsymbol{T}\{T_1, T_2, T_3\}\). For example, the orientation of facets of the unit material cube is shown in Figure (\(\PageIndex{6}\)).

    2.1.6.png
    Figure \(\PageIndex{6}\): Components of the unit normal vector on facets of a unit cube.

    The relation between the vectors of surface tractions, unit normal vector defining the surface element and the stress tensor are given by the famous Cauchy formula

    \[T_i = T_{ij}n_j\]

    or in the expanded notation,

    \[T_1 = \sigma_{1j}n_j = \sigma_{11}n_1 + \sigma_{12}n_2 + \sigma_{13}n_3\]

    \[T_2 = \sigma_{2j}n_j = \sigma_{21}n_1 + \sigma_{22}n_2 + \sigma_{23}n_3 \label{3.1.13}\]

    \[T_3 = \sigma_{3j}n_j = \sigma_{31}n_1 + \sigma_{32}n_2 + \sigma_{33}n_3\]

    To a large extent the Cauchy relation is analogous to the strain-displacement relation put in the form of Equations \ref{3.1.2} - \ref{3.1.5}.

    \[du_i = F_{ij}dx_j\]

    The displacement gradient \(F_{ij}\) transforms the increment of the length element \(dx_j\) into the increment of displacement \(du_i\). In the same way the stress tensor transforms the orientation of the surface element \(\boldsymbol{n}\) into the surface traction acting on this element.

    In order to get a physical interpretation of the concept of the stress tensor, let us see how the Cauchy formula works in the case of one and two-dimensional problems of the axially loaded bar. Consider first the normal cut of the bar with the longitudinal axis as 1-axis. The components of the surface tractions are given in Figure (\(\PageIndex{7}\)). The corresponding components of the unit normal vector were defined in Figure (\(\PageIndex{6}\)), where \(T = \frac{F}{A_o}\).

    2.1.7.png
    Figure \(\PageIndex{7}\): The unit volume element aligned with the axis of the bar.

    Substituting the values of the components of the two vectors into Equation \ref{3.1.13} one gets the following expressions:

    \[\begin{array}{c|c|c}
    \operatorname{Facet}(1,0,0) & \operatorname{Facet}(0,1,0) & \operatorname{Facet}(0,0,1) \\
    T_1 =\sigma_{11} & 0 =\sigma_{12} & 0=\sigma_{13} \\
    0 =\sigma_{21} & 0 =\sigma_{22} & 0=\sigma_{23} \\
    0=\sigma_{31} & 0=\sigma_{32} & 0=\sigma_{33}
    \end{array}\]

    Therefore the components of the stress \(3 \times 3\) matrix in the global coordinate system are

    \[\boldsymbol{\sigma} = \begin{vmatrix}
    T & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0
    \end{vmatrix} \label{3.1.17}\]

    This is the uniaxial state of stress. The two-dimensional example of the slant cut is much more interesting. This time a local coordinate system, rotated with respect to the 3-axis will be used. In this system the components \(\boldsymbol{n}\) are the same as in the global system. The components of the surface traction vector on three facets, calculated in Equation \ref{3.1.8} are defined in Figure (\(\PageIndex{8}\)).

    2.1.8.png
    Figure \(\PageIndex{8}\): Components of the surface tractions on the rotated volume element.

    Substituting the above values into the Cauchy formula we obtain

    \[\begin{array}{c|c|c}
    \operatorname{Facet}(1,0,0) & \operatorname{Facet}(0,1,0) & \operatorname{Facet}(0,0,1) \\
    T \cos ^{2} \theta=\sigma_{11} & T \sin \theta \cos \theta=\sigma_{12} & 0=\sigma_{13} \\
    T \sin \theta \cos \theta=\sigma_{21} & T \sin ^{2} \theta=\sigma_{22} & 0=\sigma_{23} \\
    0=\sigma_{31} & 0=\sigma_{32} & 0=\sigma_{33}
    \end{array}\]

    The plane stress components of the stress tensor are

    \[\boldsymbol{\sigma} = \begin{vmatrix}
    T \cos^{\theta} & T \sin \theta \cos \theta & 0 \\ T \sin \theta \cos \theta & T \sin^{2} \theta & 0 \\ 0 & 0 & 0
    \end{vmatrix} \label{3.1.19}\]

    It is interesting that the matrices Equation \ref{3.1.17} and Equation \ref{3.1.19} represent the same state of stress seen in two coordinate systems rotated with respect to one another. The transformation of the stress tensor from one coordinate system to the other is the subject Recitation 1 where the relation between Equation \ref{3.1.17} and Equation \ref{3.1.19} will be derived in a different way.

    Symmetry of the stress tensor

    It should also be noted from Equation \ref{3.1.19} that stress tensor is symmetric meaning that \(\sigma_{12} = \sigma_{21}\). The symmetry of the stress tensor comes from the moment equilibrium equation of are infinitesimal volume element. In general

    \[\sigma_{ij} = \sigma_{ji}\]

    The symmetry of the stress tensor reduce the nine components of the \(3 \times 3\) metric to only six independent components. The meaning of the two subscripts of the stress tensor is explained below

    \[\sigma_{??}\nonumber\]

    The first subscript defines the plane on which the surface tractions are acting. For example “1” denotes the surface element perpendicular to the axis \(x_1\). The second subscript indicates direction of a particular component of the surface tractions. This convention is explained in Figure (\(\PageIndex{9}\)).

    2.1.9.png
    Figure \(\PageIndex{9}\): Components of the stress tensor on three facets of the infinitesimal surface element.

    Sign convention

    The Cauchy formula can also be consistently used to determine the sign of the components of the stress tensor. The point is that the sign of the components of the vectors is known from the chosen coordinate system. For illustration, let us orient the volume element along the \(x_1\) axis. With positive direction to the right.

    2.1.10.png
    Figure \(\PageIndex{10}\): Explanation of the sign convention of the stress tensor.

    From the Cauchy formula

    \[T_1 = \sigma_{11}n_1 \label{3.1.21}\]

    On the right facet both the surface traction and the unit normal vector is positive and so must be the normal component of the stress tensor \(\sigma_{11}\). On the left facet both \(T_1\) and to the \(x_1\) axis. In order for Equation \ref{3.1.21} to hold the component \(\sigma_{11}\) must be positive, even if its visualization points out in the negative direction. in the above example the stress state is uniform along the \(x_1\) axis. This is the case of a bar under tension. In general there is a gradient of the components of the stress tensor so that stresses on both sides of the infinitesimal element differ by a small amount of \(d\sigma_{11}\). The sign convention is opening the way for deriving the equations of equilibrium for the 3-D continuum. This topic is the subject of the next section.

    Equilibrium

    The equilibrium equation for an infinitesimal volume element are derived first using two methods. Referring to Figure (\(\PageIndex{9}\)), indicated on Figure (\(\PageIndex{11}\)) are only those components of the stress tensor that are directed along \(x_2\) axis. These are \(\sigma_{12}\), \(\sigma_{22}\) and \(\sigma_{32}\).

    2.1.11.png
    Figure \(\PageIndex{11}\): All components of the stress tensor contributing to the force equilibrium in \(x_2\) direction must be in equilibrium.

    According to Newton’s law, the sum of all forces (stress times the surface area) acting along \(x_2\) must be zero

    \[\left( \sigma_{22} + \frac{\partial \sigma_{22}}{\partial x_2} dx_2 \right) dx_1 dx_3 - \sigma_2 dx_1 dx_3 + \left( \sigma_{12} + \frac{\partial \sigma_{12}}{\partial x_1}dx_1\right) dx_2 dx_3 - \sigma_{12} dx_2 dx_3 + \left( \sigma_{32} + \frac{\partial \sigma_{32}}{\partial x_3}dx_3\right) dx_1 dx_2 - \sigma_{32} dx_1 dx_2 + B_2 dx_1 dx_2 dx_3 = 0\]

    For generality, the body force (force per unit volume) was included as well. The body force represent for example gravity force \(B = \rho g\) or d’Alambert inertia force \(B = \rho \ddot{u}\) so that the derivation is valid both for static and dynamic problems. Summing up the forces one gets the first equilibrium equation

    \[\frac{\partial \sigma_{22}}{\partial x_2} + \frac{\partial \sigma_{12}}{\partial x_1} + \frac{\partial \sigma_{32}}{\partial x_3} + B_2 = 0\]

    Invoking the index notation

    \[\frac{\partial \sigma_{j2}}{\partial x_j} + B_2 = 0 \rightarrow \sigma_{j2,j} + B_2 = 0 \label{3.1.24}\]

    with the summation and coma convention. A similar procedure of summing-up forces can be repeated in the \(x_1\) and \(x_3\) direction, yielding two additional equations of equilibrium. One can immediately notice that by replacing the life subscripts “2” in Equation \ref{3.1.24} respectively by “1” and “3”, the final compact form of the equation of equilibrium reads

    \[\sigma_{ij,j} + B_i = 0 \text{ or } \frac{\partial \sigma_{ij}}{\partial x_j} + B_i = 0\]

    In the expanded notation and replacing \(x_i\) by \((x, y, z)\), the familiar form of the equilibrium equation is

    \[\color{red}{\frac{\partial \sigma_{x x}}{\partial x} + \frac{\partial \sigma_{x y}}{\partial y}} \color{black}{+ \frac{\partial \sigma_{x z}}{\partial z} + B_{x}=0}\]
    \[\color{red}{\frac{\partial \sigma_{y x}}{\partial x} + \frac{\partial \sigma_{y y}}{\partial y}} \color{black}{+ \frac{\partial \sigma_{y z}}{\partial z} + B_{y}=0}\]
    \[\frac{\partial \sigma_{z x}}{\partial x} + \frac{\partial \sigma_{z y}}{\partial y} + \frac{\partial \sigma_{z z}}{\partial z} + B_{z}=0\]

    The plane stress case, prevailing in thin plate and shells is defined by

    \[\sigma_{3j} = 0 \text{ or } \sigma_{31} = \sigma_{32} = \sigma_{33} = 0\]

    In other words all components of the stress tensor pointing out in the z-directions are zero, \(\sigma_{zz} = \sigma_{zx} = \sigma_{zy} = 0\). The components of the plane stress tensor are highlighted by the framed area, thus \(\boldsymbol{\sigma}\) is equal to

    2.1.12.png

    For plane stress, the subscripts run only over two dimensions and the Greek letters are commonly used, \(\alpha , \beta = 1, 2\). In the compact notation, the plane stress equilibrium equation reads

    \[\sigma_{\alpha \beta, \beta} + B_{\alpha} = 0\]

    In the uniaxial case only one component of the equilibrium survives, giving

    \[\frac{d \sigma_{xx}}{dx} + B = 0 \label{3.1.31}\]

    With no body force, \(B = 0\), Equation \ref{3.1.31} predicts a constant stress along the length of the bar. The addition of the d’Alambert inertia force will lead to the one-dimensional wave equation.


    This page titled 2.1: Stress Tensor is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Tomasz Wierzbicki (MIT OpenCourseWare) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.