General State of Stress |
Plane Stress |
Plane Stress Transformation |
Principal Stresses |
Maximum Shear Stress |
Mohr's Circle |
Interactive: State of Stress |
Interactive: State of Stress in a Beam |
Stress Transformation
General Stress State
The general state of stress at a point is characterized by three independent normal stress components and three independent shear stress components, and is represented by the stress tensor. The combination of the state of stress for every point in the domain is called the stress field.
\[\boldsymbol{T}= \begin{bmatrix} \sigma_{x} & \tau_{xy} & \tau_{xz}\\ \tau_{xy} & \sigma_{y} & \tau_{yz}\\ \tau_{xz} & \tau_{yz} & \sigma_{z} \end{bmatrix}\]
Note that the stress is a physical quantity and as such, it is independent of the coordinate system chosen to represent it.
Sign Convention
Plane Stress
Plane stress occurs when two faces of the cube element are stress free.
\[\sigma_{z}=\tau_{zx}=\tau_{zy}=0\]
Example:
Thin plates subject to forces acting in the mid-plane of the plate.
Plane Stress Transformation
For any surface that divides the body ( imaginary or real surface), the action of one part of the body on the other is equivalent to the system of distributed internal forces and moments and it is represented by the stress vector $t^{n}$ (also called traction), defined on the surface with normal unit vector $\boldsymbol{n}$.
The state of stress at a point in the body is defined by all the stress vectors $t^{n}$ associated with all planes (infinite in number) that pass through that point.
Cauchy’s stress theorem states that there exists a stress tensor $\boldsymbol{T}$ (which is independent of $\boldsymbol{n}$), such that $t^{n}$ is a linear function of $\boldsymbol{n}$:
\[t^n=\boldsymbol{T}\boldsymbol{n}\]
We think of stresses acting on faces, so we often associate the state of stress with a coordinate system. However, the selection of a coordinate system is arbitrary (materials don't know about coordinates - it's a mathematical construct!) and we could choose to express the steress state acting on any set of faces aligned with any coordinate system axes. Furthermore, we can relate the states of stress in each coordinate system to one another through stress transformation equations.
Sign convention:
\[\begin{align} \sigma_x' &= \frac{\sigma_x + \sigma_y}{2} + \frac{\sigma_x - \sigma_y}{2}\rm\cos(2\theta) + \tau_{xy}\sin(2\theta) \\ \sigma_y' &= \frac{\sigma_x + \sigma_y}{2} - \frac{\sigma_x - \sigma_y}{2}\rm\cos(2\theta) - \tau_{xy}\sin(2\theta) \\ \tau_{x'y'} &= -\frac{\sigma_x - \sigma_y}{2}\rm\sin(2\theta) + \tau_{xy}\cos(2\theta) \end{align}\]
Using the equations from stress in inclined planes: \[\begin{align} \sigma_{x}' &= \sigma_{x} \cos ^2(\theta) + 2 \,\tau_{xy} \sin(\theta) \cos(\theta)+\sigma_{y} \sin ^2(\theta) \\ \tau_{x'y'} &= (\sigma_{y} - \sigma_{x}) \sin(\theta) \cos(\theta) + \tau_{xy} ( \cos^2(\theta) - \sin^2(\theta) ) \end{align}\]
We use the following trigonometric relations: \[\begin{matrix} \rm\cos^2\theta = \frac{1 + \rm\cos(2\theta)}{2} & \rm\sin(2\theta) = 2\rm\sin\theta\rm\cos\theta \\ \rm\sin^2\theta = \frac{1 - \rm\cos(2\theta)}{2} & \rm\cos(2\theta) = \rm\cos^2\theta - \rm\sin^2\theta \end{matrix}\]
Plugging these in we get: \[\begin{align} \sigma_{x}' &= \sigma_{x} \frac{1 + \rm\cos(2\theta)}{2} + \tau_{xy} \sin(2\theta) + \sigma_{y} \frac{1 - \rm\cos(2\theta)}{2} \\ \tau_{x'y'} &= (\sigma_{y} - \sigma_{x}) \frac{\sin(2\theta)}{2} + \tau_{xy} (\frac{1 + \rm\cos(2\theta)}{2} - \frac{1 - \rm\cos(2\theta)}{2}) \end{align}\]
Rearranging terms: \[\begin{align} \sigma_{x}' &= \frac{\sigma_{x}}{2} + \frac{\sigma_{x}\rm\cos(2\theta)}{2} + \frac{\sigma_{y}}{2} - \frac{\sigma_{y}\rm\cos(2\theta)}{2} + \tau_{xy}\sin(2\theta) \\ \tau_{x'y'} &= -\frac{(\sigma_{x} - \sigma_{y})}{2}\sin(2\theta) + (\frac{\tau_{xy}}{2} + \frac{\tau_{xy}\rm\cos(2\theta)}{2} - \frac{\tau_{xy}}{2} + \frac{\tau_{xy}\rm\cos(2\theta)}{2}) \end{align}\]
We can then simplify to the equations above. Note that to derive $\sigma_{y}'$, we use the fact that: \[\sigma_{x}' + \sigma_{y}' = \sigma_{x} + \sigma_{y}\]
Principal Stresses
The angle at which $\sigma_{x}'$ is maximized is:
\[\rm\tan(2\theta_{p1}) = \frac{2\tau_{xy}}{\sigma_x - \sigma_y} \\ \theta_{p2} = \theta_{p1} + 90^o\]
Recall that: \[\sigma_x' = \frac{\sigma_x + \sigma_y}{2} + \frac{\sigma_x - \sigma_y}{2}\rm\cos(2\theta) + \tau_{xy}\sin(2\theta)\] To maximize an equation, we take the derivative and set it equal to zero: \[\frac{d\sigma_{x}'}{d\theta} = -2\rm\sin(2\theta)[\frac{\sigma_x-\sigma_y}{2}] + 2\rm\cos(2\theta)\tau_{xy} = 0 \\ (\sigma_x - \sigma_y)\rm\sin(2\theta) = 2\tau_{xy}\rm\cos(2\theta) \\ \rm\tan(2\theta) = \frac{2\tau_{xy}}{\sigma_x - \sigma_y}\]
The maximum/minimum nomal stress values (the principal stresses) associated with $\theta_{p1}$ and $\theta_{p2}$ are:
\[\sigma_{1,2} = \frac{\sigma_x +\sigma_y}{2} \pm \sqrt{(\frac{\sigma_x - \sigma_y}{2})^2 + \tau_{xy}^2}\]
We use the convention that $\sigma_1 > \sigma_2$
Alternative Approach
The stress tensor is a physical quantity and therefore independent of the coordinate system. There are certain invariants associated with every tensor which are also independent of the coordinate system.
Because of symmetry, the stress tensor T has real eigenvalues 𝜆 and mutually perpendicular eigenvectors v such that
From linear algebra we know that a system of linear equations $A v = 0$ has a non-zero solution $\boldsymbol{v}$ if, and only if, the determinant of the matrix $\boldsymbol{A}$ is zero, that is:
\[\det(\boldsymbol{T}-\lambda\boldsymbol{I})=0\]
Exanding this equation we get: \[\det\Biggl(\begin{bmatrix} \sigma_{x} & \tau_{xy}\\ \tau_{xy} & \sigma_{y} \end{bmatrix} - \begin{bmatrix} \lambda & 0\\ 0 & \lambda \end{bmatrix} \Biggr) = 0 \] \[\det\Biggl(\begin{bmatrix} \sigma_{x}-\lambda & \tau_{xy}\\ \tau_{xy} & \sigma_{y}-\lambda \end{bmatrix} \Biggr) = 0 \]
We evalute the determinate: \[(\sigma_{x}-\lambda)(\sigma_{y}-\lambda) - \tau_{xy}^2 = 0\] \[\sigma_{x}\sigma_{y} - \lambda\sigma_{y} -\lambda\sigma_{x} + \lambda^2 - \tau_{xy}^2 = 0\]
Rearragning we get: \[\lambda^2 - \lambda(\sigma_{y} + \sigma_{x}) + \sigma_{x}\sigma_{y} - \tau_{xy}^2 = 0\]
Now we can solve for the eigenvalues using the quadratic equation where $\rm\ a = 1$, $\rm\ b = -(\sigma_{y} + \sigma_{x})$, and $\rm\ c = \sigma_{x}\sigma_{y} - \tau_{xy}^2$ \[\begin{align} \lambda &= \frac{(\sigma_{y} + \sigma_{x}) \pm \sqrt{(\sigma_{y} + \sigma_{x})^2 - 4(\sigma_{x}\sigma_{y} - \tau_{xy}^2)}}{2} \\ &= \frac{(\sigma_{x} + \sigma_{y})}{2} \pm \frac{\sqrt{\sigma_{y}^2 + 2\sigma_{y}\sigma_{x} + \sigma_{x}^2 - 4\sigma_{x}\sigma_{y} + 4\tau_{xy}^2}}{2} \\ &= \frac{(\sigma_{x} + \sigma_{y})}{2} \pm \frac{\sqrt{\sigma_{y}^2 - 2\sigma_{y}\sigma_{x} + \sigma_{x}^2 + 4\tau_{xy}^2}}{2} \\ &= \frac{(\sigma_{x} + \sigma_{y})}{2} \pm \sqrt{\frac{(\sigma_{y} - \sigma_{x})^2 + 4\tau_{xy}^2}{4}} \\ &= \frac{(\sigma_{x} + \sigma_{y})}{2} \pm \sqrt{\biggl(\frac{\sigma_{y} - \sigma_{x}}{2}\biggr)^2 + \tau_{xy}^2} \end{align}\]
This is the same result as the geometric derivation above, thus $\lambda_{1,2} = \sigma_{1,2}$.
To find the eigenvectors, we plug our eigenvalues back into the equation $(\boldsymbol{T}-\lambda\boldsymbol{I})\boldsymbol{v} = 0$. We will start with the first eigenvalue, $\lambda_1 = \sigma_1$: \[\begin{bmatrix} \sigma_{x}-\sigma_{1} & \tau_{xy}\\ \tau_{xy} & \sigma_{y}-\sigma_{1} \end{bmatrix} \begin{bmatrix} v_{11} \\ v_{12} \end{bmatrix} = 0\]
Multiplying out gives two equations: \[\begin{align} (\sigma_{x}-\sigma_{1})v_{11} + \tau_{xy}v_{12} &= 0 \\ \tau_{xy}v_{12} + (\sigma_{x}-\sigma_{1})v_{12} &= 0 \end{align} \]
The angle of the eigenvector will be: \[\theta_{p1} = \tan^{-1}\Bigl(\frac{v_{12}}{v_{11}}\Bigr)\]
This angle can be derived from both equations, therefore: \[\theta_{p1} = \tan^{-1}\Bigl(\frac{\sigma_1 - \sigma_x}{\tau_{xy}}\Bigr) = \tan^{-1}\Bigl(\frac{\tau_{xy}}{\sigma_1 - \sigma_y}\Bigr)\]
We can repeat this procedure for the second eigenvalue, $\lambda_2 = \sigma_2$: \[\theta_{p2} = \tan^{-1}\Bigl(\frac{\sigma_2 - \sigma_x}{\tau_{xy}}\Bigr) = \tan^{-1}\Bigl(\frac{\tau_{xy}}{\sigma_2 - \sigma_y}\Bigr)\]
Maximum Shear Stress
The angle at which $\tau_{x'y'}$ is maximized is:
\[\rm\tan(2\theta_{s1}) = \frac{-(\sigma_x - \sigma_y)}{2\tau_{xy}} \\ \theta_{s2} = \theta_{s1} + 90^o\]
Recall that: \[\tau_{x'y'} = -\frac{\sigma_x - \sigma_y}{2}\rm\sin(2\theta) + \tau_{xy}\cos(2\theta)\] To maximize an equation, we take the derivative and set it equal to zero: \[\frac{d\tau_{x'y'}}{d\theta} = -\frac{\sigma_x - \sigma_y}{2}2\rm\cos(2\theta) - \tau_{xy}2\rm\sin(2\theta) = 0 \\ -2\tau_{xy}\rm\sin(2\theta) = (\sigma_x - \sigma_y)\rm\cos(2\theta) \\ \rm\tan(2\theta) = \frac{-(\sigma_x - \sigma_y)}{2\tau_{xy}}\]
The maximum/minimum in plane shear stress values associated with $\theta_{s1}$ and $\theta_{s2}$ are:
\[|\tau_{max}| = \sqrt{(\frac{\sigma_x - \sigma_y}{2})^2 + \tau_{xy}^2}\]
A maximum shear sterss element has \[\sigma_{x}' = \sigma_{y}' = \sigma_{avg} = \frac{\sigma_x + \sigma_y}{2}\] i.e. unlike with the principal stress element, the normal stresses are not zero.
The orientations for the principal stress element and max shear stress element are $45^0$ apart.
Mohr's Circle
Mohr's circle is a graphical representation of stress transformations. The equations for stress transformations actually describe a circle if we consider the normal stress $\sigma$ to be the x-coordinate and the shear stress $\tau$ to be the y-coordinate.
Circle Centroid:
\[C = \sigma_{avg} = \frac{\sigma_x +\sigma_y}{2} = \frac{\sigma_1 +\sigma_2}{2}\]
Circle Radius: \[R = \sqrt{(\frac{\sigma_x - \sigma_y}{2})^2 + \tau_{xy}^2}\]
We classify two points: \[ \begin{align} \rm\ Point \ X&: (\sigma_x, \tau_{xy}) \\ \rm\ Point \ Y&: (\sigma_y, -\tau_{xy}) \end{align}\]
State of Stress
The state of plane stress at a point on a body is represented by the element below.
Angle: | $\theta = $ {{theta}} $^o $ |
Normal stress ($\sigma_x$): | $\sigma_x = $ {{sigmax}} $MPa $ |
Normal stress ($\sigma_y$): | $\sigma_y = $ {{sigmay}} $MPa $ |
Shear stress ($\tau_{xy}$): | $\tau_{xy} = $ {{tauxy}} $MPa $ |
State of Stress - Beam
Let us analyse the stress field in this cantilever beam problem:
Choose the coordinates of the point in which you want to obtain the state of plane stress (at the plane z = 0):
x =
x = {{beam-xp}} mm
y =
y = {{beam-yp}} mm
The state of plane stress at this point is represented by the element below.
Angle:
$\theta = $ {{beam-theta}} $^o $
Let us analyse the stress field in this cantilever beam problem:
Choose the coordinates of the point in which you want to obtain the state of plane stress (at the plane z = 0):
x = | x = {{beam-xp}} mm |
y = | y = {{beam-yp}} mm |
The state of plane stress at this point is represented by the element below.
Angle: | $\theta = $ {{beam-theta}} $^o $ |