Stress Transformation - Inclined sections from other pages have moved

General State of Stress

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.

Taken from TAM251 Lecture Notes - L9S2

$$ \begin{bmatrix} \sigma_{x} & \tau_{xy} & \tau_{xz} \\ \tau_{xy} & \sigma_{y} & \tau_{yz} \\ \tau_{xz} & \tau_{yz} & \sigma_{z} \end{bmatrix} $$

Note: stress is a physical quantity and as such, it is independent of the chosen coordinate system.

Sign Convention

Sign conventions on 2D elements

Positive normal stress acts outward from all faces
Positive shear stress points towards the positive axis direction in a positive face
Positive shear stress points towards the negative axis direction in a negative face

Plane Stress

Taken from TAM251 Lecture Notes - L9S5

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.

Taken from TAM251 Lecture Notes - L9S3

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 $ 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. Cauchys stress theorem states that there exists a stress tensor $ T $ (which is independent of $ n $), such that $ t^n $ is a linear function of $ 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 stress 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.

Taken from TAM251 Lecture Notes - L9S6

Sign Convention

Both the
The orientation of an inclined plane (on which the normal and shear stress components are to be determined) will be defined using the angle

$$ \begin{align} \sigma_x' &= \frac{\sigma_x + \sigma_y}{2} + \frac{\sigma_x - \sigma_y}{2}\rm\cos(2\theta) + \tau_{xy}\sin(2\theta) \\ \sigma_y' &= \frac{\sigma_x + \sigma_y}{2} - \frac{\sigma_x - \sigma_y}{2}\rm\cos(2\theta) - \tau_{xy}\sin(2\theta) \\ \tau_{x'y'} &= -\frac{\sigma_x - \sigma_y}{2}\rm\sin(2\theta) + \tau_{xy}\cos(2\theta) \end{align} $$

**Expandable Derivation** 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:

$$ \sigma_{x}' + \sigma_{y}' = \sigma_{x} + \sigma_{y} $$

**End Derivation**

Stresses in Inclined Planes - moved from stresses page

The relations above are observed only on planes perpendicular to the axis of the member or connection

$$ \begin{align} \sigma_n &= {\bf n}\cdot\ {\bf t}^{n}={\bf n}\cdot\ {\bf T}\,{\bf n} = \sigma_{x} \cos ^2(\theta) + 2 \,\tau_{xy} \sin(\theta) \cos(\theta)+\sigma_{y} \sin ^2(\theta) \\ \tau_{ns} &= {\bf s}\cdot\ {\bf t}^{n}={\bf s}\cdot\ {\bf T}\,{\bf n} = (\sigma_y - \sigma_{x}) \sin(\theta) \cos(\theta) + \tau_{xy} ( \cos^2(\theta) - \sin^2(\theta) ) \end{align} $$

*Expandable derivation*

$$ \sum F_x: -\sigma_x (A\rm\cos\theta) - \tau_{xy}(A\rm\sin\theta) + (\sigma_{x}'\rm\cos\theta)A - (\tau_{x'y'}\rm\sin\theta)A=0\ $$

$$ \sum F_y: -\sigma_y (A\rm\sin\theta) - \tau_{xy}(A\rm\cos\theta) + (\sigma_{x}'\rm\sin\theta)A - (\tau_{x'y'}\rm\cos\theta)A=0\ $$

Rearrange terms:

$$ A[\sigma_{x}'\rm\cos\theta - \tau_{x'y'}\rm\sin\theta] = A[\sigma_{x}\rm\cos\theta + \tau_{xy}\rm\sin\theta]\ $$

$$ A[\sigma_{x}'\rm\sin\theta - \tau_{x'y'}\rm\cos\theta] = A[\sigma_{x}\rm\sin\theta + \tau_{xy}\rm\cos\theta]\ $$

Combine into a matrix:

$$ \begin{bmatrix} \rm\cos\theta & -\rm\sin\theta \\ \rm\sin\theta & \rm\cos\theta \end{bmatrix} \begin{bmatrix}\sigma_{x}' \\ \tau_{x'y'} \end{bmatrix} = \begin{bmatrix} \sigma_{x}\rm\cos\theta & \tau_{xy}\rm\sin\theta \\ \sigma_{y}\rm\sin\theta & \tau_{xy}\rm\cos\theta \end{bmatrix} $$

Multiply by the inverse:

$$ \begin{bmatrix} \sigma_{x}' \\ \tau_{x'y'} \end{bmatrix} = \begin{bmatrix} \rm\cos\theta & \rm\sin\theta \\ -\rm\sin\theta & \rm\cos\theta \end{bmatrix} \begin{bmatrix} \sigma_{x}\rm\cos\theta & \tau_{xy}\rm\sin\theta \\ \sigma_{y}\rm\sin\theta & \tau_{xy}\rm\cos\theta \end{bmatrix} $$

**End Derivation**

Torsion in Inclined Planes - moved from torsion page - should this go under the max shear stress subsection?

Recall projected forces:

$$ \sigma_n = \frac{P}{A_o}(\cos^2(\theta))\ $$

$$ \tau_{ns} = -\frac{P}{A_o}\sin(\theta)\cos(\theta)\ $$

Maximum normal stress at 90 degrees:

Max normal stress

Maximum shear stress at 45 degrees:

Max shear stress

A circular shaft under torsion develops PURE SHEAR on cross sections between longitudinal planes (the faces of element $ a $ are parallel and perpendicular to the axis of the shaft)

Taken from TAM251 Lecture Notes - L9S19

$$ \sigma_n = 2\tau_{max} \sin\theta \cos\theta = \tau_{max} \sin (2\theta)\ $$

$$ \tau_{ns} = \tau_{max}(\cos^2\theta - \sin^2\theta) = \tau_{max} \cos(2\theta)\ $$

From ref pages

Maximum normal stress at:

$$ \sigma_n = \tau_{max} = \frac{Tc}{J}\ $$

$$ \sigma_s = -\tau_{max} = -\frac{Tc}{J}\ $$

$$ \tau_{ns} = 0\ $$

Maximum shear stress at:

$$ \sigma_n = 0\ $$

$$ \sigma_s = 0\ $$

$$ \tau_{ns} = \tau_{max} = \frac{Tc}{J}\ $$

Principal Stresses

Taken from TAM251 Lecture Notes - L9S10

The angle at which $ \sigma_{x}' $ is maximized is:

$$ \rm\tan(2\theta_{p1}) = \frac{2\tau_{xy}}{\sigma_x - \sigma_y}\ $$

$$ \theta_{p2} = \theta_{p1} + 90^o\ $$

**Expandable Derivation** 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:

$$ \frac{d\sigma_{x}'}{d\theta} = -2\rm\sin(2\theta)[\frac{\sigma_x-\sigma_y}{2}] + 2\rm\cos(2\theta)\tau_{xy} = 0 \ $$

$$ (\sigma_x - \sigma_y)\rm\sin(2\theta) = 2\tau_{xy}\rm\cos(2\theta)\ $$

$$ \tan(2\theta) = \frac{2\tau_{xy}}{\sigma_x - \sigma_y}\ $$

**End Derivation** The maximum/minimum normal 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 $

from ref pages

Alternative Approach: Eigenvalues

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.

First-order tensors (vectors): magnitude is the invariant of a vector, since it is independent of the coordinate system chosen to represent the vector.
Second-order tensors (matrices): three independent invariant quantities associated with it. One set of such invariants are the eigenvalues of the stress tensor, which are called the principal stresses. The eigenvectors define the principal direction vectors.

Because of symmetry, the stress tensor T has real eigenvalues $ \lambda $ and mutually perpendicular eigenvectors $ v $ such that

$$ Tv = \lambda v \rightarrow (T-\lambda I)v = 0\ $$

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:

$$ \det(\boldsymbol{T}-\lambda\boldsymbol{I})=0\ $$

**Expandable Derivation** Expanding 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 evaluate the determinate:

$$ (\sigma_{x}-\lambda)(\sigma_{y}-\lambda) - \tau_{xy}^2 = 0\ $$

$$ \sigma_{x}\sigma_{y} - \lambda\sigma_{y} -\lambda\sigma_{x} + \lambda^2 - \tau_{xy}^2 = 0\ $$

Rearranging 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} $. **End Derivation** **Expandable Derivation** 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:

$$ (\sigma_{x}-\sigma_{1})v_{11} + \tau_{xy}v_{12} = 0\ $$

$$ \tau_{xy}v_{12} + (\sigma_{x}-\sigma_{1})v_{12} = 0\ $$

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 $:

$$ \theta_{p2} = \tan^{-1}\Bigl(\frac{\sigma_2 - \sigma_x}{\tau_{xy}}\Bigr) = \tan^{-1}\Bigl(\frac{\tau_{xy}}{\sigma_2 - \sigma_y}\Bigr)\ $$

**End Derivation**

Maximum Shear Stress

Taken from TAM251 Lecture Notes - L9S13

The angle at which $ \tau_{x'y'} $ is maximized is:

$$ \tan(2\theta_{s1}) = \frac{-(\sigma_x - \sigma_y)}{2\tau_{xy}}\ $$

$$ \theta_{s2} = \theta_{s1} + 90^o\ $$

**Expandable Derivation** 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:

$$ \frac{d\tau_{x'y'}}{d\theta} = -\frac{\sigma_x - \sigma_y}{2}2\rm\cos(2\theta) - \tau_{xy}2\rm\sin(2\theta) = 0\ $$

$$ -2\tau_{xy}\rm\sin(2\theta) = (\sigma_x - \sigma_y)\rm\cos(2\theta)\ $$

$$ \tan(2\theta) = \frac{-(\sigma_x - \sigma_y)}{2\tau_{xy}}\ $$

**End Derivation** 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 stress 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.

From ref pages

The orientations for the principal stress element and max shear stress element are $ 45^o $ apart.

Mohr's Circle

From ref pages

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 $ \pi $ 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:

$$ \begin{align} \rm\ Point \ X&: (\sigma_x, \tau_{xy}) \\ \rm\ Point \ Y&: (\sigma_y, -\tau_{xy}) \end{align} $$

Interactive: State of Stress

The state of plane stress at a point on a body is represented by the element below. **Interactive element broken**

Interactive: State of Stress in a Beam

Let us analyse the stress field in this cantilever beam problem:

From ref pages

Choose the coordinates of the point in which you want to obtain the state of plane stress (at the plane z = 0): The state of plane stress at this point is represented by the element below. **Interactive element broken**

Solid Mechanics Reference

Stress Transformation - Inclined sections from other pages have moved