Stress and Stress Transformation

Stress

Internal resistance offered by a unit area of a material to an externally applied load

Normal Stress

$\sigma$:

• Tension (+)
• Compression (-)

Tangential (Transverse) Shear Stress

$\tau$:
$\tau =\frac{VQ}{Ib}$
Where Q is the first moment of area (the area above the point of interest)
Q can be calculated using:
$Q=\overline{y}A$
Where y bar is the CoM of the area in question.
Or formally:
${\int }_{x_1}^{x_2}xdA$

Where this can be interpreted as a weighted area where the weight is given by the distance from the x-axis in this case.

Hollow shaft Shear Stress

The shear stress for a hallow shaft section is given by
$\tau =\frac{T}{2tA}$
For an equilateral triangle, area is given by
$A=\frac{1}{2}b\cdot \frac{\sqrt{3}}{2}b$

Cartesian Stress Components

• Single small cube within the analyzed member
• Goal is to identify the stress element (mini cube) with the highest stress
• Nine stress elements typically

${\sigma }_x,{\sigma }_y,{\sigma }_z$
${\tau }_{xy},{\tau }_{xz},{\tau }_{yx},{\tau }_{yz},{\tau }_{zx},{\tau }_{zy}$

Stress Tensors

For a 3D stress tensor
$\sigma =\left[\begin{array}{ccc}{\sigma }_{xx}& {\tau }_{xy}& {\tau }_{xz}\\ {\tau }_{yx}& {\sigma }_{yy}& {\tau }_{yz}\\ {\tau }_{zx}& {\tau }_{zy}& {\sigma }_{zz}\end{array}\right]$
The cross shears are equal
${\tau }_{yz}={\tau }_{xy}$ etc…
This concept can be applied to 2D tensors as well.

Plane Stresses

Plane Stress

The state of stress in which two faces of a cubic element are free of stress.

This occurs in a thin plate subjected to forces acting in the midplane of the plate.

Transformation of Plane Stress

Quote

When a 3D stress element is cut with an oblique plane with a normal (n) at an arbitrary angle (phi) counter-clockwise from the x-axis.

The transformation equations associated with plane stress are as follows for CCW cases:

\\ \tau' &= -\frac{\sigma_x - \sigma_y}{2} \sin(2\phi) + \tau_{xy} \cos(2\phi) \tag{2} \end{align}$$ Also for sigma_y, the second two terms are negative. ### Principle Stresses #### Max and Min Normal Stress To maximize sigma (normal stress), you can differentiate normal plane-stress and set it = to 0. ![[Pasted image 20240516083800 1.png|400]] $$\tan\left(2\phi_{p}\right)=\frac{2\tau_{xy}}{\sigma_{x}-\sigma_{y}}$$ This yields two principle directions associated with two principle stresses where tau = 0 in these principle directions. $$\sigma_1,\sigma_2=\frac{\sigma_{x}+\sigma_{y}}{2}_{}\pm\sqrt{\left(\frac{\sigma_{x}-\sigma_{y}}{2}_{}\right)^2}^{}+\tau_{xy}^2$$ #### Max Shear Stress The same thing can be done for shear stress ![[Pasted image 20240516084120 1.png|400]] $$\tan\left(2\phi_{s}\right)=\frac{\sigma_{x}-\sigma_{y}}{2\tau_{xy}}$$ Yielding: $$\tau_1,\tau_2=\pm\sqrt{\left(\frac{\sigma_{x}-\sigma_{y}}{2}_{}\right)^2+\tau_{xy}^2}^{}$$ In these directions: $$\sigma=\sigma_{avg}=\frac{\sigma_{x}+\sigma_{y}}{2}$$ Also note that $$\phi_{s}=\phi_{p}-45^{degrees}$$