Stress
The internal forces and moments generally vary from point to point. Obtaining this distribution is of primary importance in mechanics of materials. The total force in a cross-section, divided by the cross-sectional area, is the stress. We use stress to normalize forces with respect to the size of the geometry.
Units
The units of stress can be found in the table below.
| Measure | SI system | US system |
|---|---|---|
| Force | $[N]$ | $[lb]$ |
| Area | $[m^2]$ | $[in^2]$ |
| Stress | Pascal = $[Pa]$=$[N/m^2]$ | pound per square inch = $[psi]$ = $[lb/in^2]$ |
It is also important to recall some of the most common prefixes used to denote quantity.
| kiloPascal [kPa] | $10^3$ [Pa] |
| MegaPascal [MPa] | $10^6$ [Pa] |
| GigaPascal [GPa] | $10^9$ [Pa] |
| kilopound per square inch [ksi] | $10^3$ [psi] |
Stress under general loading conditions
We consider a homogeneous distribution of the internal force $\Delta F$ over an infinitesimal area $\Delta A$. The stress is defined by the infinitesimal force divided by the infinitesimal area
- Normal Stress: Defined by the intensity of the force acting NORMAL to $\Delta A$
- Shear Stress: Defined by the intensity of the force acting TANGENT to $\Delta A$
Average Normal Stress - Axial Loading
Here we assume that the distribution of normal stresses in an axially loaded member is uniform. Stress is calculated away from the points of application of the concentrated loads. Uniform distribution of stress is possible only if the line of action of the concentrated load $P$ passes through the centroid of the section considered
\[\sigma_{ave} = \frac{F}{A}\]
where $F$ is the internal resultant normal force and $A$ is the cross-sectional area of the bar where the normal stress $\sigma_{ave}$ is calculated
Average Shear Stress - Axial Loading
Obtained when transverse forces are applied to a member. The distribution of shear stresses cannot be assumed uniform. Common in bolts, pins and rivets used to connect various structural members
\[\tau_{ave} = \frac{V}{A}\]
where $V$ is the internal resultant shear force and $A$ is the cross-sectional area of the bar where the shear stress $\tau_{ave}$ is calculated
Stress Tensor
The components of normal and shear stress can be combined into the stress tensor.
Stresses in Inclined Planes
The relations above are observed only on planes perpendicular to the axis of the member or connection
\[\begin{align} \sigma_n &= {\bf n}\cdot\ {\bf t}^{n}={\bf n}\cdot\ {\bf T}\,{\bf n} = \sigma_{x} \cos ^2(\theta) + 2 \,\tau_{xy} \sin(\theta) \cos(\theta)+\sigma_{y} \sin ^2(\theta) \\ \tau_{ns} &= {\bf s}\cdot\ {\bf t}^{n}={\bf s}\cdot\ {\bf T}\,{\bf n} = (\sigma_y - \sigma_{x}) \sin(\theta) \cos(\theta) + \tau_{xy} ( \cos^2(\theta) - \sin^2(\theta) ) \end{align}\]
\[\sum F_x: -\sigma_x (A\rm\cos\theta) - \tau_{xy}(A\rm\sin\theta) + (\sigma_{x}'\rm\cos\theta)A - (\tau_{x'y'}\rm\sin\theta)A=0\] \[\sum F_y: -\sigma_y (A\rm\sin\theta) - \tau_{xy}(A\rm\cos\theta) + (\sigma_{x}'\rm\sin\theta)A - (\tau_{x'y'}\rm\cos\theta)A=0\]
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: \[\begin{bmatrix} \rm\cos\theta & -\rm\sin\theta \\ \rm\sin\theta & \rm\cos\theta \end{bmatrix} \begin{bmatrix} \sigma_{x}' \\ \tau_{x'y'} \end{bmatrix} = \begin{bmatrix} \sigma_{x}\rm\cos\theta & \tau_{xy}\rm\sin\theta \\ \sigma_{y}\rm\sin\theta & \tau_{xy}\rm\cos\theta \end{bmatrix} \]
Multiple by the inverse: \[\begin{bmatrix} \sigma_{x}' \\ \tau_{x'y'} \end{bmatrix} = \begin{bmatrix} \rm\cos\theta & \rm\sin\theta \\ -\rm\sin\theta & \rm\cos\theta \end{bmatrix} \begin{bmatrix} \sigma_{x}\rm\cos\theta & \tau_{xy}\rm\sin\theta \\ \sigma_{y}\rm\sin\theta & \tau_{xy}\rm\cos\theta \end{bmatrix} \]
Allowable Stress Design
Design Requirement: A structural design is intended to support and/or transmit loads while maintaining safety and utility: don't break. Strength of a structure reflects its ability to resist failure.
- Ultimate Load ($P_u$): force when specimen fails (breaks).
- Ultimate normal stress ($\sigma_u$): \[\sigma_u = \frac{P_u}{A}\]
Factor of Safety (FS): Ratio of structural strength to maximum (allowed) applied load ($P_{all}$)
\[P_{all} = \frac{P_u}{FS}\] Similarly: \[\sigma_{all} = \frac{\sigma_u}{FS}\]