# Engineering Stress ## Axial Stress Load applied on an area direct along the axis of the object: $ \sigma=\frac{P}{A_0}\:\left[\frac{\text{N}}{\text{m}^2}\right]=[\text{Pa}] $ - $P$ - load - $A_0$ - cross-sectional area of the specimen before deformation $ P=\int dF=\int_A\sigma\,dA $ Axial stress distribution assumed to be uniform across area, acts through the centroid of the object. ## Shear Stress Load applied on an area perpendicular to the axis of the object: $ \tau_\text{ave}=\frac{P}{A} $ Cannot be assumed to be uniformly distributed. ## Axial Loading on an Oblique Plane Decompose $P$ into normal component $F$ and tangential component $V$: $ F=P\cos\theta\quad V=P\sin\theta $ $ \sigma=\frac{F}{A_\theta}\quad\tau=\frac{V}{A_\theta} $ $ A_\theta=A_0/\cos\theta $ $ \sigma=\frac{P}{A_0}\cos^2\theta\quad\tau=\frac{P}{A_0}\sin\theta\cos\theta $ Naming convention: $ \begin{array}{c} \tau_{xy}\\ \;\;\swarrow\;\searrow\\ \mbox{perpedicular surface}\quad \mbox{direction of component} \end{array} $