Principal Stresses, Principal Planes & Mohr's Circle

Why stresses on inclined planes matter

The stress on a plane depends on the orientation of that plane. A bar in simple tension has no shear stress on its cross-section, yet it has shear on planes inclined to the axis — which is why ductile bars can fail by shear along 45° lines. Design against failure therefore needs the largest normal and shear stresses at a point, whatever planes they act on.

Sign convention used here: tensile normal stress positive; shear stress positive when it tends to rotate the element clockwise on the $x$-face as drawn in most Indian textbooks. The plane angle $\theta$ is measured from the plane on which ${\sigma }_x$ acts (equivalently, the normal of the inclined plane is at $\theta$ from the $x$-axis). Examiners accept any consistent convention; the magnitudes do not change.

Case 1 — uniaxial stress

A bar under direct stress $\sigma$. On a plane whose normal makes angle $\theta$ with the axis:

• ${\sigma }_n$ is maximum ($=\sigma$) at $\theta =0$ — the cross-section.
• $\tau$ is maximum ($=\sigma /2$) at $\theta =45°$, where ${\sigma }_n=\sigma /2$ as well.

Case 2 — two perpendicular normal stresses (biaxial)

${\sigma }_x$ and ${\sigma }_y$ act without shear:

• Principal stresses are ${\sigma }_x$ and ${\sigma }_y$ themselves.
• ${\tau }_{max}=\frac{{\sigma }_x-{\sigma }_y}{2}$ at 45°.
• If ${\sigma }_x={\sigma }_y$ (equal biaxial, like a thin sphere), $\tau =0$ on every plane.

Case 3 — pure shear

Only ${\tau }_{xy}$ acts. Then ${\sigma }_n={\tau }_{xy}\sin 2\theta$ and the principal stresses are $\pm {\tau }_{xy}$ on planes at 45°. A shaft in torsion is in this state.

Case 4 — general two-dimensional stress

With ${\sigma }_x$, ${\sigma }_y$ and ${\tau }_{xy}$:

Principal planes

Setting ${\tau }_t=0$:

This gives two values of ${\theta }_p$, 90° apart — the two principal planes.

Principal stresses