Principal Stresses and Mohr’s Circle

Stress on a cut depends on the cut’s orientation. Among all orientations, some planes see zero shear — the principal planes — and the normal stresses there are the principal stresses. Finding them (analytically or by Mohr’s circle) is the gateway to failure theories (Bansal Ch. 3).

What we cover
- Oblique stresses under uniaxial and biaxial loading
- Analytical principal stress formulas
- Mohr’s circle construction and reading
- Max shear and exam pitfalls

**Uniaxial tension $\sigma$**, plane at angle $\theta$ to the cross-section (Bansal’s $\theta$ from the bar axis may differ — match the figure):

Maximum shear is $\frac{\sigma}{2}$ at $45^\circ$.

General plane stress $(\sigma_x,\sigma_y,\tau_{xy})$ on a plane at $\theta$:

Set $\tau_n=0$ (or extremise $\sigma_n$) to get

and

Order them algebraically $\sigma_1\ge\sigma_2$. In plane stress the third principal is often $0$ — check whether $|\sigma_1-0|$ or $|\sigma_2-0|$ governs absolute max shear for failure theories.

Plot points $(\sigma_x,\tau_{xy})$ and $(\sigma_y,-\tau_{xy})$ (with your sign convention for $\tau$). The circle through them has:
- Centre $C=(\frac{\sigma_x+\sigma_y}{2},\,0)$
- Radius

Horizontal intercepts are $\sigma_1$ and $\sigma_2$. Top/bottom of the circle give $\pm\tau_{\max}$ (in-plane). Angles on the circle are **$2\theta$** relative to the element.

Graphical method is equivalent to the analytical formulas — use whichever is faster under exam time pressure.

- Forgetting the factor $\frac{1}{2}$ on $(\sigma_x-\sigma_y)$ inside the square root.
- Reporting only in-plane $\tau_{\max}$ when $\sigma_3=0$ gives a larger shear.
- Mixing element angle $\theta$ with Mohr double-angle $2\theta$.
- Applying Rankine/Tresca to $\sigma_x$ without finding principals.