Min Principal Stress⚠ unverified

Understanding the Parameters

• Average minus radius. Where $\sigma_1$ is the Mohr-circle centre plus the radius, $\sigma_2$ is the centre minus the radius — the two principal stresses straddle the average $\tfrac12(\sigma_x+\sigma_y)$ symmetrically (Mohr Circle Center, Mohr's Circle Radius).

• The shear and difference terms — a bigger shear $\tau_{xy}$ or normal-stress difference pushes $\sigma_2$ more negative (more compressive) just as it pushes $\sigma_1$ more positive. Under pure shear ($\sigma_x=\sigma_y=0$) the principals are $\pm\tau_{xy}$: equal tension and compression at $45^\circ$.

• The output $\sigma_2$ — the least tensile (or most compressive) normal stress, on the principal plane perpendicular to $\sigma_1$'s. When it is strongly negative it drives compression-side failures.

• Maximum shear from the pair. The greatest in-plane shear is $\tau_{max} = \tfrac12(\sigma_1-\sigma_2)$ = the Mohr radius, and the Tresca equivalent stress is $\sigma_1-\sigma_2$ (in plane stress, accounting also for the zero out-of-plane principal). So $\sigma_2$ is essential to any shear-based check.

Derivation (Approaching a Proof)

The derivation is identical to that for the maximum principal stress. Stationarising the transformed normal stress $\sigma_\theta = \tfrac12(\sigma_x+\sigma_y) + \tfrac12(\sigma_x-\sigma_y)\cos2\theta + \tau_{xy}\sin2\theta$ over the plane angle $\theta$ gives the principal-plane condition $\tan2\theta_p = 2\tau_{xy}/(\sigma_x-\sigma_y)$, and back-substitution yields both extreme values at once:

$$\sigma_{1,2} = \frac{\sigma_x+\sigma_y}{2} \pm \sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2 + \tau_{xy}^2}.$$

The minus root is the minimum principal stress $\sigma_2$. $\blacksquare$ Equivalently, $\sigma_1$ and $\sigma_2$ are the two eigenvalues of the symmetric stress tensor; their sum equals the invariant $\sigma_x+\sigma_y$ (trace) and their product equals $\sigma_x\sigma_y-\tau_{xy}^2$ (determinant) — the two in-plane stress invariants, unchanged by any rotation of axes.

Dimensional check. Average and radius are both stresses ($\text{Pa}$), so $\sigma_2$ is $\text{Pa}$. $\checkmark$

History and Development

• The invariant pair. That $\sigma_1+\sigma_2$ and $\sigma_1\sigma_2$ are rotation-invariant is the practical power of principal stresses: the physical state does not care how we orient our axes. This follows from Cauchy's stress-tensor formulation (1820s).

• Compression failures. Aerospace thin structures rarely fail by tensile overload; they fail by instability under compression — buckling (Buckling Stress), crippling (Crippling Stress), and, in composites, delamination and matrix crushing. The minimum principal stress is the flag for these, complementing the tension-focused $\sigma_1$.

• Together they close the state. Only both principals (or, equivalently, the Mohr circle) fully specify the in-plane state — which is why the calculator ships them as a matched pair.

Related Concepts: Max Principal Stress, Mohr Circle Center, Mohr's Circle Radius, Principal Stresses, Tresca Stress, Von Mises Stress 2D, Buckling Stress

Notes: Registry calculator min-principal-stress (unverified). Correct as shipped. $\sigma_2 = $ centre $-$ radius of Mohr's circle. Maximum in-plane shear $\tau_{max}=\tfrac12(\sigma_1-\sigma_2)$. Governs compression-side failure modes. All defaults $1.0$.