--- title: "Principal stresses, stress invariant, equivalent stress" subject: "Strength of Materials" chapter: 4 key: "semester_3/strength_of_materials:stress-invariant-equivalent-stress" source: https://www.bctengineer.com/q/semester_3/strength_of_materials/stress-invariant-equivalent-stress site: BCT Engineering --- # Principal stresses, stress invariant, equivalent stress Strength of Materials · Chapter 4: Stress and Strain Analysis - Asked in: 2079 Bhadra, 10 marks - Video: [Stress invariants and the equivalent stress](https://www.youtube.com/watch?v=u-xclHeuwos) — SEM LAB TV - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/strength_of_materials/stress-invariant-equivalent-stress - Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-4 ## Question The tensile stress at a point on two perpendicular planes along X-axis and Y-axis are $120\ \text{MN/m}^2$ and $60\ \text{MN/m}^2$ respectively. Find principle stresses and their direction. Based on obtained data, verify stress invariant concept. What will be the intensity of stress which acting alone can produce same maximum strain? Take poisson's ratio = 1/4. ## Short answer Since τ=0, principal stresses are σ1=120, σ2=60 MN/m². Invariant I1 = σx+σy = 180 MN/m², verified for any perpendicular planes. Maximum strain εmax = (σ1 − νσ2)/E = (120−0.25×60)/E = 105/E, so equivalent stress σeq = 105 MN/m². ## Answer **Given:** $\sigma_x = 120$ and $\sigma_y = 60\ \text{MN/m}^2$ (tensile), no shear stress ($\tau_{xy} = 0$), $\nu = 1/4$. #### Principal stresses and directions Since there is no shear on the X and Y planes, these planes are already principal planes. $$ \sigma_{1,2} = \frac{120+60}{2} \pm \sqrt{\left(\frac{120-60}{2}\right)^2 + 0} = 90 \pm 30 $$ $$ \sigma_1 = 120\ \text{MN/m}^2\ \text{(on the plane normal to the X-axis, }\theta_p = 0^\circ), \qquad \sigma_2 = 60\ \text{MN/m}^2\ \text{(}\theta_p = 90^\circ) $$ $\tan 2\theta_p = \dfrac{2\tau_{xy}}{\sigma_x-\sigma_y} = 0 \Rightarrow \theta_p = 0^\circ$ and $90^\circ$. #### Stress invariant The sum of normal stresses on any two perpendicular planes is constant (an invariant of plane stress). Check with a plane at, say, $\theta = 30^\circ$ and its perpendicular ($120^\circ$): $$ \sigma_{30} = 90 + 30\cos60^\circ = 105, \qquad \sigma_{120} = 90 + 30\cos240^\circ = 75 $$ $$ \sigma_x + \sigma_y = 120 + 60 = 180, \qquad \sigma_1 + \sigma_2 = 120 + 60 = 180, \qquad \sigma_{30} + \sigma_{120} = 105 + 75 = 180 $$ So $\sigma_x + \sigma_y = \sigma_1 + \sigma_2 = 180\ \text{MN/m}^2$ is verified. #### Equivalent stress for the same maximum strain Maximum strain (in the direction of $\sigma_1$): $$ \varepsilon_{max} = \frac{1}{E}(\sigma_1 - \nu\sigma_2) = \frac{1}{E}(120 - 0.25\times60) = \frac{105}{E} $$ A single stress $\sigma_e$ acting alone gives the strain $\sigma_e/E$. Equating: $$ \sigma_e = 105\ \text{MN/m}^2 $$ **Answer:** $\sigma_1 = 120$, $\sigma_2 = 60$ MN/m² (at $0^\circ$ and $90^\circ$); the invariant is $180$ MN/m²; equivalent stress $= 105$ MN/m². ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why are X and Y planes the principal planes? A. There is no shear stress on them B. The stresses are equal C. The stresses are compressive D. Poisson's ratio is 1/4 Answer: A. Zero shear means tan 2 theta_p = 0, so theta_p = 0 and 90 degrees. (Key point: Principal planes) 2. What is invariant? A. Difference of normal stresses B. Product of the two stresses C. Sum of normal stresses on any two perpendicular planes (180 MN/m^2) D. Shear stress on every plane Answer: C. sigma_x + sigma_y = sigma_1 + sigma_2 = sigma_30 + sigma_120 = 180. (Key point: Stress invariant) 3. Which expression is the maximum strain? A. (sigma_1 - nu sigma_2)/E B. (sigma_1 + nu sigma_2)/E C. sigma_1 sigma_2 /E D. (sigma_1 - sigma_2)/E Answer: A. Lateral stress reduces the strain through Poisson's effect: (120 - 0.25 x 60)/E. (Key point: Maximum strain) 4. What is the equivalent stress? A. 90 MN/m^2 B. 120 MN/m^2 C. 105 MN/m^2 D. 135 MN/m^2 Answer: C. sigma_e/E = 105/E. 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