--- title: "Principal stresses and planes (12, 8, 6 kN/cm2)" subject: "Strength of Materials" chapter: 4 key: "semester_3/strength_of_materials:principal-stresses-6-8-12-kncm2" source: https://www.bctengineer.com/q/semester_3/strength_of_materials/principal-stresses-6-8-12-kncm2 site: BCT Engineering --- # Principal stresses and planes (12, 8, 6 kN/cm2) Strength of Materials · Chapter 4: Stress and Strain Analysis - Asked in: 2069 Chaitra, 8 marks - Video: [How to draw Mohrs Circle (Shear strength), Mumbai University Solved Example.](https://www.youtube.com/watch?v=fHez2xUjrBQ) — civil engineering - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/strength_of_materials/principal-stresses-6-8-12-kncm2 - Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-4 ## Question Figure below shows the state of stress of point in a two dimensional stressed body. Determine the values of principal stresses and orientation of principal planes. [Figure: element with $12\ \text{kN/cm}^2$ tensile on the left and right faces, $8\ \text{kN/cm}^2$ compressive on the top and bottom faces, and shear stress $6\ \text{kN/cm}^2$.] ## Short answer Given σx=12 kN/cm² tensile, σy=-8 kN/cm² compressive, τxy=6, principal stresses σ1=13.66 kN/cm², σ2=-9.66 kN/cm², with planes at 15.48° and 105.48° from x-axis. ## Answer **Reading of the figure:** $\sigma_x = 12$ kN/cm² (tensile, left/right faces), $\sigma_y = -8$ kN/cm² (compressive, top/bottom faces), $\tau_{xy} = 6$ kN/cm². #### Principal stresses $$ \sigma_{1,2} = \frac{\sigma_x+\sigma_y}{2} \pm \sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2} = 2.00 \pm \sqrt{(10.00)^2+(6.00)^2} = 2.00 \pm 11.66 $$ $\sigma_1 = 13.66\ \text{kN/cm}^2$ and $\sigma_2 = -9.66\ \text{kN/cm}^2$. #### Principal planes $$ \tan 2\theta_p = \frac{2\tau_{xy}}{\sigma_x-\sigma_y} = \frac{2(6.00)}{20.00} \Rightarrow 2\theta_p = 30.96^\circ $$ So $\theta_p = 15.48^\circ$ (normal of the plane carrying $\sigma_1$) and $105.48^\circ$ (normal of the plane carrying $\sigma_2$), both measured anticlockwise from the $x$-axis. There is no shear stress on these planes. **Answer:** $\sigma_1 = 13.66$ kN/cm², $\sigma_2 = -9.66$ kN/cm²; principal planes at $15.48^\circ$ and $105.48^\circ$ from the $x$-axis. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which values are used for the element? A. sx = 12, sy = -8, txy = 6 kN/cm2 B. sx = 12, sy = 8, txy = 6 C. sx = -12, sy = 8, txy = 6 D. sx = 12, sy = -8, txy = 0 Answer: A. The 8 kN/cm2 on the top and bottom is compressive and therefore negative. (Key point: Sign convention for given stresses) 2. What are the principal stresses? A. 11.66 and -11.66 kN/cm2 B. 13.66 and -9.66 kN/cm2 C. 13.66 and 9.66 kN/cm2 D. 2 and 11.66 kN/cm2 Answer: B. Centre = (12-8)/2 = 2, radius = sqrt(10^2 + 6^2) = 11.66, so 2 +/- 11.66. (Key point: Principal stresses) 3. Which relation and result give the principal planes? A. tan2theta = 2txy/(sx + sy) = 12/4, theta_p = 35.3° B. tan2theta = (sx - sy)/2txy = 20/12, theta_p = 29.5° C. tan2theta = 2txy/(sx - sy) = 12/20, theta_p = 15.48° and 105.48° D. tan2theta = 2txy/(sx - sy), theta_p = 30.96° and 120.96° Answer: C. 2theta_p = 30.96°, so theta_p = 15.48° for sigma_1 and 90° more for sigma_2. 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