--- title: "Principal stresses and planes (90, 120, 50 MPa)" subject: "Strength of Materials" chapter: 4 key: "semester_3/strength_of_materials:principal-stresses-90-50-120" source: https://www.bctengineer.com/q/semester_3/strength_of_materials/principal-stresses-90-50-120 site: BCT Engineering --- # Principal stresses and planes (90, 120, 50 MPa) Strength of Materials · Chapter 4: Stress and Strain Analysis - Asked in: 2072 Chaitra, 8 marks - Video: [Strength of Materials | How to draw Mohr's circle? | Determination of Principal stresses and Plane](https://www.youtube.com/watch?v=qU_Fngzb-Vs) — Michael Thomas Rex F - 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-90-50-120 - Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-4 ## Question The state of stress in a two dimensional stress system is as shown in figure below. Determine the principal stresses and orientation of principal planes. [Figure: element with $90\ \text{MPa}$ tensile on the top and bottom faces, $120\ \text{MPa}$ tensile on the left and right faces, shear stress $50\ \text{MPa}$.] ## Short answer For σx=120, σy=90, τxy=50 MPa, principal stresses are σ1=157.20 MPa, σ2=52.80 MPa. Principal planes are at 36.65° and 126.65° from the x-axis. ## Answer **Reading of the figure:** $\sigma_x = 120$ MPa (left/right faces), $\sigma_y = 90$ MPa (top/bottom faces), $\tau_{xy} = 50$ MPa, all tensile. #### Principal stresses $$ \sigma_{1,2} = \frac{\sigma_x+\sigma_y}{2} \pm \sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2} = 105.00 \pm \sqrt{(15.00)^2+(50.00)^2} = 105.00 \pm 52.20 $$ $\sigma_1 = 157.20\ \text{MPa}$ and $\sigma_2 = 52.80\ \text{MPa}$. #### Principal planes $$ \tan 2\theta_p = \frac{2\tau_{xy}}{\sigma_x-\sigma_y} = \frac{2(50.00)}{30.00} \Rightarrow 2\theta_p = 73.30^\circ $$ So $\theta_p = 36.65^\circ$ (normal of the plane carrying $\sigma_1$) and $126.65^\circ$ (normal of the plane carrying $\sigma_2$), both measured anticlockwise from the $x$-axis. There is no shear stress on these planes. **Answer:** $\sigma_1 = 157.20$, $\sigma_2 = 52.80$ at $\theta_p = 36.65^\circ$ and $126.65^\circ$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. From the figure, which values are used in the principal stress formula? A. sigma_x = 120, sigma_y = 90, tau_xy = 50 MPa B. sigma_x = 90, sigma_y = 120, tau_xy = 50 MPa C. sigma_x = 120, sigma_y = 90, tau_xy = 0 D. sigma_x = 120, sigma_y = -90, tau_xy = 50 MPa Answer: A. The 120 MPa acts on the left/right faces (x), 90 MPa on the top/bottom (y), all tensile, with shear 50 MPa. (Key point: Reading the element) 2. Which expression gives the principal stresses? A. (sx+sy)/2 +/- sqrt((sx+sy)^2/4 + txy^2) B. (sx-sy)/2 +/- sqrt(((sx+sy)/2)^2 + txy^2) C. (sx+sy)/2 +/- sqrt(((sx-sy)/2)^2 + txy^2) D. (sx+sy)/2 +/- (sx-sy)/2 +/- txy Answer: C. The centre is the average normal stress and the radius is the root of (half difference)^2 plus shear^2. (Key point: Principal stress formula) 3. What are sigma_1 and sigma_2 here? A. 105 MPa and 52.2 MPa B. 157.20 MPa and -52.80 MPa C. 120 MPa and 90 MPa D. 157.20 MPa and 52.80 MPa Answer: D. 105 +/- sqrt(15^2 + 50^2) = 105 +/- 52.20. 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