--- title: "Principal stresses, max shear: 80, 60, 20 MPa" subject: "Strength of Materials" chapter: 4 key: "semester_3/strength_of_materials:principal-stresses-60-20-80" source: https://www.bctengineer.com/q/semester_3/strength_of_materials/principal-stresses-60-20-80 site: BCT Engineering --- # Principal stresses, max shear: 80, 60, 20 MPa Strength of Materials · Chapter 4: Stress and Strain Analysis - Asked in: 2075 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-60-20-80 - 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 shown in figure. Determine the principal stresses and their direction, maximum shear and associated normal stress. [Figure: element with $60\ \text{MPa}$ vertical stress on the top and bottom faces, $80\ \text{MPa}$ horizontal stress on the left and right faces, and shear stress $20\ \text{MPa}$.] ## Short answer Principal stresses are 92.36 MPa and 47.64 MPa, acting on planes at 31.72° and 121.72° from x. Maximum shear stress is 22.36 MPa on planes with 70.00 MPa normal stress. ## Answer **Reading of the figure:** $\sigma_x = 80$ MPa (horizontal, on left/right faces), $\sigma_y = 60$ MPa (vertical, on top/bottom faces), shear $\tau_{xy} = 20$ MPa; all tensile. #### Principal stresses $$ \sigma_{1,2} = \frac{\sigma_x+\sigma_y}{2} \pm \sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2} = 70.00 \pm \sqrt{(10.00)^2+(20.00)^2} = 70.00 \pm 22.36 $$ $\sigma_1 = 92.36\ \text{MPa}$ and $\sigma_2 = 47.64\ \text{MPa}$. #### Principal planes $$ \tan 2\theta_p = \frac{2\tau_{xy}}{\sigma_x-\sigma_y} = \frac{2(20.00)}{20.00} \Rightarrow 2\theta_p = 63.43^\circ $$ So $\theta_p = 31.72^\circ$ (normal of the plane carrying $\sigma_1$) and $121.72^\circ$ (normal of the plane carrying $\sigma_2$), both measured anticlockwise from the $x$-axis. There is no shear stress on these planes. #### Maximum shear stress $$ \tau_{max} = \frac{\sigma_1-\sigma_2}{2} = \sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2} = 22.36\ \text{MPa} $$ It acts on planes at $45^\circ$ to the principal planes, i.e. at $\theta_s = 76.72^\circ$ and $166.72^\circ$. The normal stress on these planes is $\frac{\sigma_x+\sigma_y}{2} = 70.00\ \text{MPa}$. **Answer:** $\sigma_1 = 92.36$, $\sigma_2 = 47.64$ at $\theta_p = 31.72^\circ$ and $121.72^\circ$; $\tau_{max} = 22.36$ with normal stress $70.00$ (MPa). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What values are used? A. sigma_x = 60, sigma_y = 80, tau = 20 MPa, plane at 90 B. sigma_x = 80, sigma_y = 60, tau = 20 MPa C. sigma_x = 80, sigma_y = 20, tau = 60 D. sigma_x = 70, sigma_y = 10, tau = 20 Answer: B. 80 acts on left/right faces and 60 on top/bottom. (Key point: Reading the figure) 2. What are they? A. 100 and 40 B. 80 and 60 C. 70 and 22.36 D. 92.36 and 47.64 MPa Answer: D. 70 +/- 22.36. (Key point: Principal stresses) 3. What is theta_p? A. 63.43 deg B. 31.72 deg C. 45 deg D. 22.36 deg Answer: B. tan 2 theta_p = 40/20, 2 theta_p = 63.43. (Key point: Principal plane) 4. What are tau_max and associated normal stress? A. 20 and 70 B. 92.36 and 47.64 C. 70 and 22.36 D. 22.36 MPa and 70 MPa Answer: D. Radius and centre. 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