--- title: "Principal stresses and orientation: 80 and 60 N/mm2" subject: "Strength of Materials" chapter: 4 key: "semester_3/strength_of_materials:principal-stresses-80-60-sketch" source: https://www.bctengineer.com/q/semester_3/strength_of_materials/principal-stresses-80-60-sketch site: BCT Engineering --- # Principal stresses and orientation: 80 and 60 N/mm2 Strength of Materials · Chapter 4: Stress and Strain Analysis - Asked in: 2067 Asar (old course), 12 marks - Video: [MOHR'S CIRCLE Problems MOM |VTU @mohsin_ali_14](https://www.youtube.com/watch?v=BQQOSirgAvg) — Mohsin Ali 14 - 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-80-60-sketch - Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-4 ## Question For the following stress condition of an element, obtain principal stresses and their orientation. Show the results in a neat sketch. [Figure: element with $80\ \text{N/mm}^2$ on the top and bottom faces, $80\ \text{N/mm}^2$ on the left and right faces and shear stresses $60\ \text{N/mm}^2$ on all four faces.] ## Short answer For σx=σy=80 N/mm² and τxy=60, principal stresses are 140 and 20 N/mm², acting on planes at 45° and 135° to the x-axis. Maximum shear stress is 60 N/mm² on planes with normal stress 80. ## Answer **Given:** $\sigma_x = 80$ N/mm², $\sigma_y = 80$ N/mm² (tensile) and $\tau_{xy} = 60$ N/mm² on all four faces. #### Principal stresses $$ \sigma_{1,2} = \frac{80+80}{2} \pm \sqrt{\left(\frac{80-80}{2}\right)^2 + 60^2} = 80 \pm 60 $$ $$ \sigma_1 = 140\ \text{N/mm}^2, \qquad \sigma_2 = 20\ \text{N/mm}^2 $$ #### Orientation $$ \tan 2\theta_p = \frac{2\tau_{xy}}{\sigma_x-\sigma_y} = \frac{120}{0} = \infty \Rightarrow 2\theta_p = 90^\circ,\ \theta_p = 45^\circ $$ So $\sigma_1 = 140$ N/mm² acts on the plane whose normal is at $45^\circ$ anticlockwise from the $x$-axis (the shear stresses are positive, so the major principal direction lies between the +x and +y axes). $\sigma_2 = 20$ N/mm² acts on the plane whose normal is at $135^\circ$ (perpendicular to it). #### Maximum shear $$ \tau_{max} = \frac{140-20}{2} = 60\ \text{N/mm}^2\ \text{(on planes at 0° and 90°, with normal stress 80 N/mm}^2) $$ #### Sketch of the principal element ```text y ^ sigma_2 = 20 | sigma_1 = 140 (135 deg) \ | / (45 deg) \ /\ / \ / \ / / \ \ / element rotated 45 deg \ / anticlockwise; \/ no shear on its faces ------------------+-------------------> x ``` Check: $\sigma_x+\sigma_y = 160 = \sigma_1+\sigma_2$. **Answer:** $\sigma_1 = 140$ N/mm² and $\sigma_2 = 20$ N/mm², the principal planes being at $45^\circ$ and $135^\circ$ to the $x$-axis. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. With sx = sy = 80 and txy = 60, what are the principal stresses? A. 80 and 60 N/mm2 B. 140 and 20 N/mm2 C. 140 and -20 N/mm2 D. 160 and 0 N/mm2 Answer: B. The first term in the root is zero so sigma = 80 +/- 60. (Key point: Principal stresses when sx = sy) 2. Why is theta_p = 45°? A. tan2theta = 2txy/(sx - sy) is infinite, so 2theta = 90° B. tan2theta = 0 so 2theta = 0 and plane is rotated C. tan2theta = 1 so theta = 22.5° D. Because shear stress is 60 N/mm2 Answer: A. With zero denominator, tan 2theta is infinity, 2theta = 90°, theta = 45°. (Key point: Orientation when sx = sy) 3. Which check confirms the principal stresses? A. sigma_1 - sigma_2 = sigma_x + sigma_y B. sigma_1 x sigma_2 = txy C. sigma_1 + sigma_2 = sigma_x + sigma_y = 160 D. sigma_1 + sigma_2 = txy Answer: C. The sum of normal stresses is invariant: 140 + 20 = 80 + 80. 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