--- title: "Principal stresses: 50 T, 20 C, 30 shear" subject: "Strength of Materials" chapter: 4 key: "semester_3/strength_of_materials:plane-element-50t-20c-30shear" source: https://www.bctengineer.com/q/semester_3/strength_of_materials/plane-element-50t-20c-30shear site: BCT Engineering --- # Principal stresses: 50 T, 20 C, 30 shear Strength of Materials · Chapter 4: Stress and Strain Analysis - Asked in: 2066 Chaitra (old course), 10 marks - Video: [Principal Stresses and MOHR'S CIRCLE in 12 Minutes!!](https://www.youtube.com/watch?v=aumN-Vl0OhQ) — Less Boring Lectures - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/strength_of_materials/plane-element-50t-20c-30shear - Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-4 ## Question A plane element is subjected to the stresses as shown in figure. Determine analytically or graphically, the principal stresses and their directions. Tensile stress is 50 MPa, Compressive stress is 20 MPa, Shear stress is 30 MPa. ## Short answer Given σ_x=50, σ_y=–20, τ_xy=30 MPa, principal stresses are σ1=61.10 MPa and σ2=–31.10 MPa, acting on planes at 20.30° and 110.30° from the x-axis, with zero shear. ## Answer **Given:** $\sigma_x = 50$ MPa (tension), $\sigma_y = -20$ MPa (compression), $\tau_{xy} = 30$ MPa. #### Principal stresses $$ \sigma_{1,2} = \frac{\sigma_x+\sigma_y}{2} \pm \sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2} = 15.00 \pm \sqrt{(35.00)^2+(30.00)^2} = 15.00 \pm 46.10 $$ $\sigma_1 = 61.10\ \text{MPa}$ and $\sigma_2 = -31.10\ \text{MPa}$. #### Principal planes $$ \tan 2\theta_p = \frac{2\tau_{xy}}{\sigma_x-\sigma_y} = \frac{2(30.00)}{70.00} \Rightarrow 2\theta_p = 40.60^\circ $$ So $\theta_p = 20.30^\circ$ (normal of the plane carrying $\sigma_1$) and $110.30^\circ$ (normal of the plane carrying $\sigma_2$), both measured anticlockwise from the $x$-axis. There is no shear stress on these planes. #### Graphical check (Mohr's circle) - Plot $X(50, 30)$ and $Y(-20, -30)$ and join them; the centre C is at $\sigma = 15$ MPa. - Radius $CX = 46.10$ MPa; the circle meets the $\sigma$-axis at $P_1 = 61.10$ and $P_2 = -31.10$. - $\angle XCP_1 = 2\theta_p = 40.60^\circ$, so the principal plane for $\sigma_1$ is at $20.30^\circ$ from the plane on which $\sigma_x$ acts. **Answer:** $\sigma_1 = 61.10$ MPa and $\sigma_2 = -31.10$ MPa, acting on planes whose normals are at $20.30^\circ$ and $110.30^\circ$ from the $x$-axis. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For sx = 50, sy = -20, txy = 30, what are the centre and radius of the stress circle? A. Centre 15 MPa, radius 46.10 MPa B. Centre 35 MPa, radius 46.10 MPa C. Centre 15 MPa, radius 35 MPa D. Centre 15 MPa, radius 30 MPa Answer: A. Centre = (50-20)/2 = 15, radius = sqrt(35^2 + 30^2) = 46.10. (Key point: Centre and radius) 2. What are sigma_1 and sigma_2? A. 46.10 and -46.10 MPa B. 61.10 and -31.10 MPa C. 50 and -20 MPa D. 61.10 and 31.10 MPa Answer: B. 15 +/- 46.10. (Key point: Principal stresses) 3. What is the angle of the plane carrying sigma_1? A. 40.60° from the x-axis B. 69.70° from the x-axis C. 20.30° from the x-axis, from tan2theta = 60/70 D. 20.30° from the y-axis Answer: C. 2theta_p = 40.60°, so theta_p = 20.30°; sigma_2 is on the plane at 110.30°. (Key point: Direction of principal plane) ## Related in Strength of Materials - [Normal/shear stress on 25 degree plane, principal stresses](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Ainclined-plane-30-25deg-mohr-verify) — Strength of Materials, chapter 4, asked 3× - [Principal stresses, max shear: 80, 60, 20 MPa](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aprincipal-stresses-60-20-80) — Strength of Materials, chapter 4 - [Principal and max shear stresses: 10, 7, 10 MPa](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Atransform-stress-10t-7c-10shear) — Strength of Materials, chapter 4 - [Max shear for 150 MPa stress limit; principal stresses](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Adirect-stresses-100t-60c-shear-limit) — Strength of Materials, chapter 4 - [Stresses on 20 degree plane; principal and max shear](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Ainclined-plane-20deg-50-40-80) — Strength of Materials, chapter 4, asked 4× - [Principal stresses and planes (90, 120, 50 MPa)](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aprincipal-stresses-90-50-120) — Strength of Materials, chapter 4 ## Related in other subjects - [Stresses on inclined plane and principal stresses](https://www.bctengineer.com/question.md?key=semester_4%2Fstrength_of_materials_mechanical%3Ainclined-plane-stresses-principal-stresses) — Strength of Materials, chapter 6 - [Plane stress: principal stresses and oblique plane](https://www.bctengineer.com/question.md?key=semester_5%2Fmechanics_of_solids%3Aplane-stress-principal-numerical) — Mechanics of Solids, chapter 2 - [Mohr's circle for a plane stress element](https://www.bctengineer.com/question.md?key=semester_4%2Fstrength_of_materials_mechanical%3Amohr-circle-plane-stress-construction) — Strength of Materials, chapter 6 - [Principal stresses from a 3D stress tensor](https://www.bctengineer.com/question.md?key=semester_5%2Fmechanics_of_solids%3Athree-d-stress-invariants-numerical) — Mechanics of Solids, chapter 2