--- title: "Stresses on 40 degree plane; principal stresses" subject: "Strength of Materials" chapter: 4 key: "semester_3/strength_of_materials:inclined-plane-40deg-50-30" source: https://www.bctengineer.com/q/semester_3/strength_of_materials/inclined-plane-40deg-50-30 site: BCT Engineering --- # Stresses on 40 degree plane; principal stresses Strength of Materials · Chapter 4: Stress and Strain Analysis - Asked in: 2076 Asoj, 8 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/inclined-plane-40deg-50-30 - Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-4 ## Question Determine the normal and shearing stress on the inclined plane at the angle of $40^\circ$ to the vertical. Also calculate principal stresses and their planes. [Figure: element with $30\ \text{MPa}$ acting vertically on the top and bottom faces, $50\ \text{MPa}$ on the left and right faces, and shear stress $30\ \text{MPa}$.] ## Short answer On the 40° inclined plane: normal stress 71.28 MPa, shear stress -4.64 MPa. Principal stresses are 71.62 MPa and 8.38 MPa at 35.78° and 125.78°; maximum shear 31.62 MPa with average normal 40.00 MPa. ## Answer **Reading of the figure:** both normal stresses are tensile: $\sigma_x = 50$ MPa (left/right faces), $\sigma_y = 30$ MPa (top/bottom faces), and the shear stress is $\tau_{xy} = 30$ MPa. The plane is at $40^\circ$ to the vertical, so its normal is at $40^\circ$ to the $x$-axis. #### Stresses on the inclined plane ($\theta = 40^\circ$) $$ \begin{aligned} \sigma_n &= \frac{\sigma_x+\sigma_y}{2} + \frac{\sigma_x-\sigma_y}{2}\cos 2\theta + \tau_{xy}\sin 2\theta \\ &= 40.00 + (10.00)\cos 80^\circ + (30.00)\sin 80^\circ = 71.28\ \text{MPa} \\ \tau_\theta &= -\frac{\sigma_x-\sigma_y}{2}\sin 2\theta + \tau_{xy}\cos 2\theta \\ &= -(10.00)\sin 80^\circ + (30.00)\cos 80^\circ = -4.64\ \text{MPa} \end{aligned} $$ Resultant stress on the plane $= \sqrt{71.28^2+(-4.64)^2} = 71.43\ \text{MPa}$, inclined at $3.72^\circ$ to the normal of the plane. #### Principal stresses $$ \sigma_{1,2} = \frac{\sigma_x+\sigma_y}{2} \pm \sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2} = 40.00 \pm \sqrt{(10.00)^2+(30.00)^2} = 40.00 \pm 31.62 $$ $\sigma_1 = 71.62\ \text{MPa}$ and $\sigma_2 = 8.38\ \text{MPa}$. #### Principal planes $$ \tan 2\theta_p = \frac{2\tau_{xy}}{\sigma_x-\sigma_y} = \frac{2(30.00)}{20.00} \Rightarrow 2\theta_p = 71.57^\circ $$ So $\theta_p = 35.78^\circ$ (normal of the plane carrying $\sigma_1$) and $125.78^\circ$ (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 (extra) $$ \tau_{max} = \frac{\sigma_1-\sigma_2}{2} = \sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2} = 31.62\ \text{MPa} $$ It acts on planes at $45^\circ$ to the principal planes, i.e. at $\theta_s = 80.78^\circ$ and $170.78^\circ$. The normal stress on these planes is $\frac{\sigma_x+\sigma_y}{2} = 40.00\ \text{MPa}$. **Answer:** $\sigma_n = 71.28$, $\tau = -4.64$ on the inclined plane; $\sigma_1 = 71.62$, $\sigma_2 = 8.38$ at $\theta_p = 35.78^\circ$ and $125.78^\circ$; $\tau_{max} = 31.62$ with normal stress $40.00$ (MPa). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What are sigma_x, sigma_y, tau and theta? A. 50, 30, 30 MPa and 40 deg B. 30, 50, 30 and 50 deg C. 50, 30, 40 and 30 deg D. 50, 0, 30 and 40 deg Answer: A. The plane is 40 deg from vertical so its normal is 40 deg from x. (Key point: Stress inputs) 2. What are sigma_n and tau on the plane? A. 40 and 30 MPa B. 71.62 and 8.38 MPa C. 71.28 MPa and -4.64 MPa D. 60.5 and 4.64 MPa Answer: C. 40 + 10 cos80 + 30 sin80 = 71.28 and -10 sin80 + 30 cos80 = -4.64. (Key point: Normal and shear) 3. What are sigma_1 and sigma_2? A. 71.62 and 8.38 MPa B. 80 and 0 C. 71.28 and -4.64 D. 40 and 31.62 Answer: A. 40 +/- 31.62. (Key point: Principal stresses) 4. What are the principal plane angles? A. 71.57 and 161.57 B. 40 and 130 C. 35.78 deg and 125.78 deg D. 26.57 and 116.57 Answer: C. tan 2theta_p = 60/20, 2 theta_p = 71.57 deg. (Key point: Principal planes) ## Close variants set in other exams - [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× ## 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× - [Plane stress: principal planes and max shear](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aplane-stress-15-30-60-principal) — Strength of Materials, chapter 4, asked 2× - [Plane stress: principal, max shear, normal](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aplane-stress-80-60-50-mpa) — Strength of Materials, chapter 4, asked 2× - [Stresses on 30 degree plane; Mohr's circle](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Ainclined-plane-30deg-50-30-mohr) — Strength of Materials, chapter 4, asked 3× - [Principal planes, stresses on plane AC](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aprincipal-planes-max-shear-120-180) — Strength of Materials, chapter 4 - [Resultant stress on planes EF and GE, principal stresses](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aresultant-stress-plane-ef-ge) — 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