--- title: "Principal stresses and Mohr's circle (30, 18, 75)" subject: "Strength of Materials" chapter: 4 key: "semester_3/strength_of_materials:principal-stresses-mohr-30-18-75" source: https://www.bctengineer.com/q/semester_3/strength_of_materials/principal-stresses-mohr-30-18-75 site: BCT Engineering --- # Principal stresses and Mohr's circle (30, 18, 75) Strength of Materials · Chapter 4: Stress and Strain Analysis - Asked in: 2078 Kartik, 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/principal-stresses-mohr-30-18-75 - Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-4 ## Question Determine the principal stresses, orientation of principal planes, maximum shearing and normal stress on the plane of maximum shear stress. Verify the results by drawing Mohr's Circle. [Figure: element with $30\ \text{N/mm}^2$ compressive on the top and bottom faces, $75\ \text{N/mm}^2$ tensile on the left and right faces, and shear stress $18\ \text{N/mm}^2$.] ## Short answer Principal stresses are 78.00 N/mm² and -33.00 N/mm², with planes at 9.46° and 99.46° anticlockwise from x. Maximum shear stress is 55.50 N/mm², acting with a normal stress of 22.50 N/mm². ## Answer **Reading of the figure:** $\sigma_x = 75$ N/mm² tensile (left and right faces), $\sigma_y = -30$ N/mm² (compressive, top and bottom faces), $\tau_{xy} = 18$ N/mm². #### Principal stresses $$ \sigma_{1,2} = \frac{\sigma_x+\sigma_y}{2} \pm \sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2} = 22.50 \pm \sqrt{(52.50)^2+(18.00)^2} = 22.50 \pm 55.50 $$ $\sigma_1 = 78.00\ \text{N/mm}^2$ and $\sigma_2 = -33.00\ \text{N/mm}^2$. #### Principal planes $$ \tan 2\theta_p = \frac{2\tau_{xy}}{\sigma_x-\sigma_y} = \frac{2(18.00)}{105.00} \Rightarrow 2\theta_p = 18.92^\circ $$ So $\theta_p = 9.46^\circ$ (normal of the plane carrying $\sigma_1$) and $99.46^\circ$ (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 $$ \tau_{max} = \frac{\sigma_1-\sigma_2}{2} = \sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2} = 55.50\ \text{N/mm}^2 $$ It acts on planes at $45^\circ$ to the principal planes, i.e. at $\theta_s = 54.46^\circ$ and $144.46^\circ$. The normal stress on these planes is $\frac{\sigma_x+\sigma_y}{2} = 22.50\ \text{N/mm}^2$. #### Verification by Mohr's circle - Draw the $\sigma$-axis (tension to the right) and the $\tau$-axis to a convenient scale. - Plot $X(75, 18)$ and $Y(-30, -18)$ (the two shear stresses go on opposite sides of the $\sigma$-axis); join XY. It cuts the $\sigma$-axis at the centre $C = (22.50, 0)$. - Draw the circle with centre C and radius $CX = 55.50$. - It cuts the $\sigma$-axis at $\sigma_1 = 78.00$ and $\sigma_2 = -33.00$; $2\theta_p = \angle XCP_1 = 18.92^\circ$ (scaled by protractor). - The highest and lowest points of the circle give $\tau_{max} = \pm55.50$ with $\sigma = 22.50$. **Answer:** $\sigma_1 = 78.00$, $\sigma_2 = -33.00$ at $\theta_p = 9.46^\circ$ and $99.46^\circ$; $\tau_{max} = 55.50$ with normal stress $22.50$ (N/mm²). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the compressive 30 N/mm^2 treated? A. sigma_y = +30 B. sigma_y = 0 C. sigma_y = -30 N/mm^2 D. sigma_y = -75 Answer: C. Tension is positive, compression negative. (Key point: Sign of stresses) 2. What are they? A. 78.00 and -33.00 N/mm^2 B. 105 and -30 C. 22.5 and 55.5 D. 75 and -30 Answer: A. 22.50 +/- sqrt(52.5^2 + 18^2) = 22.50 +/- 55.50. (Key point: Principal stresses) 3. What is theta_p of sigma_1? A. 18.92 deg B. 54.46 deg C. 9.46 deg from the x-axis D. 45 deg Answer: C. tan 2 theta_p = 36/105 gives 2 theta_p = 18.92 deg; theta_p is half. (Key point: Principal plane) 4. What is tau_max and the normal stress on its plane? A. 55.50 with 22.50 N/mm^2 B. 55.50 with 0 C. 22.50 with 55.50 D. 111 with 22.5 Answer: A. tau_max equals the radius and normal stress equals the centre. 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