--- title: "Principal and max shear stresses: 10, 7, 10 MPa" subject: "Strength of Materials" chapter: 4 key: "semester_3/strength_of_materials:transform-stress-10t-7c-10shear" source: https://www.bctengineer.com/q/semester_3/strength_of_materials/transform-stress-10t-7c-10shear site: BCT Engineering --- # Principal and max shear stresses: 10, 7, 10 MPa Strength of Materials · Chapter 4: Stress and Strain Analysis - Asked in: 2066 Jestha (old course), 12 marks - Video: [Understanding Stress Transformation and Mohr's Circle](https://www.youtube.com/watch?v=_DH3546mSCM) — The Efficient Engineer - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/strength_of_materials/transform-stress-10t-7c-10shear - Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-4 ## Question Transform the state of stress into the principal stresses and the maximum shearing stresses and the associated normal stresses. Show the results for both cases on properly oriented elements. Tensile stress is 10 MPa, Compressive stress is 7 MPa and shearing stress is 10 MPa. [Figure: element with $10\ \text{MPa}$ tensile on the left and right faces, $7\ \text{MPa}$ compressive on the top and bottom faces and $10\ \text{MPa}$ shear stress on all faces.] ## Short answer For σ_x=10 MPa, σ_y=–7 MPa, τ_xy=10 MPa, principal stresses are 14.62 and –11.62 MPa at 24.82° from x. Maximum shear is 13.12 MPa with associated normal stress 1.50 MPa on planes at 69.82° and 159.82°. ## Answer **Given:** $\sigma_x = 10$ MPa (tension), $\sigma_y = -7$ MPa (compression), $\tau_{xy} = 10$ MPa. #### Case 1: principal stresses $$ \sigma_{1,2} = \frac{10-7}{2} \pm \sqrt{\left(\frac{10+7}{2}\right)^2 + 10^2} = 1.50 \pm 13.12 $$ $$ \sigma_1 = 14.62\ \text{MPa}, \qquad \sigma_2 = -11.62\ \text{MPa} $$ $$ \tan 2\theta_p = \frac{2(10)}{10-(-7)} = 1.1765 \Rightarrow 2\theta_p = 49.64^\circ,\quad \theta_p = 24.82^\circ\ \text{and}\ 114.82^\circ $$ #### Case 2: maximum shear stress $$ \tau_{max} = \frac{\sigma_1-\sigma_2}{2} = 13.12\ \text{MPa}, \qquad \sigma' = \frac{\sigma_x+\sigma_y}{2} = 1.50\ \text{MPa} $$ on planes at $\theta_s = \theta_p \pm 45^\circ$, i.e. $69.82^\circ$ and $-20.18^\circ$ (the same plane as $159.82^\circ$). #### Elements (not to scale) ```text Principal element Max-shear element (rotated 24.8 deg) (rotated 69.8 deg) s2=-11.62 s'=1.50 ^ ^ tau=13.12 | s1=14.62 | /-> <--+--> (no shear) s'<-+-> (shear on | | all faces) v v s2=-11.62 s'=1.50 ``` **Answer:** (1) $\sigma_1 = 14.62$ MPa, $\sigma_2 = -11.62$ MPa at $24.82^\circ$ from the $x$-axis (no shear). (2) $\tau_{max} = 13.12$ MPa with associated normal stress $1.50$ MPa on planes at $69.82^\circ$ and $159.82^\circ$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which input values are used? A. sx = 10, sy = -7, txy = 10 MPa B. sx = 10, sy = 7, txy = 10 MPa C. sx = -10, sy = 7, txy = 10 MPa D. sx = 10, sy = -7, txy = 0 Answer: A. Compressive stress is taken negative. (Key point: Sign of stresses) 2. What are the principal stresses? A. 17 and -10 MPa B. 1.50 and 13.12 MPa C. 14.62 and -11.62 MPa D. 14.62 and 11.62 MPa Answer: C. 1.5 +/- sqrt(8.5^2 + 10^2) = 1.5 +/- 13.12. (Key point: Principal stresses) 3. What are the maximum shear stress and the associated normal stress? A. 13.12 MPa and 1.50 MPa B. 13.12 MPa and 14.62 MPa C. 10 MPa and 1.50 MPa D. 14.62 MPa and 13.12 MPa Answer: A. tau_max is half the difference of principal stresses; the normal stress on its planes is (sx+sy)/2 = 1.5. (Key point: Maximum shear and normal stress) 4. Where do the maximum shear planes lie relative to the principal planes? A. Parallel to them B. At 45° to them, theta_s = theta_p +/- 45° C. At 90° to them D. At 22.5° to them Answer: B. Maximum shear planes are 45° from principal planes, so the two elements differ by 45° rotation. 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