--- title: "Principal stresses, max shear: 80, 60 N/mm2" subject: "Strength of Materials" chapter: 4 key: "semester_3/strength_of_materials:principal-stresses-80-60-30deg-element" source: https://www.bctengineer.com/q/semester_3/strength_of_materials/principal-stresses-80-60-30deg-element site: BCT Engineering --- # Principal stresses, max shear: 80, 60 N/mm2 Strength of Materials · Chapter 4: Stress and Strain Analysis - Asked in: 2071 Chaitra, 8 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-80-60-30deg-element - Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-4 ## Question Determine the orientation of principal axes and principal stresses for the element loaded as shown in figure below. Also calculate maximum shear stress and orientation of their plane. [Figure: element on a wall with a $80\ \text{N/mm}^2$ stress acting at $30^\circ$ to the wall and a $60\ \text{N/mm}^2$ stress acting horizontally.] ## Short answer Resolve the 80 N/mm² oblique stress: σy=40, τxy=69.28, with σx=60 N/mm². Principal stresses = 120.0 and -20.0 N/mm², axes at 40.89° and 130.89° from horizontal. Maximum shear stress = 70.0 N/mm². ## Answer **Reading of the figure:** the face parallel to the wall carries a stress of 80 N/mm² acting at $30^\circ$ to the wall (an oblique resultant), and the face at right angles to it carries a horizontal normal stress of 60 N/mm². Resolving the 80 N/mm² stress into components: $$ \sigma_y = 80\sin 30^\circ = 40\ \text{N/mm}^2\ \text{(normal)}, \qquad \tau_{xy} = 80\cos 30^\circ = 69.28\ \text{N/mm}^2\ \text{(shear)} $$ So $\sigma_x = 60$, $\sigma_y = 40$, $\tau_{xy} = 69.28$ (all in N/mm²). #### Principal stresses $$ \sigma_{1,2} = \frac{60+40}{2} \pm \sqrt{\left(\frac{60-40}{2}\right)^2 + 69.28^2} = 50 \pm 70.00 $$ $$ \sigma_1 = 120.00\ \text{N/mm}^2, \qquad \sigma_2 = -20.00\ \text{N/mm}^2 $$ #### Orientation of principal axes $$ \tan 2\theta_p = \frac{2\tau_{xy}}{\sigma_x-\sigma_y} = \frac{2(69.28)}{20} = 6.928 \Rightarrow 2\theta_p = 81.79^\circ $$ $\theta_p = 40.89^\circ$ for $\sigma_1$ and $130.89^\circ$ for $\sigma_2$, measured from the horizontal (the direction of the 60 N/mm² stress). #### Maximum shear stress $$ \tau_{max} = \frac{\sigma_1-\sigma_2}{2} = 70.00\ \text{N/mm}^2 $$ It acts on planes at $\theta_s = 85.89^\circ$ and $175.89^\circ$ from the horizontal, i.e. $45^\circ$ from the principal planes, with a normal stress of $50$ N/mm² on them. **Answer:** $\sigma_1 = 120.0$, $\sigma_2 = -20.0$ N/mm² at $40.9^\circ$ from the horizontal; $\tau_{max} = 70.0$ N/mm² at $85.9^\circ$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the 80 N/mm2 stress acting at 30° to the wall resolved? A. Normal = 80 sin30° = 40, shear = 80 cos30° = 69.28 B. Normal = 80 cos30° = 69.28, shear = 80 sin30° = 40 C. Normal = 80, shear = 0 D. Normal = 80 sin30° = 40, shear = 80 sin30° = 40 Answer: A. The component perpendicular to the wall is the normal stress (80 sin30°); the component along the wall is the shear (80 cos30°). (Key point: Resolving the oblique stress) 2. With sx = 60, sy = 40, txy = 69.28, what are the principal stresses? A. 50 and 70 N/mm2 B. 120 and -20 N/mm2 C. 120 and 20 N/mm2 D. 100 and 69.28 N/mm2 Answer: B. 50 +/- sqrt(10^2 + 69.28^2) = 50 +/- 70, giving 120 and -20. (Key point: Principal stresses) 3. How is the maximum shear stress found and what is it? A. (s1 + s2)/2 = 50 N/mm2 B. s1 - s2 = 140 N/mm2 C. (s1 - s2)/2 = 70 N/mm2 D. txy = 69.28 N/mm2 Answer: C. tau_max is the radius of Mohr's circle, half the difference of principal stresses. (Key point: Maximum shear stress) 4. Where do the maximum shear planes lie and what normal stress acts on them? A. 45° from principal planes, normal stress 50 N/mm2 B. Along the principal planes, normal stress 120 N/mm2 C. 45° from principal planes, normal stress 0 D. Along the principal planes, normal stress 50 N/mm2 Answer: A. Maximum shear occurs 45° from principal planes with normal stress equal to the average (s1+s2)/2 = 50. 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