--- title: "Principal stresses, plane BE, Mohr's circle" subject: "Strength of Materials" chapter: 4 key: "semester_3/strength_of_materials:plane-stress-40-50-30-plane" source: https://www.bctengineer.com/q/semester_3/strength_of_materials/plane-stress-40-50-30-plane site: BCT Engineering --- # Principal stresses, plane BE, Mohr's circle Strength of Materials · Chapter 4: Stress and Strain Analysis - Asked in: 2066 Bhadra (old course), 16 marks - Video: [Principal Stresses and MOHR'S CIRCLE in 12 Minutes!!](https://www.youtube.com/watch?v=aumN-Vl0OhQ) — Less Boring Lectures - Page: https://www.bctengineer.com/q/semester_3/strength_of_materials/plane-stress-40-50-30-plane - Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-4 ## Question The state of stress in a two dimensionally stressed body is as shown in figure below. Determine principal stresses, principal planes and maximum shear. Determine also the normal and tangential stresses on plane BE and verify the results by drawing Mohr's circle. [Figure: rectangle ABCD; plane BE is a diagonal from B (bottom right) up to E on the top face; stresses: $40\ \text{N/mm}^2$ on the top face, $50\ \text{N/mm}^2$ on the right face, $30\ \text{N/mm}^2$ on the bottom face, with shear stresses as drawn.] ## Short answer With σ_x=50, σ_y=40, τ_xy=30 N/mm², principal stresses are σ1=75.41 and σ2=14.59 N/mm² at θ=40.27° and 130.27°, maximum shear 30.41 N/mm². On plane BE at 30°, normal stress 73.48, shear 10.67 N/mm². ## Answer **Reading of the figure:** the normal stresses are $\sigma_x = 50$ N/mm² (right/left faces) and $\sigma_y = 40$ N/mm² (top face); the stress 30 N/mm² is the shear stress $\tau_{xy}$ on the faces. The plane BE is assumed to make $60^\circ$ with the horizontal (rising to the left from B), so its normal is at $\theta = 30^\circ$ from the $x$-axis. If the figure gives another angle, replace $\theta$ in the formulas below. #### Stresses on the inclined plane ($\theta = 30^\circ$) $$ \begin{aligned} \sigma_n &= \frac{\sigma_x+\sigma_y}{2} + \frac{\sigma_x-\sigma_y}{2}\cos 2\theta + \tau_{xy}\sin 2\theta \\ &= 45.00 + (5.00)\cos 60^\circ + (30.00)\sin 60^\circ = 73.48\ \text{N/mm}^2 \\ \tau_\theta &= -\frac{\sigma_x-\sigma_y}{2}\sin 2\theta + \tau_{xy}\cos 2\theta \\ &= -(5.00)\sin 60^\circ + (30.00)\cos 60^\circ = 10.67\ \text{N/mm}^2 \end{aligned} $$ Resultant stress on the plane $= \sqrt{73.48^2+10.67^2} = 74.25\ \text{N/mm}^2$, inclined at $8.26^\circ$ to the normal of the plane. #### Principal stresses $$ \sigma_{1,2} = \frac{\sigma_x+\sigma_y}{2} \pm \sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2} = 45.00 \pm \sqrt{(5.00)^2+(30.00)^2} = 45.00 \pm 30.41 $$ $\sigma_1 = 75.41\ \text{N/mm}^2$ and $\sigma_2 = 14.59\ \text{N/mm}^2$. #### Principal planes $$ \tan 2\theta_p = \frac{2\tau_{xy}}{\sigma_x-\sigma_y} = \frac{2(30.00)}{10.00} \Rightarrow 2\theta_p = 80.54^\circ $$ So $\theta_p = 40.27^\circ$ (normal of the plane carrying $\sigma_1$) and $130.27^\circ$ (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 $$ \tau_{max} = \frac{\sigma_1-\sigma_2}{2} = \sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2} = 30.41\ \text{N/mm}^2 $$ It acts on planes at $45^\circ$ to the principal planes, i.e. at $\theta_s = 85.27^\circ$ and $175.27^\circ$. The normal stress on these planes is $\frac{\sigma_x+\sigma_y}{2} = 45.00\ \text{N/mm}^2$. #### Verification by Mohr's circle - Draw the $\sigma$-axis (tension to the right) and the $\tau$-axis to a convenient scale. - Plot $X(50, 30)$ and $Y(40, -30)$ (the two shear stresses go on opposite sides of the $\sigma$-axis); join XY. It cuts the $\sigma$-axis at the centre $C = (45.00, 0)$. - Draw the circle with centre C and radius $CX = 30.41$. - It cuts the $\sigma$-axis at $\sigma_1 = 75.41$ and $\sigma_2 = 14.59$; $2\theta_p = \angle XCP_1 = 80.54^\circ$ (scaled by protractor). - The highest and lowest points of the circle give $\tau_{max} = \pm30.41$ with $\sigma = 45.00$. - Turn CX through $2\theta = 60^\circ$ in the same sense as $\theta$ to reach point Q; its coordinates read $\sigma_n \approx 73.48$ and $\tau \approx 10.67$, which match the analytical values. **Answer:** $\sigma_n = 73.48$, $\tau = 10.67$ on the inclined plane; $\sigma_1 = 75.41$, $\sigma_2 = 14.59$ at $\theta_p = 40.27^\circ$ and $130.27^\circ$; $\tau_{max} = 30.41$ with normal stress $45.00$ (N/mm²). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which stress values are taken for the element? A. sx = 50, sy = 40, txy = 30 N/mm2 B. sx = 50, sy = 30, txy = 40 N/mm2 C. sx = 40, sy = 50, txy = 30 N/mm2 D. sx = 50, sy = 40, txy = 0 Answer: A. 30 N/mm2 is read as the shear stress on all faces, while 50 and 40 are the direct stresses. (Key point: Reading of the figure) 2. What are the principal stresses? A. 80 and 10 N/mm2 B. 75.41 and 14.59 N/mm2 C. 50 and 40 N/mm2 D. 45.00 and 30.41 N/mm2 Answer: B. 45 +/- sqrt(5^2 + 30^2) = 45 +/- 30.41. (Key point: Principal stresses) 3. What are tau_max and the associated normal stress? A. 30.41 N/mm2 and 75.41 N/mm2 B. 30.41 N/mm2 and 45 N/mm2 C. 30 N/mm2 and 50 N/mm2 D. 5 N/mm2 and 45 N/mm2 Answer: B. tau_max equals the radius 30.41; the normal stress on those planes is the average (50+40)/2 = 45. (Key point: Maximum shear stress) 4. With normal at 30°, what are sigma_n and tau on BE? A. sigma_n = 45, tau = 30.41 N/mm2 B. sigma_n = 10.67, tau = 73.48 N/mm2 C. sigma_n = 75.41, tau = 0 D. sigma_n = 73.48, tau = 10.67 N/mm2 Answer: D. sigma_n = 45 + 5cos60° + 30 sin60° = 73.48; tau = -5 sin60° + 30 cos60° = 10.67. 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