--- title: "Stresses on inclined plane and principal stresses" subject: "Strength of Materials" chapter: 6 key: "semester_4/strength_of_materials_mechanical:inclined-plane-stresses-principal-stresses" source: https://www.bctengineer.com/q/semester_4/strength_of_materials_mechanical/inclined-plane-stresses-principal-stresses site: BCT Engineering --- # Stresses on inclined plane and principal stresses Strength of Materials · Chapter 6: Transformation of stress and strain - Asked in: year not given, 8 marks - Video: [Stress on an Inclined Plane.MP4](https://www.youtube.com/watch?v=qDzE2KkMkE0) — purdueMET - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/strength_of_materials_mechanical/inclined-plane-stresses-principal-stresses - Whole chapter: https://www.bctengineer.com/md/semester_4/strength_of_materials_mechanical/chapter-6 ## Question At a point in a stressed body, the state of plane stress is sigma_x = 80 MPa (tensile), sigma_y = 40 MPa (compressive) and tau_xy = 30 MPa, where tau_xy acts in the +y direction on the face with outward normal +x. Determine (a) the normal and shear stresses on a plane whose normal is inclined at 30 degrees anticlockwise from the x axis, (b) the principal stresses and the position of the principal planes, (c) the maximum shear stress. ## Short answer Given σx=80, σy=-40, τxy=30 MPa: on plane at θ=30°, σθ=76.0 MPa, τθ=-37.0 MPa. Principal stresses: σ1=87.1 MPa, σ2=-47.1 MPa at θp=13.28°. Maximum shear stress τmax=67.1 MPa. ## Answer Use $\sigma_x = 80$, $\sigma_y = -40$, $\tau_{xy} = 30$ MPa. $$ \frac{\sigma_x+\sigma_y}{2} = 20\ \text{MPa}, \qquad \frac{\sigma_x-\sigma_y}{2} = 60\ \text{MPa} $$ #### (a) Stresses on the 30 degree plane $2\theta = 60^\circ$, $\cos60^\circ = 0.5$, $\sin60^\circ = 0.866$. $$ \sigma_\theta = 20 + 60(0.5) + 30(0.866) = 20 + 30 + 25.98 = 76.0\ \text{MPa} $$ $$ \tau_\theta = -60(0.866) + 30(0.5) = -51.96 + 15 = -37.0\ \text{MPa} $$ The negative sign means the shear acts opposite to the positive direction (clockwise sense on this face). Check on the perpendicular plane ($\theta = 120^\circ$): $\sigma = 20 - 30 - 25.98 = -36.0$ MPa. The sum $76.0 - 36.0 = 40 = \sigma_x+\sigma_y$ confirms the calculation. #### (b) Principal stresses $$ R = \sqrt{60^2 + 30^2} = \sqrt{4500} = 67.08\ \text{MPa} $$ $$ \sigma_1 = 20 + 67.08 = 87.1\ \text{MPa (tension)}, \qquad \sigma_2 = 20 - 67.08 = -47.1\ \text{MPa (compression)} $$ Principal planes: $$ \tan2\theta_p = \frac{2\tau_{xy}}{\sigma_x-\sigma_y} = \frac{60}{120} = 0.5 \;\Rightarrow\; 2\theta_p = 26.57^\circ,\ \theta_p = 13.28^\circ $$ The plane at $13.28^\circ$ anticlockwise from the $x$ axis carries $\sigma_1 = 87.1$ MPa (check: $20+60\cos26.57^\circ+30\sin26.57^\circ = 87.1$). The other principal plane is at $13.28^\circ+90^\circ = 103.28^\circ$ and carries $\sigma_2$. #### (c) Maximum shear stress $$ \tau_{max} = R = 67.1\ \text{MPa} $$ acting on planes at $\theta_p \pm 45^\circ$, i.e. $58.3^\circ$ and $148.3^\circ$, with normal stress $20$ MPa. **Answer: (a) $\sigma_\theta = 76.0$ MPa, $\tau_\theta = -37.0$ MPa; (b) $\sigma_1 = 87.1$ MPa, $\sigma_2 = -47.1$ MPa at $13.28^\circ$; (c) $\tau_{max} = 67.1$ MPa.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. With σx = 80 MPa tensile and σy = 40 MPa compressive, which values go into the transformation equations? A. σx = 80, σy = 40, so (σx+σy)/2 = 60 and (σx-σy)/2 = 20 B. σx = 80, σy = -40, so (σx+σy)/2 = 20 and (σx-σy)/2 = 60 C. σx = 80, σy = -40, so (σx+σy)/2 = 60 and (σx-σy)/2 = 20 D. σx = -80, σy = 40, so (σx+σy)/2 = -20 Answer: B. Tension is positive, so compressive 40 MPa enters as -40, giving average 20 and half-difference 60. (Key point: Sign of σy) 2. What is the normal stress on the 30° plane? A. 56.0 MPa, from 20 + 60cos60° - 30sin60° B. 87.1 MPa, the major principal stress C. 76.0 MPa, from 20 + 60cos60° + 30sin60° D. 50.0 MPa, from 20 + 30 Answer: C. σθ = 20 + 60(0.5) + 30(0.866) = 76.0 MPa. (Key point: Stresses on 30° plane) 3. Which set correctly gives the principal stresses and the first principal plane? A. 100 MPa and -60 MPa; θp = 26.57° B. 87.1 MPa and -47.1 MPa; θp = 26.57° C. 80 MPa and -40 MPa; θp = 0° D. 87.1 MPa and -47.1 MPa; θp = 13.28° from tan2θp = 60/120 Answer: D. R = sqrt(60^2+30^2) = 67.08 gives 20 ± 67.08, and tan2θp = 0.5 gives 2θp = 26.57°, θp = 13.28°. (Key point: Principal stresses and plane) 4. What is the maximum shear stress and the normal stress on its planes? A. 67.1 MPa with normal stress 20 MPa B. 67.1 MPa with normal stress 0 C. 60 MPa with normal stress 20 MPa D. 67.1 MPa with normal stress 87.1 MPa Answer: A. τmax = R = 67.08 MPa, acting on planes with the average normal stress 20 MPa. 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