--- title: "Stresses on inclined plane, bi-axial loading" subject: "Strength of Materials" chapter: 4 key: "semester_3/strength_of_materials:inclined-plane-biaxial-derive" source: https://www.bctengineer.com/q/semester_3/strength_of_materials/inclined-plane-biaxial-derive site: BCT Engineering --- # Stresses on inclined plane, bi-axial loading Strength of Materials · Chapter 4: Stress and Strain Analysis - Asked in: 2081 Bhadra, 6+2 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_3/strength_of_materials/inclined-plane-biaxial-derive - Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-4 ## Question Derive an expression for stresses on an inclined plane of an element subjected to axial loads (like stress) on two perpendicular direction (Bi-axial loading). Also explain how Mohr suggests finding it graphically. ## Short answer Consider a wedge with area dA, normal at angle θ. Force equilibrium gives σn = σx cos²θ + σy sin²θ = (σx+σy)/2 + ((σx−σy)/2) cos2θ, shear τ = ((σx−σy)/2) sin2θ. Graphically, plot OA=σy, OB=σx on σ-axis, draw circle with diameter AB, centre C; radius CA. Rotate CB by 2θ to point Q: its horizontal coordinate is σn, vertical is τ. ## Answer #### Expression for stresses on an inclined plane (bi-axial loading) Consider a small element subjected to tensile stresses $\sigma_x$ and $\sigma_y$ on two perpendicular faces (no shear stress). Cut it by a plane AB whose **normal makes angle $\theta$ with the $x$-axis** (the plane makes $\theta$ with the vertical face). Let the area of AB be $dA$ (unit thickness). ```text sy B | | | |\ v v v | \ sx | \ AB (area dA) ----> |___\ A (BA vertical face = dA cos(theta)) ``` Areas: vertical face $= dA\cos\theta$; horizontal face $= dA\sin\theta$. Forces on the wedge: $\sigma_x\,dA\cos\theta$ (horizontal) and $\sigma_y\,dA\sin\theta$ (vertical). **Normal stress $\sigma_n$** (resolve along the normal to AB): $$ \sigma_n\,dA = \sigma_x\,dA\cos\theta\cdot\cos\theta + \sigma_y\,dA\sin\theta\cdot\sin\theta $$ $$ \sigma_n = \sigma_x\cos^2\theta + \sigma_y\sin^2\theta = \frac{\sigma_x+\sigma_y}{2} + \frac{\sigma_x-\sigma_y}{2}\cos 2\theta $$ **Shear (tangential) stress $\tau$** (resolve along AB): $$ \tau\,dA = \sigma_x\,dA\cos\theta\sin\theta - \sigma_y\,dA\sin\theta\cos\theta $$ $$ \tau = \frac{\sigma_x-\sigma_y}{2}\sin 2\theta $$ **Resultant stress:** $\sigma_R = \sqrt{\sigma_n^2 + \tau^2}$, at obliquity $\tan\phi = \tau/\sigma_n$ to the normal. Maximum $\sigma_n = \sigma_x$ at $\theta = 0^\circ$; maximum shear $= \dfrac{\sigma_x-\sigma_y}{2}$ at $\theta = 45^\circ$. #### Mohr's graphical method - Draw a horizontal axis for normal stress. Mark $OA = \sigma_y$ and $OB = \sigma_x$ to scale. - Bisect AB at C and draw a circle with diameter AB (centre C, radius $\frac{\sigma_x-\sigma_y}{2}$). - From C draw a line making angle $2\theta$ with CB, cutting the circle at Q. - The horizontal distance of Q from O is $\sigma_n$ and the vertical distance of Q from the axis is $\tau$, because $OC + CQ\cos2\theta = \sigma_n$ and $CQ\sin2\theta = \tau$. - The line OQ is the resultant stress; $\angle QOC$ is its obliquity. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. If AB has area dA and its normal is at theta to the x-axis, what are the other face areas? A. Vertical dA sin(theta), horizontal dA cos(theta) B. Both equal dA C. Vertical dA/cos(theta), horizontal dA/sin(theta) D. Vertical face dA cos(theta), horizontal face dA sin(theta) Answer: D. Projection of AB gives dA cos(theta) vertically and dA sin(theta) horizontally. (Key point: Areas of faces) 2. What is sigma_n on the inclined plane? A. sigma_x sin^2(theta) + sigma_y cos^2(theta) B. sigma_x cos^2(theta) + sigma_y sin^2(theta) C. (sigma_x - sigma_y) sin(2 theta)/2 D. sigma_x + sigma_y Answer: B. Resolving forces along the normal gives sigma_x cos^2 + sigma_y sin^2. (Key point: Normal stress) 3. What is the tangential stress on the plane? A. (sigma_x + sigma_y)/2 x sin(2 theta) B. (sigma_x - sigma_y) cos(2 theta) C. (sigma_x - sigma_y)/2 x cos(2 theta) D. (sigma_x - sigma_y)/2 x sin(2 theta) Answer: D. Resolving along AB gives (sigma_x - sigma_y) sin(theta)cos(theta). (Key point: Shear stress) 4. In Mohr's graphical method, how is Q located? A. Circle with radius sigma_x; draw OQ at theta B. Circle with diameter AB (OA = sigma_y, OB = sigma_x); draw CQ at 2 theta from CB C. Circle with diameter OB; draw CQ at theta D. Circle with diameter AB; draw CQ at 90 - theta Answer: B. Coordinates of Q: OC + CQ cos 2theta = sigma_n and CQ sin 2theta = tau. 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