--- title: "Stress on oblique section with direct and shear stress" subject: "Strength of Materials" chapter: 4 key: "semester_3/strength_of_materials:oblique-section-stress-derive" source: https://www.bctengineer.com/q/semester_3/strength_of_materials/oblique-section-stress-derive site: BCT Engineering --- # Stress on oblique section with direct and shear stress Strength of Materials · Chapter 4: Stress and Strain Analysis - Asked in: 2069 Asar, 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_3/strength_of_materials/oblique-section-stress-derive - Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-4 ## Question Derive an expression for the normal stress and shear stress on an oblique section of a rectangular strained body when it is subjected to direct stresses in two perpendicular directions accompanied with simple shear stress. ## Short answer For a biaxial stress with shear, on a plane with normal at angle θ, normal stress σn = (σx+σy)/2 + {(σx-σy)/2}cos2θ + τxy sin2θ; shear stress τ = -{(σx-σy)/2}sin2θ + τxy cos2θ. ## Answer **Problem:** a rectangular element carries direct stresses $\sigma_x$ and $\sigma_y$ (tensile) on two perpendicular faces, together with a shear stress $\tau_{xy}$ on all four faces. Find the normal stress $\sigma_n$ and shear stress $\tau$ on an oblique section whose normal makes angle $\theta$ with the $x$-axis (the section makes $\theta$ with the face on which $\sigma_x$ acts). ```text sy B ------- |\ txy <- | \ AB = oblique sx <-| \ section |___\ A (wedge ABC, unit thickness) ``` Let the oblique face AB have area $dA$. Then the vertical face has area $dA\cos\theta$ and the horizontal face has area $dA\sin\theta$. #### Forces on the wedge - On the vertical face: horizontal force $\sigma_x\,dA\cos\theta$ and vertical force $\tau_{xy}\,dA\cos\theta$. - On the horizontal face: vertical force $\sigma_y\,dA\sin\theta$ and horizontal force $\tau_{xy}\,dA\sin\theta$. The shear stress $\tau_{xy}$ is taken so that the shear forces on the two faces act in the senses required by complementary shear. #### Normal stress Resolve all forces along the normal to AB (direction $\theta$ from the $x$-axis): $$ \sigma_n\,dA = \sigma_x\,dA\cos\theta\cos\theta + \sigma_y\,dA\sin\theta\sin\theta + \tau_{xy}\,dA\cos\theta\sin\theta + \tau_{xy}\,dA\sin\theta\cos\theta $$ $$ \sigma_n = \sigma_x\cos^2\theta + \sigma_y\sin^2\theta + 2\tau_{xy}\sin\theta\cos\theta $$ Using $\cos^2\theta = \frac{1+\cos2\theta}{2}$, $\sin^2\theta = \frac{1-\cos2\theta}{2}$ and $2\sin\theta\cos\theta = \sin2\theta$: $$ \sigma_n = \frac{\sigma_x+\sigma_y}{2} + \frac{\sigma_x-\sigma_y}{2}\cos2\theta + \tau_{xy}\sin2\theta $$ #### Shear stress Resolve along the plane AB (perpendicular to the normal): $$ \tau\,dA = -\sigma_x\,dA\cos\theta\sin\theta + \sigma_y\,dA\sin\theta\cos\theta + \tau_{xy}\,dA\left(\cos^2\theta - \sin^2\theta\right) $$ $$ \tau = -\frac{\sigma_x-\sigma_y}{2}\sin2\theta + \tau_{xy}\cos2\theta $$ (the sign of $\tau$ depends on the direction chosen as positive along AB; reversing that direction only changes the sign, not the magnitude). #### Result $$ \sigma_n = \frac{\sigma_x+\sigma_y}{2} + \frac{\sigma_x-\sigma_y}{2}\cos2\theta + \tau_{xy}\sin2\theta, \qquad \tau = -\frac{\sigma_x-\sigma_y}{2}\sin2\theta + \tau_{xy}\cos2\theta $$ **Principal stresses:** setting $\tau = 0$ gives $\tan2\theta_p = \dfrac{2\tau_{xy}}{\sigma_x-\sigma_y}$ and $\sigma_{1,2} = \dfrac{\sigma_x+\sigma_y}{2} \pm \sqrt{\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2 + \tau_{xy}^2}$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. If the oblique face AB has area dA and its normal makes theta with the x-axis, what are the areas of the vertical and horizontal faces? A. Vertical = dA cos(theta), horizontal = dA sin(theta) B. Vertical = dA sin(theta), horizontal = dA cos(theta) C. Both equal dA D. Vertical = dA cos(2theta), horizontal = dA sin(2theta) Answer: A. Projecting the oblique face onto the vertical and horizontal gives cos and sin components of dA. (Key point: Areas of faces of the wedge) 2. How is the normal stress on the plane obtained? A. Taking moments about point A only B. Resolving all forces on the wedge along the normal to AB C. Resolving all forces along the plane AB D. Equating strain energies of the faces Answer: B. Force equilibrium along the normal gives sigma_n dA = sum of components of the face forces. (Key point: Equilibrium used) 3. Which expression for the normal stress results? A. (sx+sy)/2 - (sx-sy)/2 cos2theta + txy sin2theta B. (sx+sy)/2 + (sx-sy)/2 sin2theta + txy cos2theta C. (sx+sy)/2 + (sx-sy)/2 cos2theta + txy sin2theta D. (sx-sy)/2 + (sx+sy)/2 cos2theta + txy sin2theta Answer: C. Using the double-angle identities, the two shear terms combine into 2txy sin theta cos theta = txy sin2theta. (Key point: Result for normal stress) 4. How are principal planes obtained from the shear stress expression? A. Set sigma_n = 0, giving tan2theta_p = (sx - sy)/2txy B. Set tau = 0, giving tan2theta_p = (sx + sy)/2txy C. Set tau maximum, giving tan2theta_p = 2txy/(sx - sy) D. Set tau = 0, giving tan2theta_p = 2txy/(sx - sy) Answer: D. Principal planes carry no shear; setting tau = 0 gives that tan 2theta_p. 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