--- title: "Plane stress/strain; 3D equilibrium equation" subject: "Computational Techniques in Civil Engineering" chapter: 3 key: "semester_8/computational_techniques_in_civil_engineering:define-plane-stress-3d-equilibrium" source: https://www.bctengineer.com/q/semester_8/computational_techniques_in_civil_engineering/define-plane-stress-3d-equilibrium site: BCT Engineering --- # Plane stress/strain; 3D equilibrium equation Computational Techniques in Civil Engineering · Chapter 3: Elasticity in solids - Asked in: 2073 Magh, 3+7 marks - Video: [FEM: Derivation for 3D equilibrium equation](https://www.youtube.com/watch?v=otU013wNv-M) — Love Mechanical - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_8/computational_techniques_in_civil_engineering/define-plane-stress-3d-equilibrium - Whole chapter: https://www.bctengineer.com/md/semester_8/computational_techniques_in_civil_engineering/chapter-3 ## Question Define plane stress and plane strain problems. Derive the differential equation of equilibrium for three-dimensional problems. ## Short answer Plane stress: thin plate with in-plane load, sigma_z=tau_xz=tau_yz=0. Plane strain: long uniform prism with constrained ends, eps_z=gamma_xz=gamma_yz=0. Three-dimensional equilibrium: dsigma_x/dx+dtau_xy/dy+dtau_xz/dz+X=0; dtau_xy/dx+dsigma_y/dy+dtau_yz/dz+Y=0; dtau_xz/dx+dtau_yz/dy+dsigma_z/dz+Z=0, or sigma_ij,j+bi=0. ## Answer #### Plane stress and plane strain **Plane stress:** a thin plate loaded only by forces in its own plane, so the stresses through the thickness are zero: $\sigma_z=\tau_{xz}=\tau_{yz}=0$. The stress state is two-dimensional, $\{\sigma\}=\{\sigma_x,\sigma_y,\tau_{xy}\}^T$, although the strain $\varepsilon_z=-\dfrac{\nu}{E}(\sigma_x+\sigma_y)\neq0$. **Plane strain:** a long prism (length much larger than cross-section) with uniform cross-section and load that does not vary along its length and is perpendicular to it. Movement along the axis is prevented, so $\varepsilon_z=\gamma_{xz}=\gamma_{yz}=0$, but $\sigma_z=\nu(\sigma_x+\sigma_y)\neq0$. Examples: thin plate with a hole under in-plane load (plane stress); long gravity dam or retaining wall of uniform section (plane strain). #### Differential equations of equilibrium in 3D Consider an infinitesimal element $dx\,dy\,dz$ in equilibrium with body forces per unit volume $X, Y, Z$ (e.g. self-weight). Stresses vary from face to face, e.g. the normal stress on the face at $x+dx$ is $\sigma_x+\dfrac{\partial\sigma_x}{\partial x}dx$. ```text y ^ sy + dsy | +----------+ txy | | | sx + dsx <-----| | dx dy |----> sx +---+----------+---> x sy ``` #### Equilibrium of forces in the $x$ direction Forces = stress $\times$ area. Faces normal to $x$ have area $dy\,dz$, those normal to $y$ have area $dx\,dz$ and those normal to $z$ have area $dx\,dy$: $$ \Big(\frac{\partial\sigma_x}{\partial x}dx\Big)dy\,dz+\Big(\frac{\partial\tau_{yx}}{\partial y}dy\Big)dx\,dz+\Big(\frac{\partial\tau_{zx}}{\partial z}dz\Big)dx\,dy+X\,dx\,dy\,dz=0 $$ (the terms $\sigma_x$, $\tau_{yx}$, $\tau_{zx}$ on opposite faces cancel). Dividing by $dx\,dy\,dz$: $$ \frac{\partial\sigma_x}{\partial x}+\frac{\partial\tau_{xy}}{\partial y}+\frac{\partial\tau_{xz}}{\partial z}+X=0 $$ using $\tau_{yx}=\tau_{xy}$ and $\tau_{zx}=\tau_{xz}$ (from moment equilibrium). #### Similarly for $y$ and $z$ $$ \begin{aligned} \frac{\partial\tau_{xy}}{\partial x}+\frac{\partial\sigma_y}{\partial y}+\frac{\partial\tau_{yz}}{\partial z}+Y&=0\\ \frac{\partial\tau_{xz}}{\partial x}+\frac{\partial\tau_{yz}}{\partial y}+\frac{\partial\sigma_z}{\partial z}+Z&=0 \end{aligned} $$ Moment equilibrium about the three axes gives $\tau_{xy}=\tau_{yx}$, $\tau_{yz}=\tau_{zy}$, $\tau_{zx}=\tau_{xz}$ (symmetry of the stress tensor). In compact form: $\ \sigma_{ij,j}+b_i=0$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which is the equilibrium equation in the x direction for a 3D element with body force X? A. d sigma_x/dx + d tau_xy/dy + d tau_xz/dz + X = 0 B. d sigma_x/dx + d sigma_y/dy + d sigma_z/dz = X C. d sigma_x/dx + d tau_xy/dx + d tau_xz/dx + X = 0 D. sigma_x + tau_xy + tau_xz = X Answer: A. Summing forces in x gives the stress gradients on faces normal to x, y and z plus the body force X. (Key point: Equilibrium in x) 2. Why do only the derivative terms (e.g. (d sigma_x/dx) dx) remain in the force balance? A. The stresses are zero on the faces B. The terms sigma_x, tau_yx, tau_zx on opposite faces act in opposite directions and cancel C. Body forces cancel them D. Moment equilibrium removes them Answer: B. The stress on one face is balanced by the equal stress on the opposite face; only the increments over dx, dy, dz remain. (Key point: Cancelling terms) 3. What does moment equilibrium of the element give? A. sigma_x = sigma_y B. tau_xy = tau_yx, tau_yz = tau_zy, tau_zx = tau_xz C. All shear stresses are zero D. tau_xy = -tau_yx Answer: B. Moment equilibrium shows that complementary shear stresses are equal, making the stress tensor symmetric. (Key point: Symmetry of shear stress) 4. Which describes plane stress? A. Thin plate loaded in its plane with sigma_z = tau_xz = tau_yz = 0 B. Long prism with epsilon_z = 0 C. Body of revolution D. Plate loaded perpendicular to its plane Answer: A. Plane stress is a thin plate with in-plane loading, so through-thickness stresses are zero. 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