--- title: "Equilibrium equations for 3D stress" subject: "Computational Techniques in Civil Engineering" chapter: 3 key: "semester_8/computational_techniques_in_civil_engineering:equilibrium-3d-stress" source: https://www.bctengineer.com/q/semester_8/computational_techniques_in_civil_engineering/equilibrium-3d-stress site: BCT Engineering --- # Equilibrium equations for 3D stress Computational Techniques in Civil Engineering · Chapter 3: Elasticity in solids - Asked in: 2070 Magh, 5 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/equilibrium-3d-stress - Whole chapter: https://www.bctengineer.com/md/semester_8/computational_techniques_in_civil_engineering/chapter-3 ## Question Derive equilibrium equations for the 3D state of stress in a solid. ## Short answer Consider an infinitesimal dx dy dz element with body forces X,Y,Z. Force equilibrium gives 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; in index form sigma_ij,j+bi=0. ## Answer 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. In the x-direction force balance of the element, why do the terms sigma_x, tau_yx, tau_zx on opposite faces not appear in the final equation? A. They are zero for a body in equilibrium B. They act on opposite faces with opposite sense and cancel, leaving only the increments d(sigma)/dx dx etc. C. They are replaced by the body force X D. They only act in the y and z directions Answer: B. The stress on one face is sigma and on the opposite face sigma + (d sigma/dx)dx, so only the increment contributes a net force. (Key point: Net force from stress change) 2. What does the term X dx dy dz in the x-equilibrium equation represent? A. The shear force on the y faces B. The moment of the stresses about the x-axis C. The body force (e.g. self-weight) per unit volume times the element volume D. The strain energy of the element Answer: C. X is the body force per unit volume in x; multiplied by the volume dx dy dz it gives the force on the element. (Key point: Role of body force) 3. Which is the equilibrium equation in the x direction after dividing by dx dy dz? 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 = 0 C. d(sigma_x)/dx + d(tau_yz)/dy + d(tau_xz)/dz = X^2 D. d(tau_xy)/dx + d(sigma_y)/dy + d(tau_yz)/dz + X = 0 Answer: A. Forces in x come from sigma_x on x-faces and tau_xy, tau_xz on y- and z-faces, plus the body force X. (Key point: Final x-equilibrium equation) 4. Which condition lets tau_yx be replaced by tau_xy in the equations? A. Force equilibrium in the z direction B. Compatibility of strains C. Moment equilibrium of the element about the axes D. Hooke's law Answer: C. Moment equilibrium about the three axes gives tau_xy = tau_yx etc. (symmetry of the stress tensor). 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