--- title: "Strain-displacement relations and strain tensor" subject: "Mechanics of Solids" chapter: 3 key: "semester_5/mechanics_of_solids:strain-displacement-small-displacement" source: https://www.bctengineer.com/q/semester_5/mechanics_of_solids/strain-displacement-small-displacement site: BCT Engineering --- # Strain-displacement relations and strain tensor Mechanics of Solids · Chapter 3: Deformable Body and Strain Tensor - Asked in: year not given, 6 marks - Video: [STRAIN DISPLACEMENT RELATIONS](https://www.youtube.com/watch?v=V1GLf4yd1LM) — EDUCATION CHANNEL - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/mechanics_of_solids/strain-displacement-small-displacement - Whole chapter: https://www.bctengineer.com/md/semester_5/mechanics_of_solids/chapter-3 ## Question Define normal and shear strain at a point. Using the small displacement theory, derive the strain-displacement relations in two dimensions and write the strain tensor. Differentiate between engineering shear strain and tensor shear strain. ## Short answer Normal strain ε = change in length/original length; shear strain γ = change in right angle. ε_x = ∂u/∂x, ε_y = ∂v/∂y, γ_xy = ∂u/∂y+∂v/∂x. Strain tensor ε_ij = (∂u_i/∂x_j + ∂u_j/∂x_i)/2. Engineering shear γ is the total angle change; tensor shear ε_xy = γ/2. ## Answer #### Definitions - **Normal strain** $\varepsilon$: change in length per unit original length of a line element through the point, $\varepsilon = \lim \dfrac{\Delta L}{L}$. - **Shear strain** $\gamma$: change in the right angle between two initially perpendicular line elements (radians). Sign: elongation is positive; the shear strain is positive when the angle between the positive axes decreases. #### Small displacement theory The displacements $u, v$ (along $x, y$) and their gradients are very small compared with unity, so products of derivatives can be neglected and the geometry before and after deformation is taken as the same. #### Derivation (2D) Take a small element $PQRS$ with $PQ = dx$, $PS = dy$. Point $P$ moves by $(u, v)$. ```text S' R' .-------------. | v+dv/dy*dy | S+--------------+R | | P--------------Q u u + du/dx dx ``` Point $Q$ has displacement $u + \dfrac{\partial u}{\partial x}dx$ along $x$, so $$ \varepsilon_x = \frac{\left(dx + \dfrac{\partial u}{\partial x}dx\right) - dx}{dx} = \frac{\partial u}{\partial x},\qquad \varepsilon_y = \frac{\partial v}{\partial y} $$ Line $PQ$ rotates by $\dfrac{\partial v}{\partial x}$ and $PS$ rotates by $\dfrac{\partial u}{\partial y}$ (small angles equal their tangents), so the change in the right angle is $$ \gamma_{xy} = \frac{\partial u}{\partial y} + \frac{\partial v}{\partial x} $$ In three dimensions also $\varepsilon_z = \partial w/\partial z$, $\gamma_{yz} = \partial v/\partial z + \partial w/\partial y$, $\gamma_{zx} = \partial w/\partial x + \partial u/\partial z$. #### Strain tensor $$ [\varepsilon] = \begin{bmatrix} \varepsilon_x & \tfrac12\gamma_{xy} & \tfrac12\gamma_{xz} \\ \tfrac12\gamma_{xy} & \varepsilon_y & \tfrac12\gamma_{yz} \\ \tfrac12\gamma_{xz} & \tfrac12\gamma_{yz} & \varepsilon_z \end{bmatrix},\qquad \varepsilon_{ij} = \tfrac12\left(\frac{\partial u_i}{\partial x_j} + \frac{\partial u_j}{\partial x_i}\right) $$ #### Engineering and tensor shear strain | Point | Engineering $\gamma_{xy}$ | Tensor $\varepsilon_{xy}$ | |---|---|---| | Meaning | Total change in right angle | Half of it (average rotation of the two lines) | | Relation | $\gamma_{xy} = 2\varepsilon_{xy}$ | $\varepsilon_{xy} = \gamma_{xy}/2$ | | Use | Hooke's law $\tau = G\gamma$, strain gauges | Transformation laws, Mohr's circle for strain, tensor algebra | The tensor form obeys the same transformation law as the stress tensor, which is why $\gamma/2$ is used in the off-diagonal terms. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What does small displacement theory allow? A. Ignoring all strains B. Treating stress as constant C. Neglecting products of displacement derivatives and using undeformed geometry D. Assuming rigid body motion only Answer: C. Displacement gradients are very small, so products are neglected and geometry before and after is taken as the same. (Key point: Small displacement theory) 2. What are the 2D normal strains? A. epsilon_x = du/dy, epsilon_y = dv/dx B. epsilon_x = du/dx, epsilon_y = dv/dy C. epsilon_x = dv/dx, epsilon_y = du/dy D. epsilon_x = u/x, epsilon_y = v/y Answer: B. The change in length of dx is (du/dx) dx, giving epsilon_x = du/dx; similarly epsilon_y = dv/dy. (Key point: Normal strain relation) 3. What is the engineering shear strain gamma_xy? A. du/dx + dv/dy B. du/dx - dv/dy C. (du/dy + dv/dx)/2 D. du/dy + dv/dx Answer: D. The right angle changes by the rotations of both lines, dv/dx and du/dy, which add. (Key point: Shear strain relation) 4. How are tensor shear strain and engineering shear strain related? A. epsilon_xy = gamma_xy/2 B. epsilon_xy = 2 gamma_xy C. They are equal D. epsilon_xy = gamma_xy^2 Answer: A. Tensor shear strain is half the engineering value, and only the tensor form obeys the same transformation law as stress. 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