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Strain-displacement relations and strain tensor

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.

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.

Definitions

  • Normal strain ε\varepsilon: change in length per unit original length of a line element through the point, ε=lim⁡ΔLL\varepsilon = \lim \dfrac{\Delta L}{L}.
  • Shear strain γ\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,vu, v (along x,yx, 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 PQRSPQRS with PQ=dxPQ = dx, PS=dyPS = dy. Point PP moves by (u,v)(u, v).

 S'               R'
  .-------------.
  |  v+dv/dy*dy |
 S+--------------+R
  |              |
  P--------------Q
   u            u + du/dx dx

Point QQ has displacement u+∂u∂xdxu + \dfrac{\partial u}{\partial x}dx along xx, so

εx=(dx+∂u∂xdx)−dxdx=∂u∂x,εy=∂v∂y\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 PQPQ rotates by ∂v∂x\dfrac{\partial v}{\partial x} and PSPS rotates by ∂u∂y\dfrac{\partial u}{\partial y} (small angles equal their tangents), so the change in the right angle is

γxy=∂u∂y+∂v∂x\gamma_{xy} = \frac{\partial u}{\partial y} + \frac{\partial v}{\partial x}

In three dimensions also εz=∂w/∂z\varepsilon_z = \partial w/\partial z, γyz=∂v/∂z+∂w/∂y\gamma_{yz} = \partial v/\partial z + \partial w/\partial y, γzx=∂w/∂x+∂u/∂z\gamma_{zx} = \partial w/\partial x + \partial u/\partial z.

Strain tensor

[ε]=[εx12γxy12γxz12γxyεy12γyz12γxz12γyzεz],εij=12(∂ui∂xj+∂uj∂xi)[\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

PointEngineering γxy\gamma_{xy}Tensor εxy\varepsilon_{xy}
MeaningTotal change in right angleHalf of it (average rotation of the two lines)
Relationγxy=2εxy\gamma_{xy} = 2\varepsilon_{xy}εxy=γxy/2\varepsilon_{xy} = \gamma_{xy}/2
UseHooke's law τ=Gγ\tau = G\gamma, strain gaugesTransformation laws, Mohr's circle for strain, tensor algebra

The tensor form obeys the same transformation law as the stress tensor, which is why γ/2\gamma/2 is used in the off-diagonal terms.

✓ Answer checked by Claude Sonnet 5.5

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