Deformable Body and Strain Tensor

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 $(u, v)$.

 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

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

(a) Strain components

Gauge $a$ along x (0$^\circ$), $b$ at 45$^\circ$, $c$ along y (90$^\circ$). The strain in direction $\theta$:

$$\varepsilon_\theta = \varepsilon_x\cos^2\theta + \varepsilon_y\sin^2\theta + \gamma_{xy}\sin\theta\cos\theta$$ $$\begin{aligned} \varepsilon_x &= \varepsilon_0 = 800\times10^{-6} \\ \varepsilon_y &= \varepsilon_{90} = -200\times10^{-6} \\ \varepsilon_{45} &= \tfrac12(\varepsilon_x+\varepsilon_y) + \tfrac12\gamma_{xy} \\ \gamma_{xy} &= 2\varepsilon_{45} - \varepsilon_x - \varepsilon_y = 800 - 800 + 200 = 200\times10^{-6} \end{aligned}$$

(b) Principal strains (units $10^{-6}$)

$$\begin{aligned} \varepsilon_{avg} &= \frac{800 - 200}{2} = 300 \\ R &= \sqrt{\left(\frac{\varepsilon_x-\varepsilon_y}{2}\right)^2 + \left(\frac{\gamma_{xy}}{2}\right)^2} = \sqrt{500^2+100^2} = 509.9 \\ \varepsilon_{1,2} &= \varepsilon_{avg} \pm R = 809.9,\ -209.9 \\ \gamma_{max} &= 2R = 1019.8 \\ \tan 2\theta_p &= \frac{\gamma_{xy}}{\varepsilon_x-\varepsilon_y} = \frac{200}{1000} \Rightarrow \theta_p = 5.65^\circ \end{aligned}$$

$\varepsilon_1$ acts at $5.65^\circ$ anticlockwise from the x-axis; $\varepsilon_2$ at $95.65^\circ$.

(c) Principal stresses (plane stress, $\sigma_z=0$)

$$\begin{aligned} \sigma_1 &= \frac{E}{1-\nu^2}(\varepsilon_1 + \nu\varepsilon_2) = \frac{200\times10^3}{0.91}(809.9 - 0.3\times209.9)\times10^{-6} = 164.2\ \text{MPa} \\ \sigma_2 &= \frac{E}{1-\nu^2}(\varepsilon_2 + \nu\varepsilon_1) = \frac{200\times10^3}{0.91}(-209.9 + 0.3\times809.9)\times10^{-6} = 7.3\ \text{MPa} \end{aligned}$$

The out-of-plane strain is $\varepsilon_z = -\dfrac{\nu}{1-\nu}(\varepsilon_1+\varepsilon_2) = -257\times10^{-6}$ (not measured by the rosette, but present).

Answer: $\varepsilon_x = 800\times10^{-6}$, $\varepsilon_y = -200\times10^{-6}$, $\gamma_{xy} = 200\times10^{-6}$; $\varepsilon_1 = 810\times10^{-6}$, $\varepsilon_2 = -210\times10^{-6}$, $\gamma_{max} = 1020\times10^{-6}$ at $5.65^\circ$; $\sigma_1 = 164.2$ MPa, $\sigma_2 = 7.3$ MPa.

Compatibility conditions

The three strain components $\varepsilon_x, \varepsilon_y, \gamma_{xy}$ in a plane are obtained from only two displacement functions $u(x,y)$ and $v(x,y)$. The strains are therefore not independent: they must satisfy a condition so that a continuous single-valued displacement field exists (the deformed body has no gaps or overlaps). This is the compatibility condition.

Derivation

From the small-displacement relations:

$$\varepsilon_x = \frac{\partial u}{\partial x},\quad \varepsilon_y = \frac{\partial v}{\partial y},\quad \gamma_{xy} = \frac{\partial u}{\partial y} + \frac{\partial v}{\partial x}$$

Differentiate:

$$\begin{aligned} \frac{\partial^2 \varepsilon_x}{\partial y^2} &= \frac{\partial^3 u}{\partial x\,\partial y^2} \\ \frac{\partial^2 \varepsilon_y}{\partial x^2} &= \frac{\partial^3 v}{\partial y\,\partial x^2} \\ \frac{\partial^2 \gamma_{xy}}{\partial x\,\partial y} &= \frac{\partial^3 u}{\partial x\,\partial y^2} + \frac{\partial^3 v}{\partial x^2\,\partial y} \end{aligned}$$

Adding the first two and comparing with the third (the order of differentiation can be interchanged):

$$\frac{\partial^2 \varepsilon_x}{\partial y^2} + \frac{\partial^2 \varepsilon_y}{\partial x^2} = \frac{\partial^2 \gamma_{xy}}{\partial x\,\partial y}$$

In 3D there are six such equations. For a simply connected body, they are necessary and sufficient.

Check of the given field

$$\begin{aligned} \frac{\partial^2 \varepsilon_x}{\partial y^2} &= 2\times 2\times10^{-6} = 4\times10^{-6} \\ \frac{\partial^2 \varepsilon_y}{\partial x^2} &= 2\times 3\times10^{-6} = 6\times10^{-6} \\ \text{LHS} &= 10\times10^{-6} \\ \frac{\partial^2 \gamma_{xy}}{\partial x\,\partial y} &= 8\times10^{-6} = \text{RHS} \end{aligned}$$

LHS $\neq$ RHS ($10\times10^{-6} \neq 8\times10^{-6}$), so the strain field is not compatible: it cannot arise from any continuous displacement field. It would need $\gamma_{xy} = 10\times10^{-6}xy$ to be compatible.

Answer: The field is not possible, since $10\times10^{-6}\neq 8\times10^{-6}$.