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\[ \epsilon = {\Delta L \over L_o} \]

where the quantities are defined in the sketch. This is also known as

The definition arises from the fact that if a 1 m long rope is pulled and fails after it stretches 0.015 m, then we would expect a 10 m rope to stretch 0.15 m before it fails. In each case, the strain is \(\epsilon\) = 0.015, or 1.5%, and is a constant value independent of the rope's length, even though the \(\Delta L's\) are different values in the two cases. Likewise, the force required to stretch a rope by a given amount would be found to depend only on the strain in the rope. It is this foundational concept of strain that makes this definition a useful choice.

\[ \gamma = {D \over T} \]

This is the shear-version of engineering strain. Note that this situation does include some rigid body rotation because the square tends to rotate counter-clockwise here, but we will ignore this complication for now.

So a better, but slightly more complex definition of shear strain, is

\[ \gamma = { \Delta x + \Delta y \over T} \]

where it is assumed that the starting point is also a square. It should be noted that the two definitions lead to the same results when the displacements and strains are small. In other words

\[ \Delta x = \Delta y = { D \over 2} \qquad \text{(small strains)} \]

This permits one to think in terms of the first definition while using the second.

The answer to this dilemma is... calculus. The approach is to define the various strains in terms of partial derivatives of the displacement field, \({\bf u({\bf X})}\), in such a way that the above definitions are preserved for the simple cases.

\[ \epsilon_x = {\partial \, u_x \over \partial X} \qquad \epsilon_y = {\partial \, u_y \over \partial Y} \qquad \epsilon_z = {\partial \, u_z \over \partial Z} \]

The simple case of uniaxial stretching can be described as

\[ x = \left( {X \over L_o} \right) L_f \]

and since \({\bf u} = {\bf x} - {\bf X}\), a little algebra can be applied to give

\[ u_x = \left( {X \over L_o} \right) \left( L_f - L_o \right) \]

So

\[ \epsilon_x \quad = \quad {\partial \, u_x \over \partial X} \quad = \quad {L_f - L_o \over L_o} \quad = \quad {\Delta L \over L_o} \]

which reproduces the "delta L over L" definition as desired.

\[ \gamma_{xy} = {\partial \, u_y \over \partial X} + {\partial \, u_x \over \partial Y} \]

The coordinate mapping equation for the shear example is

\[ \begin{eqnarray} x & = & X \\ y & = & Y + X D / T \end{eqnarray} \]

And the displacement field is

\[ \begin{eqnarray} u_x & = & 0 \\ u_y & = & X D / T \end{eqnarray} \]

The shear strain is

\[ \gamma_{xy} \quad = \quad {\partial \, u_y \over \partial X} + {\partial \, u_x \over \partial Y} \quad = \quad {\partial \, (X D / T) \over \partial X} + {\partial \, (0) \over \partial Y} \quad = \quad {D \over T} \]

This reproduces the desired result for this simple case: \(\gamma_{xy} = D / T\).

The symmetry of the equation also ensures that the computed shear value also satisfies the no-net-rotation criterion. The coordinate mapping equations for this example are

\[ \begin{eqnarray} x & = & X & + & Y \Delta x / T \\ y & = & Y & + & X \Delta y / T \end{eqnarray} \]

and they lead to

\[ \gamma_{xy} = {\Delta x + \Delta y \over T} \]

which again produces the desired result.

\[ \boldsymbol{\epsilon} = \left[ \matrix{ \epsilon_{11} & \epsilon_{12} \\ \epsilon_{21} & \epsilon_{22} } \right] = \left[ \matrix{ \epsilon_{xx} & \epsilon_{xy} \\ \epsilon_{yx} & \epsilon_{yy} } \right] = \left[ \matrix{ \epsilon_{xx} & \gamma_{xy}/2 \\ \gamma_{yx}/2 & \epsilon_{yy} } \right] \]

But since \(\gamma_{xy} = \gamma_{yx}\), all the tensors can also be written as

\[ \boldsymbol{\epsilon} = \left[ \matrix{ \epsilon_{11} & \epsilon_{12} \\ \epsilon_{12} & \epsilon_{22} } \right] = \left[ \matrix{ \epsilon_{xx} & \epsilon_{xy} \\ \epsilon_{xy} & \epsilon_{yy} } \right] = \left[ \matrix{ \epsilon_{xx} & \gamma_{xy}/2 \\ \gamma_{xy}/2 & \epsilon_{yy} } \right] \]

Setting \(\gamma_{xy} = \gamma_{yx}\) has the effect of making (requiring in fact) the strain tensors symmetric.

\[ \gamma_{ij} = 2 \epsilon_{ij} \]

It is always, always, always the case that if \(\gamma_{xy} = D / T = 0.10\), then the strain tensor will contain

\[ \boldsymbol{\epsilon} = \left[ \matrix{ ... & 0.05 & ... \\ 0.05 & ... & ... \\ ... & ... & ... } \right] \]

Alternatively, if the strain tensor is

\[ \boldsymbol{\epsilon} = \left[ \matrix{ ... & 0.02 & ... \\ 0.02 & ... & ... \\ ... & ... & ... } \right] \]

then \(\gamma_{xy} = D / T = 0.04\).

\[ \boldsymbol{\epsilon} = \left[ \matrix{ \epsilon_{11} & \epsilon_{12} & \epsilon_{13} \\ \epsilon_{21} & \epsilon_{22} & \epsilon_{23} \\ \epsilon_{31} & \epsilon_{32} & \epsilon_{33} } \right] = \left[ \matrix{ \epsilon_{xx} & \epsilon_{xy} & \epsilon_{xz} \\ \epsilon_{yx} & \epsilon_{yy} & \epsilon_{yz} \\ \epsilon_{zx} & \epsilon_{zy} & \epsilon_{zz} } \right] = \left[ \matrix{ \epsilon_{xx} & \gamma_{xy}/2 & \gamma_{xz}/2 \\ \gamma_{yx}/2 & \epsilon_{yy} & \gamma_{yz}/2 \\ \gamma_{zx}/2 & \gamma_{zy}/2 & \epsilon_{zz} } \right] \]

But rotational equilibrium requires that \(\gamma_{xy} = \gamma_{yx}\), \(\gamma_{xz} = \gamma_{zx}\), and \(\gamma_{yz} = \gamma_{zy}\). This also produces symmetric tensors.

\[ \boldsymbol{\epsilon} = \left[ \matrix{ \epsilon_{11} & \epsilon_{12} & \epsilon_{13} \\ \epsilon_{12} & \epsilon_{22} & \epsilon_{23} \\ \epsilon_{13} & \epsilon_{23} & \epsilon_{33} } \right] = \left[ \matrix{ \epsilon_{xx} & \epsilon_{xy} & \epsilon_{xz} \\ \epsilon_{xy} & \epsilon_{yy} & \epsilon_{yz} \\ \epsilon_{xz} & \epsilon_{yz} & \epsilon_{zz} } \right] = \left[ \matrix{ \epsilon_{xx} & \gamma_{xy}/2 & \gamma_{xz}/2 \\ \gamma_{xy}/2 & \epsilon_{yy} & \gamma_{yz}/2 \\ \gamma_{xz}/2 & \gamma_{yz}/2 & \epsilon_{zz} } \right] \]