• The concept of stress was examined in chapter @ref(stress).
• We have not yet discussed displacement, nor the relationship between stress and displacement.
• Therefore, we next examine the concepts of deformation, displacement, and strain in three dimensions.

Small deformation definition of strain

In an introduction to engineering course, the following description of strain was likely presented:

\[\begin{equation} {\varepsilon}= \lim_{L \rightarrow 0} \frac{\Delta L}{L} \end{equation}\]

• However, this equation is a linear approximation, is one dimensional, and makes a critical assumption of small strains.
• Hence it is inadequate for some important material we will cover this semester.

Normal strain for larger deformations

Strain is derived from deformation. Therefore we must take a closer look at deformation in order to fully understand strain.

Examine a deformation vector in a body under strain.

In this derivation, the vector is assumed to be originally aligned with the \(x\) direction, and the normal strain (\({\varepsilon_{xx}}\)) is computed.

(This assumption is not necessary, however it simplifies the calculation.)

It is noteworthy that we are looking a quadratic equation with respect to vector length, i.e., this definition examines a higher order strain equation than the linear strain encountered in many undergraduate textbooks.

• Examine the displacement vector:

\[\begin{equation} \begin{split} O'A' =& [{\delta x}{\vec{i}}] + [(u + {\delta u}){\vec{i}}+ (v+{\delta v}){\vec{j}}+(w+{\delta w}){\vec{k}}] - [u {\vec{i}}+ v {\vec{j}}+w {\vec{k}}]\\ =& ({\delta x}+{\delta u}) {\vec{i}}+ {\delta v}{\vec{j}}+ {\delta w}{\vec{k}}\\ (O'A')^2 =& ({\delta x}+{\delta u})^2 + {\delta v}^2 + {\delta w}^2 \\ =& {\delta x}^2 +2 {\delta x}{\delta u}+ {\delta u}^2 + {\delta v}^2 + {\delta w}^2 \\ \end{split} \end{equation}\]

\[\begin{equation} \begin{split} \frac{(O'A')^2}{(OA)^2}=& 1 +2 \frac{{\delta u}}{{\delta x}} + \left(\frac{{\delta u}}{{\delta x}}\right)^2 + \left(\frac{{\delta v}}{{\delta x}}\right)^2 + \left(\frac{{\delta w}}{{\delta x}}\right)^2 \end{split} \end{equation}\]

\[\begin{equation} \begin{split} \frac{(O'A')^2}{(OA)^2}-1 =& 2 \frac{{\delta u}}{{\delta x}} + \left[\left(\frac{{\delta u}}{{\delta x}}\right)^2 + \left(\frac{{\delta v}}{{\delta x}}\right)^2 + \left(\frac{{\delta w}}{{\delta x}}\right)^2 \right] \end{split} \end{equation}\]

Recall our quadratic definition of strain.

\[\begin{equation} \begin{split} {\varepsilon_{xx}}=& \lim_{OA \rightarrow 0} \left(\frac{1}{2}\right)\left[ \frac{(O'A')^2}{(OA)^2} -1 \right] \\ \end{split} \end{equation}\]

Therefore: \[\begin{equation} \begin{split} {\varepsilon_{xx}}=& {\frac{\partial u}{\partial x}} + \frac{1}{2} \left[\left({\frac{\partial u}{\partial x}}\right)^2 + \left({\frac{\partial v}{\partial x}}\right)^2 + \left({\frac{\partial w}{\partial x}}\right)^2 \right]\\ \approx & {\frac{\partial u}{\partial x}} \end{split} \end{equation}\]

Similarly, we can derive:

\[\begin{equation} \begin{split} {\varepsilon_{yy}}=& {\frac{\partial v}{\partial y}} + \frac{1}{2} \left[\left({\frac{\partial u}{\partial y}}\right)^2 + \left({\frac{\partial v}{\partial y}}\right)^2 + \left({\frac{\partial w}{\partial y}}\right)^2 \right]\\ \approx & {\frac{\partial v}{\partial y}} \end{split} \end{equation}\]

\[\begin{equation} \begin{split} {\varepsilon_{zz}}=& {\frac{\partial w}{\partial z}} + \frac{1}{2} \left[\left({\frac{\partial u}{\partial z}}\right)^2 + \left({\frac{\partial v}{\partial z}}\right)^2 + \left({\frac{\partial w}{\partial z}}\right)^2 \right]\\ \approx & {\frac{\partial w}{\partial z}} \end{split} \end{equation}\]

Definition of shear strain and its relation to linear strain

Shear strain for larger deformations

• Examiner initial and final configuration for two vectors of a strained body. In the initial configuration, the angle defined is 90 degrees.

\[\begin{equation} \angle BOA = 90 {^\circ} \end{equation}\]

We will define the shear strain as: \[\begin{equation} {\gamma_{xy}}= 90 {^\circ}- \angle B'O'A' \end{equation}\]

Manipulating our definition of \({\gamma_{xy}}\): \[\begin{equation} \begin{split} \sin {\gamma_{xy}}=& \sin \left(90 {^\circ}- \angle B'O'A' \right) \\ =& \cos \left( \angle B'O'A' \right) \end{split} \end{equation}\]

• The latter of these two can be attributed to the phase angle between sine and cosine.

• Separately, we can take the scalar product of the two vectors: \[\begin{equation} \begin{split} \vec{O'B'} \cdot \vec{O'A'} =& ||O'B'|| \; ||O'A'|| \cos \left(\angle B'O'A'\right) \\ \cos \left(\angle B'O'A'\right) =& \frac{\vec{O'B'} \cdot \vec{O'A'}}{||O'B'|| \; ||O'A'||} \\ \end{split} \end{equation}\]

• Therefore, we have: \[\begin{equation} \sin {\gamma_{xy}}= \frac{\vec{O'B'} \cdot \vec{O'A'}}{||O'B'|| \; ||O'A'||} \end{equation}\]

• We need calculate the scalar product: \[\begin{equation} \begin{split} \vec{O'A'} =& ({\delta x}+ {\delta u}) {\vec{i}}+ {\delta v}{\vec{j}}+ {\delta w}{\vec{k}}\\ \vec{O'B'} =& {\delta u}{\vec{i}}+ ({\delta y}+ {\delta v}) {\vec{j}}+ {\delta w}{\vec{k}}\\ \end{split} \end{equation}\]

\[\begin{equation} \vec{O'B'} \cdot \vec{O'A'} = ({\delta x}+ {\delta u}) {\delta u}+ ({\delta y}+ {\delta v}) {\delta v}+ {\delta w}{\delta w} \end{equation}\]

• Now, for the purposes of the change in angle, we can assume that the length of the vectors does not change significantly. Therefore: \[\begin{equation} \begin{split} O'A' \approx& \; {\delta x}\\ O'B' \approx& \; {\delta y}\\ \end{split} \end{equation}\]

Therefore:

\[\begin{equation} \begin{split} \sin {\gamma_{xy}}=& \, \frac{({\delta x}+ {\delta u}) {\delta u}+ ({\delta y}+ {\delta v}) {\delta v}+ {\delta w}{\delta w}}{{\delta x}{\delta y}} \\ =& \, (1 + {u {}_{,x}}){u {}_{,y}} + (1 + {v {}_{,y}}){v {}_{,x}} + {w {}_{,x}} {w {}_{,y}}\\ =& \, {u {}_{,y}} + {v {}_{,x}} + {u {}_{,x}} {u {}_{,y}} + {v {}_{,x}} {v {}_{,y}} + {w {}_{,x}} {w {}_{,y}} \\ \end{split} \end{equation}\]

Finally, we can also know that for small shear angles:

\[\begin{equation} \sin {\gamma_{xy}}\approx {\gamma_{xy}} \end{equation}\]

Therefore: \[\begin{equation} {\gamma_{xy}}= {u {}_{,y}} + {v {}_{,x}} + {u {}_{,x}} {u {}_{,y}} + {v {}_{,x}} {v {}_{,y}} + {w {}_{,x}} {w {}_{,y}} \end{equation}\]

Summary of the engineering strain-deformation equations

\[\begin{equation} \begin{split} {\varepsilon_{xx}}=& {u {}_{,x}} + \frac{1}{2} \left[\left({u {}_{,x}}\right)^2 + \left({v {}_{,x}}\right)^2 + \left({w {}_{,x}}\right)^2 \right]\\ {\varepsilon_{yy}}=& {v {}_{,y}} + \frac{1}{2} \left[\left({u {}_{,y}}\right)^2 + \left({v {}_{,y}}\right)^2 + \left({w {}_{,y}}\right)^2 \right]\\ {\varepsilon_{zz}}=& {w {}_{,z}} + \frac{1}{2} \left[\left({u {}_{,z}}\right)^2 + \left({v {}_{,z}}\right)^2 + \left({w {}_{,z}}\right)^2 \right]\\ {\gamma_{xy}}=& {u {}_{,y}} + {v {}_{,x}} + {u {}_{,x}} {u {}_{,y}} + {v {}_{,x}} {v {}_{,y}} + {w {}_{,x}} {w {}_{,y}}\\ {\gamma_{xz}}=& {u {}_{,z}} + {w {}_{,x}} + {u {}_{,x}} {u {}_{,z}} + {v {}_{,x}} {v {}_{,z}} + {w {}_{,x}} {w {}_{,z}}\\ {\gamma_{yz}}=& {v {}_{,z}} + {w {}_{,y}} + {u {}_{,y}} {u {}_{,z}} + {v {}_{,y}} {v {}_{,z}} + {w {}_{,y}} {w {}_{,z}}\\ \end{split} \end{equation}\]

Engineering strain vs tensor strain

Lets compare the following equations: \[\begin{equation} \begin{split} {\varepsilon_{xx}}=& {u {}_{,x}} + \frac{1}{2} \left( ({u {}_{,x}})^2 + ({v {}_{,x}})^2 + ({w {}_{,x}})^2 \right) \\ {\gamma_{xy}}=& {u {}_{,y}} + {v {}_{,x}} + {u {}_{,x}} {u {}_{,y}} + {v {}_{,x}} {v {}_{,y}} + {w {}_{,x}} {w {}_{,y}} \\ \end{split} \end{equation}\]

\[\begin{equation} \begin{split} 2 \, {\varepsilon_{xx}}=&{u {}_{,x}} + {u {}_{,x}} + {u {}_{,x}} {u {}_{,x}} + {v {}_{,x}} {v {}_{,x}} + {w {}_{,x}} {w {}_{,x}} \\ {\gamma_{xy}}= 2 \, {\varepsilon_{xy}}=&{u {}_{,y}} + {v {}_{,x}} + {u {}_{,x}} {u {}_{,y}} + {v {}_{,x}} {v {}_{,y}} + {w {}_{,x}} {w {}_{,y}} \end{split} \end{equation}\]

From the above equations we observe: \[\begin{equation} {\varepsilon_{xy}}= \frac{1}{2} {\gamma_{xy}} \end{equation}\]

Similarly: \[\begin{equation} \begin{split} {\gamma_{xy}}= 2 \, {\varepsilon_{xy}}=&{u {}_{,z}} + {w {}_{,x}} + {u {}_{,x}} {u {}_{,y}} + {v {}_{,x}} {v {}_{,y}} + {w {}_{,x}} {w {}_{,y}} \\ {\gamma_{xz}}= 2 \, {\varepsilon_{xz}}=&{u {}_{,z}} + {w {}_{,x}} + {u {}_{,x}} {u {}_{,z}} + {v {}_{,x}} {v {}_{,z}} + {w {}_{,x}} {w {}_{,z}} \\ {\gamma_{yz}}= 2 \, {\varepsilon_{yz}}=&{v {}_{,z}} + {w {}_{,y}} + {u {}_{,y}} {u {}_{,z}} + {v {}_{,y}} {v {}_{,z}} + {w {}_{,y}} {w {}_{,z}} \end{split} \end{equation}\]

Summary of the tensor strain-deformation equations

\[\begin{equation} \begin{split} {\varepsilon_{xx}}=& {u {}_{,x}} + \frac{1}{2} \left[({u {}_{,x}})^2 + ({v {}_{,x}})^2 + ({w {}_{,x}})^2 \right]\\ {\varepsilon_{yy}}=& {v {}_{,y}} + \frac{1}{2} \left[({u {}_{,y}})^2 + ({v {}_{,y}})^2 + ({w {}_{,y}})^2 \right]\\ {\varepsilon_{zz}}=& {w {}_{,z}} + \frac{1}{2} \left[({u {}_{,z}})^2 + ({v {}_{,z}})^2 + ({w {}_{,z}})^2 \right]\\ {\varepsilon_{xy}}=& \frac{1}{2} \left({u {}_{,y}} + {v {}_{,x}}\right) + \frac{1}{2} \left[{u {}_{,x}} {u {}_{,y}} + {v {}_{,x}} {v {}_{,y}} + {w {}_{,x}} {w {}_{,y}} \right]\\ {\varepsilon_{xz}}=& \frac{1}{2} \left({u {}_{,z}} + {w {}_{,x}}\right) + \frac{1}{2} \left[{u {}_{,x}} {u {}_{,z}} + {v {}_{,x}} {v {}_{,z}} + {w {}_{,x}} {w {}_{,z}} \right]\\ {\varepsilon_{yz}}=& \frac{1}{2} \left({v {}_{,z}} + {w {}_{,y}}\right) + \frac{1}{2} \left[{u {}_{,y}} {u {}_{,z}} + {v {}_{,y}} {v {}_{,z}} + {w {}_{,y}} {w {}_{,z}} \right]\\ \end{split} \end{equation}\]

• It is notable that these higher order strain equations are completely consistent with the linearized versions that are typically used in linear analysis.
  • There are many circumstances where the linear versions of these equations are perfectly adequate.
• However, in some circumstances such as stability analysis or calculations where the deformations are relatively large, it will be necessary to carry the higher-order terms.

The principal strains

\[ \left[ \begin{array}{ccc} {\varepsilon_{xx}}- {\varepsilon_{\mathrm{P}}}& {\varepsilon_{xy}}& {\varepsilon_{xz}}\\ {\varepsilon_{xy}}& {\varepsilon_{yy}}- {\varepsilon_{\mathrm{P}}}& {\varepsilon_{yz}}\\ {\varepsilon_{xz}}& {\varepsilon_{yz}}& {\varepsilon_{zz}}- {\varepsilon_{\mathrm{P}}}\\ \end{array} \right] \left\{ \begin{array}{c} l \\ m \\ n \end{array} \right\} = \left\{ \begin{array}{c} 0 \\ 0 \\ 0 \\ \end{array} \right\} \]

• The strain tensor, like the stress tensor, has principal values and principal directions.
  • The computation is identical to the computation for stress.
  • The principal strains exist in a coordinate frame where the shear strain components are zero
• It is notable that the principal stress and strain directions are not necessarily the same.
  • In general, they will be the same for linear elastic materials but may not be so for anisotropic materials.

Note that there are only 3 displacements, but 6 strains.

• To get strain from displacement, we must differentiate. Differentiation loses information, so is not difficult to stomach
• To get displacement from strain, we must integrate. While we could get 6 integration constants, we don’t have enough unique displacements to establish them. (Note, these are pointwise equations, so at any one point we do not have 6 displacements.
  • We are either overdetermined (not physically possible for a real system), or we have redundancy in the equations.
  • Consequently, there must be additional constraints on strain:

Compatibility relationships

Consider first 2D. We have 2 displacements and 3 strains. \[\begin{equation} \begin{split} {\varepsilon_{xx}}= \frac{\partial u}{\partial x} \\ {\varepsilon_{yy}}= \frac{\partial v}{\partial y} \\ 2 {\varepsilon_{xy}}= \frac{\partial u}{\partial y} + \frac{\partial v}{\partial x} \\ \end{split} \end{equation}\]

• Taking derivatives of the third equations (with respect to \(x\), then \(y\)): \[\begin{equation} \begin{split} \frac{2 \, \partial {\varepsilon_{xy}}}{\partial x} = \frac{\partial}{\partial x} \frac{\partial u}{\partial y} + \frac{\partial}{\partial x} \frac{\partial v}{\partial x} \\ \frac{2 \, \partial^2 {\varepsilon_{xy}}}{\partial x \partial y} = \frac{\partial^2 \partial u}{\partial y^2 \partial x} + \frac{\partial^2 \partial v}{\partial x^2 \partial y} \\ \end{split} \end{equation}\]

Thus: \[\begin{equation} \begin{split} \frac{2 \, \partial^2 {\varepsilon_{xy}}}{\partial y \partial x} = \frac{\partial {\varepsilon_{xx}}}{\partial y^2} + \frac{\partial {\varepsilon_{yy}}}{\partial x^2} \\ \end{split} \end{equation}\]

• This establishes a relationship between the three equations such that they are not independent.