If a pressure is applied to an object, would the object change shape, or deform? How much would it deform, and what terminology would we use to describe this? Well, if an object deforms in shape, this can be described using what's called strain; strain is often denoted by a Greek epsilon ($ \epsilon $). With all materials, they have their own measure of elasticity, or Young's Modulus (E), and this is part of the relation between stresses and strains. If a stress is applied, then a strain can be measured. We will talk in great lengths about this in later tutorials, but for now, let's learn how strain is derived.
It all starts with a shape in 3D space. This shape starts with a position of $ (x, y, z) $ and two points marked within the object, points P and Q. The shape is then deformed and in that process, point P is moved to a new position, $ (\bar{x}, \bar{y}, \bar{z}) $. What also happens is the distance between P and Q went from $ d\vec{s} $ to $ d\vec{\bar{s}} $. Let's see what is happening here with a good ol'fashioned image.
Fig. 1 - Image of an object deformed in 3D space with internal points P and Q.
Figure 1 shows what I was describing before with an object being deformed in some way such that two points internal to the blob change position and distance between one another. Notice the vectors defined as well; there is vector $ \vec{r} $, which is the position vector for $ P $ before deformation, and vector $ \vec{\bar{r}} $, which is the position vector for point $ \bar{P} $ after deformation. These vectors can be written as such:
$$ \vec{r} = x \hat{i} + y \hat{j} + z \hat{k} $$
$$ \vec{\bar{r}} = \bar{x} \hat{i} + \bar{y} \hat{j} + \bar{z} \hat{k} $$
With both of these vectors now defined, $ \vec{u} $ can be defined as the vector between the points $ P $ and $ \bar{P} $, where
$$ \vec{u} = \vec{\bar{r}} - \vec{r} $$
$$ = u \hat{i} + v \hat{j} + w \hat{k} $$
I should note that u, v, and w are not velocities like in Flight Vehicle Dynamics. In Flight Vehicle Structures, they are displacements instead, and yes, I understand the confusion in multi-use terminology for variables. It just takes some time for things to click and go from dynamics mode to structures mode and making the connection between the different uses.
The next step that can be done is defining the distance vectors between $ P $ and $ Q $ as well as $ \bar{P} $ and $ \bar{Q} $; these vectors are $ d\vec{s} $ and $ d\vec{\bar{s}} $, respectively. $ d\vec{s} $ can be defined as
$$ d\vec{s} = dx \hat{i} + dy \hat{j} + dz \hat{k} $$
And $ d\vec{\bar{s}} $ can be defined as
$$ d\vec{\bar{s}} = d\bar{x} \hat{i} + d\bar{y} \hat{j} + d\bar{z} \hat{k} $$
Both of these vectors are describing the difference between the points $ P $ and $ Q $ as well as the points $ \bar{P} $ and $ \bar{Q} $, respectively. Now, both of these can be squared to give
[ 1 ]
$$ (ds)^2 = dx^2 + dy^2 + dz^2 $$
[ 2 ]
$$ (d\bar{s})^2 = d\bar{x}^2 + d\bar{y}^2 + d\bar{z}^2 $$
The next step in the derivation is to define what $ \bar{x} $, $ \bar{y} $, and $ \bar{z} $ are. Well, these three variables are simply the change of the original position plus $ u $, $ v $, and $ w $, respectively. So we get that
$$ \bar{x} = x + u $$
$$ \bar{y} = y + v $$
$$ \bar{z} = z + w $$
Now, since $ (d\bar{s})^2 - (ds)^2 $ is the measure of distortion, we can write that
$$ (d\bar{s})^2 = (dx+du)^2 + (dy+dv)^2 + (dz+dw)^2 $$
$$ = dx^2 + dy^2 + dz^2 + 2(dudx + dvdy + dzdw) + du^2 + dv^2 + dw^2 $$
Notice that $ (ds)^2 = dx^2 + dy^2 + dz^2 $; $ (d\bar{s})^2 - (ds)^2 $ can then be written as
[ 3 ]
$$ (d\bar{s})^2 - (ds)^2 = 2(dudx + dvdy + dzdw) + du^2 + dv^2 + dw^2 $$
The next thing we need to work on is writing u, v, and w in terms of x, y, and z. After all, they are functions of x, y, and z! Let's write them out and then talk through them.
[ 4 ]
$$ du = \frac{\partial u}{\partial x} dx + \frac{\partial u}{\partial y} dy + \frac{\partial u}{\partial z} dz $$
[ 5 ]
$$ dv = \frac{\partial v}{\partial x} dx + \frac{\partial v}{\partial y} dy + \frac{\partial v}{\partial z} dz $$
[ 6 ]
$$ dw = \frac{\partial w}{\partial x} dx + \frac{\partial w}{\partial y} dy + \frac{\partial w}{\partial z} dz $$
For all three displacements, u, v, and w, the partial derivative of them are taken with respect to x, y, and z, and then they are multiplied by the infinitesimal change in x, y, and z. Now you may be wondering why u has a $ dy $ component, or why z has a $ dx $ component? This is because for a strain in one dimension, there could be a change in displacement in the other two dimensions as well. Eqs. (4-6) will account for this movement!
This can all be substituted back into $ (d\bar{s})^2 - (ds)^2 $ to get the non-linear strain-displacement relations.
[ 7 ]
$$ \epsilon_{xx} = \frac{\partial u}{\partial x} + \frac{1}{2} \begin{bmatrix} (\frac{\partial u}{\partial x})^2 + (\frac{\partial v}{\partial x})^2 + (\frac{\partial w}{\partial x})^2 \end{bmatrix} $$ $$ \epsilon_{yy} = \frac{\partial v}{\partial y} + \frac{1}{2} \begin{bmatrix} (\frac{\partial u}{\partial y})^2 + (\frac{\partial v}{\partial y})^2 + (\frac{\partial w}{\partial y})^2 \end{bmatrix} $$ $$ \epsilon_{zz} = \frac{\partial w}{\partial z} + \frac{1}{2} \begin{bmatrix} (\frac{\partial u}{\partial z})^2 + (\frac{\partial v}{\partial z})^2 + (\frac{\partial w}{\partial z})^2 \end{bmatrix} $$ $$ \gamma_{xy} = \frac{\partial u}{\partial y} + \frac{\partial v}{\partial x} + \begin{bmatrix} \frac{\partial u}{\partial x}\frac{\partial u}{\partial y}+ \frac{\partial v}{\partial x}\frac{\partial v}{\partial y} + \frac{\partial w}{\partial x}\frac{\partial w}{\partial y} \end{bmatrix} $$ $$ \gamma_{yz} = \frac{\partial w}{\partial y} + \frac{\partial v}{\partial z} + \begin{bmatrix} \frac{\partial u}{\partial y}\frac{\partial u}{\partial z}+ \frac{\partial v}{\partial y}\frac{\partial v}{\partial z} + \frac{\partial w}{\partial y}\frac{\partial w}{\partial z} \end{bmatrix} $$ $$ \gamma_{zx} = \frac{\partial u}{\partial z} + \frac{\partial w}{\partial x} + \begin{bmatrix} \frac{\partial u}{\partial z}\frac{\partial u}{\partial x}+ \frac{\partial v}{\partial z}\frac{\partial v}{\partial x} + \frac{\partial w}{\partial z}\frac{\partial w}{\partial x} \end{bmatrix} $$
[ 8 ]
These equations look like a lot, because non-linear systems are tougher to solve! However, in this specific study, we will be focusing on linear strain-displacements; these look like the following:
[ 9 ]
$$ \epsilon_{xx} = \frac{\partial u}{\partial x} $$ $$ \epsilon_{yy} = \frac{\partial v}{\partial y} $$ $$ \epsilon_{zz} = \frac{\partial w}{\partial z} $$ $$ \gamma_{xy} = \frac{\partial u}{\partial y} + \frac{\partial v}{\partial x} $$ $$ \gamma_{yz} = \frac{\partial w}{\partial y} + \frac{\partial v}{\partial z} $$ $$ \gamma_{zx} = \frac{\partial u}{\partial z} + \frac{\partial w}{\partial x} $$
These look much cleaner, right! Now you know how to derive the strain-displacement equations. These will be used throughout our studies in great lengths, so getting familiar with them is key.