Linear Deformation

Consider the same element ABCD again. If the element has to undergo a linear deformation it is necessary that u velocity change in the x-direction and v velocity in the y-direction. Let the velocities at A be u and v. Then at B the u-velocity will be $u+{\partial u \over \partial x}dx$ and at D it will be $v+{\partial v \over \partial y}dy$. As a result the element stretches both in x and y directions and assumes a shape A`B`C`D` shown in Fig. 4.8.

Figure 4.8: Linear Deformation of a Fluid Element

Now, the stretch in x-direction is given by distance BB' which is equal to

$$BB' = {\partial u \over \partial x} dx  dt$$ (4.30)

Corresponding change in the volume of the element (for a unit depth normal to the paper) is given by

$$d\forall_x = {\partial u \over \partial x} dx  dy dt$$ (4.31)

Similarly the change in the volume of the element in the y-direction is given by,

$$d\forall_y = {\partial v \over \partial y} dy  dx dt$$ (4.32)

Neglecting the change in volume due to CC```C`C``, The total change in volume is

$$d\forall = \left \{{{\partial u \over \partial x} + {\partial v \over \partial y}}\right\} dx  dy dt$$ (4.33)

Thus the rate of change of volume expressed as a fraction of the initial volume is given by,

$${1 \over \forall}{d\forall \over {dt}} = {{\partial u \over \partial x} + {\partial v \over \partial y}}$$ (4.34)

The left hand side is called the Volume Dilatation rate of the element. We have seen that the Right hand side of the equation is zero for all incompressible flows.

(c) Aerospace, Mechanical & Mechatronic Engg. 2005
University of Sydney