Consider the steady, incompressible, two-dimensional flow field of Prob. 4‒69. Using the results of Prob. 4‒69(a). do the following:

(a) From the fundamental definition of shear strain rate (half of the rate of decrease of the angle between two initially perpendicular lines that intersect at a point), calculate shear strain rate εxy in the xy-plane.(Hint: Use the lower edge and the left edge of the fluid particle, which intersect at 90° at the lower-left corner of the particle at the initial time.)

(b) Compare your results with those obtained from the equation for Ɛxy in Cartesian coordinates, i.e.,

PROBLEM: Consider steady, incompressible, two-dimensional shear flow for which the velocity field iswhere a and b are constants. Sketched in Fig. P4‒69 is a small rectangular fluid particle of dimensions dx and dy at time t. The fluid particle moves and deforms with the flow such that at a later time (t + dt). the particle is no longer rectangular, as also shown in the figure. The initial location of each corner of the fluid particle is labeled in Fig. P4‒69. The lower-left corner is at (x. y) at time t. where the x-component of velocity is u = a + by. At the later time, this corner moves to (x + u dt, y), or

(a) In similar fashion, calculate the location of each of the other three corners of the fluid particle at time t ‒ dt.

(b) From the fundamental definition of linear strain rate (the rate of increase in length per unit length). calculate linear strain rates Ɛxx and Ɛyy.

(c) Compare your results with those obtained from the equations for Ɛxx and Ɛyy in Cartesian coordinates, i.e..

FIGURE P4‒69

Then we have to compare the result we obtained using the equation

Part (a)

Step 2</p>

The fundamental definition of the shear strain given by the half of the rate of decrease of angle between the two initially perpendicular line in the given object.

The following figure shows a rectangular particle in the fluid and how it deforms due to the velocity gradient in the fluid.

The initial position of the lower left corner is given by and the upper left corner was . Now after time , the positions are

The position of the lower left corner

And the position of upper left corner is

Hence, the change in the position of the top left corner is

Hence, the angle is given by

(since is very small)

So the angle is given by

Hence, the rate of change of the angel is given by

So the shear strain is given by

So the calculated shear strain from the fundamental definition is .

Part (b)