Basics of Fluid Mechanics, 2014a

8.5. DERIVATIONS OF THE MOMENTUM EQUATION 249
y
a
b
y’
x’
45 ◦
y
a
b
c+b
d+a
c
d
y’
x’
45 ◦
(a) Deformations of the isosceles
triangular.
x
(b) Deformation of the straight angle
triangle.
x
Fig. -8.11. Different triangles deformation for the calculations of the normal stress.
constructed so it equals to dy. The forces acting in the direction of x’ on the element
are combination of several terms. For example, on the “x” surface (lower surface) and
the “y” (left) surface, the shear stresses are acting in this direction. It can be noticed
that “dx’” surface is √ 2 times larger than dx and dy surfaces. The force balance in the
x’ is
cos θ x
cos θ y
cos θ y
cos θ y
A {}}{
x {}}{ A y {}}{ A y {}}{ A
1 {}}{ 1 {}}{
x {}}{
A x ’
1 {}}{
{ }} {
1
dy τ xx √ + dx τ yy √ + dx τ yx √ + dy τ xy √ = dx √ 2 τ x’x’ (8.73)
2 2 2 2
dividing by dx and after some rearrangements utilizing the identity τ xy = τ yx results in
τ xx + τ yy
+ τ yx = τ x’x’ (8.74)
2
Setting the similar analysis in the y’ results in
τ xx + τ yy
2
− τ yx = τ y’y’ (8.75)
Subtracting (8.75) from (8.74) results in
or dividing by 2 equation (8.76) becomes
2 τ yx = τ x’x’ − τ y’y’ (8.76)
τ yx = 1 2 (τ x’x’ − τ y’y’ ) (8.77)
Equation (8.76) relates the difference between the normal shear stress and the
normal shear stresses in x’, y’ coordinates) and the angular strain rate in the regular (x, y