Basics of Fluid Mechanics, 2014a

4.5. FLUID FORCES ON SURFACES 105
The moment of inertia of the triangle around x is made of two triangles (as shown
in the Figure (4.24) for triangle 1 and 2). Triangle 1 can be calculated as the moment of
inertia around its center which is l 0 +2∗(l 1 −l 0 )/3. The height of triangle 1 is (l 1 −l 0 )
and its width b and thus, moment of inertia about its center is I xx = b(l 1 − l 0 ) 3 /36.
The moment of inertia for triangle 1 about y is
I xx1 = b(l 1−l 0 ) 3
36
+
A 1
{ }} {
b(l 1 −l 0 )
3
Δx 1
2
{
(
}} {
) 2
l 0 + 2(l 1−l 0 )
3
The height of the triangle 2 is a − (l 1 − l 0 ) and its width b and thus, the moment of
inertia about its center is
I xx2 = b[a−(l1−l0)]3
36
+
A 2
{ }} {
b[a−(l 1−l 0)]
3
Δx 2
2
{
(
}} {
) 2
l 1 + [a−(l1−l0)]
3
and the total moment of inertia
I xx = I xx1 + I xx2
The product of inertia of the triangle can
be obtain by integration. It can be noticed
that upper line of the triangle is
y = (l 1−l 0 )x
b
+ l 0 . The lower line of the
triangle is y = (l 1−l 0 −a)x
b
+ l 0 + a.
⎡
∫ b ∫ (l1−l0−a)x
b
+l 0+a
I xy = ⎣
xydx
+l 0
0
(l 1−l 0)x
b
The solution of this set equations is
A
y
l 1 b
F 3
1
2
F 1
F 2
Fig. -4.24. The general forces acting on a non
⎤ symmetrical straight area.
⎦ dy = 2 ab2 l 1 +2 ab 2 l 0 +a 2 b 2
24
{[ }} ]{
ab (g (6 l1 +3a)+6gl 0 ) ρ sin β +8P atmos
F 1 =
,
3
24
F
[ 2
ab
3
F
[ 3
ab
3
] = −
] =
„
„ « «
12 l1
(3 l 1 −14 a)−l 0 a −27 12 l02
+
a
gρ sin β
72
−
„„
a−
„„
24 l1
a −24 «
+
„
15 l1
a
«+l 0 27−
+
«
48 l0
a
P atmos
72
,
12 l1 12 l02
a
«+
a
72
„„ «
24 l1
a +24 + 48 l «
0
a
P atmos
72
End Solution
«
gρ sin β
x
l 0
a