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60 Chapter 2 ■ Fluid Statics<br />

The resultant <strong>fluid</strong><br />

force does not pass<br />

through the centroid<br />

of the area.<br />

and, therefore,<br />

x R <br />

xy dA<br />

A<br />

I xy<br />

y c A y c A<br />

where I xy is the product of inertia with respect to the x and y axes. Again, using the parallel axis<br />

theorem, 1 we can write<br />

x R I xyc<br />

y c A x c<br />

(2.20)<br />

F Rleft<br />

c<br />

Gate<br />

F Rright<br />

where I xyc is the product of inertia with respect to an orthogonal coordinate system passing through<br />

the centroid of the area and formed by a translation of the x–y coordinate system. If the submerged<br />

area is symmetrical with respect to an axis passing through the centroid and parallel to either the<br />

x or y axes, the resultant force must lie along the line x x c , since I xyc is identically zero in this<br />

case. The point through which the resultant force acts is called the center of pressure. It is to be<br />

noted from Eqs. 2.19 and 2.20 that as y c increases the center of pressure moves closer to the centroid<br />

of the area. Since y c h csin u, the distance y c will increase if the depth of submergence, h c ,<br />

increases, or, for a given depth, the area is rotated so that the angle, u, decreases. Thus, the hydrostatic<br />

force on the right-hand side of the gate shown in the margin figure acts closer to the centroid<br />

of the gate than the force on the left-hand side. Centroidal coordinates and moments of inertia<br />

for some common areas are given in Fig. 2.18.<br />

––<br />

b<br />

2<br />

––2<br />

b<br />

c<br />

y<br />

x<br />

––2<br />

a<br />

––2<br />

a<br />

A = ba<br />

I –––<br />

1 xc = ba 3<br />

12<br />

I yc = –––<br />

1<br />

ab 3<br />

12<br />

I xyc = 0<br />

R<br />

c<br />

y<br />

x<br />

A = π R 2<br />

I –––––<br />

π R 4<br />

xc = I yc =<br />

4<br />

I xyc = 0<br />

(a) Rectangle<br />

(b) Circle<br />

R<br />

c<br />

y<br />

x<br />

R<br />

4R<br />

–––<br />

3π<br />

A = –––––<br />

π R2<br />

2<br />

I xc = 0.1098R 4<br />

I yc = 0.3927R 4<br />

I xyc = 0<br />

a<br />

d<br />

b + d<br />

–––––––<br />

3<br />

c<br />

y<br />

b<br />

x<br />

A = –––<br />

ab I 2 xc = –––-–<br />

ba 3<br />

36<br />

I xyc = –––––<br />

ba 2<br />

(b – 2d)<br />

72<br />

a<br />

–– 3<br />

(c) Semicircle<br />

(d) Triangle<br />

4R<br />

–––<br />

3π<br />

c<br />

R<br />

y<br />

4R<br />

–––<br />

3π<br />

x<br />

A = –––––<br />

π R2<br />

4<br />

I xc = I yc = 0.05488R 4<br />

I xyc = –0.01647R 4<br />

F I G U R E 2.18<br />

(e) Quarter circle<br />

Geometric properties of some common shapes.<br />

1 Recall that the parallel axis theorem for the product of inertia of an area states that the product of inertia with respect to an orthogonal<br />

set of axes 1x–y coordinate system2 is equal to the product of inertia with respect to an orthogonal set of axes parallel to the original set<br />

and passing through the centroid of the area, plus the product of the area and the x and y coordinates of the centroid of the area. Thus,<br />

I xy I xyc Ax c y c .

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