Chapter 35 : Hydrostatics „„„„„ 733

35.13.CENTRE OF PRESSURE OF A COMPOSITE SECTION

The centre of pressure of a composite section (i.e., a section with cut out hole or other compos-
ite section) is obtained as discussed below :

1. First of all, split up the composite section into convenient sections (i.e., rectangles, triangles
   or circles).


2. Calculate the total pressures, P 1 , P 2 ..... on all the sections.


3. Now calculate the resultant pressure P on the whole section by the algebraic sum of the
   different pressures.


4. Then calculate the depths of centres of pressures h 1 , h 2 ... for all the sections from the
   liquid surface,


5. Finally equate Ph=+ +P h 11 P h2 2 .....

where (^) h = Depth of centre of pressure of the section from the liquid level.
Example 35.12. A trapezoidal plate, having its parallel sides (2a) and (a) at a distance (h)
apart, is immersed vertically in water with (2a) side uppermost (horizontal) at a depth of (h) below
the water surface. Find the total thrust on the surface and the centre of pressure.
Solution. Given: Area of the triangular section 1,
A 1 =
1
2
2
××=ah ah
and area of triangular section 2,
A 2 =
1
22
ah
××=ah
Total thrust on the surface
We know that the depth of centre of gravity of
triangular section 1 from the water surface,
x 1 =
4
33
hh
h+=
Similarly, x 2 =
25
33
hh
h+=
We also know that pressure on triangular portion 1,
P 1 =
2
11
44
33
hwah
wA x =× × =w ah ...(i)
Similarly, P 2 =
2
22
55
23 6
ah h wah
wA x =× × =w ...(ii)
∴ Total pressure, P=
22
12
45
36
wah wah
PP+= + =
13 2
6
wah (^) Ans.
Centre of pressure
Let h = Depth of centre of pressure of the plate, from water surface.
We know that the moment of inertia of triangular section 1 about its centre of gravity and
paralled to the base,
IG 1 =
(^333) 2()
36 36 18
bh a h ah

Similarly, moment of inertia of triangular section 2 about its centre of gravity,
IG 2 =
33
36 36
bh ah

Fig. 35.16.