Engineering Mechanics

Chapter 35 : Hydrostatics „„„„„ 727

M=wxbdx∫^2

But ∫xbdx^2 =I 0

∴ M=wI 0 ...(i)

where I 0 = Moment of inertia of the pressure about the liquid level.

We know that the sum of the moments of the pressure

=Ph× ...(ii)

where P= Total pressure on the surface, and

h = Depth of centre of pressure from the liquid surface.

Equating equations (i) and (ii),

Ph× =wI 0

wA x×h =wI 0 ...(∴=PwAx)

h =

I 0

Ax ...(iii)

We know from the Theorem of Parallel Axis that

I 0 =IG + Ah^2

where IG= Moment of inertia of the figure, about horizontal axis through

its centre of gravity.

h= Distance between the liquid surface and the centre of gravity

of the figure (x in this case)

Thus rearranging the equation (iii),

h =

2

IAx IGGx

Ax Ax

+

=+

Thus the centre of pressure is always below the centre of gravity of the area by a distance

equal to

IG

Ax

.

Example 35.8. A rectangular sluice gate is situated on the vertical wall of a lock. The
vertical side of the sluice is (d) metres in length and depth of centroid of the area is (p) metres below
the water surface. Prove that the depth of centre of pressure is equal to
2
12

d

p

p

⎛⎞
⎜⎟+
⎜⎟
⎝⎠
Solution. Given: Depth of sluice = d; Depth of centroid, xp=.
Let b= Width of sluice
∴ Area of sluice, A=bd
We know that moment of inertia of a rectangular section
about its centre of gravity and parallel to the base,

IG=

3

12

bd

and depth of centre of pressure from the water surface,

h =

32

G 12 12

bd d

I bd

xp p

Ax bd p bd p

×

+= += +

××

=

2

12

d

p

p

+ Ans.

Fig. 35.10.