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ME 111 Dr. Nikos J. Mourtos Hydrostatics 1

Example: Calculation of Hydrostatic Forces on Submerged Surfaces

A plane rectangular gate is submerged in water as shown in figure (a). What is themagnitude of the reaction force at A? For the cylindrical gate in figure (b), will themagnitude of the reaction at A be greater than, less than, or the same as that for the plane

gate?

Solution

Follow the steps for calculating the hydrostatic force on a submerged surface:

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ME 111 Dr. Nikos J. Mourtos Hydrostatics 2

1. The vertical projection of the gate AB is the rectangle BC, which has the same width(w) as the gate itself.

2. The centroid of BC is located at a depth

2

2

24

l BC

BC l ycv

=

+=

Combine these two eqs. to get ( )214

+=l

ycv

3. The hydrostatic pressure at the centroid of the vertical projection (as well as at thecentroid of the gate itself) is

( )214

+==l

y p cvc ! !

4. The area of the vertical projection of the gate is:22

)( lww BC AV ==

5. The horizontal component of the hydrostatic force can now be calculated:

F x = pc AV = 0.43 ! l2 w

6. The volume of the water directly over the surface of the gate can be calculated by

splitting it into convenient shapes, such as the prism ABD and the orthogonal parallelepiped EFBD:

( )2282

2

2

1

2

2

42

1 22

+=! ! " #

$$% &

+! ! " #

$$% &

='+'=' wl

wl wl l

BDAC EFBD LA

7. The vertical component of the hydrostatic force is equal in magnitude to the weight ofthe liquid directly over the surface of the gate:

wl W F LA LA y243.0 ! ! ="==

8. The total force on the gate is, of course, the vector sum of its components:

222 6.0 l w F F F y x ! =+=

9. Now we must find the 2 nd moment of inertia of the vertical projection of the gate(BC) about a horizontal axis through its centroid:

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ME 111 Dr. Nikos J. Mourtos Hydrostatics 3

482

12)( 33 l w BC w

I cv

==

10. The depth of the center of pressure can now be calculated from:

l A y

I y y

V cv

cvcvcp

67.0=+=

11. To find the line of action of the vertical component (F y) we must find the c.g. of thevolume of the water directly over the gate, since it is the weight of this water whichcauses the vertical force on the gate. To find the c.g. of the water directly over thegate we take the moments of the weights of the convenient shapes from step 6.

cp ycp LA AE x F DGW l

W xW M =+==!22)(

22

2

21

Now323

222

232

)()(32

)(32

22

22 l l l l AJ DA DJ DG =! " # $

% &

=! " # $

% & '! ! "

# $$% &

='==

wl wl

l W EFBD

2

1 18.0

42

2! ! ! ="

# $ %

& '

=(=

wl wl W ABC

2

2

2 25.0

2

2

2

1! ! ! =" " #

$ %%& '

=(=

Substituting into the moment eq.

l x xwl l wl l wl cpcp 29.0)43.0()24.0)(25.0()35.0)(18.0( 222 =!=+ " " "

12. To find the reaction at A lets take the moments about a horizontal axis through B:

wl Rl wl l wl l R

l y F xl F l R M cp xcp y B

222 51.00)42.0(43.0)42.0(43.022

042

2220

! ! ! ="=##

"=$ % & '( ) ##$ $ % & ''(

) ##$ $ % & ''(

) "=*

13. Regarding the cylindrical gate in figure (b) we can tell just by looking at it that thereaction will be less because:

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ME 111 Dr. Nikos J. Mourtos Hydrostatics 4

ya yb

xa xb

F F

F F !

=

since both gates have the same vertical projection (BC) but the volume of

the water over gate (b) is less than the volume of the water over gate (a).