3695 Sutardi Fluid Statics 02

8/12/2019 3695 Sutardi Fluid Statics 02

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HYDROSTATIC FORCE ON SUBMERGEDHYDROSTATIC FORCE ON SUBMERGED

SURFACESSURFACES


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HYDROSTATIC FORCE ON SUBMERGEDHYDROSTATIC FORCE ON SUBMERGED

SURFACESSURFACES

Need to determine: (1) Force magnitude

(2) Force direction(3) The location of the acting force

rr

'rr


4/19

HYDROSTATIC FORCE ON SUBMERGED SURFACESHYDROSTATIC FORCE ON SUBMERGED SURFACES

ApdFdrr

=

Static Fluids NO Shear stress

Against the surface

= AR ApdF

rr

g

dh

dp= p = p0 +g h

Hence: p and A must be expressed in terms of (x,y)

FORCE MAGNITUDE (FORCE MAGNITUDE (FFRR))


5/19

MOMENTS ABOUT O (0,0)MOMENTS ABOUT O (0,0)

== AAR ApdrFdrF'rr

r

r

r

r

r

rr

'rr

= AR ypdAF'y

=

AR xpdAF'x

Maka didapat:

THE LOCATION OF THE ACTING FORCE

dimana: ;'zk'yj'xi'r ++=r

;dAkAd =r

;FkFRR

=r

( ) ( ) ( ) +=+=+AA

R kpdAyjxiFdyjxiFk'yj'xi

r

maka:


6/19

= AR

ypdAF

'y1

= AR

xpdAF

'x1

Ap

hdAApF

atm

A

atmR

+=

THE LOCATION OF THE ACTING FORCE


7/19

EXAMPLE PROBLEM 3.5EXAMPLE PROBLEM 3.5

Resultant force,

GIVEN: Rectangular gate, hinged alongA, w = 5m

FIND:

SOLUTION:

, of the water on the gateRFr

Does the resultant force pass through the center

of gravity?

NO!


8/19

= AR ApdF

rr

g

dh

dp=

Basic Eqs.:

== AAR pwdApdF rr

k wdAd =r

k

( ) +== L

AR wdsinDgApdF

0

030 rr

k

588=RFr

k kN {Force acts in negative z direction}

h

Fdr

From the diagram:

h = D + sin 300

p = g(D + sin 300) (p = gh)

Atmospheric

pressure is neglected


9/19

Finding the line of action of the resultant force, .FRr

= AR pdAF'

( ) +=== L L

RRA

R

dsinDFgwpwd

FpdA

F'

0 0

03011

+=

+= 0

32

0

032

3032

3032

sinLDL

F

gwsin

D

F

gw'

R

L

R

= 2.22 m; and y = m.m.sin

m

sin

D ' 22622230

2

30 00 =+=+

m.FF

wpdA

F

wpdA

w

FxpdA

F'x R

RA

RA

RA

R

52222

11=====

j.i.'r 22652 +=r

m



10/19


==sss AAA

ydAdAyhdA sinsin



11/19

From solid mechanics, the location of the center of

gravity (centroid of the area) measured from the free

surface is:

=A

c ydAA

y1

AhAysinhdA ccA

==


ApF cR =



12/19


Center of pressure on an inclined plane surface:

AyI

Ay

dAy

'yc

xx

c

A==

2

Ay

Iy'y

c

xx

c+=

Ay

Ix'x

c

yx

c

+=

where:

y = y-location of acting force

x = x-location of acting force

yc = y-location of surface center of

gravity

xc = x-location of surface center of

gravity

= surface moment of inertiaaround an axis parallel to x-axis

passing through CG

= cross moment of inertiaaround axes passing through CG



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16/19

Centroidal moments of inertia for various cross sections: (a) rectangle,

(b) circle, (c) triangle, and (d) semicircle.

BUOYANCY AND STABILITYBUOYANCY AND STABILITY


17/19

Buoyancy is the net pressure force acting on a submerged

body.

The magnitude of the buoyant force is equal to the weight of

the fluid displaced by the body.

bbb VgVF ==

Buoyant force passes through the center of buoyancy, which

is at the centroid of the displaced fluid.

BUOYANCY AND STABILITYBUOYANCY AND STABILITY


18/19

Stability of floating bodies

EXAMPLE PROBLEM 3.8EXAMPLE PROBLEM 3.8


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GIVEN: Crown volume, V = 18.9 in.3

FIND: The average material density of the crown

SOLUTION:

Net force on the string, Fnet = 4.7 lbf

netF

r

buoyancyF

r

+ gravityFr

=

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3695 Sutardi Fluid Statics 02