TOTAL HYDROSTATIC FORCE ON PLANE SURFACES
F
I
I
✗ h ff
of
Moment
:
'
12
h
Inertia
I=bh3
-
12
42
b
I.
h
£
①
F
: VERTICAL height from liquid surface to center of object
e
: found between F and center of the object
2/3
r
( m)
.
I=ñr
4
r
1m27
ñ=y
ñ
: if the OBJECT IS VERTICAL
I
⑨
f-
(m )
SLANTING height from liquid surface to center of object
Area
Force is located below the
center of the object
NA
yzh
43
"
e==
AY
Cgeccentricity
bh3
'
Moment of Inertia
T
=
36
43h
SAMPLE PROBLEMS
.
Problem #1
A vertical rectangular gate 1.5 m wide and 3 m high is submerged in water with its top edge 2 m below the water surface.
Find the a.) Total Pressure acting on one side of the gate and its b.) Location from the bottom.
a.) F=8ñA
I
c.
=/ 9.81¥ )(
2m
b-
f-
m
=
1.5Mt
2m
)(
3m
)(
1.5m
)
m
te
=
21-1.5
°
e
<
1. 5m
F
2
b.)
2
;
z=
bh3
.
-5
-
→
c
e=
12
bhh
(3
1. b-
3
12
=
)
154.508kW
m=
3m
;
m=ñ
1.5
-
2=1.286
31.5
m
1.5+2
7
e- 0.214m
3.714m
1- 0.214
"
Problem #2
A vertical triangular gate with top base horizontal and 1.5 m wide is 3 m high. It is submerged in oil having sg of 0.83 with its top base
submerged to a depth of 2 m. Determine the a.) magnitude and b.) location of the total hydrostatic pressure acting on one side of the gate.
a.) F=8ñA
2m
1. 5m
oil
sg=0 -82
5=5
f-
•
e
3m
7
F
0.82/(9.81%-3)
=
2m +
¥13m) (E) (1. 5) (3)
54.298kW
=
.
2
b.)
2
2
3- (
=
213h
=
•
z=
13h
'
~e=¥y=
;
1.
8)
1. b- (3)
te
3
36
-311.8) -10.167
=
2- ( i. 5) (3) [21-1-3137]
1.833M
e.
=
0.167m
Problem #3
A 30 m long dam retains 9 m of water as shown in the figure. Find the a.) total resultant force acting on the dam and the b.) location of
the center of pressure from the location.
if
f- 30m
1
W.S.
µ
g,
.
in
/
4m
e
go.gg
Sino
h
sin
h
9m
;
60°
:/
.
( 30)( 10.39272
12
=
b.)
F=8ÑA
=
f-
=
9.811¥
30110.392)
2
;
→
h
4.5m
13,762.645kW
30m
10.392m
z=
-
e
e
2
2=10.392
-
2
z
=
3.464m
1.732
=
÷
60=91
h=1o.3qzm
30N
a.
=
1.732m
10.392
2
,
Problem #4
The isosceles triangular gate shown in the figure is hinged at A and weighs 1500 N. What is the total hydrostatic force acting on one side
of the hate in kN?
F=8ñA
a.)
nm
3.5m
9-811%-3/0.83) # (2) 1- 3. 5) [21-(1712.6/1)]
=
•
F= 44.291 kN
É
gsin5o=2_
Sino
h
Oil
zm
Sg
2m
-0.83
=
-
50°
h=
2.611
My
•3
50°
.
Problem #5
A triangular gate with a horizontal base 1.2 m long and altitude of 1.8 m is inclined 45D from the vertical with the vertex pointing upward.
The hinged horizontal base of the gate is 2.7 m below the water surface.
.*
A. Calculate the total force of water on the gate;
B. How far is the said force from the vertex measured along the gate?
hz.fm
g-
C. What normal force must be applied at the vertex of the gate to open it?
i
sin45=÷
1.8M
0
☐ = 1.273M
-11
45°
"
2.7-1.273=1.42 > my
.
*
1.427
w
f-
=
sinus
w
=
=l
2.018m
,,
c.)
a.) t=8hA
=
.
(9.81%-3)
1.427Mt }- (1-273)
24.110 kN
f- (1-2)/1.81
EMo=0
f- ( b- h
-
e)
-
24.11/31-11.8)
.
N(1. 8)
-
=O
] N( I. 8)
0.056
-
.
.
N -7.287 kN
-
'
1. 2
b.)
9
;
→
9=}- (
1.
8)
36
=
d-
te
e
=
3-
9=1
.
( 1. 8) +0.056
256m
( 1. 8)
=
1.411.8)
3-11.811-2.018
0.056m
=o
1133440¥ # NVEYY
Principle
Archimedes
Buoyant y
BF
Law
Hydrostatics
of
unit
w
liquid
displaced by
I
Force
of
volume
8 ,=V☐→
=
w
body
the
H
=
☐
volume
Total
=
of the
D
body
0
p
⑧
weight
of fluid
-
80
=
object
W=BFz
liquid
80
=
o
OF
D
^
displaced (nakalubog )
Volume
BF
•
Examples
upward
force
:
#1
F
✓
given :
305mm
go
305mm
=
6288.46m¥
W
Fy :O
v
:
3m
Ftw
~
Ft
BF
Sov
=
BF
-
=
0
81=4--0
6288.46¥
Ft
f-
-
982.774 N
0.305Mt
I
9810¥
0.305Mt (3Mt
#2
EFy=o
yvu=
1000ms
given
sgo
sg ,
Tv
.
0.305A
(0.3051m)(3mf=O
0.983 kN
4-
:
:
W
:O -92
:
=
-
BF
=
Vp
0
-8,4=0
ios
sort
?
0.921981k¥ ) VT
1.
-
,
VT
=
=Y
=
VT
-
Vu
0319.811m¥) (Vt 1000m¥
-
9363.636 ms
BF
#3
w
pi
g=o
'
"
given
w=
~
.co
water
EFy=0
:
35000 lbs
h= ?
~
*
sg=io
BF ,
W
-
BF
,
:
-
131=2--0
35000 lbs
.
-
0.8162 4) (12 ) ( R)
h= 1.495ft
^
n
.
131=2
.
.
-
.
-
l
( 62.4 ) (12) ( (2) h
=
0
#4
We
we
01m
given
"
01m
:
v
Wc= 3825 N
0.5m
0.5m
FL
=
?
2m
2m
.
.
1. 5m
1.5m
^
^
¥
BE
✓
WL
BE
~
BE
we
=
8,
Ñ
EFy=o
,
:
?
=/ 10
We 1- We
KN
m
3
Wet
8th
3. 825+110
EFy=o
Wct 8 ,
-
L
Btc
-
3. 825 t 110th
W
=
w
ff
-
0.0772m
-
☐
,
9.81
}
-8,4
=
110
( 0.0772)
8.492 kN
BF ,
-
:
=
-
Vc
=O
8fVDc=O
'
-
9.81
:
we t WL
vi.
BF
-
=
Sf
V1 = 0.0702 m3
0
DL
=
We
0
11-(0-5)>11.5?
=
=
-
9.814=0
We
=
8th
110
( 0.0702)
7. 732
kN
11-(0-5) (1-5)
=
0
APOGEE #E$#Ée
0
We
045m
given :
0.64
sgw
=
Sgs
=
0.05m
05M
.
v
7.85
2m
2. 55m
r
É
steel
sg= 7.85
w
EFy=o
-
BFW
Swvwt 8s Vs
Ws
Ws
4
=
.
=
-
?
Ws
Bf,
0
8fVDw
-
255 ✗ 10
=
8s Vs
=
7.85
=
BFS
-
-
8fVDs
( 9.81 ) ( 0.05) ( 0.05) (3)
0.64
BE
:
Wwtws
Vs
=
%
t
=
0
7.85-(9-81) Vs
0.819.81)( 0.05 ) ( 0.05) (2.55)
-
-5m 3
( 98101¥ ) ( 4.255×10-5mF)
3.277 N
w
EFy=o :
W
-
BF
gwvw
0.924
BF
given
:
sgib
sgsw
Vt
✓☐
0.924
=
=
=
1.03
1000 m3
=
?
Up
=
.
=
0
g. ✓ ☐
=
o
( 9.81 ) (1000 )
897.087ms
-
1.03
(9-81) Up
=
0
-
0.8
( 9.81 )
Vs
=o
§
✗
STEPS
OF
SOLUTION
①
consider
②
Determine all
of dam
length
1m
forces
(a) Hydrostatic
W
water
the
.
W
of
the
.
U
reduce the stability
sliding
in
the upstream
permanent
structures
on
side
( if any )
£¥¥¥r
*
,
dam
1%8
1-0%7
upstream
-
downstream
(little
to
no
water)
Forces
Hydrostatic Force
Total
↳ ( vertical projection
-
Wind pressure
.
Wave
-
Floating
-
against overturning
and
( w/ water)
-
against
dam
)
( hydrostatic uplift)
Horizontal
the
#
-
b.)
of
MWL
dam
of
steerage
to
"
the
W
due
:
of
.
pressure
→
→
a.) vertical Forces
-
Pressure
of the
submerged portion
of the
dam )
Fu
W'
~
Action
N2
"
Bodies
Earthquake load
/ ¥:/
¥
*
'
t
.
③
solve
for
-
Reactions
a.) Vertical
Ry
Rx
b.) Horizontal
^
EFV
=
-
Heel
Ry
→
EFX
Rx
→
W,
=
=
twz + W ,
Fa
-
,
-
~
Moment
Sha
8h
Faz
a.
b.
*
Righting
Moment
Overturning
Left side TOE
→
side
Location
FOUNDATION PRESSURE
2
rotation
change
of moment
e
I
=
Ry (E)
of
RM
-
OM
-
coefficient
Against sliding Cs )
Max
=YzR÷
intensity
pressure
@
the
base
@
of the
from
=
> I
e
s
E-
e
>
¥
of
RM_ >
1
Fit to B
0M
of pressure
base
☒
:
=
@
9ma×=}R¥
:
safe_
Rx
=
B
TOTAL FORCE
safe
( DAM )
-
TOE
TRIANGLE
F-
=
is
twice
PRESSURE
average
DIAGRAM
R=
12×2
TEAL
FORCE
F-
=
FAZ
t
'
Ry
(WATER)
t F- v2
B-
-
I
2
¥
9=731 I ±¥
:
HORIZONTAL SHEARING
-
I
-
=
dam
( O)
overturning
Fso
eccentricity
,
friction
=
Against
=
if :
FACTORS OF SAFETY
Fss
=
* distance
Ry
.
Area
9
Moment
B
⑤
=
=¥( 8h 1- 8ha) (b) (1)
)
about the TOE ( right
~
~
~
y
U
④
Toe
UPLIFT
STRESS
9max= t
amin
=
9min = 0
-
¥É
?⃝
eexxaammpieeo.si
⑨
.
② FH
←
#^
given
Sgc
u
=
Fss
2.4
:
i.
V
"
0.4
=
FH
1.5
①
Ft
4.
]
0M
:
372.513
:
99.326
( %)
2-
( 1. 5)
=
=
491.158kW
-
9.81
=
W
③
⑨
-
Eso
141.264
I
R×=F*= 99.326
W
=
of
Location
m
2.637m
=
(2.637 )
-
KN
141.2646 kN
=
④
(5)
Ry
RM
☒
OM
By
Ry
=
491.158-148.989
=
0.919m
PRESSURE
FOUNDATION
=
=
2.663£
=
0.440
372.513
☒
IF
=
99.326
( 2. 4) (6) (b) (1)
141.2646 kN
=
↳ 1- =4
②
0.4/141.264 b)
=
N'-372.513 kN
m
148.989 KN
1.5
W=8Vol
Ry
DRM
RX
99.326 kN
=
WRY
=
5) (1)
b.
Im
④
Fss
9.8114¥)(
=
N
:
8517
=
⑤
eccentricity
491.158kW m
¥
> e
q=R÷( it %-)
→
:
=
-
=
e=
148.989kW
Fso
3.297
=
-
I
??};§B_(i±%?)
m
SAFE !
> 1
E-
-
=
9ma×
2.637-0.919
=
269.832
kPa
2
e=
9min
0.4m
=
12.696 kPa
#2
T
w
given
,
fc
W,
23.5K¥
=
"
⇐µ
①
im
.
µ
85A
=
9.81
=
FH
③
(3) (6) (1)
,
23.5
W
,
=
(2) (8) (1)
376 KN
Wz=
8 V01
=
23.5
Wz= 188 kN
e=
0M
Ry=
=
Fss
Fss
M¥f-
=
=
3761-188
=
Fso
1.916
R¥g
=
=
0.615-64)
=
f-
1=1+2
=
+
1
=
3.903
/
=
42--1.818
'
(176.58721-0)
f-
=
176.58 kN
SAFE
376
>
1
①
T=R¥= 1764-5-8
!
(21-1)+188 (3- C2))
176.58 (E)
SAFE !
⑨
kM-oµ_
☒
=
I
=
Ry
1.818M
=
0.667
→
¥
> e
q=R¥( II %-)
Fv2
5=44.415 kPa
176.58
>
I
'
①
564 kN
=
-
0.182m
2- ¥
12=590.996 kN
②
Wit Wz
E-
m
176.58 kN
=
⑤ Fso
[E(2) (8) (1) ]
-
(176.58721-1564)
=
⑨
KN
R×2tRy2
R=
=
=8Vol
=
353.16
=
④
=
Ry
w
176.58¥)
:
e-
⑨
R×= Fit
Rx
176.58 kN
=
ON
b) Fso
c)
0M
0.6
=
a) Fss
-4m
② FH
②
:
5644-(1+-610.4822)
9ma×=
179.493
kPa
9min
102.507
kPa
=
#3
El 50m
Fu
.
Yo =F-z
m :
p5m
m=
I
u
2
n
¥
:
~
50m
FH
w
w3
,
n=
P
n
m
5. 2m
n
=
①
0.75
Iz
Wz
El O
26m
7m
-
p
=
=
=
2.
4(9.81 ) (7)
2.
4(9.81 ) E-
)( )
( 52
( 2-
=
f- ( 490.5) ( 23
Fr
=
9.81
(E) (5) ( 50) ( t )
=
9.81
( 25) ( 50) (1)
-
t
6) (
y
26m
:
2.419.81 )tz( 5. 2) ( 52) ( t )
=
,
Wz=
V
~
,
5.2m
=
VF :
W
I
Wz
10
②
:
rnp
¥1.52m
52
→
→
)
2) (1)
W,
Wz
→
wz
→
u
→
3183.149 kN
=
=
=
15915.744 kN
5689.8 kN
:
=
Fu
8570.016kW
1226.25 KN
5m
m[Drain
"m
( seepage)
HF
:
r
8h
U
③
Ry
Ry
Rx
23205.359
=
FH
-
⑨
Ut Fv
RM
3183.149
kN
RM
( f- (5.27+33)+8510.016 (3.5+26)
684048.743
KN
-
1- 15915.744
m
0M :
12262.5
( %-) +5689.8/38.2 f- ( 23.2 ))
-
014=377724.24
⑨
⑤
I
I
②
Fss
Fss
⑨
RM-Ry0M_
=
Fso
=
=
=
13.201
h¥
1.419
=
684048.743
-
KN
M
B
e
=
-
2-
I
=
3%2--13.201
23205.359
e-
-
5.899m
¥=3%2_
0.75/23205.359)
=
6.367
→
¥
> e
12262.5
> 1
SAFE !
⑤
q=R¥( II %-)
=
__R÷=§§?%¥}
1=50--1.811
-
377724.24
m
=
> 1
SAFE !
Fit
=
12262.5 KN
:
12262.5 KN
=
→
490.5 kPa
FH
=
Rx
)
9.8 / ( 50
1- Ws
Wit Wz
=
=
=
23205.359
38.2
9
max
9min
(1+-615.899) )
=
1170.318 kPa
=
44.622 kPa
38.2
[3-(6)]
t
1226.25
[
38.5
-
f-(5)
]
?⃝
$¥¥¥¥¥¥i¥"¥i¥¥¥⇐ss
"
;
T.fm
gravity
:\ center
B
w
w
"
" ""
of
buoyancy
metacenter
-
µ
g.
,
BF
BF
n
Righting
moment
G
:
overturning
STABLE
BF
moment :D
UNSTABLE
$1T#☒KE
⑥☒ 66!!
Problem # 1
A
plastic
Is the
of
cube
cube
L
side
and
specific
gravity
0.82
is
placed
vertically
in water
.
stable ?
?
W
MBO
BF
=
=L
1+0.5 tarico)
"
V,
Vobj
=
I V2
-
-
Up
,z ,
✗
obj
✓
①¥
/
_ •
"
sg= 0.82
✓ obj
=
8W VD
o.ua#)e--a.eiv
Vp
=
L2
L2
=
→
1213
.
MBO
'
0.82L
¥361)
0.089L
=
o%=÷ E
-
=
÷
Vobj
hobj
=
3
ha
v☐
÷X=¥
D= 0.936L
-
6130=0.032
BF
i
Meta centric
No
=
=
Ms
=
MBO
0
.
-
089L
MBO > GBO
s
heights :
GBO
-
0.057L
0.032L
relationship
rectangular I
132
MBo=I☐[
general ;
MBO
1+0.5
I
=
✓ ☐ cost
tariff
,
=
hi
he
V2
relationship
A,
-
Az
heights
3
U
-
of volume
of
area
=L
he
2
to
height ;
;