FLUID STATIC

PRINCIPLE OF HYDROSTATIC
PRESSURE IS THE FORCE PER UNIT AREA EXERTED BY A
LIQUID OR GAS ON A BODY OR SURFACE, WITH THE FORCE
ACTING AT RIGHT ANGLE TO THE SURFACE UNIFORMLY IN ALL
DIRECTIONS

Force F dF
PRESSURE =
Area
= A
= dA

PRINCIPLES OF HYDROSTATICS
1. THE LIQUID AT REST CANNOT RESIST SHEARING STRESS.
2. THE TOTAL FORCE IS ALWAYS NORMAL TO THE PLANE
WHERE IT ACTS. (PERPENDICULAR).

3. PRESSURE EXISTS AT EVERY POINT.

4. AT ANY POINT IN A LIQUID AT REST. THE PRESSURE IS
EQUAL IN ALL DIRECTIONS (PASCAL'S LAW).
5. THE PRESSURE OF ALL POINTS LYING ON A PLANE
PARALLEL TO THE LIQUID SURFACE IS EQUAL.
6. THE PRESSURE VARIES LINEARLY WITH DEPTH.

GAGE AND ABSOLUTE PRESSURE

GAGE PRESSURE (RELATIVE PRESSURE) ARE PRESSURES
ABOVE OR BELOW ATMOSPHERE AND CAN BE MEASURED BY
PRESSURE GAGES OR MANOMETER’S.
ATMOSPHERIC PRESSURE IS THE PRESSURE AT ONE POINT ON
THE EARTH’S SURFACE FROM THE WEIGHT OF AIR ABOVE IT.
VACUUM A REGION OR SPACE HAVING PRESSURE LESS THAN
ATMOSPHERIC PRESSURE.
RELATIONSHIP BETWEEN PRESSURES

ATMOSPHERIC PRESSURE UNDER NORMAL CONDITIONS AT

SEA LEVEL
PATM = 2166 LB/FT2
= 14.7 PSI
= 29.9 INCHES OF MERCURY (HG)
= 760MM HG
= 101.325 KPA = 1ATM = 1 BAR

ABSOLUTE PRESSURE IS THE PRESSURE ABOVE ABSOLUTE

ZERO (VACUUM)
PABS = PGAGE+ PATM
NOTE:
 ABSOLUTE ZERO IS ATTAINED IF ALL AIR IS REMOVED
 ABSOLUTE PRESSURE CAN NEVER BE NEGATIVE

VARIATION IN PRESSURE
 W=ϒV

W = ϒ (AL) [ Σ FX = 0 ]

F2 – F1 = W SINΘ

P2A – P1A= ϒ(AL) SINΘ

P2– P1= ϒL SINΘ

BUT L SINΘ = H

P2 – P1= ϒH

THE DIFFERENCE IN PRESSURE BETWEEN ANY TWO POINTS IN

A HOMOGENEOUS FLUID AT REST IS EQUAL TO THE PRODUCT
OF THE UNIT WEIGHT OF THE FLUID (ϒ) TO THE VERTICAL
DISTANCE (H) BETWEEN THE POINTS

P2 = P1 + ϒH A PRESSURE AT POINT IS THE PRESSURE AT

POINT 1 AND THE PRODUCT OF THE VERTICAL DISTANCE
BETWEEN THEM AND THE UNIT WEIGHT OF FLUID.

WHEN POINT 1 LIE ON THE FREE SURFACE: P = ϒH THIS

MEANS THE PRESSURE AT ANY POINT “H” BELOW A FREE
LIQUID SURFACE IS EQUAL TO THE PRODUCT OF THE UNIT
WEIGHT OF THE FLUID (ϒ) AND H

IF POINTS 1 AND 2 LIE ON THE SAME ELEVATION, H=0. P 1 = P2

THIS MEANS THAT THE PRESSURE ALONG THE SAME
HORIZONTAL PLANE IN A HOMOGENEOUS FLUID AT RESTS
ARE EQUAL.
PRESSURE HEAD - IS THE HEIGHT “H” OF A COLUMN OF
HOMOGENEOUS LIQUID OF UNIT WEIGHT ϒ THAT WILL
PRODUCE AN INTENSITY OF PRESSURE P H = P/ϒ

TO CONVERT PRESSURE HEAD(HEIGHT) OF LIQUID A TO

LIQUID B HB = HA SA / SB OR HB = HA ΡA / ΡB OR HB = HA ϒA / ϒB
TO CONVERT PRESSURE HEAD (HEIGHT) OF ANY LIQUID TO
WATER, JUST MULTIPLY ITS HEIGHT BY ITS SPECIFIC GRAVITY
HWATER= H LIQUID X S LIQUID

PASCAL’S LAW
“THE PRESSURE AT A POINT IN A FLUID IS EQUAL IN
DIRECTIONS”.

∑ 𝐹𝑥= 0 𝐹1- 𝐹3 SIN 𝜃 = 0

𝑝1𝐴1- 𝑝3𝐴3 SIN 𝜃 =0 ⟶ ①
∑ 𝐹𝑦= 0 𝐹2- 𝐹3 COS 𝜃 = 0
𝑝2𝐴2- 𝑝3𝐴3 COS 𝜃 =0 ⟶ ②

SIN 𝜃 = ; 𝐴1 = 𝐴3 SIN 𝜃 SUB IN ①

A1
A3
𝑝1 𝐴3 SIN 𝜃 - 𝑝3𝐴3 SIN 𝜃 =0

𝑝1 = 𝑝3

COS 𝜃 = ; 𝐴2 = 𝐴3 COS 𝜃 SUB IN

A2
A3

𝑝2𝐴3 COS 𝜃 - 𝑝3𝐴3 COS 𝜃 =0

②

𝑝2 = 𝑝3
∴ 𝑝1 = 𝑝2 = 𝑝3
APPLICATION OF PASCAL’S LAW
 HYDRAULIC JACK
 HYDRAULIC BRAKES
 HYDRAULIC PUMPS
 HYDRAULIC LIFT

F1 F1
P= =
A1 A1

HOWEVER, VOLUME OF THE LIQUID REMAINS THE SAME

 V1 – V 2
A1 = V 2
A1∆X1 = A2∆X2
PROBLEMS IN FLUID STATICS
PROBLEM 1: GRAVITY OF THE SEA WATER IS 1.03?. WHAT IS
THE PRESSURE AT A DEPTH OF 16M BELOW THE SEA FREE
SURFACE IF THE SPECIFIC
GIVEN: H=16M S=1.03
REQUIRED: PRESSURE P

kN
P16 = YSWH = (1.03)((9.81 3
)(16 m) = 161.669 KN/M2
m
= 161.669 KPA

PROBLEM 2. OIL (S=0.80) FLOWS IN AN OIL PIPE TAPPED WITH

PIEZOMETER. THE PRESSURE IN THE PIPE IS 40 KPA. WHAT IS
THE EQUIVALENT COLUMN HEIGHT IN THE PIEZOMETER?

GIVEN: S= 0.80 P= 40 KPA

REQ. : H

kN
40
p m2
H= = =5.097 M
y kN
0.8 x 9.81 3
m

PROBLEM 3. IF THE ABSOLUTE PRESSURE IN TANK IS 140 KPA,

DETERMINE THE PRESSURE HEAD IN MM OF MERCURY
(S=13.55). THE ATMOSPHERIC PRESSURE IS 100 KPA.

GIVEN: PABS = 140 KPA PATM = 100 KPA

REQ. : HHG

PABS = PG + PATM
140KPA = PG + 100 KPA
PG = 40 KPA
 PG = YHGHHG
kN
40 KPA = 13.55 (9.81 3
¿(h Hg)
m
HHG = 0.301 M = 301MM

MEASUREMENT OF PRESSURE

- MANOMETER IS A DEVICE USED FOR MEASURING

THE PRESSURE AT A POINT IN A FLUID BY
BALANCING THE COLUMN OF FLUID WITH THE SAME
COLUMN OR ANOTHER OF THE FLUID
CLASSIFICATION OF MANOMETERS

(1) SIMPLE MANOMETER:

- PIEZOMETER
- U-TUBE MANOMETER
- SINGLE COLUMN MANOMETER
- VERTICAL SINGLE COLUMN MANOMETER
- INCLINED SINGLE COLUMN MANOMETER

(2) DIFFERENTIAL MANOMETER:

- U-TUBE DIFFERENTIAL MANOMETER

- INVERTED U-TUBE DIFFERENTIAL
MANOMETER

1. PIEZOMETER : A PIEZOMETER IS THE SIMPLEST FORM OF

THE MANOMETER. IT MEASURES GAUGE PRESSURE ONLY.

THE PRESSURE AT ANY POINT IN THE LIQUID IS INDICATED BY

THE HEIGHT OF THE LIQUID IN THE TUBE ABOVE THAT POINT,
WHICH CAN READ ON THE CALIBRATED SCALE ON GLASS
TUBE.
THE PRESSURE AT POINT A IS GIVEN BY;
 P = PGH = WH

∴ H=
p
PIEZOMETER HEAD
pg
2. U-TUBE MANOMETER : IT CAN BE MEASURE
LARGE PRESSURE OR VACUUM PRESSURE AND
GAS PRESSURE.

∴ PRESSURE AT XX IN LEFT COLUMN

= PRESSURE AT X IN LEFT COLUMN

∴ P + P1GH1 = P2GH2

∴ P = P2GH2 – P1GH1
NOW, P=PGH, H HEAD IN TERMS OF WATER COLUMN,
PGH = P2GH2 – P1GH1

∴H=
P2 p1
h2− h1
p p
∴ H = S2H2 – S1H1

3. SINGLE COLUMN MANOMETER:

(A) VERTICAL SINGLE COLUMN MANOMETER
ONE OF THE LIMBS IN DOUBLE COLUMN
MANOMETER IS CONVERTED INTO A RESERVOIR
HAVING LARGE CROSS SECTIONAL AREA (ABOUT
100 TIMES) WITH RESPECT TO THE OTHER LIMB.

∴ VOLUME OF HEAVY LIQUID FALL IN RESERVOIR

= VOLUME OF HEAVY LIQUID RISE IN RIGHT COLUMN

∴ A X ∆H = A X H2
a x h2
∆H =
A
PRESSURE IN LEFT COL. = PRESSURE IN RIGHT COL.
 ∴P=
ah2
A 2
[ p g− p1 g ] + p 2 g h2− p 1 g h1 … … … ..(i)
∴ P = P2GH2 – P1GH1

(B) VERTICAL SINGLE COLUMN

MANOMETER. IT IS MODIFIED OF
VERTICAL COLUMN MANOMETER. THIS
MANOMETER IS USEFUL FOR THE
MEASUREMENT OF SMALL PRESSURE.

HERE, HEIGHT H2 = L SINΘ AND PUTTING IN ABOVE EQ. (I);

SINCE, A << A, NEGLECTING FIRST TERM;

∴P = P2GLSINΘ - P1GH1

∴H = S2GLSINΘ - S1H1

4. U-TUBE DIFFERENTIAL MANOMETER

IT IS USED TO MEASURE PRESSURE DIFFERENCE AT TWO
POINTS IN A PIPE OR BETWEEN TWO PIPES AT DIFFERENT
LEVELS.
CASE 1 – U-TUBE UPRIGHT DIFFERENTIAL
MANOMETER CONNECTED AT TWO POINTS IN A
PIPE AT SAME LEVEL
PRESSURE DIFFERENCE AT TWO POINTS IN A PIPE,
LEFT LIMB EQ. HA + (H + H1)S……………..(I)
RIGHT LIMB EQ. HS + H1S + HS1………………..(II)
* PRESSURE IN SAME AT THE DATUM LINE :
HA +(H1+ H)S = HS + H1 + HS1
HA – HS = -H1S – HS + -H1S – HS1
 HA – HS = H(S1-S)
CASE 2 – U-TUBE UPRIGHT DIFFERENTIAL
MANOMETER CONNECTED BETWEEN TWO PIPES
AT DIFFERENT LEVELS AND CARRYING
DIFFERENT FLUIDS

LEFT LIMB EQ : HA + H1S1

RIGHT LIMB EQ : HS + H2S2 + HS
* PRESSURE IS SAME AT THE DATUM LINE :
HA + H1S1 = HS + H2S2 + HS
HA – HS = H2S2 – H1S1 + HS

5. INVERTED U-TUBE DIFFERENTIAL

MANOMETER
IT IS USED FOR LOW PRESSURE DIFFERENCE.

LEFT LIMB EQ : HA - H1S1………………..(I)

RIGHT LIMB EQ : HS - H2S2 – HS………………(II)
* PRESSURE IS SAME AT THE DATUM LINE :
HA + H1S1 = HS - H2S2 – HS
HA – HS = H1S1 – H2S2 - HS

MECHANICAL GAUGES:
THESE ARE THE DEVICES IN WHICH THE PRESSURE IS
MEASURED BY BALANCING THE FLUID COLUMN BY SPRING
(ELASTIC ELEMENT) OR DEAD WEIGHT. GENERALLY, THESE
GAUGES ARE USED FOR MEASURING HIGH PRESSURE AND
WHERE HIGH PRECISION IS NOT REQUIRED. SOME COMMONLY
USED MECHANICAL GAUGES ARE:
(i) BOURDON TUBE PRESSURE GAUGE,
(ii) (II) DIAPHRAGM PRESSURE GAUGE,
(iii) (III) BELLOW PRESSURE GAUGE, AND
(iv) (IV) DEAD-WEIGHT PRESSURE GAUGE.

BOURDON TUBE PRESSURE GAUGES ARE USED

FOR THE MEASUREMENT OF RELATIVE
PRESSURES FROM 0.6 ... 7,000 BAR. THEY ARE
CLASSIFIED AS MECHANICAL PRESSURE
MEASURING INSTRUMENTS, AND THUS OPERATE
WITHOUT ANY ELECTRICAL POWER. BOURDON
TUBES ARE RADIALLY FORMED TUBES WITH AN
OVAL CROSS-SECTION.

DIAPHRAGM PRESSURE GAUGE IS A DEVICE

THAT USES A DIAPHRAGM WITH A KNOWN
PRESSURE TO MEASURE PRESSURE IN A FLUID.
IT HAS MANY DIFFERENT USES, SUCH AS
MONITORING THE PRESSURE OF A CANISTER OF
GAS, MEASURING ATMOSPHERIC PRESSURE, OR
RECORDING THE STRENGTH OF THE VACUUM IN
A VACUUM PUMP.

BELLOWS GAUGE CONTAINS AN ELASTIC

ELEMENT THAT IS A CONVOLUTED UNIT THAT
EXPANDS AND CONTRACTS AXIALLY WITH
CHANGES IN PRESSURE. THE PRESSURE TO BE
MEASURED CAN BE APPLIED TO THE OUTSIDE
OR INSIDE OF THE BELLOWS
DEAD-WEIGHT PRESSURE GAGE IS GAGE
CALIBRATED BY DEAD WEIGHT TESTER
APPARATUS USES KNOWN TRACEABLE
WEIGHTS TO APPLY PRESSURE TO A FLUID
FOR CHECKING THE ACCURACY OF READINGS
FROM A PRESSURE GAUGE. TYPICALLY,
DEADWEIGHT TESTERS ARE USED IN
CALIBRATION LABORATORIES TO CALIBRATE
PRESSURE TRANSFER STANDARDS LIKE
ELECTRONIC PRESSURE MEASURING
DEVICES.

MANOMETER PROBLEMS
STEPS IN SOLVING MANOMETER PROBLEMS:

1.DRAW A SKETCH OF THE MANOMETER APPROXIMATELY TO

SCALE.
2.DECIDE ON THE FLUID IN FEET OR METER OF WHICH THE
HEADS ARE TO BE EXPRESSED.
3.STARTING WITH ATMOSPHERIC SURFACE IN THE
MANOMETER, NUMBER IN ORDER THE LEVELS OF CONTACT
OF FLUIDS OF DIFFERENT SPECIFIC GRAVITY.
4.PROCEED FROM LEVEL TO LEVEL, ADDING (IF GOING
DOWN) OR SUBTRACTING (IF GOING UP) PRESSURE HEADS AS
THE ELEVATION DECREASES OR INCREASES, RESPECTIVELY
WITH DUE REGARD FOR THE SPECIFIC GRAVITY OF THE
FLUIDS.
MANOMETER RULE
START A POINT IN THE FLUID WHERE THE
PRESSURE IS TO BE DETERMINED, AND
PROCEED TO ADD TO IT THE PRESSURES
ALGEBRAICALLY FROM ONE VERTICAL FLUID
INTERFACE TO THE NEXT UNTIL YOU REACH
THE LIQUID SURFACE AT THE OTHER END OF THE
MANOMETER
7. A U-TUBE MANOMETER IS USED TO
MEASURE THE PRESSURE OF OIL OF SPECIFIC
GRAVITY 0.85 FLOWING IN A PIPE LINE. ITS
LEFT END IS CONNECTED TO THE PIPE AND
THE RIGHT-LIMB IS OPEN TO THE
ATMOSPHERE. THE CENTRE OF THE PIPE IS
100 MM BELOW THE LEVEL OF MERCURY
(SPECIFIC GRAVITY = 13.6) IN THE RIGHT
LIMB. IF THE DIFFERENCE OF MERCURY LEVEL
IN THE TWO LIMBS IS 160 MM, DETERMINE
THE ABSOLUTE PRESSURE OF THE OIL IN THE PIPE.

LET H1 = THE HEIGHT OF OIL IN THE LEFT LIMB = 160MM –

100MM = 60MM = 0.06M. STARTING AT FREE SURFACE,
EXPRESSING PRESSURE HEAD IN TERMS OF WATER.

PA
0 + H2S2 – H1S2 =
Yw
PA
0 + .16M (13.6) X 0.06M (0.85) = 9.81 kN
m ¿
3

¿
kg kg
PA = 1000 P =¿ 23550 3
3 Hg
m m

kg
GIVEN: H=0.8M P0 = 900 3 PW=1000
m
kg
PHG = 13550 3
m
REQUIRED : H’
REFERRING TO LEFT FIGURE
HCD = (0.2M + H’ + 0.4) -0.8M = H’ – 0.2M

STARTING FROM A, WRITING PRESSURE HEAD IN TERMS OF

WATER:

kg kg
900 3
1330 3
m m
0 + H’ + 0.4M – (H’ – 0.2M) =0
kg kg
1000 3 1000 3
m m
H = 0.246 M

9. THE TWO PIPES CONTAIN HEXYLENE GLYCOL

WHICH CAUSES THE LEVEL OF MERCURY IN
THE MANOMETER TO BE AT H= 0.3M.
DETERMINE THE DIFFERENTIAL PRESSURE IN
THE PIPES, PA – PB TAKES SHGL = 0.923, SHG =
13.55. NEGLECT THE DIAMETER OF THE PIPES.
GIVEN: H=0.3M SHGL = 0.923, SHG = 13.55
REQUIRED = PA – PB

IN TERMS OF PRESSURE OF HEXYLENE GLYCOL STARTING

FROM A
PA + SHGLYW HAC - SHGYW HCD - SHGLYW HBD = PB

kN kN
PA + (0.923)(9.81 3 )(0.1M)-(13.55)(
9.81 3 )(0.3M)- (0.923)
m m
kN
(9.81 3 )(0.1M) = PB
m
PA – PB = 39.9 KPA
10. THE INVERTED U-TUBE MANOMETER IS
USED TO MEASURE THE DIFFERENCE IN
PRESSURE BETWEEN WATER FLOWING IN THE
PIPES A AND B. IF THE TOP SEGMENT IS FILLED
WITH AIR, THE WATER LEVELS IN EACH
SEGMENT ARE INDICATED, DETERMINE THE
PRESSURE DIFFERENCE BETWEEN B AND A.
(HIBBELER, 2015) GIVEN: REFER TO THE FIGURE
REQUIRED: PB – PA
NOTICE THAT THE PRESSURE THROUGHOUT THE AIR IN THE
TUBE IS CONSTANT
PA = PW1 + PAIR = YW(HW)1 + PAIR
PB= PW2 + PAIR = YW(HW)2 + PAIR
PB – PA = YW(HW)2 + PAIR - YW(HW)1 + PAIR
PB – PA = YW(HW)2 - YW(HW)1

kN
PB – PA = (9.81 3 )(0.3M – 0.225M) = 0.73575 KPA
m
HYDROSTATIC FORCES ON PLANE AND INCLINED SURFACES

TOTAL HYDROSTATIC FORCE ON PLANE SURFACES

•PLANE SURFACE INSIDE A GAS CHAMBER
•HORIZONTAL PLANE SURFACE
•VERTICAL PLANE SURFACE (BY FORMULA & BY INTEGRATION)
•INCLINED PLANE SURFACE
IN GENERAL, THE TOTAL HYDROSTATIC PRESSURE ON ANY
PLANE SURFACE IS EQUAL TO THE PRODUCT OF THE AREA OF
THE SURFACE AND THE UNIT PRESSURE AT ITS CENTER OF
GRAVITY.
 F = PCGA
WHERE, PCG IS THE PRESSURE AT THE CENTER OF GRAVITY.
FOR HOMOGENEOUS FREE LIQUID AT REST, THE EQUATION
CAN BE EXPRESSED IN TERMS OF UNIT WEIGHT Y OF THE
LIQUID.
F = YHA

WHERE H IS THE DEPTH OF LIQUID ABOVE THE CENTROID OF

THE SUBMERGED AREA.
FOR HORIZONTAL PLANE SURFACE SUBMERGED IN LIQUID, OR
PLANE SURFACE INSIDE A GAS CHAMBER, OR ANY PLANE
SURFACE UNDER THE ACTION OF UNIFORM HYDROSTATIC
PRESSURE, THE TOTAL HYDROSTATIC FORCE IS GIVEN BY
F = PA

WHERE P IS THE UNIFORM PRESSURE AND A IS THE AREA.

DERIVATION OF FORMULAS THE FIGURE
SHOWN BELOW IS AN INCLINED PLANE
SURFACE SUBMERGED IN A LIQUID. THE
TOTAL AREA OF THE PLANE SURFACE IS GIVEN
BY A, C_G IS THE CENTER OF GRAVITY, AND
C_P IS THE CENTER OF PRESSURE

THE DIFFERENTIAL FORCE DF ACTING ON THE ELEMENT DA IS

DF=PDA SINCE P=𝛾ℎ

DF= 𝛾ℎ DA

FROM THE FIGURE H=YSINѲ DF= 𝛾 (YSINѲ)DA

INTEGRATE BOTH SIDES AND NOTE THAT 𝛾 AND Ѳ ARE
CONSTANTS,

F= 𝛾 SIN Ѳ ∫YDA

F=(ΓSIN Ѳ )A 𝑦
F= 𝛾(𝑦 SIN Ѳ )A
RECALL FROM CALCULUS THAT ∫YDA=A𝑦

FROM THE FIGURE, Y SIN Ѳ = H THUS,

THE PRODUCT YH IS A UNIT PRESSURE AT THE CENTROID AT

THE PLANE AREA, THUS, THE FORMULA CAN BE EXPRESSED IN
A MORE GENERAL TERM BELOW.
F = PCGA

LOCATION OF TOTAL HYDROSTATIC FORCE (ECCENTRICITY)

FROM THE FIGURE ABOVE, S IS THE INTERSECTION OF THE
PROLONGATION OF THE SUBMERGED AREA TO THE FREE
LIQUID SURFACE. TAKING MOMENT ABOUT POINT S.

FYP = ∫ ydF
WHERE:

DF = Y ( 𝑦 SIN Ѳ ) DA

F = Y ( 𝑦 SIN Ѳ ) A

[ Y ( 𝑦 SIN Ѳ ) A]YP = ∫ y [ y ( ysinθ ) dA ]

( 𝑦 SIN Ѳ )AY YP = (YSIN Ѳ) ∫ y 2 dA
AY YP = ∫ y 2 dA
AGAIN FROM CALCULUS, ∫ y dA IS CALLED MOMENT OF
2

INERTIA DENOTED BY I SINCE OUR REFERENCE POINS IS S,

AY YP = IS
THUS,

Is
YP =
Ay
BY TRANSFER FORMULA FOR MOMENT OF INERTIA IS = IG +
2
I g+ A y
AY2, THE FORMULA FOR YP WILL BECOME YP =
Ay
Ig
YP = Y +
Ay

YP = Y + E, THUS, THE DISTANCE

FROM THE FIGURE ABOVE,
BETWEEN CG AND CP IS

Ig
ECCENTRICITY, E =
Ay
PLANE SURFACE INSIDE THE GAS CHAMBER
12. THE FIGURE BELOW SHOWS A VERTICAL
CIRCULAR GATE IN A 3-M DIAMETER TUNNEL
WITH WATER ON ONE SIDE AND AIR ON THE
OTHER SIDE. SOLVE
(A) THE HORIZONTAL REACTION AT THE
HINGE
(B) THE LOCATION OF HYDROSTATIC FORCE
ACTING FROM THE INVERT
(C) LOCATIONS OF THE HINGE ( MEASURED FROM THE
INVERT) TO HOLD THE GATE IN POSITIONS?

GIVEN: REFER TO FIGURE

REQUIRED = R0, Y AND Z

FORCE ON THE GATE DUE TO AIR PRESSURE

2
π (3 m)
FAIR = PAIRA = 45 KPA [ ]
4
FAIR = 318.09 KN
FORCE ON THE GATE DUE TO WATER

( kN
)
2
FW = YW H A = 9.81 3
( 10.5 m )[ π (3 m) ]
m 4
FW = 728.10 KN
HORIZONTAL FORCE AT THE HINGE SUPPORT
∑FX = 0
R0 = FW - FAIR
 R0 = 728.10 KN – 318.09 KN
R0 = 410.01KN
LOCATION OF FW FROM THE INVERT

π 4
(3 )
Ig 64
E= =
Ay
( )
π 2
4
3 10.5

E = 0.0536 M
Y = 1.5 – E = 1.5-0 – 0536
Y = 1.4464 M
LOCATION OF HINGE SUPPORT
∑M0 = 0
(Y-Z) FW = (1.5 – Z)FAIR
(1.4464 – Z)728.10 KN = (1.5 – Z) 318.09 KN
Z= 1.4048 M

HORIZONTAL PLANE SURFACE

13. A RECTANGULAR PLATE 3 M LONG AND 1 M WIDE IS
IMMERSED VERTICALLY IN WATER IN SUCH A WAY THAT ITS 3
M SIDE IS PARALLEL TO THE WATER SURFACE AND IS 1 M
BELOW IT.

FIND: (A) TOTAL FORCE ON THE PLATE, AND

(B) POSITION OF CENTER OF PRESSURE.

GIVEN: B= 3 M D = 1M
 DEPTH FROM F.S. TO EDGE = 1M
REQUIRED: F, YP
A = BD = (3M)(1M) = 3M2
LOCATION OF CENTER OF GRAVITY X

1
X = 1M + (1M) = 1.5M
2
TOTAL FORCE F = PCGA

kN
3 )(1.5M)(3M ) = 44.145 KPA
2
F = (9.81
m

CENTER OF PRESSURE YP
(THE CENTER OF PRESSURE IN FIGURE IS H)
YP = H = E + X

Ig
YP = +X
Ax
3 3
IG =
b d = 3 m ( 1 m ) = 0.25M4
12 12
4
0.25 m
YP = H = 2 + 1.5M = 1.556 M
3 m (1.5 m)

14. AN ISOSCELES TRIANGULAR PLATE OF BASE 3M AND

ALTITUDE 3M IS IMMERSED VERTICALLY IN AN OIL OF SPECIFIC
GRAVITY 0.8. THE BASE OF THE PLATE COINCIDES WITH THE
FREE SURFACE OF OIL.
DETERMINE:
(a) TOTAL PRESSURE ON THE PLATE;
(b) CENTRE OF PRESSURE

GIVEN: B= 3M, D= 3M S= 0.8

REQUIRED: F, YP

A = 1/2 BD = 1/2 (3M)(3M) = 4.5M2

LOCATION OF CENTER OF GRAVITY X
X = 1/3 (3M) = 1.0M

kN 2
TOTAL FORCE F = PCGA = 0.8 (9.81 3 )(1.0M)(4.5M )= 35.32
m
KPA.

CENTER OF PRESSURE YP
(THE CENTER OF PRESSURE IN FIGURE IS H)

YP = H = E + X

Ig
YP = +X
Ax
3 3
IG =
b d = 3 m ( 3 m ) = 2.25M4
36 36
4
2.25 m
YP = H = 2 + 1.0M = 1.5 M
4.5 m (1.0 m)

PRESSURE DIAGRAM
ANOTHER APPROACH TO DETERMINE THE TOTAL
HYDROSTATIC THRUST (FORCE) AND ITS LOCATION BY THE
CONCEPT OF PRESSURE DISTRIBUTION OVER THE SURFACE IS
THE PRESSURE DIAGRAM:

PROB: A TANK CONTAINING WATER AND LIQUID (S=0.9) UP TO

HEIGHT 0.25 M AND 0.5 M RESPECTIVELY:
CALCULATE:
A. THE TOTAL PRESSURE ON THE SIDE OF THE TANK; AND
B. THE POSITION OF CENTER OF PRESSURE FROM ONE SIDE
OF THE TANK WHICH IS 1.5 WIDE.

GIVEN: DEPTH OF WATER ℎ1 = 0.25 M DEPTH OF LIQUID ℎ2 =

0.50 M S = 0.9 WIDTH OF TANK B= 1.5 M

REQUIRED: 𝑃𝑇 , ℎP

A.USING THE PRESSURE DIAGRAM SOLVE THE TOTAL

PRESSURE IN THE SIDE OF THE TANK:

𝑃𝑇= 𝑃1 + 𝑃2 + 𝑃3

𝑝𝑇𝑆= 𝛾𝑠 ℎ 1= (0.9)(9.81)(. 5)= 4.41 KPA

𝑝𝑀𝑁= 𝛾𝑠 ℎ 1 + 𝛾𝑤 ℎ 2

𝑝𝑀𝑁= 4.41 + 9.81 (.25) = 6.86 KPA

 𝑃1= ½ (𝛾𝑠ℎ1)(ℎ1B)= ½ 𝛾𝑠ℎ12B

𝑃1= ½ (4.41)(.5X1.5) = 1.65 KN

𝑃2= ℎ2(𝛾𝑠ℎ1)(B)=.25(4.41)(1.5)

𝑃2= 1.65 KN 𝑃3=½ (𝛾𝑤ℎ2)(ℎ2B)= ½ 𝛾𝑤ℎ22B

𝑃3= ½ (9.81)(0.25)2(1.5)= 0.46 KN

𝑃𝑇= 1.65 + 1.65 + 0.46 𝑃𝑇= 3.76 KN

B. FOR THE CENTER OF PRESSURE ℎ𝑝 TAKING MOMENTS OF

ℎ𝑝)= 𝑃1 (2/3 LT) + 𝑃2 (LT +1/2 TM) + 3 (LT + 2/3 TM)

ALL FORCES ABOUT L, USING THE VARIGNON’S THEOREM; P (

3.76 ( ℎ𝑝)= 1.65 (2/3)(0.5) + 1.65( 0.5 + (1/2)(.25) + 0.46 (.5

+ 2/3{.25})

ℎ𝑝= 0.5016 M FROM THE LIQUID SURFACE

HYDROSTATIC FORCES ON CURVED SURFACES

IN THE CASE OF CURVED SURFACE SUBMERGED IN LIQUID AT
REST, IT IS MORE CONVENIENT TO DEAL WITH THE
HORIZONTAL AND VERTICAL COMPONENTS OF THE TOTAL
FORCE ACTING ON THE SURFACE.
NOTE: THE DISCUSSION HERE IS ALSO APPLICABLE TO PLANE
SURFACES.
HORIZONTAL COMPONENT
THE HORIZONTAL COMPONENT OF THE HYDROSTAT FORCE
ON ANY SURFACE IS EQUAL TO THE PRESSURE O THE
VERTICAL PROJECTION OF THAT SURFACE.
FH = PCGA
VERTICAL COMPONENT
THE VERTICAL COMPONENT OF THE TOTAL HYDROSTATIC
FORCE ON ANY SURFACE IS EQUAL TO THE WEIGHT OF EITHER
REAL OR IMAGINARY LIQUID ABOVE IT.
FV= YV
TOTAL HYDROSTATIC FORCE

F= √ F +F v
2
h
2

DIRECTION OF F

Fv
TAN ΘX =
FH

CASE1: LIQUID IS ABOVE THE CURVE SURFACE

THE VERTICAL COMPONENT OF THE
HYDROSTATIC FORCE IS DOWNWARD AND
EQUAL TO THE VOLUME OF THE REAL LIQUID
ABOVE THE SUBMERGED SURFACE

CASE 2: LIQUID IS BELOW THE CURVE

SURFACE
THE VERTICAL COMPONENT OF THE
HYDROSTATIC FORCE IS GOING UPWARD
AND EQUAL TO THE VOLUME OF THE
IMAGINARY LIQUID ABOVE THE SURFACE.

18. THE QUARTER-CIRCULAR ARCHED GATE IS

3 FT WIDE, IS PINNED AT A, AND RESTS ON
THE SMOOTH SUPPORT AT B. DETERMINE THE
REACTIONS AT THESE SUPPORTS DUE TO THE
WATER PRESSURE.A
GIVEN: R = 3 FT OF QUARTER CIRCULAR GATE
REQUIRED: AX, AY