Lesson 3: Hydrostatic Force
Total Hydrostatic Force or Total Pressure:
F=(Uniform Pressure )(Area) = pA
Source: Fluid Mechanics Demystified
From the figure, the Total Hydrostatic Force acting on the inclined plane:
𝑭 = 𝜸𝒉𝑨 = 𝒑𝒄𝒈 . 𝑨
Location of the Total Hydrostatic Force:
𝒚𝒑 =
𝑰𝒔
𝑨𝒚 = 𝒚 +
𝑰𝒈
𝑨𝒚 = 𝒚 + 𝒆
Value of eccentricity, e (distance between the center of pressure (cp) and the
center of gravity (C or cg):
𝒆=
𝑰𝒈
𝑨𝒚
Geometric Properties of Common Shapes.
Source: Fundamentals of Fluid Mechanics by Munson, Young, et al.
Sample Problem No. 1:
A 60-cm square gate has its top edge 12 m
below the water surface. It is on a 45° angle
and its bottom edge is hinged as shown in
(a). What force P is needed to just open the
gate?
Solution:
Draw the FBD
Compute the Total Hydrostatic Force:
F = γhA
F = 𝟗𝟖𝟏𝟎 (𝟏𝟐 + 𝟎. 𝟑 𝒔𝒊𝒏𝟒𝟓 )(𝟎. 𝟔)( 𝟎. 𝟔) = 43,130 N or 43.13KN
Compute for the location, d of F from the bottom of the submerged object:
y=
ℎ
12 + 0.3𝑠𝑖𝑛45°
=
= 17.27𝑚
sin 45°
𝑠𝑖𝑛45°
𝑦 =y+
I
Ay = 17.27 +
(0.6x0.6 )/12
0.6(0.6)(17.27) = 17.272𝑚
d = y + 0.3 − 𝑦𝑝 = 0.3𝑚
Take moment at the hinge to compute for the value of P:
P(0.6)=0.3(43.13)
P=21.6 KN
Total Hydrostatic Force on Curved Surfaces
Case 1: Liquid is above the curve surface
Case 1: Liquid is below the curve surface
Source: www.Mathalino.com
For both cases, the value of the Total Hydrostatic Force is the resultant of the horizontal
and vertical components acting on curved surfaces.
𝐹=
𝐹
+𝐹
𝐹 = 𝛾ℎ𝐴,
𝐹 = 𝛾𝑉
tan 𝜃 =
Note that the area is the vertical projection of the submerged curve.
Sample Problem No. 2:
The 6 ft. diameter cylinder weighs 5000 lb.
and is 5 ft. long.
 Determine the upward force due to
the effect of oil in the left side.
 Compute the horizontal reaction at A.
 Compute the vertical reaction at B.
𝐹𝑉
𝐹𝐻
Solution:
Draw first the FBD and the projected area and volume.

Upward force due to the effect of oil in the left side:
P = γV
π(3)
P = 62.4(0.8)(
)(5)
2
𝐏𝐕 = 𝟑𝟓𝟐𝟗 𝐥𝐛 (𝐮𝐩𝐰𝐚𝐫𝐝)

Horizontal reaction at A:
P = R = γhA
R = (62.4)(0.80)(3)(6)(5)
𝐑 𝑨 = 𝟒𝟒𝟗𝟑 𝐥𝐛 (𝐭𝐨 𝐭𝐡𝐞 𝐥𝐞𝐟𝐭)

Vertical reaction at A:
ΣFy=0
R + P = 5000
R = 5000 − 3529
𝐑 𝑩 = 𝟏𝟒𝟕𝟏 𝒍𝒃 (upward)
Dams
Dams are structures whose purpose is to raise the water level on the upstream side of river,
stream, or other waterway. The rising water will cause hydrostatic force which will tend the
dam to slide horizontally and overturn about its downstream edge or toe. The raised water level
on the upstream edge or heel will also cause the water to seep under the dam. The pressure due
to this seepage is commonly called hydrostatic uplift and will reduce the stability of the dam
against sliding and against overturning.
Types of Dams:
GRAVITY DAMS
ARCH DAMS
EMBANKMENT DAMS
BUTTRESS DAMS
Analysis of Gravity Dams
Factor of Safety Against Sliding:
𝐹𝑆 =
𝜇𝑅
>1
𝑅
Factor of Safety Against Overturning:
𝐹𝑆 =
𝑅𝑀
>1
𝑂𝑀
Foundation Pressure
If 𝑒 ≤
𝑞=−
If 𝑒 >
𝑞 =
2𝑅
𝑥̅
1±
Sample Problem:
A dam triangular in shape has a height of 24m. and a base of 12m. Density of masonry is 2,500
kg/cu.m. If it is supporting was at a depth of 20m. Where is the location of the resultant vertical
force from the heel of the dam.
Solution:
Compute for the weight of the dam:
𝑊 = 𝛾𝑉
1
𝑊 = (24)(12)(1)(2500)
2
𝑊 =360,000 kg
R = 360,000 kg
Compute for the Total Hydrostatic Force P:
𝑃 = 𝛾ℎ𝐴
𝑃 = (1000)(10)(20)(1)
𝑃 = 200,000 𝑘𝑔
Compute for the Overturning Moment (OM)
OM= P (20/3)
OM= 200,000 (20/3)
OM= 1,333,333.33 kg-m
Compute for the Righting Moment (RM)
RM= W (8)
RM= 360,000 (8)
RM= 2,880,000 kg-m
Considering Moment at the heel:
R(12-x)= RM-OM
360,000 (12-x) = 2,880,00-1,333,333.33
(12-x) = 7.7m