1. 9.6 Fluid Pressure
According to Pascal’s law, a fluid at rest creates a
pressure ρ at a point that is the same in all
directions
Magnitude of ρ measured as a force per unit
area, depends on the specific weight γ or mass
density ρ of the fluid and the depth z of the point
from the fluid surface
ρ = γz = ρgz
Valid for incompressible fluids
Gas are compressible fluids and thus the above
equation cannot be used

2. 9.6 Fluid Pressure
Consider the submerged plate
3 points have been specified

3. 9.6 Fluid Pressure
Since point B is at depth z1 from the liquid
surface, the pressure at this point has a
magnitude of ρ1= γz1
Likewise, points C and D are both at depth z2
and hence ρ2 = γz2
In all cases, pressure acts normal to the
surface area dA located at specified point
Possible to determine the resultant force
caused by a fluid distribution and specify its
location on the surface of a submerged plate

4. 9.6 Fluid Pressure
Flat Plate of Constant Width
Consider flat rectangular plate of constant
width submerged in a liquid having a specific
weight γ
Plane of the plate makes an angle with the
horizontal as shown

5. 9.6 Fluid Pressure
Flat Plate of Constant Width
Since pressure varies linearly with depth,
the distribution of pressure over the plate’s
surface is represented by a trapezoidal
volume having an
intensity of ρ1= γz1
at depth z1 and
ρ2 = γz2 at depth z2

6. 9.6 Fluid Pressure
Magnitude of the resultant force FR = volume of
this loading diagram and FR has a line of action
that passes through the volume’s centroid, C
FR does not act at the centroid of the plate but at
point P called the center of
pressure
Since plate has a constant
width, the loading diagram
can be viewed in 2D

7. 9.6 Fluid Pressure
Flat Plate of Constant Width
Loading intensity is measured as force/length
and varies linearly from
w1 = bρ1= bγz1 to w 2 = bρ2 = bγz2
Magnitude of FR = trapezoidal area
FR has a line of action that passes through
the area’s centroid C

8. 9.6 Fluid Pressure
Curved Plate of Constant Width
When the submerged plate is curved, the pressure acting
normal to the plate continuously changes direction
For 2D and 3D view of the loading distribution,
Integration can be used to determine FR and location of
center of centroid C or pressure P

9. 9.6 Fluid Pressure
Curved Plate of Constant Width
Example
Consider distributed loading acting on the
curved plate DB

10. 9.6 Fluid Pressure
Curved Plate of Constant Width
Example
For equivalent loading

11. 9.6 Fluid Pressure
Curved Plate of Constant Width
The plate supports the weight of the liquid Wf
contained within the block BDA
This force has a magnitude of
Wf = (γb)(areaBDA)
and acts through the centroid of BDA
Pressure distributions caused by the liquid acting
along the vertical and horizontal sides of the
block
Along vertical side AD, force FAD’s magnitude =
area under trapezoid and acts through centroid
CAD of this area

12. 9.6 Fluid Pressure
Curved Plate of Constant Width
The distributed loading along horizontal side AB
is constant since all points lying on this plane are
at the same depth from the surface of the liquid
Magnitude of FAB is simply the area of the
rectangle
This force acts through the area centroid CAB or
the midpoint of AB
Summing three forces,
FR = ∑F = FAB + FAD + Wf

13. 9.6 Fluid Pressure
Curved Plate of Constant Width
Location of the center of pressure on the plate is
determined by applying
MRo = ∑MO
which states that the moment of the resultant
force about a convenient reference point O, such
as D or B = sum of the moments of the 3 forces
about the same point

14. 9.6 Fluid Pressure
Flat Plate of Variable Width
Consider the pressure distribution acting on
the surface of a submerged plate having a
variable width

15. 9.6 Fluid Pressure
Flat Plate of Variable Width
Resultant force of this loading = volume
described by the plate area as its base and
linearly varying pressure distribution as its
altitude
The shaded element may be used if integration
is chosen to determine the volume
Element consists of a rectangular strip of area dA
= x dy’ located at depth z below the liquid
surface
Since uniform pressure ρ = γz (force/area) acts
on dA, the magnitude of the differential force dF
dF = dV = ρ dA = γz(xdy’)

16. 9.6 Fluid Pressure
Flat Plate of Variable Width
FR = ∫ ρdA = ∫ dV = V
A V
Centroid V defines the point which FR acts
The center of pressure which lies on the surface
of the plate just below C has the coordinates P
defined by the equations
x=
∫ ~dV y ' = ∫ ~' dV
V
x
V
y
∫ dV
V ∫ dVV
This point should not be mistaken for centroid of
the plate’s area

17. 9.6 Fluid Pressure
Example 9.13
Determine the magnitude and location of the
resultant hydrostatic force acting on the submerged
rectangular plate AB. The
plate has a width of 1.5m;
ρw = 1000kg/m3.

18. 9.6 Fluid Pressure
Solution
The water pressures at depth A and B are
ρ A = ρ w gz A = (1000kg / m 3 )(9.81m / s 2 )(2m) = 19.62kPa
ρ B = ρ w gz B = (1000kg / m 3 )(9.81m / s 2 )(5m) = 49.05kPa
Since the plate has constant
width, distributed loading
can be viewed in 2D
For intensities of the load at
A and B,
wA = bρ A = (1.5m)(19.62kPa) = 29.43kN / m
wB = bρ B = (1.5m)(49.05kPa) = 73.58kN / m

19. 9.6 Fluid Pressure
Solution
For magnitude of the resultant force FR created by the
distributed load
FR = area of trapezoid
1
= (3)(29.4 + 73.6) = 154.5 N
2
This force acts through the
centroid of the area
1  2(29.43) + 73.58 
h=  (3) = 1.29m
3  29.43 + 73.58 
measured upwards from B

20. 9.6 Fluid Pressure
Solution
Same results can be obtained by considering
two components of FR defined by the triangle
and rectangle
Each force acts through its associated
centroid and has a magnitude of
FRe = (29.43kN / m)(3m) = 88.3kN
Ft = (44.15kN / m)(3m) = 66.2kN
Hence
FR = FRe + FR = 88.3kN + 66.2kN = 154.5kN

21. 9.6 Fluid Pressure
Solution
Location of FR is determined by summing
moments about B
∑(M R )B = ∑ M B ;
(154.5)h = 88.3(1.5) + 66.2(1)
h = 1.29m

22. 9.6 Fluid Pressure
Example 9.14
Determine the magnitude of the resultant
hydrostatic force acting on the surface of a seawall
shaped in the form of a parabola. The wall is 5m
long and ρw = 1020kg/m2.

23. 9.6 Fluid Pressure
Solution
The horizontal and vertical components of the
resultant force will be calculated since
ρ B = ρ w gz B = (1020kg / m 2 )(9.81m / s 2 )(3m) = 30.02kPa
Then
wB = bρ B = 5m(30.02kPa) = 150.1kN / m
Thus
1
Fx = (3m)(150.1kN / m) = 225.1kN
3

24. 9.6 Fluid Pressure
Solution
Area of the parabolic sector ABC can be
determined
For weight of the wafer within this region
Fy = ( ρ w gb)(area ABC )
1
= (1020kg / m )(9.81m / s )(5m)[ (1m)(3m)] = 50.0kN
2 2
3
For resultant force
FR = Fx2 + Fy2 = (225.1) 2 + (50.0) 2
= 231kN

25. 9.6 Fluid Pressure
Example 9.15
Determine the magnitude and location of
the resultant force acting on the triangular
end plates of the wafer of the water trough.
ρw = 1000 kg/m3

26. 9.6 Fluid Pressure
Solution
Magnitude of the resultant force F = volume of
the loading distribution
Choosing the differential volume element,
dF = dV = ρdA = ρ w gz (2 xdz ) = 19620 zxdz
For equation of line AB
x = 0.5(1 − z )
Integrating
1
F = V = ∫ dV = ∫ (19620) z[0.5(1 − z )]dz
V 0
1
= 9810 ∫ ( z − z 2 )dz = 1635 N = 1.64kN
0

27. 9.6 Fluid Pressure
Solution
Resultant passes through the centroid of the
volume
Because of symmetry
x =0
For volume element
1
z=
∫z
~dV
V =
∫0 z (19620) z[0.5(1 − z )]dz
∫VdV 1635
1
9810∫ ( z 2 − z 3 ) dz
= 0 = 0.5m
1635