Static Fluid Forces on Structures

03 Forces Due To Static Fluids
Water Resources

Dave Morgan

Fall 2007

Forces Due To Static Fluids WRI 1/48

Readings for “Forces Due To Static Fluids”

All readings for this topic are from Applied Fluid Mechanics by Mott:

Read sections:
4.4 - 4.8
Study Example Problems: 4.2 - 4.7


Forces on Horizontal Plane Areas

Pressure on a horizontal ﬂat plane area due to a static ﬂuid is
uniform over the plane area since the whole plane area is at the
same depth



Pressure on a horizontal flat plane area due to a static fluid is
uniform over the plane area since the whole plane area is at the
same depth

The force on the horizontal flat plane area is given by F = pA, where
p is the uniform pressure and A is the area of the plane area (this
follows from p = F /A, the definition for pressure).


Forces on Horizontal Plane Areas (Example)

Oil (sg=0.93) 0.50 m

Water 1.0 m

1.3 m

Example:
Determine the force exerted by the oil and water upon the bottom plane
area of the barrel



Solution:
Pressure at oil-water boundary:
PO −W = γ · h
Oil (sg=0.93) 0.50 m = (0.93)(9.81 kN/m3 )(0.50 m)
= 4.5617 kPa
Water 1.0 m

1.3 m



Solution:
Pressure at oil-water boundary:
PO −W = γ · h
Oil (sg=0.93) 0.50 m = (0.93)(9.81 kN/m3 )(0.50 m)
= 4.5617 kPa
Water 1.0 m
Pressure at bottom of the barrel:
PBB = PO −W +(9.81 kN/m3 )(1.0 m)
= 4.5617 kPa + 9.81 kPa
1.3 m
= 14.372 kPa



Solution:
Force on the bottom of the barrel:

PBB = 14.372 kPa

Oil (sg=0.93) 0.50 m
F = pA
πd 2
= p·
Water 1.0 m 4
π(1.3 m)2
= (14.372 kN/m2 ) ·
4
= 19.076 kN
1.3 m

F ≈ 19.1 kN



Question:
In the previous example, we took the
pressure at the surface as 0, assuming that
it was gauge pressure.
Oil (sg=0.93) 0.50 m
But, in reality, there is an atmospheric
pressure of around 101.3 kPa at the surface.
Water 1.0 m
Should we add atmospheric pressure to the
gauge pressure at the bottom of the barrel
and increase the force accordingly?
1.3 m If so, what is the force on the bottom of the
barrel? If not, why not?


Forces on Horizontal Plane Areas (Exercise)

133 kPa

A 3.0 m

sg = 1.59 3.37 m

B
Exercise:
A pressurized tank contains liquid with a speciﬁc gravity of 1.59. The
inspection hatch at A has dimensions 400 mm x 250 mm. The access
hatch at B has dimensions 500 mm x 750 mm.
Determine the force exerted by the ﬂuid on the hatch at A and on the
hatch at B



Solution:

PA = 133 kPa
133 kPa
+ (1.59)(9.81 kN/m3 )(3.0 m)
= 133 kPa + 46.794 kPa
3.0 m
= 179.79 kPa
A

sg = 1.59 3.37 m
FA = PA · AA
B = (179.79 kN/m2 )(0.4 m × 0.25 m)
= 17.979 kN

FA ≈ 18.0 kN



Solution:

PB = 133 kPa
133 kPa
+ (1.59)(9.81 kN/m3 )(6.37 m)
= 133 kPa + 99.358 kPa
3.0 m
= 232.36 kPa
A

sg = 1.59 3.37 m
FB = PB · AB
B = (232.36 kN/m2 )(0.5 m × 0.75 m)
= 87.134 kN

FB ≈ 87.1 kN


Forces on Vertical Rectangular Plane Areas

Consider a static volume of water
retained by a vertical wall or dam



0

Water pressure on the wall is 0 at the
surface



0

surface
h
Water pressure is at a maximum at the
bottom of the volume of water and can
be calculated from ∆p = γ · h
γh



0

surface
h
Water pressure is at a maximum at the
bottom of the volume of water and can
be calculated from ∆p = γ · h
γh
Water pressure is proportional to the
depth (since ∆p = γ · h, it follows that
∆p ∝ h)



The average pressure is at half-depth
0
h
Pavg = γ ·
2

h pavg=γh/2

γh



The average pressure is at half-depth
0
h
Pavg = γ ·
2

The magnitude of the resultant force
h pavg=γh/2 exerted on the wall by the water is

FR = Pavg · A
γh h
= γ · ·A
2
where A is the area of the rectangular
plane area


Forces on Inclined Rectangular Plane Areas

0

pavg=γh/2
h
γh

The same argument used for vertical rectangular plane areas applies to
inclined regular plane areas and, again,
h h
pavg = γ · , FR = γ · ·A
2 2
The direction of the pressure remains perpendicular to the plane area


Forces on Rectangular Plane Areas (Exercise)

Water

5.25 m

60O

Exercise:
The wall has a rectangular plane area in contact with the water, is
inclined at 60◦ to the horizontal and is 17 m long.
Determine the force exerted on the dam plane area by the water.


Solution:
The average pressure is at half-depth:
h
Pavg = γ ·
2
= (9.81 kN/m3 ) · (2.625 m)
= 25.751 kN/m2

Water

5.25 m

60O


Solution:
h
Pavg = γ ·
2
= (9.81 kN/m3 ) · (2.625 m)
= 25.751 kN/m2

Water The area A of the dam wall:
5.25
5.25 m A = m · (17 m)
sin 60◦
= 103.06 m2
60O


Solution:
h
Pavg = γ ·
2
= (9.81 kN/m3 ) · (2.625 m)
= 25.751 kN/m2

Water The area A of the dam wall:
5.25
5.25 m A = m · (17 m)
sin 60◦
= 103.06 m2
60O The resultant force on the wall is:
FR = pavg · A
= (25.751 kN/m2 ) · (103.06 m2 )
= 2653.8 kN

FR ≈ 2650 kN


Forces on Submerged Rectangles

Find the force on a submerged
rectangular gate



rectangular gate
0 The pressure triangle is as before,
except the pressure isn’t 0 at the
top of the rectangle so our previous
reasoning no longer holds (and
neither does FR = γ · h · A)
2



rectangular gate

The pressure triangle is as before,
2

We are only interested in the
pressure that the ﬂuid exerts on the
gate



rectangular gate

The pressure triangle is as before,
pavg
2

We are only interested in the
pressure that the ﬂuid exerts on the
gate

The average pressure on the gate is
at the gate’s mid-height, which can
be easily calculated if the gate
dimensions and location are known


Forces on Submerged Rectangle (Exercise)

Exercise:
A vertical retaining wall supports water to a depth of 4.75 m. There are
two rectangular hatches in the wall.

1 The top of the ﬁrst hatch is at a depth of 1.25 m; the hatch is 2.25
m wide × 1.5 m high. What is the magnitude of the force exerted
upon the hatch by the water?
2 A second hatch has dimensions 3.75 m wide × 1.6 m high. At what
depth can the top of this hatch be placed below the surface if the
maximum allowable force for the hatch is 128 kN?



Solution (Part 1):
Exercise sketch....

1.25 m

1.5 m



Solution (Part 1):
Exercise sketch....

The average pressure on the hatch is at
mid-height of the hatch
1.25 m 2.0 m
(h = 1.25 m + 0.75 m = 2.0 m)
pavg
1.5 m



Solution (Part 1):
Exercise sketch....

2.0 m
(h = 1.25 m + 0.75 m = 2.0 m)
pavg
pavg = γ · h
= (9.81 kN/m3 )(2.0 m)
= 19.62 kN/m2



Solution (Part 1):
Exercise sketch....

(h = 1.25 m + 0.75 m = 2.0 m)

F pavg = γ · h
= (9.81 kN/m3 )(2.0 m)
= 19.62 kN/m2

The force on the hatch is:
F = pavg · A
= (19.62 kN/m2 )(3.375 m2 )
= 66.218 kN


Solution (Part 2):
Let d be the depth of the top of the
hatch

d

1.6 m


Solution (Part 2):
hatch

Average pressure on the hatch is at
d d+0.8 m h = (d +0.8 m)

pavg
1.6 m


Solution (Part 2):
hatch

Average pressure on the hatch is at
d h = (d +0.8 m)

The force on the hatch is:
F = pavg · A
128 = (9.81)(d + 0.8)(1.6 × 3.75)
128
d +0.8 =
(9.81)(1.6 × 3.75)
= 2.1747
d = 1.3747

d ≈ 1.37 m


Centre of Pressure
The liquid exerts a force all over
the plane area; the force increases
with depth.

h


Centre of Pressure
with depth.

The resultant of the force, FR , acts
through the centroid of the pressure
triangle at a depth of 2h/3.
(Like the resultant of uniformly
2h/3 varying loads in statics....)

FR
h/3


Centre of Pressure
with depth.


FR For most problems we will consider,
the total force can be thought of as
h/3 acting at this point, called the
centre of pressure


Centre of Pressure
with depth.

pavg=γh/2
FR For most problems we will consider,
the total force can be thought of as
h/3 acting at this point, called the
centre of pressure

Use pavg to ﬁnd the average
pressure and magnitude of the
resultant force, FR = pavg A, and
the centre of pressure to ﬁnd the
position and direction of FR

Centre of Pressure

pavg=γh/2
2h/3 FR

h/3

The result is identical for an inclined rectangular plane area


Flumes

A flume is a man-made
channel used to transport
liquid, often in an elevated
wooden box-like structure.
This one was built in 1909
near the Sandy River,
Portland, Oregon

(http://www.portlandgeneral.com/community_and_env/hydropower_and_fish/phototour/sandy/little_sandy_flume.asp,
accessed 11th September, 2007)


Flumes

This ﬂume is 35 miles long.
It was completed in 1888 to supply water to San Diego.
(http://www.sandiegohistory.org/timeline/timeline2.htm,

accessed 11th September, 2007)


Flumes

(http://www.co.yamhill.or.us/pics/Sheridan/, accessed 11th September, 2007)


Flumes

Hanging Flume, Montrose County, Colorado, US
This flume is on the World Monuments Foundation’s list of the 100 Most
Endangered Sites 2006. Built during the Gold Rush, this 21 kilometre
flume transported more than 30 million litres of water a day for use in
hydraulic gold mining.
(http://wmf.org/resources/sitepages/united_states_hanging_flume.html, accessed 11th September, 2007)


Centre of Pressure (Example)

1.6 m

0.35 m
Tie
rods

Posts, spaced 2.05 m
1.75 m on centre

Example:
Water ﬂows slowly through a ﬂume, assumed to have a pinned
connection at the bottom of the sidewalls and held in place against water
pressure by tie-rods through posts every 1.75 m.
Determine the tension in the tie rods.



Solution:
h
pavg = γ ·
2
1.6 m
= (9.81 kN/m3 )(1.025 m)
0.35 m = 10.055 kPa

2.05 m



Solution:
h
pavg = γ ·
2
1.6 m
= (9.81 kN/m3 )(1.025 m)
0.35 m = 10.055 kPa

FR = pavg · A
2.05 m = (10.055 kN/m2 )
×(2.05 m × 1.75 m)
= 36.072 kN



Solution:
h
pavg = γ ·
2
1.6 m
= (9.81 kN/m3 )(1.025 m)
0.35 m = 10.055 kPa

FR = pavg · A
2.05 m = (10.055 kN/m2 )
×(2.05 m × 1.75 m)
= 36.072 kN

36.072 kN is the force exerted on
the sidewall for each 1.75 m span of
the ﬂume; this is the force restrained
by the tie-rods.



T T
0.35 m

2.05 m

FR
36.072 kN
0.68333 m

A

Solution (cont’d):
Draw an FBD, showing the forces acting on one side of the ﬂume.
Take moments about A



T
0.35 m ΣMA = T · (2.05 m + 0.35 m)
−(36.072 kN × 0.68333 m)
= 0
2.05 m
(36.072 kN × 0.68333 m)
T =
36.072 kN (2.05 m + 0.35 m)
0.68333 m
= 10.270 kN
A



T
0.35 m ΣMA = T · (2.05 m + 0.35 m)
−(36.072 kN × 0.68333 m)
= 0
2.05 m
(36.072 kN × 0.68333 m)
T =
36.072 kN (2.05 m + 0.35 m)
0.68333 m
= 10.270 kN
A
The tension in the tie-rods is 10.3 kN.


Centre of Pressure, Submerged Rectangular Plane Area

Where is the centre of pressure for a
submerged rectangular plane area?




The pressure area is a trapezoid, not a
triangle, so the centroid is not at
one-third height




one-third height

In statics, this problem was handled by
separating the trapezoid into a triangle
and a rectangle




one-third height

and a rectangle

The centres of pressure for these
simple shapes are known




one-third height

and a rectangle

The centres of pressure for these
simple shapes are known

The centres of pressure and the
magnitudes of the forces can be
combined to solve a variety of problems


Centre of Pressure, Submerged Plane Area

With the method described, the
resultant force can be calculated to
ﬁnd the tension in the bolts holding
the inspection hatch onto the tank



There is a formula that locates the
Lp centre of pressure, Lp more readily...

FR



Lp Lc centre of pressure, Lp more readily...
Find Lc , the depth of the centroid of
the plane area

FR



the plane area
Find A, the area of the surface, and IC ,
FR the moment of inertia of the plane area
about its horizontal centroidal axis



the plane area
Find A, the area of the surface, and IC ,
FR the moment of inertia of the plane area
about its horizontal centroidal axis
Then,
Ic
Lp = Lc +
Lc · A


Submerged Plane Area (Example)
Example:
A tank containing castor oil has a
1.0 m wide × 600 mm high inspection
hatch.
Castor Oil The top of the hatch is at 2.5 m below the
sg = 0.96 surface of the castor oil. The hatch cover is
attached to the tank by eight bolts, four at
2.5 m the top of the hatch and four at the bottom
of the hatch.
The bolts are oﬀset from the hatch opening
by 100 mm, as shown.
100 mm 600 mm Calculate the tension in each of the top and
in each of the bottom bolts.
(Assume that all of the top bolts have the
same tension and that all of the bottom
bolts have the same tension.)



Solution:
Calculate the area of the hatch:

Castor Oil A = 1.0 m × 0.6 m = 0.6 m2
sg = 0.96

2.5 m

100 mm 600 mm



Solution:

Castor Oil A = 1.0 m × 0.6 m = 0.6 m2
sg = 0.96
Calculate the moment of inertia of the
2.5 m hatch:
bh3 (1.0 m)(0.6 m)3
Ic = = = 0.018 m4
12 12

100 mm 600 mm



Solution:

Castor Oil A = 1.0 m × 0.6 m = 0.6 m2
sg = 0.96
Calculate the moment of inertia of the
2.8 m hatch:
bh3 (1.0 m)(0.6 m)3
Ic = = = 0.018 m4
12 12

100 mm Lc Find Lc , the location (depth) of the
centroid of the hatch area:
0.6 m
Lc = 2.5 m + = 2.8 m
2



Castor Oil Solution:
sg = 0.96 A = 0.6 m2 , Ic = 0.018 m4 , Lc = 2.8 m

2.8 m

100 mm Lc



sg = 0.96 A = 0.6 m2 , Ic = 0.018 m4 , Lc = 2.8 m

2.8 m Ic
Lp = Lc +
Lc · A
0.018 m4
= 2.8 m +
2.8 m × 0.6 m2
Lc = 2.8107 m
100 mm



sg = 0.96 A = 0.6 m2 , Ic = 0.018 m4 , Lc = 2.8 m

2.8 m Ic
Lp = Lc +
2.8107 m Lc · A
0.018 m4
= 2.8 m +
2.8 m × 0.6 m2
Lc = 2.8107 m
100 mm Lp



Solution:
Find the average pressure on the hatch:
Castor Oil
sg = 0.96 pavg = γ · Lc
= (0.96)(9.81 kN/m3 )(2.8 m)
2.8 m = 26.369 kN/m2

pavg



Solution:
Find the average pressure on the hatch:
Castor Oil
sg = 0.96 pavg = γ · Lc
= (0.96)(9.81 kN/m3 )(2.8 m)
2.8 m = 26.369 kN/m2
2.8107 m
Find the magnitude of the resultant force:

FR = pavg · A
pavg
FR = (26.369 kN/m2 )(1.0 m × 0.6 m)
= 15.822 kN



Solution:

Castor Oil The upper row of bolts is at a depth of
2.4 m and the lower row is at 3.2 m
sg = 0.96

2.4 m

pavg m
0.4
FR
0.4 m



Solution:

sg = 0.96

2.4 m The upper row of bolts is 0.4107 m above
the centre of pressure

0.4107 m pavg m
0.4
FR
0.4 m



Solution:

sg = 0.96

2.4 m The upper row of bolts is 0.4107 m above

The lower row of bolts is 0.3893 m below
0.4107 m pavg m
0.4
FR
0.3893 m 0.4 m



Solution:

sg = 0.96
The upper row of bolts is 0.4107 m above

The lower row of bolts is 0.3893 m below
0.4107 m
FR
0.3893 m We now have the values needed to calculate
the tensions in the bolts



S
A
0.4107 m 15.822 kN
FR
0.3893 m T
B

Solution:
Let S be the sum of the tensions in
the upper bolts and T be the sum of
the tensions in the lower bolts.



S
A
0.4107 m 15.822 kN
FR
0.3893 m T
B

Solution:
Let S be the sum of the tensions in
ΣMB = S × (0.8 m)
−(0.3892 m) × (15.822 kN)
=0
S = 7.6974 kN


S
A
0.4107 m 15.822 kN
FR
0.3893 m T
B

Solution: ΣMA = −T × (0.8 m)
Let S be the sum of the tensions in +(0.4107 m) × (15.822 kN)
= 0
T = 8.1226 kN
ΣMB = S × (0.8 m)
−(0.3892 m) × (15.822 kN)
=0
S = 7.6974 kN


S
A
0.4107 m 15.822 kN
FR
0.3893 m T
B

Solution: ΣMA = −T × (0.8 m)
Let S be the sum of the tensions in +(0.4107 m) × (15.822 kN)
= 0
T = 8.1226 kN
ΣMB = S × (0.8 m) The tension in each upper bolt is
−(0.3892 m) × (15.822 kN) 7.6974/4 ≈ 1.92 kN
=0 The tension in each lower bolt is
S = 7.6974 kN 8.1226/4 ≈ 2.03 kN

Submerged Plane Area (Exercise)

This is an example of a
d “self-levelling” gate. It is hinged
along its top edge.
When the water exceeds a certain
1.0 m
specified height, the hydrostatic
force on the gate is just sufficient to
10 kN 0.9 m open the gate. Water drains until it
0.75 m is again at the specified height and
the gate closes.
Find the value d for which the gate
opens.



d
A

1.0 m

B 10 kN 0.9 m
0.75 m



d
A
Cont’d
1.0 m
pavg = γ ·h = 9.81 · Lc
B 10 kN 0.9 m (kN/m2 )
0.75 m
FR = (9.81Lc )(0.9) = 8.829Lc
(kN)
Solution:
IC 0.075
Lp − Lc = Lc ·A = Lc ·(0.9)
A = 0.9 m × 1.0 m = 0.9 m2
0.083333
(0.9 m)(1.0 m)3 = Lc
Ic = 12 = 0.075 m4 (m)
IC 4
0.075 m
Lp − Lc = Lc ·A = Lc ·(0.9 m2 )

0.083333
= Lc



Cont’d (3)

d ΣMA = FR (0.5 + (Lp − Lc )) − 10(0.75)
A = 8.829Lc (0.5 + 0.08333/LC ) − 7.5
1.0 m = 4.4145Lc + 0.73572 − 7.5
= 0
B 10 kN 0.9 m
0.75 m Lc = 1.5322 m

Cont’d (2) The centroid of the gate is 1.5322 m from
the surface when the gate opens.
When the gate opens, the reaction Therefore
at B is 0 so the only two moments
to consider are the moment due to d = 1.5322 m − 0.5 m
FR and the moment due to the d ≈ 1.03 m
10 kN weight


Inclined vs. Vertical Plane Areas

With a vertical submerged plane
area, we have
Lp Lc Ic
Lp = Lc +
Lc · A
What is the situation when the
FR plane area under investigation is not
vertical?



Lp Lc
Ic
Lp = Lc +
FR Lc · A

The formula still applies if Lc and Lp are
taken parallel to the plane area under
investigation. (A vertical plane area is just
a special case of the general formula.)



Lp Lc
Ic
Lp = Lc +
FR Lc · A

The formula still applies if Lc and Lp are
taken parallel to the plane area under
investigation. (A vertical plane area is just
a special case of the general formula.)

Lc is the distance along the slope of the plane area from the
centroid of the plane area to the surface of the liquid

Lp is the distance along the slope of the plane area from the
centre of pressure to the surface of the liquid

(The formula doesn’t hold for horizontal plane areas. Why not?)

Submerged Inclined Plane Area (Example)

sg = 0.823

2.05 m

50O
m
1.0

1.0 m
1.0
m

Example:
Find the magnitude, direction and centre of pressure (point of
application) of the resultant force on the triangular hatch in the
V-shaped vessel illustrated.


Solution:
2.05 m The length AB is the “height” of the hatch:
AB
= sin 60◦
1.0
AB = 0.86603 m
B
m
1.0
A
1.0 m

1.0
m


Solution:
AB
= sin 60◦
1.0
AB = 0.86603 m
B
The area of the hatch:
m bh (1.0 m)(0.86603 m)
1.0 A= = = 0.43302 m2
A 2 2
1.0 m

1.0
m


Solution:
AB
= sin 60◦
1.0
AB = 0.86603 m
B
m bh (1.0 m)(0.86603 m)
1.0 A= = = 0.43302 m2
A 2 2
1.0 m

1.0 The moment of inertia of the hatch:
m bh3 (1.0 m)(0.86603 m)3
Ic = = = 0.18042 m4
36 36


Solution:
AB
= sin 60◦
1.0
AB = 0.86603 m
B

C 1.0 m A=
bh
=
(1.0 m)(0.86603 m)
= 0.43302 m2
A 2 2
1.0 m

1.0 The moment of inertia of the hatch:
m bh3 (1.0 m)(0.86603 m)3
Ic = = = 0.18042 m4
36 36
The centroidal axis of the triangle is at one-third
height from the base (two-thirds down from the
top):
2
BC = × AB= 0.57735 m
3



Cont’d:
Lc , the distance from the surface to the
D centroidal axis of the hatch along the slope of
the hatch:

Lc Lc = DC
2.05 m = DB + BC
50 O 2.05
= m + 0.57735 m
sin 50◦
= 3.2534 m
B
C 1.0 m
A
1.0 m

1.0
m


Frame Submerged Inclined Surface (Example)

Cont’d:
Lc , the distance from the surface to the
D centroidal axis of the hatch along the slope of
the hatch:

Lc Lc = DC
2.4922 m = DB + BC
50 O 2.05
= m + 0.57735 m
sin 50◦
= 3.2534 m
B L_{c}, the distance from the surface to the centroidal axis
The depth at the centroid of the triangular
C 1.0 m hatch: _ _ _ _ _ _ _ _ _ _ _
___
A
1.0 m

◦
1.0 hc has c · average
The depth of L_{c} = Lthecos 40 pressure for the hatch
m
= 2.4922 m
______________

p_{avg}



Cont’d:
D The average pressure on the hatch is the
pressure at the centroid of the hatch:

pavg = γ · hc
2.4922 m
= (0.823)(9.81 kN/m3 )(2.4922 m)
50O = 20.121 kPa

L_{c}, the distance from the surface to the centroidal axis
C
______________

The depth of L_{c} has the average pressure for the hatch

______________

p_{avg}



Cont’d:
D The average pressure on the hatch is the
pressure at the centroid of the hatch:

pavg = γ · hc
= (0.823)(9.81 kN/m3 )(2.4922 m)
50O = 20.121 kPa

FR The resultant force on the hatch is:

C FR = pavg · A
= (20.121 kN/m2 )(0.43302 m2 )
= 8.7128 kN



Cont’d:

D The resultant force, FR , acts at the centre of
pressure which is at a distance Lp , along the
slope of the hatch, from the surface where
Lp IC
Lp = Lc +
Lc · A
50O 0.18042 m4
= 3.2534 m +
(3.2534 m)(0.43301 m2 )
FR
= 3.3815 m
C



Cont’d:
The resultant force, FR , acts at the centre of
Lp IC
2.5904 m Lp = Lc +
Lc · A
50O 0.18042 m4
= 3.2534 m +
(3.2534 m)(0.43301 m2 )
FR
= 3.3815 m

The depth of the centre of pressure is

Lph = 3.3815 cos 40◦ = 2.5904 m



Cont’d:
The resultant force, FR , acts at the centre of
Lp IC
2.5904 m Lp = Lc +
Lc · A
50O 0.18042 m4
= 3.2534 m +
(3.2534 m)(0.43301 m2 )
FR
= 3.3815 m

The depth of the centre of pressure is

Lph = 3.3815 cos 40◦ = 2.5904 m
FR has direction 320◦ (go ﬁgure... ;-)



2.59 m

FR

Solution:

FR is 8.712 kN at 320◦ , acting on the hatch at a depth of 2.59 m


Static Fluid Forces on Structures

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