A Submerged Isosceles Triangle: Center of Pressure and Area Moment of Inertia

Linda Fahlberg-Stojanovska 1
Center of Pressure - Integral vs. Formula

for a Submerged Gate as Isosceles Triangle with Base Parallel to Liquid Surface

We are checking that the formula works and that it works for both an inverted triangle (base at top) and
a triangle with base at the bottom. The proofs are here; videos on youtube.
There are two possibilities for the triangle: a. Base is at top (inverted) or b. Base is at bottom.

Formula: The formula for
p
h as the (depth of the) center of pressure for areas symmetric with respect
to x-axis is:
c
p c
c
I
h h
h A
= +
where
3
36
c
W L
I

=
is the area moment of inertia for a triangle with
respect to the (depth of the) centroid
c
h .
Using formulas:
3
1 1
36
p c c c
c c
W L
h h I h
h A h A

= + = +

On the onther hand, the integral
1
for the (depth of the) center of pressure
p
h is:

1
p
r
A
h D pdA
F
= + q
}
, where is depth from top of submerged surface.
Recall that:
( )
c
Fr g h A =
for areas symmetric with respect to x-axis so:
0
1 1
( ) ( )
L
p
r c
A
h D g D dA D D w d
F h A
q
= + q + q = + q + q q
} }

The centroid
c
h and the width w
q
are DIFFERENT for the two possible isosceles triangles.

1
Fox, R.W. (2005) Introduction to Fluid Mechanics

Surface of liquid
D
L
gate
gate b
The triangles have different
centroids h
c
, but the area moment
for the inertia I
C
is the same. So using
the appropriate centroid, one can
find the center of pressure h
p
.
W

gate a
W

a. Base is at top
The (depth of the) centroid of an INVERTED triangle is:
1
3
c
h D L = +

Using integral:
0
1 1
( )
L
p
r c
A
h D pdA D D w d
F h A
q
= + q = + q + q q
} }

The width of the triangle w
q
at depth measured from top of triangle is:
( )
W
w L
L
q
= q
.
0
0
2 2 3
0
2 3 3 4
2 3
1
( ) ( )
1
( ) ( )
1
( )
1
2 3 3 4
1
6 12
L
p
c
L
c
L
c
c
c
W
h D D L d
h A L
W
D D L d
h A L
W
D DL D L d
h A L
W L L L L
D DL D L
h A L
DL L
D W
h A
| |
= + q + q q q
|
\ .
= + q + q q q
= + q q + q q q
| |
= + + |
|
\ .
| |
= + + |
|
\ .
}
}
}

Conjecture For
1
3
c
h D L = + , is:
3 2 3
1 1
36 6 12
c
c c
W L DL L
h D W
h A h A
| |
+ + + |
|

\ .
= .
Proof: Triangle means:
2
W L
A

= so
(3 )
( )
3 2 6
c
L WL D L WL
h A D
+
= + =
Left =
, )
3 2
6 2 6 (2 )
3 (3 ) 36 3 6(3 ) 6 3 2(3 )
D L L L W L L L L L D L
D D D D
D L WL D L D L D L
+ + + | |
+ + = + + = + = +
|
+ + + +
\ .

Right =
2 3
6 (2 )
(3 ) 6 12 3 2 2(3 )
DL L L L L D L
D W D D D
D L WL D L D L
| |
+ | | | |
+ + = + + = + |
| |
|
+ + +
\ . \ .
\ .

b. Base is at bottom
The (depth of the) centroid of a triangle with base at bottom is:
2
3
c
h D L = +

Using integral:
0
1 1
( )
L
p
r c
dA
h D pdA D D w d
F h A
q
= + q = + q + q q
} }


The width of the triangle w
q
at depth measured from top of triangle is:
W
w
L
q
= q
.
0
0
2 3
0
3 4
2 3
1
( )
1
( )
1
( )
1
3 4
1
3 4
L
p
c
L
c
L
c
c
c
W
h D D d
h A L
W
D D d
h A L
W
D D d
h A L
W L L
D D
h A L
DL L
D W
h A
| |
= + q + q q q
|
\ .
= + q + q q q
= + q +q q
| |
= + + |
|
\ .
| |
= + + |
|
\ .
}
}
}

Conjecture: For
2
3
c
h D L = + , is:
3 2 3
1 1
36 3 4
c
c c
W L DL L
h D W
h A h A
| |
+ + + |
|

\ .
= .
Proof: Triangle means:
2
W L
A

= so
(3 )
3
2
(
2 6
2
)
c
L WL D L WL
h A D
+
= + =
Left =
, )
3 2
12 8 6 (4 3 )
3 (3 2 ) 36 3 6(3 2 ) 6 3 2 2(3 2 )
2 D L L L W L L L L L D L
D D D D
D L WL D L D L D L
+ + + | |
+ + = + + = + = +
|
+ + + +
\ .

Right =
2 3
6 6 (4 3 )
(3 2 ) 3 4 3 2 3 4 2(3 2 )
DL L L D L L D L
D W D D
D L WL D L D L
| |
+ | | | |
+ + = + + = + |
| |
|
+ + +
\ . \ .
\ .

Application: A gate is in the shape of an inverted isosceles triangle. The top is 6m under water. The
height of the gate is 3m and the width is 2m. Find the resultant force on the gate and the (depth of the)
center of pressure.

Solution using formulas:
( )
c
Fr g h A =
,
1
3
c
h D L = +
,
c
p c
c
I
h h
h A
= +
,
3
36
c
W L
I

=

Surface of water
6m
3m
Side view
of gate
2m
D=6m, L=3m, W=2m, Liquid is H
2
O
Atmospheric pressure above water
and behind gate means we can use
gage pressure.
The gate is symmetric with respect to
the x-axis.
gate

1
6 3 7
3
1
3
c
h D L = + = + = ,
2 3
3
2 2
WL
A

= = =
2
3 2
( ) 1000 9.81 7 3 206
c
kg m
Fr g h A m m kN
m s
= = = ,
3 3
2 3
1.5
36 36
c
W L
I

= = =
1.5
7 7.07
7 3
c
p c
c
I
h h
h A
= + = + =

, 7.07
p
h m =
Answer: 206
r
F kN = , 7.07
p
h m = .
Solution using integrals:
A
Fr pdA =
}
and
1
p
r
A
h D pdA
F
= + q
}

1.
A A
Fr pdA g hdA = =
} }

We have many too many variables in our integral.
( sin( ) )
A A A
Fr pdA g hdA g D dA = = = + u q
} } }

where is the distance along the plate from the top of the plate, is angle of inclination of plate.
Here there is no inclination of the plate so =90 and sin()=1.
Also, because the area (triangle) is symmetric with respect to the x-axis, we can take horizontal
rectangular pieces of the triangle as our dA. Each of the pieces is at a depth of . Each piece dA has have
height d. The width of the dA at depends on so we call it w
.
We see that dA w d
q
= q and goes from 0 to L (interval of integration).
Examining similar triangles, we see: ( )
W
w L
L
q
= q
So:
2
0 0
( 1 ) ( ) ( ( ) )
L L
W W
Fr g D L d g DL D L d
L L
= + q q q = q q q
} }

2 3 2 3
2 3 3
( ( ) ) 1000 9.81 (6 3 3 (6 3) )
2 3 3 2 3
W L L
Fr g DL L D L
L
= =
Fr=10009.8121=206kN as above.
2.
0
1
( )( )
L
p
r
A
g W
h D pdA D D L d
F Fr L
= + q = + q +q q q
} }
=
2 3 4
1 2 3 3 3
6 6 3 (6 3)
21 3 2 3 4
p
h
| |
= + |
|
\ .
=6+1.7=7.7m as above.
L
W

w

dA

A Submerged Isosceles Triangle: Center of Pressure and Area Moment of Inertia

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Linda Fahlberg-Stojanovska 1

Center of Pressure - Integral vs. Formula

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