Fluid Mechanics: Topic 4.2 – Center of pressure on a plane surface


In this lesson, we will derive equations for
determining the location of the center of pressure on a plane surface. The tank on the right contains a liquid which
is exposed to the ambient pressure. We focus on a section of wall of arbitrary
shape and want to determine the location where the resultant force, FR, acts. This location is called the center of pressure. Knowledge of the location of the center of
pressure is important in situations where a gate may swing open or a retaining wall
may tip over. The origin of the coordinate system is shown
on the left diagram. The x-coordinate comes out of the screen and
is oriented along the free surface. The y-coordinate is oriented along the section
of wall we wish to examine. The coordinates of the center of pressure
are (xR, yR). The centroid of the surface is labeled as
C and its coordinates are (xC, yC) The location of some arbitrary area dA on
the surface is located at coordinate (x, y). We first will determine the y-coordinate of
the center of pressure, yR. The net hydrostatic pressure force at some
small area dA is gamma times y sin(theta) times dA. The moment this force produces about the x-axis
is the moment arm y times dFnet. Integrating over the entire surface gives
us the total moment produced by the hydrostatic pressure force. yR is the moment arm needed for the resultant
force, FR, to produce the same moment as this integral. In other words, the resultant force FR and
center of pressure location yR produce an equivalent moment on the wall as the original
pressure field. The resultant force FR is equal to gamma times
sin(theta) yC times A. We plug in the expression for FR and dFnet
into the equation, and pull gamma and sin(theta) out of the integral because they are constant. We are left with the integral of y-squared
dA. This quantity is the second moment of area
about the x-axis, and we will denote this as Ixx,O. So the right side of the equation becomes
gamma sin(theta) times Ixx,O. It can be difficult to calculate the second
moment of area about an arbitrary x-axis, but for common shapes it is easy to calculate
the second moment of area about a parallel axis that passes through the centroid, which
we will call Ixx,C. The parallel axis theorem relates Ixx,O and
Ixx,C. Plugging in the expression for Ixx,O, we find
that yR is equal to yC plus Ixx,C divided by yC times A. The second term, Ixx,C divided by yC times
A, is the distance between the centroid and center of pressure along the y-axis. We can follow a similar procedure to find
the x-coordinate of the center of pressure, xR. The equation for the net force exerted on
area dA is the same, gamma times y sin(theta) dA. But now we want to know the moment produced
by dFnet about the y-axis. The moment arm for dFnet is x, and the moment
produced by dFnet is x times dFnet. We can find the total moment about the y-axis
by integrating x dFnet over the entire wall. xR is the moment arm needed for the resultant
force, FR, to produce the same moment about the y-axis as the entire pressure field. Plug in the expression for FR and dFnet, and
pull gamma and sin(theta) out of the integral on the right because they are constant. The integral of x y dA is the product moment
of area with respect to the x and y axes, and is denoted by Ixy,O. So the right side of the equation becomes
gamma sin(theta) times Ixy,O. It can be difficult to calculate the product
moment of area about an arbitrary set of x-y axes. However, for common shapes it is easy to calculate
this quantity about a parallel set of orthogonal axes that pass through the centroid. We will call this quantity Ixy,C. The parallel axis theorem relates Ixy,O and
Ixy,C. Plugging in the expression for Ixy,O, we find
that xR is equal to xC plus Ixy,C divided by yC A. The second term, Ixy,C divided by yC A, is
the distance between the centroid and center of pressure along the x-axis. For many problems in fluid mechanics, it is
not necessary to calculate xR. Although not derived in this video, for closed
tanks that are pressurized to a gage pressure P0, the equations for xR and yR need to be
modified slightly. xR becomes xC plus Ixy,C divided by the quantity
yC plus P0 divided by gamma sin(theta), times A. yR becomes yC plus Ixx,C divided by the quantity
yC plus P0 divided by gamma sin(theta), times A. We now will list the equations for the area,
Ixx,C, Ixy,C, and the location of the centroid for common shapes. For rectangular walls, the area is the height
times the base. Ixx,C is 1/12 times the height to the third
power times the base. Ixy,C is 0 since the shape is symmetric about
the y-axis. The centroid is located at the midpoint between
the left and right sides, and the midpoint between the top and bottom sides. For triangular walls, the area is the 1/2
times the height times the base. Ixx,C is the height to the third power times
the base, divided by 36. Ixy,C is one divided by 72, times the base,
times the height squared, times the quantity base minus two times d, where the distance
from the vertex to the left side of the triangle, as measured along the x-axis. If the triangle is symmetric about the y-axis,
Ixy,C becomes 0. The distance from the base to the centroid
is one-third times the height. The distance from the left side of the triangle
to the centroid is one-third base plus the distance d. For circular walls, the area is pi times the
radius squared. Ixx,C is 1/4 pi times the radius to the fourth
power. Ixy,C is 0 since the shape is symmetric about
the y-axis. The centroid is located at the center of the
disk. For semi-circular walls, the area is one-half
pi times the radius squared. Ixx,C is 0.1098 times the radius to the fourth
power. Ixy,C is 0 since the shape is symmetric about
the y-axis. The centroid is located at a distance four-thirds
times the radius divided by pi away from the base of the semi-circle. For quarter circle walls, the area is one-quarter
pi times the radius squared. Ixx,C is 0.05488 times the radius to the fourth
power. Ixy,C is -0.01647 times the radius to the
fourth power. The centroid is located at a distance four-thirds
times the radius divided by pi away from the two flat ends of the quarter-circle.

Leave a Reply

Your email address will not be published. Required fields are marked *