Chapter 11: Equivalent Systems, Distributed Loads, Centers of Mass, and Centroids
11-1
Chapter 11
Equivalent Systems, Distributed
Loads, Centers of Mass, and
Centroids
“To be truly ignorant, be content with your own knowledge.”
∼ Chuang Tzu (c. 360 BC - c. 275 BC)
11.1
Overview
Thus far we have covered a wide range of topics, but they all have one common theme in
the sense that you are learning to apply physics and mathematics to more and more realistic
problems in mechanics. We began this course with some preliminary vector concepts and
moved quickly into the particle equilibrium. After solving some interesting, but relatively
simple problems, we extended our capabilities by studying moments and distributed loads.
That allowed us to consider the equilibrium of rigid bodies and solve a much wider range
of problems. We applied the concepts of particle and rigid body equilibrium to analyze
structures including trusses, frames, and machines. Then friction, where we were able to
incorporate slipping/tipping, wedges, and belts. Now we will consider equivalent systems,
distributed loads, centers of mass and centroids. A proper understanding of these concepts
will make it possible for the student to apply their knowledge of statics and structures to
complicated shapes that are subjected to complicated loadings. After completing this part
of the course, the student should be able to,
1. Replace one set of forces and moments with an equivalent single force and single mo-
11-2
Chapter 11: Equivalent Systems, Distributed Loads, Centers of Mass, and Centroids
ment.
2. Use integral formulations to transform distributed loads into a single equivalent force.
3. Use composite parts to transform simple distributed loads into a single equivalent force.
4. Calculate centers of mass and centroids using integral formulations.
5. Calculate the centers of mass and centroids using the method of composite parts.
11.2
Important Points
1. Equivalent systems form the basis for replacing multiple forces and moments or distributed loads with a simplified set of reactions.
2. For a distributed load described by the function w(x), the force, F , is,
Z
F =
L
w(x)dx,
0
and the location of that force, X̄,
RL
F =
0
RL
xw(x)dx
xw(x)dx
= R0 L
.
F
w(x)dx
0
Please note that you must integrate the numerator and denominator separately.
3. The center of mass is a physical property, the centroid is a geometric property.
4. The center of mass is based on the concept of equivalent systems. What we would
like to do is replace the complicated distribution of material with a single force so that
the single force generates the same total force and same total moment as the original
material distribution,
R
xdm
x̃ = R
dm
R
ydm
ỹ = R
.
dm
Chapter 11: Equivalent Systems, Distributed Loads, Centers of Mass, and Centroids
11-3
5. The centroid effectively defines the geometric center of an object,
R
xdA
x̄ = R
dA
R
ydA
ȳ = R
.
dA
6. When your object has constant thickness and uniform density, the center of mass and
the centroid occupy the same point. This equivalency can be exploited to break objects
down into relatively simple pieces and make calculations much more straight forward.
11.3
Equivalent Systems
The concept of equivalent systems is a very powerful one in mechanics. Moreover, introducing
it here should pose few difficulties for students because expertise in working with vectors and
calculating moments has already been established. The concept is quite simple. We consider
one set of forces and moments and replace them with a second set of forces and moments
that has the same net effect. The only real difficulty is distinguishing between equivalent
systems and equilibrium. It is a subtle distinction, but for equivalent systems, we just want
to simplify a system of forces and moments. We may not consider all the reactions acting on
the object and the net effect is often different from zero. For equilibrium, we must consider
all the forces and moments and they must add up to zero.
In truth, we have already touched on this concept back when we discussed force couples.
There we replaced two equal and opposite forces offset by some distance, d with a single
concentrated moment. One thing to keep in mind is that you must draw a separate diagram
for each system. That is very important because trying to put both systems on the same
diagram can only lead to confusion and catastrophe.
Let’s begin with an illustration of equivalent systems.
11-4
Chapter 11: Equivalent Systems, Distributed Loads, Centers of Mass, and Centroids
Example Consider the example below. If System I consists of four forces and a concentrated moment1 and System II consists of a force located at point B and a different
concentrated moment. If the force in System II has a magnitude of 300 lbs. and the concentrated moment has a magnitude of 600 ft.*lbs., determine the magnitudes of F1 , F2 , and
F3 .
For system I we have,
X
Fx = F − F1 = 200lbs. − F1
X
Fy = F3 − F2
X
M/B = (2f t.)F1 + (4f t.)F3 + (2f t.)F2 − M = (2f t.)F1 + (4f t.)F3 − 500f t.lbs.
For system II we have,
1
Recall that a concentrated moment is really a force couple.
Chapter 11: Equivalent Systems, Distributed Loads, Centers of Mass, and Centroids
X
Fx = Rcos(53.13o )
X
Fy = Rsin(53.13o )
X
11-5
M/B = 600f t.lbs.
In order to make the two systems equivalent, the forces in the x and y direction must be
the same2 and the net moments must be the same. This leads us to,
2
X
F1 = 20lbs.
X
F3 = 265lbs.
X
F2 = 25lbs.
N.B. the system may or may not be in equilibrium
11-6
Chapter 11: Equivalent Systems, Distributed Loads, Centers of Mass, and Centroids
Example Here is a cantilevered beam with a system of forces and moments on it. What
force and moment must act at point A in order to produce an equivalent system of forces
(not including the reactions)? What about points B, C, and D?
The beam has a length, L = 3.5m, a = 0.25m, b = 0.75m, c = 1.5m, and d = 0.5m. The
value of F1 = 100N, F2 = 200N, F3 = 300N.
y
F1
b
a
d
a
F3
O
x
B
A
c
C
D
F1
F2
Chapter 11: Equivalent Systems, Distributed Loads, Centers of Mass, and Centroids
11-7
Example Here is our wind turbine again. Replace the wind forces on the blades with an
equivalent force and concentrated moment at the origin (the center of the actual turbine).
We may assume that the thrust force on each blade (F1 ) is approximately 35 N (in the −x
direction). The lift force (F2 ) causing each blade to rotate is approximately 200N. You may
assume that the forces act at the center of each 20 m blade. These forces are all in the y − z
plane. In addition, d1 = 65 m, d2 = 5 m.
z
FL
y
FD
FD
FL
x
FD
FL
d1
d2
11-8
11.4
Chapter 11: Equivalent Systems, Distributed Loads, Centers of Mass, and Centroids
Distributed Loads
Wind and water loads, cars on a bridge, and people on a crowded walkway often generate
loads that are approximated as a pressure (force per unit area) or a distributed load (force
per unit length). In order to utilize our equilibrium equations, however, we need forces and
moments. Consequently, it is often necessary to replace a pressure or distributed load with
a single force. First, consider a simple example. We will apply a uniform load to a beam
that is 3 m long and the space, a between the wall and the beginning of the applied load is
0.5 m. It should be easy to see that, if we want to replace this with a single force, it must
be a 250 N load placed in the middle of the loaded region (i.e. X̄ = 1.75 m).
100 N/m
a
L
F
X
Chapter 11: Equivalent Systems, Distributed Loads, Centers of Mass, and Centroids
11-9
Next, take the system shown below, a cantilevered beam with an increasing, triangular
distributed load which peaks at w0 .
w0
F
X
The distributed load has units of force per unit length (N/m or lbs./ft.) and, in this case,
can be written as,
w(x) =
w0
x.
L
(11.1)
In order to ensure that the forces in the two systems are equivalent, we require
Z
F =
w(x)dx,
0
which yields,
L
(11.2)
11-10 Chapter 11: Equivalent Systems, Distributed Loads, Centers of Mass, and Centroids
L
Z
w(x)dx
F =
0
w0 2 L
x )|0
2L
w0 L
=
.
2
= (
We also want it to have the same moment about the origin. In order to do this we require
that,
L
Z
M/O = X̄F =
xw(x)dx,
0
which yields,
Z
L
xw(x)dx
X̄F =
0
Z
L
w0 2
x dx
L
0
w0 L2
=
3
=
and the position of the force, X̄, must be 23 L.
(11.3)
Chapter 11: Equivalent Systems, Distributed Loads, Centers of Mass, and Centroids 11-11
Example Here is another distributed load acting on a beam. Replace the distributed load
with a single force.
y
w(x)=A(x 2 +1) kN/m
x
L=2m
11-12 Chapter 11: Equivalent Systems, Distributed Loads, Centers of Mass, and Centroids
11.5
Centers of Mass and Centroids
Right up front we should note that the center of mass is a physical property while the centroid
is a geometric property.Having said that, the concepts of center of mass and centroid are
based on the notion of equivalent loads. The center of mass provides the location of the
single gravitational force that we usually use in our equilibrium equations and is defined by,
R
xdm
x̃ = R
dm
R
ydm
ỹ = R
dm
R
zdm
z̃ = R
.
dm
The differential element dm is usually broken down in the following way, dm = ρdv.
Consequently, if the object has a uniform density, these equations simplify,
R
xdv
x̄ = R
dv
R
ydv
ȳ = R
dv
R
zdv
z̄ = R
,
dv
which effectively defines the centroid of the volume in x, y, and z coordinates. We should
note that the tilde is used to denote the coordinates of the center of mass while the over
bar is used for the coordinates of the centroid. For two dimensional problems (constant
thickness) it suffices to specify the x and y coordinates,
R
xdA
x̄ = R
dA
R
ydA
ȳ = R
.
dA
Chapter 11: Equivalent Systems, Distributed Loads, Centers of Mass, and Centroids 11-13
Depending on the geometry some integrals may be especially difficult to evaluate analytically. In that case, we can use numerical methods to estimate the integrals or simplify the
geometry and approximate it with simple shapes. This technique is very useful to engineers
and is referred to as the the method of composite parts. Probably the best way to proceed
is to do a few examples.
11-14 Chapter 11: Equivalent Systems, Distributed Loads, Centers of Mass, and Centroids
Example - Integral Formulation
Determine the x and y coordinates of the centroid for this particular shape.
y
y=x 2
x
2m
Chapter 11: Equivalent Systems, Distributed Loads, Centers of Mass, and Centroids 11-15
Example - Integral Formulation Calculate the centroid and center of mass of this
particular structure. You may assume that the density, ρ varies as 2000kg/m3 + 100kg/m4 y.
y
y=(x 2 +1)
x
2m
11-16 Chapter 11: Equivalent Systems, Distributed Loads, Centers of Mass, and Centroids
Example - Integral Formulation One more centroid calculation. In this case, the x
coordinate should be obvious, but the y coordinate is a little more challenging.
y
y=x 2
x
2m
2m
Chapter 11: Equivalent Systems, Distributed Loads, Centers of Mass, and Centroids 11-17
Example - Composite Parts
y
5m
1m
x
2m
11-18 Chapter 11: Equivalent Systems, Distributed Loads, Centers of Mass, and Centroids
Example - Composite Parts
Use composite parts to find the centroid (and coincidentally the center of mass) of this
homogeneous steel plate (with uniform thickness).
Chapter 11: Equivalent Systems, Distributed Loads, Centers of Mass, and Centroids 11-19
Example - Composite Parts
Use composite parts to find the centroid (and coincidentally the center of mass) of this
homogeneous steel plate (with uniform thickness). Be careful with the circular hole.
y
0.5 m
0.4 m
1m
x
3m
11-20 Chapter 11: Equivalent Systems, Distributed Loads, Centers of Mass, and Centroids
Chapter 12: Fluid Statics
12-1
Chapter 12
Fluid Statics
“Eureka!”
∼ Archimedes, after his discovery of buoyancy1 .
12.1
Overview
In fluid mechanics we usually worry about moving fluids, especially flows through pipes,
around airfoils, and fluid jets. One application of moving fluids that fits into our discussion
of statics is air drag which was very important to the survival of Mr. Sulu and Mr. Kirk in
an earlier chapter. Here we will consider what happens in fluids that are not moving.
Beginning with buoyancy, we will investigate the freedom ship, submarines, and offshore
oil rigs. After that, the effects of fluid pressure on dams, gates, and submerged structures
will be explored as well as wind loads on frames and machines. As you may have guessed
the problems will continue to get more realistic.
At the end of this chapter, the student will be able to,
1
There are many interesting aspects to this story. The first is that the Greeks had such a single, short word
to express the idea, “I have made a leap in understanding,” suggesting that they were accustomed to such
things. The other is that, after having figured out that the density of an objected could be determined by
comparing its weight to the amount of water it displaced, Archimedes was so overjoyed that he failed to grab
a towel and ran from the public bath back to his home, naked. Lastly, many people remember Archimedes
only for his role in elucidating buoyancy, but his contributions extended to other areas of mechanics (the
discovery of the principle of the lever, design of a screw pump) and mathematics (the “method of exhaustion”
to compute infinite series and the approximation of π).
12-2
Chapter 12: Fluid Statics
1. Apply the concept of buoyancy to model the forces acting on submerged or partially
submerged structures.
2. Model distributed loads caused by pressure and quantify the forces (and their location)
that act on submerged structures.
12.2
Buoyancy
Everyone knows the story of Archimedes and whatever you think of celebrating your discoveries by running naked down the street, it represented a tremendous leap in understanding.
In simple terms, his discovery led us o the definition for the buoyancy force acting on a
submerged or partially submerged object,
Fb = ρf luid gV,
(12.1)
where Fb is the buoyancy force, ρf luid is the density of the fluid in which the object is
submerged, and V is the volume of fluid displaced2 . Essentially, the buoyancy force is equal
to the weight of the fluid displaced by your object. Keep in mind that an object may not
always be completely submerged. In that case, only the submerged portion contributes to
the buoyancy force.
2
Some useful numbers to remember: The density of air is approximately 1.28kg/m3 , the density of water
is approximately 1, 000kg/m3 , and ρg = 62.4lbs./f t3
Chapter 12: Fluid Statics
12-3
Example: The Freedom Ship
For an example, we will use the Freedom Ship which, once completed, will be the largest
ship in history and will support an entire floating city. It will continuously circle the globe,
visiting most of the earths inhabited coastal regions every two years. The Freedom Ship will
be approximately one mile long, 750 ft. wide, and 25 stories high ( 340 ft.) and its top-most
deck will function as an airfield for small passenger aircraft. Perhaps the most impressive
statistic is the proposed weight of the ship, approximately 2.7 million tons. It will be able
to house between 50,000 and 200,000 passengers with a crew of 15,000. Our goal today is to
figure out how deep this boat is going to sit in the water. A very simple analysis you might
say. But a good starting point. The most difficult aspect is the actual construction. It is
likely that completely new manufacturing techniques will have to be developed before the
ship ever leaves port.
Figure 12.1: An aerial view of the proposed Freedom Ship
12-4
Chapter 12: Fluid Statics
Example: The Hindenburg Disaster
The Hindenburg (technically the LZ 129 Hindenburg) was a German-made dirigible (sometimes called an airship) that was famously destroyed on May 6, 1937. As it was engulfed in
flame, Herbert Morrison’s famous narration of the disaster accentuated the feeling of helplessness and tragedy felt by those on the ground. There were 97 people on board the giant
ship (36 passengers and 61 crew members) and, given the magnitude and swiftness of the
fire, it is remarkable that only 35 of them died (13 passengers and 22 crew members) in
addition to 1 person on the ground.
This disaster was a key moment in engineering history because it was one of the first
really well publicized disasters and it shook the public’s faith in these types of airships. To
some extent, it also damaged the reputation of engineers in general. While it is unlikely
that airships would have been able to compete with passenger planes in the long term, the
Hindenburg disaster dramatically decreased their utility.
Chapter 12: Fluid Statics
12-5
Figure 12.2: Transcript of Herbert Morrison’s commentary: It’s practically standing still
now they’ve dropped ropes out of the nose of the ship; and (uh) they’ve been taken ahold of
down on the field by a number of men. It’s starting to rain again; it’s... the rain had (uh)
slacked up a little bit. The back motors of the ship are just holding it (uh) just enough to
keep it from...It’s burst into flames! It’s burst into flames and it’s falling it’s crashing! Watch
it; watch it! Get out of the way; Get out of the way! Get this, Charlie; get this, Charlie!
It’s fire... and it’s crashing! It’s crashing terrible! Oh, my! Get out of the way, please! It’s
burning and bursting into flames and the... and it’s falling on the mooring mast. And all
the folks agree that this is terrible; this is the one of the worst catastrophes in the world.
[indecipherable] its flames... Crashing, oh! Four- or five-hundred feet into the sky and it...
it’s a terrific crash, ladies and gentlemen. It’s smoke, and it’s in flames now; and the frame
is crashing to the ground, not quite to the mooring mast. Oh, the humanity! And all the
passengers screaming around here. I told you; it - I can’t even talk to people, their friends
are out there! Ah! It’s... it... it’s a... ah! I... I can’t talk, ladies and gentlemen. Honest: it’s
just laying there, mass of smoking wreckage. Ah! And everybody can hardly breathe and
talk and the screaming. Lady, I... I... I’m sorry. Honest: I... I can hardly breathe. I... I’m
going to step inside, where I cannot see it. Charlie, that’s terrible. Ah, ah... I can’t. Listen,
folks; I... I’m gonna have to stop for a minute because [indecipherable] I’ve lost my voice.
This is the worst thing I’ve ever witnessed.
12-6
Chapter 12: Fluid Statics
Here we will estimate how much water ballast was required to keep the Hindenburg at a
fixed height. First, treat the Hindenburg as a particle. The Hindenburg had a volume of
200,000 m3 , a weight of 130 tons (you will need to convert this number to metric units) and
a maximum speed of 135 km/h. If the density of hydrogen is 0.089 kg/m3 , that of air is 1.28
kg/m3 , and the density of water is 1,000 kg/m3 , determine the volume of water required to
maintain equilibrium. What if you had used helium (0.18 kg/m3 ) instead of hydrogen?
Chapter 12: Fluid Statics
12.3
12-7
Fluid Statics and Distributed Loads
In fluids the pressure increases linearly with depth below the water’s surface. In fact, fluid
pressure only depends on atmospheric pressure (which is often negligible) and the depth
below the surface. To illustrate how this works, we will consider a Sea Wall of the type used
to protect cities (such as New Orleans) that lie below sea level. It’s very tricky to build these
types of structures because cities that lie below sea level tend to have soft soil. Consequently,
they cannot support very large moments.
x
z
dy
y
y
w
Sea
Wall
H
Water
Sea
Wall
Figure 12.3: A simple sea wall with width, w, and a water depth, H.
Right now we are primarily concerned with the force that the water exerts on the Sea Wall
and its location. Since the pressure, p, can be written,
p = ρgy,
(provided that we put the origin at the surface of the water and measure y positive downward,
and neglect atmospheric pressure), we can integrate the pressure over the area to obtain the
total force. First, we draw a differential area with width, w and height, dy. Since the
pressure is constant across the width, we do not have to deal with differential areas in the z
direction. The total force on this differential area is,
dF = (ρgw)ydy,
and the total force is obtained by integrating over the surface (in this case by integrating
through the depth),
12-8
Chapter 12: Fluid Statics
Z
F =
H
(ρgw)ydy = (ρgw)
0
H2
.
2
The next challenge is figuring out where the force is located. Keep in mind that we have
already derived the proper location for a force caused by a triangular load distribution. In
this case, we have just rotated the problem 90o .
Chapter 12: Fluid Statics
12-9
Our next example involves Hoover Dam, one of the greatest monuments to industrial
strength and human audacity. The dam itself rises approximately 726 feet above the floor
of the Black Canyon between Nevada and Arizona. It is 1,244 feet long, 660 feet thick at its
base, 45 feet thick at its crest and weighs 5,500,000 tons. The lake it created, Lake Mead,
backs up 110 miles behind the dam and is nearly 500 feet deep. It contains enough water
to submerge the state of Connecticut under 10 feet of water or to supply 5,000 gallons to
every person in the world. Perhaps even more impressively, the concrete dam was cast as a
single solid piece. Pouring started in June of 1933 and continued 24 hours a day, seven days
a week until May of 1935.
Figure 12.4: An aerial view of Hoover Dam
The engineer responsible for this feat was Frank Crowe.He studied civil engineering at
the University of Maine and, after a summer internship studying the drainage basin of the
Yellowstone River, joined the U.S. Governments Reclamation Service. Crowe was widely
considered the best construction engineer in the United States and was eventually hired by
the Six Companies to spearhead their efforts to obtain the Hoover Dam contract. He was the
first to use cable ways to transport people and supplies to the worksite. In many instances
the cables were required to support loads of several tons. In addition, because falling rocks
were always a hazard during the construction of the dam, the first hard hats were developed
12-10
Chapter 12: Fluid Statics
by workers using baseball caps and tar as raw materials. It is worth noting that, after
painstaking calculations and recalculations and four years of construction, nobody knew for
sure that the dam would work. The only way to be certain was seal the diversion tunnels
and hope for the best.
Figure 12.5: Transporting equipment and materials at Hoover Dam
Model the dam as a simple cantilever that rises up out of the bottom of Black Canyon.
Given the weight of the dam and the depth of the water behind it, calculate the reaction
forces and moment at the base of the dam.
Chapter 12: Fluid Statics
12-11
One of the key aspects of this type of analysis is that the force that the water exerts on the
dam increases with the depth squared and the moment increases with the depth cubed. That
is crucial. Using these methods, will make it possible to analyze a wide array of relevant
problems from the failure of the Teton Dam in southeastern Idaho on June 5th, 1976 to
the failure of the levees in New Orleans after Hurricane Katrina. While these disasters are
probably the best known to American students, some of the biggest failures have occurred in
China and Hungary 3 . The most dangerous dam in the world is in just upriver from Mosul,
Iraq.
3
The Banqiao Dam failed in 1975 killing approximately 26,000 people during the resulting flood and
causing nearly 145,000 more deaths as a result of disease. The Kolontar red mud dam failure did not kill
nearly as many people but contaminated huge tracts of land to the point that humans cannot return for
many years.
12-12
Chapter 12: Fluid Statics
A parabolic bridge. One elegant trick that Civil Engineers use to improve the stability of
their dams is to make them parabolic. This allows the weight of the water to press the dam
into the ground making it harder to tip over and easier to create a seal between the bottom
of the dam and the river bed. At first glance, this looks like a very challenging problem
because you know that the pressure always acts normal to the face of the dam.
y
D
y = x²
Water
H
x
Chapter 12: Fluid Statics
12-13
It turns out that this is a challenging problem, but not as bad as it looks. You can turn a
very difficult integral into an easy problem by including part of the water in your free body
diagram.
y
D
y = x²
Water
H
x
The best place to start is by calculating the weight of the water and the distributed
load caused by the water that you sectioned through. While this may seem like a strange
approach, it works quite well. Assume that D = 5m and determine the forces acting on the
dam.
12-14
Chapter 12: Fluid Statics