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Chapter 11: Equivalent Systems, Distributed Loads, Centers of Mass, and Centroids 11-
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,
- 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.
- Use integral formulations to transform distributed loads into a single equivalent force.
- Use composite parts to transform simple distributed loads into a single equivalent force.
- Calculate centers of mass and centroids using integral formulations.
- Calculate the centers of mass and centroids using the method of composite parts.
11.2 Important Points
- Equivalent systems form the basis for replacing multiple forces and moments or dis- tributed loads with a simplified set of reactions.
- For a distributed load described by the function w(x), the force, F , is,
F =
∫ L
0
w(x)dx,
and the location of that force, X¯,
F =
∫ L
0 xw(x)dx F
∫ L
∫^0 xw(x)dx L 0 w(x)dx
Please note that you must integrate the numerator and denominator separately.
- The center of mass is a physical property, the centroid is a geometric property.
- 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,
x˜ =
xdm ∫ dm
˜y =
ydm ∫ dm
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 con- centrated moment^1 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 con- centrated moment has a magnitude of 600 ft.*lbs., determine the magnitudes of F 1 , F 2 , and F 3.
For system I we have,
Fx = F − F 1 = 200lbs. − F 1 ∑ Fy = F 3 − F 2 ∑ M/B = (2f t.)F 1 + (4f t.)F 3 + (2f t.)F 2 − M = (2f t.)F 1 + (4f t.)F 3 − 500 f 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 11-
Fx = Rcos(53. 13 o) ∑ Fy = Rsin(53. 13 o) ∑ M/B = 600 f t.lbs.
In order to make the two systems equivalent, the forces in the x and y direction must be the same^2 and the net moments must be the same. This leads us to,
F 1 = 20 lbs. ∑ F 3 = 265 lbs. ∑ F 2 = 25 lbs.
(^2) N.B. the system may or may not be in equilibrium
Chapter 11: Equivalent Systems, Distributed Loads, Centers of Mass, and Centroids 11-
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 (F 1 ) is approximately 35 N (in the −x direction). The lift force (F 2 ) 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, d 1 = 65 m, d 2 = 5 m.
FD
x
y
z
FL
FD FD
FL
FL
d 2
d 1
11-8 Chapter 11: Equivalent Systems, Distributed Loads, Centers of Mass, and Centroids
11.4 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. X¯ = 1.75 m).
F
X
a
L
100 N/m
11-10 Chapter 11: Equivalent Systems, Distributed Loads, Centers of Mass, and Centroids
F =
∫ L
0
w(x)dx
= (
w 0 2 L
x^2 )|L 0
=
w 0 L 2
We also want it to have the same moment about the origin. In order to do this we require that,
M/O = XF¯ =
∫ L
0
xw(x)dx, (11.3)
which yields,
XF¯ =
∫ L
0
xw(x)dx
∫ L
0
w 0 L
x^2 dx
w 0 L^2 3
and the position of the force, X¯, must be 23 L.
Chapter 11: Equivalent Systems, Distributed Loads, Centers of Mass, and Centroids 11-
Example Here is another distributed load acting on a beam. Replace the distributed load with a single force.
w(x)=A(x +1) kN/m 2
x
y
L = 2 m
Chapter 11: Equivalent Systems, Distributed Loads, Centers of Mass, and Centroids 11-
Depending on the geometry some integrals may be especially difficult to evaluate analyti- cally. 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=x
x
y
2 m
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=x 2
x
y
2 m 2 m
Chapter 11: Equivalent Systems, Distributed Loads, Centers of Mass, and Centroids 11-
Example - Composite Parts
x
y
2 m
1 m
5 m
Chapter 11: Equivalent Systems, Distributed Loads, Centers of Mass, and Centroids 11-
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.
x
y
3 m
1 m
0.5 m
0.4 m
11-20 Chapter 11: Equivalent Systems, Distributed Loads, Centers of Mass, and Centroids
