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This of course is an idealization. Even the tires of a car are in contact with the supporting road not at a single point but over a small area which becomes larger when the air pressure inside the tire is low. Assuming that the force of the road onto the car is a concentrated load will reflect reality quite well as long as the size of the contact area remains small in comparison to the distances among the forces involved (i.e. the distance between the wheels), which is usually the case.
In cases for which such an idealization might not lead to accurate enough solutions we speak of a body or structure being exposed to a continuous or distributed load. The weight (meaning the force which gravity exerts) of the structure itself is an obvious example (although in many case the weight of the structure might not be important in itself because it might be small in comparison to the other forces). In addition to the weight of the structure itself continuous or distributed loads come about in basically two ways :
FOOT NOTE : When gravity is involved, all our calculations will be
done assuming that the
gravitational field is constant over the extent of the bodies
we are investigating. In this case the location of the Center of Mass
and the Center of Gravity of a body coincide.
For us normal earthlings this is a very
good assumption because practically all structures we are concerned with
are small in comparison to the distance over which the earth's gravitational
field changes appreciably. But the few space faring individuals among us
might get into trouble with this assumption.
| In Fig. 7.2a we have a simple example of a body (the triangular load) lying on top of a beam. Our goal is to determine the support forces FA and FB for a given weight W of the load. |
An elegant way of finding these support forces is to apply the
Principle of the Center of Gravity. According to this principle
we replace the action of the body onto the beam by a single force equal
to its weight acting at its center of gravity.
(in Figure 7.2a) indicates the
location of the center of
gravity, here measured as distance from the left end of the triangular
load. It's value is a characteristic of the body at hand and does
not change as the body is moved (which would change the distance "a"
in Figure 7.2a) nor does it change with the strength of the gravitational field.
The equilibrium equations for this case would be then
which we can solve for the two support forces.
Of course, we have to find first the location of the center of gravity
(the distance 
in Fig. 7.2a)
which depends on the way how the mass of the
body (the triangular load) is exactly distributed.
One worry you might have when applying Equations 7.2a, is that it implies that the details of contact between the triangular load and the beam underneath do not matter. Have a look at Problem 7.8a which investigates this concern.
Assuming for the moment that the location of the center of gravity of some common bodies is known, the method can be expanded to find the support forces even if we have a composite body lying on top of our beam.
| Here the composite body consists of three "elementary" or "simple" bodies, the triangular load of weight W1, the rectangular body of weight W2, and the second triangular load of weight W3. |
Again, applying the Principle of Center of Gravity, we can write down the equilibrium equation :
which allow us to determine the two support forces uniquely.
|
The next question to be asked, in preparation for the next section
in this chapter, is : How can we determine the location of the center
of gravity of the entire composite body ? Or in other words
at which location
(the distance a+ in Fig. 7.2c)
must I place the total weight
of the composite body in order to obtain the same support forces as with the equations 7.2b. |
|
Here is the secret as to how you can arrive at the result yourself. Write out the equilibrium equations ( sum of the forces and sum of the moments equal to zero ) for the beam of Figure 7.2c. Equate the solutions to the results from Equat. 7.2b and after a little bit of algebra you find :
Equation 7.2c can be used to determine the location of the center of
gravity of a composite body if the locations of the center of gravitity
of its individual parts with respect to a common point (here the left
corner of the composite body) are known.
Equation 7.2c is more than just a prescription to determine the location
of the center of mass of a composite body. It is also the validation
of the Principle of Center of Gravity : if the distance
which indicates the location of the center
of gravity from the left corner of the composite body,
would depend on where the composite body is located ( given by the
distance "a" in Fig. 7.2c ) the entire concept would not make
much sense. Right ?
Finally, equation 7.2c can be expanded to take care of composite
bodies which consist of any number, say "n", of parts.
Mathematicians would write it this way :
|
meaning, that you should first add up the all the weights, Wi,
of all the individual parts which are numbered 1 through n in order
to find the total weight, Wtotal.
Then proceed in summing up the moments,
i Wi ,
of all the individual parts
with respect to a common reference point
and divide by the total weight. That will give you the distance of the
center of gravity of the composite body from the same reference point.
In my example above, the chosen reference point was the left bottom corner
of the composite body, but any other point would have been fine as well.
Problems
Prob. 7.2a :
Details of Contact
,
throughout. That means that each cubic
inch (or whatever other measure of volume) of the body has the same
weight regardless of its location inside the body.
|
Fig. 7.3a is an example of such a body. If you like an underlying beam
is represented by the x-axis. The body extends from x=a to x=b
and its shape above the x-axis is assumed to be given by some
mathematical function f(x). For example :
y = 3 x - 5 x2 = f(x) for a ≤ x ≤ b and zero otherwise. |
We use now the concept of composite bodies to determine the weight of such a body and the location of its center of gravity.
|
In Fig. 7.3b I approximate the original body by a sum of n=8
vertical strips each having the same width
 =(b-a)/n and the thickness t. The
center of gravity of each individual strip is located at its geometric
center, xi in
Fig. 7.3b, halfway inbetween its left and right side.
The height of each these strips is chosen to
be equal to the value of f(x) at this mid-point.
|
The approximate weight of the composite body is now determined using Equat. 7.2d :
Here "A" is used to denote the approximate area of the composite body.
And using Equat. 7.2d we obtain for the x-location of the center of gravity :
Because of its constant thickness and uniform specific weight the
location of the center of gravity,
, is independent of these quantities and
is therefor just a property of the displayed area under the curve f(x). In this
context we speak of as describing the
x-location of the centroid of the area under the curve f(x).
Of course the equations 7.3a through 7.3c give you only approximations to the correct values of the area A and the location of the centroid. The quality of the approximation can be increased if the number of strips is increased ( a short computer program may come in handy here ). In the limit of infinitely many strips of vanishingly small width we make the transition to integration and the above equations become :
(7.3d) 
|
In Equat. 7.3d the integral over x*f(x)*dx is often referred to as
the 1. moment of the area w.r.t. the y-axis. Such quantities or properties
of areas including the location of the centroid and higher order
moments ( x2*f(x)*dx ) are of importance when discussing
bending of beams (their cross-section is the area whose properties
are of interest) on one side and in discussing the dynamic properties
of rigd bodies on the other side. ( EMch 13 and EMch 12, respectively ).
| We also can determine the y-location of the center of gravity by rotating Fig.3b by 90o in clockwise direction as shown in Fig. 7.3c. The weight of the individual (now horizontal) strips is now directed parallel to the x-axis and the center of gravity of a single strip is located at a distance 1/2f(xi) away from the x-axis. |
The resulting equation for the distance of the centroid (center of gravity) from the x-axis is :
(7.3e) 
|
Problems
Problem 7.3a : Area and location of centroid for f(x) = sin(x)
The St. Louis Gateway Arch (if it has a constant cross-section) or the main cables of the Golden Gate Bridge in San Francisco (Diameter = 0.92 m, length = 2332m) might be some good examples but you might as well think of the cables of a powerline leaping from tower to tower or maybe a gas pipeline going across mountain ranges of known profile (The center of gravity might - admittedly - not be of interest in this case).
|
Fig. 7.4a depicts a generic example. A solid line is stretching between x=a
and x=b according to some given function f(x). If this line represents some
cable ( for example ) the weight of the cable iwe could determine its weight
be dividing it into small straight pieces, determine the weight of each
piece and summing over all the pieces making up the entire cable.
The weight of such a small piece, see the insert in Fig. 7.4a, is given by :
where A = cross-section of cable, For all intend and purposes the small line segment is parallel to the slope f'(x) the line has at the location of the small segment which allows us to rearrange above equation :
|
Adding up the contributions of all line segments gives us the total weight :
|
Notice that the appearing integral contains just information about the shape of the line in terms of f'(x) and can be solved once and for all independent of the shape and size of the cross-section and the specific weight. The integral itself merely represents the length of the line between the points x=a and x=b.
Sorry, none available.
Prob. 7.2a :
Details of Contact
Prob. 7.3a : Area and location of centroid for f(x) = sin(x)
=
specific weight [g/cm3].
