In Chapter 10, I describe how to find the resultant (combined) force of a variety of distributed loads by simply determining the areas under the load diagram. This calculation provides you with two of the three pieces of information required to fully define a force vector — namely, the magnitude (the vector's length) and the sense (the vector's direction). However, you also need to determine a force vector's point of application in order to properly define the vector. (Check out Chapter 4 for details on these vector properties.)
For concentrated loads (single loads applied at a point — see Chapter 9), you can determine the point of application almost by inspection. If a small object hits a wall, a concentrated force from the ball is located at the point of impact. However, distributed loads (loads spread over a line or area — see Chapter 10) are different.
To find the point of application of a resultant of a distributed load, you have to calculate the center of area or the centers of mass and gravity for the load or object. In this chapter, I show you how to perform these calculations.
Depending on the type of distributed loads you encounter along your statics journeys, the resultant force of each of those loads must act at a specific location. As you understand how to specify the locations where distributed loads and self weight (the force resulting from the gravitational effects of the mass of an object) are positioned, you encounter several terms to define these positions: centroid, center of mass, and center of gravity.
Centroid: The centroid or center of area of a geometric region is the geometric center of an object's shape. For most external distributed loads (where the force acts on the object from the outside), the resultant force acts at a location known as the centroid of the load distribution. See the following section for more on centroids.
Center of mass: The center of mass is the location at which the resultant mass is assumed to act.
Center of gravity: The center of gravity is the average location at which the self weight of the object is assumed to act. Usually, the center of gravity and center of mass are assumed to be the same location; I explain why a little later in this chapter.
Chapter 9 gives you the lowdown on lumped self weight, and Chapter 10 describes distributed self weight.
The centroid is actually a set of coordinate values (x,y,z) measured relative to a specific reference point. Usually, the origin — or coordinates (0,0,0) — of the Cartesian coordinate system that you implement is a convenient reference point. (See Chapter 5 for more on Cartesian coordinates.) For many shapes, this location often occurs inside the boundaries of the region. However, in some situations you may actually compute the centroid coordinates at a position outside the boundary.
Determining the location for the resultant of a distributed load involves calculating the centroid of the load region, which I show you how to do in the following sections.
When you compute a centroid location, your first step is to always determine which equation you should use. To make this decision, you must first classify whether a region is discrete or continuous.
Discrete region: A discrete region is an area that can be broken up into several subregions composed of simple shapes, such as rectangles, circles, triangles, and parabolic segments, with known or easily determined areas. You can also easily express the centroids of these regions based on the dimensions of the region.
Continuous regions: A continuous region is any region that isn't classified as a discrete region. This region is normally enclosed by a complex or irregular-shaped boundary. To determine the centroid of continuous regions, you have to define the boundaries by using mathematical functions and then employ basic calculus and integration techniques.
A discrete region is a type of region made up of a combination of shapes referred to as subregions. Each subregion has its own individual area and centroid calculation that is usually fairly simple to compute. You can then combine these subregion properties to compute a single centroid location by using the following equations:
xi is the distance from the origin to the centroid of subregion i measured parallel to the Cartesian x-axis.
yi is the distance from the origin to the centroid of subregion i measured parallel to the Cartesian y-axis.
Ai is the area of subregion i.
n is the number of subregions that make up the discrete region.
Tip
If the sigma notation in this equation looks foreign to you, flip to Chapter 2.
Figure 11-1 shows six simple shapes that allow you to handle the centroidal calculations for the vast majority of discrete regions. With these six basic shapes, you can construct many more-complex discrete regions.
Tip
Pay special attention to the location of the origin from which the centroid distances xi and yi are determined. Most statics books and other reference sources include similar graphics for properties of areas (usually inside the front or back cover), but the authors of these texts may base their measurements on completely different origins.
When calculating the centroid of discrete shapes, such as the one in Figure 11-2, I find that constructing a simple table makes the solution process much easier and more straightforward.
To calculate a centroid coordinate, you need a separate table for each x- and y-centroid dimension. To determine the x-centroid location, you start by creating a table with the column headings shown in Table 11-1.
Table 11.1. X-Centroid Coordinate Table
Region # | xi (in) | Ai (in2) | xi Ai (in3) |
|---|---|---|---|
1 (triangle) | 0.67(8) = 5.36 | 0.5(8)(8) = 32.00 | (5.36)(32.00) = 171.52 |
2 (rectangle) | 4 + 0.5(4) = 6.00 | (4)(4) = 16.00 | (6.00)(16.00) = 96.00 |
3 (circle hole) | 8 – 2 = 6.00 | −π(0.5)2 = −0.785 | (6.00)(–0.785) = −4.71 |
TOTAL | ------------------ |
 |
 |
Next, use the following simple steps to help you complete the table. At the end, the calculation in Step 7 is the actual x-coordinate location.
In the first column, list each of the areas that make up the discrete region, including any holes or subtracted regions.
In the second column, calculate the distance from the origin of the combined region to the centroid of each shape.
For example, Region #1 (i = 1) is a triangle; Figure 11-1 earlier in the chapter shows you that the x-distance to the centroid of a triangle is
Calculate the area for each region and fill the results into the third column.
For Region #1, you can calculate the area of a triangle from
Tip
For regions that are missing or subtracted from other regions (such as holes), you calculate the area of the region containing the hole (Region #2) as though the hole doesn't exist and then calculate and subtract the area of the hole (Region #3). See "Including holes in discrete regions" later in the chapter for more information about this process.
Multiply the values in the second and third columns and put the product in the fourth column.
For Region #1:
x1A1 = (5.33)(32.00) = 171.52 cubic inches
Notice that this product for Region #3 contains a negative value, because the hole creates a negative area, as I note in Step 3.
Add all the values in the third column and record this value on the bottom row for the TOTAL of that column.
Repeat Step 5 for the values in the fourth column.
Compute the
Based on the result of Step 7, you can now locate the x-coordinate of the centroid, which is located 5.57 inches from the origin in the positive x-direction. You measure this distance from the same origin you use in all the calculations.
To locate the y-coordinate of the centroid, you need to create a table similar to Table 1-1, using y in place of x.
When you're working with a strangely shaped discrete region, sometimes it's convenient to overestimate an area with a simple shape and then subtract another simple shape from your calculations. Using Figure 11-3a, you can overestimate the total region by drawing a rectangle with a horizontal dimension of b1 + b2 and a vertical dimension of h1 + h2. However, you're overestimating the total area of the actual region (see Figure 11-3b). To correct this estimation, you can then subtract a region having a horizontal dimension of b2 and a vertical dimension of h2 (see Figure 11-3c). The area of this subtracted region is computed as a negative number and included in Table 11-1 earlier in the chapter.
Tip
You can add or subtract regions from your estimated shape as long as you apply the correct sign to the area of the region when you calculate it. Areas being subtracted are always negative.
Note
You must measure the distance to the centroid of each simple area, including subtracted regions relative to the same reference point.
One of the more common discrete regions you come across is the trapezoidal region, which shows up frequently in submerged surface calculations (which I discuss in Chapter 23) and linearly varying load distributions (head to Chapter 10). Just like with other discrete regions, you can separate the trapezoidal region into smaller subregions. Figure 11-4 shows two possibilities for this division.
The first option is to break the trapezoid in Figure 11-4a into a rectangle and a triangle (see Figure 11-4b). The second option is to break the region into two triangles (see Figure 11-4c). Regardless of which method you choose, your centroid calculations produce the same resulting centroid location (assuming you do the math right, of course!).
Finding the centroidal coordinates for a continuous region is usually more mathematically complex than the discrete centroidal calculations that I describe in the preceding section because you need to use your calculus skills to perform the integrations to find the centroidal coordinates. The equations that you need to use when working with continuous regions are
If you examine these formulas, the continuous region formulas are actually very similar to the discrete region formulas. To illustrate how these equations are used, consider the continuous region in Figure 11-5a, which is bounded by two functions, f1(x) = x2 on the lower bound edge and f2(x) = x on the upper bound edge.
To use these integral equations, you first need to develop expressions for the incremental area dA. Start by examining the shaded incremental slice shown. This rectangular area can be computed from
The distance x in this centroidal calculation is the distance from the origin to the centroid of the rectangular incremental area. It's just the same variable x that you use when you perform your integration calculation.
Note
Because you've now transformed the area integration into a linear integration calculation (as indicated by the dx), you need to change the limits of integrations as well. The upper limit of the linear integrations along the x-axis is 1 and the lower limit is 0. To compute the x-direction centroidal coordinate as shown in Figure 11-5b, you then substitute into the formula and perform the integration as follows:
Calculating Figure 11-5c's centroid location in the y-direction works much the same way as the x-direction calculation does, but with a few added issues. As in the x-direction, you start by modifying the incremental area calculation to become a linear integration calculation. The same dA expression you use for the x calculation still works. However, the y-distance requires a bit of work. In Figure 11-5c, notice that the variable yi is actually different for each value of x. In this case, you need to transform that expression as well.
Note
Remember, when you integrate with respect to a variable (in this case, dx), all variables in the expression must be in terms of x. In this example, the y location of the centroid is also a variable, but you know its relationship to the x variable because you know the equations of the upper and lower boundaries of the region.
Next, you substitute the expression for y into the centroidal equation:
Notice here that you have to complete a multiplication of two polynomials before you can perform the integration. Now, if those polynomials are reasonably simple, this multiplication may not be that big an issue.
Tip
If you choose a slicing direction and the algebra and boundary-defining functions seem complex, try developing the incremental area calculations by slicing the region in the opposite direction as I describe in the nearby "Slicing a centroidal calculation differently" sidebar. After all, you still end up getting the same numeric results no matter how you slice it.
In some cases, you have to find the centroids of objects that have one or more axes of symmetry. An object is said to be symmetrical about an axis if the part of the object on one side of that axis is a reflection, or mirror image, of the part on the other side. Many shapes in engineering are symmetrical.
Some objects have one axis of symmetry, such as the images shown in Figure 11-6a and b. Other objects can have multiple axes of symmetry, such as the object in Figure 11-6c. And yet other objects can have an axis of symmetry that is neither horizontal nor vertical but rather oblique as in Figure 11-6d.
Note
If you know that an object has an axis of symmetry, you can assume (and rightly so!) that the centroid location in the opposite direction must be located on that axis of symmetry. For example, Figure 11-6a has a vertical axis of symmetry. If you identify that this axis is located 5 millimeters to the right of the origin, you also know that the horizontal centroid distance is 5 millimeters to the right of the origin as well. You've just found one of the centroid locations without ever having to write a single equation!
Slicing a centroidal calculation differently
If the polynomials you need to multiply become lengthy, complex, or full of trigonometric functions, you may consider using a process similar to the following:
Sometimes you can simplify the math by changing the direction of the area slices. Reexamine the example in Figure 11-5 by using a horizontal incremental area as shown in the following figure.
For example, if you repeat the calculation for the y-centroid coordinate but instead slice the incremental area dA horizontally, you get a different expression:
For the horizontal slice, notice that the parameters x1 and x2 are actually dependent on the functions f2(x) = y = x2 and f1(x) = y = (x1)2. Substituting into the same continuous centroid equation and evaluating the integral,
Note that the result is identical to the traditionally calculated result in "Finding centroids of continuous regions." However, the polynomial multiplication for this method is a bit easier to work with. Unfortunately, depending on how you slice the continuous region (horizontal or vertical) to obtain your incremental area, you can either simplify your calculations or make them significantly more complicated. And you won't know which until you actually try.
Although you can use the calculations for centroids and centers of area in the previous sections with internal forces and external loads, self weight has its own special location requirements. To locate self weight, you first need to understand the difference between an object's center of mass and its center of gravity.
Center of mass: An object's center of mass is the single location where its total mass can be applied as a single lumped value.
Center of gravity: An object's center of gravity is the location on the object where resultant force due to gravity is acting. Self weight is a significant gravitational force on any object and is always located at the center of gravity.
Centers of mass and centers of gravity don't necessarily coincide with the centroids of geometric areas, although very often they do (check out "Getting to the Center of Centroids" earlier in the chapter for more on centroids). The center of gravity and center of mass also don't necessarily have to be contained within the boundary of the region.
The center of mass isn't necessarily tied to the geometric dimensions of the object but rather to the distribution of the mass within the object. For example, engineers often want a racing vehicle's center of mass as low as possible in order to ensure its stability at high speeds. However, the centroid of the vehicle is usually located at a position much higher up in the vehicle as a result of the physical dimensions of the automobile (see Figure 11-7).
For discrete regions, you can calculate the center of mass from the following expression:
For continuous regions, you have to fall back to the integral form (similar to the centroid calculations for continuous regions that I discuss earlier in the chapter), as follows.
where m is the total mass of the object, and dm is the mass of an incremental slice of the object. The variable x represents the distance from the reference point to the center of the incremental mass, dm.
If you think this formula looks very similar to the centroid equations I discuss earlier in the chapter, you're right. The only difference is that instead of using the region's geometric area in your calculation, you're now using its mass. Consider the system of two masses (m1 and m2) shown in Figure 11-8.
Note
The center of mass is directly related to the location of the mass of each subregion and its position relative to a reference location, so always include a reference or origin when measuring the center of mass.
As long as the gravitational pull on an object is uniform, the centers of gravity and mass share the same position. On Earth, you can reasonably assume this scenario to be the case because the Earth's gravitational field typically doesn't fluctuate very much over short distances.
In this book, I assume that the gravitational field is constant for the object, and thus that the center of mass and the center of gravity occur at the same location.