When you throw a small object such as a golf ball at a wall, the force the ball makes onto the wall (and of course the force that the wall makes onto the ball) acts on a very small area. In many cases, you idealize this force as a concentrated load (which I cover in Chapter 9) because the force is concentrated onto a small area. But what happens when you throw a larger object (such as your malfunctioning TV) with the same force at the same wall? In this case, the area of the force when it meets the wall is spread out and is therefore no longer concentrated. This type of force is known as a distributed force and has several unique properties that you need to remember.
In this chapter, I discuss distributed loads in detail and show you how to calculate their total combined effect. I also reveal how you can consider self weight (see Chapter 9) as a distributed load as well.
A distributed load is a load that doesn't act at a point but rather is spread out over a specified length. Because distributed forces are spread out over the entire section, they have some properties you want to be aware of: intensity, start point, and end point.
Intensity: The intensity of a distributed load is actually the magnitude of the load (flip to Chapter 4 for more on magnitude). For a distributed load, intensity can be any shape. It's often a polynomial of zero, first, or second order (see Chapter 2); however, you can use any function to describe the intensity depending on the loading you're describing. In the following section, I describe several common shapes of intensity functions that you may encounter.
Start point: The start point of a distributed load indicates where the intensity load application actually begins.
End point: The end point of a distributed load indicates where the intensity loading actually ends.
Figure 10-1 shows a depiction of a fully distributed load (where the start point is at one end of the object and the end point is at the opposite end), a partially distributed load, and a concentrated load applied to a beam.
Distributed loads come in many shapes and varieties depending on how the load was created and how it's applied on the object. Distributed loads can be linear, surface, or volumetrically distributed; I deal with these varieties in more detail in the following section.
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You can typically classify distributed loads based on the dimensions in which they're acting. The units of a load intensity always give you some insight into the type of load you're dealing with and can help you choose an appropriate method for working with distributed loads. Check out the following sections for more on the types of loads and the units they use.
For distributed forces that act over a linear distance, you measure in force-per-distance measurements: Newton per meter (N/m) for SI units and pounds per foot (lb/ft or plf) in U.S. customary units. Figure 10-2 shows several different distributed forces. By far, the two most common of the distributed forces are known as uniformly distributed and linearly distributed (though they're not the only options).
Uniformly distributed: A uniformly distributed load is a load that has a constant intensity over the length of the load. Figure 10-2a illustrates a uniformly distributed load with a constant magnitude of wo. A uniform load is also a zero-order load distribution because the order of a polynomial of zero order (n = 0) for a polynomial is of the form w(x) = wox0 = wo, which indicates a zero-order (or constant) function.
Linearly distributed: A linearly distributed load is a load with an intensity that varies linearly over the length of the load. The lowest intensity occurs at one end of the load, and the maximum occurs at the other. You can establish all intensity values in between from a linear function, as shown in Figure 10-2b. In a linearly varying function, the function is first order because it has the form w(x) = ax+b, which defines a linear function.
Higher order distributed: A higher order distributed load is a distributed load that has an intensity that can be determined by a polynomial of order greater than one (or n > 1) or by completely different functions altogether (such as trigonometric functions). Higher-order distributions such as the one shown in Figure 10-2c can be quite complex and may require special calculations.
Surface distributed loads (also known as surface pressures) are loads that act over a prescribed area. Over that area, the intensity of a surface load can vary greatly. You measure a surface distributed load as a force per area, so you express the units as Newton per square meter (N/m2) or the pascal (Pa.) in SI and pounds per square foot (lb/ft2 or psf) in U.S. customary.
Forces spread over an area are also called pressures. Pressures always act over a two-dimensional area (or surface) and can vary in intensity in multiple directions. Examples of surface pressures include the pressure of water acting on a dam (pressures are very common when dealing with fluids), and the weight of snow on the roof of your house. Figure 10-3 shows a pressure on a surface; the pressure intensity has the form w(x,y) because it can have changing intensities in the x- and y-directions.
Volumetric distributed loads are loads that act over a volume. The most common volumetric load is the specific weight, a useful value for calculating the self weight of an object. (Take a look at Chapter 9 for details on specific weight and self weight.) You measure a volumetric distributed load as a force per volume, so you use the units Newton per cubic meter (N/m3) for SI and pounds per cubic foot (lb/ft3) in the U.S. customary system. Volumetric loads require all three dimensions to calculate (versus the two dimensions pressures require; see the preceding section).
Sometimes in statics work, you want to be able to consolidate a distributed load into a single value that acts at a single, specific location. This combined or consolidated force is known as the resultant. The resultant of a distributed load is similar to the resultants of concentrated loads I discuss in Chapter 7. With distributed loads, the resultant is a single combined force that represents the entire effect of the distribution.
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After calculating the resultant, you still need to find the unique point of application where it's acting. I show you how to compute this location in Chapter 11.
The simplest resultant forces to calculate are those forces that occur in two dimensions. In two dimensions, you calculate the resultant of a distribution by evaluating the following relationship: resultant = area under loading distribution.
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To start the calculating process, I like to replace the load diagram with a dashed line to represent the boundary as shown in Figure 10-4. In addition, I place a concentrated force to represent the resultant of the distributed load. The dimension x helps determine the specific location of the resultant, which I show you how to calculate in Chapter 11. I use this type of sketch for two reasons. First, it helps determine the necessary dimensions of the area calculation. Second, it also preserves a reminder of the original load distribution that I use to compute the resultant magnitude.
The uniform distribution in Figure 10-4a is one of the simplest resultants to compute. By looking at the distribution boundary line, you see that the resultant area that you need to compute is actually the area of the rectangular shaded region.
Recall from your basic geometry class (remember that?) that the formula for the area of a rectangle is area = base · height, where base and height are the dimensions of the rectangle. You can expand this formula to apply to uniform distributions:
For the uniform distribution, the base is the length of the length of the uniform distribution, or L. The height of the distribution is actually the intensity, wo, of the distribution.
For example, if you stack 300 pounds over a one-foot area and then copy this loading repeatedly for the entire length of the beam, you've just defined a uniform load of 300 pounds per foot. If the beam is 12 feet long, the resultant load of this uniform distribution is 300 pounds per foot · 12 feet = 3,600 pounds.
Thus, the resultant of this uniform load is 3,600 pounds total acting on the beam.
The linearly varying distribution shown in Figure 10-4b is another common distribution that you often encounter.
To compute the magnitude of the resultant, you need to be able to compute the area of a trapezoid (the shaded area of the distribution's boundary):
where length of the distribution is given by the dimension L, the maximum intensity is given by wMAX, and the minimum intensity is given by wMIN. Geometrically, you can represent this calculation as a sum of a rectangle and triangle or as a sum of two triangles as shown in Figure 10-5 in the following section.
Other two-dimensional distributions that you may encounter can be described by higher-order continuous or trigonometric functions and distributions composed of combinations of simple distributions.
Figure 10-5 illustrates a more complex two-dimensional distribution that has a varying intensity, w(x), acting over a specified length L.
If you remember from calculus, you can easily use an integral to determine the area bound by a continuous function. You start by taking an incremental slice that has a differential width, at some arbitrary position x. The height of the slice is actually the value of the intensity function, w(x), evaluated at that point. That is, h = w(x). Using integration, you can then compute the area from the following integral:
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As long as you can define the intensity of the distribution as a function w(x), or a series of functions w1(x), w2(x), and so on, you can compute the magnitude of the resultant of the distribution.
For working with complex or combinations of distributions, you can always break up a larger distribution into smaller (and often simpler) pieces. Figure 10-6 illustrates a distribution that's actually a combination of two simpler distributions.
In this example, you can break this seemingly complex distribution into a combination of a linearly varying distribution shown as Area #1 and a uniform distribution shown as Area #2, and you can compute the resultant for each of those two distributions separately. After you determine these resultants, you can then calculate the resultant of the two smaller resultants (Resultant #1 and Resultant #2). This final resultant then has its own position, as indicated by xRES. I discuss this topic more in Chapter 11.
In concept, calculating the resultant of a surface load is similar to how you treat linear distributed loads. Instead of calculating the area under a linear distribution, for surface loads, you actually calculate the volume inside the pressure distribution. Figure 10-7 shows an example of a three-dimensional pressure diagram.
For a given Cartesian axis, you know (or can determine from calculus) the function that describes the pressure distribution. In Figure 10-7, the pressure varies in two directions (both x and y) — that is, a pressure distribution can vary in two directions. To find the resultant, you have to use some basic calculus involving a double integral for this evaluation:
Now, before you completely panic about all the calculus you may have forgotten, here's how you can handle a pressure distribution defined by a geometric prism. A geometric prism is a three-dimensional shape that has equal areas on opposite ends of the shape and a constant length. In Figure 10-7 earlier in the section, if PTOP1 = PTOP2 and PBOT1 = PBOT2, the distribution shown is a geometric prism. To calculate the volume of this prism, you can make use of a basic equation for calculating the volume of a prism:
If you redraw the distribution of Figure 10-7 as a two-dimensional linear distribution that is constant along its length, you can create a new, simpler distribution as shown in Figure 10-8. If you use the following basic steps, you can avoid the double integral and calculus.
Determine the resultant (Resultant #1) of the distribution for the cross-sectional area of the geometric prism, using the techniques I discuss in the section "Uniform and linearly varying forces in two dimensions."
When you calculate this resultant, you end up with a force per distance, so your units are Newton per meter or pounds per foot.
Create a new distribution in the direction of the length of the original distribution.
As shown in Figure 10-8, you evaluate the second distribution in the Cartesian x-direction. The intensity of this new distribution is equal to Resultant #1 and is applied uniformly along length L.
Use those same techniques to resolve this second distribution into a final resultant (Resultant #2).
Resultant #2 is actually the resultant of the surface pressure. The units on this final resultant are either Newton or pounds.
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This technique only works on distributions that can be defined as geometric prisms. If you can't define the shape as a geometric prism, you have no choice but to use the double integrals technique.
As I state in Chapter 9, treating mass and weight as a single lumped value is valid for prismatic objects, or simple objects having constant dimensions in each direction. However, for cases where the object shape is irregular (nonprismatic), or the density of the object isn't constant throughout, the mass may have a distribution of its own. For example, if you take a prismatic bar that has a constant density and cut it into two pieces, each piece still has a weight associated with it, which means each piece must have mass as you can see in Figure 10-9.
Figure 10-9 shows a prismatic bar with a length L that has been broken into n equal-length pieces. That means that the length of each piece is given by
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If the cross section of the bar is not prismatic, you have to assume that the mass is constant for any given piece and calculate the mass (as I describe in Chapter 9) for each piece separately. A distributed mass problem is basically a lot of lumped mass problems combined into a single problem.
After you determine the mass of each piece, you can calculate the self-weight (W)i of each piece by using the following formula:
where g is the acceleration due to gravity.
Each piece of the bar actually has its own self weight. Usually you can assume that the mass is constant over very small increments (or on a piece-by-piece basis), but from one piece to the next, the distributed masses can be vastly different.
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Here's a handy guide to the units and constants you need (in both SI and U.S. customary) to work with distributed mass:
Measurement | SI | U.S. Customary |
|---|---|---|
Mass | kg/m | slugs/ft |
Gravity | 9.81 m/sec2 | 32.2 ft /sec2 |
Weight | N/m | plf |
To verify the piecewise values, you can still determine the total weight of the bar by adding the weights of all the individual pieces:
The final units of WTOTAL give you the same units (Newton for SI and pounds for U.S. customary) as the lumped mass system in Chapter 9.