7.8: Distributed Loads
- Page ID
- 70258
Key Questions
- What is a distributed load?
- Given a distributed load, how do we find the magnitude of the equivalent concentrated force?
- Given a distributed load, how do we find the location of the equivalent concentrated force?
Distributed loads are forces which are spread out over a length, area, or volume. Most real-world loads are distributed, including the weight of building materials and the force of wind, water, or earth pushing on a surface. Pressure, load, weight density and stress are all names commonly used for distributed loads. Distributed load is a force per unit length or force per unit area depicted with a series of force vectors joined together at the top, and will be designated as \(w(x)\) to indicate that the distributed loading is a function of \(x\text{.}\)
For example, although a shelf of books could be treated as a collection of individual forces, it is more common and convenient to represent the weight of the books as as a uniformly distributed load. A uniformly distributed load is a load which has the same value everywhere, i.e. \(w(x) = C\text{,}\) a constant
(a) A shelf of books with various weights.
(b) Each book represented as an individual weight
(c) All the books represented as a distributed load.
Figure 7.8.1.
We can use the computational tools discussed in the previous chapters to handle distributed loads if we first convert them to equivalent point forces. This equivalent replacement must be the resultant of the distributed loading, as discussed in Section 4.7. Recall that this resultant force has the identical effect on the object as the original system of forces had.
To be equivalent, the point force must have a:
- Magnitude equal to the the area or volume under the distributed load function.
- Line of action that passes through the centroid of the distributed load distribution.
The next two sections will explore how to find the magnitude and location of the equivalent point force for a distributed load.
Equivalent Magnitude
The magnitude of the distributed load of the books is the total weight of the books divided by the length of the shelf
\[ w(x) = \frac{\Sigma W_i}{\ell}\text{.} \nonumber \]
It represents the average book weight per unit length. Similarly, the total weight of the books is equal to the value of the distributed load times the length of the shelf or
\begin{align*} W \amp = w(x) \ell\\ \text{total weight} \amp = \frac{\text{weight}}{\text{length}} \times\ \text{length of shelf} \end{align*}
This total load is simply the area under the curve \(w(x)\text{,}\) and has units of force. If the loading function is not uniform, integration may be necessary to find the area.
Example 7.8.2. Bookshelf.
A common paperback is about \(\cm{3}\) thick and weighs approximately \(\N{3}\text{.}\)
What is the loading function \(w(x)\) for a shelf full of paperbacks and what is the total weight of paperback books on a \(\m{6}\) shelf?
- Answer
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\begin{align*} w(x) \amp = \Nperm{100}\\ W \amp = \N{600} \end{align*}
- Solution
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The weight of one paperback over its thickness is the load intensity \(w(x)\text{,}\) so
\[ w(x) = \frac{\N{3}}{\cm{3}}= \Nperm{100}\text{.} \nonumber \]
The total weight is the the area under the load intensity diagram, which in this case is a rectangle. So, a \(\m{6}\) bookshelf covered with paperbacks would have to support
\[ W = w(x) \ell = (\Nperm{100})(\m{6}) = \N{600}\text{.} \nonumber \]
The line of action of this equivalent load passes through the centroid of the rectangular loading, so it acts at \(x = \m{3}\text{.}\)
Equivalent Location
To use a distributed load in an equilibrium problem, you must know the equivalent magnitude to sum the forces, and also know the position or line of action to sum the moments.
The line of action of the equivalent force acts through the centroid of area under the load intensity curve. For a rectangular loading, the centroid is in the center. We know the vertical and horizontal coordinates of this centroid, but since the equivalent point force’s line of action is vertical and we can slide a force along its line of action, the vertical coordinate of the centroid not important in this context.
Similarly, for a triangular distributed load — also called a uniformly varying load — the magnitude of the equivalent force is the area of the triangle, \(bh/2\) and the line of action passes through the centroid of the triangle. The horizontal distance from the larger end of the triangle to the centroid is \(\bar{x} = b/3\text{.}\)
Essentially, we’re finding the balance point so that the moment of the force to the left of the centroid is the same as the moment of the force to the right.
The examples below will illustrate how you can combine the computation of both the magnitude and location of the equivalent point force for a series of distributed loads.
Example 7.8.3. Uniformly Varying Load.
Find the equivalent point force and its point of application for the distributed load shown.
- Answer
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The equivalent load is \(\lb{30}\) downward force acting \(\ft{4}\) from the left end.
- Solution 1
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The equivalent load is the ‘area’ under the triangular load intensity curve and it acts straight down at the centroid of the triangle. This triangular loading has a \(\ft{6}\) base and a\(\lbperft{10}\) height so
\[ W = \frac{1}{2} b h =\frac{1}{2}(\ft{6})(\lbperft{10}) =\lb{30}. \nonumber \]
and the centroid is located \(2/3\) of the way from the left end so,
\[ \bar{x} = \ft{4}\text{.} \nonumber \]
- Solution 2
Distributed loads may be any geometric shape or defined by a mathematical function. If the load is a combination of common shapes, use the properties of the shapes to find the magnitude and location of the equivalent point force using the methods of Section 7.5. If the distributed load is defined by a mathematical function, integrate to find their area using the methods of Section 7.7.
A few things to note:
- You can include the distributed load or the equivalent point force on your free body diagram, but not both!
- Since you’re calculating an area, you can divide the area up into any shapes you find convenient. So, if you don't recall the area of a trapezoid off the top of your head, break it up into a rectangle and a triangle.
Distributed Load Applications
Once you convert distributed loads to the resultant point force, you can solve problem in the same manner that you have other problems in previous chapters of this book. Note that while the resultant forces are externally equivalent to the distributed loads, they are not internally equivalent, as will be shown Chapter 8.
Example 7.8.4. Cantilever Beam.
Find the reactions at the fixed connection at \(A\text{.}\)
- Answer
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\begin{align*} A_x\amp = 0\\ A_y \amp = \N(16)\\ M \amp = \Nm{64} \end{align*}
- Solution
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Draw a free body diagram with the distributed load replaced with an equivalent concentrated load, then apply the equations of equilibrium.
\begin{align*} \Sigma F_x \amp = 0 \amp \amp \rightarrow \amp A_x \amp = 0\\ \Sigma F_y \amp = 0 \amp \amp \rightarrow \amp A_y \amp = \N{16}\\ \Sigma M_A \amp = 0 \amp \amp \rightarrow \amp M_A \amp = (\N{16})(\m{4}) \\ \amp \amp \amp \amp \amp = \Nm{64} \end{align*}
Example 7.8.5. Beam Reactions.
Find the reactions at the supports for the beam shown.
- Answer
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\[ B_y = F_y = \lb{295}, B_x = 0 \nonumber \]
- Solution 1
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\begin{align*} \sum M_B \amp = 0\\ +(\lbperin{12})(\inch{10}) (\inch{5}) -(\lb{100}) (\inch{6})\\ -(\lb{150})(\inch{12}) -(\lb{100}) ( \inch{18})\\ +(F_y) (\inch{24}) - (\lbperin{12}) (\inch{10}) (\inch{29})\amp = 0 \rightarrow \amp F_y \amp= \lb{295}\\ \\ \sum F_y\amp = 0\\ -(\lbperin{12}) (\inch{10}) + B_y - \lb{100} - \lb{150} \\ - \lb{100} +F_y - (\lbperin{12})( \inch{10})\amp = 0 \rightarrow \amp B_y\amp= \lb{295}\\ \\ \sum F_x \amp = 0 \rightarrow \amp B_x \amp = 0 \end{align*}
- Solution 2
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1. The two distributed loads are \((\inch{10}) (\lbperin{12}) = \lb{120}\) each.
2. The total downward force is
\[ W = (2 \times \lb{120}) + (2 \times \lb{100}) + \lb{150} = \lb{590} \nonumber \]
3. Since the beam and loading are symmetrical supports \(B\) and \(F\) share the load equally, so
\begin{gather*} B_y = F_y = \frac{\lb{590}}{2} = \lb{295} \end{gather*}
4. There are no horizontal loads acting on the beam, so
\begin{gather*} B_x =0 \end{gather*}