For many engineers, the ability to analyze the internal forces of bending members is probably one of the most important skills that you can develop, partly because many structures and objects in the real world aren't just axially loaded. In fact, the beams in the floor you're sitting on (assuming it's not the ground floor) or even the rafters in your roof are all examples of bending members.
In this chapter, I help define what a bending member is, and I show you what to look for when you identify it. When you know you're dealing with a bending member, you're ready to actually compute the magnitudes of internal bending forces, and I show you how. I also decipher yet another sign convention (assumption) and show you the steps to develop generalized equations for internal loads. Finally, I show you one of the most important tools for engineers: the ability to quickly draw diagrams of these internal forces.
In Chapter 19, I show you how to handle truss systems, which are comprised of members subjected to only axial internal forces, which cause a member to become longer in tension and shorter in compression (see Figure 20-1).
But what happens when a member isn't loaded with just an axial load? Figure 20-1 also shows a point load that is acting perpendicular to the member's longitudinal axis. When this type of member (also known as a bending member) is loaded, it wants to deflect in the direction of the applied load, but not in the direction of the longitudinal axis of the member. As a result, axial tension or compression are no longer the only forces that appear internally.
To start the investigation, you need to cut the bending member to expose the juicy goodness inside, as I show in Figure 20-2 at an arbitrary distance x from the left support. When you cut a bending member, instead of just an axial force, three internal loads will appear to help maintain equilibrium.
Axial: By inspecting Figure 20-2a, to balance the horizontal reaction Ax, there must be an internal axial force Nx that acts parallel to the longitudinal axis to ensure translational equilibrium in this direction.
Shear: By inspecting Figure 20-2b, you can see that without the presence of the force Vx, the free-body diagram (F.B.D. — see Part IV) of the object can't be in translational equilibrium in that direction. This internal force Vx is also known as a shear force, and it acts perpendicular to the longitudinal axis of the member.
Moment: From Figure 20-2c, you can see that without the presence of the moment Mx the F.B.D. can't be in rotational equilibrium about Point X. The effect of the vertical support reaction and the shear force actually causes a rotational behavior that must be balanced by the internal moment. The contribution of the axial force Nx to the internal moment is often neglected because the perpendicular distance, Δ, is generally very, very small.
As I show in Figure 20-3, in three dimensions, six internal loads appear on every exposed surface (or section) — one for each equation of equilibrium (flip to Chapter 16).
To balance the exposed internal loads in a three-dimensional F.B.D., you need three noncoplanar translational forces to provide the necessary forces for equilibrium. You always have an axial force Nx that acts along the longitudinal axis of the member, and you also have a vertical shear force, labeled Vy. The third internal force Vz also acts perpendicular to the longitudinal axis — which makes it an additional shear force (refer to Figure 20-3).
There must also be three rotational equilibrium requirements in a three-dimensional free-body diagram (refer to Figure 20-3). Just as with two- dimensional problems, a moment about the z-axis Mz is necessary. There must also be a torsional moment (sometimes referred to as a twisting moment) about the longitudinal axis Mx. The third moment that is required is an additional bending moment, My.
When beams are designed to be used in buildings, these internal loads are what engineers must design for. Engineers must follow local building codes and guidelines and have a thorough understanding of the materials that they are working with. At this point, I just stick with showing you how to crunch the numbers.
To calculate internal loads at a point, you simply cut your structure at the point of interest, draw the F.B.D. (including the newly exposed internal forces), and then apply the equations of equilibrium (sound familiar?). However, with internal loads, you want to be very mindful of the sign convention that you select.
When you expose an axial force on an axial-only member, the sign of the force depends on which side of the cut (or exposed face) you're working with. An axial force that is positive in one direction for one F.B.D. would be negative for the F.B.D. on the other side of the cut line (or exposed surface). Figure 20-4 shows the sign convention for two-dimensional bending members.
Axial forces: The sign of the internal axial forces depends on whether the object of the F.B.D. is subject to a tensile or compressive load. The sign convention for axial forces from bending effects is the same. Tension is positive and compression is negative.
Shear forces: If the force is acting on the left side of an interface or exposed face, it's positive if it acts upward. If the force is acting on the right side of the interface or exposed face, it's positive if it acts downward.
Bending moments: A positive bending moment on each end of a section causes a beam to flex downward in the middle. A clockwise moment on the left side of the interface (or cut line) is considered positive, while a counterclockwise moment on the right side of the cut line is also considered positive.
Now, if you picture the positive moment deflected shape as the mouth, and the (+) indicator as a nose, all you need are a couple of eyes on your sketch, and you'll have created a happy face. That's why you can say, "Positive moments make you happy!"
Often, you don't know the direction of an internal load until after you perform your equilibrium calculations. Just make sure to include the internal loads on your F.B.D.s at the beginning, and let your equilibrium calculations confirm your assumed directions. It usually helps to be consistent, so normally you want to assign the directions of your internal forces based on the positive sign convention that I discuss in the preceding section.
Consider the basic portal frame of Figure 20-5 which experiences a 50-kiloNewton (kN) force, a 2 kN per-meter distributed force, and a 40 kN per-meter concentrated moment. Suppose you want to compute the internal forces acting at Point E of the beam BC. In this example, the support reactions and directions at Points A and D have already been shown and are included on the drawing. Self weight (gravitational effects on the system) isn't a concern for this problem.
To calculate internal forces, the first step is to cut the beam at the location of interest (Point E, for this example), and draw a free-body diagram of one of the pieces that remains. F.B.D. #1 in Figure 20-6 shows the left half of the structure with all the applied external loads, support reactions, and appropriate dimensions. Remember, you also need to include the internal forces -— an axial force, a shear force, and an internal moment.
The assumed direction of the internal loads is determined by looking at the cut line. From the F.B.D. #1 of Figure 20-6, because all of the internal loads are acting to the right of the cut line, you look at the "right" side of the sign convention diagram in Figure 20-4 for each of the internal forces.
The axial force is assumed to be positive if it's pulling (or acting tensile) on the longitudinal axis of the cut member. In this case, NE represents the axial force on F.B.D. #1 and is acting to the right. In order for the internal shear force VE (which is acting on the right side of the cut line) to be positive, it must be acting downward (which is perpendicular to the longitudinal axis). The bending moment ME (also on the right side of the exposed face) must be acting in a counterclockwise direction to be positive.
For reference, F.B.D. #2 in Figure 20-6 shows the other half of the structure, with the appropriate loads, moments, support reactions, and dimensions already included on that F.B.D. For this diagram, the same internal loads have been revealed, but this time they're acting on the left side of the cut line which means they're applied with their senses reversed from those of F.B.D. #1. On F.B.D. #2 of Figure 20-6, the axial load NE, is in tension when it acts to the left, the internal shear VE is assumed positive when it acts upward, and the bending moment ME is now acting positively in a clockwise direction.
Tip
Notice that the internal loads drawn on a F.B.D. on one side of a cut line are always equal and opposite to the internal loads drawn on the F.B.D. on the other side of the cut line. This ensures that the equilibrium is maintained at the cut line.
With the internal forces drawn, you're now ready to employ the equations of equilibrium that I show you in Part V. For now, I use F.B.D. #1 of Figure 20-6 for my calculations.
Note
It really doesn't matter which of the two free-body diagrams of Figure 20-6 you choose to work with. Assuming that you've calculated the support reactions correctly and that you have all the applied loads on both free-body diagrams and at their proper locations, the numerical calculations you perform produce the same results.
I start by summing forces in the x-direction (assuming to the right as positive).
At Point E, the axial load, NE= 0 kN. Don't worry that the axial value is computed to be zero. Numerically, internal forces are permitted to be zero. Next, you compute the internal shear force at Point E:
The negative value on the magnitude of the shear VE indicates that the direction that was assumed on the free-body diagram. Finally, you compute the internal moment at Point E.
The magnitude of the internal moment, ME, is computed to be +161.33 kN-meters. Because this value is positive, the assumed direction is correct.
I can also compute the magnitude of the internal moment, ME, using F.B.D. #2:
Thus, the internal bending moment calculations produce the same result, regardless of which F.B.D. you decide to work with. The only difference is that the directions (or the senses) of the forces are equal and opposite.
If you want to calculate the magnitudes of internal forces at specific points, the procedure I outline in the preceding section is more than adequate. Unfortunately, if you want to properly design a beam or a column, you also need to determine the maximum and minimum values of axial, shear, and moment that your design will safely support. The problem is, you may not actually know where these locations are. So what are your options?
One option you may choose is to cut the same member at a hundred different locations and perform the same calculations over and over and over. . . . (This task isn't exactly the way I would want to spend a Saturday afternoon.) A better option is to try to formulate a more general set of expressions for internal loads along the length of the bending member.
Generalized internal load equations are typically created for shear and moment at all locations along the length of the member. In the following section, I show you how to create a set of algebraic functions as a function of location in the member.
The generalized equations are valid over distinct regions that occur between critical points, which are locations on a beam where a change to the shear, the moment, or the relationships between them occurs. Critical points may occur at the following locations:
Start and end of the bending member: Both ends of the member are considered as critical points.
Support reactions and internal hinges: All points where support reactions and internal hinges occur are critical points.
All concentrated forces: All points where concentrated forces occur are critical points.
All concentrated moments: All points where concentrated moments occur are critical points.
Beginning and end of all distributed loads: All points where a distributed load starts or stops are considered critical points.
Tip
Multiple critical points can (and often do) occur at the same location on your beam.
Consider the simply supported beam shown in Figure 20-7. It's subjected to a linearly distributed load with a maximum intensity of 50 pounds per linear foot at the right end. The slope of this distribution is 50 pounds per linear foot ÷ 20 feet = 2.5 pounds per square foot. Also acting on this beam is an 800-pound concentrated force at 6 feet from Point A and a concentrated moment of 2,000 pound-feet acting at 5 feet from Point B.
Figure 20-7 also shows the F.B.D. of this example, with all loads and moments already included at their appropriate locations on the beam. The support reactions have already been calculated.
The critical points for this structure are indicated by dashed lines on the F.B.D. of Figure 20-7. One critical point occurs at Point A because that's both the left end of the beam and the location of a support reaction. A second critical point occurs at the location of the 800-pound concentrated load. A third critical point occurs at the 2,000-pound-foot concentrated moment. The last critical point occurs at Point B because that is both the right end of the structure and the location of a support reaction.
The generalized equations for internal forces of bending members are valid within specific regions which are shown in the Figure 20-7 example, between those dashed lines that you created for the critical points in the last section. I list them in that figure as Region #1, Region #2, and Region #3. What these regions allow you to do is create a generic F.B.D. that is valid for all points within those regions.
First you need to establish a reference location that remains unchanged for each region. I usually choose to take this reference point as the left end of the structure, but you can choose any point you want — just don't move it! You can then draw F.B.D.s specific for each region, as I've done in Figure 20-8.
For Region #1, you cut an arbitrary location at an arbitrary distance, x, from your reference. This F.B.D. is valid for all values of x between 0 feet and 6 feet. Whether x = 1 foot or x = 3.697 feet, this F.B.D. looks exactly the same. The only parameter that changes is the x dimension or the location of the generalized section. Knowing this, you can then employ the equation of equilibrium and write expressions for the shear and moment:
This expression allows you to compute the value of the shear for every location value of x within the region. Similarly, you can compute the generalized moment equations as
Repeating the process of writing equilibrium equations for each F.B.D. of Region #2 and Region #3, you can easily develop the generalized equations. For Region #2,
And for Region #3,
These equations represent the generalized equations for internal bending forces. Instead of having to perform a ridiculous number of internal load calculations with a large number of free-body diagrams, you can now create a much smaller number of algebraic expressions that fully define the values of the internal loads within the specific regions.
The best part about developing mathematical expressions for the generalized equations (which are actually mathematical functions) is that you can now utilize basic calculus principles to find maximum and minimum values for each of those functions.
With calculus, you can also discover an interesting relationship between the function for shear and the function for moment. If you examine the generalized equations for shear and moment for each region, for Region #1 of Figure 20-8, you can compute the first derivative of the internal bending moment Mx.
If you recognize the result of this operation, it's because this derivative actually ends up being the exact same equation as the internal shear function Vx for the same region. You can verify that this relationship is valid for all regions and can thus be defined as
Note
The first derivative of the generalized moment function is equal to the generalized shear function over the same interval. Although it may not seem like a big deal at this point, this equation is fundamentally crucial in the advanced study of structural analysis and mechanics/strength of materials classes. This equation also serves as your graphical basis for establishing a shortcut method for creating shear and moment diagrams in the next section.
Another neat feature with the generalized equations is that you can now also calculate the minimum and maximum values for shear and moment over a given region. You've gone to all the work to create nice continuous algebraic functions for shear and moment. You can now apply the principles of calculus that I discuss in Chapter 2 to find the maximum and minimum values of the shear and moment function, as well as their locations.
Another useful result of having these algebraic expressions for the generalized shear and moment equations is that you can now plot their functions on a computer (or by hand) and create a complete shear and moment diagram that shows you every value of the internal shear and moments along the length of the bending member. These shear and moment diagrams are among the more important tools an engineer has available for design.
When you compute the generalized internal force equations in the preceding section, you see that those calculations aren't terribly difficult, but they can take a good amount of time (especially if your problem has a lot of critical regions). A faster method is definitely worth investigating.
To help explain a new method based on area calculations, check out Figure 20-9, which contains a simply supported beam with a cantilever at the left end and is loaded as shown. The reactions have already been provided and are indicated on the figure.
To use the area methods, load diagrams must be fairly simple and consist of only point loads, concentrated moments, and uniformly distributed loads. Any higher-order distributed loads, and the geometry becomes less than friendly and you're better off just creating the generalized equations.
Note
When you're working with area methods, keep in mind the following:
Construct the framework for the graphs first and align them vertically. Place the load diagram on top, with all reactions shown as point loads or concentrated moments. Next, place the shear diagram directly below the load diagram and then place the moment diagram directly below that.
Obey the sign convention, and locate all critical points.You use the sign convention for the left end of the beam, as shown in Figure 20-4. Locate all critical points as I outline in "Defining the critical points" earlier in this chapter.
Finish the shear diagram first.You must complete the shear diagram before you can start the moment diagram.
Start from the left and work to the right.This method allows you to use the sign convention as described in the preceding bullet. Place points at V = 0 and M = 0 on the left end of each diagram.
The shear and moment diagrams you draw, and the calculations you perform, must compute to zero at the right end of the diagram. If you don't calculate a zero when you reach the end, you've made an error somewhere along the way. What's worse is that if your shear diagram doesn't equal zero at the end, you're guaranteed to have an incorrect moment diagram.
The first diagram that you construct after drawing the loading diagram (or free-body diagram) is always the shear diagram. The shear diagram is built exclusively from the free-body diagram of the entire system, including external applied loads as well as the computed support reactions. If you make a mistake in finding the support reaction calculations, chances are your shear diagram will be incorrect as well.
Using the example in Figure 20-9, you can compute the shear diagram in the following basic steps. Check out Figure 20-10 for a visual.
Starting at Point 1, place a point at V = 0 and then examine the first critical region (between Point 1 and Point 2).
If there are no loads acting in this region, the shear remains unchanged, so because the start point is at V1 = 0, the unchanged end point will be at the same value, or V2 = 0. There is a concentrated moment at Point 1, but remember that moments don't affect the shear diagram, so just leave it alone for now.
At Point 2 place a point at a shear value of V2 = 0; draw a line to connect Points 1 and 2.
At Point B, a reaction of 13.5 kN is acting upward on the beam. This reaction is the same as a concentrated load, which will cause the shear diagram to jump instantly.
Starting at the value of V2 = 0 at Point 2, add another +13.5 kN to that value (for V3 = 13.5 kN total) and place a point for V3 = 13.5 kN on the graph.
This becomes the new value for Point 3.
Beginning at Point 3, which has a value of V3 = +13.5 kN, you can see that this region is subjected to a uniformly distributed load. The area under this load (or the resultant) is equal to the change in shear value and helps you calculate the shear at Point 4. For this example, the resultant of the distributed load on this region is (−4 kN per meter)(8 meters) = −32 kN. Because this uniform load is acting downward, the resultant must also be acting downward. This fact means that the total change from Point 3 to Point 4 must be −32 kN. The value of shear at point 4 is then (+13.5 kN – 32 kN) V4 = −18.5 kN.
But be careful: The shape of the shear function between Points 3 and 4 is dependent on the shape of the load between those points. The order of the shear function is always one order higher than the load function for the same interval. Thus, a uniform distributed load (with an order of zero) results in a linear (or first order shear function). So a straight line connecting Point 3 and Point 4 is correct. If this load had been linear (first order), the shear function would have been quadratic (or second order).
At this point, you're standing at the end of the beam at Point 4, and sitting on a value of −18.5 kN. But the previous section says that you must end at a value of zero. So what happened? Even though you've reached the end of the beam, remember that there is also a vertical reaction, Cy =+18.5 kN. So the value of Point 5 is equal to the value of Point 4 plus the effect of the concentrated point load. Thus, the shear at Point 5, V5 = −18.5 kN + 18.5 kN = 0, which means the shear diagram ends on a zero value. Good news!
Denote any areas of positive shear with a plus (+) sign inside the region and areas of negative shear with a negative (–) sign.
This notation helps with the moment calculations I discuss in the next section. I also like to shade the areas to make them a bit more visible.
Look for a secondary critical point that occurs at locations of zero shear.
At any locations where the shear diagram has a value of zero, you need to add a new critical point if there isn't one there already. In this diagram, Point 1, Point 2, and Point 5 are already critical points. However, there is another place where V = 0, somewhere between Point 3 and Point 4. At this location, add a dotted or dashed line and draw it down into the moment diagram, as shown in Figure 20-10. These critical points are locations of maximum or minimum moments.
To calculate the location of this point, you need to employ a bit of geometry. In this example, because the shear function is linear at this location, you can use the concept of proportions or similar triangles to determine the distance of the new critical point from Point 3. Using a variable x to denote this location:
Note
You use this dimension when you calculate the points on the moment curve in the following section. If the functions aren't linear, you need to formulate a generalized shear equation, set the equation equal to zero, and then solve for the x-dimension that satisfies that condition.
When the shear diagram is complete, you're ready to begin drawing the moment diagram. The moment diagram is based directly off the shear diagram, and all your calculations come from the shear diagram that you previously created. The only exception is the presence of concentrated moments, which cause a jump in the moment diagram. You need to look at the original load diagram to find them!
Using the example earlier in Figure 20-9, follow these steps to create a moment diagram:
Place a point at the left end of the moment diagram at a value of M6 = 0.
This becomes Point 6 in this example. At this critical point, the beam also has a concentrated moment in the amount of 20 kN-meters acting clockwise.
Note
A clockwise moment is a positive moment. This means that Point 7 has a value of M7 = (0) + 20 kN-meters = +20 kN-meters.
Place a new Point 7 at M7 = +20 kN-meters.
Compute the change in moment between Point 7 and Point 8 by calculating the area under the shear diagram between the same two critical points (in this case it's the area under the shear diagram between Point 1 and Point 2).
Because the shear is zero in this region, the change in moment between Point 7 and Point 8 is (0 kN-meters)(2 meters) = 0 kN-meters. The value at Point 8 is then M8 = 20 kN-meters + 0 kN-meters = +20 kN-meters. Because there is no shear between Point 7 and Point 8, the moment must remain constant.
Starting at the value of M8 = +20 kN-meters, compute the change in moment by calculating the area under the shear diagram for the next region.
The final value of Point 9 is then M9= +20 kN-meters + 22.78 kN-meters = +42.78 kN-meters. The positive area under the shear diagram means that the moment will increase. So now you have the two endpoints declared, but what does the moment function between them look like?
At this point, you need to do a bit of detective work. The first clue is in the shape of the shear diagram. Remember that moment diagrams are always one order higher than the shear diagram in the same region. So, if the shear diagram is linear (first order), as in this case, the moment diagram must be second order (or one order higher).
But that leaves you with another dilemma. It takes a minimum of three points to define a second-order curve, and you only have two so far! This discrepancy means there are two second-order curves that can possibly fit between Point 8 and Point 9 (see Figure 20-11).
To deduce which second-order curve actually fits, you need to look at the slope of the moment diagram at each point. Recall that the slope of the moment diagram is equal to the value of the shear at that point. Thus, the slope of the moment diagram at Point 8 is +13.5. The slope at Point 9 is 0. Laying these two slopes on the two possible second-order curves reveals that curve #1 is the right shape and is the only one that is possible. It has the highest positive slope on the left end of the region, where the shear was the highest positive value.
Be careful, though — this second-order curve won't necessarily work for every linear shear function. You have to look at the slopes to make the decision!
Starting at a moment value, M9 = 42.78 kN-meters, compute the change in moment from Point 9 to Point 10 as the area under the shear diagram in this region.
Because the area under the shear curve is negative, you should expect that the change in moment will also be negative.
Thus, the moment value at Point 10 is (M10 = +42.78 kN-meters – 42.78 kN-meters) = 0 kN-meters. Hence, the moment diagram ends at a value of zero. You can deduce the shape of the second-order curve in the same manner as in the preceding step. Because the final value was zero, this indicates that the work you did is most likely correct. Be sure to label and shade your positive and negative moment regions as a useful reminder when you're done.
Look at your diagrams and declare the maximum and minimum shear and moment values.
By looking at the shear diagram, you can see that the maximum positive shear VMAX+ = +13.5 kN at Point B (or Point 3) and the maximum negative shear VMAX− = −18.5 kN at Point D (or Point 4). Similarly, the maximum positive moment, MMAX+ = +42.78 kN-meters at the new critical point (or Point 9) and the maximum negative moment, MMAX− = 0 kN at both Point"6 and Point 10. With these values determined, you're ready to begin designing this beam — but I'll save that discussion for another book.
