So far you have learned very important and fundamental concepts in statics. Now lets go one step further and introduce another fundamental subject. The force and moment diagrams.
In the example in the previous post of this series, we had a beam, which was shown with its supports and an external force acting on it, and then we calculated the reaction forces, at the support locations. But this is usually not enough for us. What if we wanted to know the value of forces and moments not only at supports, but anywhere on the beam? We can do so by showing these on force and moment diagrams.
Look at the figure statics 2c in the previous post again.
Now, we know that this beam is stable and is not moving. So, any part of this beam is also not moving and stable. What if we were to take only a portion of the beam, right of point A? Since that portion is also not moving, we could show the forces and moments only for that portion, so that it will not move. Let’s take a 1 meter portion of the beam, to the right of point A. It would look like this:
From now, assume x = 0 at point A.
At x = 1 meter:
At 1 meter right of point A, the horizontal force must be equal to horizontal force at A and opposite in direction so we have 17.3 N in opposite direction. Similarly, the vertical force must be equal and opposite direction for satisfying equilibrium. So, that is 6 N and pointing downwards.
But we are not done yet. Although we cancelled out forces in x and y directions, there is a turning effect now. and we need to insert a moment at this point, to counter that effect. So why did we need the moment M here? Because although the forces in x and y directions cancel each other out, there is still the turning effect, since forces in y direction are not on the same line but they have distance between each other. When there is force and distance it means moment. So they will try to rotate the beam. So we must counter that and insert a moment at this new location.
Then, if we take moment with respect to point A for instance, both horizontal forces pass through A, so they have no effect, the Fy at point A has no effect as it also passed through point A, so all we have remaining is the force Fy at the point 1 meter right of A. So that force creates a moment of 6N x 1 meter = 6 Newton-meter = 6Nm. This moment happens at that point 1 meter right of point A.
We could have simply reached this 6N.m answer, just by multiplying the value of Fy force with distance between them, because they are a moment couple (equal value with opposite direction, but with a distance between them), so we could have simply said 6×1 = 6 N.m, without taking moment with respect to anywhere, because nothing else affects the moment here anyway.
As you can see, although in the final result of previous example, we saw that there was no external moment on the beam, now we see that there is internal moment inside the beam, just because of these external forces.
At x = 2 meters:
Now assume we take a 2 meter portion of beam. With the same logic, all forces in x and y directions will cancel out, and since reaction A does not change, forces Fx and Fy is again the same. But now, we have more distance between the y direction forces. This means, we have more moment. The moment value of the couple is now increased, as they have now 2 meters between each other.
So, M = 6×2 = 12 N.m this time, for the point 2 meters right of the beam. Again, note that Fx and Fy are still the same.
At x = 3 meters:
Now lets move on to 3 meters right of point A. Now the moment will be M = 6×3 = 18 N.m. And the forces Fx and Fy still did not change.
At x = 4 meters (left of B):
Now let’s move on to 4 meters right of point A, but let’s stop, just before reaching to point B. Now we are at very slightly left of point B. Now the moment will be M = 6×4 = 24 N.m. And the forces Fx and Fy still did not change.
Now, let’s move just a little bit more towards right, a very, very small distance, so we are still at 4 meters, but now we are to the right of point B. Now the shape will look like this:
At x = 4 meters (right of B):
Here, we now include the external force that was being applied to point B. This causes a sudden change in forces Fx and Fy at the right face of the beam. Let’s take a look…
In x direction, the reaction force A was 17.3 N, to the left. Now that we have force Fb at point B, the x component of that force was also 17.3 N, but to the right. So these two cancel each other, and there remains zero force, for the internal force in x direction at the right face of the beam. So, suddenly, Fx became zero. And since we will have no other horizontal force until we reach the end, to point C, it means, the force in x direction will remain zero until the end. So, we can see that only the portion AB of the beam is in tension, with a force of 17.3 N, and the rest of the beam has zero axial force (axial means, in the direction of main beam axis, which is x direction here, and the graph they produce is called the axial force graph). This is logical because, remember, the roller support at point C could not produce any horizontal reaction force. So if there is no reaction force to hold it, there can be no action force to push it either, to satisfy equilibrium.
In y direction, just to the left of point B, just before we included the external force at B, Fy at the right end of the beam was 6N pointing down, to cancel out the upwards force of 6 N at point A. Now, when we moved slightly to the right of B, we suddenly include the downwards force of 10N at B. Now the balance changed. So now in y direction we have 6N pointing up at A, and 10 N pointing down at B. So we must have 4 N pointing up, to satisfy zero net force in y direction. So, when we moved from slightly left of B to slightly right of B, the Fy suddenly changed from 6N pointing down, to 4N pointing up. So there is a sudden jump here, and the amount of jump is equal to the external force at B in y direction. The forces in y direction is called the shear force. Because they are perpendicular to the main axis of the beam, as if trying to slide the two sections of the beam along each other (remember the definition of shear force). And the graph they produce is called the shear force graph.
And the moment? That remains the same for now. We moved slightly to the right of B, but the net force at B is still 6 N (as in 10-4), and therefore the couple moment is still 6×4=24N.m. Or, as we described before, another way to calculate the moment is to take moment with respect to point A from here, and you will reach to same result. So far, the internal moment had increased, until we reached point B. Now, it will start to decrease, as we continue to move towards right.
At x = 5 meters:
Lets continue towards right, to see how. Now let’s move 1 meter to the right of point B. Now when we take moment with respect to this point, the force at point A and force at point B are trying to turn this 5 meter long piece piece is in opposite directions. So for example, you will have 6N x 5 meters from point A, trying to turn it in clockwise direction, and you will have 10N x 1 meter from point B, trying to turn it in counter clockwise direction, so, 6×5 – 10×1 = 30-10 = 20 N.m, at a distance of x = 5 meters.
Fx will not change anymore, until the end which is point C, and remain as zero, as we explained above.
Fy will remain the same, as 4N pointing up, until it reaches point C, which only will then change, by a value equal to the force at C.
At x = 6 meters:
Again, for this 6 meter portion of beam to remain static, the net moment with respect to any point must be zero. So lets take a moment with respect to this point at x=6m. So from point A, we have 6N x 6 meters, trying to turn it in clockwise direction, and from point B, we have 10N x 2 meters, trying to turn it in counterclockwise direction, which makes, 36-20 = 16 N.m. As you see, the moment keeps decreasing. Fy and Fx remain just the same, as we met no more external effect.
At x = 8 meters:
Again, for this 8 meter portion of beam to remain static, the net moment with respect to any point must be zero. So lets take a moment with respect to this point at x=8m. So from point A, we have 6N x 8 meters, trying to turn it in clockwise direction, and from point B, we have 10N x 4 meters, trying to turn it in counterclockwise direction, which makes, 48-40 = 8 N.m. The moment decreased further, and we can also see that it is decreasing linearly, with constant slope (we will see it better when we draw the graph).
At x = 10 meters, which is also point C:
When we take moment to this point, from point A we have 6N x 10 meters, trying to turn in clockwise direction and fro point B, we have 10N x 6 meters, trying to turn it in counterclockwise direction, which makes, 60-60 = 0 N.m, which is what we expected, because point C is only a roller support point and can not resist any moment. Since it cannot resist any moment, no internal moment in the beam can exist there either. So we confirmed the accuracy of our solution of reactions this way too.
And as for Fx, it is still zero, as the roller support at C did not produce any horizontal reaction, and as for Fy, which was 4 N upwards, now we have 4N upwards reaction at point C, so now there remains 4-4=0 for Fy. In other words, while we still had the 4N upwards internal beam shear force, just before we reached point C, when we reach exactly to point C, there is already a 4N upwards reaction, so the shear force equals zero now as well, and the graph is closed.
Now lets put all these results into the graph form, which we call axial force, shear force and bending moment diagrams. Try to study this graph, in comparison to what we have described in previous paragraphs, and it should be self explanatory. See how each graph changes when there is an external effect, and what it becomes at supports. For a civil engineer, these were elementary but if you are not an engineer, you have just learned a very, very important concept in civil engineering, if you understood how these graphs are drawn. If not, try to find another example from a book or on the internet, there are millions of it, as these are the most basics, but what we described here should be self sufficient, with a fairly detailed explanation at each step. Even if you did not fully understand these examples and graphs, there is certainly nothing to worry about as far as understanding the rest of this post series, and you can just keep going.
Also notice in this graph, the following, which are very helpful when drawing these graphs:
- the area under shear diagram equals change in moment diagram
- the slope of moment diagram equals value of shear diagram (i.e. if shear is +, moment will increase or if shear is zero, moment slope will be zero such as at max. or min. moments)
- the change in shear equal area of load on beam (not applicable here, as we only have point loads, so shear never changed, except when we had point load as listed next item
- a point load on beam decreases or increases value of shear equal to load’s value
The positive side on these graphs are taken by convention, according to a beam segment and the forces acting on it as below being positive.
In other words, the internal forces Fx and Fy on the right face of our beam we kept finding for different x distance are positive, if they have the same direction of forces on the right face of these beam segment above. Similarly for the internal moment we kept finding for different x values, if they agree with the direction of the moment on the right face of this beam segment above, our moment is positive. Why right face? That is just how we chose to proceed in or example and that is all. We could have taken point C as x=0 and proceeded to the left, and this time we would have compared the internal forces we found on the left side of beam segment to the left face of the beam segment below, to determine positive or negative.
There is a simpler way to memorize this among engineers as, the moments on both sides of this beam segment above, make it bend down, as if it can hold water. Indeed, the bent shape of our beam looks like it is bent down, everywhere, which indicates a positive moment everywhere. (This certainly might not have been the case though. In a certain frame, a beam might bent up, from the effects of other members in a frame for example, and those bent up sections are said to have negative moment). As you can simply infer from here, where positive and where the negative bending takes place also affects our steel reinforcing in a concrete beam for instance, as concrete is weak in tension and those sections must be reinforced.
That is why are these graphs very important, because we can design all structural members such as beams or columns or slabs etc… only if we know the forces and moments acting on them and the forces and moments within them. Of course, in real life structures, we have far more complicated frames that the simple beam here, which require much more complicated methods to solve the reactions, but the main idea is as what you already learned, which is, finding the external reactions on the frame, and then, finding the internal forces and moments at any given point of any given member. We will not cover any of those more advanced situations or methods in this book, but at least you now know about the fundamental logic behind all, which is very important to know, for a civil engineer.
In the next post, we will introduce the term “Determinate and Indeterminate Structures”