Example for solving reactions:
Now let’s do a simple example, to better digest everything we have learned so far… Let’s calculate reaction forces on a beam, based on given external forces. Assuming you are not an engineer, if you do not understand this example, do not spend too much time on it, although it is written for a non engineer, in a way that a person who knows basic high school math, and someone who read the previous posts attentively would probably be able to understand. Just be sure you grasped whatever you could from the previous posts, and that will be fine for this section.
In the figure below, a beam is shown, with a force acting on it, and has two supports at its ends.
The support at point A is a pin (hinge) support and the support at point C is a roller support. This beam is simply supported. Simply supported means, a hinge connection on one end and a roller connection on the other end, which is the minimum way we can support a member, without any redundant reaction, but just enough to resist forces in all directions and moments (we say moments too because altough neither of support can resist moment, the vertical reactions they create cause a net moment to cancel out the moment caused by the external force).
In our example the magnitude of the force is given as 20 N, and it makes 30 degrees angle with the horizontal x axis, and it acts 4 meters from point A and 6 meters from point C, as shown in figure.
As we have learned in previous pages, in order for this beam to be at rest, the net force and moment on it must be zero, in all directions.
forces in x direction (horizontal) must be zero, Fx = 0,
forces in y direction (vertical) must be zero, Fy=0,
and the moment must be zero Mz=0.
By the way, why did we write z for moment? Because the moment M is as if rotating around z axis, so that it can fall on our paper plane. Note that this is a 2 dimensional problem, so we have x and y directions on paper, and, a z direction perpendicular to the paper, which would make 3rd dimension. And only if the moment rotates around that z axis, it can fall on our xy plane. Try to visualise it in your mind.
Now let’s determine the reaction forces. To determine the reaction forces, let’s draw the free body diagram of this beam.
Here we have removed the external supports, and instead, we have shown the reaction forces they would create. As we explained above, the pin (hinge) support will create both x and y reactions, and the roller support can only create a reaction in y direction.
Here you also see that we have reduced the inclined force of 20 N, to its perpendicular components, from simple geometry of a right angled triangle, so that we only deal with forces in x and y directions. We normally do not show the original inclined force here, and only show its components, but here we showed as grayed out, so that you can understand, where these perpendicular forces at point B are coming from.
Now, first let’s calculate the x and y components at point B.
From simple trigonometry, Fbx must be (Fbx means, force F at point B in x direction):
Fbx = F . cos30 = 20 x 0.87 = 17.3 N
and Fby must be
Fby = F . sin30 = 20 x 0.5 = 10 N
Again, for the body to be at rest, the force components on all axes must be zero.
Fx Total = 0 and Fy Total = 0
For forces in x direction, we have only Fbx and Fax.
Therefore Fbx + Fax = 0
We already know Fbx, as we found above. So,
Fax + 17.3 = 0
Fax = – 17.3 N
Note that the sign is minus, which indicates it is in opposite direction to what we considered as plus. It doesn’t matter which direction you choose as plus (here we had taken direction to the right as plus, so this means horizontal reaction at A is acting towards left, since it came as minus)
Now let’s try to find Fay and Fcy. All forces on y direction must also be equal to zero. So,
Fay + Fby + Fcy = 0
We already know Fby, which is 10 N.
so we get
Fay + 10 + Fcy = 0
We have only one equation but two unknowns. We must have one more equation.
That additional equation comes from moment. Remember that we said for a body to be at rest, not only forces, but moments must also cancel out and the resultant moment must be zero. So, if we were to take moments of forces with respect to any desired point on this beam, and make it equal to zero, we will have our additional equation.
Let’s consider the moment with respect to point C for simplicity. With respect to point C, the force at C does not create a turning effect. Because it acts directly on C. So we can leave that one out of our new equation. Furthermore, we can also leave out Fax and Fbx, because their lines of action, pass through point C, and they have no distance with point C, so they also cannot create any moment with respect to C. So we leave those out too. Then, the only forces that can cause moment around point C, is, Fay and Fby and their total moment must add to zero.
Remember, we did not have to choose point C, to calculate the turning effect. We could have chosen anywhere on the beam, since no part of beam is turning, and we will get same results, it would be just with more equations, if we chose another point. We always want to solve in the simplest way possible.
So let’s write our new equation:
Fay . 10 + Fby . 6 = 0
Remember that for moment we multiplied Fay by its distance to C, which is 10, and Fby by 6 meters.
We already knew Fby, which was 10 N.
Fay.10 + 10×6 = 0
This gives us
Fay = -6 N
Note that it is minus, which indicates that it is in opposite direction to where we took as plus, which was downwards (we wrote Fby which was downwards as plus 10, not minus), so it means Fay is upwards. If we put this in the other equation we solve Fcy as:
-6 + 10 + Fcy = 0,
Fcy = -4 N and which means it is upwards. (If we show it upwards as below, no need to write minus)
So we can show the result as in the figure below:
If you missed some of the details above no problem, just compare figure statics 2a with 2c now, and you will see the big picture.
In the next post, we will introduce the term “Axial force, shear force, bending moment diagrams”