Before proceeding with any calculations, we should chose in which coordinate system we will work. This is very important, because from what coordinate system we adopt depends on whether the acting force, for example on the beam, will have a positive (+) or negative (-) value. Let’s see how it works on an example.
We are currently accepting such a coordinate system.
Then, on the example of a simple cantilever beam, I will try to explain how it works. We have such a beam.
I will calculate support reactions, you don’t have to understand it for now, as you will see them in the next steps. Look at the forces’ returns, our coordinate system and the value they will have when calculating.
Let’s start with projecting forces on the X axis, wchich are horizontal forces.
Remember! In the coordinate system, the X axis is directed to the right, the equation will look as follows.
∑X = 0
Ha + 6kN = 0, what gives us
Let’s see the force of Ha is directed to the right (that’s how I took it), so the value is positive (+). The force (6 kN) acting on the beam is also directed to the right, which gives us a positive (+) value and all this should be equally zero.
Unknown to the left, known to the right, and we receive Ha of -6kN, which is balanced by horizontal forces. We will not go into it for now, because we are entering the topic of calculating supporting reactions, and here it is about understanding what turns of forces have to the coordinate system we have adopted.
If you do not understand that yet, let’s take up vertical forces.
Remember! In the coordinate system, the Y axis is directed downwards, the equation will look as follows.
∑Y = 0
–Va + 5kN = 0, what gives us
–Va = –5kN /*(-1)
Va = 5kN
Now verbally, the vertical support reaction Va is directed up (that’s how I took it), it has a negative value (-), the vertical force 5kN is directed downwards, which gives it a positive (+) value because our Y axis is pointing down and all this should be equal to zero. Again unknown to the left, known to the right, and get a supporting force Va of 5 kN.
In the beam there is still the bending moment Ma. We will not calculate it because it works on a different basis, what we will do in the next steps of introduction.
Great, can we go to next?