Deriving internal forces: Segmenting the beam
Since, as we noted earlier, both f and M are internal, they will not appear in any equations that deal with the whole beam. We find expressions for Fa and Fb without reference to these internal forces.
To calculate the sizes of the internal forces, we need an equation which features them, alongside Fa and Fb etc. We must therefore consider a segment of the beam, hypothetically split from the rest. At this split end, there will be the internal force resulting from the effect of the rest of the beam on the segment under consideration.
For example, we may first consider the situation in a segment AP, where the point P lies somewhere within the left half of the beam. Thus AP will involve the external support force Fa, and some proportion of the beam's weight, but not Fb. As we will discover later, the situation is somewhat different in the right half of the beam, due to the opposite 'sense' (anti-clockwise versus clockwise) of the bending moment in that half.
At this first stage, we will assume that the moveable 'point' load V is positioned at the centre of the beam's length (i.e. d = 5). This can be changed later.
We haven’t yet included the weight of the beam segment AP, which will be a proportion of the overall beam weight, according to its length x. If the beam has a weight distribution of 1 per unit length, as we have taken it so far, then an x unit length will simply weigh x.
This segmental weight will act at its centre of mass, a distance x/2 from both A and P. Note that the moveable load V doesn't appear, as we haven't got to that part of the beam yet.
If we take the situation at the end of section 6 as our initial ’loaded beam’ setup, then W = 10, and V = 20. The formulas gave Fa = 25 – 2d, and Fb = 5 + 2d.
In our present case, we have the load in the centre, so d = 5, and Fa = Fb = 15.
Resolving forces vertically:
Fa + f = 0
So, f = x – Fa = x - 15
Taking clockwise moments about P:
M + (Fa . x) – [x. (x/2)] = 0
So, M = 0.5x2 – 15x