The scheme below represents a simply supported beam loaded by an evenly distributed transverse load across the entire length l of the beam. The distributed load is represented graphically by the light blue area and has the physical meaning of a force per length:
|p| = (F/l)
Its resultant, pil, is represented by the dark blue vector oriented downwards and applied at the midpoint of the beam.

We will take a horizontal unit vector u with positive orientation to the right side. The vertical vector can then be obtained by multiplying the unit vector by the unit bivector (ui); the opposite vector to this is then iu; A transverse force is in the same direction of this second---vertical---vector; this leads to the following result:
p = F (iu) / lu = (F/l) i
which means that in this case the distributed load is in fact a pure bivector; we will write it as (pi).

By enforcing equilibrium equations and because of the symmetric nature of the problem we find that the two reactions at the endpoints of the beam are equal:
2A + pil = 0 (vertical equilibrium) => A = - 1/2 pil

We observe that:
pil = p i |l| u = p |l| iu --> this is a vertical vector oriented downwards

We can therefore write:
A = - 1/2 pil = 1/2 pli --> this is a vertical vector oriented upwards

In the scheme we show the value of the bending moment M(x) computed at a generic location x from the left tip of the beam. This location is identified by the dark green point. The bending moment, according to its definition, is computed by applying the zero torque condition to one of the two partitions of the beam that we obtain by setting point x. We will consider the partition at the left side: on its right end will then be the point x (dark green).

M(x) + dA^A + dF^F = 0 => M(x) = -dA^A - dF^F,
where A is the reaction of the support at the left tip (red vector) and F is the resultant of the distributed load (orange vector) on the partition of the beam considered, whose length is x (that is, F = pix). The arms of the two forces, dA and dF, are represented, respectively, by the horizontal light and dark green vectors, their origin being located at point x. Their expression is:
dA = - x
dF = - 1/2 x

The light yellow areas represent the two contributions |dA^A| and |dF^F|. They appear to opposite sides of the beam to account for the fact that the absolute value of the bending moment can be computed by subtracting |dF^F| from |dA^A|.

Finally, the light red rectangle represents the magnitude of the bending moment M(x):
M(x) = - (-x) ^ (1/2 pli) - (-1/2 x) ^ (pix)
M(x) = x ^ (1/2 pli) + 1/2 x ^ (pix)
M(x) = 1/2 {x ^ (pli) + x ^ (pix)}
M(x) = 1/2 x ^ (pli + pix) = 1/2 x ^ (pli - pxi)
M(x) = 1/2 px ^ {(l-x)i}
and due to the fact that these two vector factors are always perpendicular, we can substitute the outer product with the full geometric product:
M(x) = 1/2 px(l-x)i = 1/2 (plx - pxx) i = 1/2 pll (x/l - xx/ll)

As a final note, we observe that the two sides of the light red rectangle, that is the grey and purple vectors, have a magnitude of |pxi| and |(l-x)/2|, respectively.

By dragging the red and green interactive point it is possible to change the magnitude of the load p and the position x where to compute the bending moment. Note that the maximum moment is located at midpoint of the beam (black point), as is obvious by maximizing the bending moment function. Note also that the bending moment vanishes at the endpoints of the beam, as is expected considering the boundary conditions provided.

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Created with Cinderella by Luca Redaelli