A beam is a member that can resist bending. Beams are often statically indeterminate, which means they have too many supports for them to be analysed using simple statics techniques. However, such a beam will not be covered here.

To find the internal forces in a beam, cut the beam in two and draw a free body diagram of one part. At the break, the forces will be tension (T) and shear (F), and there will also be a bending moment (M).

As we are interested in the distribution of forces along the length of a beam, we can plot graphs of the shear force and bending moment. These are best placed underneath the sketch of the beam.

The variations in the shear force and bending moment can be derived either analytically, by deriving equations for each section of the beam, or graphically, by remembering the effect of each type of load and constructing the graphs manually.

By convention, a downward shear force is positive if we are working from the left end of the beam and negative if we are working from the right end. An anticlockwise bending moment is positive if we are working from the left end and negative if we are working from the right end. A positive bending moment will therefore result in the beam taking the shape of a dish.

Consider a simply supported beam, fixed at one end and on rollers at the other. A point load L is placed on the beam a distance *a* from the left end and a distance *b* from the right end. The reaction at the left end is R_{A}, and that at the right end is R_{B}.

Because the beam is in equilibrium, summing the moments about the right end gives ΣM_{B}=R_{A}(a+b)-Lb=0; therefore, R_{A}=Lb/(a+b). Summing the moments about the left end gives ΣM_{A}=La-R_{B}(a+b)=0; therefore, R_{B}=La/(a+b).

Because we are analysing two sections of the beam - on either side of the point load - there will be two sets of bending moment and shear force equations. For section 1, 0<x<a, where x is the distance from the left end of the beam. Summing the vertical forces acting on that section gives ΣF_{Y}=R_{A}-F=0; therefore, F=R_{A}. Summing the moments about the break gives ΣM=R_{A}x-M=0; therefore, M=R_{A}x.

For section 2, a<x<(a+b). Summing the vertical forces gives ΣF_{Y}=R_{A}-L-F=0; therefore, F=R_{A}-L. Taking moments about the break gives ΣM=R_{A}x-L(x-a)-M=0; this leads to M=R_{A}x-L(x-a), or M=(R_{A}-L)x+La.

It will be noticed that at a free or simply supported end of a beam, the bending moment is 0Nm. This is because there is nothing to exert a reaction moment at such a point.

It will also be noticed that at a point load, there is a step in the shear force diagram equal to the magnitude of the applied force; working from left to right, an upward force increases the shear force, while a downward force decreases it. In addition, the gradient of the bending moment diagram changes abruptly. Between point loads, the shear force is constant and the bending moment varies linearly. Incidentally, the shear force is equal to dM/dx, the rate of change of the bending moment; this is a general rule.

A cantilever beam is one that is unsupported at one end. To take an example, consider a beam that is fixed to a wall at its left end, with the other end loose. A point load L_{1} is placed on the beam a distance b from the left end, and a point load L_{2} is placed on the beam a distance a+b from the left end. The wall exerts a reaction force A_{Y} and a reaction moment M_{o}.

Because the vertical forces must balance, A_{Y}=L_{1}+L_{2}. And because the moments about the left end of the beam must also balance, M_{o}=L_{1}b+L_{2}(a+b); this can be rewritten as M_{o}=b(L_{1}+L_{2})+L_{2}a.

Another type of beam problem features a simply supported beam subjected to a couple. A clockwise couple c takes effect a distance *a* from the left end and a distance *b* from the right end. R_{A}, R_{B} and x have the same meanings as in the first problem.

Taking moments about the left end leads to R_{B}=c/(a+b). And taking moments about the right end leads to R_{A}=-c/(a+b).

To the left of the couple, F=R_{A} and M=R_{A}x. To the right of the couple, F=R_{A} and M=c+R_{A}x.

This shows that at a couple, there is a step in the bending moment diagram equal to the magnitude of the couple; working from left to right, a clockwise couple increases the bending moment, while an anticlockwise couple decreases it. It also shows that a couple has no effect on the shear force diagram; this is because a couple has a tendency to twist the beam, not to cause cross-sections to slip past each other.

Because R_{A} is negative, the bending moment to the left of a clockwise couple is also negative. Substituting for R_{A} in the right hand side bending moment equation gives M=c{1-x/(a+b)}; because x≤a+b, the bending moment to the right of a clockwise couple is positive, falling to 0Nm at the free end.

A distributed load is one that is spread over a finite area. The simplest such load is a uniform load of w N/m spread over a distance l; its resultant force is wl. Two other examples are a triangular load, which increases linearly along its length from 0 to w N/m and whose resultant is 0.5wl, and a trapezoidal load, which increases linearly from w_{1} to w_{2} N/m and whose resultant is 0.5(w_{1}+w_{2})l.

Consider a simply supported beam, of length l, subjected to a uniform load of w N/m along its length. To find the moment exerted by the load, consider the force exerted on a small section dx of the beam. If this section is a distance x from an axis, the moment exerted about that axis is w(dx)x. The moment exerted by a uniform load is therefore ∫ wx dx. If we want to find the moment exerted by the entire load about one end of its range, the integral becomes ∫_{0}^{l} wx dx, which gives 0.5wl².

Taking moments about each end shows that R_{A}=R_{B}=0.5wl. Note that R_{A} and R_{B} have the same meanings as before.

Because the vertical forces must balance, F=R_{A}-wx, where x is again the distance from the left end of the beam. Substituting for R_{A} and simplifying gives F=w(0.5l-x). This shows that between the ends of a uniform distributed load, the shear force varies linearly.

Because the moments about any point must balance, M=R_{A}x-0.5wx². Substituting for R_{A} gives M=0.5wlx-0.5wx². Differentiating this gives dM/dx=0.5wl-wx. Between the ends of a distributed load, therefore, the gradient of the bending moment diagram changes smoothly.