Superposition: stepwise procedure for bending moment

Principle of superposition: for a linear elastic structure under small deformation, the response (reaction, shear force, bending moment, deflection) caused by several loads acting together equals the algebraic sum of the responses caused by the loads acting one at a time.

Stepwise procedure for the bending moment of a beam

1. Check the conditions: the material is linearly elastic and deflections are small, so superposition is valid.
2. Break the loading into simple cases: split the given loading into separate cases, each containing a single load or a standard load (point load, UDL, triangular load, couple, support settlement) on the same beam with the same supports.
3. Find the reactions of each case using $\Sigma F_y = 0$ and $\Sigma M = 0$ (and the hinge condition, if any).
4. Write or draw the bending moment of each case as a function of $x$ (or as a diagram), using one sign convention for all cases (sagging +ve, hogging −ve).
5. Add the moments at each section (or add the ordinates of the diagrams): $M(x) = M_1(x) + M_2(x) + \dots$
6. Draw the resultant BMD, find where $V = dM/dx = 0$ for the maximum moment, and check that $M = 0$ at the free ends and simple supports.

Example: for a simply supported beam with a UDL $w$ and a central point load $P$, the mid-span moment is $\dfrac{wL^2}{8} + \dfrac{PL}{4}$, the sum of the two separate BMDs (parabolic and triangular).