Beam Reaction by Moment Equilibrium — Two Point Loads

This is the calculation $\sum M = 0$ is actually taught on. Two supports, two point loads, and one rule: take moments about one support and the other support's reaction drops out of the equation because its moment arm is zero. Taking moments about B, $R_A L = P_1(L - a_1) + P_2(L - a_2)$, so each load is weighted by its distance from the far support. That weighting is the entire content of the method, and it is the part beginners reverse.

A 10 m span with 20 kN at 3 m and 30 kN at 7 m from A: $R_A = [20(7) + 30(3)]/10 = 230/10 = 23$ kN. The far reaction is the remainder, $50 - 23 = 27$ kN — and you should always check it independently by taking moments about A instead: $R_B \times 10 = 20(3) + 30(7) = 270$, giving 27 kN. Two routes, same answer. If they disagree, you have mixed up which distance goes with which load, which is by far the commonest error and is silent, because the wrong answer is still a plausible force.

Two supports and any number of vertical loads is statically determinate: three equations of equilibrium, and after the two vertical reactions and the horizontal one there is nothing left unknown. Add a third support and it stops being determinate — equilibrium no longer has enough equations, the answer depends on the beam's stiffness and on how level the supports are, and you need moment distribution, slope-deflection or a stiffness analysis. That boundary is worth knowing precisely, because a continuous beam looks like this problem and is not this problem. The same weighting extends to any number of loads on the determinate case: $R_A = \sum P_i(L - a_i)/L$, one term per load.