drawn the next step is to find the value of the shear force and moment at any given point along the element. For a horizontal beam one way to perform this is at any point to "chop off" the right end of the beam. The example below includes a point load, a distributed load, and an applied moment. The supports include both hinged supports and a fixed end support. The first drawing shows the beam with the applied forces and displacement constraints. The second drawing is the loading diagram with the reaction values given without the calculations shown or what most people call a free body diagram. The third drawing is the shear force diagram and the fourth drawing is the bending moment diagram. For the bending moment diagram the normal sign convention was used. Below the moment diagram are the stepwise functions for the shear force and bending moment with the functions expanded to show the effects of each load on the shear and bending functions. The example is illustrated using United States customary units. Point loads are expressed in kips (1 kip = 1000 lbf = 4.45 kN), distributed loads are expressed in k/ft (1 k/ft = 1 kip/ft = 14.6 kN/m), moments are expressed in ft-k (1 ft-k = 1 ft-kip = 1.356 kNm), and lengths are in ft (1 ft = 0.3048 m). Step 1: Compute the reaction forces and moments The first step obtaining the bending moment and shear force equations is to determine the reaction forces. This is done using a free body diagram of the entire beam. The beam has three reaction forces, Ra, Rb at the two supports and Rc at the clamped end. The clamped end also has a reaction couple Mc. These four quantities have to be determined using two equations, the balance of forces in the beam and the balance of moments in the beam. Four unknowns cannot be found given two independent equations in these unknown variables and hence the beam is statically indeterminate. One way of solving this problem is to use the principle of linear superposition and break the problem up into the superposition of a number of statically determinate problems. The extra boundary conditions at the supports have to be incorporated into the superposed solution so that the deformation of the entire beam is compatible. From the free-body diagram of the entire beam we have