will be multiplied by the distance after 10 ft. i.e. (x-10) the moment location is defined in the middle of the distributed force, which is also changing. This is where (x+10)/2 is derived from. Alternatively, we can take moments about the cross-section to get Again, in this case, Step 5: Compute shear forces and moments - third piece Taking the third segment, and summing forces, we have and summing moments about the cross-section, we get Therefore, and Notice that the distributed force can now be considered one force of 15 kips acting in the middle of where it is positioned. Step 6: Compute shear forces and moments - fourth piece Taking the fourth and final segment, a balance of forces gives and a balance of moments around the cross-section leads to Solving for V4 and M4, we have and By plotting each of these equations on their intended intervals, you get the bending moment and shear force diagrams for this beam. In particular, at the clamped end of the beam, x = 50 and we have Step 7: Compute deflections of the four segments We now use the Euler–Bernoulli beam theory to compute the deflections of the four segments. The differential equation that relates the beam deflection (w) to the bending moment (M) is where E is the Young's modulus and I is the area moment of inertia of the beam cross-section. Substituting the expressions for M1, M2, M3, M4 into the beam equation and solving for the deflection gives us Step 8: Apply boundary conditions Now we will apply displacement boundary conditions for the four segments to determine the integration constants. For the fourth segment of the beam, we consider the boundary conditions at the clamped end where w4 = dw/dx = 0 at x = 50. Solving for C7 and C8 gives Therefore, we can express w4 as Now, w4 = w3 at x = 37.5 (the point of application of the external couple). Also, the slopes of the deflection curves at this point are the same, i.e., dw4/dx = dw3/dx. Using these boundary conditions and solving for C5 and C6, we get Substitution of these constants into the expression for w3 gives us Similarly, at the support between segments 2 and 3 where x = 25, w3 = w2 and dw3/dx = dw2/dx. Using these and solving for C3 and C4 gives Therefore, At the support between