a) Since the forces are to be applied successively, I'll start by computing the deflection due to the force P on the right. Taking P acting at E (force 1 at position1, no force at D), and using Appendix D, case 5,

The subscript "11" refers to position1 and force 1. The deflection at D (position 2) because of this single force P acting at E (P₁) is:

Now suppose P is applied at D (force 2, position 2) - without the other force P at E:

The subscripts mean: "22 = position 2, force 2_." Of course, P₁ and P₂ are simply P. Without going through the same algebra, it should be clear from the symmetry that

I'll drop the minus signs since all of the deflections are down. If P₁ is applied first, load-displacement diagrams can be drawn (see also Fig. 11.40 on page 710):
 

Explanation -- P₁ is applied first (it could have just as well have been the other way around). As it is applied slowly (and increased to its maximum value), the displacement y₁ also increases until the curve reaches the point C'₁ and the plot is a straight line of slope , or as the text would write (Figure 11.39 and since the authors begin with P₁), P₁/x₁₁. Now we apply P₂ at D, slowly increasing its magnitude until it has reached the value of P, during which we get the triangle in the second plot on the right. But at the same time, P₁ (plot on the left) does work, since it moves through an additional displacement from C'₁ to C₁. But its work is represented by a rectangle, since P₁ is constant while P₂ is slowly increasing.

Now I can use the areas under the graphs to compute the energies:

To see how Castigliano's Theorem is used to find deflections, please look at this example.

b) To use the method of §11.4, I need the moment as a function of x. Each reaction force is equal to P, by symmetry.