Question 1.4.11: For the wire form of diameter d shown in Fig. 4–11a, determi...

Figure 4–11b shows free body diagrams where the body has been broken in each section,and internal balancing forces and moments are shown. The sign convention for the force and moment variables is positive in the directions shown. With energy methods, sign conventions are arbitrary, so use a convenient one. In each section, the variable x is defined with its origin as shown. The variable x is used as a variable of integration for each section independently, so it is acceptable to reuse the same variable for each section. For completeness, the transverse shear forces are included, but the effects of transverse shear on the strain energy (and deflection) will be neglected.
Element BC is in bending only so from Eq. (4–31)^5

\delta _i=\frac{∂U}{∂F_i}=\int{\frac{1}{EI} \left(M\frac{∂M}{∂F_i}\right)dx}    bending

\frac{∂U_{BC}}{∂F}=\frac{1}{EI}\int_{0}^{a}{\left(Fx\right) } \left(x\right) dx =\frac{Fa^3}{3EI}     (1)

Element CD is in bending and in torsion. The torsion is constant so Eq. (4–30)can be written as

\theta _i=\frac{∂U}{∂M_i}=\int{\frac{1}{GJ} \left(T\frac{∂T}{∂M_i}\right)dx}        torsion    (4–30)

where l is the length of the member. So for the torsion in member  CD \ \ , F_i= F, \ T = Fa, \ and \ l = b. Thus,

\left(\frac{∂U_{CD} }{∂F}\right)_{torsion} =\left(Fa\right) \left(a\right) \frac{b}{GJ} =\frac{Fa^2b}{GJ}     (2)

For the bending in CD,

\left(\frac{∂U_{CD} }{∂F}\right)_{bending} =\frac{1}{EI}\int_{0}^{b}({Fx})\left(x\right)dx =\frac{Fb^3}{3EI}     (3)

Member DG is axially loaded and is bending in two planes. The axial loading is constant, so Eq. (4–29)can be written as

\delta _i=\frac{∂U}{∂F_i} = \int_{0}^{b}{\frac{1}{EI}}\left(F\frac{∂F}{∂F_i} \right)dx      tension and compression     (4–29)

where l is the length of the member. Thus, for the axial loading of DG \ \ , F_i= F,  l = c. and,

\left(\frac{∂U_{DG}}{∂F}\right) _{axial}=\frac{Fc}{AE}   (4)

The bending moments in each plane of DG are constant along the length, with M_{DG2}= Fb \ and \ M_{DG1}= Fa . Considering each one separately in the form of
Eq. (4–31) gives

\delta _i=\frac{∂U}{∂F_i}=\int{\frac{1}{EI} \left(M\frac{∂M}{∂F_i}\right)dx}    bending       (4–31)

Adding Eqs. (1) to (5), noting that  I = {\pi d^4}/{64} \ , \ J = 2I \ , \ A = {\pi d^2}/{4}, \ and \ G={E}/{\left[2\left(1+\nu \right) \right] }  , we find that the deflection of B in the direction of F is

Now that we have completed the solution, see if you can physically account for each term in the result using an independent method such as superposition.

^5 It is very tempting to mix techniques and try to use superposition also, for example. However, some subtle things can occur that you may visually miss. It is highly recommended that if you are using Castigliano’s theorem on a problem, you use it for all parts of the problem