Question 9.16: A cantilever beam AB (Fig. 9-34) is subjected to three diffe......

Use a four-step problem-solving approach. Combine steps as needed for an efficient solution.
Part (a): Beam with concentrated load P.
1, 2. Conceptualize, Categorize: The bending moment in the beam (Fig. 9-34a) at distance x from the free end is M = -Px.
Substituting this expression for M into Eq. (9-95a) leads to the expression for the strain energy of the beam:

U=\int{\frac{M^{2}d x}{2E I}}\qquad (9-95a)\\
\quad\quad\quad\quad U=\int_{o}^{L}{\frac{M^{2}d x}{2E I}}=\int_{o}^{L}{\frac{(-P x)^{2}d x}{2E I}}={\frac{P^{2}L^{3}}{6E I}}\quad\quad\quad(9-99)
3. Analyze: To obtain the vertical deflection \delta_A under the load P, equate the work done by the load to the strain energy:
\quad\quad\quad\quad W=U\quad\mathrm{or}\quad\frac{P\delta_{A}}{2}=\frac{P^{2}L^{3}}{6E I}
from which
\quad\quad\quad\quad \delta_{A}={\frac{P{{L}}^{3}}{3E I}}
The deflection \delta_A is the only deflection that can be found by this procedure, because it is the only deflection that corresponds to the load P.
Part (b): Beam with moment M_0.
1, 2. Conceptualize, Categorize: In this case, the bending moment (Fig. 9-34b) is constant and equal to -M_0. Therefore, the strain energy [from Eq. (9-95a)] is
\quad\quad\quad\quad U = \int_{0}^{L}\frac{M^{2}d x}{2E I}=\int_{0}^{L}\frac{(-M_{0})^{2}d x}{2E I}=\frac{M_{0}^{2}L}{2E I}\quad\quad(9-100)
3. Analyze: The work W done by the couple M_0 during loading of the beam is M_0\theta_A / 2, where \theta_A is the angle of rotation at end A. Therefore,
\quad\quad\quad\quad W\,=\,U\quad\mathrm{or}\quad\frac{M_{o}\theta_{A}}{2}=\frac{M_{o}^{2}L}{2E I}
and
\quad\quad\quad\quad {{\theta}}_{A}={\frac{M_{0}L}{E I}}
The angle of rotation has the same sense as the moment (counterclockwise in this example).
Part (c): Beam with both loads acting simultaneously.
1, 2. Conceptualize, Categorize: When both loads act on the beam (Fig. 9-34c), the bending moment in the beam is
\quad\quad\quad\quad M = -P x-M_{0}
3. Analyze: Therefore, the strain energy is
\quad\quad\quad\quad U\;\;=\;\;\int_{0}^{L}\frac{M^{2}d x}{2E I}=\frac{1}{2E I}\;\;\int_{0}^{L}(-P x-M_{0})^{2}d x \\ \quad\quad\quad\quad = \frac{P^2L^3}{6EI} + \frac{PM_0L^2}{2EI} + \frac{M_0^2L}{2EI}\quad\quad\quad\quad (9-101)
The first term in this result gives the strain energy due to P acting alone [Eq. (9-99)], and the last term gives the strain energy due to M_0 alone [Eq. (9-100)]. However, when both loads act simultaneously, an additional term appears in the expression for strain energy.
4. Finalize: Therefore, the strain energy in a structure due to two or more loads acting simultaneously cannot be obtained by adding the strain energies due to the loads acting separately. The reason is that strain energy is a quadratic function of the loads, not a linear function. Therefore, the principle of superposition does not apply to strain energy.
Also observe that a deflection for a beam with two or more loads cannot be calculated by equating the work done by the loads to the strain energy. For instance, equating work and energy for the beam of Fig. 9-34c gives
\quad\quad\quad\quad {W} = U \quad\mathrm{or}\quad \frac{{P}\delta_{A2}}{2}+\frac{M_{o}\theta_{A2}}{2} = \frac{P^{2}{L}^{3}}{6E{I}}+\frac{{P}M_{o}L^{2}}{2E{I}}+\frac{M_{o}^{2}{L}}{2E\bar{I}}\quad\quad\quad (a)
in which \delta_{A2} ~ and ~ \theta_{A2} represent the deflection and angle of rotation at end A of the beam with two loads acting simultaneously (Fig. 9-34c). Although the work done by the two loads is indeed equal to the strain energy, and Eq. (a) is correct, a solution for either \delta_{A2} ~ or ~ \theta_{A2} cannot be obtained because there are two unknowns and only one equation.