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

(a) Beam with concentrated load P (Fig. 9-34a). The bending moment in the beam at distance x from the free end is M = -Px. Substituting this expression for M into Eq. (9-80a), we get the following expression for the strain energy of the beam:

U=\int_{0}^{L}{\frac{M^2dx}{2EI}}=\int_{0}^{L}{\frac{(-PX)^2dx}{2EI}}=\frac{P^2L^3}{6EI}                  (9-84)

To obtain the vertical deflection δ_A under the load P, we equate the work done by the load to the strain energy:

W = U        or         \frac{Pδ_A}{2}=\frac{P^2L^3}{6EI}

from which

δ_A=\frac{PL^3}{3EI}

The deflection δ_A is the only deflection we can find by this procedure.

(b) Beam with moment M_0 (Fig. 9-34b). In this case the bending moment is constant and equal to -M_0. Therefore, the strain energy (from Eq. 9-80a) is

U=\int_{0}^{L}{\frac{M^2dx}{2EI}}=\int_{0}^{L}{\frac{(-M_0)^2dx}{2EI}}=\frac{M_0^2L}{2EI}                  (9-85)

The work W done by the couple M_0 during loading of the beam is M_0θ_A/2, where θ_A is the angle of rotation at end A. Therefore,

W = U        or         \frac{M_0θ_A}{2}=\frac{M_0^2L}{2EI}

and                                                 θ_A=\frac{M_0L}{EI}

The angle of rotation has the same sense as the moment (counterclockwise in this example).
(c) Beam with both loads acting simultaneously (Fig. 9-34c). When both loads act on the beam, the bending moment in the beam is

M = -Px – M_0

Therefore, the strain energy is

U=\int_{0}^{L}{\frac{M^2dx}{2EI}}=\frac{1}{2EI}\int_{0}^{L}{(-Px\ -\ M_0)^2dx}\\ =\frac{P^2L^3}{6EI}+\frac{PM_0L^2}{2EI}+\frac{M_0^2L}{2EI}                  (9-86)

The first term in this result gives the strain energy due to P acting alone (Eq. 9-84), and the last term gives the strain energy due to M_0, alone (Eq. 9-85). However, when both loads act simultaneously, an additional term appears in the expression for strain energy.
Therefore, we conclude that 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.
We also observe that we cannot calculate a deflection for a beam with two or more loads by equating the work done by the loads to the strain energy. For instance, if we equate work and energy for the beam of Fig. 9-34c, we get

W = U        or         \frac{Pδ_{A2}}{2}+\frac{M_0θ_{A2}}{2}=\frac{P^2L^3}{6EI}+\frac{PM_0L^2}{2EI}+\frac{M_0^2L}{2EI}             (h)

in which δ_{A2} and θ_{A2} represent the deflection and angle of rotation at end A of the beam with two loads (Fig. 9-34c). Although the work done by the loads is indeed equal to the strain energy, and Eq. (h) is quite correct, we cannot solve for either δ_{A2} or θ_{A2} because there are two unknowns and only one equation.