Question 9.16: A cantilever beam AB (Fig. 9-34) is subjected to three diffe...
A cantilever beam AB (Fig. 9-34) is subjected to three different loading conditions: (a) a concentrated load P at its free end, (b) a couple M_{0} at its free end, and (c) both loads acting simultaneously.
For each loading condition, determine the strain energy of the beam. Also, determine the vertical deflection \delta_{A} at end A of the beam due to the load P acting alone (Fig. 9-34a), and determine the angle of rotation \theta_{A} at end A due to the moment M_{0} acting alone (Fig. 9-34b). (Note: The beam has constant flexural rigidity EI.)
(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^{2}dx}{2EI}}=\int_{0}^{L}{\frac{(-Px)^{2}dx}{2EI}}=\frac{P^{2}L^{3}}{6EI} (9-84)
To obtain the vertical deflection \delta_{A} under the load P, we equate the work done by the load to the strain energy:
W = U or \frac{P\delta_{A}}{2}=\frac{P^{2}L^{3}}{6EI}
from which
\delta_{A}=\frac{PL^{3}}{3EI}The deflection \delta_{A} is the only deflection we can find by this procedure, because it is the only deflection that corresponds to the load P.
(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^{2}dx}{2EI}}=\int_{0}^{L}{\frac{(-M_{0})^{2}dx}{2EI}}=\frac{M^{2}_{0}L}{2EI} (9-85)
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,
W = U or \frac{M_{0}\theta_{A}}{2}=\frac{M^{2}_{0}L}{2EI}
and \theta_{A}=\frac{M_{0}L}{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^{2}dx}{2EI}}=\frac{1}{2EI}\int_{0}^{L}{(-Px-M_{0})^{2}}dx
=\frac{P^{2}L^{3}}{6EI}+\frac{PM_{0}L^{2}}{2EI}+\frac{M^{2}_{0}L}{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{P\delta_{A2}}{2}+\frac{M_{0}\theta_{A2}}{2}=\frac{P^{2}L^{3}}{6EI}+\frac{PM_{0}L^{2}}{2EI}+\frac{M^{2}_{0}L}{2EI} (h)
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. (h) is quite correct, we cannot solve for either \delta_{A2} or \theta_{A2} because there are two unknowns and only one equation.