Question 11.13: The cantilever beam AB supports a uniformly distributed load...
The cantilever beam AB supports a uniformly distributed load w (Fig. 11.44). Determine the deflection and slope at A.
Deflection at A. We apply a dummy downward load Q _{A} at A (Fig. 11.45) and write
y_{A}=\frac{\partial U}{\partial Q_{A}}=\int_{0}^{L} \frac{M}{E I} \frac{\partial M}{\partial Q_{A}} d x (11.77)
The bending moment M at a distance x from A is
M=-Q_{A} x-\frac{1}{2} w x^{2} (11.78)
and its derivative with respect to Q _{A} is
\frac{\partial M}{\partial Q_{A}}=-x (11.79)
Substituting for M and \partial M / \partial Q_{A} from (11.78) and (11.79) into (11.77), and making Q _{A} = 0, we obtain the deflection at A for the given loading:
y_{A}=\frac{1}{E I} \int_{0}^{L}\left(-\frac{1}{2} w x^{2}\right)(-x) d x=+\frac{w L^{4}}{8 E I}
Since the dummy load was directed downward, the positive sign indicates that
y_{A}=\frac{w L^{4}}{8 E I} \downarrow
Slope at A. We apply a dummy counterclockwise couple M _{A} at A (Fig. 11.46) and write
u _{A}=\frac{\partial U}{\partial M_{A}}
Recalling Eq. (11.17), we have
U=\int_{0}^{L} \frac{M^{2}}{2 E I} d x (11.17)
u _{A}=\frac{\partial}{\partial M_{A}} \int_{0}^{L} \frac{M^{2}}{2 E I} d x=\int_{0}^{L} \frac{M}{E I} \frac{\partial M}{\partial M_{A}} d x (11.80)
The bending moment M at a distance x from A is
M=-M_{A}-\frac{1}{2} w x^{2} (11.81)
and its derivative with respect to M _{A} is
\frac{\partial M}{\partial M_{A}}=-1 (11.82)
Substituting for M and \partial M / \partial M_{A} from (11.81) and (11.82) into (11.80), and making M_{A}=0, we obtain the slope at A for the given loading:
u _{A}=\frac{1}{E I} \int_{0}^{L}\left(-\frac{1}{2} w x^{2}\right)(-1) d x=+\frac{w L^{3}}{6 E I}
Since the dummy couple was counterclockwise, the positive sign indicates that the angle u _{A} is also counterclockwise:
u _{A}=\frac{w L^{3}}{6 E I} a