Question 11.13: The cantilever beam AB supports a uniformly distributed load...
The cantilever beam A B supports a uniformly distributed load w (Fig. 11.41). Determine the deflection and slope at A.
Deflection at \boldsymbol{A}. We apply a dummy downward load \mathbf{Q}_{A} at A (Fig. 11.42) 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 \boldsymbol{A}. We apply a dummy counterclockwise couple \mathbf{M}_{A} at A (Fig. 11.43) and write
\theta_{A}=\frac{\partial U}{\partial M_{A}}
Recalling Eq. (11.17),
U=\int_{0}^{L}{\frac{M^{2}}{2E I}}\,d x (11.17)
we have
\theta_{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:
\theta_{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 \theta_{A} is also counterclockwise:
\theta_{A}=\frac{w L^{3}}{6 E I} ⦨
