Question 11.CA.13: The cantilever beam AB supports a uniformly distributed load......
The cantilever beam AB supports a uniformly distributed load w (Fig. 11.39a). Determine the deflection and slope at A.
Deflection at A. Apply a dummy downward load Q _A at A (Fig. 11.39b) 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 (1)
The bending moment M at a distance x from A is
M=-Q_A x-\frac{1}{2} w x^2 (2)
and its derivative with respect to Q _A is
\frac{\partial M}{\partial Q_A}=-x (3)
Substituting for M and \partial M / \partial Q_A from Eqs. (2) and (3) into Eq. (1) and making Q_A=0 , the deflection at A for the given load is:
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. Apply a dummy counterclockwise couple M _A at A (Fig. 11.39c) and write
\theta_A=\frac{\partial U}{\partial M_A}
Recalling Eq. (11.15),
U=\int_0^L \frac{M^2}{2 E I} d x (11.15)
\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 (4)
The bending moment M at a distance x from A is
M=-M_A-\frac{1}{2} w x^2 (5)
and its derivative with respect to M_A is
\frac{\partial M}{\partial M_A}=-1 (6)
Substituting for M and \partial M / \partial M_A from Eqs. (5) and (6) into Eq. (4) and making M_A = 0, the slope at A for the given load is
\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} ⦨