Question 11.12: The cantilever beam AB supports a uniformly distributed load...
The cantilever beam A B supports a uniformly distributed load w and a concentrated load \mathbf{P} as shown (Fig. 11.40). Knowing that L=2 \mathrm{~m}, w=4 \mathrm{kN} / \mathrm{m}, P=6 \mathrm{kN}, and E I=5 \mathrm{MN} \cdot \mathrm{m}^{2}, determine the deflection at A.
The deflection y_{A} of the point A where the load \mathbf{P} is applied is obtained from Eq. (11.70).
x_{j}={\frac{\partial U}{\partial P_{j}}}=\;\int_{0}^{L}{\frac{M}{E I}}{\frac{\partial M}{\partial P_{j}}}\,d x (11.70)
Since \mathbf{P} is vertical and directed downward, y_{A} represents a vertical deflection and is positive downward. We have
y_{A}=\frac{\partial U}{\partial P}=\int_{0}^{L} \frac{M}{E I} \frac{\partial M}{\partial P} d x (11.73)
The bending moment M at a distance x from A is
M=-\left(P x+\frac{1}{2} w x^{2}\right) (11.74)
and its derivative with respect to P is
\frac{\partial M}{\partial P}=-x
Substituting for M and \partial M / \partial P into Eq. (11.73), we write
\begin{gathered} y_{A}=\frac{1}{E I} \int_{0}^{L}\left(P x^{2}+\frac{1}{2} w x^{3}\right) d x \\ y_{A}=\frac{1}{E I}\left(\frac{P L^{3}}{3}+\frac{w L^{4}}{8}\right) & (11.75) \end{gathered}
Substituting the given data, we have
\begin{gathered} y_{A}=\frac{1}{5 \times 10^{6} \mathrm{~N} \cdot \mathrm{m}^{2}}\left[\frac{\left(6 \times 10^{3} \mathrm{~N}\right)(2 \mathrm{~m})^{3}}{3}+\frac{\left(4 \times 10^{3} \mathrm{~N} / \mathrm{m}\right)(2 \mathrm{~m})^{4}}{8}\right] \\ y_{A}=4.8 \times 10^{-3} \mathrm{~m} \quad y_{A}=4.8 \mathrm{~mm} \downarrow \end{gathered}
We note that the computation of the partial derivative \partial M / \partial P could not have been carried out if the numerical value of P had been substituted for P in the expression (11.74) for the bending moment.