Question 9.CA.12: For the prismatic beam and loading shown (Fig. 9.37a), deter......
For the prismatic beam and loading shown (Fig. 9.37a), determine the slope and deflection at point D.
Reference Tangent at Support A. Compute the reactions at the supports and draw the M∕EI diagram (Fig. 9.37b). The tangential deviation t_{B / A} of support B with respect to support A is found by applying the second moment-area theorem and computing the moments about a vertical axis through B of the areas A_1 \text { and } A_2 .
\begin{gathered} A_1=\frac{1}{2} \frac{L}{4} \frac{3 P L}{16 E I}=\frac{3 P L^2}{128 E I} \quad A_2=\frac{1}{2} \frac{3 L}{4} \frac{3 P L}{16 E I}=\frac{9 P L^2}{128 E I} \\ t_{B / A}=A_1\left(\frac{L}{12}+\frac{3 L}{4}\right)+A_2\left(\frac{L}{2}\right) \\ =\frac{3 P L^2}{128 E I} \frac{10 L}{12}+\frac{9 P L^2}{128 E I} \frac{L}{2}=\frac{7 P L^3}{128 E I} \end{gathered}
The slope of the reference tangent at A (Fig. 9.37c) is
\theta_A=-\frac{t_{B / A}}{L}=-\frac{7 P L^2}{128 E I}
Slope at D. Applying the first moment-area theorem from A to D,
\theta_{D / A}=A_1=\frac{3 P L^2}{128 E I}
Thus, the slope at D is
\theta_D=\theta_A+\theta_{D / A}=-\frac{7 P L^2}{128 E I}+\frac{3 P L^2}{128 E I}=-\frac{P L^2}{32 E I}
Deflection at D. The tangential deviation D E=t_{D / A} is found by computing the moment of the area A_1 about a vertical axis through D:
D E=t_{D / A}=A_1\left(\frac{L}{12}\right)=\frac{3 P L^2}{128 E I} \frac{L}{12}=\frac{P L^3}{512 E I}
The deflection at D is equal to the difference between the segments DE and EF (Fig. 9.37c). Thus,
\begin{aligned} y_D & =D E-E F=t_{D / A}-\frac{1}{4} t_{B / A} \\ & =\frac{P L^3}{512 E I}-\frac{1}{4} \frac{7 P L^3}{128 E I} \\ & =-\frac{3 P L^3}{256 E I}=-0.01172 P L^3 / E I \end{aligned}
