Question 9.CA.13: Determine the maximum deflection of the beam of Concept Appl......

Determination of Point K Where Slope Is Zero. Recall that the slope at point D, where the load is applied, is negative. It follows that point K, where the slope is zero, is located between D and the support B (Fig. 9.39b). Our computations are simplified if the slope at K is related to the slope at B, rather than to the slope at A.
Since the slope at A has already been determined in Concept Application 9.12, the slope at B is obtained by

\begin{aligned} \theta_B & =\theta_A+\theta_{B / A}=\theta_A+A_1+A_2 \\ & =-\frac{7 P L^2}{128 E I}+\frac{3 P L^2}{128 E I}+\frac{9 P L^2}{128 E I}=\frac{5 P L^2}{128 E I} \end{aligned}

Observing that the bending moment at a distance u from end B is M=\frac{1}{4} P u (Fig. 9.39c), the area A′ located between K and B under the M/EI diagram (Fig. 9.39d) is expressed as

A^{\prime}=\frac{1}{2} \frac{P u}{4 E I} u=\frac{P u^2}{8 E I}

Use the first moment-area theorem to obtain

\theta_{B / K}=\theta_B-\theta_K=A^{\prime}

and since \theta_K=0, \quad \theta_B=A^{\prime}

Substituting the values obtained for \theta_B \text { and } A^{\prime} ,

\frac{5 P L^2}{128 E I}=\frac{P u^2}{8 E I}

and solving for u,

u=\frac{\sqrt{5}}{4} L=0.559 L

Thus, the distance from the support A to point K is

AK = L − 0.559L = 0.441L

Maximum Deflection. The maximum deflection |y|_{\max } is equal to the tangential deviation t_{B / K} and thus to the first moment of area A′ about a vertical axis through B (Fig. 9.39d).

|y|_{\max }=t_{B / K}=A^{\prime}\left(\frac{2 u}{3}\right)=\frac{P u^2}{8 E I}\left(\frac{2 u}{3}\right)=\frac{P u^3}{12 E I}

Substituting the value obtained for u,

|y|_{\max }=\frac{P}{12 E I}\left(\frac{\sqrt{5}}{4} L\right)^3=0.01456 P L^3 / E I