Question 9.13: Determine the maximum deflection of the beam of Example 9.12...
Determine the maximum deflection of the beam of Example 9.12.
Determination of Point K Where Slope Is Zero. We recall from Example 9.12 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. 9.69). Our computations, therefore, will be simplified if we relate the slope at K to the slope at B, rather than to the slope at A.
Since the slope at A has already been determined in Example 9.12, the slope at B is obtained by writing
u _{B}= u _{A}+ u _{B / A}= u _{A}+A_{1}+A_{2}u _{B}=-\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}
Observing that the bending moment at a distance u from end B is M=\frac{1}{4} P u (Fig. 9.70a), we express the area A^{\prime} located between K and B under the (M/EI) diagram (Fig. 9.70b) as
A^{\prime}=\frac{1}{2} \frac{P u}{4 E I} u=\frac{P u^{2}}{8 E I}
By the first moment-area theorem, we have
u _{B / K}= u _{B}- u _{K}=A^{\prime}
and, since u _{K}=0, u_{B}=A^{\prime}
Substituting the values obtained for u _{B} and A^{\prime}, we write
\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 the area A^{\prime} about a vertical axis through B (Fig. 9.70b). We write
|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, we have
|y|_{\max }=\frac{P}{12 E I}\left(\frac{\sqrt{5}}{4} L\right)^{3}=0.01456 P L^{3} / E I
