Question 7.10: A simply supported beam AB carries a concentrated load P at ...

Using Theorem II, we get
Tangential deviation of A with respect to B:

$t_{ A / B }= AA ^{\prime}=\frac{P L}{4 E I} \cdot \frac{L}{4} \cdot \frac{L}{3}+\frac{P L}{8 E I} \frac{L}{4} \cdot \frac{2 L}{3}=\frac{P L^3}{24 E I}$

Now $\theta_{ B } \text { is slope at } B = AA ^{\prime} / AB$

or        $\theta_{ B }=\frac{P L^3}{24 E I} \times \frac{1}{L}=\frac{P L^2}{24 E I}$           (1)

Similarly, tangential deviation of B with respect to A:

$t_{ B / A }= BB ^{\prime}=\frac{P L}{8 E I} \cdot \frac{L}{4} \cdot \frac{L}{3}+\frac{P L}{4 E I} \cdot \frac{L}{4} \cdot \frac{2 L}{3}=\frac{5 P L^3}{96 E I}$

and also,

$\theta_{ A }=\frac{ BB ^{\prime}}{ AB }=\frac{5 P L^3}{96 E I} \times \frac{1}{L}=\frac{5 P L^2}{96 E I}$            (2)

From Eqs. (1) and (2), we get

$\frac{\theta_{ A }}{\theta_{ B }}=\frac{5}{4}$

Now to get the deflection $\delta_{ C } \text { at point } C \text {, we first find } t_{ C / B }$ ,  that is, tangential deviation of C with respect to B:

$t_{ C / B }=\frac{P L}{8 E I} \cdot \frac{L}{4} \cdot \frac{L}{6}=\frac{P L^3}{192 E I}$

From Figure 7.15(c) and (d):

$\delta_{ C }=\frac{ AA ^{\prime}}{2}-t_{ C / B }=\frac{P L^3}{48 E I}-\frac{P L^3}{192 E I}=\frac{P L^3}{64 E I}$

so the slopes are

$\theta_{ A }=\frac{5 P L^2}{96 E I} \quad \text { and } \quad \theta_{ B }=\frac{P L^2}{24 E I}$

and deflection at C is

$\delta_{ C }=\frac{P L^3}{64 E I}$