Question 7.9: For a simply supported beam as shown in Figure 7.14 find out...

Let us complete the moment divided by EI diagram first and then from Figure 7.14(c), we can write slope at Q as

\left|\theta_{ Q }\right|=\left|\theta_{ P }\right|-\left|\theta_{ Q / P }\right|=\left|\frac{t_{ S / P }}{8}\right|-\left|\theta_{ Q / P }\right|

This can be written because t_{ S / P } is substantially small such that we can write:

\left|\theta_{ P }\right|=\left|\frac{t_{ S / P }}{L_{ PS }}\right|=\left|\frac{t_{ S / P }}{8}\right|

where L_{ PS } is the length between points P and S.

Using Theorem I, \left|\theta_{ Q / P }\right| is the area under moment diagram between P and Q, and is given by

\left|\theta_{ Q / P }\right|=\frac{1}{2} \times 2 \times \frac{8}{E I}=\frac{8}{E I}  kNm ^2

Using Theorem II, t_{ S / P } is the moment of the area under M/EI diagram from P to S about S, and is given by

t_{ S / P }=\left\lgroup 2+\frac{1}{3} \times 6 \right\rgroup\left[\frac{1}{2} \times 6 \times \frac{24}{E I}\right]+\frac{2}{3} \cdot 2\left[\frac{1}{2} \times 2 \times \frac{24}{E I}\right]=\frac{320}{E I}  kN m ^3

So, \left|\theta_{ Q }\right|=\frac{320}{8 E I}-\frac{8}{E I}=\frac{32}{E I}  kN m ^2

Putting E=200 \times 10^6  kN / m ^2 \text { and } I=17 \times 10^{-6}  m ^4 \text {, we get }\left|\theta_{ Q }\right| \text { slope at } Q \text { as } 0.00941  rad