Question 7.11: For the beam shown in Figure 7.19, compute (a) the slope at ...

We draw the free-body diagram of the beam as shown in Figure 7.20 and compute the support reactions to write the expression of bending moment at any section at a distance x from the left-end support of the beam.

We know

M_x=\frac{P(L-a)}{L} x-P\langle x-a\rangle             (1)

But from our flexure equation, we get by substituting M_{ x } from Eq. (1):

(E I) \frac{ d ^2 y}{ d x^2}=-M_x=-\frac{P(L-a)}{L} x+P\langle x-a\rangle

Integrating twice, we obtain

(E I) \frac{ d y}{ d x}=-\frac{P(L-a)}{2 L} x^2+\frac{P}{2}\langle x-a\rangle^2+C_1

and      (E I) y=-\frac{P(L-a)}{6 L} x^3+\frac{P}{6}\langle x-a\rangle^3+C_1 x+C_2           (3)

Now, at x = 0 and at x = L, y = 0. Therefore, putting x = 0 in the Eq. (3), we get

C_2=0 \quad\left(\text { as } x=0<a,\langle x-a\rangle^3=0\right)

Similarly, putting x = L(> a) and noting \langle x-a\rangle^3=(x-a)^3 \text { for } x>a , in Eq. (3) we get

C_1 L+\frac{P}{6}(L-a)^3-\frac{P(L-a) L^2}{6}=0

or      C_1=\frac{P b\left(L^2-b^2\right)}{6 L}

Assuming L – a = b, we get

(E I) y=-\frac{P b x^3}{6 L}+\frac{P}{6}\langle x-a\rangle^3+\frac{P b\left(L^2-b^2\right)}{6 L} x             (4)

and        \text { (EI) } \frac{ d y}{ d x}=-\frac{P b x^2}{2 L}+\frac{P}{2}\langle x-a\rangle^2+\frac{P b\left(L^2-b^2\right)}{6 L}               (5)

Thus, to obtain slope at A, we put x = 0 in Eq. (5)

Therefore, slope at A is

\left.\theta_{ A } \approx \tan \theta_{ A } \approx \frac{ d y}{ d x}\right|_{ A }=\frac{P b\left(L^2-b^2\right)}{6 E I L} ⦪

Similarly, to obtain deflection at C, we need to put x = a in Eq. (4) to get

Therefore,

\begin{aligned} \left.(E I) y\right|_{x=a} & =\frac{P b a}{6 L}\left\{L^2-L^2+2 L a-2 a^2\right\} \\ & =\frac{P b a^2}{3 L}(L-a)=\frac{P a^2 b^2}{3 L} \end{aligned}

or      \left.y\right|_{x=a}=\frac{P a^2 b^2}{3 E I L}(\downarrow)