Question 7.13: For the beam with EI constant shown in Figure 7.23, obtain t...

Let us draw the free-body diagram of the beam as shown in Figure 7.24(a) to determine the support reactions.

\sum F_y=0 \Rightarrow R_{ A }+R_{ B }=\frac{w_{ o } L}{6}, \sum M_{ A }=0 \Rightarrow R_{ B }=\frac{5 w_{ o } L}{54} \Rightarrow R_{ A }=\frac{4}{54} w_{ o } L(\uparrow)

The equivalent loading diagram is shown in Figure 7.24(b) to write the bending moment at a distance x from end A of the beam. Therefore,

M_x=R_{ A } x-\frac{2 w_{ o }}{6(2 L / 3)}\left\langle x-\frac{L}{3}\right\rangle^3+\frac{w_{ o }}{2}\left\langle x-\frac{2 L}{3}\right\rangle^2+\frac{w_{ o }}{6(L / 3)}\left\langle x-\frac{2 L}{3}\right\rangle^3

Putting the expression for R_{ A } , we get

M_x=\left\lgroup \frac{4 w_0 L}{54} \right\rgroup x-\frac{w_0}{2 L}\left\langle x-\frac{L}{3}\right\rangle^3+\frac{w_0}{2}\left\langle x-\frac{2 L}{3}\right\rangle^2+\frac{w_0}{2 L}\left\langle x-\frac{2 L}{3}\right\rangle^3            (1)

From flexure equation, we get

(E I) \frac{ d ^2 y}{ d x^2}=-M_x

Substituting M_x from Eq. (1):

\text { (EI) } \frac{ d ^2 y}{ d x^2}=\left\lgroup -\frac{4 w_{ o } L}{54} \right\rgroup x+\frac{w_{ o }}{2 L}\left\langle x-\frac{L}{3}\right\rangle^3-\frac{w_{ o }}{2}\left\langle x-\frac{2 L}{3}\right\rangle^2-\frac{w_{ o }}{2 L}\left\langle x-\frac{2 L}{3}\right\rangle^3

Integrating twice, we get

(E I) \frac{ d y}{ d x}=\left\lgroup -\frac{2 w_{ o } L}{54} \right\rgroup x^2+\frac{w_{ o }}{8 L}\left\langle x-\frac{L}{3}\right\rangle^4-\frac{w_{ o }}{6}\left\langle x-\frac{2 L}{3}\right\rangle^3-\frac{w_{ o }}{8 L}\left\langle x-\frac{2 L}{3}\right\rangle^4+C_1               (2)

and      (E I) y=\left\lgroup -\frac{2 w_0 L}{162} \right\rgroup x^3+\frac{w_0}{40 L}\left\langle x-\frac{L}{3}\right\rangle^5-\frac{w_{ o }}{24}\left\langle x-\frac{2 L}{3}\right\rangle^4-\frac{w_0}{40 L}\left\langle x-\frac{2 L}{3}\right\rangle^5+C_1 x+C_2             (3)

Now, at x = 0, y = 0, so we get from Eq. (3) C_2=0 and at x = L, y = 0, thus we get

-\frac{w_{ o } L^4}{81}+\frac{w_{ o }}{40 L}\left\lgroup \frac{2 L}{3} \right\rgroup^5-\frac{w_{ o }}{24}\left\lgroup \frac{L}{3} \right\rgroup^4-\frac{w_{ o }}{40 L}\left\lgroup \frac{2 L}{3} \right\rgroup^5+C_1 L=0           [from Eq. (3)]

or        C_1=\frac{47 w_{ o } L^3}{4860}

Thus, the equation of the elastic line is given by

and the slope is found as

We note the negative sign in the expression of slope. Hence the tangent at end B to the elastic line rotates anticlockwise with respect to the x-axis. Therefore,

\theta_{ B }=\frac{101 w_{ o } L^3}{9720 E I} ⦨