Question 7.12: Determine the angle of rotation θB and deflection δB at the ...
Determine the angle of rotation \theta_B \text { and deflection } \delta_B at the free end of a cantilever beam AB with uniform load intensity w_o as shown in Figure 7.21. Consider EI to be constant.
We draw the free-body diagram of the beam first as shown in Figure 7.22 to determine the support reactions as follows:
Clearly from Figure 7.22(a), considering equilibrium of the entire beam we note that
\sum F_y=0 \Rightarrow R_{ A }=\frac{w_{ o } L}{3}(\uparrow) \text { and } \sum M_{ A }=0 \Rightarrow M_{ A }=\frac{w_{ o } L^2}{3}
Now, we note from Figure 7.22(a) that the distributed load does not continue till the end point of the beam, we draw its equivalent loading diagram as shown in Figure 7.22(b). Therefore the bending-moment at a section at a distance x from the end A of the beam is given by
M_x=R_{ A } x-M_{ A }-\frac{w_{ o }}{2}\left\langle x-\frac{L}{3}\right\rangle^2+\frac{w_{ o }}{2}\left\langle x-\frac{2 L}{3}\right\rangle^2
Putting expression for R_{ A }
M_x=\left\lgroup \frac{w_0 L}{3}\right\rgroup x-\left\lgroup \frac{w_0 L^2}{6}\right\rgroup-\frac{w_0}{2}\left\langle x-\frac{L}{3}\right\rangle^2+\frac{w_0}{2}\left\langle x-\frac{2 L}{3}\right\rangle^2 (1)
From the flexure equation we get by substituting the expression for M_x from Eq. (1):
(E I) \frac{ d ^2 y}{ d x^2}=-M_x=-\left\lgroup \frac{w_{ o } L}{3} \right\rgroup x+\left\lgroup \frac{w_{ o } L^2}{6}\right\rgroup +\frac{w_{ o }}{2}\left\langle x-\frac{L}{3}\right\rangle^2-\frac{w_{ o }}{2}\left\langle x-\frac{2 L}{3}\right\rangle^2
Integrating,
\text { (EI) } \frac{ d y}{ d x}=-\frac{w_{ o } L x^2}{6}+\frac{w_{ o } L^2 x}{6}+\frac{w_{ o }}{6}\left\langle x-\frac{L}{3}\right\rangle^3-\frac{w_{ o }}{6}\left\langle x-\frac{2 L}{3}\right\rangle^3+C_1
But at x = L, dy/dx = 0 and C_1=0 since Macaulay’s functions vanish therefore,
(E I) \frac{ d y}{ d x}=-\frac{w_{ o } L x^2}{6}+\frac{w_{ o } L^2 x}{6}+\frac{w_{ o }}{6}\left\langle x-\frac{L}{3}\right\rangle^3-\frac{w_{ o }}{6}\left\langle x-\frac{2 L}{3}\right\rangle^3 (2)
Again integrating Eq. (2), we get
(E I) y=-\frac{w_0 L x^3}{18}+\frac{w_0 L^2 x^2}{12}+\frac{w_0}{24}\left\langle x-\frac{L}{3}\right\rangle^4-\frac{w_0}{24}\left\langle x-\frac{2 L}{3}\right\rangle^4+C_2
Again at x = 0, y = 0 so from the above expression C_2=0 , since Macaulay’s functions vanish so
(E I) y=-\frac{w_{ o } L x^3}{18}+\frac{w_{ o } L^2 x^2}{12}+\frac{w_{ o }}{24}\left\langle x-\frac{L}{3}\right\rangle^4-\frac{w_{ o }}{24}\left\langle x-\frac{2 L}{3}\right\rangle^4 (3)
Putting x = L in Eqs. (2) and (3) and noting Macaulay’s functions, we get
\left.(E I) \frac{ d y}{ d x}\right|_{x=L}=-\frac{w_{ o } L^3}{6}+\frac{w_{ o } L^3}{6}+\frac{w_0}{6}\left\lgroup \frac{2 L}{3} \right\rgroup^3-\frac{w_{ o }}{6}\left\lgroup \frac{L}{3} \right\rgroup^3=\frac{7 w_{ o } L^3}{112}
Thus, the slope at B is
\left.\frac{ d y}{ d x}\right|_{x=L}=\tan \theta_{ B } \approx \theta_{ B }=\frac{7 w_{ o } L^3}{162 E I} ⦪
and
\begin{aligned} \left.(E I) y\right|_{x=L} & =-\frac{w_{ o } L^4}{18}+\frac{w_{ o } L^4}{12}+\frac{w_{ o }}{24}\left\{\left\lgroup \frac{2 L}{3} \right\rgroup^4-\left\lgroup \frac{L}{3} \right\rgroup^4\right\} \\ & =-\frac{w_{ o } L^4}{18}+\frac{w_{ o } L^4}{12}+\frac{5 w_{ o } L^4}{648}=\frac{23 w_{ o } L^4}{648} \end{aligned}
Thus, deflection at B is
\left.y\right|_{x=L}=\delta_{ B }=\frac{23 w_{ o } L^4}{648 E I}(\downarrow)
