Question 7.15: Calculate the deflections at points A and C of the beam show...

Let us first draw the free-body diagram of the beam as shown in Figure 7.28(a) to determine the support reactions.

Consider

\sum M_{ B }=0 \Rightarrow R_{ D } L+\frac{w_{ o } L^2}{8}=\frac{3 w_{ o } L^2}{8}

Therefore,

R_{ D }=\frac{w_{ o } L}{4}(\uparrow)

But,

\sum F_y=0 \Rightarrow R_{ B }=w_{ o } L-\frac{w_{ o } L}{4}=\frac{3}{4} w_{ o } L(\uparrow)

In Figure 7.28(b), we represent the equivalent loading diagram. We can now write the bending moment, M_x at any arbitrary section at a distance x from end A as shown:

M_x=-\frac{w_{ o } x^2}{2}+R_{ B }\left\langle x-\frac{L}{2}\right\rangle+\frac{w_{ o }}{2}\left\langle x-\frac{L}{2}\right\rangle^2-\frac{w_{ o }}{2}\langle x-L\rangle^2

Putting R_{ B }=3 w_{ o } L / 4 , we get

M_x=-\frac{w_{ o } x^2}{2}+\frac{3}{4} w_{ o } L\left\langle x-\frac{L}{2}\right\rangle+\frac{w_{ o }}{2}\left\langle x-\frac{L}{2}\right\rangle^2-\frac{w_{ o }}{2}\langle x-L\rangle^2

Now, from flexure equation, we get from the above expression

(E I) \frac{ d ^2 y}{ d x^2}=-\frac{w_0 x^2}{2}-\frac{3}{4} w_{ o } L\left\langle x-\frac{L}{2}\right\rangle-\frac{w_0}{2}\left\langle x-\frac{L}{2}\right\rangle^2+\frac{w_0}{2}\langle x-L\rangle^2

Integrating successively the above equation, we get

(E I) \frac{ d y}{ d x}=\frac{w_{ o } x^3}{6}-\frac{3}{8} w_{ o } L\left\langle x-\frac{L}{2}\right\rangle^2-\frac{w_{ o }}{6}\left\langle x-\frac{L}{2}\right\rangle^3+\frac{w_{ o }}{6}\langle x-L\rangle^3+C_1             (1)

and      (E I) y=\frac{w_{ o } x^4}{24}-\frac{w_{ o } L}{8}\left\langle x-\frac{L}{2}\right\rangle^3-\frac{w_{ o }}{24}\left\langle x-\frac{L}{2}\right\rangle^4+\frac{w_{ o }}{24}\langle x-L\rangle^4+C_1 x+C_2         (2)

Now, the boundary conditions are: (a) at x = L/2, y = 0 and (b) at x = 3L/2, y = 0.
From Eq. (2) by putting the boundary conditions, we get

\frac{w_{ o }}{24}\left\lgroup \frac{L}{2} \right\rgroup^4+\left\lgroup \frac{L}{2} \right\rgroup C_1+C_2=0

or          \frac{L C_1}{2}+C_2=-\frac{w_{ o } L^4}{384}           (3)

and        \frac{w_{ o }}{24}\left\lgroup \frac{3 L}{2} \right\rgroup^4-\frac{w_{ o } L^4}{8}-\frac{w_{ o } L^4}{24}+\frac{w_{ o }}{24}\left\lgroup \frac{L}{2} \right\rgroup^4+\frac{3 L C_1}{2}+C_2=0

or            \left\lgroup \frac{3 L}{2} \right\rgroup C_1+C_2=-\frac{3 w_0 L^4}{64}            (4)

From Eqs. (3) and (4),

C_1=-\frac{17 w_{ o } L^3}{384} \text { and } C_2=\frac{5 w_{ o } L^4}{256}

After substituting C_1 \text { and } C_2 , Eqs. (1) and (2) become

(E I) \frac{ d y}{ d x}=\frac{w_{ o } x^3}{6}-\frac{3}{8} w_{ o } L\left\langle x-\frac{L}{2}\right\rangle^2-\frac{w_{ o }}{6}\left\langle x-\frac{L}{2}\right\rangle^3+\frac{w_{ o }}{6}\langle x-L\rangle^3-\frac{17 w_{ o } L}{384}              (5)

and        (E I) y=\frac{w_{ o } x^4}{24}-\frac{w_{ o } L}{8}\left\langle x-\frac{L}{2}\right\rangle^3-\frac{w_{ o }}{24}\left\langle x-\frac{L}{2}\right\rangle^4+\frac{w_{ o }}{24}\langle x-L\rangle^4-\frac{17 w_{ o } L^3 x}{384}+\frac{5 w_{ o } L^4}{256}               (6)

Now to obtain deflection at A and C, we need to put x = 0 and x = L in Eq. (6)

\left.(E I) y\right|_{x=0}=(E I) \delta_{ A }=\frac{5 w_{ o } L^4}{256}

or        \delta_{ A }=\frac{5 w_{ o } L^4}{256 E I}(\downarrow)

and        \left.(E I) y\right|_{x=l}=(E I) \delta_{ C }=\frac{w_{ o } L^4}{24}-\frac{w_{ o } L^4}{64}-\frac{w_{ o } L^4}{384}+\frac{5 w_{ o } L^4}{256}-\frac{17 w_{ o } L^4}{384}=-\frac{w_{ o } L^4}{768 E I}

or        \delta_{ C }=-\frac{w_{ o } L^4}{768 E I}

Since deflection at point C is negative, it must be in the upward direction. Hence,

\delta_{ C }=\frac{w_{ o } L^4}{768 E I}(\uparrow)