Calculate the deflections at points A and C of the beam shown in Figure 7.27. Assume EI to be constant.

Question

Accepted Answer

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

Consider

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

Therefore,

[latex] R_{ D }=\frac{w_{ o } L}{4}(\uparrow) [/latex]

But,

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

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

[latex] M_x=-\frac{w_{ o } x^2}{2}+R_{ B }\left\langle x-\frac{L}{2} ight angle+\frac{w_{ o }}{2}\left\langle x-\frac{L}{2} ight angle^2-\frac{w_{ o }}{2}\langle x-L angle^2 [/latex]

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

[latex] M_x=-\frac{w_{ o } x^2}{2}+\frac{3}{4} w_{ o } L\left\langle x-\frac{L}{2} ight angle+\frac{w_{ o }}{2}\left\langle x-\frac{L}{2} ight angle^2-\frac{w_{ o }}{2}\langle x-L angle^2 [/latex]

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

[latex] (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} ight angle-\frac{w_0}{2}\left\langle x-\frac{L}{2} ight angle^2+\frac{w_0}{2}\langle x-L angle^2 [/latex]

Integrating successively the above equation, we get

[latex] (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} ight angle^2-\frac{w_{ o }}{6}\left\langle x-\frac{L}{2} ight angle^3+\frac{w_{ o }}{6}\langle x-L angle^3+C_1 [/latex] (1)

and [latex] (E I) y=\frac{w_{ o } x^4}{24}-\frac{w_{ o } L}{8}\left\langle x-\frac{L}{2} ight angle^3-\frac{w_{ o }}{24}\left\langle x-\frac{L}{2} ight angle^4+\frac{w_{ o }}{24}\langle x-L angle^4+C_1 x+C_2 [/latex] (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

[latex] \frac{w_{ o }}{24}\left\lgroup \frac{L}{2} ight group^4+\left\lgroup \frac{L}{2} ight group C_1+C_2=0 [/latex]

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

and [latex] \frac{w_{ o }}{24}\left\lgroup \frac{3 L}{2} ight group^4-\frac{w_{ o } L^4}{8}-\frac{w_{ o } L^4}{24}+\frac{w_{ o }}{24}\left\lgroup \frac{L}{2} ight group^4+\frac{3 L C_1}{2}+C_2=0 [/latex]

or [latex] \left\lgroup \frac{3 L}{2} ight group C_1+C_2=-\frac{3 w_0 L^4}{64} [/latex] (4)

From Eqs. (3) and (4),

[latex] C_1=-\frac{17 w_{ o } L^3}{384} ext { and } C_2=\frac{5 w_{ o } L^4}{256} [/latex]

After substituting [latex] C_1 ext { and } C_2 [/latex], Eqs. (1) and (2) become

[latex] (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} ight angle^2-\frac{w_{ o }}{6}\left\langle x-\frac{L}{2} ight angle^3+\frac{w_{ o }}{6}\langle x-L angle^3-\frac{17 w_{ o } L}{384} [/latex] (5)

and [latex] (E I) y=\frac{w_{ o } x^4}{24}-\frac{w_{ o } L}{8}\left\langle x-\frac{L}{2} ight angle^3-\frac{w_{ o }}{24}\left\langle x-\frac{L}{2} ight angle^4+\frac{w_{ o }}{24}\langle x-L angle^4-\frac{17 w_{ o } L^3 x}{384}+\frac{5 w_{ o } L^4}{256} [/latex] (6)

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

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

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

and [latex] \left.(E I) y ight|_{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} [/latex]

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

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

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

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

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