Question 7.17: For the beam shown in Figure 7.31, find the support reaction...

We note from the given beam loading that it is a statically indeterminate system. Let us now draw the free-body diagram of the beam as shown in Figure 7.32(a).

We note from Figure 7.32(a) that

\sum F_y=0 \Rightarrow R_{ A }+R_{ B }=\frac{w_{ o } L}{2}             (1)

\sum M_{ A }=0 \Rightarrow R_{ B } L-M_{ A }=\frac{3 w_{ o } L^2}{8}        (2)

Now, from Figure 7.32(b) bending moment expression at any section at a distance x from end A is given by

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

Now from the flexure equation, we get

(E I) \frac{ d ^2 y}{ d x^2}=-M_x=-R_{ A } x-M_{ A }+\frac{w_{ o }}{2}\left\langle x-\frac{L}{2}\right\rangle^2

Integrating once, we get

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

Now, at x = 0, dy/dx = 0, so C_1=0 . Substituting in the above expression, we get

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

Again integrating, we get

(E I) y=-\frac{R_{ A } x^3}{6}-\frac{M_{ A } x^2}{2}+\frac{w_{ o }}{24}\left\langle x-\frac{L}{2}\right\rangle^4+C_2

Again we note that at x = 0, y = 0, so C_2=0 . Thus, we have

(E I) y=-\frac{R_{ A } x^3}{6}-\frac{M_{ A } x^2}{2}+\frac{w_{ o }}{24}\left\langle x-\frac{L}{2}\right\rangle^4

We observe that at x = L, y = 0. Therefore, from the above expression we note that

-\frac{R_{ A } L^3}{6}-\frac{M_{ A } L^2}{2}+\frac{w_{ o } L^4}{384}=0

or          R_{ A } L+3 M_{ A }=\frac{w_{ o } L^2}{64}           (3)

Now solving Eqs. (1)–(3) simultaneously,

\left\lgroup \frac{w_{ o } L}{2}-R_{ A } \right\rgroup L-M_{ A }=\frac{3 w_{ o } L^2}{8}

or          -R_{ A } L-M_{ A }=\frac{3 w_{ o } L^2}{8}-\frac{w_{ o } L^2}{2}=-\frac{w_{ o } L^2}{8}            (4)

Adding Eqs. (3) and (4),

2 M_{ A }=\frac{w_{ o } L^2}{64}-\frac{w_{ o } L^2}{8}=-\frac{7 w_{ o } L^2}{64}

Therefore,

M_{ A }=-\frac{7 w_{ o } L^2}{128}

Thus from Eq. (2),

R_{ B }=\frac{1}{L}\left\{\frac{3 w_{ o } L^2}{8}-\frac{7 w_{ o } L^2}{128}\right\}=\frac{41 w_{ o } L}{128} \Rightarrow R_{ A }=\frac{23 w_{ o } L}{128}

Thus, we have the support reactions as

R_{ A }=\frac{23 w_{ o } L}{128}(\uparrow), \quad M_{ A }=\frac{7 w_{ o } L^2}{128}(\uparrow) \quad \text { and } \quad R_{ B }=\frac{41 w_{ o } L}{128}(\uparrow)