Question 15.CA.5: Determine the reactions at the supports for the prismatic be...

Equilibrium Equations. From the free-body diagram of Fig. 15.14b,

\begin{array}{lll}\stackrel{+}{\rightarrow} F_x=0: & & A_x=0 \\+\uparrow \Sigma F_y=0: & & A_y+B-w L=0 \\+\circlearrowleft \Sigma M_A=0: & & M_A+B L-\frac{1}{2} w L^2=0\end{array}     (1)

Equation of Elastic Curve. Draw the free-body diagram of a portion of beam AC (Fig. 15.16) to obtain

+\circlearrowleft \Sigma M_C=0: \quad M+\frac{1}{2} w x^2+M_A-A_y x=0     (2)

Solving Eq. (2) for M and carrying into Eq. (15.4),

\frac{d^2 y}{d x^2}=\frac{M(x)}{E I}     (15.4)

Integrating in x gives

EI  \theta=E I \frac{d y}{d x}=-\frac{1}{6} w x^3+\frac{1}{2} A_y x^2-M_A x+C_1     (3)

EI  y=-\frac{1}{24} w x^4+\frac{1}{6} A_y x^3-\frac{1}{2} M_A x^2+C_1 x+C_2     (4)

Referring to the boundary conditions indicated in Fig. 15.15, we set x = 0, θ = 0 in Eq. (3); x = 0, y = 0 in Eq. (4); and conclude that C_1 = C_2 = 0. Thus, Eq. (4) is rewritten as

EI  y=-\frac{1}{24} w x^4+\frac{1}{6} A_y x^3-\frac{1}{2} M_A x^2     (5)

But the third boundary condition requires that y = 0 for x = L. Carrying these values into Eq. (5),

or

3 M_A-A_y L+\frac{1}{4} w L^2=0     (6)

Solving this equation simultaneously with the three equilibrium equations of Eq. (1), the reactions at the supports are