Question D.3: Determine the reactions for the continuous beam shown in Fig...

Since support A of the beam is fixed, we replace it with an imaginary interior roller support with an adjoining end span of zero length, as shown in Fig. D.5(b).

Redundants. From Fig. D.5(b), we can see that the bending moments M_A and M_B at the supports A and B, respectively, are the redundants.

Three-Moment Equation at Joint A. By using Eq. (D.10)

M_{\ell} L_{\ell}+2 M_{c}\left(L_{\ell}+L_{r}\right)+M_{r} L_{r}=-\sum P_{\ell} L_{\ell}^{2} k_{\ell}\left(1-k_{\ell}^{2}\right)-\sum P_{r} L_{r}^{2} k_{r}\left(1-k_{r}^{2}\right)-\frac{1}{4}\left(w_{\ell} L_{\ell}^{3}+w_{r} L_{r}^{3}\right)-6 E I\left(\frac{\Delta_{\ell}-\Delta_{c}}{L_{\ell}}+\frac{\Delta_{r}-\Delta_{c}}{L_{r}}\right)                    (D.10)
for supports A′, A, and B, we obtain

or

2 M_{A}+M_{B}=-337.5                              (1)

Three-Moment Equation at Joint B. Similarly, applying Eq. (D.10) for supports A,B, and C, we write

M_{A}(20)+2 M_{B}(20+30)+M_{C}(30)=-45(20)^{2}(1 / 2)\left[1-(1 / 2)^{2}\right]-(1 / 4)(1.8)(30)^{3}
The bending moment at end C of the cantilever overhang CD is computed as

By substituting M_{C}=-90 k – ft into the foregoing three-moment equation and simplifying, we obtain

M_{A}+5 M_{B}=-810                          (2)

Support Bending Moments. Solving Eqs. (1) and (2), we obtain

Span End Shears and Reactions. See Figs. D.5(c) and (d).