Question D.1: Determine the reactions and draw the shear and bending momen...
Determine the reactions and draw the shear and bending moment diagrams for the beam shown in Fig. D.3(a) by using the three-moment equation.
Redundant. The beam has one degree of indeterminacy. The bending moment M_B, at the interior support B, is the redundant.
Three-Moment Equation at Joint B. Considering the supports, A, B, and C as \ell, c, and r, respectively, and substituting L_{\ell}=24 ft , L_{r}=20 ft , I_{\ell}=2 I, I_{r}=I, P_{\ell 1}=30 k , k_{\ell 1}=1 / 3, P_{\ell 2}=20 k , k_{\ell 2}=2 / 3, w_{r}=2.5 k / ft , \text { and } P_{r}=w_{\ell}=\Delta_{\ell}=\Delta_{c}=\Delta_{r}=0 Eq. (D.9), we obtain
\frac{M_{\ell } L_{\ell}}{I_{\ell}} + 2M_c \left( \frac{L_{\ell}}{I_{\ell}} + \frac{L_r}{I_r}\right) + \frac{M_r L_r}{I_r} \\ = -\sum{\frac{P_{\ell}L_{\ell}^2 k_{\ell}}{I_{\ell}}} (1-k^2_{\ell}) – \sum{\frac{P_r L_r^2k_r}{I_r}}(1-k^2_r) – \frac{w_\ell L^3 _\ell}{4I_\ell} – \frac{w_r L_r^3}{4I_r} -6E \left( \frac{\Delta _\ell -\Delta _c}{L_\ell} + \frac{\Delta _r – \Delta _c}{L_r}\right) (D.9)
\frac{M_{A}(24)}{2 I}+2 M_{B}\left(\frac{24}{2 I}+\frac{20}{I}\right)+\frac{M_{C}(20)}{I}=-\frac{30(24)^{2}(1 / 3)}{2 I}\left[1-(1 / 3)^{2}\right]-\frac{20(24)^{2}(2 / 3)}{2 I}\left[1-(2 / 3)^{2}\right]-\frac{2.5(20)^{3}}{4 I}
Since A and C are simple end supports, we have by inspection
M_{A}=M_{C}=0Thus, the three-moment equation becomes
64 M_{B}=-9,693.33from which we obtain the redundant bending moment to be
M_{B}=-151.5 k – ftSpan End Shears and Reactions. The shears at the ends of the spans AB and BC of the continuous beam can now be determined by applying the equations of equilibrium to the free bodies of the spans shown in Fig. D.3(b). Note that the negative bending moment M_B is applied at the ends B of spans AB and BC so that it causes tension in the upper fibers and compression in the lower fibers of the beam. By considering the equilibrium of span AB, we obtain
+\circlearrowleft \sum M_{B}=0 \quad-A_{y}(24)+30(16)+20(8)-151.5=0A_{y}=20.4 k \uparrow +\uparrow \sum F_{y}=0 \quad 20.4-30-20+B_{y}^{A B}=0
B_{y}^{A B}=29.6 k \uparrow
Similarly, for span BC,
+\circlearrowleft \sum M_{C}=0 \quad-B_{y}^{B C}(20)+151.5+2.5(20)(10)=0B_{y}^{B C}=32.6 k \uparrow
+\uparrow \sum F_{y}=0 \quad 32.6-2.5(20)+C_{y}=0
C_{y}=17.4 k \uparrow
By considering the equilibrium of joint B in the vertical direction, we obtain
B_{y}=B_{y}^{A B}+B_{y}^{B C}=29.6+32.6=62.2 k \uparrowThe reactions are shown in Fig. D.3(c).
Shear and Bending Moment Diagrams. See Fig. D.3(d).