Question D.2: Determine the reactions for the continuous beam shown in Fig...
Determine the reactions for the continuous beam shown in Fig. D.4(a) due to the uniformly distributed load and due to the support settlements of 10 mm at A, 50 mm at B, 20 mm at C, and 40 mm at D. Use the three-moment equation.
Redundants. The bending moments M_B and M_C, at the interior supports B and C, respectively, are the redundants.
Three-Moment Equation at Joint B. By considering the supports A, B, and C as \ell, c and r, respectively, and substituting L=10 m , E=200 GPa =200\left(10^{6}\right) kN / m ^{2}, I=700\left(10^{6}\right) mm ^{4}=700\left(10^{-6}\right) m ^{4}, w_{\ell}=w_{r}=30 kN / m , \Delta_{\ell}=\Delta_{A}=10 mm =0.01 m , \Delta_{c}=\Delta_{B}=50 mm =0.05 m , \Delta_{r}=\Delta_{C}=20 mm =0.02 m and P_{\ell}=P_{r}=0 into Eq. (D.11), we write
M_\ell + 4M_c +M_r \\ = -\sum{P_\ell L k _\ell (1-k^2_\ell)}-\sum{P_rL k_r (1-K^2_r)} – \frac{L^2}{4}(w_\ell + w_r) – \frac{6EI}{L^2}(\Delta _\ell – 2\Delta _c +\Delta _r) (D.11)
M_{A}+4 M_{B}+M_{C}=-\frac{(10)^{2}}{4}(30+30)-\frac{6(200)(700)}{(10)^{2}}[0.01-2(0.05)+0.02]
Since A is a simple end support, M_{A}=0 The foregoing equation thus simplifies to
4 M_{B}+M_{C}=-912 (1)
Three-Moment Equation at Joint C. Similarly, by considering the supports B, C, and D as \ell, c and r, respectively, and
by substituting the appropriate numerical values in Eq. (D.11), we obtain
Since D is a simple end support, M_{D}=0 Thus, the foregoing equation becomes
M_{B}+4 M_{C}=-1,920 (2)
Support Bending Moments. Solving Eqs. (1) and (2) simultaneously for M_B and M_C, we obtain
M_{B}=-115.2 kN \cdot mM_{C}=-451.2 kN \cdot m
Span End Shears and Reactions. With the redundants MB and MC known, the span end shears and the support reactions can be determined by considering the equilibrium of the free bodies of the spans AB;BC, and CD, and joints B and C, as shown in Fig. D.4(b). The reactions are shown in Fig. D.4(c).