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.

Question

Accepted Answer

Redundant. The beam has one degree of indeterminacy. The bending moment [latex]M_B[/latex], at the interior support B, is the redundant.

Three-Moment Equation at Joint B. Considering the supports, A, B, and C as [latex]\ell, c[/latex], and r, respectively, and substituting [latex]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 , ext { and } P_{r}=w_{\ell}=\Delta_{\ell}=\Delta_{c}=\Delta_{r}=0[/latex] Eq. (D.9), we obtain

[latex] \frac{M_{\ell } L_{\ell}}{I_{\ell}} + 2M_c \left( \frac{L_{\ell}}{I_{\ell}} + \frac{L_r}{I_r} ight) + \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} ight) [/latex] (D.9)

[latex]\frac{M_{A}(24)}{2 I}+2 M_{B}\left(\frac{24}{2 I}+\frac{20}{I} ight)+\frac{M_{C}(20)}{I}=-\frac{30(24)^{2}(1 / 3)}{2 I}\left[1-(1 / 3)^{2} ight]-\frac{20(24)^{2}(2 / 3)}{2 I}\left[1-(2 / 3)^{2} ight]-\frac{2.5(20)^{3}}{4 I}[/latex]

Since A and C are simple end supports, we have by inspection

[latex]M_{A}=M_{C}=0[/latex]

Thus, the three-moment equation becomes

[latex]64 M_{B}=-9,693.33[/latex]

from which we obtain the redundant bending moment to be

[latex]M_{B}=-151.5 k - ft[/latex]

Span 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 [latex]M_B[/latex] 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

[latex]+\circlearrowleft \sum M_{B}=0 \quad-A_{y}(24)+30(16)+20(8)-151.5=0[/latex]

[latex]A_{y}=20.4 k \uparrow[/latex]

[latex]+\uparrow \sum F_{y}=0 \quad 20.4-30-20+B_{y}^{A B}=0[/latex]

[latex]B_{y}^{A B}=29.6 k \uparrow[/latex]

Similarly, for span BC,

[latex]+\circlearrowleft \sum M_{C}=0 \quad-B_{y}^{B C}(20)+151.5+2.5(20)(10)=0[/latex]

[latex]B_{y}^{B C}=32.6 k \uparrow[/latex]

[latex]+\uparrow \sum F_{y}=0 \quad 32.6-2.5(20)+C_{y}=0[/latex]

[latex]C_{y}=17.4 k \uparrow[/latex]

By considering the equilibrium of joint B in the vertical direction, we obtain

[latex]B_{y}=B_{y}^{A B}+B_{y}^{B C}=29.6+32.6=62.2 k \uparrow[/latex]

The reactions are shown in Fig. D.3(c).

Shear and Bending Moment Diagrams. See Fig. D.3(d).

Question D.1: Determine the reactions and draw the shear and bending momen...

Related Answered Questions

Determine the reactions for the continuous beam shown in Fig. D.5(a) by the three-moment equation. ...

Verified Answer:

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. ...

Verified Answer: