Determine the reactions and draw the shear and bending moment diagrams for the two-span continuous beam shown in Fig. 13.5(a) using the method of consistent deformations.

Question

Determine the reactions and draw the shear and bending moment diagrams for the two-span continuous beam shown in [latex]Fig. 13.5(a)[/latex] using the method of consistent deformations.

Accepted Answer

Degree of Indeterminacy The beam is supported by four reactions, so its
degree of indeterminacy is equal to [latex]4 - 3 = 1[/latex].

Primary Beam The vertical reaction [latex]B_y[/latex] at the roller support [latex]B[/latex] is selected to be the redundant, and the primary beam is obtained by removing the roller support [latex]B[/latex] from the given indeterminate beam, as shown in Fig. 13.5(b). Next, the primary beam is subjected separately to the external loading and a unit value of the unknown redundant [latex]B_y[/latex], as shown in Fig. 13.5(b) and (c), respectively. As shown in these figures, [latex]Delta_{BO}[/latex] denotes the deflection at [latex]B[/latex] due to the external loading, whereas [latex]f_{BB}[/latex] denotes the flexibility coefficient representing the deflection at [latex]B[/latex] due to the unit value of the redundant [latex]B_y[/latex].

Compatibility Equation Because the deflection at support [latex]B[/latex] of the actual indeterminate beam is zero, the algebraic sum of the deflections of the primary beam at [latex]B[/latex] due to the external loading and the redundant [latex]B_y[/latex] must also be zero. Thus, the compatibility equation can be written as

[latex]\Delta_{BO} + f_{BB} B_y = 0[/latex] (1)

Deflections of Primary Beam The flexural rigidity [latex]EI[/latex] of the primary beam is not constant (since the moment of inertia of the right half of the beam, [latex]BD[/latex], is twice the moment of inertia of the left half, [latex]AB[/latex]), so we cannot use the formulas given inside the front cover of the book for computing deflections. Therefore, we will use the conjugate-beam method, discussed in Chapter 6, for determining the deflections of the primary beam.

To determine the deflection [latex]\Delta_{BO}[/latex] due to the external loading, we draw the conjugate beams for the [latex]15-kN/m[/latex] uniformly distributed load and the [latex]60-kN[/latex] concentrated load, as shown in Fig. 13.5(d) and (e), respectively. Recalling that the deflection at a point on a real beam is equal to the bending moment at that point in the corresponding conjugate beam, we determine the deflection [latex]\Delta_{BO}[/latex] due to the combined effect of the distributed and concentrated loads as

[latex]EI \Delta_{BO}=[-4,218.75(10) +(\frac{2}{3})(10)(750)(\frac{30}{8})][/latex]

[latex]+ [-718.75(10) + (\frac{1}{2})(10)(150)(\frac{10}{3})][/latex]

[latex]\Delta_{BO}=\frac{28,125 kN . m^3}{EI}[/latex]

in which the negative sign indicates that the deflection occurs in the downward direction. Note that although the numerical values of [latex]E[/latex] and [latex]I[/latex] are given, it is usually convenient to carry out the analysis in terms of [latex]EI[/latex]. The flexibility coefficient [latex]f_{BB}[/latex] can be computed similarly by using the conjugate beam shown in Fig. 13.5(f ). Thus

[latex]EIf_{BB} = 20.833(10) -(\frac{1}{2})(10)(5)(\frac{10}{3}) = 125 kN . m^3/kN[/latex]

[latex]f_{BB}=\frac{125 kN .m^3/kN}{EI}[/latex]

Magnitude of the Redundant By substituting the values of [latex]\Delta_{BO}[/latex] and [latex]f_{BB}[/latex] into the compatibility equation (Eq. (1)), we obtain

[latex]-\frac{28,125}{EI} + (\frac{125}{EI})B_y=0[/latex] [latex]B_y = 225 kN \uparrow [/latex] Ans.

Reactions To determine the remaining reactions of the indeterminate beam, we apply the equilibrium equations (Fig. 13.5(g)):

[latex]+ \longrightarrow \sum{F_x}=0[/latex] [latex]A_x=0[/latex] Ans.

[latex]+\circlearrowleft \sum{M_D}=0[/latex] [latex]-A_y(20) - 225(10) + 15(20)(10) + 60(5) = 0[/latex]

[latex]A_y= 52.5 kN \uparrow[/latex] Ans.

[latex]+ \uparrow \sum{F_y}=0[/latex] [latex]52.5 + 225 - 15(20) - 60 + D_y = 0[/latex]

[latex]D_y = 82.5 kN \uparrow[/latex] Ans.

Shear and Bending Moment Diagrams See Fig. 13.5(h). Ans.

Question 13.3: Determine the reactions and draw the shear and bending momen...

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