Question 10.3: A two-span continuous beam ABC supports a uniform load of in......

Use a four-step problem-solving approach.
1. Conceptualize: This beam has three unknown reactions (R_A, R_B, ~ and ~ R_C). Since there are two equations of equilibrium for the beam as a whole, it is statically
indeterminate to the first degree. For convenience, select the reaction R_B at the middle support as the redundant.
2. Categorize:
Equations of equilibrium: Express the reactions R_A ~ and ~ R_C in terms of the redundant R_B by means of two equations of equilibrium. The first equation, which is for equilibrium of moments about point B, shows that R_A ~ and ~ R_C are equal. The second equation, which is for equilibrium in the vertical direction, yields
\quad\quad\quad\quad R_{A}=R_{C}=q L-{\frac{R_{B}}{2}}\quad\quad(a)
Equation of compatibility: Because the reaction R_B is selected as the redundant, the released structure is a simple beam with supports at A and C (Fig. 10-14b).
The deflections at point B in the released structure due to the uniform load q and the redundant R_B are shown in Figs. 10-14c and d, respectively. Note that the deflections are denoted (\delta_B)_1 ~ and ~ (\delta_B)_2. The superposition of these deflections must produce the deflection \delta_B in the original beam at point B. Since the latter deflection is equal to zero, the equation of compatibility is
\quad\quad\quad\quad \delta_{B}=(\delta_{B})_{1}-(\delta_{B})_{2}=0\quad\quad(b)
in which the deflection (\delta_B)_1 is positive downward, and the deflection (\delta_B)_2 is positive upward.
Force-displacement relations: The deflection (\delta_B)_1 caused by the uniform load acting on the released structure (Fig. 10-14c) is obtained from Table H-2, Case 1 as
\quad\quad\quad\quad (\delta_{B})_{1}=\frac{5 q(2{L})^{4}}{384E I}=\frac{5 q{L}^{4}}{24{L}}
where 2L is the length of the released structure. The deflection (\delta_B)_2 produced by the redundant (Fig. 10-14d) is
\quad\quad\quad\quad (\delta_{B})_{2}=\frac{{R}_{B}(2{L})^{3}}{48E{I}}=\frac{{R}_{B}{L}^3}{6E{I}}
as obtained from Table H-2, Case 4.
3. Analyze:
Reactions: The equation of compatibility pertaining to the vertical deflection at point B [Eq. (b)] now becomes
\quad\quad\quad\quad \delta_{B}=\frac{5 q L^{4}}{24E I}-\frac{R_{B}L^{3}}{6E I}=0\quad\quad(c)
from which the reaction at the middle support is
\quad\quad\quad\quad R_{B}={\frac{5q L}{4}}\quad\quad(10-28)
The other reactions are obtained from Eq. (a):
\quad\quad\quad\quad R_{A}=R_{C}={\frac{3q L}{8}}\quad\quad(10-29)
With the reactions known, the shear forces, bending moments, stresses, and deflections can be found without difficulty.
4. Finalize: The purpose of this example is to demonstrate the method of superposition, so all steps were described in the analysis. However, this particular beam (Fig. 10-14a) can be analyzed by inspection because of the symmetry of the beam and its loading.
From symmetry, the slope of the beam at the middle support must be zero; therefore, each half of the beam is in the same condition as a propped cantilever beam with a uniform load (see, for instance, Fig. 10-6). Consequently, all of the previous results for a propped cantilever beam with a uniform load [Eqs. (10-1) to (10-12)] can be adapted immediately to the continuous beam of Fig. 10-14a.