Question 10.3: A two-span continuous beam ABC supports a uniform load of in......
A two-span continuous beam ABC supports a uniform load of intensity q, as shown in Fig. 10-14a. Each span of the beam has length L. Using the method of superposition, determine all reactions for this beam.
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, let us select the reaction R_B at the middle support as the redundant.
Equations of equilibrium. We can 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 the following result:
R_A=R_C=qL\ -\ \frac{R_B}{2} (k)
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 Rg are shown in Figs. 10-14c and d, respectively. Note that the deflections are denoted (δ_B)_1 and (δ_B)_2. The superposition of these deflections must produce the deflection δ_B in the original beam at point B. Since the latter deflection is equal to zero, the equation of compatibility is
δ_B=(δ_B)_1\ -\ (δ_B)_2 = 0 (l)
in which the deflection (δ_B)_1 is positive downward and the deflection (δ_B)_2 is positive upward.
Force-displacement relations. The deflection (δ_B)_1 caused by the uniform load acting on the released structure (Fig. 10-14c) is obtained from Table G-2, Case 1, as follows:
(δ_B)_1=\frac{5q(2L)^4}{384EI}=\frac{5qL^4}{24EI}
where 2L is the length of the released structure. The deflection (δ_B)_2 produced by the redundant (Fig. 10-14d) is
(δ_B)_2=\frac{R_B(2L)^3}{48EI}=\frac{R_BL^3}{6EI}
as obtained from Table G-2. Case 4.
The equation of compatibility pertaining to the vertical deflection at point B (Eq. 1) now becomes
δ_B=\frac{5qL^4}{24EI}\ -\ \frac{R_BL^3}{6EI} = 0 (m)
from which we find the reaction at the middle support:
R_B=\frac{5qL}{4} (10-23)
The other reactions are obtained from Eq. (k):
R_A=R_C=\frac{3qL}{8} (10-24)
With the reactions known, we can find the shear forces, bending moments, stresses, and deflections without difficulty.