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

This beam has three unknown reactions \left(R_{A,} R_{B,} \text { and } R_{C}\right) \text {. }. Since there are two equations of equilibrium for the beam as a whole, it is statically indetermi-nate 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}=q L-\frac{R_{B}}{2}   (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 \left(\delta_{B}\right)_{1} \text { and }\left(\delta_{B}\right)_{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

\delta_{B}=\left(\delta_{B}\right)_{1}-\left(\delta_{B}\right)_{2}=0   (b)

in which the deflection \left(\delta_{B}\right)_{1} is positive downward and the deflection \left(\delta_{B}\right)_{2} is positive upward.
Force-displacement relations. The deflection \left(\delta_{B}\right)_{1} caused by the uniform load acting on the released structure (Fig. 10-14c) is obtained from Table G-2, Case 1 as

Table G-2
Deflections and Slopes of Simple Beams

	v = deflection in the y direction (positive upward) v^{\prime} = dv/dx = slope of the deflection curve \delta_{c} = -v(L/2) = deflection at midpoint C of the beam (positive downward) x_{1} = distance from support A to point of maximum deflection \delta_{\max }=-v_{\max } = maximum deflection (positive downward) \theta_{A}=-v^{\prime}(0)= angle of rotation at left-hand end of the beam (positive clockwise) \theta_{B}=v^{\prime}(L) =angle of rotation at right-hand end of the beam (positive counterclockwise) EI = constant
	\begin{aligned}&v=-\frac{q x}{24 E I}\left(L^{3}-2 L x^{2}+x^{3}\right) \\&v^{\prime}=-\frac{q}{24 EL}\left(L^{3}-6 L x^{2}+4 x^{3}\right) \\&\delta_{C}=\delta_{\max }=\frac{5 q L^{4}}{384 E I} \quad \theta_{A}=\theta_{B}=\frac{q L^{3}}{24 E I}\end{aligned}
	\begin{aligned}&v=-\frac{q x}{384 E I}\left(9 L^{3}-24 L x^{2}+16 x^{3}\right) \quad\left(0 \leq x \leq \frac{L}{2}\right) \\&v^{\prime}=-\frac{q}{384 E I}\left(9 L^{3}-72 L x^{2}+64 x^{3}\right) \quad\left(0 \leq x \leq \frac{L}{2}\right) \\&v=-\frac{q L}{384 E I}\left(8 x^{3}-24 L x^{2}+17 L^{2} x-L^{3}\right) \quad\left(\frac{L}{2} \leq x \leq L\right) \\&v^{\prime}=-\frac{q L}{384 E I}\left(24 x^{2}-48 L x+17 L^{2}\right) \quad\left(\frac{L}{2} \leq x \leq L\right) \\&\delta_{C}=\frac{5 q L^{4}}{768 E I} \quad \theta_{A}=\frac{3 q L^{3}}{128 E I} \quad \theta_{B}=\frac{7 q L^{3}}{384 E I}\end{aligned}
	\begin{aligned}&v=-\frac{q x}{24 L E I}\left(a^{4}-4 a^{3} L+4 a^{2} L^{2}+2 a^{2} x^{2}-4 a L x^{2}+L x^{3}\right) \quad(0 \leq x \leq a) \\&v^{\prime}=-\frac{q}{24 L E I}\left(a^{4}-4 a^{3} L+4 a^{2} L^{2}+6 a^{2} x^{2}-12 a L x^{2}+4 L x^{3}\right) \quad(0 \leq x \leq a) \\&v=-\frac{q a^{2}}{24 L E I}\left(-a^{2} L+4 L^{2} x+a^{2} x-6 L x^{2}+2 x^{3}\right) \quad(a \leq x \leq L) \\&v^{\prime}=-\frac{q a^{2}}{24 L E I}\left(4 L^{2}+a^{2}-12 L x+6 x^{2}\right) \quad(a \leq x \leq L) \\&\theta_{A}=\frac{q a^{2}}{24 L E I}(2 L-a)^{2} \quad \theta_{B}=\frac{q a^{2}}{24 L E I}\left(2 L^{2}-a^{2}\right)\end{aligned}
	\begin{aligned}&v=-\frac{P X}{48 E I}\left(3 L^{2}-4 x^{2}\right) \quad v^{\prime}=-\frac{P}{16 EI}\left(L^{2}-4 x^{2}\right) \quad\left(0 \leq x \leq \frac{L}{2}\right) \\&\delta_{C}=\delta_{\max }=\frac{P L^{3}}{48 E I} \quad \theta_{A}=\theta_{B}=\frac{P L^{2}}{16 E I}\end{aligned}
	\begin{aligned}&v=-\frac{P b x}{6 L E I}\left(L^{2}-b^{2}-x^{2}\right) \quad v^{\prime}=-\frac{P b}{6 L E I}\left(L^{2}-b^{2}-3 x^{2}\right) \quad(0 \leq x \leq a) \\&\theta_{A}=\frac{P a b(L+b)}{6 L E I} \quad \theta_{B}=\frac{P a b(L+a)}{6 L E I} \\&\text { If } a \geq b, \quad \delta_{C}=\frac{P b\left(3 L^{2}-4 b^{2}\right)}{48 E I} \text { If } a \leq b, \quad \delta_{C}=\frac{P a\left(3 L^{2}-4 a^{2}\right)}{48 E I} \\&\text { If } a \geq b, x_{1}=\sqrt{\frac{L^{2}-b^{2}}{3}} \text { and } \delta_{\max }=\frac{P b\left(L^{2}-b^{2}\right)^{3 / 2}}{9 \sqrt{3} L E I}\end{aligned}
	\begin{array}{ll}v=-\frac{P X}{6 E I}\left(3 a L-3 a^{2}-x^{2}\right) & v^{\prime}=-\frac{P}{2 E I}\left(a L-a^{2}-x^{2}\right) \quad(0 \leq x \leq a) \\v=-\frac{P a}{6 E I}\left(3 L x-3 x^{2}-a^{2}\right) & v^{\prime}=-\frac{P a}{2 E I}(L-2 x) \quad(a \leq x \leq L-a) \\\delta_{C}=\delta_{\max }=\frac{P a}{24 E I}\left(3 L^{2}-4 a^{2}\right) \quad \theta_{A}=\theta_{B}=\frac{P a(L-a)}{2 E I}\end{array}
	\begin{aligned}&v=-\frac{M_{0} x}{6 L E I}\left(2 L^{2}-3 L x+x^{2}\right) \quad v^{\prime}=-\frac{M_{0}}{6 L E I}\left(2 L^{2}-6 L x+3 x^{2}\right) \\&\delta_{C}=\frac{M_{0} L^{2}}{16 E I} \quad \theta_{A}=\frac{M_{0} L}{3 E I} \quad \theta_{B}=\frac{M_{0} L}{6 E I} \\&x_{1}=L\left(1-\frac{\sqrt{3}}{3}\right) \quad \text { and } \quad \delta_{\max }=\frac{M_{0} L^{2}}{9 \sqrt{3} E I}\end{aligned}
	\begin{aligned}&v=-\frac{M_{0} X}{24 L E I}\left(L^{2}-4 x^{2}\right) \quad v^{\prime}=-\frac{M_{0}}{24 L E I}\left(L^{2}-12 x^{2}\right) \quad\left(0 \leq x \leq \frac{L}{2}\right) \\&\delta_{C}=0 \quad \theta_{A}=\frac{M_{0} L}{24 E I} \quad \theta_{B}=-\frac{M_{0} L}{24 E I}\end{aligned}
	\begin{aligned}&v=-\frac{M_{0} x}{6 L E I}\left(6 a L-3 a^{2}-2 L^{2}-x^{2}\right) \quad(0 \leq x \leq a) \\&v^{\prime}=-\frac{M_{0}}{6 L E I}\left(6 a L-3 a^{2}-2 L^{2}-3 x^{2}\right) \quad(0 \leq x \leq a) \\&\text { At } x=a: v=\frac{M_{0} a b}{3 L EI}(2 a-L) \quad v^{\prime}=-\frac{M_{0}}{3 L E I}\left(3 a L-3 a^{2}-L^{2}\right) \\&\theta_{A}=\frac{M_{0}}{6 L E I}\left(6 a L-3 a^{2}-2 L^{2}\right) \quad \theta_{B}=\frac{M_{0}}{6 L E I}\left(3 a^{2}-L^{2}\right)\end{aligned}
	\begin{aligned}&v=-\frac{M_{0} x}{2 E I}(L-x) \quad v^{\prime}=-\frac{M_{0}}{2 E I}(L-2 x) \\&\delta_{C}=\delta_{\max }=\frac{M_{0} L^{2}}{8 EI} \quad \theta_{A}=\theta_{B}=\frac{M_{0} L}{2 EI}\end{aligned}
	\begin{aligned}&v=-\frac{q_{0} x}{360 L E I}\left(7 L^{4}-10 L^{2} x^{2}+3 x^{4}\right) \\&v^{\prime}=-\frac{q_{0}}{360 L E I}\left(7 L^{4}-30 L^{2} x^{2}+15 x^{4}\right) \\&\delta_{C}=\frac{5 q_{0} L^{4}}{768 E I} \quad \theta_{A}=\frac{7 q_{0} L^{3}}{360 E I} \quad \theta_{B}=\frac{q_{0} L^{3}}{45 EI} \\&x_{1}=0.5193 L \quad \delta_{\max }=0.00652 \frac{q_{0} L^{4}}{E I}\end{aligned}
	\begin{aligned}&v=-\frac{q_{0} x}{960 L E I}\left(5 L^{2}-4 x^{2}\right)^{2} \quad\left(0 \leq x \leq \frac{L}{2}\right) \\&\left.v^{\prime}=-\frac{q_{0}}{192 L E I}\left(5 L^{2}-4 x^{2}\right) L^{2}-4 x^{2}\right) \quad\left(0 \leq x \leq \frac{L}{2}\right) \\&\delta_{C}=\delta_{\max }=\frac{q_{0} L^{4}}{120 E I} \quad \theta_{A}=\theta_{B}=\frac{5 q_{0} L^{3}}{192 E I}\end{aligned}
	\begin{aligned}&v=-\frac{q_{0} L^{4}}{\pi^{4} E I} \sin \frac{\pi x}{L} \quad v^{\prime}=-\frac{q_{0} L^{3}}{\pi^{3} E I} \cos \frac{\pi x}{L} \\&\delta_{C}=\delta_{\max }=\frac{q_{0} L^{4}}{\pi^{4} E I} \quad \theta_{A}=\theta_{B}=\frac{q_{0} L^{3}}{\pi^{3} EI}\end{aligned}

where 2L is the length of the released structure. The deflection \left(\delta_{B}\right)_{2}
produced by the redundant (Fig. 10-14d) is

as obtained from Table G-2, Case 4.
Reactions. The equation of compatibility pertaining to the vertical deflection at point B [Eq. (b)] now becomes

\delta_{B}=\frac{5 q L^{4}}{24 E I}-\frac{R_{B} L^{3}}{6 E I}=0   (c)

from which we find the reaction at the middle support:

R_{B}=\frac{5 q L}{4}   (10-28)

The other reactions are obtained from Eq. (a):

R_{A}=R_{C}=\frac{3 q L}{8}  (10-29)

With the reactions known, we can find the shear forces, bending moments, stresses, and deflections without difficulty.

Note: The purpose of this example is to provide an illustration of the method of superposition, and therefore we have described all steps 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 we know that the slope of the beam at the middle sup-port must be zero, and 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 our previous results for a propped cantilever beam with a uniform load [Eqs. (10-1) to (10-12)]

R_{B}=\frac{3 q L}{8}    (10-1)

R_{A}=\frac{5 q L}{8} \quad M_{A}=\frac{q L^{2}}{8}    (10-2a,b)

V=R_{A}-q x=\frac{5 q L}{8}-q x  (10-3)

M=R_{A} x-M_{A}-\frac{q x^{2}}{2}=\frac{5 q L x}{8}-\frac{q L^{2}}{8}-\frac{q x^{2}}{2}    (10-4)

V_{\max }=\frac{5 q L}{8}    (10-5)

M_{\text {pos }}=\frac{9 q L^{2}}{128} \quad M_{\text {neg }}=-\frac{q L^{2}}{8}    (10-6 a,b)

v^{\prime}=\frac{q x}{48 E I}\left(-6 L^{2}+15 L x-8 x^{2}\right)    (10-7)

v=-\frac{q x^{2}}{48 E I}\left(3 L^{2}-5 L x+2 x^{2}\right)    (10-8)

x_{1}=\frac{15-\sqrt{33}}{16} L=0.5785 L    (10-9)

\begin{aligned}\delta_{\max } &=-(v)_{x=x_{1}}=\frac{q L^{4}}{65,536 E I}(39+55 \sqrt{33}) \\&=\frac{q L^{4}}{184.6 E I}=0.005416 \frac{q L^{4}}{E I}\end{aligned}  (10-10)

\delta_{0}=-(v)_{x=L 4}=\frac{5 q L^{4}}{2048 E I}=0.002441 \frac{q L^{4}}{EI}    (10-11)

\theta_{B}=\left(v^{\prime}\right)_{x-L}=\frac{q L^{3}}{48 E_{j}}    (10-12)

can be adapted immedi-ately to the continuous beam of Fig. 10-14a.