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 a length L. Using the method of superposition, determine all reactions for this beam.

Question

Accepted Answer

Use a four-step problem-solving approach.
1. Conceptualize: This beam has three unknown reactions [latex](R_{A} , R_{B} , and R_{C})[/latex]. 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 [latex]R_{B}[/latex] at the middle support as the redundant.

2. Categorize:
Equations of equilibrium: Express the reactions [latex]R_{A} and R_{C}[/latex] in terms of the redundant [latex]R_{B}[/latex] by means of two equations of quilibrium. The first equation, which is for equilibrium of moments about point B, shows that [latex]R_{A} and R_{C}[/latex] are equal. The second equation, which is for equilibrium in the vertical direction, yields

[latex]R_{A}=R_{C}=q L-\frac{R_{B}}{2}[/latex] (a)

Equation of compatibility: Because the reaction [latex]R_{B}[/latex] 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 [latex]R_{B}[/latex] are shown in Figs. 10-14c and d, respectively. Note that the deflections are denoted [latex](\delta_{B} )_{1} and (\delta_{B} )_{2}[/latex]. The superposition of these deflections must produce the deflection [latex]\delta_{B}[/latex] in the original beam at point B. Since the latter deflection is equal to zero, the equation of compatibility is

[latex]\delta_{B}=\left(\delta_{B} ight)_{1}-\left(\delta_{B} ight)_{2}=0[/latex] (b)

in which the deflection [latex](\delta_{B} )_{1}[/latex] is positive downward, and the deflection [latex](\delta_{B} )_{2}[/latex] is positive upward.

Force-displacement relations: The deflection [latex](\delta_{B} )_{1}[/latex] caused by the uniform load acting on the released structure (Fig. 10-14c) is obtained from Table H-2, Case 1 as

[latex]\left(\delta_{B} ight)_{1}=\frac{5 q(2 L)^{4}}{384 E I}=\frac{5 q L^{4}}{24 E I}[/latex]

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

[latex]\left(\delta_{B} ight)_{2}=\frac{R_{B}(2 L)^{3}}{48 E I}=\frac{R_{B} L^{3}}{6 E I}[/latex]

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

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

from which the reaction at the middle support is

[latex]R_{B}=\frac{5 q L}{4}[/latex] (10-28)

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

[latex]R_{A}=R_{C}=\frac{3 q L}{8}[/latex] (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.

[latex]R_{B}=\frac{3 q L}{8}[/latex] (10-1)

[latex] heta_{B}=v^{\prime}(L)=\frac{q L^{3}}{48 E I}[/latex] (10-12)

Table H-2
Deflections and Slopes of Simple Beams
		Notation:
		v = deflection in the y direction (positive upward)
		v′ = dv/dx = slope of the deflection curve
		[latex]\delta_{C}=-v(L / 2)=[/latex] deflection at midpoint C of the beam (positive downward)
		[latex]x_{1}[/latex] = distance from support A to point of maximum deflection
		[latex]\delta_{\max }=-v_{\max }=[/latex] maximum deflection (positive downward)
		[latex] heta_{A}=-v^{\prime}(0)=[/latex] angle of rotation at left-hand end of the beam (positive clockwise)
		[latex] heta_{B}=v^{\prime}(L)=[/latex] angle of rotation at right-hand end of the beam (positive counterclockwise)
		EI = constant
		[latex]v=-\frac{q x}{24 E I}\left(L^{3}-2 L x^{2}+x^{3} ight)[/latex]
		[latex]v^{\prime}=-\frac{q}{24 E I}\left(L^{3}-6 L x^{2}+4 x^{3} ight)[/latex]
		[latex]\delta_{C}=\delta_{\max }=\frac{5 q L^{4}}{384 E I} \quad heta_{A}= heta_{B}=\frac{q L^{3}}{24 E I}[/latex]
		[latex]v=-\frac{q x}{384 E I}\left(9 L^{3}-24 L x^{2}+16 x^{3} ight) \quad\left(0 \leq x \leq \frac{L}{2} ight)[/latex]
		[latex]v^{\prime}=-\frac{q}{384 E I}\left(9 L^{3}-72 L x^{2}+64 x^{3} ight) \quad\left(0 \leq x \leq \frac{L}{2} ight)[/latex]
		[latex]v=-\frac{q L}{384 E I}\left(8 x^{3}-24 L x^{2}+17 L^{2} x-L^{3} ight) \quad\left(\frac{L}{2} \leq x \leq L ight)[/latex]
		[latex]v^{\prime}=-\frac{q L}{384 E I}\left(24 x^{2}-48 L x+17 L^{2} ight) \quad\left(\frac{L}{2} \leq x \leq L ight)[/latex]
		[latex]\delta_{C}=\frac{5 q L^{4}}{768 E I} \quad heta_{A}=\frac{3 q L^{3}}{128 E I} \quad heta_{B}=\frac{7 q L^{3}}{384 E I}[/latex]
		[latex]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} ight) \quad(0 \leq x \leq a)[/latex]
		[latex]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} ight) \quad(0 \leq x \leq a)[/latex]
		[latex]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} ight)[/latex] [latex](a \leq x \leq L)[/latex]
		[latex]v^{\prime}=-\frac{q a^{2}}{24 L E I}\left(4 L^{2}+a^{2}-12 L x+6 x^{2} ight)[/latex] [latex](a \leq x \leq L)[/latex]
		[latex] heta_{A}=\frac{q a^{2}}{24 L E I}(2 L-a)^{2} \quad heta_{B}=\frac{q a^{2}}{24 L E I}\left(2 L^{2}-a^{2} ight)[/latex]
		[latex]v=-\frac{P x}{48 E I}\left(3 L^{2}-4 x^{2} ight) \quad v^{\prime}=-\frac{P}{16 E I}\left(L^{2}-4 x^{2} ight) \quad\left(0 \leq x \leq \frac{L}{2} ight)[/latex]
		[latex]\delta_{C}=\delta_{\max }=\frac{P L^{3}}{48 E I} \quad heta_{A}= heta_{B}=\frac{P L^{2}}{16 E I}[/latex]
		[latex]v=-\frac{P b x}{6 L E I}\left(L^{2}-b^{2}-x^{2} ight) \quad v^{\prime}=-\frac{P b}{6 L E I}\left(L^{2}-b^{2}-3 x^{2} ight) \quad(0 \leq x \leq a)[/latex]
		[latex] heta_{A}=\frac{P a b(L+b)}{6 L E I} \quad heta_{B}=\frac{P a b(L+a)}{6 L E I}[/latex]
		[latex] ext { If } a \geq b, \quad \delta_{C}=\frac{P b\left(3 L^{2}-4 b^{2} ight)}{48 E I} \quad ext { If } a \leq b, \quad \delta_{C}=\frac{P a\left(3 L^{2}-4 a^{2} ight)}{48 E I}[/latex]
		[latex] ext { If } a \geq b, \quad x_{1}=\sqrt{\frac{L^{2}-b^{2}}{3}} \quad ext { and } \quad \delta_{\max }=\frac{P b\left(L^{2}-b^{2} ight)^{3 / 2}}{9 \sqrt{3} L E I}[/latex]
		[latex]v=-\frac{P x}{6 E I}\left(3 a L-3 a^{2}-x^{2} ight) \quad v^{\prime}=-\frac{P}{2 E I}\left(a L-a^{2}-x^{2} ight) \quad(0 \leq x \leq a)[/latex]
		[latex]v=-\frac{P a}{6 E I}\left(3 L x-3 x^{2}-a^{2} ight) \quad v^{\prime}=-\frac{P a}{2 E I}(L-2 x) \quad(a \leq x \leq L-a)[/latex]
		[latex]\delta_{C}=\delta_{\max }=\frac{P a}{24 E I}\left(3 L^{2}-4 a^{2} ight) \quad heta_{A}= heta_{B}=\frac{P a(L-a)}{2 E I}[/latex]
		[latex]v=-\frac{M_{0} x}{6 L E I}\left(2 L^{2}-3 L x+x^{2} ight) \quad v^{\prime}=-\frac{M_{0}}{6 L E I}\left(2 L^{2}-6 L x+3 x^{2} ight)[/latex]
		[latex]\delta_{C}=\frac{M_{0} L^{2}}{16 E I} \quad heta_{A}=\frac{M_{0} L}{3 E I} \quad heta_{B}=\frac{M_{0} L}{6 E I}[/latex]
		[latex]x_{1}=L\left(1-\frac{\sqrt{3}}{3} ight) ext { and } \delta_{\max }=\frac{M_{0} L^{2}}{9 \sqrt{3} E I}[/latex]
		[latex]v=-\frac{M_{0} x}{24 L E I}\left(L^{2}-4 x^{2} ight) \quad v^{\prime}=-\frac{M_{0}}{24 L E I}\left(L^{2}-12 x^{2} ight) \quad\left(0 \leq x \leq \frac{L}{2} ight)[/latex]
		[latex]\delta_{C}=0 \quad heta_{A}=\frac{M_{0} L}{24 E I} \quad heta_{B}=-\frac{M_{0} L}{24 E I}[/latex]
		[latex]v=-\frac{M_{0} x}{6 L E I}\left(6 a L-3 a^{2}-2 L^{2}-x^{2} ight) \quad(0 \leq x \leq a)[/latex]
		[latex]v^{\prime}=-\frac{M_{0}}{6 L E I}\left(6 a L-3 a^{2}-2 L^{2}-3 x^{2} ight) \quad(0 \leq x \leq a)[/latex]
		[latex] ext { At } x=a: \quad v=-\frac{M_{0} a b}{3 L E I}(2 a-L) \quad v^{\prime}=-\frac{M_{0}}{3 L E I}\left(3 a L-3 a^{2}-L^{2} ight)[/latex]
		[latex] heta_{A}=\frac{M_{0}}{6 L E I}\left(6 a L-3 a^{2}-2 L^{2} ight) \quad heta_{B}=\frac{M_{0}}{6 L E I}\left(3 a^{2}-L^{2} ight)[/latex]
		[latex]v=-\frac{M_{0} x}{2 E I}(L-x) \quad v^{\prime}=-\frac{M_{0}}{2 E I}(L-2 x)[/latex]
		[latex]\delta_{C}=\delta_{\max }=\frac{M_{0} L^{2}}{8 E I} \quad heta_{A}= heta_{B}=\frac{M_{0} L}{2 E I}[/latex]
		[latex]v=-\frac{q_{0} x}{360 L E I}\left(7 L^{4}-10 L^{2} x^{2}+3 x^{4} ight)[/latex]
		[latex]v^{\prime}=-\frac{q_{0}}{360 L E I}\left(7 L^{4}-30 L^{2} x^{2}+15 x^{4} ight)[/latex]
		[latex]\delta_{C}=\frac{5 q_{0} L^{4}}{768 E I} \quad heta_{A}=\frac{7 q_{0} L^{3}}{360 E I} \quad heta_{B}=\frac{q_{0} L^{3}}{45 E I}[/latex]
		[latex]x_{1}=0.5193 L \quad \delta_{\max }=0.00652 \frac{q_{0} L^{4}}{E I} [/latex]
		[latex]v=-\frac{q_{0} x}{960 L E I}\left(5 L^{2}-4 x^{2} ight)^{2} \quad\left(0 \leq x \leq \frac{L}{2} ight)[/latex]
		[latex]v^{\prime}=-\frac{q_{0}}{192 L E I}\left(5 L^{2}-4 x^{2} ight)\left(L^{2}-4 x^{2} ight) \quad\left(0 \leq x \leq \frac{L}{2} ight)[/latex]
		[latex]\delta_{C}=\delta_{ max }=\frac{q_{0} L^{4}}{120 E I} \quad heta_{A}= heta_{B}=\frac{5 q_{0} L^{3}}{192 E I}[/latex]
		[latex]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}[/latex]
		[latex]\delta_{C}=\delta_{ max }=\frac{q_{0} L^{4}}{\pi^{4} E I} \quad heta_{A}= heta_{B}=\frac{q_{0} L^{3}}{\pi^{3} E I}[/latex]

Table H-2
Deflections and Slopes of Simple Beams
		Notation:
		v = deflection in the y direction (positive upward)
		v′ = 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
		$v=-\frac{q x}{24 E I}\left(L^{3}-2 L x^{2}+x^{3}\right)$
		$v^{\prime}=-\frac{q}{24 E I}\left(L^{3}-6 L x^{2}+4 x^{3}\right)$
		$\delta_{C}=\delta_{\max }=\frac{5 q L^{4}}{384 E I} \quad\theta_{A}=\theta_{B}=\frac{q L^{3}}{24 E I}$
		$v=-\frac{q x}{384 E I}\left(9 L^{3}-24 L x^{2}+16 x^{3}\right) \quad\left(0 \leq x \leq \frac{L}{2}\right)$
		$v^{\prime}=-\frac{q}{384 E I}\left(9 L^{3}-72 L x^{2}+64 x^{3}\right) \quad\left(0 \leq x \leq \frac{L}{2}\right)$
		$v=-\frac{q L}{384 E I}\left(8 x^{3}-24 L x^{2}+17 L^{2} x-L^{3}\right) \quad\left(\frac{L}{2} \leq x \leq L\right)$
		$v^{\prime}=-\frac{q L}{384 E I}\left(24 x^{2}-48 L x+17 L^{2}\right) \quad\left(\frac{L}{2} \leq x \leq L\right)$
		$\delta_{C}=\frac{5 q L^{4}}{768 E I} \quad \theta_{A}=\frac{3 q L^{3}}{128 E I} \quad \theta_{B}=\frac{7 q L^{3}}{384 E I}$
		$v=-\frac{q x}{24 L E I}\left(a^{4}-4 a^{3} L+4 a^{2} L^{2}+2 a^{2} x^{2}-4 a L x^{2}+L x^{3}\right) \quad(0 \leq x \leq a)$
		$v^{\prime}=-\frac{q}{24 L E I}\left(a^{4}-4 a^{3} L+4 a^{2} L^{2}+6 a^{2} x^{2}-12 a L x^{2}-4 L x^{3}\right) \quad(0 \leq x \leq a)$
		$v=-\frac{q a^{2}}{24 L E I}\left(-a^{2} L+4 L^{2} x+a^{2} x-6 L x^{2}+2 x^{3}\right)$ $(a \leq x \leq L)$
		$v^{\prime}=-\frac{q a^{2}}{24 L E I}\left(4 L^{2}+a^{2}-12 L x+6 x^{2}\right)$ $(a \leq x \leq L)$
		$\theta_{A}=\frac{q a^{2}}{24 L E I}(2 L-a)^{2} \quad \theta_{B}=\frac{q a^{2}}{24 L E I}\left(2 L^{2}-a^{2}\right)$
		$v=-\frac{P x}{48 E I}\left(3 L^{2}-4 x^{2}\right) \quad v^{\prime}=-\frac{P}{16 E I}\left(L^{2}-4 x^{2}\right) \quad\left(0 \leq x \leq \frac{L}{2}\right)$
		$\delta_{C}=\delta_{\max }=\frac{P L^{3}}{48 E I} \quad \theta_{A}=\theta_{B}=\frac{P L^{2}}{16 E I}$
		$v=-\frac{P b x}{6 L E I}\left(L^{2}-b^{2}-x^{2}\right) \quad v^{\prime}=-\frac{P b}{6 L E I}\left(L^{2}-b^{2}-3 x^{2}\right) \quad(0 \leq x \leq a)$
		$\theta_{A}=\frac{P a b(L+b)}{6 L E I} \quad \theta_{B}=\frac{P a b(L+a)}{6 L E I}$
		$\text { If } a \geq b, \quad \delta_{C}=\frac{P b\left(3 L^{2}-4 b^{2}\right)}{48 E I} \quad \text { If } a \leq b, \quad \delta_{C}=\frac{P a\left(3 L^{2}-4 a^{2}\right)}{48 E I}$
		$\text { If } a \geq b, \quad x_{1}=\sqrt{\frac{L^{2}-b^{2}}{3}} \quad \text { and } \quad \delta_{\max }=\frac{P b\left(L^{2}-b^{2}\right)^{3 / 2}}{9 \sqrt{3} L E I}$
		$v=-\frac{P x}{6 E I}\left(3 a L-3 a^{2}-x^{2}\right) \quad v^{\prime}=-\frac{P}{2 E I}\left(a L-a^{2}-x^{2}\right) \quad(0 \leq x \leq a)$
		$v=-\frac{P a}{6 E I}\left(3 L x-3 x^{2}-a^{2}\right) \quad v^{\prime}=-\frac{P a}{2 E I}(L-2 x) \quad(a \leq x \leq L-a)$
		$\delta_{C}=\delta_{\max }=\frac{P a}{24 E I}\left(3 L^{2}-4 a^{2}\right) \quad \theta_{A}=\theta_{B}=\frac{P a(L-a)}{2 E I}$
		$v=-\frac{M_{0} x}{6 L E I}\left(2 L^{2}-3 L x+x^{2}\right) \quad v^{\prime}=-\frac{M_{0}}{6 L E I}\left(2 L^{2}-6 L x+3 x^{2}\right)$
		$\delta_{C}=\frac{M_{0} L^{2}}{16 E I} \quad \theta_{A}=\frac{M_{0} L}{3 E I} \quad \theta_{B}=\frac{M_{0} L}{6 E I}$
		$x_{1}=L\left(1-\frac{\sqrt{3}}{3}\right) \text { and } \delta_{\max }=\frac{M_{0} L^{2}}{9 \sqrt{3} E I}$
		$v=-\frac{M_{0} x}{24 L E I}\left(L^{2}-4 x^{2}\right) \quad v^{\prime}=-\frac{M_{0}}{24 L E I}\left(L^{2}-12 x^{2}\right) \quad\left(0 \leq x \leq \frac{L}{2}\right)$
		$\delta_{C}=0 \quad \theta_{A}=\frac{M_{0} L}{24 E I} \quad \theta_{B}=-\frac{M_{0} L}{24 E I}$
		$v=-\frac{M_{0} x}{6 L E I}\left(6 a L-3 a^{2}-2 L^{2}-x^{2}\right) \quad(0 \leq x \leq a)$
		$v^{\prime}=-\frac{M_{0}}{6 L E I}\left(6 a L-3 a^{2}-2 L^{2}-3 x^{2}\right) \quad(0 \leq x \leq a)$
		$\text { At } x=a: \quad v=-\frac{M_{0} a b}{3 L E I}(2 a-L) \quad v^{\prime}=-\frac{M_{0}}{3 L E I}\left(3 a L-3 a^{2}-L^{2}\right)$
		$\theta_{A}=\frac{M_{0}}{6 L E I}\left(6 a L-3 a^{2}-2 L^{2}\right) \quad \theta_{B}=\frac{M_{0}}{6 L E I}\left(3 a^{2}-L^{2}\right)$
		$v=-\frac{M_{0} x}{2 E I}(L-x) \quad v^{\prime}=-\frac{M_{0}}{2 E I}(L-2 x)$
		$\delta_{C}=\delta_{\max }=\frac{M_{0} L^{2}}{8 E I} \quad \theta_{A}=\theta_{B}=\frac{M_{0} L}{2 E I}$
		$v=-\frac{q_{0} x}{360 L E I}\left(7 L^{4}-10 L^{2} x^{2}+3 x^{4}\right)$
		$v^{\prime}=-\frac{q_{0}}{360 L E I}\left(7 L^{4}-30 L^{2} x^{2}+15 x^{4}\right)$
		$\delta_{C}=\frac{5 q_{0} L^{4}}{768 E I} \quad \theta_{A}=\frac{7 q_{0} L^{3}}{360 E I} \quad \theta_{B}=\frac{q_{0} L^{3}}{45 E I}$
		$x_{1}=0.5193 L \quad \delta_{\max }=0.00652 \frac{q_{0} L^{4}}{E I}$
		$v=-\frac{q_{0} x}{960 L E I}\left(5 L^{2}-4 x^{2}\right)^{2} \quad\left(0 \leq x \leq \frac{L}{2}\right)$
		$v^{\prime}=-\frac{q_{0}}{192 L E I}\left(5 L^{2}-4 x^{2}\right)\left(L^{2}-4 x^{2}\right) \quad\left(0 \leq x \leq \frac{L}{2}\right)$
		$\delta_{C}=\delta_{ max }=\frac{q_{0} L^{4}}{120 E I} \quad \theta_{A}=\theta_{B}=\frac{5 q_{0} L^{3}}{192 E I}$
		$v=-\frac{q_{0} L^{4}}{\pi^{4} E I} \sin \frac{\pi x}{L} \quad v^{\prime}=-\frac{q_{0} L^{3}}{\pi^{3} E I} \cos \frac{\pi x}{L}$
		$\delta_{C}=\delta_{ max }=\frac{q_{0} L^{4}}{\pi^{4} E I} \quad \theta_{A}=\theta_{B}=\frac{q_{0} L^{3}}{\pi^{3} E I}$

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