Question 9.8: A compound beam ABC has a roller support at A, an internal h...

Use a four-step problem-solving approach.
1. Conceptualize: For purposes of analysis, consider the compound beam to consist of two individual beams: a simple beam AB of length a and a cantilever beam BC of length b. The two beams are linked together by a pin connection at B.

2. Categorize: If you separate beam AB from the rest of the structure (Fig. 9-20b), there is a vertical force F at end B equal to 2P/3. This same force acts downward at end B of the cantilever (Fig. 9-20c). Consequently, the cantilever beam BC is subjected to two loads: a uniform load and a concentrated load.

3. Analyze: The deflection at the end of this cantilever (which is the same as the deflection $\delta_{B}$ of the hinge) is readily found from Cases 1 and 4 of Table H-1, Appendix H:

or, since F = 2P / 3,

$\delta_{B}=\frac{q b^{4}}{8 E I}+\frac{2 P b^{3}}{9 E I}$            (9-65)

The angle of rotation $\theta_{A}$ at support A (Fig. 9-20d) consists of two parts: an angle BAB′ produced by the downward displacement of the hinge and an additional angle of rotation produced by the bending of beam AB (or beam AB′) as a simple beam. The angle BAB′ is

The angle of rotation at the end of a simple beam with a concentrated load is obtained from Case 5 of Table H-2. The formula given there is

in which L is the length of the simple beam, a is the distance from the left-hand support to the load, and b is the distance from the right-hand support to the load. Thus, in the notation of this example (Fig. 9-20a), the angle of rotation is

Combine the two angles to obtain the total angle of rotation at support A:

$\theta_{A}=\left(\theta_{A}\right)_{1}+\left(\theta_{A}\right)_{2}=\frac{q b^{4}}{8 a E I}+\frac{2 P b^{3}}{9 a E I}+\frac{4 P a^{2}}{81 E I}$          (9-66)

4. Finalize: This example illustrates how the method of superposition can be adapted to handle a seemingly complex situation in a relatively simple manner.

Table H-1
Deflections and Slopes of Cantilever Beams
	Notation:
	v = deflection in the y direction (positive upward)
	v′ = dv/dx = slope of the deflection curve
	$\delta_{B}=-v(L)=$ deflection at end B of the beam (positive downward)
	$\theta_{B}=-v^{\prime}(L)=$ angle of rotation at end B of the beam (positive clockwise)
	EI = constant
	$v=-\frac{q x^{2}}{24 E I}\left(6 L^{2}-4 L x+x^{2}\right)   \quad v^{\prime}=\frac{q x}{6 E I}\left(3 L^{2}-3 L x+x^{2}\right)$
	$\delta_{B}=\frac{q L^{4}}{8 E I}   \quad \theta_{B}=\frac{q L^{3}}{6 E I}$
	$v=-\frac{q x^{2}}{24 E I}\left(6 a^{2}-4 a x+x^{2}\right)   \quad(0 \leq x \leq a)$
	$v^{\prime}=-\frac{q x}{6 E I}\left(3 a^{2}-3 a x+x^{2}\right)   \quad(0 \leq x \leq a)$
	$v=-\frac{q a^{3}}{24 E I}(4 x-a) \quad v^{\prime}=-\frac{q a^{3}}{6 E I}   \quad(a \leq x \leq L)$
	At  $x=a: v=-\frac{q a^{4}}{8 E I}   \quad v^{\prime}=-\frac{q a^{3}}{6 E I}$
	$\delta_{B}=\frac{q a^{3}}{24 E I}(4 L-a)   \quad \theta_{B}=\frac{q a^{3}}{6 E I}$
	$v=-\frac{q b x^{2}}{12 E I}(3 L+3 a-2 x)$          $(0 \leq x \leq a)$
	$v^{\prime}=-\frac{q b x}{2 E I}(L+a-x)$            $(0 \leq x \leq a)$
	$v=-\frac{q}{24 E I}\left(x^{4}-4 L x^{3}+6 L^{2} x^{2}-4 a^{3} x+a^{4}\right)   \quad(a \leq x \leq L)$
	$v^{\prime}=-\frac{q}{6 E I}\left(x^{3}-3 L x^{2}+3 L^{2} x-a^{3}\right)  \quad(a \leq x \leq L)$
	At $x=a:   v=-\frac{q a^{2} b}{12 E I}(3 L+a)  \quad v^{\prime}=-\frac{q a b L}{2 E I}$
	$\delta_{B}=\frac{q}{24 E I}\left(3 L^{4}-4 a^{3} L+a^{4}\right)  \quad \theta_{B}=\frac{q}{6 E I}\left(L^{3}-a^{3}\right)$
	$v=-\frac{P x^{2}}{6 E I}(3 L-x)  \quad v^{\prime}=-\frac{P x}{2 E I}(2 L-x)$
	$\delta_{B}=\frac{P L^{3}}{3 E I}  \quad \theta_{B}=\frac{P L^{2}}{2 E I}$
	$v=-\frac{P x^{2}}{6 E I}(3 a-x) \quad v^{\prime}=-\frac{P x}{2 E I}(2 a-x)  \quad(0 \leq x \leq a)$
	$v=-\frac{P a^{2}}{6 E I}(3 x-a)  \quad v^{\prime}=-\frac{P a^{2}}{2 E I}  \quad(a \leq x \leq L)$
	At $x=a:  \quad v=-\frac{P a^{3}}{3 E I}  \quad v^{\prime}=-\frac{P a^{2}}{2 E I}$
	$\delta_{B}=\frac{P a^{2}}{6 E I}(3 L-a)  \quad \theta_{B}=\frac{P a^{2}}{2 E I}$
	$v=-\frac{M_{0} x^{2}}{2 E I} \quad v^{\prime}=-\frac{M_{0} x}{E I}$
	$\delta_{B}=\frac{M_{0} L^{2}}{2 E I}  \quad \theta_{B}=\frac{M_{0} L}{E I}$
	$v=-\frac{M_{0} x^{2}}{2 E I}  \quad v^{\prime}=-\frac{M_{0} x}{E I} \quad(0 \leq x \leq a)$
	$v=-\frac{M_{0} a}{2 E I}(2 x-a)  \quad v^{\prime}=-\frac{M_{0} a}{E I} \quad(a \leq x \leq L)$
	At $x=a:  \quad v=-\frac{M_{0} a^{2}}{2 E I}  \quad v^{\prime}=-\frac{M_{0} a}{E I}$
	$\delta_{B}=\frac{M_{0} a}{2 E I}(2 L-a)  \quad \theta_{B}=\frac{M_{0} a}{E I}$
	$v=-\frac{q_{0} x^{2}}{120 L E I}\left(10 L^{3}-10 L^{2} x+5 L x^{2}-x^{3}\right)$
	$v^{\prime}=-\frac{q_{0} x}{24 L E I}\left(4 L^{3}-6 L^{2} x+4 L x^{2}-x^{3}\right)$
	$\delta_{B}=\frac{q_{0} L^{4}}{30 E I} \quad \theta_{B}=\frac{q_{0} L^{3}}{24 E I}$
	$v=-\frac{q_{0} x^{2}}{120 L E I}\left(20 L^{3}-10 L^{2} x+x^{3}\right)$
	$v^{\prime}=-\frac{q_{0} x}{24 L E I}\left(8 L^{3}-6 L^{2} x+x^{3}\right)$
	$\delta_{B}=\frac{11 q_{0} L^{4}}{120 E I}  \quad \theta_{B}=\frac{q_{0} L^{3}}{8 E I}$
	$v=-\frac{q_{0} L}{3 \pi^{4} E I}\left(48 L^{3} \cos \frac{\pi x}{2 L}-48 L^{3}+3 \pi^{3} L x^{2}-\pi^{3} x^{3}\right)$
	$v^{\prime}=-\frac{q_{0} L}{\pi^{3} E I}\left(2 \pi^{2} L x-\pi^{2} x^{2}-8L^{2} \sin \frac{\pi x}{2 L}\right)$
	$\delta_{B}=\frac{2 q_{0} L^{4}}{3 \pi^{4} E I}\left(\pi^{3}-24\right)   \quad\theta_{B}=\frac{q_{0} L^{3}}{\pi^{3} E I}\left(\pi^{2}-8\right)$

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