Question 9.12: A simple beam ADB supports a concentrated load P acting at t...

Use a four-step problem-solving approach.
1. Conceptualize: The deflection curve, showing the angle of rotation \theta_{A} and the deflection \delta_{D}, is sketched in Fig. 9-26b. The directions of \theta_{A}   and   \delta_{D} are determined by inspection, so write the moment-area expressions using only absolute values.

2. Categorize:
M/EI diagram: The bending-moment diagram is triangular, with the maximum moment (equal to Pab/L) occurring under the load. Since EI is constant, the M/EI diagram has the same shape as the moment diagram (see Fig. 9-26c).

3. Analyze:
Angle of rotation at support A: To find this angle, construct the tangent AB_{1} at support A. Note that the distance BB_{1} is the tangential deviation t_{B/A} of point B from the tangent at A. Calculate this distance by evaluating the first moment of the area of the M/EI diagram with respect to point B and then applying the second moment-area theorem.
The area of the entire M/EI diagram is

The centroid C_{1} of this area is at distance \bar{x}_{1} from point B (see Fig. 9-26c). This distance, obtained from Case 3 of Appendix E, is

Therefore, the tangential deviation is

The angle \theta_{A} is equal to the tangential deviation divided by the length of the beam:

\theta_{A}=\frac{t_{B / A}}{L}=\frac{P a b}{6 L E I}(L+b)            (9-82)

Thus, the angle of rotation at support A has been found.
Deflection under the load: As shown in Fig. 9-26b, the deflection \delta_{D} under the load P is equal to the distance DD_{1} minus the distance D_{2}D_{1}. The distance DD_{1} is equal to the angle of rotation \theta_{A} times the distance a; thus,

D D_{1}=a \theta_{A}=\frac{P a^{2} b}{6 L E I}(L+b)            (a)

The distance D_{2}D_{1} is the tangential deviation t_{D/A} at point D; that is, it is the deviation of point D from the tangent at A. This distance can be found from the second moment-area theorem by taking the first moment of the area of the M/EI diagram between points A and D with respect to D (see Fig. 9-26c). The area of this part of the M/EI diagram is

and its centroidal distance from point D is

Thus, the first moment of this area with respect to point D is

t_{D / A}=A_{2} \bar{x}_{2}=\left(\frac{P a^{2} b}{2 L E I}\right)\left(\frac{a}{3}\right)=\frac{P a^{3} b}{6 L E I}              (b)

The deflection at point D is

Substitute from Eqs. (a) and (b) to find

\delta_{D}=\frac{P a^{2} b}{6 L E I}(L+b)-\frac{P a^{3} b}{6 L E I}=\frac{P a^{2} b^{2}}{3 L E I}           (9-83)

4. Finalize: The preceding formulas for \theta_{A} and \delta_{D} in Eqs. (9-82) and (9-83) can be verified by using the formulas of Case 5, Table H-2, Appendix H.

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}