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 θ_A and the deflection δ_D, is sketched in Fig. 9-26b. The directions of θ_A and δ_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
\quad\quad\quad\quad{\mathit{A}}_{\mathrm{i}}={\frac{1}{2}}(L){{\Bigg\lgroup}}{\frac{P a b}{L E I}}{{\Bigg\rgroup}}={\frac{P a b}{2E I}}
The centroid C_1 of this area is at distance \overline{x_1} from point B (see Fig. 9-26c). This distance, obtained from Case 3 of Appendix E, is
\quad\quad\quad\quad {\overline{{x}}}_{1}={\frac{L+b}{3}}
Therefore, the tangential deviation is
\quad\quad\quad\quad t_{B/A}=A_{1}\overline{{{x}}}_{1}=\frac{{{{P}}}a b}{2E{{{I}}}}\bigg(\frac{L+b}{3}\bigg)=\frac{{{{P}}}a b}{6E{{{I}}}}\left(L+b\right)
The angle θ_A is equal to the tangential deviation divided by the length of the beam:
\quad\quad\quad\quad\theta_{A}=\frac{t_{B/ A}}{L}=\frac{P a b}{6L E I}(L+b)\quad\quad(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_2D_1. The distance DD_1 is equal to the angle of rotation θ_A times the distance a; thus,
\quad\quad\quad\quad{{D}}{{{D}}}_{1}=a\theta_{A}={\frac{{{{P}}}a^{2}b}{6L E I}}(L+b)\quad\quad(a)
The distance D_2D_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
\quad\quad\quad\quad A_{2}={\frac{1}{2}}(a){\Bigg\lgroup}{\frac{P a b}{L E I}}{\Bigg\rgroup}={\frac{P a^{2}b}{2L E I}}
and its centroidal distance from point D is
\quad\quad\quad\quad \overline{{x}}_{2}={\frac{a}{3}}
Thus, the first moment of this area with respect to point D is
\quad\quad\quad\quad t_{D/A}=A_{2}\overline{{{x}}}_{2}={\Bigg\lgroup}\frac{P a^{2}b}{2L E I}{\Bigg\rgroup}{\Bigg\lgroup}\frac{a}{3}{\Bigg\rgroup}=\frac{P a^{3}b}{6L E I}\quad\quad(b)
The deflection at point D is
\quad\quad\delta_{D} = {D}{D}_{1}-{D}_{2}D_{1}={D}{D}_{1}-t_{D/A}
Substitute from Eqs. (a) and (b) to find
\quad\quad\quad\quad \delta_{ D}=\frac{P a^{2}b}{6L E I}(L+b)-\frac{P a^{3}b}{6L E I}=\frac{P a^{2}b^{2}}{3L E I}\quad\quad (9-83)
4. Finalize: The preceding formulas for θ_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.