Question 9.12: A simple beam ADB supports a concentrated load P acting at t...
A simple beam ADB supports a concentrated load P acting at the position shown in Fig. 9-26. Determine the angle of rotation \theta_{A} at support A and the deflection \delta_{D} under the load P. (Note: The beam has length L and constant flexural rigidity EI.)
The deflection curve, showing the angle of rotation \theta_{A} and the deflection \delta_{D}, is sketched in the second part of Fig. 9-26. Because we can determine the directions of \theta_{A} and \delta_{D} by inspection, we can write the moment-area expressions using only absolute values.
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 the third part of Fig. 9-26).
Angle of rotation at support A. To find this angle, we construct the tangent AB_{1} at support A. We then note that the distance BB_{1} is the tangential deviation t_{B/A} of point B from the tangent at A. We can 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 \overline{x}_{1} from point B (see Fig. 9-26). This distance, obtained from Case 3 of Appendix D, is
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Triangle (Origin of axes at centroid)
A=\frac{bh}{2} \overline{x}=\frac{b+c}{3} \overline{y}=\frac{h}{3}
I_{x}=\frac{bh^{3}}{36} I_{y}=\frac{bh}{36}(b^{2}-bc+c^{2})
I_{xy}=\frac{bh^{2}}{72}(b-2c) I_{P}=\frac{bh}{36}(h^{2}+b^{2}-bc+c^{2}) |
Therefore, the tangential deviation is
t_{{B}/{A}}=A_{1}\overline{x}_{1}=\frac{Pab}{2EI}\left(\frac{L+b}{3}\right)=\frac{Pab}{6EI}(L+b)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{Pab}{6LEI}(L+b) (9-68)
Thus, the angle of rotation at support A has been found.
Deflection under the load. As shown in the second part of Fig. 9-26, 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,
DD_{1}=a\theta_{A}=\frac{Pa^{2}b}{6LEI}(L+b) (h)
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 the last part of Fig. 9-26). The area of this part of the M/EI diagram is
A_{2}=\frac{1}{2}(a)\left(\frac{Pab}{LEI}\right)=\frac{Pa^{2}b}{2LEI}and its centroidal distance from point D is
\overline{x}_{2}=\frac{a}{3}Thus, the first moment of this area with respect to point D is
t_{{D}/{A}}=A_{2}\overline{x}_{2}=\left(\frac{Pa^{2}b}{2LEI}\right)\left(\frac{a}{3}\right)=\frac{Pa^{3}b}{6LEI} (i)
The deflection at point D is
\delta_{D}=DD_{1}-D_{2}D_{1}=DD_{1}-t_{{D}/{A}}Upon substituting from Eqs. (h) and (i), we find
\delta_{D}=\frac{Pa^{2}b}{6LEI}(L+b)-\frac{Pa^{3}b}{6LEI}=\frac{Pa^{2}b^{2}}{3LEI} (9-69)
The preceding formulas for \theta_{A} and \delta_{D} (Eqs. 9-68 and 9-69) can be verified by using the formulas of Case 5, Table G-2, Appendix G.
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\nu=-\frac{Pbx}{6LEI}(L^{2}-b^{2}-x^{2}) \nu^{\prime}=-\frac{Pb}{6LEI}(L^{2}-b^{2}-3x^{2}) (0 ≤ x ≤ a)
\theta_{A}=\frac{Pab(L+b)}{6LEI} \theta_{B}=\frac{Pab(L+a)}{6LEI}
If a ≥ b, \delta_{C}=\frac{Pb(3L^{2}-4b^{2})}{48EI} If a ≤ b, \delta_{C}=\frac{Pa(3L^{2}-4a^{2})}{48EI}
If a ≥ b, x_{1}=\sqrt{\frac{L^{2}-b^{2}}{3}} and \delta_{\max}=\frac{Pb(L^{2}-b^{2})^{{3}/{2}}}{} |