A simple beam ADB supports a concentrated load P acting at the position shown in Fig. 9-26. Determine the angle of rotation θA at support A and the deflection δD under the load P. (Note: The beam has length L and constant flexural rigidity EI.)

Question

A simple beam ADB supports a concentrated load P acting at the position shown in Fig. 9-26. Determine the angle of rotation [latex] heta_{A}[/latex] at support A and the deflection [latex]\delta_{D}[/latex] under the load P. (Note: The beam has length L and constant flexural rigidity EI.)

Accepted Answer

The deflection curve, showing the angle of rotation [latex] heta_{A}[/latex] and the deflection [latex]\delta_{D}[/latex], is sketched in the second part of Fig. 9-26. Because we can determine the directions of [latex] heta_{A}[/latex] and [latex]\delta_{D}[/latex] 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 [latex]AB_{1}[/latex] at support A. We then note that the distance [latex]BB_{1}[/latex] is the tangential deviation [latex]t_{B/A}[/latex] 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

[latex]A_{1}=\frac{1}{2}(L)\left(\frac{Pab}{LEI} ight)=\frac{Pab}{2EI}[/latex]

The centroid [latex]C_{1}[/latex] of this area is at distance [latex]\overline{x}_{1}[/latex] from point B (see Fig. 9-26). This distance, obtained from Case 3 of Appendix D, is

Triangle (Origin of axes at centroid)

[latex]A=\frac{bh}{2}[/latex] [latex]\overline{x}=\frac{b+c}{3}[/latex] [latex]\overline{y}=\frac{h}{3}[/latex]

[latex]I_{x}=\frac{bh^{3}}{36}[/latex] [latex]I_{y}=\frac{bh}{36}(b^{2}-bc+c^{2})[/latex]

[latex]I_{xy}=\frac{bh^{2}}{72}(b-2c)[/latex] [latex]I_{P}=\frac{bh}{36}(h^{2}+b^{2}-bc+c^{2})[/latex]

[latex]\overline{x}_{1}=\frac{L+b}{3}[/latex]

Therefore, the tangential deviation is

[latex]t_{{B}/{A}}=A_{1}\overline{x}_{1}=\frac{Pab}{2EI}\left(\frac{L+b}{3} ight)=\frac{Pab}{6EI}(L+b)[/latex]

The angle [latex] heta_{A}[/latex] is equal to the tangential deviation divided by the length of the beam:

[latex] heta_{A}=\frac{t_{{B}/{A}}}{L}=\frac{Pab}{6LEI}(L+b)[/latex] (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 [latex]\delta_{D}[/latex] under the load P is equal to the distance [latex]DD_{1}[/latex] minus the distance [latex]D_{2}D_{1}[/latex]. The distance [latex]DD_{1}[/latex] is equal to the angle of rotation [latex] heta_{A}[/latex] times the distance a; thus,

[latex]DD_{1}=a heta_{A}=\frac{Pa^{2}b}{6LEI}(L+b)[/latex] (h)

The distance [latex]D_{2}D_{1}[/latex] is the tangential deviation [latex]t_{D/A}[/latex] 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

[latex]A_{2}=\frac{1}{2}(a)\left(\frac{Pab}{LEI} ight)=\frac{Pa^{2}b}{2LEI}[/latex]

and its centroidal distance from point D is

[latex]\overline{x}_{2}=\frac{a}{3}[/latex]

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

[latex]t_{{D}/{A}}=A_{2}\overline{x}_{2}=\left(\frac{Pa^{2}b}{2LEI} ight)\left(\frac{a}{3} ight)=\frac{Pa^{3}b}{6LEI}[/latex] (i)

The deflection at point D is

[latex]\delta_{D}=DD_{1}-D_{2}D_{1}=DD_{1}-t_{{D}/{A}}[/latex]

Upon substituting from Eqs. (h) and (i), we find

[latex]\delta_{D}=\frac{Pa^{2}b}{6LEI}(L+b)-\frac{Pa^{3}b}{6LEI}=\frac{Pa^{2}b^{2}}{3LEI}[/latex] (9-69)

The preceding formulas for [latex] heta_{A}[/latex] and [latex]\delta_{D}[/latex] (Eqs. 9-68 and 9-69) can be verified by using the formulas of Case 5, Table G-2, Appendix G.

[latex] u=-\frac{Pbx}{6LEI}(L^{2}-b^{2}-x^{2})[/latex] [latex] u^{\prime}=-\frac{Pb}{6LEI}(L^{2}-b^{2}-3x^{2}) [/latex] (0 ≤ x ≤ a)

[latex] heta_{A}=\frac{Pab(L+b)}{6LEI}[/latex] [latex] heta_{B}=\frac{Pab(L+a)}{6LEI}[/latex]

If a ≥ b, [latex]\delta_{C}=\frac{Pb(3L^{2}-4b^{2})}{48EI}[/latex] If a ≤ b, [latex]\delta_{C}=\frac{Pa(3L^{2}-4a^{2})}{48EI}[/latex]

If a ≥ b, [latex]x_{1}=\sqrt{\frac{L^{2}-b^{2}}{3}}[/latex] and [latex]\delta_{\max}=\frac{Pb(L^{2}-b^{2})^{{3}/{2}}}{}[/latex]

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

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