Question 9.11: Find the angle of rotation θB and deflection δB at the free ...
Find the angle of rotation \theta_{B} and deflection \delta_{B} at the free end B of a cantilever beam ACB supporting a uniform load of intensity q acting over the right-hand half of the beam (Fig. 9-25). (Note: The beam has length L and constant flexural rigidity EI.)
The deflection and angle of rotation at end B of the beam have the directions shown in Fig. 9-25. Since we know these directions in advance, we can write the moment-area expressions using only absolute values.
M/EI diagram. The bending-moment diagram consists of a parabolic curve in the region of the uniform load and a straight line in the left-hand half of the beam. Since EI is constant, the M/EI diagram has the same shape (see the last part of Fig. 9-25). The values of M/EI at points A and C are −3qL²/8EI and −qL²/8EI, respectively.
Angle of rotation. For the purpose of evaluating the area of the M/EI diagram, it is convenient to divide the diagram into three parts: (1) a parabolic spandrel of area A_{1}, (2) a rectangle of area A_{2}, and (3) a triangle of area A_{3}. These areas are
A_{1}=\frac{1}{3}\left(\frac{L}{2}\right)\left(\frac{qL^{2}}{8EI}\right)=\frac{qL^{3}}{48EI} A_{2}=\frac{L}{2}\left(\frac{qL^{2}}{8EI}\right)=\frac{qL^{3}}{16EI}
A_{3}=\frac{1}{2}\left(\frac{L}{2}\right)\left(\frac{3qL^{2}}{8EI}-\frac{qL^{2}}{8EI}\right)=\frac{qL^{3}}{16EI}
According to the first moment-area theorem, the angle between the tangents at points A and B is equal to the area of the M/EI diagram between those points. Since the angle at A is zero, it follows that the angle of rotation \theta_{B} is equal to the area of the diagram; thus,
\theta_{B}=A_{1}+A_{2}+A_{3}=\frac{7qL^{3}}{48EI} (9-66)
Deflection. The deflection \delta_{B} is the tangential deviation of point B with respect to a tangent at point A (Fig. 9-25). Therefore, from the second moment-area theorem, \delta_{B} is equal to the first moment of the M/EI diagram, evaluated with respect to point B:
\delta_{B}=A_{1}\overline{x}_{1}+A_{2}\overline{x}_{2}+A_{3}\overline{x}_{3} (g)
in which \overline{x}_{1}, \overline{x}_{2}, and \overline{x}_{3}, are the distances from point B to the centroids of the respective areas. These distances are
\overline{x}_{1}=\frac{3}{4}\left(\frac{L}{2}\right)=\frac{3L}{8} \overline{x}_{2}=\frac{L}{2}+\frac{L}{4}=\frac{3L}{4} \overline{x}_{3}=\frac{L}{2}+\frac{2}{3}\left(\frac{L}{2}\right)=\frac{5L}{6}
Substituting into Eq. (g), we find
\delta_{B}=\frac{qL^{3}}{48EI}\left(\frac{3L}{8}\right)+\frac{qL^{3}}{16EI}\left(\frac{3L}{4}\right)+\frac{qL^{3}}{16EI}\left(\frac{5L}{6}\right)=\frac{41qL^{4}}{384EI} (9-67)
This example illustrates how the area and first moment of a complex M/EI diagram can be determined by dividing the area into parts having known properties. The results of this analysis (Eqs. 9-66 and 9-67) can be verified by using the formulas of Case 3, Table G-1, Appendix G, and substituting a = b = L/2.
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\mathrm{v}=-\frac{qbx^{2}}{12EI}(3L+3a-2x) (0 ≤ x ≤ a)
\mathrm{v}^{\prime} =-\frac{qbx}{2EI}(L+a-x) (0 ≤ x ≤ a)
\mathrm{v}=-\frac{q}{24EI}(x^{4}-4Lx^{3}+6L^{2}x^{2}-4a^{3}x+a^{4}) (a ≤ x ≤ L)
\mathrm{v}^{\prime} =-\frac{q}{6EI}(x^{3}-3Lx^{2}+3L^{2}x-a^{3}) (a ≤ x ≤ L)
At x = a: \mathrm{v}=-\frac{qa^{2}b}{12EI}(3L+a) \mathrm{v}^{\prime} =-\frac{qabL}{2EI}
\delta_{B}=\frac{q}{24EI}\left(3L^{4}-4a^{3}L+a^{4}\right) \theta_{B}=\frac{q}{6EI}\left(L^{3}-a^{3}\right) |