Question 9.9: A simple beam AB of a span length L has an overhang BC of le......

Use a four-step problem-solving approach.
1. Conceptualize: Find the deflection of point C by imagining the overhang BC (Fig. 9-21a) to be a cantilever beam subjected to two actions. The first action is the rotation of the support of the cantilever through an angle $\theta_B$, which is the angle of rotation of beam ABC at support B (Fig. 9-21c). (Assume that a clockwise angle $\theta_B$ is positive.) This angle of rotation causes a rigid-body rotation of the overhang BC, resulting in a downward displacement $\delta_1$ of point C.
The second action is the bending of BC as a cantilever beam supporting a uniform load. This bending produces an additional downward displacement $\delta_2$ (Fig. 9-21c). The superposition of these two displacements gives the total displacement $\delta_C$ at point C.
2. Categorize:
Deflection $\delta_1$: Begin by finding the deflection $\delta_1$ caused by the angle of rotation $\theta_B$ at point B. To find this angle, observe that part AB of the beam is in the same condition as a simple beam (Fig. 9-21b) subjected to the following loads: a uniform load of intensity q, a couple $M_B$ (equal to $qa^2 / 2$), and a vertical load P (equal to qa). Only the loads q and $M_B$ produce angles of rotation at end B of this simple beam. These angles are found from Cases 1 and 7 of Table H-2, Appendix H. Thus, the angle $\theta_B$ is
$\quad\quad\quad\quad\theta_{B}=-{\frac{q L^{3}}{24E I}}+{\frac{M_{B}L}{3E I}}=-{\frac{q{L}^{3}}{24{E}I}}+{\frac{q a^{2}{L}}{6{E}I}}={\frac{q{L}(4 a^{2}-{L}^{2})}{24{E}{I}}}\quad\quad$(9-67)
in which a clockwise angle is positive, as shown in Fig. 9-21c.
The downward deflection $\delta_1$ of point C, due solely to the angle of rotation $\theta_B$, is equal to the length of the overhang times the angle (Fig. 9-21c):
$\quad\quad\quad\quad\delta_{1}=a\theta_{_{ B}}\,=\,\frac{q a{L}(4a^{2}-{L}^{2})}{24{ E}{ L}}\quad\quad$(a)
Deflection $δ_2$: Bending of the overhang BC produces an additional downward deflection $δ_2$ at point C. This deflection is equal to the deflection of a cantilever beam of length a subjected to a uniform load of intensity q (see Case 1 of Table H-1):
$\quad\quad\quad\quad{\delta}_{2}={\frac{q a^{4}}{8E I}}\quad\quad\quad\quad$(b)
3. Analyze:
Deflection $\delta_C$: The total downward deflection of point C is the algebraic sum of $\delta_1 ~ and ~ \delta_2$:
$\quad\quad\quad\quad\delta_{C}=\delta_{1}+\delta_{2}={\frac{q a{L}(4a^{2}-{L}^{2})}{24E I}}+{\frac{q a^{4}}{8E I}}={\frac{q a}{24E I}}[{L}(4a^{2} – {L}^{2})+3 a^{3}]$
or
$\quad\quad\quad\quad\delta_{C}={\frac{q a}{24E I}}(a+{L})(3a^{2}+a L- {L}^{2})\quad\quad$(9-68)
4. Finalize: In the preceding equation, the deflection $\delta_C$ may be upward or downward, depending upon the relative magnitudes of the lengths L and a. If a is relatively large, the last term in the equation (the three-term expression in parentheses) is positive, and the deflection $\delta_C$ is downward. If a is relatively small, the last term is negative, and the deflection is upward. The deflection is zero when the last term is equal to zero:
$\quad\quad\quad\quad 3a^{2}+a {L} – {L}^{2} = {0}$
or
$\quad\quad\quad\quad a={\frac{{L}\Big(\sqrt{13}\,-\,1\Big)}{6}}=0.434{{3L}}\quad\quad$(c)
From this result, note that if a is greater than 0.4343L, the deflection of point C is downward; if a is less than 0.4343L, the deflection is upward.
Deflection curve: The shape of the deflection curve for the beam in this example is shown in Fig. 9-21c for the case where a is large enough (a > 0.4343L) to produce a downward deflection at C and small enough (a < L) to ensure that the reaction at A is upward. Under these conditions, the beam has a positive bending moment between support A and a point such as D. The deflection curve in region AD is concave upward (positive curvature). From D to C, the bending moment is negative; therefore, the deflection curve is concave downward (negative curvature).
Point of inflection: At point D, the curvature of the deflection curve is zero because the bending moment is zero. A point such as D, where the curvature and bending moment change signs, is called a point of inflection (or point of contraflexure). The bending moment M and the second derivative $d^2v / dx^2$ always vanish at an inflection point.
However, a point where M and $d^2v / dx^2$ equal zero is not necessarily an inflection point because it is possible for those quantities to be zero without changing signs at that point; for example, they could have maximum or minimum values.