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

We can 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 θ_B, which is the angle of rotation of the original beam at support B (Fig. 9-21c). This angle of rotation causes a rigid-body rotation of the overhang BC. resulting in a downward displacement δ_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 δ_2 (Fig. 9-21c). The superposition of these two displacements gives the total displacement δ_C at point C.

Let us begin by finding the deflection δ_1 caused by the angle of rotation θ_B at point B. To find this angle, we observe that part AB of the beam is in the same condition as a simple beam (Fig. 9-21b) subjected to the following loads: (1) a uniform load of intensity q. (2) a couple M_B (equal to qa²/2). and (3) 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 G-2, Appendix G. Thus, the angle θ_B is

θ_B=-\frac{qL^3}{24EI}+\frac{M_BL}{3EI}=-\frac{qL^3}{24EI}+\frac{qa^2L}{6EI}=\frac{qL(4a^2\ -\ L^2)}{24EI}             (9-58)

in which a clockwise angle is positive, as shown in Fig. 9-21c.
The downward deflection δ_1 of point C. due solely to the angle of rotation θ_B. is equal to the length of the overhang times the angle (Fig. 9-21c):

δ_1=aθ_B=\frac{qaL(4a^2\ -\ L^2)}{24EI}                        (e)

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 G-1):

δ_2=\frac{qa^4}{8EI}                      (f)

Therefore, the total downward deflection of point C is

δ_C=δ_1+δ_2=\frac{qaL(4a^2\ -\ L^2)}{24EI}+\frac{qa^4}{8EI}=\frac{qa}{24EI}[L(4a^2\ -\ L^2)+3a^3]

or                                               δ_C=\frac{qa}{24EI}(a+L)(3a^2+aL\ -\ L^2)                  (9-59)

From the preceding equation we see that the deflection δ_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 expressioin in parentheses) is positive and the deflection δ_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:

3a² + aL – L² = 0

or                                                    a=\frac{L\sqrt{13}\ -\ 1}{6} = 0.4343L                          (g)

From this result, we see 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.
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, and therefore the deflection curve is concave downward (negative contrflexure).
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 d²v/dx² always vanish at an inflection point. (However, a point where M and d²v/dx² equal zero is not always 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.)