Question 8.9: 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. 8-21a) 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 beam ABC at support B (Fig. 8-21c). (We assume that a clockwise angle  θ_B  is positive.) 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. 8-21c). The superposition of these two displacements gives the total displacement  δ_C  at point C.

Deflection   δ_1.  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. 8-21b) subjected to the following loads: (1) a uniform load of intensity q, (2) a couple  M_B  (equal to  qa^2/2),  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 H-2, Appendix H (available online). 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}                            (8-58)

in which a clockwise angle is positive, as shown in Fig. 8-21c.

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. 8-21c):

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

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, available online):

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

Deflection  δ_C.  The total downward deflection of point C is the algebraic sum of  δ_1  and  δ_2:

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)                            (8-59)

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 expression 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:

3a^2+aL-L^2=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.

Deflection curve. The shape of the deflection curve for the beam in this example is shown in Fig. 8-21c for the case where a is large enough  (a\gt 0.4343L)  to produce a downward deflection at C and small enough  (a\lt 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 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.