Question 8.9: A simple beam AB of span length L has an overhang BC of leng...
A simple beam AB of span length L has an overhang BC of length a (Fig. 8-21a). The beam supports a uniform load of intensity q throughout its length.
Obtain a formula for the deflection \delta_C at the end of the overhang (Fig. 8-21c). (Note: The beam has constant flexural rigidity EI.)
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 \theta_B , which is the angle of rotation of beam ABC at support B (Fig. 8-21c). (We 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. 8-21c). The superposition of these two displacements gives the total displacement \delta _C at point C.
Deflection \delta _1. Let us begin by finding the deflection \delta _1 caused by the angle of rotation \theta_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 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 (available online). Thus, the angle \theta_B is
\theta _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 \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. 8-21c):
\delta _1=a\theta _B=\frac{qaL(4a^2-L^2)}{24EI} (e)
Deflection \delta _2. Bending of the overhang BC produces an additional downward deflection \delta _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, available online):
\delta _2=\frac{qa^4}{8EI} (f)
Deflection \delta _C. The total downward deflection of point C is the algebraic sum of \delta _1 and \delta _2:
\delta _C=\delta _1+\delta _2=\frac{qaL(4a^2 – L^2)}{24EI}+\frac{qa^4}{8EI} = \frac{qa}{24EI} [L(4a^2 – L^2)+3a^3]
or
\delta _C= \frac{qa}{24EI} (a + L)(3a^2 + aL – L^2) (8-59)
From the preceding equation we see that 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:
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 > 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 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 d²v/dx² always vanish at an inflection point.
However, a point where M and d²v/dx² 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.