Question 9.9: 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. 9-21a). 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. 9-21c). (Note: The beam has constant flexural rigidity EI.)
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 \theta_{B}, which is the angle of rotation of beam ABC at support B (Fig. 9-21c). (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. 9-21c). 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. 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.
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\nu=-\frac{qx}{24EI}\left(L^{3}-2Lx^{2}+x^{3}\right)
\nu^{\prime}=-\frac{q}{24EI}\left(L^{3}-6Lx^{2}-4x^{3}\right)
\delta_{C}=\delta_{\max}=\frac{5qL^{4}}{384EI} \theta_{A}=\theta_{B}=\frac{qL^{3}}{24EI} |
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\nu=-\frac{M_{0}x }{6LEI}(2L^{2}-3Lx+x^{2}) \nu^{\prime}=-\frac{M_{0}}{6LEI}(2L^{2}-6Lx+3x^{2})
\delta_{C}=\frac{M_{0}L^{2}}{16EI} \theta_{A}= \frac{M_{0}L}{3EI} \theta_{B}= \frac{M_{0}L}{6EI}
x_{1}=L\left(1-\frac{\sqrt{3}}{3}\right) and \delta_{\max}=\frac{M_{0}L^{2}}{9\sqrt{3}EI} |
Thus, the angle \theta_{B} is
\theta_{B} = -\frac{qL^{3}}{24EI}+\frac{M_{B}L}{3EI} = -\frac{qL^{3}}{24EI}+\frac{qa^{2}L}{6EI} = \frac{qL\left(4a^{2}-L^{2}\right)}{24EI} (9-58)
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):
\delta_{1} = a\theta_{B} = \frac{qaL\left(4a^{2}-L^{2}\right)}{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):
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\nu=-\frac{qx^{2}}{24EI}(6L^{2}-4Lx+x^{2}) \nu^{\prime}=-\frac{qx}{6EI}(3L^{2}-3Lx+x^{2})
\delta_{B}=\frac{qL^{4}}{8EI} \theta_{B}=\frac{qL^{3}}{6EI} |
\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\left(4a^{2}-L^{2}\right)}{24EI}+\frac{qa^{4}}{8EI} = \frac{qa}{24EI}\left[L\left(4a^{2}-L^{2}\right)+3a^{2}\right]or
\delta_{C} = \frac{qa}{24EI}\left(a+L\right)\left(3a^2+aL-L^{2}\right) (9-59)
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:
\left(3a^2+aL-L^{2}\right) = 0or
a = \frac{L\left(\sqrt{13}-1\right)}{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. 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 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.
