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.)

Question

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 [latex]\delta_{C}[/latex] at the end of the overhang (Fig. 9-21c). (Note: The beam has constant flexural rigidity EI.)

Accepted Answer

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 [latex] heta_{B}[/latex], which is the angle of rotation of beam ABC at support B (Fig. 9-21c). (We assume that a clockwise angle [latex] heta_{B}[/latex] is positive.) This angle of rotation causes a rigid-body rotation of the overhang BC, resulting in a downward displacement [latex]\delta_{1}[/latex] 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 [latex]\delta_{2}[/latex] (Fig. 9-21c). The superposition of these two displacements gives the total displacement [latex]\delta_{C}[/latex] at point C.
Deflection [latex]\delta_{1}[/latex]. Let us begin by finding the deflection [latex]\delta_{1}[/latex] caused by the angle of rotation [latex] heta_{B}[/latex] 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 [latex]M_{B}[/latex] (equal to qa²/2), and (3) a vertical load P (equal to qa). Only the loads q and [latex]M_{B}[/latex] 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.

[latex] u=-\frac{qx}{24EI}\left(L^{3}-2Lx^{2}+x^{3} ight)[/latex]

[latex] u^{\prime}=-\frac{q}{24EI}\left(L^{3}-6Lx^{2}-4x^{3} ight)[/latex]

[latex]\delta_{C}=\delta_{\max}=\frac{5qL^{4}}{384EI}[/latex] [latex] heta_{A}= heta_{B}=\frac{qL^{3}}{24EI}[/latex]

[latex] u=-\frac{M_{0}x }{6LEI}(2L^{2}-3Lx+x^{2}) u^{\prime}=-\frac{M_{0}}{6LEI}(2L^{2}-6Lx+3x^{2})[/latex]

[latex]\delta_{C}=\frac{M_{0}L^{2}}{16EI} heta_{A}= \frac{M_{0}L}{3EI} heta_{B}= \frac{M_{0}L}{6EI}[/latex]

[latex]x_{1}=L\left(1-\frac{\sqrt{3}}{3} ight)[/latex] and [latex]\delta_{\max}=\frac{M_{0}L^{2}}{9\sqrt{3}EI}[/latex]

Thus, the angle [latex] heta_{B}[/latex] is

[latex] heta_{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} ight)}{24EI}[/latex] (9-58)

in which a clockwise angle is positive, as shown in Fig. 9-21c.
The downward deflection [latex]\delta_{1}[/latex] of point C, due solely to the angle of rotation [latex] heta_{B}[/latex], is equal to the length of the overhang times the angle (Fig. 9-21c):

[latex]\delta_{1} = a heta_{B} = \frac{qaL\left(4a^{2}-L^{2} ight)}{24EI}[/latex] (e)

Deflection [latex]\delta_{2}[/latex]. Bending of the overhang BC produces an additional downward deflection [latex]\delta_{2}[/latex] 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):

[latex] u=-\frac{qx^{2}}{24EI}(6L^{2}-4Lx+x^{2})[/latex] [latex] u^{\prime}=-\frac{qx}{6EI}(3L^{2}-3Lx+x^{2})[/latex]

[latex]\delta_{B}=\frac{qL^{4}}{8EI}[/latex] [latex] heta_{B}=\frac{qL^{3}}{6EI}[/latex]

[latex]\delta_{2} = \frac{qa^{4}}{8EI}[/latex] (f)

Deflection [latex]\delta_{C}[/latex]. The total downward deflection of point C is the algebraic sum of [latex]\delta_{1}[/latex] and [latex]\delta_{2}[/latex]:

[latex]\delta_{C} = \delta_{1}+\delta_{2} = \frac{qaL\left(4a^{2}-L^{2} ight)}{24EI}+\frac{qa^{4}}{8EI} = \frac{qa}{24EI}\left[L\left(4a^{2}-L^{2} ight)+3a^{2} ight] [/latex]

or

[latex]\delta_{C} = \frac{qa}{24EI}\left(a+L ight)\left(3a^2+aL-L^{2} ight)[/latex] (9-59)

From the preceding equation we see that the deflection [latex]\delta_{C}[/latex] 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 [latex]\delta_{C}[/latex] 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:

[latex]\left(3a^2+aL-L^{2} ight) = 0[/latex]

or

a = [latex]\frac{L\left(\sqrt{13}-1 ight)}{6} [/latex] = 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.

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