The cantilever beam AB supports a uniformly distributed load w (Fig.11.41). Determine the

Question

The cantilever beam [latex]A B[/latex] supports a uniformly distributed load [latex]w[/latex] (Fig. 11.41). Determine the deflection and slope at [latex]A[/latex].

Accepted Answer

Deflection at [latex]\boldsymbol{A}[/latex]. We apply a dummy downward load [latex]\mathbf{Q}_{A}[/latex] at [latex]A[/latex] (Fig. 11.42) and write

[latex]y_{A}=\frac{\partial U}{\partial Q_{A}}=\int_{0}^{L} \frac{M}{E I} \frac{\partial M}{\partial Q_{A}} d x[/latex] (11.77)

The bending moment [latex]M[/latex] at a distance [latex]x[/latex] from [latex]A[/latex] is

[latex]M=-Q_{A} x-\frac{1}{2} w x^{2}[/latex] (11.78)

and its derivative with respect to [latex]Q_{A}[/latex] is

[latex]\frac{\partial M}{\partial Q_{A}}=-x[/latex] (11.79)

Substituting for [latex]M[/latex] and [latex]\partial M / \partial Q_{A}[/latex] from (11.78) and (11.79) into (11.77), and making [latex]Q_{A}=0[/latex], we obtain the deflection at [latex]A[/latex] for the given loading:

[latex]y_{A}=\frac{1}{E I} \int_{0}^{L}\left(-\frac{1}{2} w x^{2} ight)(-x) d x=+\frac{w L^{4}}{8 E I}[/latex]

Since the dummy load was directed downward, the positive sign indicates that

[latex]y_{A}=\frac{w L^{4}}{8 E I} \downarrow[/latex]

Slope at [latex]\boldsymbol{A}[/latex]. We apply a dummy counterclockwise couple [latex]\mathbf{M}_{A}[/latex] at A (Fig. 11.43) and write

[latex] heta_{A}=\frac{\partial U}{\partial M_{A}}[/latex]

Recalling Eq. (11.17),

[latex]U=\int_{0}^{L}{\frac{M^{2}}{2E I}}\,d x[/latex] (11.17)

we have

[latex] heta_{A}=\frac{\partial}{\partial M_{A}} \int_{0}^{L} \frac{M^{2}}{2 E I} d x=\int_{0}^{L} \frac{M}{E I} \frac{\partial M}{\partial M_{A}} d x[/latex] (11.80)

The bending moment [latex]M[/latex] at a distance [latex]x[/latex] from [latex]A[/latex] is

[latex]M=-M_{A}-\frac{1}{2} w x^{2}[/latex] (11.81)

and its derivative with respect to [latex]M_{A}[/latex] is

[latex]\frac{\partial M}{\partial M_{A}}=-1[/latex] (11.82)

Substituting for [latex]M[/latex] and [latex]\partial M / \partial M_{A}[/latex] from (11.81) and (11.82) into (11.80), and making [latex]M_{A}=0[/latex], we obtain the slope at [latex]A[/latex] for the given loading:

[latex] heta_{A}=\frac{1}{E I} \int_{0}^{L}\left(-\frac{1}{2} w x^{2} ight)(-1) d x=+\frac{w L^{3}}{6 E I}[/latex]

Since the dummy couple was counterclockwise, the positive sign indicates that the angle [latex] heta_{A}[/latex] is also counterclockwise:

[latex] heta_{A}=\frac{w L^{3}}{6 E I} ⦨[/latex]

Question 11.13: The cantilever beam AB supports a uniformly distributed load...

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