Determine the deflection curve and the deflection of the free end of the cantilever shown in Fig. 15.17(a). The cantilever has a doubly symmetrical cross-section.

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Accepted Answer

The bending moment, M, at any section Z is given by

[latex]M=\frac{w}{2}(L-z)^{2}[/latex] (i)

Substituting for M in the second of Eqs. (15.31) and rearranging, we have

[latex]u^{\prime \prime}=-\frac{M_{y}}{E I_{y y}}, \quad v^{\prime \prime}=-\frac{M_{x}}{E I_{x x}}[/latex] (15.31)

[latex]E I v^{\prime \prime}=-\frac{w}{2}(L-z)^{2}=-\frac{w}{2}\left(L^{2}-2 L z+z^{2} ight)[/latex] (ii)

Integration of Eq. (ii) yields

[latex]E I v^{\prime}=-\frac{w}{2}\left(L^{2} z-L z^{2}+\frac{z^{3}}{3} ight)+C_{1}[/latex]

When z = 0 at the built-in end, [latex]v^{\prime}=0[/latex] so that [latex]C_{1}=0[/latex] and

[latex]E I v^{\prime}=-\frac{w}{2}\left(L^{2} z-L z^{2}+\frac{z^{3}}{3} ight)[/latex] (iii)

Integrating Eq. (iii), we have

[latex]E I v=-\frac{w}{2}\left(L^{2} \frac{z^{2}}{2}-\frac{L z^{3}}{3}+\frac{z^{4}}{12} ight)+C_{2}[/latex]

and since v = 0 when z = 0, [latex]C_{2}=0[/latex]. The deflection curve of the beam therefore has the equation

[latex]v=-\frac{w}{24 E I}\left(6 L^{2} z^{2}-4 L z^{3}+z^{4} ight)[/latex] (iv)

and the deflection at the free end when z = L, is

[latex]v_{ ext {tip }}=-\frac{w L^{4}}{8 E I}[/latex]

which is again negative and downward.

Question 15.6: Determine the deflection curve and the deflection of the fre...

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