A cantilever of length L and having a flexural rigidity EI carries a distributed load that varies in intensity from w/unit length at the built-in end to zero at the free end. Find the deflection of the free end.

Answer: $w L^{4}/30E I$ (downward)

Question

A cantilever of length L and having a flexural rigidity EI carries a distributed load that varies in intensity from w/unit length at the built-in end to zero at the free end. Find the deflection of the free end.

Answer: $w L^{4}/30E I$ (downward)

Answer: [latex]w L^{4}/30E I[/latex] (downward)

Accepted Answer

Take the origin of z at the free end of the cantilever. The load intensity at any section a distance z from the free end is wz/L. The bending moment at this section is given by

[latex]M_{z}=\left({\frac{z}{2}}\right)\left({\frac{w z}{L}}\right)\left({\frac{z}{3}}\right)={\frac{w z^{3}}{6L}}[/latex]

Substituting in Eqs (16.31),

[latex]u^{\prime\prime}=-{\frac{M_{y}}{E I_{y y}}},\quad ν^{\prime\prime}=-{\frac{M_{x}}{E I_{x x}}}[/latex] (16.31)

[latex]E I\left({\frac{\mathrm{d}^{2}v}{\mathrm{d}z^{2}}}\right)={\frac{-w z^{3}}{6L}}[/latex]

[latex]E I{\biggl(}{\frac{\mathrm{d}v}{\mathrm{d}z}}{\biggr)}={\frac{-w z^{4}}{24L}}+C_{1}[/latex]

[latex]E I v={\frac{-w z^{5}}{120L}}+C_{1}z+C_{2}[/latex]

When z=L, (dυ/dz) = 0 so that [latex]C_{1}=w L^{3}/24.[/latex] When z=L, υ = 0, i.e., [latex]C_{2}=-w L^{4}/30.[/latex]
The deflected shape of the beam is then given by

[latex]E I v=-\left({\frac{w}{120L}}\right)\left(z^{5}-5zL^{4}+4L^{5}\right)[/latex]

At the free end where z = 0,

[latex]\ v=-{\frac{w L^{4}}{30E I}}[/latex]

Question 16.10: A cantilever of length L and having a flexural rigidity EI c......

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