Determine the extension, due to its own weight, of the conical bar shown in Figure P5.13. The bar is made of aluminum alloy [E = 10,600 ksi and γ = 0.100 lb/in.^3]. The bar has a 2-in. radius at its upper end and a length of L = 20 ft. Assume the taper of the bar is slight enough for the assumption

Question

Determine the extension, due to its own weight, of the conical bar shown in Figure P5.13. The bar is made of aluminum alloy [E = 10,600 ksi and γ = 0.100 [latex]lb / in .{ }^{3}[/latex]]. The bar has a 2-in. radius at its upper end and a length of L = 20 ft. Assume the taper of the bar is slight enough for the assumption of a uniform axial stress distribution over a cross section to be valid.

Accepted Answer

An incremental length dy of the bar has an incremental deformation given by

[latex]d \delta=\frac{F(y)}{A E} d y[/latex]

The force in the bar can be expressed as the product of the unit density of the aluminum [latex]\left(\gamma_{ ext {alum }} ight)[/latex] and the volume of the bar below the incremental slice dy. The volume below the slice dy is a cone. At y, the cross-sectional area of the base of the cone is:

[latex]A_{y}=\pi\left(\frac{y}{L} r ight)^{2}[/latex]

and the volume of a cone is given by:

[latex]V=\frac{1}{3}[/latex] (area of base)(altitude)

The internal force at y can be expressed as

[latex]F(y)=\gamma_{ ext {alum }} imes \frac{1}{3} A_{y} y[/latex]

The incremental deformation can be expressed as

[latex]d \delta=\frac{\gamma_{ ext {alum }} A_{y} y}{3 A_{y} E} d y=\frac{\gamma_{ ext {alum }} y}{3 E} d y[/latex]

Integrate this expression over the entire length L of the bar:

[latex]\delta=\int_{0}^{L} \frac{\gamma_{ ext {alum }} y}{3 E} d y=\frac{\gamma_{ ext {alum }}}{3 E} \int_{0}^{L} y d x=\frac{\gamma_{ ext {alum }}}{3 E} \frac{1}{2}\left[y^{2} ight]_{0}^{L}=\frac{\gamma_{ ext {alum }} L^{2}}{6 E}[/latex]

The change in length of the bar due to its own weight is therefore:

[latex]\delta=\frac{\gamma_{ alum } L^{2}}{6 E}=\frac{\left(0.100 lb / in .^{3} ight)(20 ft imes 12 in. / ft )^{2}}{6(10,600,000 psi )}=9.0566 imes 10^{-5} in .=90.6 imes 10^{-6} in.[/latex]

Question 5.13: Determine the extension, due to its own weight, of the conic...

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