A 1-in.-diameter by 16-ft-long cold-rolled bronze bar [E = 15,000 ksi and γ = 0.320 lb/in^3] hangs vertically while suspended from one end. Determine the change in length of the bar due to its own weight.

Question

A 1-in.-diameter by 16-ft-long cold-rolled bronze bar [E = 15,000 ksi and γ = 0.320 [latex]lb / in ^{3}[/latex]] hangs vertically while suspended from one end. Determine the change in length of the bar due to its own weight.

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 bronze [latex]\left(\gamma_{ ext {bronze }} ight)[/latex] and the volume of the bar below the incremental slice dy:

[latex]F(y)=\gamma_{ ext {bronze }} A y[/latex]

Therefore, the incremental deformation can be expressed as

[latex]d \delta=\frac{\gamma_{ ext {bronze }} A y}{A E} d y=\frac{\gamma_{ ext {bronze }} y}{E} d y[/latex]

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

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

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

[latex]\delta=\frac{\gamma_{ ext {bronze }} L^{2}}{2 E}=\frac{\left(0.320 lb / in .{ }^{3} ight)(16 ft imes 12 in . / ft )^{2}}{2(15,000,000 psi )}=0.000393216 in .=0.000393 in .[/latex]

Question 5.11: A 1-in.-diameter by 16-ft-long cold-rolled bronze bar [E = 1...

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