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A bar of copper is subjected to a tensile load. The elastic modulus of the copper is 110 GPa, its yield strength is 240 MPa, its tensile strength is 400 MPa, and its Poisson's ratio is 0.340. If the original length of the rod is 250 mm and it has a round cross-section with an original diameter of 9.50 mm, what load (in N) would need to be applied in order for the rod's length to be 250.1 mm

Respuesta :

Answer:

The answer is "3,118 N"

Explanation:

Using the hookes laws:

[tex]y=\frac{stress}{strain}=\frac{\frac{F}{A}}{\frac{\Delta L}{L}}[/tex]

So,  

[tex]F=\frac{y \times \Delta L \times A}{L}\\\\[/tex]

    [tex]=\frac{110 \times 10^{3} \times 0.1 \times \frac{\pi}{4} \times 9.5^2 }{250}\\\\=\frac{110 \times 10^{3} \times 0.1 \times \frac{3.14}{4} \times 90.25 }{250}\\\\=3,117.235 \ N\\\\=3,118 \ N\\\\[/tex]

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