A square steel bar has a length of 5.1 ft and a 2.7 in by 2.7 in cross section and is subjected to axial tension. The final length is 5.10295 ft . The final side length is 2.69953 in . What is Poisson's ratio for the material

Respuesta :

Answer:

The Poisson's ratio for the material is 0.0134.

Step-by-step explanation:

The Poisson's ratio ([tex]\nu[/tex]), no unit, is the ratio of transversal strain ([tex]\epsilon_{t}[/tex]), in inches, to axial strain ([tex]\epsilon_{a}[/tex]), in inches:

[tex]\nu = -\frac{\epsilon_{t}}{\epsilon_{a}}[/tex] (1)

[tex]\epsilon_{a} = l_{a,f}-l_{a,o}[/tex] (2)

[tex]\epsilon_{t} = l_{t,f}-l_{t,o}[/tex] (3)

Where:

[tex]l_{a,o}[/tex] - Initial axial length, in inches.

[tex]l_{a,f}[/tex] - Final axial length, in inches.

[tex]l_{t,o}[/tex] - Initial transversal length, in inches.

[tex]l_{t,f}[/tex] - Final transversal length, in inches.

If we know that [tex]l_{a,o} = 61.2\,in[/tex], [tex]l_{a,f} = 61.235\,in[/tex], [tex]l_{t,o} = 2.7\,in[/tex] and [tex]l_{t,f} = 2.69953\,in[/tex], then the Poisson's ratio is:

[tex]\epsilon_{a} = 61.235\,in - 61.2\,in[/tex]

[tex]\epsilon_{a} = 0.035\,in[/tex]

[tex]\epsilon_{t} = 2.69953\,in - 2.7\,in[/tex]

[tex]\epsilon_{t} = -4.7\times 10^{-4}\,in[/tex]

[tex]\nu = - \frac{(-4.7\times 10^{-4}\,in)}{0.035\,in}[/tex]

[tex]\nu = 0.0134[/tex]

The Poisson's ratio for the material is 0.0134.

ACCESS MORE