# A cylindrical specimen of a brass alloy having a diameter of 40 mm and length of 80 mm elongated...

## Question:

A cylindrical specimen of a brass alloy having a diameter of 40 mm and length of 80 mm elongated during tensile test by 0.8 mm (elastic deformation). Under these circumstances, what is the diameter of the specimen (in mm) at this load level? Consider modulus of elasticity (E) = 97 GPa and Poisson's Ratio (v) = 0.34 .Use 4 significant figures 1 GPa = 1000 MPa, 1 MPa = 1 N/mm{eq}^2{/eq}

## Diameter of the Cylinder Under Load - Poisson's Ratio

Poisson's ratio is the ratio of lateral contraction strain to longitudinal extension strain in the direction of the applied force. In the simplest form the ratio of lateral strain to longitudinal strain is known as Poisson's ratio. Lateral strain is the ratio of lateral contraction to original length and linear strain is the ratio of linear extension to original length. The highest value of Poisson's ratio is 0.5.

Given points

• Original diameter of the brass cylinder {eq}d = 40 \times 10^{-3 } \ m {/eq}
• Original length of the brass cylinder {eq}L = 80 \times 10^{-3 } \ m {/eq}
• Increase in length of the cylinder under load condition {eq}\Delta L = 0.8 \times 10^{-3 } \ m {/eq}
• Poisson's ratio {eq}\sigma = 0.34 {/eq}

Let Lateral contraction in the diameter be {eq}\Delta d {/eq}

Then lateral contraction strain {eq}\epsilon_s = \dfrac { \Delta d } { d } {/eq}

Linear expansion strain {eq}\epsilon_l = \dfrac { \Delta L } { L } {/eq}

We have the equation for Poisson's ratio {eq}\sigma = \dfrac { \epsilon_s } { \epsilon_l } \\ \sigma = \dfrac { \Delta d \times L } { d \times \Delta L } {/eq}

So the contraction in the diameter {eq}\Delta d = \dfrac { \sigma \times d \times \Delta L } { L } \\ \Delta d = \dfrac { 0.34 \times 40 \times 10^{-3 } \times 0.8 \times 10^{-3 } } { 80 \times 10^{-3 } } \\ \Delta d = 1.36 \times 10^{-4 } \ m {/eq}

So the diameter of the specimen under the load condition {eq}d_1 = d - \Delta d \\ d_1 = 40 \times 10^{-3 } - 1.36 \times 10^{-4 } \\ d_1 = 39.864 \times 10^{-3 } \ m {/eq} 