# A cylindrical bar of steel 10.2 mm (0.4016 in.) in diameter is to be deformed elastically by...

## Question:

A cylindrical bar of steel 10.2 mm (0.4016 in.) in diameter is to be deformed elastically by application of a force along the bar axis. Determine the force that will produce an elastic reduction of 3.4{eq}\times {/eq}10{eq}^3 {/eq} mm (1.339{eq}\times {/eq}10{eq}^4 {/eq}in.) in the diameter. For steel, values for the elastic modulus (E) and Poisson's ratio (?) are, respectivley, 207 GPa and 0.30.

## Young's Modulus

In an elastic material like steel when deformation takes place then it follows the Hook's law.

It states that the stress ({eq}\sigma {/eq}) produced in the material is directly proportional to the strain ({eq}\epsilon {/eq}) of the material.

{eq}\begin{align*} Stress&\propto Strain \\ Stress &= Y \times Strain \\ Y&=\frac{Stress}{Strain} \\ &=\frac{\sigma}{\varepsilon } \\ &=\frac{\frac{F}{A}}{\frac{\Delta L}{L} } \\ &=\frac{FL}{A \Delta L} \end{align*} {/eq}

- F is the external force
- A is the cross-sectional area
- L is the original length
- {eq}\Delta L {/eq} is the elongation

Poisson ratio of a material is defined as the ratio of the lateral strain to the linear strain of the material.

{eq}\eta=\displaystyle \frac{Lateral \ Strain}{Linear \ Strain}= \frac{ \displaystyle \frac{\Delta D}{D}}{\displaystyle \frac{\Delta L}{L}} {/eq}

## Answer and Explanation:

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View this answerGiven:

- The diameter of a cylindrical bar of steel is {eq}d = 10.2 \ mm = 10.2 \times 10^{-3} \ m {/eq}.

- The elastic reduction in diameter produced...

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