# 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:

Given:

- 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 is {eq}\Delta D=- \rm 3.4 \times 10^{-3} \ mm {/eq}.

- The modulus of elasticity of steel is {eq}E = 207 {/eq} GPa.

- The Poisson's ratio of steel is {eq}\mu = 0.30 {/eq}.

Within the elastic limit of operation, the Poisson ratio of a material is defined as the ratio of the lateral strain to the linear strain of the material.

{eq}\begin{align*} \eta&=\displaystyle \frac{Lateral \ Strain}{Linear \ Strain} \\ 0.3& =- \frac{ \displaystyle \frac{\Delta D}{D}}{ \displaystyle \frac{\Delta L}{L}}\\ 0.3 & = - \frac{ \displaystyle \frac{-3.4 \times 10^{-3}}{10.2}}{\displaystyle \frac{\Delta L}{L}}\\ \frac{\Delta L}{L} &=1.11 \times 10^{-3} \end{align*} {/eq}

Let the force applied on the specimen be *F*.

We apply the expression of Young's Modulus

{eq}\begin{align*} E&=\frac{FL}{A \Delta L}\\\\ 207 \times 10^9 &=\frac{F }{\frac{\pi D^2}{4} } \times \frac{L}{\Delta L}\\\\ 207 \times 10^9 &=\frac{F }{\displaystyle \frac{\pi \times (10.2 \times 10^{-3} )^2}{4} } \times \frac{1}{1.11 \times 10^{-3} }\\\\ F & = 18.76 \ kN\Rightarrow(Answer) \end{align*} {/eq}

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