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}


Learn more about this topic:

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Poisson's Ratio: Definition & Equation

from Introduction to Engineering

Chapter 1 / Lesson 17
19K

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