Question

The acrylic plastic rod is 200mm long and 15mm in diameter. If an axial load of 300N is applied to it, determine the change in its length and the change in its diameter. Ep = 2.70 GPa, Vp = 0.4.

Answer 1

Concepts and reason

Stress:

Stress is defined as the resisting force per unit area. It is expressed in N/m².

Strain:

Strain is defined as the change in dimensions divided by the original dimensions. It has no units.

Modulus of Elasticity: Stress is directly proportional strain in elastic deformation. The proportionality constant between stress and strain is called as modulus of elasticity.

Deflection or Deformation:

Deflection is defined as the change in dimensions or the shape due to the application of force or forces. It is expressed in terms of units of length.

Poisson’s ratio: The ratio of the lateral strain and axial strain is called Poisson’s ratio.

Fundamentals

Normal stress due to axial force (Direct stress) is given by,

$\sigma = \frac{P}{A}$

Here, $P$ is the axial load and $A$ is the cross-sectional area.

Hooke’s law:

Stress is proportional to strain, as long as the stress induced doesn’t exceed the proportional limit of the material.

$\begin{array}{l}\\\sigma \propto \varepsilon \\\\\sigma = E\varepsilon \\\end{array}$

The above equation can be rearranged as,

$\varepsilon = \frac{\sigma }{E} = \frac{{\left( {\frac{P}{A}} \right)}}{E} = \frac{P}{{AE}}$ …… (1)

Here, $\varepsilon$ is the strain, $\sigma$ is the stress, $E$ is the young’s modulus, $P$ is the load, and $A$ is the cross sectional area

Deformation is given by,

$\begin{array}{l}\\\delta = \frac{{PL}}{{AE}}\\\\\delta = \frac{\sigma }{E}L\\\\\delta = {\varepsilon _{long}}L\\\end{array}$

Formulas for calculating strains in different direction,

$\begin{array}{l}\\{\varepsilon _x} = \frac{\sigma }{E}\\\\{\varepsilon _r} = - \nu {\varepsilon _x}\\\end{array}$

Calculate the stress:

$\begin{array}{l}\\\sigma = \frac{P}{A}\\\\\sigma = \frac{P}{{\frac{\pi }{4}{D^2}}}\\\end{array}$

Substitute 300 for P, and 0.015 for D.

$\begin{array}{c}\\\sigma = \frac{{300}}{{\frac{\pi }{4}{{\left( {0.015} \right)}^2}}}\\\\ = 1.697\;{\rm{Mpa}}\\\end{array}$

Calculate the longitudinal strain:

${\varepsilon _{long}} = \frac{\sigma }{E}$

Substitute $1.697\left( {{{10}^6}} \right)$ for $\sigma$ and $2.70\left( {{{10}^9}} \right)$ for $E$ .

$\begin{array}{c}\\{\varepsilon _{long}} = \frac{{1.697\left( {{{10}^6}} \right)}}{{2.70\left( {{{10}^9}} \right)}}\\\\ = 0.000628\\\end{array}$

The deformation (change in length) is given by:

$\delta = {\varepsilon _{long}}L$

Substitute 0.000628 for ${\varepsilon _{long}}$ and 200 for L.

$\begin{array}{c}\\\delta = 0.000628\left( {200} \right)\\\\ = 0.126{\rm{ mm}}\\\end{array}$

Calculate the lateral strain,

$\varepsilon = - V{\varepsilon _{long}}$

Substitute 0.4 for $V$ and 0.000628 for ${\varepsilon _{long}}$ .

$\begin{array}{c}\\\varepsilon = - 0.4\left( {0.000628} \right)\\\\ = - 0.000251\\\end{array}$

Change in diameter is:

$\Delta d = \varepsilon d$

Substitute 0.000251 for $\varepsilon$ and 15 for $d$ .

$\begin{array}{c}\\\Delta d = - 0.000251\left( {15} \right)\\\\\Delta d = - 0.00377\;{\rm{mm}}\\\end{array}$

Ans:

The change in length of rod is $0.126\;{\rm{mm}}$ .