Stress, strain and Young's modulus

In a nutshell

The Young's modulus defines how easy it is to deform an object based upon its composition. The Young's modulus value for a material is constant regardless of dimension. Stress is the force per unit area and is measured in pascals, $Pa$ . Strain is the ratio of extension compared to original length and is unitless.

Equations

DESCRIPTION	EQUATION
Stress	$\sigma = \dfrac {F}{A}$
Strain	$\varepsilon = \dfrac {x}{L}$
Young's Modulus	$E = \dfrac {\sigma}{\varepsilon}$

Variable definitions

QUANTITY NAME	SYMBOL	DERIVED UNIT	SI Base UNITs
$force$	$F$	$N$	$kgms^{-2}$
$cross\,sectional\,area$	$A$	$m^2$	$m^2$
$extension$	$x$	$m$	$m$
$original\, length$	$L$	$m$	$m$
$stress$	$\sigma$	$Pa$	$kg\,m^{-1} \, s^{-2}$
$strain$	$\varepsilon$	No units	No units
$Young's\,modulus$	$E$	$Pa$	$kg\,m^{-1} \, s^{-2}$

Force constant

The force constant describes how much force is required to deform an object by a certain amount of extension. The force constant depends on the dimensions of the materials being deformed.

Young's modulus is a property of a material which will be the same regardless of its dimensions.

Example

Two springs made of the same material but with a different length and thickness will have difference force constant and the same Youngs modulus.

Before looking at Young's modulus, lets first consider its constituents, stress and strain.

Stress

Stress, $\sigma$ , is defined as the force per unit area. This is the same equation as pressure and thus has the same units of pascals, $Pa$ .

Stress is given as:

$\sigma = \dfrac {F}{A}$

Example

Calculate the stress on a wire of cross-sectional area $1.7\ m^2$ when a force of $13\ N$ is applied onto it.

Firstly, write down the known values:

$A = 1.7\ m^2$ 

$F = 13\ N$

Calculate the stress using the force and the cross-sectional area:

$\sigma = \dfrac {F}{A}$

$\sigma = \dfrac{13}{1.7}$

$\sigma = 7.647\ Pa$

$\sigma = 7.6 Pa$

The stress on the wire is $\underline{7.6\ Pa}.$

Strain

Strain, $\varepsilon$ , is defined as the ratio of the extension to the length. It does not have a unit as the value is a ratio of two lengths. Since strain is a ratio, it can also be written as a percentage.

The equation for strain is:

$\varepsilon = \dfrac {x}{L}$

Example

A wire of original length $10\ m$ extends by $2.5\ m$ when a force is applied onto it. Calculate the strain on the wire.

Firstly, write down the known values:

$L=10\ m$

$x=2.5\ m$

Calculate the strain using the length and the extension:

$\varepsilon = \dfrac {x}{L}$

$\varepsilon = \dfrac {2.5}{10}$

$\varepsilon = 0.25$

The strain of the wire is $\underline{0.25}.$ 

Young's Modulus

The Young's modulus, $E$ , is defined as the stress per unit strain and has the units of pascals, $Pa$ . This is because stress has the units of pascals and strain is unitless.

Mathematically it is given as:

$E = \dfrac {\sigma}{\varepsilon}$

Note: Physicists are clever people, but be careful not to confuse Young's modulus, $E$ , with energy, $E$ , or electric field strength, $E$ !

The Young's modulus can be calculated for a material as long as it is within the limit of proportionality as beyond this, the stress is no longer proportional to the strain and the equation no longer applies. When calculating Young's modulus from a stress strain graph, the gradient must be calculated from the straight line.

Example

A spring of length $5\ m$ and a cross-sectional area of $1.2\ m^2$  is extended by $0.8\ m$ when a force of $9\ N$ is applied on it. Calculate the Young's Modulus of the material.

Firstly, write down the known values:

$L= 5\ m$

$A = 1.2\ m^2$

$x=0.8\ m$ 

 $F = 9\ N$ 

Calculate the stress using the force and the cross-sectional area:

$\sigma = \dfrac {F}{A}$

 $\sigma = \dfrac{9}{1.2}$ 

$\sigma = 7.5\ Nm^{-2}$

Calculate the strain using the length and the extension:

$\varepsilon = \dfrac {x}{L}$

 $\varepsilon = \dfrac {0.8}{5}$ 

$\varepsilon = 0.16$

Calculate the Young's Modulus using the stress and the strain:

$E = \dfrac{\sigma}{\varepsilon}$

$E = \dfrac {7.5}{0.16}$ 

$E = 46.875\ Nm^{-2}$ 

$E \approx 47\ Nm^{-2}$

The Young's Modulus of the material is $\underline{47\ Nm^{-2}}$