Elastic Constants

Describes the realtionship between the Elastic Constants, and introduces Bulk Modulus and Young's Modulus.

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Contents

Introduction - Elastic Constants
Bulk Modulus
The Relationship Between E And C
Page Comments

Introduction - Elastic Constants

In the science of materials, numbers that quantify the response of a particular material to elastic or non-elastic deformation when a stress load is applied to that material, are known as Elastic Constants.

They are the relationships that determine the deformations produced by a given Stress system acting on a particular Material, and within the limits for which Hooke's Law is obeyed, these factors are constant:

The Modulus of Elasticity, $\inline E$
The Modulus of rigidity, $\inline C$
The Bulk Modulus, $\inline K$
Poisson's Ratio $\inline \displaystyle\frac{1}{m}$ or $\inline \sigma$

An elastic modulus, or modulus of elasticity, is the mathematical description of an object or substance's tendency to be deformed elastically (i.e. non-permanently) when a force is applied to it. The elastic modulus of an object is defined as the slope of its stress-strain curve in the elastic deformation region.

Shear modulus or modulus of rigidity is defined as the ratio of shear stress to the shear strain.

Poisson's ratio is the ratio, when a sample object is stretched, of the contraction or transverse strain (perpendicular to the applied load), to the extension or axial strain (in the direction of the applied load).

Hooke's Law of elasticity is an approximation that states that the extension of a spring is in direct proportion with the load applied to it.

Bulk Modulus

If a "hydrostatic" pressure $\inline p$ (i.e. one which is equal in all directions) acting on a body of initial volume $\inline V$ , produces a reduction in the Volume equal numerically to $\inline \delta V$ , then the Bulk Modulus $\inline K$ is defined as the ratio between the fluid pressure and the Volumetric Strain, i.e.

$\;\;\;\;\;\;K=\frac{- p}{\displaystyle\frac{\delta V}{V}}=- p\;\frac{V}{\delta V}$

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The negative sign allows for the reduction in Volume.

The above diagram is of a Unit Cube of material (or fluid) which is under the action of a pressure $\inline p$ . It can be seen that the Principal Stresses are $\inline -p$ , $\inline -p$ , and $\inline -p$ , and that the linear Strain in each direction is (see Compound Stress and Strain Part 2):

$\frac{- p}{E} + \frac{p}{mE} + \frac{p}{mE}= \left(\frac{- p}{E} \right)\left(1 - \frac{2}{m} \right)$

But, Volumetric Strain = Sum of Linear Strains $\inline = \left(\displaystyle\frac{- 3p}{E} \right)\left(1 - \displaystyle\frac{2}{m} \right)$

Hence by definition, $\inline K=\displaystyle\frac{-p}{\left(\displaystyle\frac{- 3p}{E} \right)\left(1-\displaystyle\frac{2}{m} \right)}$

Or,

$E=3K\left(1-\frac{2}{m} \right)$

(2)

Strain Energy per unit volume $\inline U$ in terms of the Principal Stresses is given by:

$U=\left(\frac{1}{2E} \right)[p^2+p^2+p^2-\left(\frac{2}{m} \right)(p^2+p^2+p^2)]=\left(\frac{3p^2}{2E} \right)\left(1-\frac{2}{m} \right)$

$\therefore\;\;\;\;\;\;\;\;U=\frac{p^2}{2K}$

Example:

[imperial]

Example - Example 1

Problem

A frictionless plunger $0.25\;inches$ in diameter and weighing $2\;lbs.$ , compresses oil in steel container. A weight of $3\;lbs.$ is dropped from a height of $2\;ins.$ onto the plunger.

Calculate the maximum pressure set up in the oil if its volume is $500\;cu.\;ins.$ and the container is assumed to be rigid.

$\displaystyle K=0.4\times10^6\;lb.\;in.^{-2}$ for Water

Workings

Let $p\;lb/sq.in.$ be the additional momentary maximum pressure produced by the falling weight, if the loss of energy at impact is neglected.

The loss of the Potential energy of the falling weight = The gain in Strain energy of the water

The Volumetric Strain produced $p$ is $- \displaystyle\frac{p}{k}$ and hence the decrease in the volume of water is $\displaystyle \left(\frac{p}{K} \right)\times 500$ and this is taken up by the Plunger which will therefore sink a further distance equal to :

$\left(\frac{p}{K} \right)\times 500 \times \frac{4}{\pi\times 0.25^2 }=\left(\frac{64\;p\times 500}{\pi K} \right)$

Therefore, Loss of potential energy = $3\left\{2 + \left(\displaystyle\frac{64\;p\times 500}{\pi K} \right) \right\}$

And gain in Strain energy = $\left(\displaystyle\frac{p^2}{2K} \right)\times 500$

Equating these last two quantities and multiplying through by $\displaystyle\frac{K}{500}$ produces the quadratic

$\frac{p^2}{2} = \frac{3\times 64\;p}{\pi } + \frac{3\times 2\times K}{500}$

$\frac{p^2}{2}-61.2\;p-48,000=0$

Solving and taking the positive root gives: $p=161.1\;lb.in.^{-2}$

Adding the pressure due to the $2\;lb.$ weight gives a final maximum pressure of:

$p=161.1+2\times \frac{64}{\pi }=202\;lb.in.^{-2}$

Solution

The maximum pressure is $p=202\;lb.in.^{-2}$

The Relationship Between E And C

It is necessary to establish , first of all, the relationship between Pure Shear Stress and a pure normal Stress system at a point in an elastic material.

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In the diagram the applied Stresses are $\inline f$ tensile on $\inline AB$ and $\inline f$ compressive on $\inline BC$ . If the Stress components on a plane $\inline AC$ at $\inline \displaystyle 45^0$ to $\inline AB$ are $\inline \displaystyle f_\theta$ and $\inline s_\theta$ , then the forces acting are as shown, taking the area on $\inline AC$ as unity.

Resolving along and at right angles to $\inline AC$

$s_\theta=\left(\frac{f}{\sqrt{2}} \right)\;\sin45^0+\left(\frac{f}{\sqrt{2}} \right)\;\cos45^0$

And,

$\;\;\;\;f_\theta =\left(\frac{f}{\sqrt{2}} \right)\;\cos45^0-\left(\frac{f}{\sqrt{2}} \right)\;\sin45^0$

i.e., there is pure shear on planes at $\inline \displaystyle 45^0$ to $\inline AB$ and $\inline AC$ of magnitude equal to the applied normal Stresses.

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The square element $\inline ABCD$ has sides of unstrained length 2 units which are under the equal normal Stresses $\inline f$ both tension and compression. It has been shown the element $\inline EFGH$ is in pure shear of equal magnitude $\inline f$ .

The linear Strain $\inline e$ in the direction $\inline \displaystyle EG = \frac{f}{E} + \frac{f}{mE}$

$\therefore\;\;\;\;\;\;e = \left(\frac{f}{E} \right)\left(1 + \frac{1}{m} \right)$

(3)

The linear Strain in the direction $\inline HF\;\displaystyle = - \frac{f}{E} - \frac{f}{mE} = -e$

Hence the Strained lengths of $\inline EO$ and $\inline HO$ are $\inline 1+e$ and $\inline 1-e$ respectively.

The Shear Strain,

$\phi = \frac{f}{C}$

(4)

( See "Modulus of Rigidity" in pages on Shear Stress)

This distorts the element $\inline EFGH$ and the angle $\inline EHG$ increase to $\inline \displaystyle\frac{\pi }{2} + \phi$ . Angle $\inline EHO$ is half this i.e. $\inline \displaystyle \frac{\pi }{4} + \frac{\phi}{2}$

Consider the triangle $\inline EHO$ .

$\inline \tan EHO = \displaystyle\frac{EO}{HO}=\tan\left(\displaystyle\frac{\pi }{4} + \displaystyle\frac{\phi }{2} \right) = \displaystyle\frac{1 + e}{1 - e}$

Expanding this equation gives:

$\frac{1 + e}{1 - e} = \frac{\tan\displaystyle\frac{\pi}{4} + \tan\displaystyle\frac{\phi}{2}}{1 - \tan\displaystyle\frac{\pi}{4}.\tan\displaystyle\frac{\phi}{2}} = \frac{1 + \displaystyle\frac{\phi}{2}}{1 - \displaystyle\frac{\phi}{2}}$

Note $\inline \displaystyle \tan\frac{\pi }{4} = 1$ and for small angles it is permissible to write $\inline \displaystyle \tan\frac{\phi }{2} = \frac{\phi }{2}$

By inspection $\inline \displaystyle e = \frac{\phi }{2}$ and by substituting for $\inline e$ and $\inline \displaystyle \phi$ from equations (3) and (4)

$\left(\frac{f}{E} \right)\left(1 + \frac{1}{m} \right) = \frac{f}{2C}$

Or re-arranged into a more normal form:

$E = 2C\left(1 + \frac{1}{m} \right)$

(5)

By using equation (2) it is possible to eliminate Poisson's Ratio from equation (5) and hence it can be shown that:

$E = \frac{9\;C\;K}{C + 3K}$

In fact if any two elastic constants are known, the other two may be calculated. Experimentally however, it is not satisfactory to calculate Poisson's Ratio by determining $\inline E$ and $\inline C$ separately.

Example:

[imperial]

Example - Example 2

Problem

Show that if $E$ is assumed correct then an error of $1%$ in the determination of $C$ will involve an error of $5%$ in the calculation of Poisson's Ratio.

Workings