Elastic Constants
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.
Describes the realtionship between the Elastic Constants, and introduces Bulk Modulus and Young's Modulus.
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, $E$
- The Modulus of rigidity, $C$
- The Bulk Modulus, $K$
- Poisson's Ratio $\displaystyle\frac{1}{m}$ or $\sigma$
Bulk Modulus
If a "hydrostatic" pressure $p$ (i.e. one which is equal in all directions) acting on a body of initial volume $V$, produces a reduction in the Volume equal numerically to $\delta V$, then the Bulk Modulus $K$ is defined as the ratio between the fluid pressure and the Volumetric Strain, i.e. $\;\;\;\;\;\;K=\frac{- p}{\displaystyle\frac{\delta V}{V}}=- p\;\frac{V}{\delta V}$
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 $p$. It can be seen that the Principal Stresses are $-p$, $-p$, and $-p$, 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 $= \left(\displaystyle\frac{- 3p}{E} \right)\left(1 - \displaystyle\frac{2}{m} \right)$
Hence by definition, $K=\displaystyle\frac{-p}{\left(\displaystyle\frac{- 3p}{E} \right)\left(1-\displaystyle\frac{2}{m} \right)}$
Or,
Strain Energy per unit volume $U$ 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 1 [imperial]
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
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
$\frac{p^2}{2} = \frac{3\times 64\;p}{\pi } + \frac{3\times 2\times K}{500}$ Or $\frac{p^2}{2}-61.2\;p-48,000=0$
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: $p=161.1+2\times \frac{64}{\pi }=202\;lb.in.^{-2}$
- 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.
In the diagram the applied Stresses are $f$ tensile on $AB$ and $f$ compressive on $BC$. If the Stress components on a plane $AC$ at $\displaystyle 45^0$ to $AB$ are $\displaystyle f_\theta$ and $s_\theta$, then the forces acting are as shown, taking the area on $AC$ as unity.
Resolving along and at right angles to $AC$
$s_\theta=\left(\frac{f}{\sqrt{2}} \right)\;\sin45^0+\left(\frac{f}{\sqrt{2}} \right)\;\cos45^0$ 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 $\displaystyle 45^0$ to $AB$ and $AC$ of magnitude equal to the applied normal Stresses.
The square element $ABCD$ has sides of unstrained length 2 units which are under the equal normal Stresses $f$ both tension and compression. It has been shown the element $EFGH$ is in pure shear of equal magnitude $f$.
The linear Strain $e$ in the direction $\displaystyle EG = \frac{f}{E} + \frac{f}{mE}$
The linear Strain in the direction $HF\;\displaystyle = - \frac{f}{E} - \frac{f}{mE} = -e$
Hence the Strained lengths of $EO$ and $HO$ are $1+e$ and $1-e$ respectively.
The Shear Strain,
( See "Modulus of Rigidity" in pages on Shear Stress)
This distorts the element $EFGH$ and the angle $EHG$ increase to $\displaystyle\frac{\pi }{2} + \phi$. Angle $EHO$ is half this i.e. $\displaystyle \frac{\pi }{4} + \frac{\phi}{2}$
Consider the triangle $EHO$.
$\tan EHO = \displaystyle\frac{EO}{HO}=\tan\left(\displaystyle\frac{\pi }{4} + \displaystyle\frac{\phi }{2} \right) = \displaystyle\frac{1 + e}{1 - e}$
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 $\displaystyle \tan\frac{\pi }{4} = 1$ and for small angles it is permissible to write $\displaystyle \tan\frac{\phi }{2} = \frac{\phi }{2}$
By inspection $\displaystyle e = \frac{\phi }{2}$ and by substituting for $e$ and $\displaystyle \phi$ from equations (2) and (3)
$\left(\frac{f}{E} \right)\left(1 + \frac{1}{m} \right) = \frac{f}{2C}$
Or re-arranged into a more normal form:
By using equation (1) it is possible to eliminate Poisson's Ratio from equation (4) 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 $E$ and $C$ separately.
Example 1 [imperial]
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.
Let the correct values be $E,C$ and $\sigma$ then
If $C$ is increased to $1.01\;C$ let the calculated value of Poisson's ratio be $\sigma '$ then:
Eliminating $E$ from equations (5) and (6) $C(1=\sigma )=1.01C(1+\sigma ')$ $\therefore\;\;\;\;\;\;1+\sigma =1.01+1.01\sigma '$ Or $\sigma '-\sigma =-0.01-0.01 \sigma '$
The percentage error is given by: $\frac{\sigma ' - \sigma}{\sigma }\times 100 =-0.01\left(\frac{1+\sigma '}{\sigma } \right)\times 100$ $\approx -\frac{(1 + 0.25)}{0.25}=- 5\%$