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Engineering › Materials ›

Elastic Constants

Describes the realtionship betyween the Elastic Constants and introduces Bulk Modulus and Young's Modulus

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Contents

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

Elastic Constants

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

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

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.

$i.e.\;\;\;\;\;\;K=\frac{- p}{\delta V/V}$

(1)

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 equation (49)) :

$\frac{- p}{E} + \frac{p}{mE} + \frac{p}{mE}$

(2)

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

(3)

But using equation (98) from Compound Stress and Strain Part 2:-

$Volumetric\;Strain = Sum\;of\;Linear\;Strains$

(4)

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

(5)

Hence by definition:

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

(6)

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

(7)

Using Equation (99) of Compound Stress and Strain part 2:

Strain Energy per unit volume (U) in terms of the Principal Stresses is given by:-

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

(8)

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

(9)

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

(10)

Example 1

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 it 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 (U.L.)

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 - 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 :-

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

(11)

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

(12)

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

(13)

Equating these last two quantities and multiplying through by K/500 produces the quadratic

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

(14)

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

(15)

Solving and taking the positive root gives:

$p=161.1\;lb.in.^{-2}$

(16)

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

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

(17)

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

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

(18)

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

(19)

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 $\displaystyle EG = \frac{f}{E} + \frac{f}{mE}$

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

(20)

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.

$The\; Shear\; Strain\; \phi = \frac{f}{C}$

(21)

( 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 = EO/HO

$\tan\left(\frac{\pi }{4} + \frac{\phi }{2} \right) = \frac{1 + e}{1 - e}$

(22)

Expanding this equation gives:

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

(23)

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 (20) and (21)

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

(24)

Or re-arranged into a more normal form:-

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

(25)

By using equation (7) it is possible to eliminate Poisson's Ratio from equation (25) and hence it can be shown that:-

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

(26)

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.