LECTURE 10

RELATION AMONG ELASTIC CONSTANTS

Relation between E, G and u :

Let us establish a relation among the elastic constants E,G and u. Consider a cube of material of side Ďa' subjected to the action of the shear and complementary shear stresses as shown in the figure and producing the strained shape as shown in the figure below.

Assuming that the strains are small and the angle A C B may be taken as 450.

Therefore strain on the diagonal OA

= Change in length / original length

Since angle between OA and OB is very small hence OA @ OB therefore BC, is the change in the length of the diagonal OA

Now this shear stress system is equivalent or can be replaced by a system of direct stresses at 450 as shown below. One set will be compressive, the other tensile, and both will be equal in value to the applied shear strain.

Thus, for the direct state of stress system which applies along the diagonals:

We have introduced a total of four elastic constants, i.e E, G, K and g. It turns out that not all of these are independent of the others. Infact given any two of then, the other two can be found.

irrespective of the stresses i.e, the material is incompressible.

When g = 0.5  Value of k is infinite, rather than a zero value of E and volumetric strain is zero, or in other words, the material is incompressible.

Relation between E, K and u :

Consider a cube subjected to three equal stresses s as shown in the figure below

The total strain in one direction or along one edge due to the application of hydrostatic stress or volumetric stress s is given as

Relation between E, G and K :

The relationship between E, G and K can be easily determained by eliminating u from the already derived relations

E = 2 G ( 1 + u ) and E = 3 K ( 1 - u )

Thus, the following relationship may be obtained

Relation between E, K and g :

From the already derived relations, E can be eliminated

Engineering Brief about the elastic constants :

We have introduced a total of four elastic constants i.e E, G, K and u. It may be seen that not all of these are independent of the others. Infact given any two of them, the other two can be determined. Futher, it may be noted that

hence if u = 0.5, the value of K becomes infinite, rather than a zero value of E and the volumetric strain is zero or in otherwords, the material becomes incompressible

Futher, it may be noted that under condition of simple tension and simple shear, all real materials tend to experience displacements in the directions of the applied forces and Under hydrostatic loading they tend to increase in volume. In otherwords the value of the elastic constants E, G and K cannot be negative

Therefore, the relations

E = 2 G ( 1 + u )

E = 3 K ( 1 - u )

Yields

In actual practice no real material has value of Poisson's ratio negative . Thus, the value of u cannot be greater than 0.5, if however u > 0.5 than Îv = -ve, which is physically unlikely because when the material is stretched its volume would always increase.

Determination of Poisson's ratio: Poisson's ratio can be determined easily by simultaneous use of two strain gauges on a test specimen subjected to uniaxial tensile or compressive load. One gage is mounted parallel to the longitudnal axis of the specimen and other is mounted perpendicular to the longitudnal axis as shown below:

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