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
„a' 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. Further, 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 other words, the
material becomes incompressible
Further, 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 other words 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.
Related Topics
Privacy Policy, Terms and Conditions, DMCA Policy and Compliant
Copyright © 2018-2023 BrainKart.com; All Rights Reserved. Developed by Therithal info, Chennai.