Question

what are elastic constant? establish the relation between them.

Answer 1

heya here Is ua answer.
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elastic constant. : one of the constants that express the elastic behavior of a given material — compare bulk modulus, elasticlimit, poisson's ratio , shear modulus, yield point, young's modulus.

Although only 2 independent elastic constants are required to completely the properties of an isotropic, elastic medium, there are many different constants from which to choose. Most of these constants arise as constants of proportionality between stress and strain for various loading conditions. For example, the ratio of hydrostatic stress (pressure) to volumetric strain (dilatation) is , which is given the name bulk modulus and the symbol . The constants referred to below are:

Has no name, as far as I know, so I call it "Lam's constant" shear modulus; sometimes G is used instead of Poisson's ratio Young's modulus Bulk modulus

To these we may add the following combinations that appear frequently.

mark it as brainlist. .

Relationship between elastic constants
The elastic constants describing a solid are closely related to each other. Here we
prove the relation between the modulus of rigidity, Young’s modulus and Poisson’s
ratio
G =
Y
2(1 + ν)
. (1)
Another such relation is the subject of Problem 3.1.
To prove (1) we proceed in two steps. Consider first a cube of side length a that
is exposed to shear stress. A side view of this cube is given in Fig. 1(a). We
are interested in how much the length of the two diagonals d1 and d2 changes.
To obtain this, we first determine the length of the distortion to the right (red
arrow). This must be a tan α ≈ aα for a small distortion. With this we calculate
the diagonal length d1 to be
d1 =
p
(a + aα)
2 + a
2 ≈
p
2a
2 + 2a
2α, (2)
neglecting the quadratic term in α. We now expand d1 for small α as
d1 =
√
2a +
2a
2
2
√
2a
2 + 2a
2α

α=0
α =
√
2a +
1
√
2
aα (3)
The extension of d1 is thus aα/√
2 and the strain along this direction is
∆d1
d1
=
αa
√
2a
√
2
=
α
2
=
τ
2G
. In the last step, we have used the definition of modulus of rigidity G = τ/α.
Following exactly the same strategy for d2, we find that this contracts by the same
amount as d1 expands.
We can now construct a situation in which we obtain the same deformation by
combined application of conventional stress in two orthogonal directions. Since the
deformation of the cube is symmetric (same expansion of d1 as contraction of d2)
what we have to do is pull along the d1 axis and push along the d2 axis using
the same force. In order to do this, we imagine the strained cube inside a bigger
cuboid as shown in Fig. 1(b). The forces F acting on the sides all have the same
magnitude. When we compare the force F to the forces FS used to obtain the
shear distortion, we see that F = 2FS/
√
2 = √
2FS. From this we obtain
F
√
2a
2
=
√
2FS
√
2a
2
=
FS
a
2
. (5)
On the left hand side of this equation we recognize the magnitude of the stress σ
needed to achieve the strained situation in Fig. 1(b). On the right hand side we
have the magnitude of shear stress τ in Fig. 1(a). We therefore see that σ = τ .
We now require that applying conventional stress gives rise to the same strain of
the cube along the axis d1. The strain must be
=
σ
Y
+ ν
σ
Y
. (6)
The first term is the usual strain and the second originates from the fact that we
also push in the horizontal direction (and it is therefore positive). We can now set
this stain equal to the one obtained for a shear strain (4)
τ
2G
=
σ
Y
(1 + ν). (7)
Since we have also shown that σ = τ , this gives the desired relation (1).



 

Answer 2

Explanation:

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