Isotropic Material

40 An isotropic material is symmetric with respect to every plane and every axis, that is the elastic properties are identical in all directions.

41 To mathematically characterize an isotropic material, we require coordinate transformation with rotation about x2 and xi axes in addition to all previous coordinate transformations. This process will enforce symmetry about all planes and all axes.

42 The rotation about the x2 axis is obtained through cos 6 0 — sin 6 0 1 0 sin 6 0 cos 6

we follow a similar procedure to the case of transversely isotropic material to obtain

Ciiii = C3333 1

43 next we perform a rotation about the x1 axis it follows that which will finally give

cos 6 sin 6

sin 6 cos 6

C1133 1 2

Cijkm

C1122 C1133 C2222 C2233 C3333

SYM.

c with a — 1 (C1111 — C1122), b — 2(c2222 — C2233), and c — 1 (C3333 — C1

44 If we denote C1122 = C1133 = C2233 = A and C1212 = C2323 = C3131 = p then from the previous relations we determine that c1111 = c2222 = c3333 = A + or

m and we are thus left with only two independent non zero coefficients A and j which are called Lame's constants.

45 Substituting the last equation into Eq. 7.29,

Or in terms of A and j, Hooke's Law for an isotropic body is written as

Sij Tkk

Tij 3A

2j or or

46 It should be emphasized that Eq. 7.47 is written in terms of the Engineering strains (Eq. 7.29) that is Yij = 2Eij for i = j. On the other hand the preceding equations are written in terms of the tensorial strains Ej

7.3.5.1 Engineering Constants

47 The stress-strain relations were expressed in terms of Lame's parameters which can not be readily measured experimentally. As such, in the following sections we will reformulate those relations in terms of "engineering constants" (Young's and the bulk's modulus). This will be done for both the isotropic and transversely isotropic cases.

7.3.5.1.1 Isotropic Case

7.3.5.1.1.1 Young's Modulus

48 In order to avoid certain confusion between the strain E and the elastic constant E, we adopt the usual engineering notation Tij ^ aij and Eij ^ eij

49 If we consider a simple uniaxial state of stress in the direction, then from Eq. 7.51

50 Yet we have the elementary relations in terms engineering constants E Young's modulus and v Poisson's ratio

then it follows that then it follows that

53 When the strain equation is expanded in 3D cartesian coordinates it would yield:

1

—v

—v

0

0

0

axx

£yy

—v

1

—v

0

0

0

ayy

<

1

—v

—v

1

0

0

0

<

azz

Ixy(2&xy)

f = E

0

0

0

1 + v

0

0

Txy

f

7yz(2^yz)

0

0

0

0

1 + v

0

Tyz

7zx(2^zx)

0

0

0

0

0

1 + v

. Tzx .

54 If we invert this equation, we obtain

axx

1 — v v v

&xx

ayy

(l+v)(l-2v)

v 1 — v v

0

£yy

azz

v v 1 — v

<

£ zz

Txy

=

1 0

0

Yxy(2£xy)

Tyz

0

G

0 1

0

7yz(2£yz)

Tzx

0 0

1

Yzx(2£zx)

7.3.5.1.1.2 Bulk's Modulus; Volumetric and Deviatoric Strains

7.3.5.1.1.2 Bulk's Modulus; Volumetric and Deviatoric Strains

51 Similarly in the case of pure shear in the x\x3 and x2x3 planes, we have

and the j is equal to the shear modulus G.

52 Hooke's law for isotropic material in terms of engineering constants becomes

55 We can express the trace of the stress Ia in terms of the volumetric strain IE From Eq. 7.50

an = XSii £kk + 2/xejj = (3A + 2fj,)£ii = 3Keu (7.61)

56 We can provide a complement to the volumetric part of the constitutive equations by substracting the trace of the stress from the stress tensor, hence we define the deviatoric stress and strains as as and the corresponding constitutive relation will be a = Kel + 2pe' (7.65)

0 0

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  • chantelle wilson
    How to use elasticity tensor for transversely isotropic material?
    8 years ago

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