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 stressstrain 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)(l2v) 
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)
Responses

chantelle wilson8 years ago
 Reply