K a
where p = 1 tr (a) is the pressure, and a' = a — pI is the stress deviator.
7.3.5.1.1.3 Restriction Imposed on the Isotropic Elastic Moduli
but since dW is a scalar invariant (energy), it can be expressed in terms of volumetric (hydrostatic) and deviatoric components as dW = pde + oj dEj (7.68)
substituting p = Ke and oj = 2GEj, and integrating, we obtain the following expression for the isotropic strain energy
W = 1 Ke2 + GEj Ej and since positive work is required to cause any deformation W > 0 thus
58 The isotropic strain energy function can be alternatively expressed as
59 From Table 7.2, we observe that v = 1 implies G = E, and K = 0 or elastic incompressibility
E,v 
M,v 
e,m 
K,v 
vE 
2ßv 
M(E2M) 
3Kv 
ßE 
1+v 3K (12v)  
K  
3(12v) 
3(12v) 
3(3ßE)  
E 
2m(1 + v) 
E 
3K(1  2v) 
2ß L 
v 
Table 7.2: Conversion of Constants for an Isotropic Elastic Material
Table 7.2: Conversion of Constants for an Isotropic Elastic Material
60 The elastic properties of selected materials is shown in Table 7.3. 7.3.5.1.2 Transversly Isotropic Case
6i For transversely isotropic, we can express the stressstrain relation in tems of
£zz Ixy Yyz a 1 1axx + a 1 2ayy + a 13a zz a 12 &xx + a i i Oyy + a 13a zz a 1 3 (axx + ayy) + a33a zz 2(an  a12)rxy a44 Txy
Material 
E (MPa) 
v 
A316 Stainless Steel 
196,000 
0.3 
A5 Aluminum 
68,000 
0.33 
Bronze 
61,000 
0.34 
Plexiglass 
2,900 
0.4 
Rubber 
2 
^0.5 
Concrete 
60,000 
0.2 
Granite 
60,000 
0.27 
Table 7.3: Elastic Properties of Selected Materials at 200c
Table 7.3: Elastic Properties of Selected Materials at 200c and
au = e; °12 =_E' ai3 =_e; a33 =_E; a44 =_î (7.74j where E is the Young's modulus in the plane of isotropy and E' the one in the plane normal to it. v corresponds to the transverse contraction in the plane of isotropy when tension is applied in the plane; v' corresponding to the transverse contraction in the plane of isotropy when tension is applied normal to the plane; ¡i' corresponding to the shear moduli for the plane of isotropy and any plane normal to it, and ¡i is shear moduli for the plane of isotropy.
7.3.5.2 Special 2D Cases
62 Often times one can make simplifying assumptions to reduce a 3D problem into a 2D one.
7.3.5.2.1 Plane Strain
For problems involving a long body in the z direction with no variation in load or geometry, then
Oxx ayy
0. Thus, replacing into Eq. 5.2 we obtain
7.3.5.2.2 Axisymmetry
v 
0 
1 &xx  
 v) 
0  
v 
0 
^ &yy  
2 J 
64 In solids of revolution, we can use a polar coordinate sytem and du 65 The constitutive relation is again analogous to 3D/plane strain
7.3.5.2.3 Plane Stress If the longitudinal dimension in z direction is much smaller than in the x and y directions, then Yxz = Yyz = 0 throughout the thickness. Again, substituting into Eq. 5.2 we obtain:

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