9780203009512.ch6

9780203009512.ch6 - 0749_Frame_C06 Page 95 Wednesday 5:06...

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95 Stress-Strain Relation and the Tangent-Modulus Tensor 6.1 STRESS-STRAIN BEHAVIOR: CLASSICAL LINEAR ELASTICITY Under the assumption of linear strain, the distinction between the Cauchy and Piola- Kirchhoff stresses vanishes. The stress is assumed to be given as a linear function of linear strain by the relation (6.1) in which c ijkl are constants and are the entries of a 3 × 3 × 3 × 3 fourth-order tensor, C . If T and E L were not symmetrical, C might have as many as 81 distinct entries. However, due to the symmetry of T and E L there are no more than 36 distinct entries. Thermodynamic arguments in subsequent sections will provide a rationale for the Maxwell relations: (6.2)± It follows that c ijkl = c klij , which implies that there are, at most, 21 distinct coefﬁ- cients. There are no further arguments from general principles for fewer coefﬁcients. Instead, the number of distinct coefﬁcients is speciﬁc to a material, and reﬂects the degree of symmetry in the material. The smallest number of distinct coefﬁcients is achieved in the case of isotropy, which can be explained physically as follows. Suppose a thin plate of elastic material is tested such that thin strips are removed at several angles and then subjected to uniaxial tension. If the measured stress-strain curves are the same and independent of the orientation at which they are cut, the material is isotropic. Otherwise, it exhibits anisotropy, but may still exhibit limited types of symmetry, such as transverse isotropy or orthotropy. The notion of isotropy is illustrated in Figure 6.1. In isotropic, linear-elastic materials (which implies linear strain), the number of distinct coefﬁcients can be reduced to two, m and l , as illustrated by Lame’s equation, (6.3) 6 TcE ij ijkl kl L = () , = T E T E ij kl kl ij . TEE ij ij L kk L ij =+ 2 µλ δ . © 2003 by CRC CRC Press LLC

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96 Finite Element Analysis: Thermomechanics of Solids This can be inverted to furnish (6.4) The classical elastic modulus E 0 and Poisson’s ratio n represent response under uniaxial tension only , provided that T 11 = T , T ij = 0. Otherwise, (6.5) It is readily veriﬁed that (6.6)± from which it is immediate that (6.7) Leaving the case of uniaxial tension for the normal (diagonal) stresses and strains, we can write (6.8)± (6.9) FIGURE 6.1 Illustration of isotropy. 1 2 3 4 T 21 E ET T ij L ij kk ij () . =− + 1 22 3 µ λ µλ δ E == = T E E E E E L L L L L 11 11 22 11 33 11 . ν 11 2 1 23 1 22 3 E E + = + , 1 2 1 = + E , T T TT T L 11 11 22 33 11 22 33 1 2 1 1 1 [ ( )], + + + + E T T L 22 22 33 11 1 [ ( )], + E © 2003 by CRC CRC Press LLC
Stress-Strain Relation and the Tangent-Modulus Tensor 97 and (6.10) and the off-diagonal terms satisfy (6.11)± 6.2 ISOTHERMAL TANGENT-MODULUS TENSOR 6.2.1 C LASSICAL E LASTICITY Under small deformation, the fourth-order tangent-modulus tensor D in linear elas- ticity is deﬁned by (6.12) In linear isotropic elasticity, the stress-strain relations are written in the Lame’s form as (6.13)± Using Kronecker Product notation from Chapter 2, this can be rewritten as (6.14)

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9780203009512.ch6 - 0749_Frame_C06 Page 95 Wednesday 5:06...

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