9780203009512.ch6

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

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
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
Background image of page 2
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)
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Page1 / 12

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

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online