Page 260 - Intro to Tensor Calculus

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Assume that
F
 
A 0 0
σ ij =  0 0 0 
0 0 0
where F is the constant applied force and A is the cross sectional area of the cylinder. Consequently, the
generalized Hooke’s law (2.4.21) produces the nonzero strains

1+ ν ν σ 11
e 11 = σ 11 − (σ 11 + σ 22 + σ 33 )=
E E E
−ν
e 22 = σ 11
E
−ν
e 33 = σ 11
E
From these equations we obtain:
The first part of Hooke’s law
F
σ 11 = Ee 11 or = Ee 11 .
A
The second part of Hooke’s law
lateral contraction −e 22 −e 33
= = = ν = Poisson’s ratio.
longitudinal extension e 11 e 11
This example demonstrates that the generalized Hooke’s law for homogeneous and isotropic materials
reduces to our previous one dimensional result given in (2.3.1) and (2.3.2).

Basic Equations of Elasticity

Assuming the density % is constant, the basic equations of elasticity reduce to the equations representing
conservation of linear momentum and angular momentum together with the strain-displacement relations
and constitutive equations. In these equations the body forces are assumed known. These basic equations
produce 15 equations in 15 unknowns and are a formidable set of equations to solve. Methods for solving
these simultaneous equations are: 1) Express the linear momentum equations in terms of the displacements
u i and obtain a system of partial differential equations. Solve the system of partial differential equations
for the displacements u i and then calculate the corresponding strains. The strains can be used to calculate
the stresses from the constitutive equations. 2) Solve for the stresses and from the stresses calculate the
strains and from the strains calculate the displacements. This converse problem requires some additional
considerations which will be addressed shortly.