Page 259 - Intro to Tensor Calculus

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p
∂y ∂y q
and convert equation (2.4.23) to a more generalized form. Multiply equation (2.4.23) by m n and verify
∂x ∂x
the result
q
∂y ∂y q
e
σ mn = λ m n rr + µ (e mn + e nm ) ,
∂x ∂x
which can be simplified to the form
ij
σ mn = λg mn ij g + µ (e mn + e nm ) .
e
Dropping the bar notation, we have

ij
σ mn = λg mn g e ij + µ (e mn + e nm ) .

The contravariant form of this equation is

sr ij
ns mr
σ sr = λg g e ij + µ (g ms nr + g g ) e mn .
g
Employing the equations (2.4.24) the above result can also be expressed in the form
E ms nr ns mr 2ν sr mn
rs
σ = g g + g g + g g e mn . (2.4.26)
2(1 + ν) 1 − 2ν
This is a more general form for the stress-strain constitutive equations which is valid in all coordinate systems.
Multiplying by g sk and employing the use of associative tensors, one can verify

E i ν m i
i
σ = e + e δ
j j m j
1+ ν 1 − 2ν
m i
i
i
or σ =2µe + λe δ ,
m j
j
j
are alternate forms for the equation (2.4.26). As an exercise, solve for the strains in terms of the stresses
and show that
m i
i
i
Ee =(1 + ν)σ − νσ δ .
j
j
m j
EXAMPLE 2.4-2. (Hooke’s law) Let us construct a simple example to test the results we have
developed so far. Consider the tension in a cylindrical bar illustrated in the figure 2.4-3.

Figure 2.4-3. Stress in a cylindrical bar