Page 276 - Intro to Tensor Calculus

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where
1 ∂
∂u r u r
Θ= + = (ru r )
∂r r r ∂r
is the dilatation. These stresses are found to be

2µ 2µ
σ rr =(λ + µ)c 1 − 2 c 2 σ θθ =(λ + µ)c 1 + 2 c 2 σ zz = λc 1 σ rθ = σ rz = σ zθ =0.
r r
We now apply the boundary conditions

2µ 2µ
σ rr | r=R 1 n r = − (λ + µ)c 1 − 2 c 2 =+P 1 and σ rr | r=R 0 n r = (λ + µ)c 1 − 2 c 2 = −P 0 .
R 1 R 0
Solving for the constants c 1 and c 2 we find
2 2 2 2
R P 1 − R P 0 R R (P 1 − P 0 )
0
1
0
1
c 1 = 2 2 , c 2 = 2 2 .
(λ + µ)(R − R ) 2µ(R − R )
0 1 0 1
This produces the displacement field
2 2 2 2
R P 1 r R 0 R P 0 r R 1
1
0
u r = + − + , u θ =0, u z =0,
2
2
2
2
2(R − R ) λ + µ µr 2(R − R ) λ + µ µr
1
0
0
1
and stress fields
2 2 2 2
R P 1 R 0 R P 0 R 1
1
0
σ rr = 1 − − 1 −
2
2
R − R 2 1 r 2 R − R 2 1 r 2
0
0
2 2 2 2
R P 1 R 0 R P 0 R 1
1
0
σ θθ = 1+ − 1+
2
2
R − R 2 r 2 R − R 2 r 2
0 1 0 1
2 2
λ R P 1 − R P 0
0
1
σ zz = 2 2
λ + µ R − R 1
0
σ rz = σ zθ = σ rθ =0
EXAMPLE 2.4-8. By making simplifying assumptions the Navier equations can be reduced to a more
tractable form. For example, we can reduce the Navier equations to a one dimensional problem by making
the following assumptions
1. Cartesian coordinates x 1 = x, x 2 = y, x 3 = z
2.u 1 = u 1 (x, t), u 2 = u 3 =0.
3. There are no body forces.
∂u 1(x, 0)
4. Initial conditions of u 1 (x, 0) = 0 and =0
∂t
5. Boundary conditions of the displacement type u 1(0,t)= f(t),
where f(t) is a specified function. These assumptions reduce the Navier equations to the single one dimen-
sional wave equation
2 2
∂ u 1 2 ∂ u 1 2 λ +2µ
= α , α = .
∂t 2 ∂x 2 ρ
The solution of this equation is
f(t − x/α), x ≤ αt

u 1 (x, t)= .
0, x > αt