ENGM 922 Exam prep
Hydrostatic Pressure:
•
Pressurized hollow sphere
e3
Assume that
No body forces act on the sphere
The sphere has uniform temperature
pb
pa
in spherical coord
The inner surface R=a is subjected to pressure
The outer surface R=b is subjected to pressure
a
e2
b
e1
σ ij ,i = 0
∂
(σ rr + σ rθ + σ rφ ) = 0
∂r
1 ∂
(σ θ r + σ θθ + σ θφ ) = 0
r ∂θ
1
∂
(σ φr + σ φθ + σ φφ ) = 0
r sin θ ∂φ
⎡λ ( ε rr + εθθ + ε φφ ) + 2με rr
⎢
0
σ=⎢
⎢
⎢
0
⎣
Constitutive relation:σ ij = λε kk δ ij + 2με ij
∂
Given the θ & φ symmetry of the hydrostatic
λ ε rr + εθθ + εφφ + 2με rr = 0
state only ur ≠ 0 & the 3 =librium eq’s become: ∂r
( (
)
)
0
λ ( ε rr + εθθ + ε φφ ) + 2μεθθ
0
⎤
⎥
⎥
0
⎥
λ ( ε rr + εθθ + ε φφ ) + 2με φφ ⎥⎦
0
Hydrostatic σ state
0 as well due to symmetry
1 ∂
( λ ur , r ) = 0
r ∂θ
1
∂
( λ ur , r ) = 0
r sin θ ∂φ
1 ⎛ ∂ui ∂u j ⎞ The simplistic ε – u relation to the left is not so
+
⎟
2 ⎝ ∂x j ∂xi ⎟⎠ straight forward for coord sys other than cartesian:
ε ij = ⎜
⎜
= σ 31 the
+ σ 32
= μθmomentum
x1 + x2 ==librium
μθρ eq becomes:
Applying the ε – u relations for sphericalτ coord’s
radial
2
uθ = 0
uφ = 0
2
2
2
uθ = 0
⎞ ⎞⎤
∂u
∂ 2 ur ∂ ⎡ λ
∂
∂ ⎡ λ ⎛ ⎛ ∂u
1 ⎛ ∂uφ
⎞
⎤
+
+
=
+
+ ⎢ ( 2ur ) ⎥ = 0
u
u
λ (ε rr + εθθ + εφφ ) + 2με rr = ( λ + 2μ ) 2r + ⎢ ⎜ ⎜ θ + ur ⎟ +
sin
θ
cos
θ
λ
2
μ
(
)
⎥
⎟
⎜
⎟
θ
r
2
∂r
∂r
∂r ⎢⎣ r ⎝ ⎝ ∂θ
∂r
∂r ⎣ r
⎦
⎠ sin θ ⎝ ∂φ
⎠ ⎠ ⎥⎦
2
∂u
1 ∂u u
( λ + 2μ ) 2r + 2λ ⎛⎜ r − 2r ⎞⎟ = 0
∂r
⎝ r ∂r r ⎠
(
4/25/2010
)
2
Jeff Berg
1
ENGM 922 Exam prep
Hydrostatic Pressure (con’d):
•
( λ + 2μ )
⎡
∂ 2ur
E
⎛ 1 ∂ur ur ⎞ ⎤
−
+ 2υ ⎜
− 2 ⎟⎥ = 0
1
υ
(
)
⎢
2
∂r
(1 + υ )(1 − 2υ ) ⎣
⎝ r ∂r r ⎠ ⎦
∂ 2ur
⎛ 1 ∂ur ur ⎞
+ 2λ ⎜
− 2⎟=0
2
∂r
⎝ r ∂r r ⎠
note:
http://solidmechanics.org/
In Dr. Allan Bower’s text, Applied Mechanics of Solids, from Brown Univ. he managed to get rid of the coeff terms dependent on the
Poisson’s ratio that prevent me from obtaining the simpler ordinary diff eq which would allow me to resolve u(r) via integration. There
must be some obscure assumption or relation that I am not aware of which would allow me to proceed per Dr. Bower’s direction.
(
2
⎛
Adopting the governing diff ∂ ur ⎛ 1 ∂ur ur ⎞ ∂ ⎜ 1 ∂ r ur
+
−
=
⎜
⎟
eq per Dr. Bower’s text: ∂r 2 ⎝ r ∂r r 2 ⎠ ∂r ⎜ r 2
∂r
⎝
∂u
2
Resolve A & B via σ rr ( r = a ) = ( λ + 2μ ) 2r
∂r
application of BC’s:
σ rr ( r = b ) = ( λ + 2μ )
4/25/2010
∂ ur
∂r 2
r =b
) ⎞⎟ = 0
⎟
⎠
soln via direct integration
ur =
1⎛A 3
⎞
r + B⎟
2 ⎜
r ⎝3
⎠
+
∂ ⎡λ
( 2ur ) ⎥⎤ = Pa
⎢
∂r ⎣ r
⎦ r =a
( λ + 2μ )
6B
⎡1 ⎛
B⎞ 1⎛
B ⎞⎤
+ 2λ ⎢ ⎜ A − 2 3 ⎟ + 2 ⎜ Aa + 2 ⎟ ⎥ = Pa
4
a
a ⎠ a ⎝
a ⎠⎦
⎣a ⎝
+
∂ ⎡λ
( 2ur )⎤⎥ = Pb
⎢
∂r ⎣ r
⎦ r =b
( λ + 2μ )
6B
⎡1 ⎛
B⎞ 1⎛
B ⎞⎤
+ 2λ ⎢ ⎜ A − 2 3 ⎟ + 2 ⎜ Aa + 2 ⎟ ⎥ = Pb
4
b
b ⎠ b ⎝
b ⎠⎦
⎣b ⎝
r =a
2
2
Jeff Berg
2
ENGM 922 Exam prep
4.1.3 General solution to the spherically symmetric linear elasticity problem
Our goal is to solve the equations given in Section 4.1.2 for the displacement, strain and stress in the sphere. To do so,
1. Substitute the strain-displacement relations into the stress-strain law to show that
2.
Substitute this expression for the stress into the equilibrium equation and rearrange the result to see that
Given the temperature distribution and body force this equation can easily be integrated to calculate the displacement u. Two arbitrary constants of
integration will appear when you do the integral
these must be determined from the boundary conditions at the inner and outer surface of the
sphere. Specifically, the constants must be selected so that either the displacement or the radial stress have prescribed values on the inner and outer
surface of the sphere.
In the following sections, this procedure is used to derive solutions to various boundary value problems of practical interest.
4/25/2010
Jeff Berg
3