EM 424: Exact solution for thick wall pressure vessel
Pressure loading of a thick walled cylinder is a relatively simple example where we can
demonstrate explicitly how we can satisfy all the governing equations and boundary
conditions to arrive at a complete solution of a stress analysis problem. Figures 1(a), (b)
show the geometry and loading of the problem.
σrr
σr
σzr
θ
C
A
σzθ σrr
B
σr
ri
θ
re
Fig 1(a) Stresses on the cylinder surfaces (the z-axis is
outwards from this planar view)
r
σzz
σzr
D
σrz
σrr
σrz
σzr
A
B
C
σrr
σzz
pi
pe
L
Fig. 1 (b) Stresses on the surfaces of the cylinder and the pressure loading.
z
EM 424: Exact solution for thick wall pressure vessel
It is assumed that the internal pressure, pi , and the external pressure, pe , are uniform
along the entire inner and outer surfaces as shown in Fig. 1(b) and that all the other
surfaces of the cylinder are unloaded. In this case the boundary conditions for this
problem are:
for all 0 ≤ θ ≤ 2π , 0 ≤ z ≤ L on the inner and outer surfaces we have
σ rr ( r ,θ , z ) r = r = − pi
i
σ rθ ( r ,θ , z ) r = r = σ rz ( r ,θ , z ) r = r = 0
i
i
σ rr ( r ,θ , z ) r = r = − pe
(1)
e
σ rθ ( r ,θ , z ) r = r = σ rz ( r ,θ , z ) r = r = 0
e
e
and for all 0 ≤ θ ≤ 2π , ri ≤ r ≤ re on the cylinder ends
σ zz ( r ,θ , z ) z =0 = σ zr ( r ,θ , z ) z =0 = σ zθ ( r ,θ , z ) z =0 = 0
σ zz ( r ,θ , z ) z = L = σ zr ( r ,θ , z ) z = L = σ zθ ( r ,θ , z ) z = L = 0
(2)
Given that the pressure loading is all radially directed, it is reasonable to assume that the
only stresses developed in the cylinder are the normal stresses σ rr , σ θθ and to assume
that these stresses are only a function of r, i.e.
σ rr = σ rr ( r ) , σ θθ = σ θθ ( r )
σ rθ = σ rz = σ zθ = σ zz = 0
(3)
In this case the boundary conditions of Eq. (2) are satisfied identically for the cylinder
ends and the only boundary conditions in Eq. (1) that are not satisfied identically are
σ rr ( r ) r = r = − pi
i
σ rr ( r ) r = r = − pe
(4)
e
Under these assumptions, the equations of equilibrium, which in cylindrical coordinates
are
EM 424: Exact solution for thick wall pressure vessel
∂σ rr σ rr − σ θθ 1 ∂σ rθ ∂σ rz
+
+
+
+ fr = 0
r
r ∂θ
∂r
∂z
∂σ rθ 2σ rθ 1 ∂σ θθ ∂σ θ z
+
+
+
+ fθ = 0
r
r ∂θ
∂r
∂z
∂σ zr σ zr 1 ∂σ zθ ∂σ zz
+
+
+
+ fz = 0
r
r ∂θ
∂r
∂z
(5)
reduce to only one non-trivial equation
∂σ rr σ rr − σ θθ
+
=0
r
∂r
(6)
The stress state given by Eq. (3) is one of plane stress so that the 3-D stress-strain
relations for a homogeneous, isotropic elastic solid reduce to
1
(σ rr −νσ θθ )
E
1
εθθ = (σ θθ − νσ rr )
E
−ν
ε zz =
(σ rr + σ θθ )
E
ε rr =
ε rθ =
ε zθ =
ε rz =
σ rθ
G
σ zθ
G
σ rz
G
(7)
=0
=0
=0
or, inverting these relations, we have the usual plane stress equations
E
(ε rr + νεθθ )
1 −ν 2
E
σ θθ =
(εθθ + νε rr )
1 −ν 2
−ν
ε zz =
(σ rr + σ θθ )
E
σ rr =
(8)
To guarantee that we automatically satisfy compatibility, we will work directly with the
displacements as the fundamental unknowns for this problem. Because of the symmetry
we expect that there will be no displacement component , uθ , in the cylinder and that the
other displacements will also be independent of θ. Also, because the pressure loading is
uniform in z, we expect that the radial displacement is only a function of r, i.e.
EM 424: Exact solution for thick wall pressure vessel
ur = ur ( r ) . In this case the strains in the cylinder, which are given in cylindrical
coordinates as
∂ur
1 ∂uθ ur
∂u
, εθθ =
+ , ε zz = z
∂r
∂z
r ∂θ
r
1 ⎛ 1 ∂ur ∂uθ uθ ⎞
ε rθ = ⎜
+
− ⎟
2 ⎝ r ∂θ
∂r
r ⎠
ε rr =
1 ⎛ ∂u ∂u ⎞
ε rz = ⎜ z + r ⎟
2 ⎝ ∂r
∂z ⎠
1 ⎛ 1 ∂u z ∂uθ ⎞
+
εθ z = ⎜
⎟
∂z ⎠
2 ⎝ r ∂θ
(9)
reduce to
dur
u
du
, εθθ = r , ε zz = z
dr
r
dz
ε rθ = εθ z = ε rz = 0
ε rr =
(10)
Placing the strain-displacement relations in Eq.(10) into the stress-strain relations of Eq.
(8) gives
E ⎛ dur
u ⎞
+ν r ⎟
2 ⎜
1 −ν ⎝ dr
r ⎠
E ⎛ ur
du ⎞
+ν r ⎟
σ θθ =
2 ⎜
1 −ν ⎝ r
dr ⎠
−ν
du
ε zz = z =
(σ rr + σ θθ )
dz
E
σ rr =
(11)
so that when the stresses of Eq. (11) are substituted into the equilibrium equation (Eq.
(6)), we find
d 2ur 1 dur ur
+
− =0
dr 2 r dr r 2
(12)
It is easy to solve this equation for the displacement since we can rewrite it equivalently
in the form
d ⎡1 d
( rur )⎤⎥ = 0
⎢
dr ⎣ r dr
⎦
(13)
EM 424: Exact solution for thick wall pressure vessel
Integrating once on r gives
d
( rur ) = C1′ r
dr
(14)
where C1′ is a constant of integration. Integrating Eq.(14) once more then yields
ur =
C1′
C
C
r + 2 = C1r + 2
r
r
2
(15)
where C1 = C1′ / 2, C2 are both constants of integration. If the displacement expression in
Eq. (15) is placed into the stress-strain relations of Eq. (11), one finds
E
E C2
C1 −
1 −ν
1 +ν r 2
E
E C2
C1 +
σ θθ =
1 −ν
1 +ν r 2
−2ν
C1 = constant
ε zz =
1 −ν
σ rr =
(16)
If we define two new constants, A, B as
A=
E
C1 ,
1 −ν
B=
E
C2
1 +ν
(17)
then we can rewrite Eq.(16) as
B
r2
B
σ θθ = A + 2
r
du z −2ν A
ε zz =
=
= constant
dz
E
σ rr = A −
(18)
At this point our solution for the stresses (Eq. (18)) and the displacement (Eq. (15))
satisfy equilibrium, compatibility, and the stress-strain relations and are given in terms of
two unknown constants ( C1 and C2 or A and B). To find these constants we must satisfy
the boundary conditions (Eq. (4)) which , together with Eq.(18) yield
EM 424: Exact solution for thick wall pressure vessel
A−
B
= − pi
ri 2
B
A − 2 = − pe
re
(19)
whose solution is
r2 p − r2 p
A = i 2i e2 e ,
re − ri
ri 2 re2 ( pi − pe )
B=
re2 − ri 2
(20)
which gives the stresses and displacements as:
σ rr =
pi ri 2 ⎛ re2 ⎞
pe re2 ⎛ ri 2 ⎞
−
−
1
⎜
⎟
⎜1 − ⎟
re2 − ri 2 ⎝ r 2 ⎠ re2 − ri 2 ⎝ r 2 ⎠
σ θθ =
pi ri 2 ⎛ re2 ⎞
pe re2 ⎛ ri 2 ⎞
+
−
1
⎜
⎟
⎜1 + ⎟
re2 − ri 2 ⎝ r 2 ⎠ re2 − ri 2 ⎝ r 2 ⎠
du
−2ν ⎛ ri 2 pi − re2 pe ⎞
−2ν ⎛ ri 2 pi − re2 pe ⎞
u
⇒
=
ε zz = z =
⎜
⎟
⎜
⎟z+C
z
dz
E ⎝ re2 − ri 2 ⎠
E ⎝ re2 − ri 2 ⎠
1 − ν ) ⎛ ri 2 pi − re2 pe ⎞
1 + ν ) re2 ri 2 ( pi − pe ) 1
(
(
ur =
⎜
⎟r +
E ⎝ re2 − ri 2 ⎠
E
re2 − ri 2
r
(21)
where C is an arbitrary constant displacement (translation) in the z-direction.
The solution of Eq. (21) is an exact solution to our pressurized cylinder problem
since it satisfies all the governing equations and boundary conditions. One can show that
for a pressurized cylinder subject to an internal pressure only, the highest stresses occur
on the inner wall where we have
σ rr
σ θθ
r = ri
r = ri
=
pi ri 2 ⎛ re2 ⎞
⎜1 − ⎟ = − pi
re2 − ri 2 ⎝ ri 2 ⎠
p r2 ⎛ r2 ⎞
= 2 i i 2 ⎜1 + e2 ⎟
re − ri ⎝ ri ⎠
For a thin wall cylinder these stresses become
(22)
EM 424: Exact solution for thick wall pressure vessel
σ rr
r = ri
= − pi
⎞
pi ri 2 ⎛ re2 ⎞ pi ⎛ re2 + ri 2 ⎞ pi ⎛ re + ri
t2
+
=
=
+
1
⎜
⎟
⎜
⎟
⎜
⎟
r = ri
2 ( re + ri ) ⎟⎠
re2 − ri 2 ⎝ ri 2 ⎠ t ⎝ re + ri ⎠ t ⎜⎝ 2
p ⎛r +r ⎞ pr
≈ i ⎜ e i⎟= i m
t ⎝ 2 ⎠
t
σ θθ
=
(23)
in terms of the thickness t = re − ri and the mean radius rm = ( re + ri ) / 2 . This agrees with
the result for the thin wall cylindrical pressure vessel hoop stress obtained in Strength of
Materials (where the small radial stress component is normally ignored).