Thick Walled Cylinders
Consider a thick walled cylinder with open ends as shown above. It is loaded by internal pressure Pi and external pressure Po as seen below. It has inner radius r i and outer radius r o.
Po
r o
Pi r i
σθ
σr + dσr
σθ
σr
dr
rdθ
r
Now consider and element at radius r and defined by an angle increment dθ and a radial increment dr. By circular symmetry, the stresses σθ and σr are functions of r only, not θ and the shear stress on the element must be zero. For an element of unit thickness, radial force equilibrium gives:
(σ r + d σ r )(r + dr )d θ = σ r rd θ + σ d θ dr θ
Ignoring second order terms gives:
d σ r dr
+
σ r + σ θ
r
= 0 LLL (1)
Assuming that there are no body forces. Now consider strains in the element. By symmetry there is no θ displacement v. there is only a radial displacement u given by line aa’. Point c is displaced radially by (u + du) given by line cc’. As the original radial length of the element is dr (line ac), the radial strain is: ε r
=
u + du − u dr
=
=
d
c’ b’
du
c
dr
b
Line ab has length rdθ and line a’b’ has length (r + u)dθ. The tangential strain is thus: ε θ
d’
(r + u )d θ − rd θ rd θ
=
u r
a’ a
As the ends are open, σz = σ3 =0 and we thus have plane stress conditions. From Hooke’s law we get: ε r = ε θ
=
du
=
1
dr E u 1 r
Solving for the stresses gives:
=
E du u σ r = + ν 2 1 − ν dr r
(σ r −νσ ) θ
(σ − σ r ) θ
and
σ θ
E u du = + ν 2 1 − ν r dr
Substituting into equation above yields:
d 2u 2
dr
+
1 du
r dr
−
u 2
r
=0
u = C 1r +
Which has solution :
Giving the stresses as:
C 2 r
σ r
=
E 1 −ν ( ) ν 1 + − C C LLL ( 2) 1 2 2 1 − ν 2 r
σ θ
=
E 1 − ν ν 1 ( ) + + C C LLL(3) 2 2 1 2 1 − ν r
The boundary conditions are: σ r (r i ) = − P i
σ r
and
(r o ) = − P o
This yields the integration constants:
C 1 =
1 − ν r i 2 P i − r o2 P o
1 − ν r i 2 r o2 ( P i − P o )
2 2 and C 2 = 2 E r o − r i E r o − r 2i
Giving the stresses as a function of radius: σ r
=
r i 2 P i − r o2 P o
(r
2 o
− r i ) 2
−
( P i − P o ) r i2 r o2
(r − r )r 2 o
2
2
i
These are known as Lamé’s equations.
σ θ
=
r i 2 P i − r o2 P o
(r
2 o
− r i ) 2
+
( P i − P o )r i2 r o2
(r
2 o
− r i 2 )r 2
From equations 2 and 3 above we can see that the sum of the radial and tangential stresses is constant, regardless of radius: σ r +σ θ = 2 EC 1 / (1 − ν ) Hence the longitudinal strain is also constant since: ε z
=−
ν
E
(σ r + σ ) = θ
constant. Hence we get σz = Eεz = constant = c
If the ends of the cylinder and open and free we have Fz = 0, hence: r o
∫
σ z ⋅ 2π r dr = π c
(r
2 p
− r i2 ) = 0
or c = σz = 0 as we assumed.
r i
If the cylinder has closed ends, the axial stress can be found separately using only force equilibrium considerations as was done for the thin walled cylinder. The result is then simply superimposed on the above equations. The pressure Pi acts on area given by π r i2. The pressure Po acts on area given by πr o2. The axial stress σz acts on an area given by π(r o2 – r i2) Force equilibrium then gives:
P i r i2 − P o r o2 σ z = r 2 − r 2 o i
The following is a summary of the equations used to determine the stresses found in thick walled cylindrical pressure vessels. In the most general case the vessel is subject to both internal and external pressures. Most vessels also have closed ends - this results in an axial stress component. Principal stresses at radius r : σ 1
= σ θ = − K + C / r 2
And, if the ends are closed, Where:
(a)
σ θ
: σ 3
σ 2
= σ r = − K + C / r 2
= σ axial = − K
r i 2 r o2 C = ( P o − P i ) 2 2 r 0 − r i
:
( P r K =
− P i r i 2 ) r − r i 2
2 o 0 2 0
Internal Pressure only ( Po = 0 ):
P i r i 2 r o2 = 2 2 1 − 2 : r o − r i r
P i r i 2 r o2 σ r = 2 1+ 2 : 2 r o − r i r
σ z
=
P i r i 2 r o2 − r i 2
At inside surface, r = r i:
r o2 + r i2 σ θ = P i 2 : 2 r o − r i
σ r
= − P i :
σ z
=
P i r i 2 r o2 − r i 2
At outside surface, r = r o:
2 P i r i2 σ θ = 2 : 2 r o − r i
σ r
=0 :
σ z
=
P i r i 2 r o2 − r i 2
(b) External Pressure only ( Pi = 0 ):
− P o r o2 r i2 σ θ = 2 1+ 2 : 2 r o − r i r
− P o r o2 r i2 σ r = 2 1− 2 : 2 r o − r i r
At inside surface, r = r i:
− 2 P o r o2 σ θ = 2 : 2 r o − r i
− P o r o2 σ r = 0 : σ z = 2 r o − r i 2
At outside surface, r = r o:
r o2 + r i2 σ θ = − P o 2 : 2 r o − r i
− P o r o2 σ r = − P o : σ z = 2 r o − r i 2
− P o r o2 σ z = 2 r o − r i 2
Thick cylinder stresses 2500
Circum. Radial
2000
Axial 1500 i s p s s e r t s
VonMises
1000 500 0 -500 -1000 -1500 2
2.5
3
3.5
4
Radius - in
Example of cylinder with Pi = 1000 psi, r i = 2” and r o=4”
Note that in all cases the greatest magnitude of direct stress is the tangential stress at the in-side surface. The maximum magnitude of shear stress also occurs at the inside surface. (c) Press and shrink fits When a press or shrink fit is used between 2 cylinders of the same material, an interface pressure pi is developed at the junction of the cylinders. If this pressure is calculated, the stresses in the cylinders can be found using the above equations. The pressure is:
pi =
E δ (c 2 − b 2 )(b2 − a 2 )
2b2 (c 2 − a 2 )
b
Where: E = Young’s Modulus δ = radial interference between the two cylinders a = inner radius of the inner cylinder b = outer radius of inner cylinder and inner radius of outer cylinder c = outer radius of outer cylinder It is assumed that δ is very small compared to the radius b and that there are no axial stresses. Thus we have δ = binner – bouter. Note that this small difference in the radii is ignored in the above equation.
All stresses are calculated at the inner radius and are for a cylinder with closed ends and internal pressure only.
FEA - Theory Comparison 2000 Radial FEA-radial 1500
Circum. FEA-circum
1000 i s p s s e r t S
500
0
-500
-1000 2
2.5
3
Radius - in.
3.5
4
Problems on Thick Cylinders. 1. A steel cylinder is 160 mm ID and 320 mm OD. If it is subject to an internal pressure of 150 MPa, determine the radial and tangential stress distributions and show the results on a plot (using a spreadsheet). Determine the maximum shear stress in the cylinder. Assume it has closed ends. ( σt = 250 to 100 MPa, σr = 0 to ±150 MPa, τmax = 200 MPa.) 2. A cylinder is 150 mm ID and 450 mm OD. The internal pressure is 160 MPa and the external pressure is 80 MPa. Find the maximum radial and tangential stresses and the maximum shear stress. The ends are closed. (σt = 20 to ±60 MPa, σr = ±80 to ±160 MPa, τmax = 90 MPa.) 3. A cylinder has an ID of 100 mm and an internal pressure of 50 MPa. Find the needed wall thickness if the factor of safety n is 2.0 and the yield stress is 250 MPa. Use the maximum shear stress theory, i.e. maximum shear stress = yield strength/2n. ( wall = 61.8 mm thick ) 4. A 400 mm OD steel cylinder with a nominal ID of 240 mm is shrunk onto another steel cylinder of 240 mm OD and 140 mm ID. The radial interference δ is 0.3 mm. Use Young's Modulus E = 200 GPa and Poisson's Ratio n = 0.3. Find the interface pressure pi and plot the radial and tangential stresses in both cylinders. Then find the maximum internal pressure which may be applied to the assembly if the maximum tangential stress in the inside cylinder is to be no more than 140 MPa. ( pi = ±120.3 MPa. :: inner cylinder : σt = ±365 to ±244 MPa, σr = 0 to ±120.3 MPa. :: outer cylinder : σt = 256 to 135 MPa, σr = ±120.3 to 0 MPa. :: maximum internal pressure = 395 MPa.) 5. A cylinder with closed ends has outer diameter D and a wall thickness t = 0.1D. Determine the %age error involved in using thin wall cylinder theory to calculate the maximum value of tangential stress and the maximum shear stress in the cylinder. (tangential stress ± 9.75% : max. shear stress ± 11.1%)
