INME 4011 Machine Component Design 1
Pressure Vessel Design
Group Members:
Osvaldo Caraballo
Hector Castro
Daniel Andreu
Carlos Baez
Department of Mechanical Engineering
University of Puerto Rico at Mayaguez
May/07/2007
Engineering Drawing
Objectives
1- Design a pressure vessel which maximizes total volume with a largest external
dimension of 10in.
2- Find the best geometry for both, sustaining internal stresses and increase mobility.
3- Select a material which meets the requirements of strength to weight and resist
temperature changes.
Description
Taking into consideration the designed parameters we selected a cylindrical vessel with
a radius of 5in and a height of 10in. These dimensions maximize the volume, thereby increasing
the amount of fluid that can be stored inside the reactor. This geometry is also practical for the
transportation of this light weight vessel. The material selection will depend on the magnitude
of wall stresses and also on weight considerations. This material will have to sustain thermal
gradients and cyclic loads for a period of 30min. To analyze the problem maximum stresses
have to be found and a safety factor established to prevent any crack formation and
propagation.
Design Details
Thick-Walled Cylinders
It will be assumed that there are not axial forces on the cylinder. The state of stress at
any point within the wall is biaxial. If they do exist, the stress component σl may be superposed
later. Because the cylinder is symmetrical about the center axis the plane section remains plane
when the cylinder is pressurized. The differential equation of force equilibrium may be written
by talking a vertical force resolution:
r
 2r l    
r
   r
t
r
r
2r  dr l  2 r ldr  0
d r
0
dr
To solve for the two unknowns, σr and σt requires another independent relationship. For this
we are going to use Hooks law in our analysis.
1
 r  t 
E
1
 t   t   r 
E
1
 l     r   t 
E
r 
Where εr εt εs are radial, tangential and longitudinal strain.
Rewriting the stress equation:
    2C1
t
r
Subtracting and substituting:
2C1  
E l

 2 r  r
d r
 2C1
dr
Multiplying by r and rearranging,


d 2
r  r  2rC1
dr
Integrating,
r 2 r  r 2C1  C2 ;
Solving:
σr, =
c1 
c2
r2
and
Where C2 is a constant of integration
σ t = C1 -
Using the boundary conditions of:
σr = -pi
at
r=a
σr = -po
at
r=b
Inserting the boundary conditions:
 p i  c1 
 p o  c1 
c2
a2
c2
b2
Solving the two equations simultaneously for C1 and C2 finally,
 a 2b 2
pi a 2  p o b 2   2
 r
r 
b2  a2

 p o  pi 


 a 2b 2
pi a  po b   2
 r
t 
2
b  a2
2
2

 po  pi 


For a pressure vessel subjected to internal pressure only:
a 2 pi
r  2 2
b a
 b2 
1  2 
 r 
a 2 pi
t  2 2
b a
 b2 
1  2 
 r 
For internally pressurized vessels, the peak stress magnitudes of σr and σt both occur at the
inner radius r=a;
 r   Pi
 b2  a 2 
 t  Pi  2 2 
b a 

a2


 l  Pi  2
2 
b a 
Now that we have calculated the maximum stresses due to internal pressure only in the critical
section (the inner wall) we can use them as design parameters and then find the appropriate
material for the task.
Thin Walled Cylindrical Vessels
To treat the vessel as thin walled there are three assumptions that have to be met:
 Planes remain plane.
 Bending stress is proportional to distance away from the neutral axis.
 The Hoop stress in the wall of a thin-walled pressurized cylinder is uniform across the
wall thickness.
For thin wall σr has to be negligibly small compared to σt, typically if t is 10 percent of the
diameter d, the pressure vessel may be accurately analyzed as a thin-walled vessel. To analyse
the stresses present in the cylinder walls, we first dived the cylinder cutting longitudinally and
exposing the tangential stresses. Then we calculate the vertical stress which is the resultant
from all the Pi acting on the wall of the vessel.
dθ
ds
d
2
The element of area is obtained by multiplying ds by l.
ds 
d
* d
2
dA  ds * l
dA 
d
* d * l
2
Vertical force resolution:
FPV  2( t tl )  0
The vertical force component FPV produced by the pressure Pi may be found by integrating the
vertical component of pressure-induced force, PidA, over the semicircular surface.

d


2   Pi * * d * l * sen   2 t tl  0
2

0 
2
This results in:
Pdl
 2 t tl  0
i
Where d*l is the projected area, It may be found that the tangential stress component is
t 
Pd
i
2t
And the longitudinal stress component is
l 
Pd
i
4t
It is important to notice that these equations do not apply for elements near stress
concentrations like closures, nozzle, or supports. Local stresses must be separately determined
in the vicinity of any stress concentration site.
Spherical Pressure Vessels
Consider a spherical pressure vessel with radius r and wall thickness t subjected to an internal
gage pressure p.
For reasons of symmetry, all four normal stresses on a small stress element in the wall must be
identical. Furthermore, there can be no shear stress.
Summing forces we obtain:
 * t * 2 r  P *  r 2

Pr
2t
Calculation of stresses under static loads
A. Cylindrical section
Assuming thick walled pressure vessels, now we can use the maximum stress equations for
radial, tangential, and longitudinal stresses. Assuming that the internal radius a=4.75in and
external radius b=5in we can calculate:
 r  150 psi
 t  2927 psi
 l  1388 psi
These stresses are at their maximum because they were measured at the internal wall of the
cylindrical section of the pressure vessel which is a critical section. Because no shear stresses
are present at the θ=0 plane these stresses are the principal stresses and this plane is the
principal plane.
y
 t  1
l  2
r  3
z
x
 2 0 0 


0

0
1


 0 0  
3

The principal shearing stresses are calculated using the following equations:
1 
2 
3 
1  769 psi
 2  1538.5 psi
 3  769 psi
B. Spherical section

pr
 1425 psi
2t
2 3
2
 3  1
2
1   2
2
Selection of Material
Pressure vessels are vulnerable to potential failure by many possible modes these include
yielding, ductile rupture, brittle fracture, fatigue, stress corrosion cracking, creep, combined
creep and fatigue. Candidate materials for pressurized cylinders should include good strength,
fatigue strength, high ductility, good formability, good weld ability, and low cost.
There are two ways for designing a safe pressure vessel, it has to either yield or leak before it
brakes. For small pressure vessels like this design, are usually allow to yield at a pressure low to
propagate any crack. The alteration by yielding is easy to detect and the pressure can be
released safely.
Table1 Design requirements
Function
Contain pressure
Objective
Minimize cost
Minimize weight
Constraints
Must yield before break
For pressure vessel design it’s needed to obtain stresses less than the yield strength (  y ) of the
wall,

pR
y
2t
This design is considered as a small pressure vessel then establishing that it contains no crack or defect
of diameter greater than 2 a c . The stress required to make such a crack is:
f 
CK 1c
a c
Designing a pressure vessels needs to be very safety, for this the working stresses have to be less than
the fracture stresses, but to get greater security the crack will not propagate even if the stresses reaches
the general yield stress. Then the deformation will be stable and would not be detected. We can
expressed with 
f
greater than the  y so;
K 
ac  C 2  1c 
  y 
M1 
2
K1c
y
The material indices M 1 establish above with a fracture toughness K Ic  36MPa m
and a
 y  241MPa , it can calculated M 1  .1494 m . Looking in the chart of fracture toughness K Ic
against elastic limit  y we can select the material needed in a “yield before break” problem, in this case
we choose four different materials:
1) Tough steels
2) Tough copper alloys
3) Ti-alloys
4) High Strength Al-Alloys
After knowing the safes material we can now minimized the mass with the largest values of other
material indices;
M4 
y

But for light pressure vessels require that M 4 have to be maximized. This is plotted as a bar chart in the
graph 2 (see Appendix A). The candidate materials are those with large values of M 1 and M 4 . Since,
the tough steels, Ti-alloys and Mg-alloys are greater in M 1 and M 4 that the high strength Al-alloys, we
consider this last one because it has a good performance and is less expensive that the others.
Stress Concentrations
The cylindrical part is produced in a form of a plate then it is folded by a machine and welded
along the longitudinal direction. This weld produces stress concentrations which have to be
accounted in the design. To find the value of this stress concentration we need to find the
stress concentration factor Kt, which depends on the type of weld and the zone at which it is
located. Since the design parameters established cyclic fluctuations with changes in
temperature the zone selected is classified as a heat affected zone (HAZ) and the type of weld
selected by design is a butt weld and that combination gives a Kf equal to 1.2.
From Table 13.9 Mechanical Designs of Machine Elements and Machines
K f  1.2
 max  K f  nom
In the cylindrical region
 VM  2665.7 psi
 max  3198.8 psi
The next stress concentration is located in the joint between the cylindrical and spherical sections.
These two sections are also joined by Butt welding.
 max  K f  nom
 max  1710 psi
Cyclic loads
This design includes the application of varying loads, this variation will depend on temperature
fluctuations. Because the pressure vessel serves like a chemical reactor as a result it must be
able to sustain cyclic stresses due to changes in pressure. Cyclic stresses are so called because
of their dependence in time as the parameter. The figure below shows a sinusoidal
approximation for these types of loadings.
First the maximum and minimum loads have to be established, the design requires the vessel to
support a maximum pressure of 150psi. We chose the minimum pressure to be 100psi now it
can be written that:
Pmax = 150psi
Pmin = 100psi
Next the amplitude and mean load components are calculated
Pa =
(Pmax -Pmin )
= 25psi
2
Pm =
(Pmax +Pmin )
= 125psi
2
With these two loads now the stress at the wall are found
First the stresses for amplitude load are found, there are no shear stresses so this plane θ=0 is
the principal plane:
 r = z = -Pi  25 psi
 b2 + a 2 
 t =  y =Pi  2 2   487.8 psi
 b -a 
 a2 
 l = x = Pi  2 2   231.4 psi
b -a 
Now the stress for mean load are found
 r = z = -Pi  125 psi
 b2 + a 2 
 t =  y =Pi  2 2   2, 439 psi
 b -a 
 a2 
 l = x = Pi  2 2   1,157 psi
b -a 
With the mean and amplitude stresses calculated now we need the σVM
 VM 

2
2
2
x   y    y   z    z   x   6  xy   yz   zx 
2
2
2
2
 VM,a  444.1psi
 VM,M  2, 220.5 psi
Because these stresses are cyclic they need to be corrected, to do this we need to multiply
them by a Kf which accounts for the variation of loads with time
The Kf that will be used was obtained in the stress concentration section:
For a Butt Weld:
K f  1.2
 VM ,corrected  K f  VM ,nom
 VM ,a ,corrected  532.92 psi
 VM , M ,corrected  2664.6 psi
To calculate the fatigue strength of the material we first need the SUT which depends in the
specific alloy and the different treatments giving to it. The material selected was:
Aluminum Alloy 6020-T651
SUT  38ksi
S 'f  15.2ksi
Now the correction factors will be found
Ksurf  aSUT b  1
Kload  1
A95  0.010462d 2  0.262in
dequiv 
A95
 1.85in
0.0766
Ksize  0.869(deq )0.097  0.819
Ktemp  1
K reliable  0.702@99.99%
The corrected fatigue strength
S f  8.7ksi
The factor of safety using the modified-Goodman relation depends on the material chosen and
the calculated von misses stresses.
nf 
a
Sf
1

m
S UT
 7.6
Discussion
For the pressure vessel being design we took into consideration the weight parameter.
Aluminum alloys are broadly used on industry because of their convenient properties: low
density, corrosion resistance, and high tensile strength. This material is also characterized by
being the third most abundant element in the earth’s crust for a total of 8.1wt%, which lowers
its overall cost. So to minimize weight and lower cost in this particular application our pressure
vessel will be made of Aluminum Ultr Alloy 6020-T651 which posses a density of ρ=0.0979lb/in3
and a yield strength of σy =35ksi. The maximum external dimension was restrained to 10in for
a total volume of the aluminum V=136in3. The resulting weight of the vessel is 13.31lbf without
taking into consideration the weight of the fluid. The advantage of having an aluminum
pressure vessel is the reduction of weight that makes it easier to move or transport form place
to place. This design was also made to maximize the total volume of fluid it can store at a time
by making every external dimension as close to the constrain of 10in as possible. The cost of the
material was also lowered by choosing aluminum alloy as compared to other options like
titanium and composite materials. The disadvantage of choosing aluminum over other
materials is that it has a lower melting point and so it can only work at temperatures below
600℃. Because this vessel is to serve as a chemical reactor it must withstand high temperatures
produced by heat released by the components inside the vessel.
The vessel design established produces a safety factor of n=7.6 which is relatively high from a
designer point of view. The advantage is that the inside pressure can be overload and no failure
will occur, but also a high factor of safety implies a less economical design model. To improve
this design the wall thickness should be lowered to reduce the cost of the plate. The optimum
design for a pressure vessel should be safe, manufactured at low cost, and light weight.
Conclusion
Designing a pressurized vessel requires many analysis and calculations, which evidently
changes for different applications. The design established had the purpose to be safe and light.
In conclusion by the material indices the material results to be an Aluminum alloy. Inside the
great variety of these alloys, Aluminum 6020-T651 was selected. At the moment of establishes
the geometry of the vessel the maximum volume for the restrains was taken in consideration,
the favorable geometry was a spherical vessel, but this shape makes the manufacture difficult
and expensive. Then, the considered design was a cylindrical vessel with spherical ends. For the
established geometry stresses was calculated for both sections, resulting in maximum stresses
at the critical section, in the welding of the cylindrical section. Calculations were made for the
safety factor, using Modified Goodman Criteria, resulting in η=7.6, that for a designer point of
view the vessel was over designed.
Appendix A
Graph and Tables
Graph 1.Fracture toughness K Ic against elastic limit  y .
Graph 2. Chart showing M 4 . Strong, light materials lie near the top
Table 1. Materials for light pressure vessels
Appendix B
Calculations
1. Cyclic stress
Pa =
(Pmax -Pmin )
(150-100)
=
= 25psi
2
2
Pm =
(Pmax +Pmin )
= 125psi
2
 b2 + a 2 
 52  4.752 
 t =  y =Pi  2 2   25  2
 487.8 psi
2 
b
a
5

4.75




 a2 
 4.752 
 l = x = Pi  2 2   25  2
 231.4 psi
2 
b -a 
 5  4.75 
 r = z = -Pi  125 psi
 b2 + a 2 
 52  4.752 
 t =  y =Pi  2 2   125  2
 2, 439 psi
2 
b
a
5

4.75




 a2 
 4.752 
 l = x = Pi  2 2   125  2
 1,157 psi
2 
b
a
5

4.75




 231.4  487.8   487.8  25   25  231.4 
2
 VM,a 
2
2
1157  2439   2439  125   125  1157 
2
 VM,M 
2
2
VM ,a,corrected  1.2 444.1  532.92 psi
2
 444.1 psi
2
 2, 220.5 psi
VM ,M ,corrected  1.2 2220.5  2664.6 psi
Ksurf  aSUT b  2.7(38)0.265  1
Kload  1
A95  0.010462d 2  0.010462  5   0.262in 2
2
A95
.262

 1.85in
0.0766
0.0766
dequiv 
K size  0.869(1.85) 0.097  0.819
Ktemp  1
K reliable  0.702@99.99%
S f  (15.2ksi )(1)(1)(0.819)(1)(0.702)  8.7 ksi
nf 
a
Sf
1


m
S UT
1
0.5329 2.664

8.7
38
 7.6
2. Cylinders section
 r   Pi = -150 psi
2
2
 b2  a 2 
 t  Pi  2 2  = (150)  5 2  4.752  = 2926.92 psi
b a 
 5  4.75 


a2

4.75 2

 = (150)  2
 = 1388.46 psi
 l  Pi  2
2 
2 
b a 
 5  4.75 
1 
2 3
2
=
2927  1388
= 769.5 psi
2
2 
3 
 3  1
2
1   2
2
=
1388  (150)
= 769 psi
2
=
(150)  2927
= 1538.5 psi
2
3. Spherical section
pr 150  4.75 


 1425 psi
2t
2  0.25 
4.Material Selection
Aluminum Alloys
M1 
K1c
y

36
 0.1494m0.5
241
 y 241x106
m2
M4 

 88929 2

2710
s
Steel
M1 
K1c
y

50
 0.0485m0.5
1030
 y 1030 x106
m2
M4 

 132903 2

7750
s
Titanium Alloy
M1 
K1c
y

66
 0.4714m0.5
140
 y 1030 x106
m2
M4 

 132903 2

7750
s
5. Stress concentrations
 max  1.2 2665.7 psi   3198.8 psi
 max  1.21425 psi   1710 psi