Al-Qadisiya Journal For Engineering Sciences                                          Vol. 2      No. ٢      Year 2009 
  

 

 

٣٦٨

 

 

 

 

 

 

BEHAVIOR OF THICK SPHERICAL VESSELS UNDER HIGH 

PRESSURE  
 

Maase Fasel Mailak 

College of engineering,  

University of Qadisiya  
 

Abstract 

 

The aim of this study was to investigate elasto-plastic thermal stresses in a thermoplastic of Thick 

Spherical vessels under high pressure The present study deals with spherical shells analysis, The 

elastic and plastic theory   of spherical shell is consider in this search, with a thick sphere subjected 

to different types of loading such as internal pressure, external pressure and thermal loading have 

been studies. The Tresca yield condition used in this study. When the applied pressure exceeds the 

minimum pressure required to initiate the yielding at the inner radius, plastic zone starts to be 

formed. The residual stress components also were calculated using elastic and elasto-plastic 

solution result 

 


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Nomenclature 

 

 

 

 

 

 

 

 

 

 

 

σy: Yield strength of material of 
uniaxial tension test 

σr: Radial Stress 
σθ: Hoop stress  
A and B: Constant 

r : Radius of thick sphere  

a: inner Radius 

b: outer Radius 

pe :pressure at plastic Zone 

 

cr
)(σ : The radial stress at r=c 

C: Radius of sphere at plastic zone 

Pi: inner radius 

Po: outer radius 

Tb and Ta: temperatures at the outer 

and inner radius respectively 

α : linear coefficient of expansion. 
β: Dimensionless quantity  
 



               Al-Qadisiya Journal For Engineering Sciences                                          Vol. 2      No. ٢      Year 2009 
  

 

 

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Introduction 

 

Cylindrical or spherical pressure vessels (e.g., hydraulic cylinders, sewage treatment plants, gun 

barrels, pipes, boilers and tanks) are commonly used in industry to carry both liquid and gases under 

pressure. When the pressure vessel is exposed to this pressure, the material comprising the vessel is 

subjected to pressure loading, and hence stresses, from all directions. The normal stresses resulting 

from this pressure are functions of the radius of the element under consideration; the shape of the 

pressure vessel (i.e., open ended cylinder, closed end cylinder, or sphere) as well as the applied 

pressure. A detailed study of stress analysis of spherical pressure vessel with internal pressure and 

external pressure are to be considered. Spherical pressure vessels are classified into thick sphere if 

the outer radius is larger then the inner radius by 10 %, otherwise the sphere can be considered as 

thin sphere. in the thin spheres the radial and tangential stresses are assumed to be constant along 

the thickness of the sphere but in thick  spheres the radial and tangential stresses very across the 

thickness of the sphere according to the type of loading such as internal pressure and external 

pressure . An effective numerical iterative method is proposed for the pressure loading analysis. 

Elastic stress behavior is considered first, followed by some cases of elasto-plastic stress 

distributions, and finally plastic deformation. for the spherical pressure vessel, the hoop and axial 

stresses are equal and are one half of the hoop stress in the cylindrical pressure vessel as shown in 

Figure (1). This makes the spherical pressure vessel a more “efficient” pressure vessel geometry [1-

4] 

 
Failure Criteria 
 
The purpose of failure criteria is to predict or estimate the failure/yield of structural members 

subjected biaxial or triaxial states of stress. There are more one theory for Failure Criteria, 

dependent on the nature of the material (i.e. brittle or ductile), as shown below: 

 
- Maximum shear stress criterion, Von Mises criterion for Ductile material. 
- Maximum Shear Stress Criterion, Tresca's criterion for Ductile material. 
- Maximum normal stress criterion, Mohr’s theory for Brittle material. 

 

Whether a material is brittle or ductile could be a subjective guess, and often depends on 

temperature, strain levels, and other environmental conditions. However, a 5% elongation criterion 

at break is a reasonable dividing line. Materials with a larger elongation can be considered ductile 

and those with a lower value brittle. Another distinction is a brittle material's compression strength 

is usually significantly larger than its tensile strength. 

 
All popular failure criteria rely on only a handful of basic tests (such as uniaxial tensile and/or 

compression strength), even though most machine parts and structural members are typically 

subjected to multi-axial loading. This disparity is usually driven by cost, since complete multi-axial 

failure testing requires extensive, complicated, and expensive tests. The success of all machine parts 

and structural members are not necessarily determined by their strength. Whether a part succeeds or 

fails may depend on other factors, such as stiffness, vibrational characteristics, fatigue resistance, 

and/or creep resistance. The maximum shear stress criterion, also known as Tresca's or Guest's 

criterion [4] , is often used to predict the yielding of ductile materials. Yield in ductile materials is 

usually caused by the slippage of crystal planes along the maximum shear stress surface. Therefore, 

a given point in the body is considered safe as long as the maximum shear stress at that point is 

under the yield shear stress obtained from a uniaxial tensile test. With respect to plane stress, the 

maximum shear stress is related to the difference in the two principal stresses. Therefore, the 

criterion requires the principal stress difference, along with the principal stresses themselves, to be 



               Al-Qadisiya Journal For Engineering Sciences                                          Vol. 2      No. ٢      Year 2009 
  

 

 

٣٧٠

less than the yield shear stress, this criterion has a good agreement with experimented results 

obtained for ductile material and due to simplicity of this criteria will be used in this study. 

 
|σ1|< σy, |σ2|< σy, |σ1-σ2|< σy 

 
Where , σy = Yield strength of material of uniaxial tension test 
 

As shown below, the maximum shear stress criterion requires that the two principal stresses . 

 

 

 

 

 

 

 

 

 

 

 
Mathematical Formulation 
 
The condition of spherical symmetry is assumed for the thick sphere with stress-strain relations and 

incompressibility conditions for plastic deformation. The equation is solved in the spherical 

Coordinates. The governing equation (equilibrium equation) for the thick spherical vessel is: 
 

rdr

d

rdrdr

d

drrdividingbyandsimplified

edrdrrdrrr

rdrrdrrdr

ddrrdddrrdrd

rr

rr

rrrr

rrr

rrr

)(2

0
22

)(

02)2(

02)2)((

0)2/sin(4)())(()(

2

222

22

222

σσσ

σσσ

σσσσσ

σσσσ

θθσθσσθσ

θ

θ

θ

θ

θ

−
=∴

=+−

→−−

=+++−

=+++−

=+++−

 

 

Formulation Of Tresca's 

 
This theory predicts that yielding will start when the maximum shear stress in the material becomes 

equal to the maximum shear stress at yielding in a simple tension test: 

1
2

1 21
21

±=−⇒=−
YY

Y
σσ

σσ                                                                             ……..(1) 

For thick spheres the principle stress are θσσ andr ,Where 

r
σ is compressive and θσ is tension so that: 

Y
r

=− σσ θ                                                                                                  ………….(2) 

 
 



               Al-Qadisiya Journal For Engineering Sciences                                          Vol. 2      No. ٢      Year 2009 
  

 

 

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Case 1: Internal Pressure Only 

 
Elastic Analysis: 
In the analysis of elastic and elastic-plastic case we must use some fundamental equations, such as 

equilibrium and compatibility equations. 

From Equilibrium equation for spherical coordinates has been derived in last section, and it is given 

as follows: 

 

rdr

d
rr
)(2 σσσ θ −=                                                                                                     …………….(3) 

 

Compatibility equation in terms of stresses is given by: 

0)2( =+ θσσ r
dr

d
                                                                                                     …………….(4) 

 

In the elastic solution for this case and for other, Lames equation has been used to find the solution, 

from equation (4), it can be seen that the amount )2( θσσ +r  is independent of the radius r, there 

fore: 

)2( θσσ +r =A 

Where A is a constant. 

 

By substitute Eq. (4) into the Eq. (3) and integrating, will be obtained  

3
r

B
Ar +=

σ                                                                                                    ………(5) 

3
2r

B
A −=θσ                                                                                                                    ………..(6) 

pr −=σ  at  r=a 

0=
r

σ  at r=b 

By substituting the above boundary condition in Eq.(5) and (6), we obtain: 

 

1
3

3

−

=

a

b

P
A  

1
3

3

3

−

−
=

a

b

Pb
B  

 

Thus Eq. (5) and (6) can be written as: 

)1(

)1(

3

3

3

3

−

−−
=

r

b

r

b
p

r
σ                                                                                        ……….(7) 



               Al-Qadisiya Journal For Engineering Sciences                                          Vol. 2      No. ٢      Year 2009 
  

 

 

٣٧٢

)1(

)1
2

(

3

3

3

3

−

+
=

a

b

r

b
p

θσ                                                                                …………….(8) 

 
These Equations express the radial and tangential elastic stress distribution across the thickness of 

the wall and it is noted that the redial stress is compressive while the tangential stress is tensile. 

 

 
Plastic Analysis 

If the internal pressure is increased to a critical value (pe), plastic yielding begins at the radius, 

where the yield criterion (Tresca) is first satisfied. 

By substituting Eq. (7) and (8) into Eq. (2): 

 

Y
r

b

a

b

p

r

b

a

b

p
=








−









−

+







+









−

1

1

1
2

1

3

3

3

33

3

3

3
 

 
This equation reduces to 

 

Pe = 3

3

3

3

1
3

2

b

r

a

b
Y 








−  

 
It has note that has minimum value at r=a (i.e) the difference ( )(

r
σσ θ −  has the greatest value at 

r=a, this means that the initiation of the plastic zone will begin-from the inner radius when p=pe 
 

Pe = )1(
3

2
3

3

b

a
Y −  

With further increase in the internal pressure, the plastic zone spreads out wards and the elastic-

plastic boundary is a spherical trace at each stage. 

 
To find the stress distribution at the plastic zone (at each point) in this zone the yield criterion will 

be satisfied as will as the equilibrium equation therefore by substituting equation. 

 

From Eq (1) and (2) we get: 

 

Y
rdr

d
r 2=

σ
 then integrating 

ArY
r

+= ln2σ  
Where A is a constant to be evaluated from the boundary condition, 

At r=a, p
r

−=σ  then A= (-P-2Yln a), 

And P
a

r
Y

r
−







= ln2σ                                                                                   …………… (9) 

To obtain the expression for θσ Eq. (9) is substituted into the Tresca's criterion 

Eq. (2) we get. 

 



               Al-Qadisiya Journal For Engineering Sciences                                          Vol. 2      No. ٢      Year 2009 
  

 

 

٣٧٣

P
a

r
Y −+= ))ln(21(θσ                                                                                ……… (10) 

To find the stresses in the elastic zone, Lames equation are also applied as for the elastic case, but 

the constant will be a function of the plastic radius (c), there for the sphere is assumed to be 

consisted from two spheres ,the inner sphere is completely plastic and the outer sphere is 

completely elastic, thus the boundary condition become: 

0=
r

σ  at r=b 

crr
)(σσ =  at r=c 

Where  

cr
)(σ  is the radial stress at r=c, and it is equal to: 

-(
cr

)(σ = )1(
3

2
3

3

b

c
Y −  from the first boundary condition: 

A=-B/b
3 

And from second boundary condition, B=



















−

−

)1(
1

1

3

2

3

3

3

3

3

c

b

b

b

C

Y  

 
Then Lames equations become  

)1(
3

2
3

3

3

3

−−=
r

b

b

C
Y

r
σ                                                                                     ………. (11) 

 and 

 )1
2

(
3

2
3

3

3

3

+=
r

b

b

C
Yθσ                                                                                      ……… (12) 

At the elastic-plastic boundary the stresses must satisfy the continuity condition therefore equalizing 

Eqs. (11) and (12) at r= c, will yield 

 

3

)1(

ln
2

3

3

b

c

a

c

Y

P
−

+=    

 This equation is used to determine the radius of plastic phase. 

The solution is accomplished at any value of pressure (P), by solving it by using suitable numerical 

method  
 
Residual Stress Analysis 

 
The residual stresses can be found by subtracting Eqs. (7) and (8) from Eqs (9) and (10).but in the 

elastic zone the residual stress can be found by subtracting Eqs. (7) and (8) from Eqs (11) and (12). 

 

))((
3

2
3

3

3

3

3

3

b

a

r

a

P

P

a

CY

e

r
−−−=σ    ,c≤r≤b 

))((
3

2
3

3

3

3

3

3

b

a

r

a

P

P

a

CY

e

+−−=θσ , c≤r≤b 

3

3

3

3

ln)1((
3

2

a

r

r

a

p

pY

e

r
−−−=σ  , a≤r≤c 



               Al-Qadisiya Journal For Engineering Sciences                                          Vol. 2      No. ٢      Year 2009 
  

 

 

٣٧٤

 
3

3

3

3

ln)1((
3

2

a

r

r

a

p

pY

e

−+−=θσ ,  a≤r≤c 

 
Case 2: External Pressure Only 

 
There are some case in which a sphere is subjected to external pressure, the analysis for such a case 

is similar to for the internal pressure case, but boundary condition are varied: 

  
Elastic Analysis 

 
The Lames equations after applying the boundary condition becomes : 

 

)1(

)1(
3

3

3

3
b

a

b

a

P
O

r
−

−

=σ  and  )
2

1(

)1(
3

3

3

3
r

a

b

a

P
O +

−

=θσ  

It is noted that the radial stress and as well as the tangential stress are both compressive while they 

are tensile for the case of internal pressure only 

 

Plastic Analysis 

 

To find the stress distribution at the plastic zone (r<c) Tresca yield criteria should be satisfied as 

well as the equilibrium equation. 

Thus the stresses will be: 

)
3

1(
3

2
33

3

R

b
Y

b

c
P

r
−−−=σ      

)
2

3
1(

3

2
33

3

R

b
Y

b

c
P +−−=θσ  ,Since these stresses must be continuous across the elastic-plastic 

boundary: 

So R=C    , [ ] [ ]
plasticrelasticr

σσ =   then  

-P-
c

a
Y

c

b
Y

b

c
ln2)1(

3

2
3

3

3

3

=−  

 

So P = )1(
3

2
ln2

3

3

−+=
b

C
Y

b

a
YP  

By solving the above equation numerically using Newton Raphson method the value of the radius C 

can be obtain for any value of applied pressure. 

  
Case 3: Internal and External Pressure 

 
The sphere in this case is subjected to internal and external pressure applied simultaneously. By 

superposition of the two elastic stresses for the internal and external pressure case and by applying 

the following boundary condition in to lames equation: 

ir
P−=σ  at r=a 

r
σ =-PO at r=b 

Then 

 



               Al-Qadisiya Journal For Engineering Sciences                                          Vol. 2      No. ٢      Year 2009 
  

 

 

٣٧٥

)1(

1

)1(

1

3

3

3

33

3

3

3 r

a

b

a

P

r

b

a

b

P
Oi

r
−









−

−−









−

−=σ  and  )
2

1(

1

)1
2

(

1

3

3

3

33

3

3

3 r

a

b

a

P

r

b

a

b

P
Oi +









−

−+









−

−=θσ  

 

To obtain the critical of the difference in pressures that will case the initiation of plastic zone at the 

inner radius (r=a),sub, θσσ andr  in Tresca criterion( Yr =− )( σσ θ  

 
  Case 4: Thermal Loading 

 

The thermal loading, has significant effect on the value of stresses due to temperature difference 

that may occurs between the inner and the outer radius of the sphere, this will cause a difference in 

the expansion or contraction between the layers of the thick sphere, there for this will give rise-to 

the radial and tangential stresses. The steady state temperature distribution in the case of spherical 

symmetry is give by: 

1

1

)(

−

−
×−+=

a

b
r

b

TTTT
bab

 

Where Tb &Ta are temperatures at the outer and inner radius respectively. 

The strain in the sphere is the superposition of that due to pressure loading and thermal loading, 

then the elastic stress-strain equation is: 

 

T
E

T
E

r

rr

ανσσνε

ανσσε

θθ

θ

+−−=

+−=

))1((
1

)2(
1

        

       where α is the linear coefficient of expansion. 
 

After Appling compatibility equation and substitution T with integrating we can get the 

dimensionless quantity given as: 

)1(

)(

ν

α
β

−

−
= ba

TT
  

The radial stress and as well as the tangential stress are given by the Equations below: 



















−

−
−

−

−
−=

1

1

1

1

3

3

3

3

a

b

r

b

a

b
r

b

E
r

βσ  

 



















−

+
−

−

−
−=

1

1
2

1

1
2

3

3

3

3

a

b

r

b

a

b
r

b

Eβσ θ  

 
Result And Discussion 

The Calculation show that the internal pressure required to initiate the plastic zone from the inner 

radius is Pi=14 Mpa and the state of total plasticity is reached when Pi=33.2Mpa, because of 

increasing the   loading of vessel, the volume increase, which leads to plastic deformations take 



               Al-Qadisiya Journal For Engineering Sciences                                          Vol. 2      No. ٢      Year 2009 
  

 

 

٣٧٦

place. At internal pressure Pi=8Mpa, the whole sphere is in fully elastic state. This pressure is lower 

than the pressure needed to initiate the plastic deformation at the inner surface, Figure(1) shows the 

non-dimensional radial and tangential stress distribution along the thickness of the sphere, we 

conclude from that when the external pressure increased beyond the critical value of the pressure 

the plastic zone will propagate outward that’s leads to increasing the radial stress . Figure (4) shows 

the variation of the dimensional residual radial and tangential stresses across the thickness of the 

sphere when  it is loaded with pressure P=30 Map. we see that the hoop stress increase during the 

thickness of the sphere  because of the Residual stress distribution increase due to overloads. Figure 

(5) show the non-dimensional radial and tangential stress distributions due to thermal loading ,we 

see that the hoop stress is larger then  radial stress because of  the Coefficient of thermal expansion 

mismatch between different phases. Figure (6) show the elasto-plastic state of stresses for external 

pressures of 20Mpa this figures also show the propagation of plastic zone as the pressure increases. 

due to the plastic area increase leads to yielding take place. 

 

Conclusions 

For the internal pressure case it can be observed that the plastic zone starts at the inner radius and 

spreads outward as the pressure increase, also there is a direct proportionality between elastic 

boundary radius and internal pressure. The residual stresses increases as the amount of plastic 

deformation increases. There is a direct proportionality between the internal pressure and the value 

of the residual tangential. For thermal loading it is noted that the radial stress is negative and 

reaches a maximum value between the inner and the outer radius. This maximum stress will be at 

R/B=0.64.it also observed those values of stresses increases as temperature difference increases. 

Effects of residual stress may be either beneficial or detrimental, depending upon the magnitude, 

sign, and distribution of the stress with respect to the load-induced stresses. Very commonly, the 

residual stresses are detrimental, and there are many documented cases in which these stresses were 

the predominant factor contributing to fatigue and other structural failures when the service stresses 

were superimposed on the already present residual stresses. 

            
References 

       
[1] Pavel Elesztos - Ladislav Ecsi ‘’ stress test of spherical pressure-retaining vessels with 

hydrocarbon Gases’’ 

 

[2] Dechant,K.E.: Effects of high pressure testing technique on pipelines. 1996 OMAE - Volume V, 

Pipeline Technology, ASME 1996. 

 

[3]  Hossein Mahbadi-  Mohamad R. Eslami ‘’ Elastic-plastic-creep Cyclic Loading of Thick 

Pressure Vessels based on the Frederick-Armstrong Kinematic Hardening Model’’Paper # F05-7 

 

[4] Faruk Sen and Metin Sayer ‘’ Elasto-Plastic thermal stress analysis in a thermoplastic composite 

Disc under uniform temperature using FEM’’Mathematical and Computational Applications, Vol. 

11, No. 1, pp. 31-39, 2006. 



               Al-Qadisiya Journal For Engineering Sciences                                          Vol. 2      No. ٢      Year 2009 
  

 

 

٣٧٧

0.40 0.60 0.80 1.00

-0.40

-0.20

0.00

0.20

0.40

  
 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

             

             

             

             

             

             

             

             

             

             

    

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Figure (1): Pressure and Internal Hoop and Axial Stresses. 

0.40 0.60 0.80 1.00

-0.4

-0.2

0

0.2

0.4

Y/σ

 

R/B 

θσ 

r
σ 

Figure (1): Elastic stress distribution 

for internal pressure(pi=8 Mpa) 

θσ 

r
σ 

Figure (2): Elastic stress distribution 

for internal pressure(pi= 18 Mpa) 

Y/σ

 

R/B 

0.40 0.60 0.80 1.00

-0.80

-0.40

0.00

0.40

0.80

Plastic 

zone 

R/B 

Elastic 

zone 
Y/σ

 

θσ 

r
σ 

Figure (3):Elastic-Plastic stress 

Distribution for( pi=25 Mpa) 

0.40 0.60 0.80 1.00

-1.50

-1.00

-0.50

0.00

0.50

θσ 

r
σ 

R/B 

Y/σ

 

Figure (4):Residual stress distribution for      

     internal pressure ( pi=30 Mpa) 



               Al-Qadisiya Journal For Engineering Sciences                                          Vol. 2      No. ٢      Year 2009 
  

 

 

٣٧٨

     

             

             

             

             

             

             

             

             

             

             

             

             

             

             

             

             

             

             

             

             

      
 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

  

0.40 0.60 0.80 1.00

-0.02

0.00

0.02

0.04

0.06

θσ 

r
σ 

R/B 

Y/σ

 

Figure (5):Radial and tangential stress 

Distribution for thermal loading 

( Ta=500,Tb=20 ) 

0.40 0.50 0.60 0.70 0.80 0.90

-1.60

-1.20

-0.80

-0.40

0.00

R/B 

Y/σ

 

θσ 

r
σ 

Figure (6):Elastic –plastic stress distribution 

for external pressure(po=20 Mpa)