Calculating stress and strain on Loading Systems.

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MODULE TITLE : MECHANICAL PRINCIPLES

TOPIC TITLE : LOADED BEAMS AND CYLINDERS

LESSON 2 : THIN-WALLED PRESSURE VESSELS

MP - 2 - 2

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________________________________________________________________________________________

INTRODUCTION ________________________________________________________________________________________

Pressure vessels are commonly used in industry to carry a liquid, vapour, or

gas at a pressure other than atmospheric pressure at the same elevation and can

be divided into two categories, thin- and thick-walled. You could even say that

a carbonated soda bottle is a pressure vessel, as the pressure in the bottle is

over 105 Pa (1 atmosphere). Also, any container that has the potential to

pressurize should be considered a pressure vessel. Under pressure the material

of which they are made is subjected to a loading from all directions. When the

stress state somewhere in the wall material exceeds some criterion, the

pressure vessel will fail. It is, therefore, very important to understand and

quantify stresses that occur in pressure vessels.

Two types of analysis are usually applied to pressure vessels. The most

common method is based on a simple mechanics approach and is applicable to

thin-walled pressure vessels. The second method is based on elasticity

solution and is always applicable for thick-walled pressure vessels. This

lesson contains an investigation of thin-walled pressure vessels, which have

most engineering applications.

Pressure vessels very often have the following shapes:

• spherical (e.g. LPG storage tanks) (see Photograph 1)

• cylindrical (e.g. liquid storage tanks, boiler steam drums, condensers,

gun barrels, hydraulic and pneumatic cylinders) (see Photograph 2)

• cylindrical shells with hemispherical ends (e.g. compressed-air

receivers, distillation columns, submarine hulls) (see Photograph 3).

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Photograph 1 Photograph 2

Photograph 3

(Reproduced by kind permission of Duncan Rogers (Engineering) Ltd.)

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YOUR AIMS ________________________________________________________________________________________

At the end of the lesson you should be able to:

• determine whether a pressure vessel should be classified as thin-

walled or thick-walled

• determine the principal stresses acting on a thin-walled pressure

vessel

• calculate the stresses in the wall of a thin-walled cylinder and sphere.

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________________________________________________________________________________________

THIN-WALLED PRESSURE VESSELS ________________________________________________________________________________________

These pressure vessels are the most used. In general, thin wall refers to wall

thickness, t, which is less than 10% of the inner radius, r. Another way of

saying this is a pressure vessel is thin-walled if the radius is 10-times or more

the thickness.

In most cases, . Specifically, when , the results of a thin-wall

analysis will predict a stress that is approximately 4% less than the actual

maximum stress in the vessel. For the greater r/t ratios this error will be even

smaller.

When the vessel wall is thin, the stress distribution throughout its thickness

will not vary significantly, and we can assume that it is uniform or constant.

Using this assumption, we will now analyse the state of stress in thin-walled

cylindrical and spherical vessels. In both cases, the pressure occurring on the

inside and outside of the vessel’s wall is understood to be the gauge pressure.

t = 10

t > 50

t d≥ ≥10 20or ( is inner diameter)

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THIN-WALLED CYLINDRICAL VESSELS

Consider the cylindrical vessel with a wall thickness t and inner radius r as

shown in FIGURE 1. A pressure p is developed within the vessel by a

contained fluid, which is assumed to have negligible weight.

When an internal pressure p acts on the inside of a cylinder, two types of stress

are set up within the wall to resist the bursting effect of applied p:

• circumferential or hoop stress σ1

• longitudinal or axial stress σ2.

FIG. 1 A cylindrical vessel with an inner radius r and a wall thickness t

Both these stress components exert tension on the material. The cylinder may

fail in one of two ways, depending upon the values of the stresses. If σ1 becomes excessive, the material will tear along a line parallel to the axis of

the vessel. If σ2 becomes excessive, the material will tear along a line perpendicular to the axis.

Longitudinal joint

Circumferential joint

σ 1

σ 2

σ 1

σ 2

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Hoop Stress σσ1

Consider a ring from the cylinder with a length l as shown in FIGURE 2. In

the x direction, the resultant of the forces due to the internal pressure p, which

is p(2rl), must be determined and balanced with the force in the wall, which is

called tangential force 2[σ1(tl)]. Therefore,

Then the stress is expressed by

This is the equation for the hoop stress in a thin-walled cylinder subjected to

internal pressure.

FIG. 2 A ring from the cylinder carrying pressure showing hoop stress

σ 1

p Area 2tl on which σ

1 acts

Projected area 2rl on

which p acts l

σ1 = pr

t ....................................... 1( )

2 2 01σ tl p rl( )⎡⎣ ⎤⎦ ( ) =–

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Longitudinal Stress σσ2

FIG. 3 A ring from the cylinder carrying pressure showing longitudinal

stress

FIGURE 3 shows a part of a cylinder, which is subjected to an internal

pressure p, cut perpendicularly to its axis y. Assuming that the end of the

cylinder is closed, the force due to the pressure acting on the circular area,

p(πr2), must be resisted by the force in the walls of the cylinder due to σ2, σ2(2πrt). In the y direction, we have the force balance as following

σ 2 22 0π πrt p r( ) ( ) =–

or σ 2 2 = pr

t ......................................... 2( )

σ 2

Area πr2on which p acts

Cylinder wall cross- sectional area 2πrt

r Internal pressure

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Comparing equations (1) and (2), it is found that σ1 = 2σ2, which means that the cylinder would fail by tearing along a line parallel to the axis, rather than

on a section perpendicular to the axis. Consequently, when fabricating

cylindrical vessels from rolled-formed plates, the longitudinal joints must be

designed to carry twice as much stress as the circumferential joints.

In the above equations,

σ1 and σ2 = the normal stress in the hoop and longitudinal directions, respectively, which are assumed to be constant throughout

the cylinder wall and subject the material to tension (Pa or

MPa)

p = the internal pressure developed by the fluid contained in the

cylinder (Pa or MPa)

r = the inner radius of the cylinder (m)

t = the thickness of the cylinder wall (r/t ≥ 10) (m)

Example 1

A thin-walled cylindrical pipe has an inside diameter of 150 mm and a wall

thickness of 4 mm. What will be the hoop stress and the longitudinal stress

when the pipe has an internal gauge pressure of 5 MPa?

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Solution

From the question, it is known that:

Using equation (1), the hoop stress σ1 can be calculated as follows.

The longitudinal stress σ2 is half the σ1 and thus is 46.9 MPa.

Example 2

A cylindrical tank holding oxygen at 2000 kPa pressure has an outside

diameter of 450 mm and a wall thickness of 10 mm. Calculate the hoop stress

and the longitudinal stress in the wall of the cylinder.

σ1 65 10 0 075

0 004

93 8

= = × ×

. .

. MPa

r d

= = ×

= = = =

= =

2 150

2 75

MPa 5 10 Pa

mm 0.075 m

mm 0.

0004 m

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Solution

From the question:

The inner radius can be calculated using:

Since , this is a thin-walled cylindrical tank, then

equation (1) can be used to calculate the hoop stress σ1.

The longitudinal stress .σ σ

2 1

2 21 5= = . MPa

σ1 6

2 10 0 215 0 01

43 10

= = × ×

= ×

. .

43 MPa

t = = >0 215

0 01 21 5 10

. .

r D

0 45 2

0 01

0 215

–

. – .

. m

= = ×

= =

2000

450

kPa 2 10 Pa

mm 0.45 m

mm 0.01 m

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SPHERICAL VESSELS

Using a similar method to the above, we can analyse stresses in a thin-walled

spherical vessel as shown in FIGURE 4.

FIG. 4 A spherical pressure vessel with an inner radius r and a wall thickness t

If the spherical vessel is cut into two pieces through the x-z plane, the stresses

acting on the cross-section are as shown in FIGURE 5.

FIG. 5 Stresses on a spherical vessel carrying internal pressure

Area 2πrt on which σ acts

Projected area πr2 on

which p acts

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Like the cylinder, equilibrium in the y direction requires

Then, the tensile stress to be developed can be expressed by

where σ = the tensile stress due to the tensile force acting on the sphere wall (Pa or MPa)

p = the internal pressure developed by the fluid contained in the

sphere (Pa or MPa)

r = the inner radius of the sphere (m)

t = the thickness of the sphere wall (r/t ≥ 10) (m)

Note that for the spherical pressure vessel, the hoop and longitudinal stresses

are equal and are half of the hoop stress in the cylindrical pressure. This

makes the spherical vessel a more 'efficient' pressure vessel shape, which will

be clearly shown in Example 4.

Example 3

Calculate the stress in the wall of a sphere having an inside diameter of

300 mm and a wall thickness of 1.5 mm when carrying nitrogen gas at

3500 kPa internal pressure.

σ = pr t2

......................................... 3( )

σ 2 02π πrt p r( ) ( ) =–

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Solution

From the question:

Since , this spherical vessel is thin-walled. We can

now use equation (3) to calculate the stress σ.

Example 4

A cylindrical pressure vessel has an inner diameter of 1.2 m and a wall

thickness of 12.5 mm. Determine the maximum internal pressure it can sustain

so that neither its circumferential nor its longitudinal stress component exceeds

138 MPa. Under the same conditions, what is the maximum internal pressure

that a similar-sized spherical vessel can sustain?

σ = = × × × ×

= × =

t2 3 5 10 0 15 2 1 5 10

175 10

. . . –

Pa 175 MMPa

t =

× = >0 15

1 5 10 100 10

. . –

r d

= = × =

= = = =

3500

2 300 2

150

kPa 3.5 10 Pa 3.5 MPa

mm 0

..15 m

mm 1.5 10 m3t = = ×1 5. –

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Solution

From the question:

Cylindrical pressure vessel – the maximum stress occurs in the

circumferential direction. From equation (1) we have:

Then, the maximum internal pressure pmax will be:

p t

rmax max

= ( )

= × ×

= ×

σ1

138 10 0 0125 0 6

2 88 10

. .

== 2.88 MPa

σ1 = pr

r d

= = =

= =

2 1 2 2

0 6

12 5

. .

mm 0.0125 m

the maximum sttress 138 MPa 138 10 Pa6= = ×

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Spherical pressure vessel – here the maximum stress occurs in any two

perpendicular directions on an element of the sphere, as shown in FIGURE 4.

From equation (3) we have:

Therefore, the maximum internal pressure pmax in the spherical pressure vessel

can be determined as:

Although it is more difficult to fabricate, the spherical pressure vessel will

carry twice as much internal pressure as a cylindrical vessel.

The above analysis indicates that an element of material taken from either a

cylindrical or a spherical pressure vessel is subjected to biaxial stress, which is

the normal stress existing in only two directions. Actually, there is another stress

which is a radial stress σ3 acting along a radial line. This stress has its maximum value equal to the pressure p at the interior wall and decreases through

the wall to zero at the exterior surface of the vessel since the gauge pressure

there is zero. However, for thin-walled vessels the radial stress component can

be ignored. This is because the assumption that r/t is greater than 10 results in

σ1, and σ2 or σ being at least 10 and 5 times higher than the maximum radial stress (σ3)max = p, respectively. Note that the three normal stresses are the principal stresses and can be directly used to determine failure criteria.

p t

rmax max

= × × ×

= ×

2 138 10 0 0125 0 6

5 75 10

. .

== 5.75 MPa

σ = pr t2

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It should be noted that the above equations are applied for vessels subjected to

an internal pressure. If the vessel is subjected to an external pressure, for

example, a submarine (as shown in Photograph 4), the above formulae are still

valid. However, the stresses are now negative since the wall is now in

compression instead of tension. The compressive stress developed within the

thin wall may cause the vessel to become unstable and collapse may occur by

buckling of the wall.

Photograph 4 The Jimmy Carter

(Reproduced by kind permission of General Dynamics Electric)

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SAFETY FACTOR AND JOINT EFFICIENCY

Safety Factor

By transposing equation (1), we can obtain an expression for the thickness of

the cylinder wall suitable for a particular pressure vessel as following:

There would be no point in using equation (2) for the calculation of the wall

thickness of the cylinder as it would give a value half of that obtained from

equation (4). Therefore, the required wall thickness is determined by the hoop

or circumferential stress σ1.

For steady pressure, the design stress can be based on the yield strength of the

material. The choice of the safety factor N is often dictated by industrial code

because of the danger created when a pressure vessel fails. This is particularly

true for vessels containing gases or steam under pressure since failures produce

a violent expulsion of gas as the high level of stored energy is released. In the

absence of a code, we will use N ≥ 4 (i.e. the design stress will be at most a

quarter of the allowable stress for the material).

For cycling pressure, the design stress is based on the ultimate strength, and

the safety factor should be greater than 8.

t pr pd= =

σ σ1 12 ......................................... 4( )

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Joint Efficiency

In developing the formulae for hoop and longitudinal stresses, we have ignored

the efficiency of the joints (e.g. rivets) that go to make up a pressure vessel.

The strength of the wall at any joint should be sufficient to withstand the force

due to internal pressure on the end of the cylinder. Allowance has to be made

for the joint efficiency, which is denoted by η (eta). η is often quoted as the ratio of strength of the joint to the strength of the unjoined plate. So taking this

into account for the hoop stress we may write:

For longitudinal stress we may write:

Referring to FIGURE 1, it can be seen that the efficiency of the longitudinal

joints ηl affects the hoop stress at the joint, and similarly the longitudinal stress is affected by the efficiency of the circumferential joints ηh.

Example 5

A cylindrical compressed air vessel has an internal diameter of 50 cm and is

manufactured from steel plate 4 mm thick. The longitudinal plate joints have

an efficiency of 90% and the circumferential joints have an efficiency of 55%.

Determine the maximum air pressure which may be accommodated by the

vessel if the maximum permissible tensile stress of the steel is 100 MPa.

σ η η2 2 4

= =pr t

th h ...................................... 6( )

σ η η1 2

= =pr t

tl l ......................................... 5( )

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Solution

From the question:

Using equations (5) and (6), we have:

Then, the maximum air pressure accommodated by the longitudinal joints is:

p t

r l

max max=

( )

= × × ×

σ η1

6100 10 0 004 0 9 0 25

1 44

. . .

. ××

106 Pa

1.44 MPa

σ η

σ η1 2

= =pr t

tl h and 2

r d

= = = =

= =

2 50 2

cm 0.25 m

mm 0.004 m

tthe maximum tensile stress 100 MPa 100 10 P6= = × aa

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Similarly, the maximum air pressure accommodated by the hoop joints is:

So the maximum air pressure is determined by the longitudinal joints and is

1.44 MPa.

p t

r h

max max=

( )

= × × × ×

2 100 10 0 004 0 55 0 25

σ η

. . .

11 76 106. ×

1.76 MPa

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________________________________________________________________________________________

NOTES ________________________________________________________________________________________

...................................................................................................................................................

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________________________________________________________________________________________

SELF-ASSESSMENT QUESTIONS ________________________________________________________________________________________

1. Determine the stresses set up in the shell plate of a torpedo whose

diameter is 500 mm and wall thickness 10 mm. The internal pressure is

12.5 MPa.

2. A compressed air cylinder has an internal diameter of 200 mm. What

wall thickness will be required for a gauge pressure of 15 MPa if the steel

used for the cylinder has a maximum allowable stress of 600 MPa?

Assume a safety factor N of 4.

3. A spherical pressure vessel has an internal diameter of 400 mm and a wall

thickness of 5 mm. What is the maximum pressure that can be used if the

yield stress of the material is 630 MPa and a safety factor of 6 is to be

used?

4. A cylindrical pressure vessel has an internal diameter of 1.5 m and is

made of plate of thickness 10 mm. The efficiency of the longitudinal

joint is 75% and of the circumferential joints 50%. The plate has a tensile

strength of 400 MPa. What is the maximum internal pressure to which

the vessel should be used if there is to be a safety factor of 5?

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________________________________________________________________________________________

NOTES ________________________________________________________________________________________

...................................................................................................................................................

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________________________________________________________________________________________

ANSWERS TO SELF-ASSESSMENT QUESTIONS ________________________________________________________________________________________

1. We know that:

Force tending to tear wall around the circumference pπr2 should be at least balanced with the force due to longitudinal stress σ22πrt

thus

then

p r rt

π π2 2

22 5 10 0 25 2 0 1

156 25 10

. . .

× × ×

= ×

156.25 MPa

r d

= = = =

= =

2 500 2

250

12 5

mm 0.25 m

mm 0.01 m

MPa. 112.5 10 Pa6×

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Force tending to tear wall along length of torpedo p2rl must be resisted by

the force due to hoop stress σ12tl, where l is the length of the torpedo:

2. From the question:

As we have seen, the thickness of the cylinder wall is determiend by the

hoop stress σ1. Considering the factor of safety, the allowable tensile stress will be:

σ1 6

6600 10 4

150 10= × = × Pa

r d

= = = =

= = ×

2 200 2

100

mm 0.1 m

MPa 15 10 Pa

the max

iimum allowable stress of steel 600 MPa 600 1= = × 00 Pa6

N = 4

p rl tl

2 2

12 5 10 0 25 0 01

∴ =

= × ×

. . .

112 5 106. ×

312.5 MPa

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Using equation (1):

The wall thickness must be at least 10 mm.

3. From the question:

After considering the safety factor N, the maximum allowable tensile

stress in the spherical pressure vessel will be:

σ σ

max

= ×

630 10 6

105 10

r d

= = = =

= =

2 400 2

200

630

mm 0.2 m

mm 0.005 m

MPa 6σ 330 10 Pa6×

=N 6

15 10 0 1 150 10

= × × ×

t pr

then

00 01. m 10 mm=

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Applying equation (3), we have:

Then, the maximum pressure that can be used is:

4. From the question:

Considering the safety factor N, the maximum allowable tensile stress

will be:

400 10 5

80 10 6

6× = × Pa

r d

= = =

= =

2 1 5 2

0 75

. .

mm 0.01 m

the

ttensile strength of the material 400 MPa 400= = ××

10 Pa6

N 5

p t

rmax max

= × × × ×

= ×

2 5 10 105 10 0 2

5 25 10

3 6

–

== 5.25 MPa

σ max max=

p r

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Using equations (5) and (6), we have:

Then the maximum internal pressure accommodated by the longitudinal

joints is:

The maximum internal pressure accommodated by the circumferential

joints is:

Thus, the maximum internal pressure is determined by the longitudinal

joints and is 0.8 MPa.

p t

r h

max max=

( )

= × × × ×

2 80 10 0 01 0 5 0 75

1 0

σ η

. . .

. 77 10

6×

1 MPa.

p t

r l

max max=

( )

= × × ×

= ×

σ η1

680 10 0 01 0 75 0 75

0 8 1

. . .

. 006 Pa

0.8 MPa=

σ η

σ η1 2 2

= =pr t

tl h and

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________________________________________________________________________________________

SUMMARY ________________________________________________________________________________________

In this lesson, we have applied the principle of forces to thin-walled cylindrical

and spherical pressure vessels subjected to an internal pressure. Using the

results from our analyses, the principal stresses have been determined. It

should be noted that safety factor and joint efficiency must be taken into

account when we carry out an engineering design.

In the next lesson, we will analyse the stresses occurring in thick-walled

pressure vessels and the distribution of stresses along the wall.

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/PTB 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/ITA 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setdistillerparams << /HWResolution [2400 2400] /PageSize [612.000 792.000] >> setpagedevice