ch04 section15 pressure vessel design

† Text Eq. refers to MechaR. Mischke and Richard G

MECHANICAL ENGINEERING DESIGN TUTORIAL 4 –15: PRESSURE VESSEL DESIGN

PRESSURE VESSEL DESIGN MODELS FOR CYLINDERS:

1. Thick-walled Cylinders 2. Thin-walled Cylinders

THICK-WALL THEORY

• Thick-wall theory is developed from the Theory of Elasticity which yields the state of stress as a continuous function of radius over the pressure vessel wall. The state of stress is defined relative to a convenient cylindrical coordinate system:

1. tσ — Tangential Stress 2. rσ — Radial Stress 3. lσ — Longitudinal Stress

• Stresses in a cylindrical pressure vessel depend upon the ratio of the inner radius to

the outer radius ( /o ir r ) rather than the size of the cylinder. • Principal Stresses ( 1 2 3, ,σ σ σ )

1. Determined without computation of Mohr’s Circle; 2. Equivalent to cylindrical stresses ( , ,t r lσ σ σ )

• Applicable for any wall thickness-to-radius ratio.

Cylinder under Pressure

Consider a cylinder, with capped ends, subjected to an internal pressure, pi, and an external pressure, po,

ir

ip

rσlσ

t

nica. Bu

lσ

l Engdyna

rσ

ineers; eq

tσ

ing Duatio

FI

σ

esign, 7th edition text by Joseph Edward Shigley, Charles ns and figures with the prefix T refer to the present tutorial.

GURE T4-15-1

opor

Shigley, Mischke & Budynas Machine Design Tutorial 4–15: Pressure Vessel Design 2/10

The cylinder geometry is defined by the inside radius, ,ir the outside radius, ,or and the cylinder length, l. In general, the stresses in the cylindrical pressure vessel ( , ,t r lσ σ σ ) can be computed at any radial coordinate value, r, within the wall thickness bounded by

ir and ,or and will be characterized by the ratio of radii, / .o ir rζ = These cylindrical stresses represent the principal stresses and can be computed directly using Eq. 4-50 and 4-52. Thus we do not need to use Mohr’s circle to assess the principal stresses. Tangential Stress:

22

22222 /)(

io

iooiooiit rr

rpprrrprp−

−−−=σ for i or r r≤ ≤ (Text Eq. 4-50)

Radial Stress:

22

22222 /)(

io

iooiooiir rr

rpprrrprp−

−+−=σ for i or r r≤ ≤ (Text Eq. 4-50)

Longitudinal Stress:

• Applicable to cases where the cylinder carries the longitudinal load, such as capped ends.

• Only valid far away from end caps where bending, nonlinearities and stress concentrations are not significant.

22

22

io

ooiil rr

rprp−−

=σ for i or r r≤ ≤ (Modified Text Eq. 4-52)

Two Mechanical Design Cases

1. Internal Pressure Only ( 0=op ) 2. External Pressure Only ( 0=ip ) Design Case 1: Internal Pressure Only

• Only one case to consider — the critical section which exists at irr = . • Substituting 0=op into Eqs. (4-50) and incorporating / ,o ir rζ = the

largest value of each stress component is found at the inner surface:

2 2 2

,max 2 2 21( )1

o it i t i i i ti

o i

r rr r p p p Cr r

ζσ σζ

+ += = = = =− −

(T-1)


where 2 22

2 2 2

11

o iti

o i

r rCr r

ζζ

++= =− −

is a function of cylinder geometry only.

irir prr −=== max,)( σσ Natural Boundary Condition (T-2)

• Longitudinal stress depends upon end conditions:

i lip C Capped Ends (T-3a)

lσ = 0 Uncapped Ends (T-3b)

where 21

1liCζ

=−

.

Design Case 2: External Pressure Only

• The critical section is identified by considering the state of stress at two

points on the cylinder: r = ri and r = ro. Substituting pi = 0 into Text Eqs. (4-50) for each case:

r = ri 0)( == ir rrσ Natural Boundary Condition (T-4a)

( )2 2

,max 2 2 22 2

1o

t i t o o o too i

rr r p p p Cr r

ζσ σζ

= = = − = − = −− −

(T-4b)

where, 22

2 2 2

221

oto

o i

rCr r

ζζ

= =− −

.

r = ro oror prr −=== max,)( σσ Natural Boundary Condition (T-5a)

2 2 2

2 2 21( )1

o it o o o o ti

o i

r rr r p p p Cr r

ζσζ

+ += = − = − = −− −

(T-5b)

• Longitudinal stress for a closed cylinder now depends upon external

pressure and radius while that of an open-ended cylinder remains zero:

o lop C− Capped Ends (T-6a)

lσ = 0 Uncapped Ends (T-6b)


where 2

2 1loC ζζ

=−

.

Example T4.15.1: Thick-wall Cylinder Analysis

Problem Statement: Consider a cylinder subjected to an external pressure of 150 MPa and an internal pressure of zero. The cylinder has a 25 mm ID and a 50 mm OD, respectively. Assume the cylinder is capped. Find:

1. the state of stress ( rσ , tσ , lσ ) at the inner and outer cylinder surfaces;

2. the Mohr’s Circle plot for the inside and outside cylinder surfaces; 3. the critical section based upon the estimate of maxτ .

Solution Methodology:

Since we have an external pressure case, we need to compute the state of stress ( ,rσ ,tσ lσ ) at both the inside and outside radius in order to determine the critical section.

1. As the cylinder is closed and exposed to external pressure only, Eq. (T-6a) may be applied to calculate the longitudinal stress developed. This result represents the average stress across the wall of the pressure vessel and thus may be used for both the inner and outer radii analyses.

2. Assess the radial and tangential stresses using Eqs. (T-4) and (T-5) for the inner and outer radii, respectively.

3. Assess the principal stresses for the inner and outer radii based upon the magnitudes of ( ,rσ ,tσ lσ ) at each radius.

4. Use the principal stresses to calculate the maximum shear stress at each radius.

5. Draw Mohr’s Circle for both states of stress and determine which provides the critical section.

Solution: 1. Longitudinal Stress Calculation:

OD 50 mm ID 25 mm25 mm ; 12.5 mm2 2 2 2o ir r= = = = = =

Compute the radius ratio, ζ

25 mm 2.012.5 mm

o

i

rr

ζ = = =


Then, 2 2

2 2

22

2

(2)1 (2) 1

( ) ( ) ( 150MPa)(1.3333 mm )1

lo

l i l o o o lo

C

r r r r p p C

ζζ

ζσ σζ

= = =− −

= = = = − = − = −−

21.3333 mm

MPa200−−−−====lσσσσ

2. Radial & Tangential Stress Calculations:

Inner Radius (r = ri) 2 2

2 2

2

,max 2 2

2 2(2)1 (2) 1

2( ) ( 150 MPa)(2.6667)

to

ot i t o o to

o i

C

rr r p p Cr r

ζζ

σ σ

= = =− −

= = = − = − = −−

2.6667

400 MPat iσ (r r ) Compressive= = −

0pforConditionBoundaryNatural i === 0)r(rσ ir

Outer Radius (r = ro)

2 2

2 2

2 2

,min 2 2

1 (2) 11 (2) 1

( ) ( 150 MPa)(1.6667)

ti

o it o t o o ti

o i

C

r rr r p p Cr r

ζζ

σ σ

+ += = =− −

+= = = − = − = −−

1.6667

eCompressivMPa250−−−−======== )r(rσ ot

ConditionBoundaryNaturalMPa150−−−−====−−−−======== oir p)r(rσ

3. Define Principal Stresses:

Inner Radius (r = ri ) Outer Radius (r = ro )

MPa400MPa200

MPa0

3

2

1

−==−==

==

t

l

r

σσσσσσ

MPa250MPa200MPa150

3

2

1

−==−==−==

t

l

r

σσσσσσ

4. Maximum Shear Stress Calculations:

Inner Radius (r = ri ) 1 3max

0 ( 400)( )2 2ir r σ στ − − −= = = =200 MPa

Shigley, Mischke & Budynas Machine Design Tutorial 4–15: Pressure Vessel

Outer Radius (r = ro ) 1 3max

( 150) ( 250)( )2 2or r σ στ − − − −= = = =50 MPa

5. Mohr’s Circles:

Inner Radius (r = ri ) Outer Radius (r = ro ) Critical Section

SectionCriticalrr i ⇐⇐⇐⇐======== MPa200)(maxττττ

3σ = -250 MPa

τ

2σ = -200 MPa

τ

σ

2σ = -200 MPa

3σ = -400 MPa

τ

1 0 MPaσ =

maxτ = 200 MPa

FIGURE T4-15-2

FIGURE T4-15-3

•

Design 6/10

!RadiusInsideatis

1 150 MPaσ = −

σ

max = 50 MPa


THIN-WALL THEORY

• Thin-wall theory is developed from a Strength of Materials solution which yields the state of stress as an average over the pressure vessel wall.

• Use restricted by wall thickness-to-radius ratio:

��1According to theory, Thin-wall Theory is justified for20

tr

≤

��1In practice, typically use a less conservative rule,

10tr

≤

• State of Stress Definition:

1. Hoop Stress, tσ , assumed to be uniform across wall thickness.

2. Radial Stress is insignificant compared to tangential stress, thus, 0.rσ �

3. Longitudinal Stress, lσ

��Exists for cylinders with capped ends;

��Assumed to be uniformly distributed across wall thickness;

��This approximation for the longitudinal stress is only valid far away from the end-caps.

4. These cylindrical stresses ( , , )t r lσ σ σ are principal stresses ( , , )t r lσ σ σ which

can be determined without computation of Mohr’s circle plot.

• Analysis of Cylinder Section

1

t FV

FHoop FHoop

Pressure Acting over Projected Vertical Area

di

FIGURE T4-15-4


The internal pressure exerts a vertical force, FV, on the cylinder wall which is

balanced by the tangential hoop stress, FHoop.

tpdFFF

ttAFpddppAF

tiHoopVy

ttstressedtHoop

iiprojV

σ

σσσ

� −=−==

===

===

220

)}1)({(

)}1)({(

Solving for the tangential stress,

Hoop Stress2

it

pdt

σ = (Text Eq. 4-53)

• Comparison of state of stress for cylinder under internal pressure verses external

pressure:

Internal Pressure Only

20

(Text Eq.4-55)4 2

it

r

i tl

pd Hoop Stresst

By Definitionpd Capped Case

t

σ

σσσ

=

=

= =

External Pressure Only

20

4 2

ot

r

o tl

pd Hoop Stresst

By Definitionpd Capped Case

t

σ

σσσ

=

=

= =

Example T4.15.2: Thin-wall Theory Applied to Cylinder Analysis

Problem Statement: Repeat Example T1.1 using the Thin-wall Theory (po = 150 MPa, pi = 0, ID = 25 mm, OD = 50 mm). Find: The percent difference of the maximum shear stress estimates found using the Thick-wall and Thin-wall Theories.


Solution Methodology:

1. Check t/r ratio to determine if Thin-wall Theory is applicable. 2. Use the Thin-wall Theory to compute the state of stress 3. Identify the principal stresses based upon the stress magnitudes. 4. Use the principal stresses to assess the maximum shear stress. 5. Calculate the percent difference between the maximum shear stresses

derived using the Thick-wall and Thin-wall Theories.

Solution:

1. Check t/r Ratio: 101

201

21

mm25mm5.12 or

rt

�==

The application of Thin-wall Theory to estimate the stress state of this cylinder is thus not justified.

2. Compute stresses using the Thin-wall Theory to compare with Thick-wall theory estimates.

definitionbyStress Radialb.

mm)2(12.5)mm50)(MPa150(

2

wall)acrossuniformstress,(averageStressHoopa.

r 0

MPa300

=

−=−=−=

σ

σtdp oo

t

c. Longitudinal Stress (average stress, uniform across wall)

4 2o o t

lp d

tσσ −= = = −150 MPa

3. Identify Principal Stresses in terms of “Average” Stresses:

MPa300MPa150

MPa0

3

2

1

−==−==

==

t

l

r

σσσσσσ

4. Maximum Shear Stress Calculation:

MPa1502

)MPa300(02

31max +=−−=

−=

σστ

5. Percent Difference between Thin- and Thick-wall Estimates for the

Critical Section:


max,Thin max,Thick

max,Thick

% 100%

( 150) ( 200) (100%)( 200)

Differenceτ τ

τ−

= ∗

+ − += ∗ = −+

25%

�� Thin -wall estimate is 25% low! ⇐⇐⇐⇐

ch04 section15 pressure vessel design

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