Composite Reinforcement of Cylindrical Pressure Ves sels

MAE 661

Laminated Composite Materials

1

Laminated Composite Materials

Spring 2012

Cylindrical Pressure Vessels

Cylindrical pressure vessels are in widespread use for a variety of applications

• SCBA and SCUBA tanks

• Propane tanks

• Compressed Natural Gas (CNG) and hydrogen for Alternative Fuel Vehicles

• Medical oxygen tanks

• Laboratory gas tanks

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• Laboratory gas tanks

Depending on the application, primary design consid erations include:

• Weight

• Cost

• Pressure capacity

• Storage capacity

• Safety and durability

First, consider a metallic, thin-walled cylindrical vessel

For preliminary design/analysis, and for today’s di scussion, we will restrict ourselves to the following conditions and assumptio ns:

• Vessels are thin-walled (t < R/10)

– Stresses are uniform through the wall thickness (membrane loading, no bending)

– Stress normal to the wall thickness is much less than membrane stresses

Design and Analysis Considerations

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• Material (typically steel or aluminum) is elastic-perfectly plastic

– von Mises yield criterion applies

• We will consider the cylinder portion only

– End closures (domes) are beyond today’s scope

Note that the vessel is axisymmetric about cylinder axis

Applied pressure loading is also axisymmetric

σ

σy

ε

0=− pRLtLhσR

p

Equibrium in Hoop and Axial Directions

Notation

p, internal pressure

R, radius

t, wall thickness

σ, stress

h, hoop direction t

pRh =σ

Sum forces in horizontal (vertical) direction:

σh

σh

p

Consider a slice of length L:

4

p( ) ( ) 02 2 =− RptRa ππσ

a, axial directiont

σa

t

pRa 2

=σ

Sum forces in horizontal direction:

Summary of Stresses

Biaxial state of stress

• Normal stress in the hoop direction

• Normal stress in the axial direction

• No shear stress

t

pRh =σ

t

pRa 2

=σ

σσσσh

σσσσa

σσσσa

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• No shear stress

• Therefore, σh and σa are principal stresses

0=ahτ

• von Mises yield criterion in two dimensions:

• Failure occurs when load line reaches the von Mises ellipse

222

2221

21

2

aahhy

y

σσσσσ

σσσσσ

+−=

+−=

σσσσh

σσσσa

σσσσy

σσσσy

σσσσy

σσσσy

Design Equation

+−

=

+

−

=

4

1

2

11

22

222

2

t

Rp

t

Rp

t

Rp

t

Rp

t

Rp fffffyσ

222aahhy σσσσσ +−=

Failure criterion:

Rp

Substitute for the hoop and axial stresses, set p = pf , and simplify:

6

t

Rp fy 2

3=σ

Example

Given:

• Tank Dimensions

– Diameter = 12 in.

– Length of cylinder section = 3 ft. = 36 in.

• Load

– Service pressure = 3600 psi

– Factor of safety against burst = 2.25

Pressure at failure:

• pf = (3600)2.25 = 8100 psi

( )( )6810033

2

3

==⇒

=

f

fy

Rpt

t

Rp

σ

σ

7

– Factor of safety against burst = 2.25

• Material is 4130 steel

– E = 30 x 106 psi

– Poisson’s ratio, ν = 0.25

– Yield stress, σy = 112,000 psi

– Weight density, ρ = 0.283 lb/in3

Determine the required wall thickness

000,11222==⇒

y

tσ

Weight of tank (cylinder section):

( )( )( )( ).144

36376.062283.0

2

lbW

W

tLRVW

==

==π

πρρ

.376.0 int = (< R/10)

Some Observations

t

pRh =σ

t

pRa 2

=σ

σσσσh

σσσσa

σσσσy

σσσσy

σσσσy

The hoop stress is twice as large as the axial stre ss

If we apply reinforcement in the hoop direction, ma ybe we can

• Reduce the tank load in the hoop direction

• Make hoop stress more nearly equal to the axial str ess

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σσσσy

Wrap the cylinder section with continuous fiber com posite in the hoop direction

• With all reinforcing fibers in the hoop direction, we will initially assume that the composite carries load in the hoop direction only

• Assume that the composite is linearly elastic to fa ilure

• Enable reduction in tank wall thickness

• Reduce the weight of the tank (improved structural efficiency)

Equilibrium Considerations

Notation: subscript w denotes the hoop-wrap reinfor cement

Equilibrium in the axial direction is unchangedt

pRa 2

=σ

( ) 0=−+ pRLLtt wwh σσ

⇒ tσ h + twσ w = pR

σh

σw

Equilibrium in the hoop direction:

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⇒ tσ h + twσ w = pR

The problem is now statically indeterminate

To determine how the hoop load is divided between t he tank and the wrap, we need an additional equation: hoop strain in tank and wrap must be equal

( )ahh Eυσσε −= 1

ww

w Eσε 1=

As long as the tank has not yielded, Hooke’s law ap plies:

⇒1E

σ h −νσ a( )= 1Ew

σ w

p

σhσw

Solution to the Elastic Equations

t

pRa 2

=σ tσ h + twσ w = pR 1

Eσ h −νσ a( )= 1

Ew

σ w

⇓

σ w = pR − tσ h

tw w

h

wh t

tpR

Et

pR

E

σνσ −=

−⇒

⇓

1

2

1

and σ =Et + ν

2Ewtw pRpR=σ σ = pR − tσ h

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These equations are valid as long as the wrap has n ot failed:

σw < σ fw

and the tank has not yielded:

σ vm = σ h2 −σ hσ a + σ a

2 < σ y

and σ h = 2Et + Ewtw

pR

tt

pRa 2

=σ σ w = pR − tσ h

tw

(equilibrium ) (equilibrium ) (Hooke’s Law )

Following Tank Yield

Once the tank has yielded (assuming the wrap is sti ll intact):

t

pRa 2

=σ still, from axial equilibrium

But now the tank hoop stress is obtained from the y ield criterion

σ y2 = σ h

2 −σ hσ a + σ a2 ⇒ σ h

2 −σ hσ a + σ a2 −σ y

2 = 0

Solve for the hoop stress using the quadratic formu la:

( )−−+ 4 222 σσσσ 2

1 pRpR

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And get the stress in the wrap from equilibrium in the hoop direction:

σ w = pR − tσ h

tw

Note from the equation for σσσσh, the value inside the square root must be ≥ 0:

yy

pRt

t

pR

σσ

4

30

234

22 ≥⇒≥

−

( )⇒

−−+=

2

4 222yaaa

h

σσσσσ

−+=2

2

234

22

1

t

pR

t

pRyh σσ

What’s it All Mean?

p

σσσσaσσσσw

σσσσhσσσσvm

σσσσa

σσσσh

σσσσy

σσσσy

σσσσy

σσσσy

Before wrap

After wrap

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As the pressure increases up to tank yield, hoop st ress and axial stress in the tank, as well as the stress in the wrap, increase linearly.

After the tank yields:

• Axial stress in the tank continues to increase linearly (axial equilibrium)

• Hoop stress decreases, keeping σh and σa on the von Mises ellipse (yield criterion)

• Stress in the wrap must increase at a faster rate (hoop equilibrium)

Two possible failure modes:

• Failure of the wrap (preferred)

• Failure of the tank

Example

Given:

• Tank Dimensions

– Diameter = 12 in.

– Length of cylinder section = 3 ft. = 36 in.

• Load

– Service pressure = 3600 psi

– Factor of safety against burst = 2.25

• Material is 4130 steel

Minimum tank thickness

( )( ).188.0

000,112

68100

4

3

4

3in

Rpt

y

f ==≥σ

Set t = 0.200 in.

At burst pressure, ( )( )

( ) psit

Rp fa 500,121

200.02

68100

2===σ

Tank has yielded; use post-yield equations for σσσσ and σσσσ at burst pressure:

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– E = 30 x 106 psi

– Poisson’s ratio, ν = 0.25

– Yield stress, σy = 112,000 psi

– Weight density, ρ = 0.283 lb/in3

• Hoop wrap with T300 carbon fiber/epoxy

– Ew = 22 x 106 psi

– Failure stress σfw = 270,000 psi

– Weight density, ρω = 0.056 lb/in3

Determine thickness of tank and wrap

for σσσσh and σσσσa at burst pressure:

psit

Rp

t

Rp fy

fh 121,99

234

22

12

2 =

−+= σσ

.107.0 intRp

tt

tRp

fw

hfw

w

hffw =

−=⇒

−=

σσσ

σ

Weight of tank (cylinder section):

( )( )( ) ( )( ) ( )( )( )

.85

107.0056.0200.0283.03662

2

lbW

W

ttRLW ww

=+=

+=π

ρρπ

Closing Comments

Governing equations are readily entered into a spre adsheet

• Examine the effects of different material selections, thicknesses, tank geometries

• For a given tank radius, material selection, and pressure requirement find the thicknesses that give optimum design (minimum weight)

Minimum weight design is achieved by enforcing the condition that both tank and wrap fail simultaneously at the required burst pressure (82 l b. in previous example)

Additional considerations

• Compare pressure at which tank yields to service pressure (5618 psi vs. 3600 psi in previous example)

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example)

– Autofrettage (intentionally pressurize beyond the elastic limit) to increase the elastic range and improve fatigue strength

• Thermal effects: tank and wrap have different coefficients of thermal expansion

– Processing-induced residual stresses due to elevated temperature composite cure

– Operation at elevated and reduced pressures

The next steps to reduce weight :

• Fully-wrapped with load-sharing metallic liner

• Fully-wrapped with plastic liner

Filament Winding of Cylindrical Tanks

Continous fibers under tension are applied to a mandrel

Carbon fiber bundles (tows) consist of 12k fibers provided on a spool; four tows constitute a band

Wet-winding vs. pre-preg winding

Helical layers cover cylinder plus domes

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Helical layers cover cylinder plus domes

Hoop plies cover cylinder section only

Oven curing to harden the resin follows winding

Mandrel doubles as an impermeable barrier to the stored high-pressure gas

Prototype Testing

Hydroburst test conducted remotely

Painted white with stripes for visibility

Instrumented with strain gages on overwrap and domes

Filmed with high-speed camera

Burst at 4530 psi (90% of target)

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Failure initiated in overwrap near one end

Both cells ruptured in cylinder region, extending to the dome

Examination of carcass revealed some waviness in fibers of overwrap

Safety Testing

Safety requirements dictated by industry standard A NSI-NGV2

• Operating pressure

– 3600 psi for CNG, 5000 psi for hydrogen

• Burst pressure

– 2.25 safety factor

• Cycle life

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• Bonfire

• Bullet impact

• Pendulum impact

• Drop

• Environmental exposure

• Permeation

• Flaw tolerance