GIÁO TRÌNH KHAI THÁC PHẦN mềm TRONG GIA CÔNG KHUÔN mẫu chapter x sheet forming processes

Also the thickness, the loads and stresses are axisymmetric during forming... Chapter X: Sheet Forming Membrane Theory 3Examples Examples: q z C r0 0, sin R φ , sin r φ Surface BC: Surfa

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Chapter X: Sheet Forming (Membrane Theory) 1

Content of Membrane Analysis

• Basic Assumptions

• Static Equilibrium Equations

• Strain State

• Application 1: Hole Expansion

• Application 2: Drawing

• Application 3: Flaring

P

C

z

Surface Normal

r

f f

Basic Assumption 1

The sheet metal part is a surface of revolution, so that it is symmetric about a

central axis Also the thickness, the loads and stresses are axisymmetric during

forming

q

P

C

z

Principal radii of curvature:

,

r rθ φ

meridian

curve C

center of curvatures

sin

hoop plane

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Examples Examples:

q

z

C

r0

0,

sin

R

φ

, sin

r

φ

Surface BC:

Surface AB:

q

z

dq

sf

sq

For thin plastically deforming shells, the bending moments are negligible and

because of axial symmetry the hoop (σθ) and tangential (σφ) stresses are

principal stresses The stress normal to the surface can be neglected, so that

the resulting stress state is the one of plane stress

1

2

3 0

θ

φ

σ

=

Normal pressure is assumed small enough

as compared to the flow stress of the material

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P

Normal

f

dr

P’

90o

f

P

Normal

f

df

rfdf f

dr

Surface Normal

P’

A Useful Relation

A useful geometric

relationship: cos

dr

r dφ φ

φ =

PP ′ = r dφ φ

cos

dr PP

φ

′ =

Friction forces are neglected Only uniform pressure loads normal to the

surface (although small enough wrt the flow stress) and uniform edge tensions

tangential to the surface are allowed:

Pressure Load Edge Tension Load

( sf)0

z

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Basic Assumptions 4 & 5

Assumption 4: (Not used to derive equilibrium equations)

Work hardening is compensated by thinning of the sheet, so that the product of

flow stress times current thickness is constant: σf ⋅ = t Tf= constant

Assumption 5:

The Tresca flow condition is assumed to be applicable:

sf

sq

sf0

wo

rk-hard

Tf= sft

Tf = sf0 0t = sft

Tq= sqt

Tf= sft

Tf = sf0 0t = sft

Tq= sqt

I

VI V

IV

III

II

The Flow Criterion

Region Stress State Flow Condition

f f f f f f

T T T T T

T T T T

T T T T T

φ θ φ θ

φ θ θ

θ φ φ

θ φ θ φ

> > − =

> > = −

> > − =

: Force per meridian-width in hoop direction : Force per hoop-width in meridian direction

T T

θ φ

The entity T is also called

a force-resultant

max min

f

Recall:

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Static Equilibrium Equations (1)

A typical infinitesimal

membrane element

as a free-body

dq

r

T rq fdf

( +d )( +d ) Tf Tf r r dq

T rq fdf

T rf dq

z

p

r

dq hoop-plane T rq fdf

T rq fdf dq/2

meridian plane

dq/2

( T r dθ φ ) d 2 θ ( T r dθ φ ) d 2 θ

T r d dθ φ φ θ

=

Normal resultant of hoop-forces:

Radial resultant in hoop-plane of hoop-forces

(in normal direction)

meridian-plane

T rq fd d f q df

surface normal

z

f

T rq fd d sin f q f

rf

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Tφ dTφ r dr d φ T r dθ φ

Normal resultant of meridian-forces:

T rd dθ φ θ

=

T rdφ φ dT rdφ φ

2

d

T drdφ φ θ +

2

d

dT drdφ φ θ

+

2

d

T rdφ φ θ

Equilibrium in normal direction:

p ⋅ rφ ⋅ d φ ⋅ r d ⋅ θ − Tθ ⋅ rφ ⋅ d φ ⋅ d θ ⋅ φ − Tφ ⋅ r d ⋅ θ ⋅ d φ =

Cancelling out the term dφ dθ and pulling pout yields:

sin

T T

p

φ θ φ

φ

T T p

φ θ

θ φ

Or, using the geometric relation given

in Slide 02.002:

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Equilibrium of forces in the tangential plane to the shell surface yields:

( Tφ + dTφ) ( r + dr d ) θ − Tφ⋅ ⋅ r d θ − Tθ ⋅ ( rφ ⋅ d φ ) ⋅ d θ ⋅ cos φ = 0

T r dφ⋅ ⋅ θ dT r d T dr d dT dr d

T r d

φ

θ

+ ⋅ ⋅ + ⋅ ⋅ + ⋅ ⋅

− ⋅ ⋅ − Tθ⋅ ( r dφ⋅ φ ) ⋅ d θ ⋅ cos φ = 0

dTφ⋅ + r Tφ⋅ dr + dTφ⋅ dr − Tθ ⋅ r dφ ⋅ φ ⋅ φ =

Expanding:

Deleting dθ and cancelling terms:

cos

dr

r dφ φ

φ = Using from Slide 02.005

0

dTφ⋅ + r Tφ⋅ dr + dTφ⋅ dr − Tθ ⋅ dr =

Yields:

0

dT

Hence:

REMARKS:

1) As seen from the last equilibrium equation the stress distribution is a

function of radius ronly and is independent of the shape (rφ) of the shell

2) The external axial force F delivers the boundary stress as:

( )0

0 0

2 sin

F T

r

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Review of Membrane Model

The generalized stress distribution can be obtained from the Tresca flow

condition and the static equilibrium equation in the tangential direction:

0

dT

The tool pressure can be found from the static equilibrium equation in normal

direction: T T

p

φ θ

Remark: No dependency

on shape of shell!

Remark: Dependency on shape of shell!

0 0 0 0

0

f f f f f f

T T T T

T T T T T

T T T T

T T T T T

φ θ φ θ

φ θ θ

θ φ φ

θ φ θ φ

> > =

> > − =

> > = −

> > − =

Strain State (1)

2 1

σ α σ

=

Defining the plane stress by the

three principal stress components: σ σ1; 2 and σ3 = 0

We can introduce a stress ratio α:

The hydrostatic stress is found by: 1 2 3

3

h

σ σ σ

1 1 3

h

σ = + α σ

The deviatoric stress components are: 1 1 ( ) 1

1 2 3

h

σ ′ = σ − σ = − α σ

1

3

h

σ ′ = σ − σ = α − σ

1 1 3

h

σ ′ = σ − σ = − + α σ

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The strain increments are given as: d ε1; d ε2 and d ε3 = − ( d ε1+ d ε2)

Introducing the strain ratio β: 2

1

d d

ε β ε

= d ε3 = − ( 1 + β ) d ε1

From the flow rule:

1 2 3

d

λ

σ ′ = σ ′ = σ ′ =

1 2

σ ′ = σ ′

or:

d ε β d ε

⋅

=

2

α β

α

−

=

− and

2

β α

β

+

= +

by volume constancy

Hence having found Tθand Tφ, the stress-ratio α can be determined as: α = T Tθ φ

and using this stress-ratio, the strain-ratio β can be determined So, knowing

one of the strain components, the other components can be derived Also the

equivalent strain and equivalent stress (flow stress) can be determined:

( )2 ( )2 ( )2 1 2 2 3 3 1

1 2

σ =σ = σ −σ + σ −σ + σ −σ 

f

σ =σ −α α+ =σ +β +β +β

Similarly, the equivalent plastic strain

increment can be determined as: 12 22 32

2 3

d ε = d ε + d ε + d ε

d ε = d ε   + β + β   = d ε − α α + − α

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Application 1: Hole Expansion (1)

Note:

1) Both Tθand Tφare tensile

2) At the hole rim we have Tφ= 0

and Tθ>0, from which we can

conclude that Tθ> Τφ >0

3) Hence: Tθ = T f

4) From tangential equilibrium we find:

0

dT

0

f

dT

φ

f

φ φ

=

− − ln ( Tf − Tφ) = ln r + C

with Tφ= 0 at r = r i: Tφ = Tf  −  1 ( r ri )  

f

Tθ = T

Remark 1: The stress state varies from uniaxial tension at the edge of the hole

towards equal biaxial tension at the periphery for large radii

Remark 2: As the hole radius approaches zero r i 0, almost the entire shell

is in a state of uniform biaxial tension in which

f

Tθ = T ≈ Tφ

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Remark 3: The case of r i= 0 provides an approximate solution for hydraulic bulging:

Since α= 1 we obtain β= 1 Hence:

f θ φ

2 t

ε =ε = − ε

2 θ t

ε = ε = −ε But by definition:

( 0)

ln

ε =

So: t = t e0 εt = t e0 −ε

0

n

f

Checking the assumption of constant T f:

ε

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Application 2: Drawing (1)

Note:

1) At the outer rim we have Tφ= 0

and Tθ< 0, from which we can

conclude that Tφ> Tθ

2) Hence: Tφ– Tθ= T f

3) Tθ< 0 always 4) From tangential equilibrium we find:

0

dT

dr r

φ

+ =

f

φ

= − − Tφ = Tf ln r + C

with Tφ = 0 at r = r 0: Tφ = Tf ln ( r r0 ) Tθ = Tf   ln ( r r0 ) −  1 

Application 2: Drawing (2)

Remark 1: The given relations are valid if and only if Tφ> 0 > Tθ.

Remark 2: The stress resultant at the inner boundary is:

i

Tφ = T r r

( 0 )

f

Tθ = T   r r −  ≤  ln(r r0 )≤1 (r e0 )≤ ≤r r0

( 0 )

i

r ≥ r e

Remark 3: Note again that these results are independent of the shape of the die!

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Application 3:

Flaring

Note the following:

1) At the bottom outer rim we have Tφ = 0 and Tθ>

0, from which we can conclude that Tθ> 0 > Tφ

2) Hence: Tθ – Tφ = T f

3) Tθ> 0 and Tφ≤0 always!

4) From tangential equilibrium we find:

0

dT

dr r

φ

− =

f

φ

= Tφ = Tf ln r + C

( 0)

ln

f

Tφ = T r r Tθ = Tf   ln ( r r0) + 1  

with Tφ = 0 at r = r 0:

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