GIÁO TRÌNH KHAI THÁC PHẦN mềm TRONG GIA CÔNG KHUÔN mẫu chapter III strain

Chapter III: Strain 1Content • Engineering strain versus true strain • Strain increment • Strain rates • Equivalent strain rate and equivalent strain • Examples of application Engineerin

Chapter III: Strain 1

Content

• Engineering strain versus true strain

• Strain increment

• Strain rates

• Equivalent strain rate and equivalent strain

• Examples of application

Engineering Strain & True Strain

λ0

λ

λ1

intermediate

configuration

initial

configuration

final

configuration

dλλ

Chapter III: Strain 3

Mechanical engineering strain:

0

l Elongation is based to the initial length.

1 0

1 0

ln

∫

l

l

True (logarithmic, natural) strain:

Forming technology is related to the actual length, useful to

describe the large amounts of deformation

Remark on Plastic Strains

elastic plastic

The total strain can be splitted (phân tách)into:

In metalforming, plastic strains are much larger than elastic ones:

plastic

ε ≈ ε since

Chapter III: Strain 5

Examples

Example 3.1:

a) A uniform bar of length λ0 is uniformly extended until its final

length is λ = 2 λ0 Compute the values of engineering strain and

true strain for this extension

b) To what final length, λ, must a bar of initial length λ0, be

compressed if the strains are to be the same (except for sign) as

those in part (a)?

Example 3.2:

A uniform bar of 100 mm initial length is elongated to a length of 200

mm in three stages:

Stage 1: 100 mm to 120 mm, Stage 2: 120 mm to 150 mm and

Stage 3: 150 mm to 200 mm

Calculate the engineering and true strains for each stage and

compare the sums of the three with the overall values of the strains

Engineering Strain & True Strain

0 1

ln 1

Series expansion:

εεεε 0.0009995 0.00995 0.0198 0.0487 0.0953 0.182 0.405 0.693

Chapter III: Strain 7

Advantages and disadvantages resulting from the

Large strains lead to numerical values which are not easy to deal with

(True strain ɛ=1 equals percentage strain of about 173%.)

The logarithmic strain is mechanical more sensible as the strain related to

the initial length and present calculation advantages So in procedures

with several forming steps the total strain equals the summation of the

single strains.

In contrast to (ngược với) the strain the true strain take in consideration

not only the initial and the end state but also all intermediate steps of the

total forming

In the calculation with strains only, the single strains can not be added to

the total strain, because they are related to the initial length

Definition: True Strain

The point has initial coordinations: x, y, z, After forming it has current coordinations: x’, y’, z’.

Displacements of point following x, y, z directions are:

x’ - x = u x = u y’ - y = uy= v z’ - z = uz= w

Chapter III: Strain 9

True Strain Differentials

Similarly we can define the true strain differentials:

{ }

x

y

z

xy

zx

yx

zy

xz

d d d d

d d d d

ε ε ε ε ε ε ε ε ε

where:

( )

, etc

x

du d

x

∂

1

, etc

2

xy

d

Here: u, v and w are displacements in x,

y and z directions and x, y, z are current

coordinates

True Strain Differentials

T

ε

=

































1 2

1 2 1

2

1 2 1

2

1 2

ε

ij

xx xy xz

yx yy yz

zx zy zz

=





















Tensor of strain:

ε ∂

∂

ε ∂

∂

ε ∂

∂

x x

y y

z z

u x u y u z

=

=

=





















∂

∂ +

∂

∂

= γ

∂

∂ +

∂

∂

= γ

∂

∂ +

∂

∂

= γ

z

u x u y

u z u x

u y u

x z zx

z y yz

y x xy

Where:

Chapter III: Strain 11

Strain Rates

( )

x

du d

x

∂

Sometimes it is more convenient (thuận tiện) to introduce strain rates:

Consider the strain differential:

Dividing both sides by the differential time dt :

( )

x du d

dt

=

∂

x

du dt

ε

&

&

Sometimes the strain-rate tensor is called “Rate-of-Deformation”

tensor ε&ij = D ij

Strain Rates

( )

xx xx

du dt

ε

&

&

( )

yy yy

ε

&

&

1

1 2

2

xy

ε

& &

Similar to strain, we have terms of strain rates:

Chapter III: Strain 13

Strain Rate Tensor

& & &

& & & &

& & &

xx

u x

ε = ∂

∂

&

y

ε = ∂

∂

&

z

ε = ∂

∂

&

&

1 2

xz zx

ε = ε =   ∂ + ∂  

& &

& &

1 2

xy yx

ε = ε =   ∂ + ∂  

& &

& &

1 2

ε = ε =   ∂ + ∂  

& &

& &

Normal Strain Rate Components:

Shear Strain Rate

Components:

The Strain Rate Tensor in 3-D:

particle velocity in -direction particle velocity in -direction particle velocity in -direction

&

&

&

Examples

Example 2.3:

A uniform bar with current length λ is extended at its free end by a

tool with velocity of v tool Determine the strain rate in axial

direction

λ

v tool

Example 2.4:

A sheet with dimensions shown in

the figure is sheared by a tool

with velocity v tool Determine the

shear strain rate

v tool

y

x h

Chapter III: Strain 15

Strain Rates in Cylindrical Coordinate System

z

r

θ

P(r, ,z)θ

1 1 2

1 1 2 1 2

r

z

zr rz

u

θ

θ

θ θ

ε ε

θ

ε ε

θ

ε ε

& &

& & &

& &

& &

& &

& &

,

,

1

r

z

u u z

θ θθ

θ ε

∂

∂

& & & &

&

& &

θ

θ

& & &

& & & &

& & &

Strain Rates for Axisymmetrical Problems

Conditions of axial symmetry:

uθ

θ

∂ ⊗

∂

&

Using these conditions results:

1

ε = ∂ = ε = ε = ∂ =

& & & & & &

1

2

& &

Chapter III: Strain 17

Strain Rates in Spherical Coordinate System

,

,

,

cot sin

1

cot

1 1 2

rr r r

r

r

u

r

r

r

r

r

θθ θ θ

ϕθ θϕ θ ϕ θ ϕ ϕ

ε

θ ε

θ

θ

=



& &

& & & &

& & &

& & & & &

& & & & &

& & & & &



r

θ

P(r, ϕ,θ )

ϕ

Strain Rates in Spherical Coordinate System:

Symmetry

For symmetry with: uϕ uθ 0 and

ϕ

∂ ⊗

∂

& &

,

,

,

0

2

r

r

u u

r

u r

θ

&

& & & &

& & & &

&

& &

Chapter III: Strain 19

Total Finite Strains

Total finite strains have physical meaning if and only if:

1 All shear strain rates are zero,

2 The straining path is straight,

0

ε & = ∀ ≠

constant, etc.

xx yy

ε & ε & =

λ

v tool dx

x Assuming uniform straining:

tool

x

u & = v

l

( )

{

tool xx

u x

v

u du

x dx

∂

&

& &

&

d v

dt

= l but so that xx d dt

ε & = l

l

Integration:

0

ε = ∫ ε = ∫ = ∫l ⇒

l

&

ln

xx

ε = l

l

Principal Strains

x

y z

1

2 3

element before deformation

element after deformation

ln ln ln

a a

b b

c c

ε ε ε

=

=

=

principal

strain

directions

Chapter III: Strain 21

Equivalent Plastic Strain

The hardening of metals in multidimensional strain states

is represented by the equivalent strain and its rate The

( 2 2 2 ) ( 2 2 2 )

2

2

ε & =   ε & + ε & + ε & + ε & + ε & + ε &  

The total equivalent strain is simply:

t

dt

ε = ∫ ε &

In terms of principal strains: ( 2 2 2)

2 3

Condition of Volume Constancy (1)

During plastic deformations volume of the deformed body

remains constant (experimental observations)

final initial

V = a × b × c = a × b × c = const

dt

Chapter III: Strain 23

Condition of Volume Constancy (2)

( 1 1 1)

0

a b c

1 1 1 1 1 1 1 1 1

dV

a dt b dt c dt

0 = ε & + ε & + ε & {

!

tensor

⇒ 0 = ε &xx + ε &yy + ε &zz

Expand (khai triển)

Examples (1)

Example 2.5:

Consider the frictionless

axisym-metrical upsetting process as

shown in the figure Determine the

equivalent plastic strain for the

process by assuming that there is

no bulging and the cylinder is

compressed from height h 0 to h 1

F, vtool

ur. r z D

h

lower die (fixed) upper die

(moving)

Chapter III: Strain 25

Equivalent Strain Rate in Upsetting

0

1

2

3

4

Height Reduction in %

v tool = 100 mm/s, h 0= 100 mm

0%

60%

40%

20%

Equivalent Plastic Strain in Upsetting

0

1

2

3

Height Reduction in %

v tool = 100 mm/s, h 0= 100 mm

0%

80%

60%

40%

20%

0 /

h h

∆

Chapter III: Strain 27

Examples (2)

Example 2.6:

Consider the frictionless

axisymmetrical drawing

process as shown in the

figure Determine the

equivalent plastic strain

rate for the process by

assuming that the axial

velocity component is a

function of the axial

coordinate only