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Mefp 2

The document discusses the mechanics of metal working, focusing on plastic deformation and the analysis of stress distribution. It covers concepts such as stress tensors, yield criteria, and the principles of plasticity, including the von Mises yield surface and the relationship between stress and strain. The document also includes examples and calculations related to strain and metal sheet deformation under various conditions.

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nelsonosman2023
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0% found this document useful (0 votes)
40 views44 pages

Mefp 2

The document discusses the mechanics of metal working, focusing on plastic deformation and the analysis of stress distribution. It covers concepts such as stress tensors, yield criteria, and the principles of plasticity, including the von Mises yield surface and the relationship between stress and strain. The document also includes examples and calculations related to strain and metal sheet deformation under various conditions.

Uploaded by

nelsonosman2023
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF, TXT or read online on Scribd
You are on page 1/ 44

METAL

FORMING
PROCESSES

J D.C.
J Ellard
Borniface
@MUST 2021

2018/2019

1
Mechanics of Metal
Working

2
Mechanics of Metal Working
 Metal working occurs due to plastic deformation
which is associated with analysis of complex stress
distribution.
 In order to still analyse these stresses, the following
simplifications are made:
 Only (large) plastic strain is considered while
elastic strain is very small and can be neglected.
 Strain hardening is often neglected.
 Metal is considered to be isotropic and
homogeneous.

3
Mechanics of Metal Working
 Required theory of plasticity, and for plastic
deformation, a constant-volume relationship is
required.

 In metal working, compressive stress and strain are


predominant.

4
Mechanics of Metal Working

5
Mechanics of Metal Working
 The reduction of area in metal working
deformation is given by:

From the constant vol.


relation

6
Mechanics of Metal Working
Example
Determine the engineering strain, true strain
and reduction of area for
a. A bar which is doubled in length
b. A bar which is halved in length

7
Mechanics of Metal Working
Example

Tensile tests are carried out on two metal


sheets, (A) and (B), at 0°, 45° and 90° with
respect to the rolling direction. The specimens
were deformed until just past maximum load,
unloaded and their dimensions measured. The
values obtained were:

8
Mechanics of Metal Working

The initial width was 12 mm and the initial thickness


was 1.6 mm for both specimens.
a. Calculate the r-values (and the mean planar r) of
the two sheets.
b. Calculate the strain to maximum load of both
specimens. 9
Mechanics of Metal Working
c. Identify the sheet metal that would be best suited
for a forming operation dominated by stretching,
and the sheet metal for an operation dominated
by drawing. Please justify your answer.

10
Stress Tensor and Yield
Criteria

11
Stress Tensor
Stress
The stress (s) is defined as the force (F) divided by the
area (A):

But force is a vector and, for stress, so is the area:

12
Stress Tensor
Stress
Therefore the stress can be better defined as:

In matrix notation:

which is a shorthand for:

Subscript notation:

with the summation over the common index: 13


Stress Tensor
Stress

There are 9 components of stress: this can be


reduced.
Symmetry Stress Tensor

14
Stress Tensor
Stress

So we only need 6 independent components


to define the stress:

Principal Stresses
Because the components of force and area vectors
depend on the coordinate frame, so do the
components of stress.

15
Stress Tensor
Stress
Because stress is a real and symmetrical tensor,
there will always be coordinate frames where
the shear components become zero.

But we lose the information about the relationship


between the original coordinate frame and the
principal directions. 16
Strain
Strain is a measure of deformation.
Small deformations: where the maximum length
change (‘stretch’) is 0.01 to 0.05 of the original
length.

But the change in the vector between two points in a


material is also a vector:

17
Strain
Only the symmetrical part of J is (small) strain:

These relationships can be written in a general sense


as:

So we only need 6 independent components in the


case of the strain: 3 normal strains and 3 shear
strains

18
Strain rate
The time derivative of strain in the current
geometry is the strain rate:

The strain rate is defined as the symmetrical part


of the velocity gradient, L:

The antisymmetrical part of L is called the spin,


and is the rate of rotation:

As with stress and strain, principal values of strain


rate can be found. 19
An important ‘decomposition’
Stress, (small) strain and strain rate can be
decomposed into mean normal and deviatoric
components.
This decomposition is an additive one: just add
the mean normal and deviatoric parts together
and get the original quantities.

20
An important ‘decomposition’
NOTE: p, the pressure, is the negative of the
hydrostatic stress
These values are always the same (invariant)
for a given stress, strain or strain rate!!

NOTE: this subtraction only applies to


normal (direct) components!! Shear stresses
and strains are always already deviatoric. 21
An important ‘decomposition’
So, for example for stress:

Why is this ‘decomposition’ useful?


 Mean normal quantities relate to pressure
and changes in volume
 Deviatoric quantities relate to changes in
shape at constant volume
22
Yield Criteria

Yielding in unidirectional
tension test takes place when
the stress σ = 𝐹 𝐴 reaches the
critical value.

Yielding in
multiaxial stress
states is not
dependent on a
single stress but on
a combination of
all stresses. 23
Yield Criteria

Yielding occurs when the second invariant of the


stress deviator 𝐽2 > critical value 𝑘 2

In uniaxial tension, to evaluate the constant 𝑘, note


𝝈𝟏 = 𝝈𝟎 ; 𝝈𝟐 = 𝝈𝟑 = 𝟎; where 𝝈𝟎 is the yield stress;

Therefore

Substituting for 𝑘 24
Yield Criteria

In pure shear, to evaluate the constant 𝑘, note


𝝈𝟏 = −𝝈𝟑 = 𝝉𝒚 ; 𝝈𝟐 = 𝟎; where 𝝈𝟎 is the yield
stress; when yields: 𝝉𝟐𝒚 + 𝝉𝟐𝒚 + 𝟒𝝉𝟐𝒚 = 𝟔𝒌𝟐 then
𝝉𝒚 = 𝒌

By comparing with we
then have

25
Yield Criteria

Special case : Plane stress


Let 𝝈𝟏 , 𝝈𝟐 and 𝝈𝟑 be the principal stresses 𝝈𝟑 =0

Von Mises yield criterion becomes :

In 𝝈𝟏 − 𝝈𝟐 plane, this equation represents an ellipse


26
Yield Criteria

Von Misses yield surface for plane stress problems

27
Yield Criteria

Yielding occurs when the maximum shear stress, 𝝉𝒎𝒂𝒙


reaches the value of the shear stress in the uniaxial-
tension test, 𝝉𝟎 .

where 𝜎1 = Algebraically largest


principal stress
𝜎3 = Algebraically smallest
principal stress
For uniaxial tension, 𝝈𝟏 = 𝝈𝟎 ; 𝝈𝟐 = 𝝈𝟑 = 𝟎;
𝝈𝟎
and the shearing yield stress 𝝉𝟎 = 𝟐

28
Yield Criteria

Therefore the maximum shear stress criterion


is given by:

In pure shear, 𝝈𝟏 = − 𝝈𝟑 = 𝒌; 𝝈𝟐 = 𝟎 → 𝝉𝒎𝒂𝒙 = 𝝉𝒚

29
Yield Criteria

Special case : Plane stress


Let 𝝈𝟏 , 𝝈𝟐 and 𝝈𝟑 be the principal stresses 𝝈𝟑 =0
When 𝝈𝟏 and 𝝈𝟐 are of opposite sign

The yield condition is given by :

30
Yield Criteria

Special case : Plane stress


When 𝝈𝟏 and 𝝈𝟐 carry the same sign :

31
Yield Criteria

Tresca yield surface for plane stress problems

32
Yield Criteria
Comparison of yield criteria

The differences in the maximum shear stress


prediction from both criteria lie between 1-
15%.
However experiments confirmed that the
Von Mises criterion is more accurate to
describe the actual situations. 33
Yield Criteria
Comparison of yield criteria
Tresca and Von Misses yield surfaces : 2D space

34
Yield Criteria
Comparison of yield criteria
Tresca and Von Misses yield surfaces : 3D space

35
Plasticity
Plasticity is much more complicated, both in
concepts and in the mathematics involved,
than elasticity.
The Thermodynamic Basis
Plasticity is thermodynamically irreversible. When it
occurs, the great majority of the work done is
transformed to heat in a one-way process.

36
Plasticity
Some simplifications
1. Assume an isotropic material
2. Assume no change in volume by plastic
deformation
3. Use principal values for stress and strain rate..

37
Plasticity
The von Mises surface and the Levy- Mises Plasticity
relationships

= a 'constant' which depends on the current state of the


material. Dimensions: 𝑆𝑡𝑟𝑒𝑠𝑠. 𝑇𝑖𝑚𝑒 −1 .
Useful values that are associated with the von Mises surface:

38
Plasticity
The von Mises surface and the Levy- Mises Plasticity
relationships
Useful values that are associated with the von Mises
surface:

Calculate plastic strain rate from stress assuming von


Mises behaviour:
 Calculate deviatoric stresses
 Calculate λ (given directly or derived)
 Plastic strain rates are λ x corresponding deviatoric stress.
39
Plasticity
The von Mises surface and the Levy- Mises Plasticity
Relationships

An increment of plastic strain can be derived by just


multiplying the strain rate by a time step. ADD the
elastic strain to give a total strain increment!!
BUT: the state of the material (and λ) changes
(evolves) during plastic deformation.

Change in stress during plastic deformation


ASSUMPTION: the plastic potential size only
depends on the accumulated effective plastic strain

40
Plasticity
Change in stress during plastic deformation

41
Plasticity
Change in stress during plastic deformation
True strain
Considering the path length of a simple
tensile (or compression) test:

42
Plasticity
Plastic Power and Work
The rate of plastic work or plastic power:

If the material is isotropic:

and if it is von Mises:

Work is calculated by integrating power over time.


43
Plasticity
Temperature rise
Nearly all of plastic work (usually η >95%) goes to
heat.

Adiabatic assumption & stress remains close to


constant:

44

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