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SLAB Analysis Methods for Metal Forming-1

By jiangshutong Oct 25, 2014 2035 Words
Centre for
Advanced
Composite
Materials

Part 4 - Slab Analysis (Force Balance) Method-1
Block Compression
MANUFACTURING AND INDUSTRIAL PROCESSES
(MECHENG 747)

Dr. Raj Das

Module Objectives
• Introduction of slab analysis (or force balance)
method for compression of a block under plane
strain
• Calculation of pressure distribution, average
pressure and total force
• Role of sliding and sticking friction on the
pressure and forces during compression of a
block

Slab Analysis (or Force Balance) Method



Slab analysis in forging


Compression of a block



Axi-symmetric compression



Slab Analysis in rolling



Slab analysis in drawing/extrusion


Sheet drawing



Rod drawing

Slab Analysis



Slab analysis is an improvement on the ideal work method.



It takes into account frictional work by estimating it.



This method works by doing a total force balance.





Determining redundant work is difficult, hence use of an efficiency term is still required.
Slab analysis also underestimates the force (or stress/pressure) to carry out a bulk deformation process. Hence it is also a lower bound to the actual force (or stress) required.

Slab Analysis
The slab analysis is based on making a force balance on a differentially thick slab of material. It is useful in estimating the role of friction on forces required in drawing, extrusion and rolling. The important assumptions are :

1. The direction of the applied load and planes perpendicular to this direction define principal directions, and the principal stresses do not vary on these planes.
2. Although effects of surface friction are included in the force balance, these do not influence the internal distortion of the metal or the orientation of principal directions.
3. Plane sections remain plane; thus the deformation is homogeneous in regard to the determination of induced strain.

Plane Strain Compression of a Block
Slab Analysis can also be illustrated by the following case, which is the compression of a block under plane strain deformation conditions. Figure shows such a situation, wherein it is assumed that h < b, and sliding (Coulomb) friction with a constant friction coefficient µ exists at the interfaces. The width of the block is large and is assumed to not change in dimension.

Sliding Friction Compression of a Block
A force balance on the slab of unit width in the x direction gives

 x h  2 Pdx  ( x  d x )h  0
2  Pdx  hd x

(1)
(2)

where σx and P (which is equal to – σy) are principal stresses. Now apply Tresca yield criterion, (σx - σy) = 2k (= Y) where Y is the yield stress in uniaxial tension, while k is the yield stress in pure shear. [Note that σx > σz > σy ]

Sliding Friction Compression of a Block
Note: As per the von Mises criterion applied to Flow Rules, in terms of principal stresses and strains,
1


d  3  d   3   1   2  
2



where d  

d



If dε3 = 0 (i.e. plane strain),

3 

1
 1   2 
2

Sliding Friction Compression of a Block
In the present instance, x, y and z are assumed to be principal directions, and dεz = 0

1

Therefore,  z  2  x   y 
We know that σx is +ve and σy is –ve. (σy is also the largest stress in terms of

influence, and is equal to –P).
Therefore, σz will be the intermediary stress (and will be negative). i.e.

x  z  y

Sliding Friction Compression of a Block
If we use the von Mises yield criterion, we would still get

 x   y  2k , but 2k  2Y
Hence,  x  P  2k

3

 x  2k  P

Differentiating, we get, d x   dP

(3)
(4)

Sliding Friction Compression of a Block
Therefore, Eq can be re-written as

2  Pdx   hdP

(5)

dP 2 

dx
P
h

(6)

Integrating eq. (6), with the limits as follows:
• At any distance x = x, Pressure P = P
• At x = b/2, σx = 0 and therefore P = 2k [from eq. (3)]
We get,

x 2 
dP

2k P b / 2 h dx
P

ln P  ln(2k ) 

2 
b
(x  )
2
h

Sliding Friction Compression of a Block
b  2  b
 P  2  

Therefore, ln  2k   h  x  2   h  2  x   




2  b

Re-writing,



 x
P
 e h 2 
2k

(7)

This equation applies from x = 0 to x = b/2. The maximum value of P is at the centerline, where x = 0. If we put x = 0 in the above equation, we get, b
Pmax
 e h (3.8)
2k

Note: Pmin is at x = b/2.

Pmin  2k

b

Pmax  2ke h

Friction Hill in Sliding Friction

Figure: Illustration of the friction hill in plane-strain compression for different values of the coefficient of friction.

Average Pressure in Sliding Friction
We can also determine an average pressure value.
We know that P  2ke
2  b

Therefore

2  b 
 x
h 2 



 x
P
h 2 
e
2k

Average P can be determined by integrating from 0 to b/2 and dividing by b/2.
2  b 
b /2
 x
1
2 b /2
h 2 
Pa 
Pdx   2ke
dx
b 0
(b / 2) 0
Let a = b/2 and c = 2μ/h in the following derivation.

1 a
1 a
2keca
c a  x 
Pa   Pdx   2ke
dx 
a 0
a 0
a



a

0

e  cx dx

Average Pressure in Sliding Friction
2keca
Pa =
a
Thus,

2kh  hb 
Pa 
e  1
b 


Now, e

b h

So,

2keca  e  ca 1 
 
Pa 
a  c
c

x=a

 -e-cx 
 c 

 x=0

1  1 

b
h

 b h 

2

2

 ...  1

(10)

 b

 ...
Pa  2k 1 
 2h

P a  b 
 1
2k  2h 

P a  b 
 1
2k  2h 

For small values of μb/h, higher
order terms can be ignored.
(11)

Problem 1
A piece of metal 40 mm x 62.5 mm x 150 mm, having a yield stress of 7 MPa in shear, is to be press forged between flat dies to a size of 20 mm x 125 mm x 150mm.
The co-efficient of friction between the metal and the dies is 0.20. Determine the pressure distribution and the total forging force F. Assume sliding friction exists between the dies and the workpiece.

Solution 1

Given Information:
Final dimensions: h = 20 mm, b = 125 mm, w = 150 mm
Plane strain compression (no change in w)
k = 7 MPa, μ = 0.2, sliding friction
To determine:
Pressure distribution
Total forging force

Solution 1
Pressure distribution:
2  b



2*0.2  125 
x

20  2


 x
P
 e h 2 
2k

P
e
2*7

P
 e1.250.02 x
14

P  14e1.250.02 x ( MPa )

Note: Use of correct units and consistency in units is very important. Pressure or Stress: Pascal and its multiples
Force: Newton and its multiples
1 MPa = 1 MN/m2 = 1 N/mm2

Solution 1
Maximum Pressure:
b
Pmax
eh
2k

Pmax  2*7 * e

0.2*125
20

Pmax  48.86 MPa

Average Pressure:

Minimum Pressure:

Pa
h  hb 

 e  1
2k  b 


Pmin  2k  2*7  14 MPa


20  0.2*125
20
Pa  2*7 *
e

1


0.2*125 


Pa  27.89 MPa
Forging Force:
F  Pa bw  27.89*125*150  522.972 N

Solution 1

3.49
μ=0.2

Sticking Friction Compression of a Block








There is a limit to which sliding friction can prevail at the workpiece-die interface.
At the limit, the workpiece sticks to the die, and is sheared as compression (forging) takes place.
In this case, the frictional stress μP is replaced by the yield shear strength of the workpiece metal, k.
Following a similar analysis to that used for sliding friction, with sticking friction we get,

 x h  2kdx   x  d x  h  0
From this,

(12)

2kdx  d x h  0

(13)

Sticking Friction Compression of a Block
Hence,

2kdx  hd x

As before, d x  dP
Therefore, 2kdx  hdP

dP 

2kdx
h

(14)

Once again, the limits are as follows:
At x = x, P = P
At x = b/2, σx = 0, P = 2k (= Y)
Integrating eq (14) between the above limits, we get,
P
x
2k
d
P


2 k
b / 2 h d x

 P 2 k 
P

2 k
x
 x b /2
h

Sticking Friction Compression of a Block
P  2k 

2 k 
b  2k  b

x 
  x
h 
2 h 2


 b

x

2k  b

P  2k    x   2k  1  2

h 2
h





b
x
P
2
 1
2k
h





(15)

According to this equation, P varies linearly from the outer edge to the centreline, as shown in the figure in the following slide.
The maximum value of P occurs at the centreline (x = 0), and is b 

Pmax  2k  1  
 2h 

(16)

Sticking Friction Compression of a Block
We know that P is minimum at x = b/2.
Pmin = 2k
With linear variation of P,
Average Pressure
Therefore,

Simplifying,

Pa 

Pmax  Pmin
2


b 

 2k   1  2h    2k


Pa  
2

b 

Pa  2k 1  
 4h 

(17)

Sticking Friction – Friction Hill

Pmax

Problem 2
Consider Problem 1 with sticking friction conditions.
Solution 2
The corresponding results are
P = (57.75 – 0.7x) MPa
Pmax = 57.75 MPa
Pmin = 14 MPa
Average pressure Pa = 35.88 MPa
Forging force F = 672,656.3 N
Note that both pressure and force are greater, compared to sliding conditions. This is to be expected, as the workpiece shears at its top and bottom boundaries as it deforms. There is hence a bigger pressure (and force) required to perform the same forging operation.

Mixed Sliding – Sticking Condition
In a real situation, both sliding and sticking can occur. Hence the forging would be partially under sliding and partially under sticking conditions.
When pressure is high, sliding stops and sticking starts. At the cross-over region,  P  k





Let the cross-over be at a distance xc from the centre line. Recall that for sliding friction,

P  2k e

2  b 
 x
h 2 

P  k

At x = xc,  P  k and so P 
At this distance (x = xc),

k

k



 2k e



2  b

  xc 
h 2


Mixed Sliding – Sticking Condition
2  b



  xc 
1
 e h 2 
2

 1  2  b

ln 
  xc 


 2  h  2
h  1  b

ln 
    xc 
2  2   2


 xc 

b h  1 
ln 


2 2  2 

Friction Hill



Sticking friction would exist from x = 0 to x = xc.
Sliding friction would be present from x = xc to x = b/2.

Problem 3
Figure shows a billet before and after hot forging from an initial size of 2.5 mm x 2.5 mm x 25 mm to final size of 5 mm x 1.25 mm x 25 mm. This is accomplished by using a flat-face drop hammer. Sticking friction can be assumed. For the rate of deformation and the temperature, a flow stress of 18 MPa can be assumed.

a) Find the force necessary.
b) Find the ideal work required.
c) What is the actual work and hence, what is the efficiency?

Solution 3
Given:
Final dimensions: b = 5 mm, h = 1.25 mm, l = 25 mm (no change, plane strain) Sticking friction
Flow stress σf = 18 MPa
Determine:
(a)
Forging force
(b)
Ideal work
(c)
Actual work and efficiency
Flow stress σf = 18 MPa = 2k (Tresca), where k is the yield stress in pure shear. k = 9 MPa (as per Tresca)
If we use von Mises, 2k = 2Y/√3, k = Y/√3 = 10.39 MPa
(We can use either; we will go with Tresca for convenience). Average Pressure

b 

Pa  2k 1  
 4h 

5 

Pa  2*9 1 
  36 MPa
 4*1.25 

Solution 3
Forging force F  Pa bl  36*5* 25  4500 N
Ideal Work per unit volume = wi


wi    d   
0

z  0
x  y z  0
 y   x



2
2 2
2 2
 x   y2   z2  
 x  ( x ) 2   x


3
3
3
 5 
  0.693
 2.5 

 x  ln 

   f  18MPa

Solution 3
wi     18*

2  0.693
 14.41 N .mm / mm3
3

Wi  wi  bhl  14.41 5  1.25  25 N .mm
Wi  2251.1 ( N .mm)  2.25 N .m  2.3 Joules

[Wi is the total ideal work, if we wish to determine this].
Actual work per unit volume = wa=Pa*ε
wa  Pa   36 
w

2  0.693
 28.82 N .mm / mm3
3

14.41

Efficiency:   wi  28.82  0.5  50%
a
[If we use k = 10.39 MPa (von Mises), Pa = 41.56 MPa, wi remains the same, η = 43.3%].

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