Transcript Shape Rolling - Faculty Of Engineering And Technology

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Chapter 2
Fundamentals of the
Mechanical Behavior of
Materials
Types of Strain
FIGURE 2.1 Types of strain. (a) Tensile, (b) compressive, and (c) shear. All deformation
processes in manufacturing involve strains of these types. Tensile strains are involved in
stretching sheet metal to make car bodies, compressive strains in forging metals to make
turbine disks, and shear strains in making holes by punching.
Universal testing machine
Tension Test
Figure 2.2 (a) Original and final shape of a standard tensile-test specimen. (b) Outline of a
tensile-test sequence showing stages in the elongation of the specimen.
Understanding shear
Definition
Not required
Imperfections in engineering strain
First case ln(2Lo/Lo)=ln(2)
Second case ln(Lo/2Lo)=ln(1/2)=-ln(2)
ln(2)=0.6931
ln(1.5)+ln(2/1.5)=0.6931
Stess in three dimensions
n is a vector, so tn is the stress on a plan with
n as orthogonal to it
Special case n= x,
There are other cases where n=y, n=z
Thus we have 9 stress vectors
Hydraustaic and deviatoric stress
Homework: prove the above relation
The plastic deformation, the plastic deformation
Causes zero volume change
Prove the above relation for
A unit cube
Mechanical Properties of Materials
METALS (WROUGHT)
E (GPa)
Y (MPa)
UTS (MPa)
ELONGATION
(%) in 50 mm
45-5
65-3
50-9
21-5
40-30
60-5
65-2
60-20
25-7
0
POISSONÕS
RATIO (
0.31-0.34
0.33-0.35
0.43
0.29-0.35
0.32
0.31
0.28-0.33
0.28-0.30
0.31-0.34
0.27
Alumi num and its alloys
69-79
35-550
90-600
Copper and its alloys
105-150
76-1100
140-1310
Lead and its alloys
14
14
20-55
Magnesium and its alloys
41-45
130-305
240-380
Molybdenum a nd its alloys
330-360
80-2070
90-2340
Nicke l and its alloys
180-214
105-1200
345-1450
Steels
190-200
205-1725
415-1750
Stainless steels
190-200
240-480
480-760
Titanium a nd its alloys
80-130
344-1380
415-1450
Tungsten and its alloys
350-400
550-690
620-760
NONMETALLIC MATERIALS
Ceramics
70-1000
140-2600
0
0.2
Diamond
820-1050
Glass and porcelain
70-80
140
0
0.24
Rubbers
0.01-0.1
0.5
Thermoplastics
1.4-3.4
7-80
1000-5
0.32-0.40
Thermoplastics, reinforced
2-50
20-120
10-1
Thermosets
3.5-17
35-170
0
0.34
Boron fiber
380
3500
0
Carbon fibers
275-415
2000-5300
1-2
Glass fibers (S, E)
73-85
3500-4600
5
Kevlar fibers (29, 49, 129)
70-113
3000-3400
3-4
Spectra fibers (900, 1000)
73-100
2400-2800
3
Note: In the upper table the lowest values for E, Y, and UTS and the highest values for elongation are for the
pure metals. Multiply GPa by 145,000 to ob tain psi, and MPa by 145 to obtain psi. For example, 100 GPa =
14,500 ksi, and 100 MPa = 14,500 psi.
Table 2.1 Typical mechanical properties of various materials at room temperature.
Loading and Unloading
FIGURE 2.3 Schematic illustration of
loading and unloading of a tensile-test
specimen. Note that during unloading, the
curve follows a path parallel to the
original elastic slope.
True Stress-TrueStrain Curves in
Tension
FIGURE 2.5 (a) True stress-true-strain curve in tension. Note that, unlike in an engineering
stress-strain curve, the slope is always positive, and the slope decreases with increasing strain.
Although stress and strain are proportional in the elastic range, the total curve can be
approximated by the power expression shown. On this curve, Y is the yield stress and Yf is the
flow stress. (b) True-stress true-strain curve plotted on a log-log scale. (c) True stress-true-strain
curve in tension for 1100-O aluminum plotted on a log-log scale. Note the large difference in the
slopes in the elastic and plastic ranges. Source: After R. M. Caddell and R. Sowerby.
Power Law Material Behavior
MATERIAL
Alumi num, 1100-O
2024-T4
5052-O
6061-O
6061-T6
7075-O
Brass, 70-30, annealed
85-15, cold-rolled
Bronze (phosphor), annealed
Cobalt-base alloy, heat treated
Copper, annealed
Molybdenum, annealed
Steel, low-carbon, annealed
1045 hot-rolled
1112 annealed
1112 cold-rolled
4135 annealed
4135 cold-rolled
4340 annealed
17-4 P-H annealed
52100 annealed
304 stainless, annealed
410 stainless, annealed
Note: 100 MPa = 14,500 psi.
K (MPa)
180
690
210
205
410
400
895
580
720
2070
315
725
530
965
760
760
1015
1100
640
1200
1450
1275
960

n
0.20
0.16
0.13
0.20
0.05
0.17
0.49
0.34
0.46
0.50
0.54
0.13
0.26
0.14
0.19
0.08
0.17
0.14
0.15
0.05
0.07
0.45
0.10
  K
n
K: strength coefficient
n: strain hardening coefficient
Table 2.3 Typical values of
K and n in Eq. (2.11) at
room temperature.
True Stress True Strain
Curves for
Various Metals
FIGURE 2.6 True-stress-true-strain curves in tension at room temperature for various metals.
The point of intersection of each curve at the ordinate is the yield stress Y; thus, the elastic
portions of the curves are not indicated. When the K and n values are determined from these
curves, they may not agree with those given in Table 2.3, because of the different sources from
which they were collected. Source: S. Kalpakjian.
Strain Rate Effects
m

  C
C
3
MATERIAL
TEMPERATURE,
psi x 10
MPa
Alumi num
200-500
12-2
82-14
Alumi num alloys
200-500
45-5
310-35
Copper
300-900
35-3
240-20
Copper alloys (brasses)
200-800
60-2
415-14
Lead
100-300
1.6-0.3
11-2
Magnesium
200-400
20-2
140-14
Steel
Low-carbon
900-1200
24-7
165-48
Medium-carbon
900-1200
23-7
160-48
Stainless
600-1200
60-5
415-35
Titanium
200-1000
135-2
930-14
Titanium a lloys
200-1000
130-5
900-35
*
Ti-6Al-4V
815-930
9.5-1.6
65-11
Zirconium
200-1000
120-4
830-27
-4 -1
* At a strain ra te of 2 x 10 s .
Note: As temperature increases, C decreases and m increases. As strain increases, C i ncreases
increase or decrease, or it may b ecome n egative within certain ra nges of temperature and strain.
Source: After T. Altan and F.W. Boulger.
m
0.07-0.23
0-0.20
0.06-0.17
0.02-0.3
0.1-0.2
0.07-0.43
0.08-0.22
0.07-0.24
0.02-0.4
0.04-0.3
0.02-0.3
0.50-0.80
0.04-0.4
and m ma y
Table 2.5 Approximate range of values for C and m in Eq. (2.16) for various annealed
materials at true strains ranging from 0.2 to 1.0.
C: strength coefficient
M: strain rate sensitivity exponent
Effect of temperature
Power Law Creep
One of the most common forms of plastic flow is
Power-Law Creep, given by the formula:
Strain Rate = C (Stress)n exp(-Q/RT)
Let's take each part of the formula in turn:
C is a scaling constant.
n means that the strain rate increases much faster than stress. Typically n is abou
Q is the activation energy required to get crystal dislocations moving. It's typicall
R is the Universal Gas Constant that turns up everywhere in physical chemistry. I
T is the temperature in degrees Kelvin. As T increases, Q/RT decreases and thus e
Barreling In
Compression
FIGURE 2.15 Barreling in compression of a
round solid cylindrical specimen (7075-O
aluminum) between flat dies. Barreling is caused
by interfaces, which retards the free flow of the
material. See also Figs. 6.1 and 6.2. Source: K.
M. Kulkarni and S. Kalpakjian.
Plane Strain
Compression
FIGURE 2.16 Schematic illustration of the
plane-strain compression test. The dimensional
relationships shown should by satisfied for this
test to be useful and reproducible. This test give
the yield stress of the material in plane strain, Y’.
Source: After A. Nadai and H. Ford.
Toughness
It is the Area under the stress strain curve
 d
Derive the instability point for plain stress case with equal stresses
Principle stresses
Since x, y, z are optional coordinates, then we can obtain
Certain coordinates where the shear stresses disappear
Example, the normal directions in simple tension
Two dimensional-plain stress case
Homework, (don’t submitted it, just solve it)
Find the maximum stress case for the simple tension
And
Find the principle stresses for the simple shear case
Not required
Plain strain transformation
Not required
Plane Strain
• A state of plane strain exits when the strains are confined
to a single plane, such as the x-y plane.
• This generally means that the stresses in the other
direction; eg., the z direction, are non-zero.
• Plane Strain occurs in thick sections that “constrain” out of
plane deformations
Transformations in plane strain
The state of plane strain at a point p is given by  x ,  y , g xy . Determine the
principal strains and the maximum in-plane shear strain and show the
orientations of the elements subjected to these strains. Also determine the
absolute maximum shear strain.
 x  0.003;  y  0.001; g xy  0.006;

 1 xx
 ij   2 g yx
 12 g zx

g xy
 yy
1
2 g zy
1
2
g xz 

1
2 g yz 
 zz 
1
2
=
 0.003  0.003 0
 0.003 0.001 0


 0
0
0
Solution
tan 2 q p
qp
g xy
x y
g xy
1.
atan
x y
2
q p  0.6245 rad  39.8
1.
-0.006
atan
(0.003-0.001)
2
Principal Strains
' x
' y
' x
x y
2
x y
2
0.004
2
=0.005162
’y=-0.001162
x y
.cos 2 .q
p
2
g xy
.sin 2 .q
p
2
x y
.cos 2 .q
p
2
g xy
.sin 2 .q
p
2
0.002 .cos -79.6º -0.006 sin -79.6º
2
2
Maximum Shear Strain
The absolute maximum shear strain is given by the largest of the three values
' x ' y
g test
' x
' y
0.005162 (0.001162)  0.006325


0
.
005162
= 



0.001162
gmax = 0.006325
Use diameters of circles!, Mohr’s circle plots g/2, so gis the diameter
(=2x radius).
Max In-Plane Shear Strain
x
tan 2 .q s
1
2
y
g xy
qs
1.
2
atan
y
g xy
 0.003 0.001 1
  a tan(0.3333)  0.1608rad
 0.006  2

q s  a tan 
=9.2º
g' xy
x
x y
.sin 2 .q
s
2
=-0.006325
g xy
.cos 2 .q
.2
s
2
Mohr’s Circle
• Plot normal strain on the x-axis
• Plot ½ the shear strain on the y-axis
• Solve as you would for plane stress problem
g/2

Strain energy
For the elastic region
For 3D case
For principle stress case
Plane Strain
• A state of plane strain exits when the strains are confined
to a single plane, such as the x-y plane.
• This generally means that the stresses in the other
direction; eg., the z direction, are non-zero.
• Plane Strain occurs in thick sections that “constrain” out of
plane deformations
Transformations in plane strain
The state of plane strain at a point p is given by  x ,  y , g xy . Determine the
principal strains and the maximum in-plane shear strain and show the
orientations of the elements subjected to these strains. Also determine the
absolute maximum shear strain.
 x  0.003;  y  0.001; g xy  0.006;

 1 xx
 ij   2 g yx
 12 g zx

g xy
 yy
1
2 g zy
1
2
g xz 

1
2 g yz 
 zz 
1
2
=
 0.003  0.003 0
 0.003 0.001 0


 0
0
0
Solution
tan 2 q p
qp
g xy
x y
g xy
1.
atan
x y
2
q p  0.6245 rad  39.8
1.
-0.006
atan
(0.003-0.001)
2
Principal Strains
' x
' y
' x
x y
2
x y
2
0.004
2
=0.005162
’y=-0.001162
x y
.cos 2 .q
p
2
g xy
.sin 2 .q
p
2
x y
.cos 2 .q
p
2
g xy
.sin 2 .q
p
2
0.002 .cos -79.6º -0.006 sin -79.6º
2
2
Maximum Shear Strain
The absolute maximum shear strain is given by the largest of the three values
' x ' y
g test
' x
' y
0.005162 (0.001162)  0.006325


0
.
005162
= 



0.001162
gmax = 0.006325
Use diameters of circles!, Mohr’s circle plots g/2, so gis the diameter
(=2x radius).
Max In-Plane Shear Strain
x
tan 2 .q s
1
2
y
g xy
qs
1.
2
atan
y
g xy
 0.003 0.001 1
  a tan(0.3333)  0.1608rad
 0.006  2

q s  a tan 
=9.2º
g' xy
x
x y
.sin 2 .q
s
2
=-0.006325
g xy
.cos 2 .q
.2
s
2
Mohr’s Circle
• Plot normal strain on the x-axis
• Plot ½ the shear strain on the y-axis
• Solve as you would for plane stress problem
g/2

*Prove the above relations
*prove that equation 21 and 22 are the same
(for the plain stress case)
Principle stresses in three direction
ses are easiest. Unfortunately, we can't add stresses, only forces, so we have
is c1. So F1 = S1c1. Similarly, F2 = S2c2 and F3 = S3c3. The force normal
Fn = S1c12 + S2c22 + S3c32
Furthermore, stress = force/area, but the area of the plane is one, so we have
Sn = S1c12 + S2c22 + S3c32
r stress can be a lot messier, if we do things the brute force way. Or we can d
Here we are looking in the plane of the normal and shear
an area of one and stress = force per unit area, we have F2 = Sn2 + Ss2. No
The total force F can be found from the three vectors F1, F2 and F3
F2 = F12
Sn2 + Ss2 = S12c12 + S22c22 + S
So we have:
Ss2 = F2 - Sn2 = F12 + F22 + F32 - (S1c12 + S2c22 + S3c32)2
= S12c12 + S22c22 + S32c32 - (S1c12 + S2c22 + S3c32)2
22c22 + S32c32 - S12c14 - S22c24 - S32c34 - 2S1S2c12c22 - S2S3c32c22
We regroup terms to get:
Ss2 = S12c12(1 - c12) + S22c22(1 - c22) + S32c32(1 - c32)
- 2S1S2c12c22 - S2S3c32c22 - S3S1c12c32
Now, since 1 - c12 = c22 + c32, we can rewrite the above as:
Ss2 = S12c12(c22 + c32) + S22c22(c32 + c12) + S32c32(c22 + c12)
- 2S1S2c12c22 - S2S3c32c22 - S3S1c12c32
Gathering terms, we get
2S1S2 + S22)c12c22 + (S22 - 2S2S3 + S32)c32c22 + (S32 - 2S3S1 + S12)
Ss2 = (S1 - S2)2c12c22 + (S2 - S3)2c32c22 + (S1 - S3)2c12c32
Triaxial Stress State


(+ve sense shown)
3D Principal – Triaxial Stress
 max   int   min
 3   2  1
3D Stress – Principal Stresses
The three principal stresses are obtained as the
three real roots of the following equation:
where
 3  I1 2  I 2  I3  0
I1   x   y   z
I 2   x y   x z   y z   xy2   xz2   yz2
I 3   x y z  2 xy xz yz         
2
x yz
2
y xz
I1, I2, and I3 are known as stress invariants as
they do not change in value when the axes are
rotated to new positions.
2
z xy
Stress Invariants for Principal Stress
I1   1   2   3
I 2   1 2   2 3   1 3
I 3   1 2 3
Zero shear stress on
principal planes
I1   x   y   z
I 2   x y   x z   y z   xy2   xz2   yz2
I 3   x y z  2 xy xz yz         
2
x yz
2
y xz
2
z xy
Derive the two dimensional principle stress case using the
Equations for the tri-axial stress case
Mohr’s Circle?
• There is no Mohr’s circle solution for problems of triaxial
stress state
• Solution for maximum principal stresses and maximum shear
stress is analytical
• Either closed form solution or numerical solution (or computer
program) are used to solve the eigenvalue problem.
Maximum Shear Stresses
Absolute max shear stress is the numerically larger of:
 max,3  
1   2
 max,2  
2
1   3
 max,1  
2
y’z’, abs max
1
y’z’
x’y’
2
3
Normal
Stress, 
2 3
2
3D Mohr’s Circle – Plane Stress
A Case Study – The two principal stresses are of the same sign
1


2
3
3D Mohr’s Circle – Plane Stress
A Case Study – The two principal stresses are of opposite sign
1
2
3


Example:
For the following state of stress, find the principal and critical values.
 120 50 0


 ij   50 80 0 MPa


0 0
 0
Tensor shows that:
z = 0 and  xz =  yz = 0
y
80 MPa
50 MPa
120 MPa
x
The other 2 faces:
y
80 MPa
x
120 MPa
0 MPa
0 MPa
0 MPa
z
0 MPa
z
3-D Mohr’s Circles
 max = 77 MPa
80
60
H
40
20
0
-20
-40
V
-60
-80
-25
0
25 50 75 100 125 150 175
Normal Stress (MPa)
Example: triaxial stress state, not plane stress
• Determine the maximum principal stresses and the maximum
shear stress for the following triaxial stress state. (+ve values
as defined in slide 1)
 20 40  30
  40 30 25 MPa
 30 25  10 
Solution

 x  xy  zx 


 xy  y  zy 
 zx  yz  z 



 20 40  30
 40 30 25 


 30 25  10 
MPa
  I1  I 2  I3  0
3
2
I1   x   y   z = 20 + 30 –10 = 40 MPa
I 2   x y   x z   y z   xy2   xz2   yz2 = -3025 MPa
I 3   x y z  2 xy xz yz         
2
x yz
= 89500 MPa
2
y xz
2
z xy
Solve
Results
 3  65.3MPa
 2  26.5MPa
 1  51.8MPa
 max  1 / 2(65.3  51.8)
 58.5MPa
Mohr’s circles
Shear (MPa)
y’z’, abs max=58.5
2=26.5
1= -51.8
3=63.5
Normal
Stress, (MPa
Safety Factor?
Not Required
If the stress state was determined on a steel crankshaft, made of forged
SAE1045 steel with a yield strength of 300 MPa, what is the factor
of safety against yield?
1.
Tresca Criterion: max= 58.5 MPa
FS 
Sy
2 max
300

 2.6
2(58.5)
2. Max Principal Stress Criterion: max= 63.5 MPa
FS 
Sy
2 max
300

 4.6
65.3
3. Von Mises Criterion:

NOT REQUIRED
 3  65.3MPa
 2  26.5MPa
 1  51.8MPa

1
2
2
2 1/ 2
e 
( 1   2 )  ( 2   3 )  ( 3   1 )
2
1
2
2
2 1/ 2
e 
(51.8  26.5)  (26.5  65.3)  (65.3  (51.8))
2

=103.31 MPa
Sy
300
FS 

 2.9
 e 103.3

Yield criterion
For simple tension
Tresca yield locus
Flow rules
Prove it
Ignoring the
Shear strains
Efficiency = ideal work / (total work)
Next lecture problems and solutions for two lectures
And tutorials
Using abacus, the program is a finite element software for
Modelling and analysis of Manufacturing processes
We will use it to understand the operations and its analysis