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N07-UA-TheoreticalStrength.pdf
HOMEWORK
Module #7
Estimate the theoretical
shear and cleavage
strengths for tungsten.
Elastic constants can be
found are on p. 50 of
your text.
Plasticity and the Theoretical Strength of Materials
READING LIST
DIETER: Pages 117-119 and 243-245
Pages 44-53 in Meyers and Chawla, 1st edition
Ch. 1 in Strong Solids
When exposed to external forces,
materials will “break”
deformation
[i.e., change shape]
and/or
fracture
[i.e., separate into pieces]
Ways to “break” things

• Specimen breaks suddenly with little or no
distortion; consumes little energy.
• Brittle fracture.
• The fracture path is perpendicular to a
plane.
Uniaxial Normal
Stresses

CLEAVAGE


• Specimen breaks with lots of distortion
(i.e., deformation) a specimen; consumes
energy.
• Ductile fracture.
Shear Stresses

SHEAR
• The fracture path is parallel to a plane.

What is more likely,
cleavage or shear?
Theoretical Strength
• We can answer the question posed on the previous
viewgraph by estimating estimate the theoretical strengths
required to cleave or shear a “perfect” material (i.e., a
material that contains no defects).
• Theoretical strengths are related to:
– Interatomic forces
– Temperature
– State of stress
• For now, let’s ignore temperature effects and variations in
the state of stress.
“Perfect” Crystalline Solids
• Atoms, ions, or molecules are arranged in periodic, repeating,
symmetric patterns in three dimensions (i.e., atoms, ions, or
molecules occupy specific lattice sites and exhibit specific
symmetry relationships in their arrangement).
FCC
Fig. 1.7 from Hull & Bacon, 4th ed.
HCP
Fig. 1.8 from Hull & Bacon, 4th ed.
Cleavage
• Separation of all atomic bonds on a
plane perpendicular to an applied
stress.
F
A
do
• Polanyi1 and Orowan2 devised a simple
method to determine the theoretical
strength of a perfect crystal.
F
F
A
• Critical assumptions:
– The crystal contains no defects and no
stress concentrations at the crack tip.
– All atoms separate simultaneously once they
reach a critical separation distance d.
1M.
d
Polanyi, Z. Phys., v. 7 (1921) p. 323.
2E. Orowan, Rep. Prog. Phys., v. 12 (1949) p. 185.
F
Cleavage – cont’d

max
A
do
At rest: F, σ = 0
do
d
F
A
di
Force applied:
Bonds stretch;
F, σ > 0
A
At failure: F, σ = 0
d
F
• The stress required to separate two
atomic planes varies as a function of
the distance between planes.
• Orowan assumed a sinusoidal
variance for simplicity.
F
F
x
• The area beneath the curve
represents the work required to cleave
the crystal (i.e., the work of
deformation).
Work to Cleave
• Two new surfaces are created by cleavage. Work
is needed to create these surfaces. Thus there is a
surface energy associated with these surfaces.
– Let  = surface energy/area (= E/A)

– For the two surfaces created, the
total surface energy = 2A.
• The work to cleave must be greater than the surface
energy of the newly created surfaces.
• The stress dependence on planar separation can be
expressed as:

F
2
 K sin
( x  do )
2d
A
[1]
• We can find the constant K by relating the initial slope
of the force-distance curve to Young’s modulus.
• When x is close to do [i.e., when (x-do) is small], the
material responds in accord with Hooke’s law
(i.e., σ = E).


max
F
A
do
do
F
d
x
• If we assume that elastic deformation is restricted to
the two planes shown in the adjacent figure, and
that the material is isotropic, we can define the
incremental strain (dε) as:
F
dx
d 
do
• If we substitute this expression into
Hooke’s law:
d
d
d
E
dx / d o
or
d
E
do
dx
• Where E is Young’s modulus.
A
do
F

max
do
d
x
• Recall that:
  K sin

d
( x  do )
• We can determine K by taking the derivative of this
expression with respect to x and substituting it into
our expression for E at x = do.
d


do
 K d o cos ( x  d o )  E
dx
d
d
E d
K
 do
[2]
• The value of d, the interplanar spacing, is unknown
and must be estimated.
• Polanyi and Orowan equated the area under the stressdistance curve to the energy of the surfaces created:
do  d

 dx  2 
do
do  d

do
K sin

d
( x  d o )dx
• From calculus:
1
 sin ay dy   a cos ay
• Let y = x – do; then dx = dy.
a

d
 d

K  sin ydy  2  K   cos   cos 0 
d

 

0

2d
K
 2  d 
.
K

[3]
The theoretical cleavage stress is the, max, maximum value
of σ in equation [1] where we set the sine term equal to 1.
 max
E d 
K  
  do 
If we substitute our recently derived value for d (from the
previous slide) into this expression , we get:
 max  K 
E
Kd o
which can be re-written as:
K2 
E
2
  max  since  max  K .
do
E
do
  max 
The cleavage strain is:
 max 
d


2d o 2

Ed o

  max
2 E
In this model, the surface energy term can be expressed as:
Ka
E


do

2
d 
  .
 
The critical separation distance d is of the same order of
magnitude as do (i.e., d  do). THUS we can estimate the
surface energy as:

Ed o

2
Ed o
10

 max 
E
E

do

From this derivation we can conclude that the cleavage
strength at absolute zero (i.e., 0 K) will be highest in materials
with high values of E and , and low values of do.
Some typical values for the theoretical cleavage stress as calculated
from the derived expression are reproduced below. Table 1.1 in
Chapter 1 of Strong Solids has a more complete listing.
Direction
E
(GPa)
Surface energy
(mJ m-2)
max
(GPa)
Silver
<111>
121
1130
24
Silver
<100>
44
1130
16
Gold
<111>
110
1350
27
Nickel
<100>
138
1730
37
Tungsten
<100>
390
3000
61
-Iron
<100>
132
2000
30
Diamond
<111>
1210
5400
205
Al2O3
<0001>
460
1000
46
Substance
The largest sources of error are the surface energies (), which are
very difficult to measure accurately.
More accurate methods for determining max are summarized
Chapter 1 of Strong Solids (pp. 7-24).
Shear
•
The other way to break/deform a
material is via shear.
•
Frenkel3 has devised a simple method to
estimate the theoretical shear strength
of a perfect crystal.
•
•
b
d

Consider a crystal structure consisting of
two neighboring planes with a
separation distance d and an interatomic
spacing b.
Assume the individual planes do not
distort under an applied shear stress.
Ao
δ
F
d
θ

F
3J.

Frenkel, Z. Phys., v. 37 (1926) p. 572.


d
F
Ao
 tan 
y
b
At non-equilibrium positions, 
increases.  is maximum at ½b.
A

B
C
D
d
Potential
Energy
At each equilibrium position, the
potential energy  is minimum and
 = 0.

Shear
Stress
Under an applied stress, atoms will
pass sequentially through
equilibrium positions (i.e., A, B, C,
etc…).
x
Displacement, x
Plastic deformation will occur when the applied shear stress (app) is
large enough to overcome the potential energy barriers. At that point,
atoms will move from one equilibrium position to the next one.
This process is known as “slip.”
y
b
We can express this variation as:
2 x
.
b
A

B
C
d
D
Shear
Stress
  K sin

Potential
Energy
Between equilibrium positions, 
varies cyclically. Assuming the
variation is sinusoidal.
x
Displacement, x
In this equation K is a constant that describes the amplitude of the sine
wave and b is the period of the sine wave.
Recall: for small displacements, Hooke’s law (i.e.,  = εE or  = γG for
isotropic solids) applies leading to:
 K
2 x
x
 G  G   .
b
d

Shear
Strain
Ao
δ
If we can re-arrange the equation, and solve
for K.
K
b
2 x
G
x
Gb

d 2 d
F
d
θ

If we substitute this expression back into our
idealized expression for  we get:

F


d
F
Ao
 tan 
Gb
2 x
sin
2 d
b
Since τ is essentially the derivative of the Φ-x curve, τmax occurs when
x = b/4 (i.e., when the sine term → 1)
 max 
Gb
2 d
The relationships between energy and stress for shear yielding are more
clearly illustrated on the next viewgraph.
Fig. 6.15. Energy and stress curves for
rigid shear yielding.
[Copied from G. Gottstein, Physical
Foundations of Materials Science,
(Springer-Verlag, Berlin, 2004) p. 213]
Consider a face centered cubic (fcc) crystal. The relationship between the
lattice parameter and the interatomic spacing is given by:
b  ao
2
2 2 ao
 ao

2
2 2
2
The spacing between crystal planes in a cubic
system is given by:
d hkl 
ao
ao
b
ao
ao
b is the
distance
between
atom
centers on
lines where
they touch!
h2  k 2  l 2
For fcc crystals, the close packed plane, which is the most likely plane
where shear will occur is (1 1 1).
d111 
ao
3
Substituting the expression for d111, the spacing between (111) planes, in
for d in the original equation and our new expression for b into the equation
for max:
 max 
G
2
2ao 3
G

2 ao 5.13
More refined methods of calculation are described in Chapter 1 of Strong
Solids.
Some typical values for the theoretical shear
stress (adapted from p.29 of Strong Solids).
max
G
(GPa)
max/G
(GPa)
Copper (10K)
33.2
0.039
1.29
Copper
30.8
0.039
1.2
Gold
19.0
0.039
0.74
Silver
19.7
0.039
0.77
Cobalt
84
0.039
3.49
Aluminum
23
0.039
0.9
Aluminum
23
0.114
2.62
Nickel
62
0.039
2.4
Nickel
62
0.114
7.1
Silicon*
57
0.24
13.7
α-Iron
60
0.11
6.6
Tungsten
150
0.11
16.5
Al2O3 (basal)
147
0.115
16.9
Zinc
38
0.034
2.3
Graphite
2.3
0.05
11.5  10-2
NaCl
18
0.159
2.9
Material
*For Si, the same value of max/G has been taken as for diamond.
Therefore the estimate for Si is likely high.
If we compare max and
max, we find that max is
two or more times
larger than max.
What this means is that
materials are more
likely to deform via
shear than to cleave.
This isn’t always true.
Just most of the time.
Slip bands:
Composed of
several slip
lines/traces
100’s of Å
1000’s of Å
Figure
Slip in a zinc single crystal.
[Copied from W.D.
Callister, Jr, Materials
Science and Engineering:
An Introduction, 7th Ed.,
(Wiley, New York, 2007)
p.183.]
In ductile crystalline
solids, proof that slip (i.e.,
planar shear) occurs is
provided by the
appearance of slip lines
on the surfaces of single
crystals or individual
grains.
I’ve schematically
illustrated slip lines above
and provided a picture of
the real thing. We’ll
discuss their
crystallographic nature a
little later.
How do theoretical strengths compare with real
strengths?
ESTIMATE
In most cubic crystals, b  d
(i.e., they are of the same order of magnitude).
 max
Gb
G
E
E




2 d 2 4(1   ) 17
 max 
E

Theoretical and experimental yield/shear strengths for various Materials
Theoretical Shear Stress
Experimental Shear Stress
G/2
(GPa)
G/2
(106 psi)
(MPa)
(psi)
max/exp.
Silver
12.6
1.83
0.37
55
~3  104
Aluminum
11.3
1.64
0.78
115
~1  104
Copper
19.6
2.84
0.49
70
~4  104
Nickel
32
4.64
3.2-7.35
465-1,065
~1  104
Iron
33.9
4.92
27.5
3,990
~1  103
Molybdenum
54.1
7.85
71.6
10,385
~8  102
Niobium
16.6
2.41
33.3
4,830
~5  102
Cadmium
9.9
1.44
0.57
85
~2  104
7
1.02
39.2
5,685
~2  104
7
1.02
39.2
5,685
~2  104
16.9
2.45
13.7
1,985
~1  103
49.3
7.15
1.37
200
~4  104
49.3
7.15
52
7,540
~1  103
Material
Magnesium
(basal slip)
Magnesium
(prism slip)
Titanium
(prism slip)
Beryllium
(basal slip)
Beryllium
(prism slip)
experimental << max ; also, experimental << max
The difference is attributed to the
presence of DEFECTS
• Microcracks → σexperimental << σmax
• Dislocations → τexperimental << τmax
• Etc…
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