18 November 2009 3.14/3.40 Lecture Summary
We know how small
precipitates affect
material properties...
10 nm
But what about larger inclusions?
10 μm
For particle size greater than about 1 μm, it is necessary to
take into account the stress distribution inside the particle. Bounding Case - Isostrain
1
�
Examine simplified cases to determine a
range of possible outcomes
�
Assume the following:
2
�
�
Two materials with different E and � (or G
and K)
�
Perfect bonding between zones 1 and 2
�
Constant cross-section
Given these assumptions:
�
What is E for the composite?
�
How are loads shared between the two
zones?
Bounding Case - Isostrain
�1�� 2 ��tot
� 1�
�E
E 1 �1�E 1 �tot
; � 2�
�E
E 2 �2 �
�E
E 2 �tot
P 1� A1 � 1� A1 E 1 �tot ; P 2 � A 2 � 2 � A 2 E 2 �tot
P tot �P 1�P 2��tot � A1 E 1� A 2 E 2 �
�
P tot
A1
A2
�� tot
E 1�
E2
� tot �
A1 � A 2
A1 � A 2
A1 � A2
1
2
� tot �� f 1 E 1� f 2 E 2 �� tot
Etot
=f E f E
E
tot � f 11E 11� f 22E 22
P1, P2 are the loads on 1 and 2.
f1, f2 are the volume fractions of 1 and 2. �
Bounding Case - Isostrain
�
Rule of Mixtures
E2
Etottot �
=ff 11E 11�
ff 22E 22
P 22 A22 E 22 f 22E 22
�
�
=
=
P 11 A11 E 11 f 11E 11
f 1 E1 + f 2 E2
E1
�
Load Sharing
The phase with the higher Young's
Modulus has the higher stress.
Bounding Case ­ Isostress
1
 1 = 2 = tot
 1 =E 1 1 ;  2 =E 2 2
 tot
 tot
tot = f 1 1 f 2 2 = f 1
f2
E1
E2
E=
2
 tot
tot
1
=
f1
1

f2
2
=
E1 E2
f 1 E 2 f 2 E 1
Modulus Limits
Isostrain case: E tot �f 1 E 1�f 2 E 2
E2
Isostress case: E �E 1 E 2
tot
f 1 E 1 �f 2 E 2
f 1 E 2�f 2 E 2
These two cases provide the
bounds for more complicated
microstructures.
E 1E 2
E1
0.0
f 1 E 2�f 2 E 2
f2
1.0
Modulus Limits (Hashin-Shtrikman)
E2
E1
Assumption of isotropy leads to
tighter bounding curves.
0.0
f2
1.0
Contiguity
Take two systems, each with f2 = 0.5: Which bounding line is
a particular system closest to? Look at the connectivity: 2
1
1
2
Closer to E1
Closer to E
2
Contiguity and Percolation
E
1
0.0
fcrit
1.0
fcrit (the “percolation threshold”) is a critical volume
fraction of f2 above which there is a connected network of
(randomly scattered) material 2 throughout the matrix.
Fiber-Epoxy Example
�
Compromise between strength
and ductility:
� For
better load transfer, you
want long fibers, which lead to
better long-range connectivity
� For
better ductility, you want
short fibers (a “dispersed
structure”). A broken short fiber
has less effect on the
macroscopic properties than a
broken long fiber
Fiber-Epoxy Example
�
i
i:
fiber
composite
epoxy
�
� Epoxy
��Fiber
f 1E 1�
f 2E 2�
Fiber-Epoxy Example
�
i
� Epoxy
��Fiber
i:
f 1E 1�
f 2E 2�
ii :
� 1 ���
ii
fiber
composite
epoxy
�
f 2E 2�
Fiber-Epoxy Example
�
i
� Epoxy
��Fiber
i:
f 1E 1�
f 2E 2�
ii :
� 1 ���
f 2E 2�
iii :
� 1 ���
� 2 ���
ii
iii
fiber
composite
epoxy
�
Fiber-Epoxy Example
�
i
� Epoxy
��Fiber
i:
f 1E 1�
f 2E 2�
ii :
� 1 ���
f 2E 2�
iii :
� 1 ���
� 2 ���
iv :
� 1 ���
���
ii
iii
iv
fiber
composite
epoxy
�
questions?
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3.40J / 22.71J / 3.14 Physical Metallurgy
Fall 2009
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