MME 467
Ceramics for Advanced Applications
Lecture 06
Effect of Chemical Forces on
Physical Properties of Ceramics
Ref: Barsoum, Fundamentals of Ceramics, Ch04, McGraw-Hill, 2000.
Prof. A. K. M. Bazlur Rashid
Department of MME, BUET, Dhaka
Topics to discuss....
1.
2.
3.
4.
5.
Melting Point
Coefficient of Thermal Expansion
Young’s Modulus of Elasticity
Theoretical Strength
Surface Energy
1 MELTING POINTS
Pure substance at constant pressure melts at a fixed temperature
with the absorption of heat, known as the heat of fusion, ΔHf.
ΔSf =
ΔHf
Tm
ΔSf = entropy difference between solid and liquid @ MP
ΔHf = enthalpy difference between solid and liquid @ MP
Tm = melting point in kelvin
∆Sf for most crystalline solids ≈ 10-12 J/mol K
Ceramics, in general, have high melting points (MP)
than metals or polymers.
But there is a quite bit of variability in the MPs – Why?
Factors Affecting MP of Ionic Ceramics
1. Properties from Bonding
The most important factor determining melting point of a ceramic
is the bond strength holding the ions in place.
Energy
Energy
r0 = equilibrium distance
r0
r
r
E0 = bond energy
z1z 2 e 2 % 1 (
Ebond =
'1− *
4 π ε 0 r0 & n )
Tm is larger if Eo is larger
€
2 z1z 2 e 2 % 1 (
Ebond =
'1− *
4 π ε 0 r0 & n )
Ionic charge
Most important.
€Higher z
1 and z2 results in stronger attraction between atoms.
è higher melting point (MP)
MgO (2852 C) , NaCl (800 C)
[Mg2+, O2–], [Na+, Cl–]
Atomic radius
Larger r0 results in lower MP.
NaCl (800 C) , NaF (997 C)
[rCl = 167 pm], [rF = 119 pm]
2. Covalent Character of Ionic Bond
q  Melting point is proportional to ΔHf
q  Increasing covalent character of ionic bond reduces
bond strength by stabilising discrete units in the melt
è lowers the MP
v  Covalency per se does not necessarily favour either higher
or lower MPs.
v  The important consideration is the melt structure.
v  If strong covalent bonds have to be broken during melting,
extremely high MPs can result.
TiO2 (rutile structure – 1857 C)
CdI2 (layered structure – 387 C)
CO2 (molecular structure – -58 C)
3 Extent of Covalency of Ionic Bond
Assumption:
Bond is purely ionic, then impart a covalent character to it
u  The extent of distortion and sharing of electron cloud between
two ions is a measure of covalent character of ionic bond.
Extent of Covalency Depends on 3 Factors:
1. Polarising power of cation, φ = z+ / r
MgO (+2, -2) – low covalency, high MP (2852 C)
Al2O3 (+3, -2) – high covalency, low MP (2054 C)
But based on z alone, Al2O3 should have high MP.
2. Polarising power of anion
Increasing polarising power of anion
increases covalent character of ionic bond.
LiF (848 C),
LiCl (613 C),
LiBr (547 C),
LiI (446 C)
3. Electronic configuration of Cation
d-electrons are less effective in shielding nuclear charge
than s- or p-electrons è more polarising, form covalent bond
CaCl2 (782 C),
HgCl2 (276 C)
4 Melting Points of Covalent Ceramics
q  Very high MPs due to very strong primary bond
q  Some materials do not melt, rather decompose
Example: Si3N4 (decomposes over 2000 C)
Melting Points of Glass Forming Liquids
SiO2, B2O3, GeO2, P2O5
q  Very low entropy !!!
(Signifies that @MP, liquids and solids have quite similar structures)
THERMAL EXPANSION
ΔL
L0
= α ΔT
q  Asymmetry in the energy well determines thermal expansion
!
Effect of heat on interatomic distance between atoms
5 Energy
r
larger α
o
r
Eo
smaller α
Eo
Asymmetry of energy well increases with decreasing bond strength
è Increases thermal expansion (Thus, α is larger if Ebond is smaller)
Solid Ar
Most metals and ceramics
~ 10–3 C–1
~ 10–5 C–1
Generalizations:
(a) α for ceramics typically < metals.
(b) α is a function of T.
Thus you need to specify T range
(c) Atomic packing can be crucial.
More open structures (like those of covalent
ceramics) è have lower α’s
Example: SiC and Si3N4
Atoms vibrate in the “open spaces”
MgO and NaCl (Close packed) – high α
(d) Thermal expansion anisotropy can
also be very important
Example: Quartz (high α) and fused silica (low α)
!
6 YOUNG’S MODULUS OF ELASTICITY
How ceramics behave when they
are subjected to an external force ?
Study on the shape of Energy vs. Distance
curve is necessary
What is relationship between
energy and force?
F(r) = dE(r) / dr
Many problems in science and engineering
can be solved by minimizing energy or
equating forces.
You should get the same answer at the end.
7 Stiffness and theoretical
strength of a material can be
related to the shape of the
Energy vs. Distance curve.
From the general shape of E(r) curve,
one can easily determine the shape
of Force vs. Distance curve.
General Features
of F(r) Curve
q  Net force between atoms or ions is zero at equilibrium (r = r0).
q  Pulling the atoms apart creates an attractive restoring force, and vice versa.
q  At r = r0, this response is nearly linear.
q  Beyond r = rfail, the bond fails. This represents Fmax that the bond can withstand.
Atomic View of Modulus
The Hook’s Law:
σ = Y ε
Near the vicinity of r0 :
F ≈ S0 (r – r0)
!
Dividing both sides with r02 :
F ≈ S0 (r – r0)
r02
r0
The stiffness of bond :
S0 =
dF
dr
r = r0
r0
stress = Y x strain
where Y is a measure of stiffness
Y≈
S0
r0
8 Y =
1
r0
dF
dr
=
r = r0
1
r0
d2E
dr2 r = r
0
Stiffness is directly related to the curvature of
Force–Distance or Energy–Distance curve
q  Strong bonds are stiffer than weaker bonds
q  Ceramics are stiffer solids
Theoretical Bond Strength
What is the theoretical strength
of a solid that requires to
simultaneously break all bonds?
From the diagram,
S0 ≈
!
Fmax =
S0
( rfail – r0 )
2
=
S0
( 0.25 r0 )
2
2 Fmax
rfail – r0
Typically, most bonds fail if they are
stretched by about 25%, i.e.,
rfail = 1.25 ro
9 Fmax =
S0
( 0.25 r0 )
2
Dividing both sides with r02 :
Fmax
=
r02
Fmax
r0
2
σtheo
S0
2 r0
1
8
Y
=
8
=
0.25 r0
r0
S0
r0
If you actually find the second derivative of E vs. r,
i.e., solve it exactly, you will get, for ionic bond:
σtheo =
Y
15
Actual strengths of ceramics
are much lower and close to
Y/100 – Y/1000.
Why ?
SURFACE ENERGY
Surface energy is the energy needed to create unit area of
new surface.
This equals product of the number of bonds broken per unit area
of crystal surface, NS, and the energy of bond :
γ = Ns Ebond
Since NS is a function of crystallography, surface energy is also
function of crystallography.
10 Let us determine the surface
energy required to cleave a
rocksalt structure along its
(100) plane.
Considering only the firstneighbour interactions,
!
!
Number of ions in the plane = 4 (2 anions and 2 cations)
Area of plane = (2r0)2
Total surface area created = 2 (2r0)2
Number of bonds to be broken = 4
γ = Ebond Ns = -
z1 z2 e2
4 π ε0 r0
γ = -
z1 z2 e2
8 π ε0 r03
11-
1
n
4
2 (2r0)2
1
n
SUMMARY
Strong bonds result in solids with high melting points that are
also stiff and posses high theoretical strengths and low thermal
expansion coefficients.
They also result in solids with high surface energies.
Large bond energy
large Tm
large E
small α
11 Next Class
Lecture 07
Deformation Behaviour of
Ceramic Materials
12