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Dr. Ehab Hegazy
Properties of materials
Crystal Structure
• properties of materials (especially mechanical)
are determined by the arrangement of the
atoms. This arrangement is called the material’s
crystal structure.
Atomic structure
Relates to the number of protons and neutrons
in the nucleus of an atom, as well as the number
and probability distributions of the electrons.
Crystal structure
arrangement of atoms in the crystalline solid
material.
Atoms can be arranged either in a Regular,
periodic array (i.e., long-range order) or
completely Disordered (amorphous) short
range order.
What is the Unit Cell?
unit cell is the smallest group of atoms which can
generate the entire crystal by translation.
Definition: the length of each unit cell axis is
called a lattice parameter.
Metallic crystals
Why metals are densely packed
1. -Typically, only one element is present, so all
atomic radii are the same.
2. Metallic bonding is not directional.
3. Nearest neighbor distances(IAD) tend to be
small in order to lower bond energy
• APF for a simple cubic structure = 0.52
BODY CENTERED CUBIC
STRUCTURE (BCC)
FACE CENTERED CUBIC
STRUCTURE (FCC)
FCC
3 Common Unit Cells with Cubic
Symmetry
Simple Cubic Body Centered Cubic Face Centered Cubic
(SC) (BCC) (FCC)
1- CUBIC CRYSTAL SYSTEMS
Mechanical Properties
Tablet manufacturing
Compression of tablet
Compression of tablet is reduction in the volume of the
powder owing to application of the forces .
Elasticity of soft gelatin capsules
 Application of compression force
Testing the elasticity of soft gels typically serves as an indicator whether the
product can be "released-to-pack" - providing a reliable and repeatable measure
to prevent failure during packaging.
Ultimate strength
Stress
 Stress = Force/Area
 Compressive, tensile, and
shear.
 Unit Newton /meter2
Types of Stress
tensile stress occurs when
equal and opposite forces
are directed away from
each other.
A compressive stress occurs
when equal and opposite
forces are directed toward
each other.
Shear stress
F
W
Tension
F
W
Compression
STRAIN
 Strain is the change in length per unit
length that a material undergoes when
a force is applied to it.
dL/L
 No unit
Change of length /original length
Types of Strain
 Elastic strain
Is strain that totally disappears once the
external load that caused it is removed.
 Plastic strain
is strain that permanently remains once the
external load that caused it is removed.
Example 1 (Cont.) A 10 m steel wire
stretches 3.08 mm due to the 200 N load.
What is the longitudinal strain?
L
DL
Given: L = 10 m; DL = 3.08 mm
Longitudinal Strain
3.08 x 10-4
0.00308 m
10 m
L
Srain
L
D
 
Elastic Moduli
The elastic modulus of a material characterize the
degree of its resistance to a given type of a deforming
force.
Stress is defined as the force applied per unit area
(N/m2).
Strain
Stress
modulusElastic 
N/m2
N/m2
……
The strain indicates some fractional change in a
dimension or volume.
Elastic Moduli
1- Young modulus
2- Bulk modulus
3- Shear modulus
Young's Modulus
Figure A force applied normal to the end face of a rod
causes a change in length.
DL
L0
A
FnFn
Fn Fn
A
Fn
stressTensile 
0
Strain
L
LD

0
LL
AF
E
/
/
;modulussYoung'
n
D

2 - 28
Stress-Strain Test
2 - 29
Stress-Strain Diagram: Ductile
Materials
2 - 30
Stress-Strain Diagram: Brittle
Materials
The stress-strain relationship
for a metal.
108
Stress
(N/m2)
Strain
Proportional
point
Yield
point
Break
point
Plastic range
Physics
Loading of long bones
Forces on vertebral column
Stress strain curve human bone
Shear Modulus
The shear force is applied tangentially to
the surface.
A shearing force Ft causes the
body to deform as shown.
Dx
A Ft
h
fixed
A
Ft
stressShear 
h
xD
strainShear
hx
AF
G
/
/
modulusShear t
D

Bulk Modulus
The bulk modulus of a solid or a fluid indicates its
resistance to a change in volume.
A
F
P 
VV
P
B
/
strainVolume
stressVolume
modulusBulk
D
D

 The bulk modulus is
negative because of
decrease in V.
TYPES OF PURE FORCES
 Compressive: a force that results in a decrease
in length along the direction of the force
 Tensile: a force that results in an increase in
length along the direction of the force
 Shear: a force that causes a sliding displacement
of one side of a structure relative to another
side
PROPORTIONAL LIMIT
 can be defined as the limit of
proportionality of stress to strain
 is represented on the stress-strain
diagram as the point where the
plotting converts from a straight line
to a curve. Below the proportional
limit, stress is proportional to strain
41
1. Initial 2. Small load 3. Unload
Elastic means reversible.
Elastic Deformation
42
1. Initial 2. Small load 3. Unload
Plastic means permanent.
Plastic Deformation (Metals)
DUCTILITY
 Ductility is the ability of a material to
undergo permanent tensile deformation
without fracture or rupture
MALLEABILITY
 Malleability is the ability of a material
to undergo permanent compressive
deformation without fracture .
BRITTLENESS
 Brittleness is the material behavior where a
material undergoes fracture or rupture with
little or no prior permanent deformation.
 Materials that are brittle usually have a very
ordered atomic Structure which does not permit
the easy movement of dislocations.
 A good example is the class of materials
known as ceramics. Their ordered atomic
structure does not permit easy dislocation
movement, and hence, they are brittle.
Ductile fracture
 Plastic deformation takes place before fracture
Brittle fracture
 No Plastic deformation takes place before fracture
RESILIENCE
Energy absorbed up to proportional limit
51
Toughness
Lower toughness: ceramics
Higher toughness: metals
Toughness is the ability to
absorb energy up to
fracture (energy per unit
volume of material).
A “tough” material has
strength and ductility.
Approximated by the area
under the stress-strain
curve.
FATIGUE
 Fatigue refers to the fact that under
cyclic loading a material will undergo
failure at a lower applied stress than it
normally would if it were not under
cyclic loading.
 The name "fatigue" is derived from the
fact that the materials seem to tire
under this type of repetitive loading.
WEAR
Wear is the loss of material from
one or both of two contacting
surfaces because of the
mechanical activity between
them
Questions model
1- The ability of material to absorb energy up to proportional limit is called ……..
2- Ultimate strength and breaking strength are the same in case of ………
3- fatigue occurs when material subjected to …….load
4- Elastic modulus represent ………of materials
5- The whole area under stress strain curve of dental material represent ….
6- No plastic deformation takes place in of ………….fracture.
7- The stress at which the material fracture is called -----------
9- dislocations generally move more ……….. at lower temperature.
10- Malleability is the ability of a material to undergo permanent ……….. deformation without fracture
11- Brittleness is the material behavior where a material undergoes fracture or rupture with …………….
prior permanent deformation
12-…………. is the loss of material from one or both of two contacting surfaces because of the mechanical
activity between them
13- area under stress strain including elastic and plastic deformation is known as ……..
14- area under stress curve up to fracture point is called ………….
15- area under stress strain curve up to proportional limit is called …………………..
Example 1. A steel wire 10 m long and
2 mm in diameter is attached to the
ceiling and a 200-N weight is attached
to the end. What is the applied stress?
L
DL
A
A
F
First find area of wire:
2 2
(0.002 m)
4 4
D
A
 
 
A = 3.14 x 10-6 m2
-6 2
200 N
3.14 x 10 m
F
Stress
A
  Stress
6.37 x 107 Pa
Example 2. The elastic limit for steel is
2.48 x 108 Pa. What is the maximum
weight that can be supported without
exceeding the elastic limit?
L
DL
A
A
F
8
2.48 x 10 Pa
F
Stress
A
 
Recall: A = 3.14 x 10-6 m2
F = (2.48 x 108 Pa) A
F = (2.48 x 108 Pa)(3.14 x 10-6 m2)
F = 779 N
Example 2(Cont.) The ultimate strength
for steel is 4089 x 108 Pa. What is the
maxi- mum weight that can be supported
without breaking the wire?
L
DL
A
A
F
8
4.89 x 10 Pa
F
Stress
A
 
Recall: A = 3.14 x 10-6 m2
F = (4.89 x 108 Pa) A
F = (4.89 x 108 Pa)(3.14 x 10-6 m2)
F = 1536 N
Example 3. In our previous example, the
stress applied to the steel wire was 6.37
x 107 Pa and the strain was 3.08 x 10-4. Find
the modulus of elasticity for steel.
L
DL
7
-4
6.37 x 10 Pa
3.08 x 10
Stress
Modulus
Strain
 
Modulus = 207 x 109 Pa
This longitudinal modulus of elasticity is
called Young’s Modulus and is denoted by
the symbol Y.
Example 4: Young’s modulus for brass is
8.96 x 1011Pa. A 120-N weight is
attached to an 8-m length of brass wire;
find the increase in length. The diameter
is 1.5 mm.
8 m
DL
120 NFirst find area of wire:
2 2
(0.0015 m)
4 4
D
A
 
  A = 1.77 x 10-6 m2
or
FL FL
Y L
A L AY
 D 
D
Example 4: (Continued)
8 m
DL
120 N
Y = 8.96 x 1011 Pa; F = 120 N;
L = 8 m; A = 1.77 x 10-6 m2
F = 120 N; DL = ?
or
FL FL
Y L
A L AY
 D 
D
-6 2 11
(120 N)(8.00 m)
(1.77 x 10 m )(8.96 x 10 Pa)
FL
L
AY
D  
DL = 0.605 mmIncrease in length:
Example 7. A hydrostatic press contains 5
liters of oil. Find the decrease in volume of
the oil if it is subjected to a pressure of
3000 kPa. (Assume that B = 1700 MPa.)
/
P PV
B
V V V
 
 
D D
6
9
(3 x 10 Pa)(5 L)
(1.70 x 10 Pa)
PV
V
B
 
D  
DV = -8.82 mL
Decrease in V; milliliters (mL):
Thank you

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Properties of materials

  • 2. Crystal Structure • properties of materials (especially mechanical) are determined by the arrangement of the atoms. This arrangement is called the material’s crystal structure.
  • 3. Atomic structure Relates to the number of protons and neutrons in the nucleus of an atom, as well as the number and probability distributions of the electrons. Crystal structure arrangement of atoms in the crystalline solid material.
  • 4. Atoms can be arranged either in a Regular, periodic array (i.e., long-range order) or completely Disordered (amorphous) short range order.
  • 5. What is the Unit Cell? unit cell is the smallest group of atoms which can generate the entire crystal by translation. Definition: the length of each unit cell axis is called a lattice parameter.
  • 6.
  • 7.
  • 8. Metallic crystals Why metals are densely packed 1. -Typically, only one element is present, so all atomic radii are the same. 2. Metallic bonding is not directional. 3. Nearest neighbor distances(IAD) tend to be small in order to lower bond energy
  • 9.
  • 10. • APF for a simple cubic structure = 0.52
  • 13. FCC
  • 14. 3 Common Unit Cells with Cubic Symmetry Simple Cubic Body Centered Cubic Face Centered Cubic (SC) (BCC) (FCC) 1- CUBIC CRYSTAL SYSTEMS
  • 17. Compression of tablet Compression of tablet is reduction in the volume of the powder owing to application of the forces .
  • 18. Elasticity of soft gelatin capsules  Application of compression force Testing the elasticity of soft gels typically serves as an indicator whether the product can be "released-to-pack" - providing a reliable and repeatable measure to prevent failure during packaging.
  • 20. Stress  Stress = Force/Area  Compressive, tensile, and shear.  Unit Newton /meter2
  • 21. Types of Stress tensile stress occurs when equal and opposite forces are directed away from each other. A compressive stress occurs when equal and opposite forces are directed toward each other. Shear stress F W Tension F W Compression
  • 22. STRAIN  Strain is the change in length per unit length that a material undergoes when a force is applied to it. dL/L  No unit Change of length /original length
  • 23. Types of Strain  Elastic strain Is strain that totally disappears once the external load that caused it is removed.  Plastic strain is strain that permanently remains once the external load that caused it is removed.
  • 24. Example 1 (Cont.) A 10 m steel wire stretches 3.08 mm due to the 200 N load. What is the longitudinal strain? L DL Given: L = 10 m; DL = 3.08 mm Longitudinal Strain 3.08 x 10-4 0.00308 m 10 m L Srain L D  
  • 25. Elastic Moduli The elastic modulus of a material characterize the degree of its resistance to a given type of a deforming force. Stress is defined as the force applied per unit area (N/m2). Strain Stress modulusElastic  N/m2 N/m2 …… The strain indicates some fractional change in a dimension or volume.
  • 26. Elastic Moduli 1- Young modulus 2- Bulk modulus 3- Shear modulus
  • 27. Young's Modulus Figure A force applied normal to the end face of a rod causes a change in length. DL L0 A FnFn Fn Fn A Fn stressTensile  0 Strain L LD  0 LL AF E / / ;modulussYoung' n D 
  • 29. 2 - 29 Stress-Strain Diagram: Ductile Materials
  • 30. 2 - 30 Stress-Strain Diagram: Brittle Materials
  • 31. The stress-strain relationship for a metal. 108 Stress (N/m2) Strain Proportional point Yield point Break point Plastic range
  • 35. Stress strain curve human bone
  • 36. Shear Modulus The shear force is applied tangentially to the surface. A shearing force Ft causes the body to deform as shown. Dx A Ft h fixed A Ft stressShear  h xD strainShear hx AF G / / modulusShear t D 
  • 37. Bulk Modulus The bulk modulus of a solid or a fluid indicates its resistance to a change in volume. A F P  VV P B / strainVolume stressVolume modulusBulk D D   The bulk modulus is negative because of decrease in V.
  • 38.
  • 39. TYPES OF PURE FORCES  Compressive: a force that results in a decrease in length along the direction of the force  Tensile: a force that results in an increase in length along the direction of the force  Shear: a force that causes a sliding displacement of one side of a structure relative to another side
  • 40. PROPORTIONAL LIMIT  can be defined as the limit of proportionality of stress to strain  is represented on the stress-strain diagram as the point where the plotting converts from a straight line to a curve. Below the proportional limit, stress is proportional to strain
  • 41. 41 1. Initial 2. Small load 3. Unload Elastic means reversible. Elastic Deformation
  • 42. 42 1. Initial 2. Small load 3. Unload Plastic means permanent. Plastic Deformation (Metals)
  • 43.
  • 44. DUCTILITY  Ductility is the ability of a material to undergo permanent tensile deformation without fracture or rupture
  • 45. MALLEABILITY  Malleability is the ability of a material to undergo permanent compressive deformation without fracture .
  • 46. BRITTLENESS  Brittleness is the material behavior where a material undergoes fracture or rupture with little or no prior permanent deformation.  Materials that are brittle usually have a very ordered atomic Structure which does not permit the easy movement of dislocations.  A good example is the class of materials known as ceramics. Their ordered atomic structure does not permit easy dislocation movement, and hence, they are brittle.
  • 47. Ductile fracture  Plastic deformation takes place before fracture
  • 48. Brittle fracture  No Plastic deformation takes place before fracture
  • 49.
  • 50. RESILIENCE Energy absorbed up to proportional limit
  • 51. 51 Toughness Lower toughness: ceramics Higher toughness: metals Toughness is the ability to absorb energy up to fracture (energy per unit volume of material). A “tough” material has strength and ductility. Approximated by the area under the stress-strain curve.
  • 52. FATIGUE  Fatigue refers to the fact that under cyclic loading a material will undergo failure at a lower applied stress than it normally would if it were not under cyclic loading.  The name "fatigue" is derived from the fact that the materials seem to tire under this type of repetitive loading.
  • 53. WEAR Wear is the loss of material from one or both of two contacting surfaces because of the mechanical activity between them
  • 54. Questions model 1- The ability of material to absorb energy up to proportional limit is called …….. 2- Ultimate strength and breaking strength are the same in case of ……… 3- fatigue occurs when material subjected to …….load 4- Elastic modulus represent ………of materials 5- The whole area under stress strain curve of dental material represent …. 6- No plastic deformation takes place in of ………….fracture. 7- The stress at which the material fracture is called ----------- 9- dislocations generally move more ……….. at lower temperature. 10- Malleability is the ability of a material to undergo permanent ……….. deformation without fracture 11- Brittleness is the material behavior where a material undergoes fracture or rupture with ……………. prior permanent deformation 12-…………. is the loss of material from one or both of two contacting surfaces because of the mechanical activity between them 13- area under stress strain including elastic and plastic deformation is known as …….. 14- area under stress curve up to fracture point is called …………. 15- area under stress strain curve up to proportional limit is called …………………..
  • 55. Example 1. A steel wire 10 m long and 2 mm in diameter is attached to the ceiling and a 200-N weight is attached to the end. What is the applied stress? L DL A A F First find area of wire: 2 2 (0.002 m) 4 4 D A     A = 3.14 x 10-6 m2 -6 2 200 N 3.14 x 10 m F Stress A   Stress 6.37 x 107 Pa
  • 56. Example 2. The elastic limit for steel is 2.48 x 108 Pa. What is the maximum weight that can be supported without exceeding the elastic limit? L DL A A F 8 2.48 x 10 Pa F Stress A   Recall: A = 3.14 x 10-6 m2 F = (2.48 x 108 Pa) A F = (2.48 x 108 Pa)(3.14 x 10-6 m2) F = 779 N
  • 57. Example 2(Cont.) The ultimate strength for steel is 4089 x 108 Pa. What is the maxi- mum weight that can be supported without breaking the wire? L DL A A F 8 4.89 x 10 Pa F Stress A   Recall: A = 3.14 x 10-6 m2 F = (4.89 x 108 Pa) A F = (4.89 x 108 Pa)(3.14 x 10-6 m2) F = 1536 N
  • 58. Example 3. In our previous example, the stress applied to the steel wire was 6.37 x 107 Pa and the strain was 3.08 x 10-4. Find the modulus of elasticity for steel. L DL 7 -4 6.37 x 10 Pa 3.08 x 10 Stress Modulus Strain   Modulus = 207 x 109 Pa This longitudinal modulus of elasticity is called Young’s Modulus and is denoted by the symbol Y.
  • 59. Example 4: Young’s modulus for brass is 8.96 x 1011Pa. A 120-N weight is attached to an 8-m length of brass wire; find the increase in length. The diameter is 1.5 mm. 8 m DL 120 NFirst find area of wire: 2 2 (0.0015 m) 4 4 D A     A = 1.77 x 10-6 m2 or FL FL Y L A L AY  D  D
  • 60. Example 4: (Continued) 8 m DL 120 N Y = 8.96 x 1011 Pa; F = 120 N; L = 8 m; A = 1.77 x 10-6 m2 F = 120 N; DL = ? or FL FL Y L A L AY  D  D -6 2 11 (120 N)(8.00 m) (1.77 x 10 m )(8.96 x 10 Pa) FL L AY D   DL = 0.605 mmIncrease in length:
  • 61. Example 7. A hydrostatic press contains 5 liters of oil. Find the decrease in volume of the oil if it is subjected to a pressure of 3000 kPa. (Assume that B = 1700 MPa.) / P PV B V V V     D D 6 9 (3 x 10 Pa)(5 L) (1.70 x 10 Pa) PV V B   D   DV = -8.82 mL Decrease in V; milliliters (mL):