project e - waves - trinity college dublin, the university … matter divided into three phases:...
TRANSCRIPT
Because of thermal expansion,
volume increases with increasing temperature,
Density decreases
with increasing temperature.
Density (r) is defined as mass divided by
volume: r =
Density
SI unit for density is kg.m-3
Objects of the same volume are not necessarily
the same mass.
Depends on density.
Example: block of aeroboard & block of metal
Density of object depends on temperature
m
V
A rigid (high modulus of elasticity) low
density material is used
equal performance can be achieved
considerable saving in weight.
Density
Density is important for restorations
Example:
Choosing an alloy with which to construct
components of an upper denture
Heavy alloy would result in large forces
making retention difficult.
To reduce such forces choose
lower density alloy
Dental Materials
Material Density (r) kg/m3
Enamel 3,000
Dentine 2,100
Composite ≈2,000
Amalgam 11,600
Gold 19,300
Silver 10,500
Water 1,000
Density
Density of some common materials
Density value may have significant influence on
thermal characteristics of dental materials
Solids
Matter divided into three phases:
•Solids
•Liquids
•Gases
Solids:
•definite shape and volume
•not absolutely rigid
•elastically deformed by external forces
All solids are elastic to some degree
Solid that is slightly deformed by an applied
force will return to its original shape when the
force is removed.
Shape changes are reversible
Simple model Atoms of the solid are
assumed to be held
together with
“spring-like” forces
Crystalline structure
Highly ordered arrangement of atoms
Amalgam, gold, pure ceramic materials
Non-crystalline structure
Amorphous, disordered atom positions
Dental waxes (thermoplastics)
Glasses
Solids
Materials characteristics
Simple Cubic
Properties
No definite melting temperature
Gradually softens
Solids
Simple model
The elastic properties of solids are usually
referred to in terms of stress and strain
The elasticity of the springs represent
the resilient nature of the inter-atomic
forces
Force Fstress
area A
F is the magnitude of the force applied
perpendicular to the cross sectional area A
Units: Nm-2
Compressive Tensile Shear
Solids
0
0 0
change in length l l lstrain
original length l l
Strain has no units
Solid: subjected to tensile stress
Result: strain
F F
l0l
A
Force F
area A
l
Compressive stress
F F l
0l
A
Solid: subjected to compressive stress
Result: strain
l
0
0 0
l l lstrain
l l
Solids
Stress is proportional strain
Resulting strain depends on the applied stress
)elastic modulusstress strain
Depends on nature of material
elastic modulusstress
strain
3 types of moduli associated with stress:
•Young’s Modulus; concerns change in length
•Shear Modulus; concerns change in shape
•Bulk modulus; concerns change in volume
stress strain )constantstress strain
Defines the rigidity of the material
Solids
Stress
(F/A)
Strain
L/L0)
fracture
Stress & strain
directly proportional,
known as Hooke’s law
Length :
0
/
/
F AE
L L
0
EAF L
L
: Young’s modulus E = Stress
Strain Thomas Young (1773 – 1829) England
Slope of linear portion is a measure
of the rigidity of the material
Elastic limit Greater stress
permanent deformation
F kx
Solids
Substance Young’s Modulus
(Nm-2)
Aluminium 7x1010
Bone (tension) 1.6x1010
Bone (compression) 9.3x109
Dentine (compression) 6.8x109
Nylon 7x108
Steel 20x1010
Tendon 2x107
0 1FLL
A E
1L
E
Large Young’s modulus :
Large force required to produce small
change in length
0
/
/
F AE
L L
Solids
Tensile (or compressive) strength of a material is
the amount of tensile (or compressive) stress that
causes it to break
Substance
Tensile
Strength
Compressive
Strength
Dentine 4.5x107Nm-2
2.6x108Nm-2
Enamel
2.0x107Nm-2
3.0x108Nm-2
Bone
8.3x107Nm-2
(Collagen
fibres)
1.0x108Nm-2
(Calcium Salts
within fibres)
Aluminium 2.2x108Nm-2
stress
strain
fracture Tensile (or compressive)
Strength
Forces and stresses
Biting stresses during chewing
Region Bite force (N)
Molar 400-800
Premolar 220-450
Cuspid 130-330
incisor 90-110
Cusp tip area of a molar ≈ 0.04cm2
Applied biting force = 600N
Compressive stress during chewing
8 2
2 6 2
600 6001.5 10
0.04 4 10
Force N NNm
area cm m
Energy of a bite is absorbed by
Food
Teeth
Periodontal ligament
Bone
Tooth design is such that it can absorb large
static and impact energies
Forces and stresses
Str
ess N
m-2
x10
6
strain
enamel
dentine
0.01 0.02
100
200
Compressive stress versus strain
EEnamel ≈33x109 Nm-2
EDentine ≈12x109 Nm-2
Enamel:
Relatively high elastic modulus
Brittle material
Dentine:
More flexible
Tougher
J.W. Stanford et. al.
J Am Dent Assoc
60, 746,1960
Amalgams, ceramics, composites: Brittle
>>small strain before fracture
stress
strain
fracture
Solids
Resilience of a material
Characteristic of a material to absorb energy
when elastically stressed such that the energy
is recovered when the stress is removed
Resilience of a material
Recoverable energy
Quantitatively, modulus of resilience Represented graphically by area under the linear part
of the stress- strain graph
syield
)2
2
yieldarea
E
s Modulus of resilience Units Jm-3
Large modulus of resilience required for
orthodontic wires. Store large amount of energy
and release it over long period of time
Exercise Determine the force required to extend a person’s
femur by 0.01% when in horizontal traction.
Assume the bone is of circular cross section
with a radius of 2.0cm and a Young’s modulus
E =1.6x1010Nm-2
0
/
/
F AE
L L
4
0/ 0.01% 1 10L L
10 21.6 10E Nm
2 2(0.02 )A r m
)0/F E L L A
) )10 2 4 21.6 10 1 10 (0.02 )F Nm m
32.0 10F N
Tension
What is the resultant force (vertically downward)
exerted on the tooth by the braces in the
figure if T = 25 N?
10
0
1
16
16 25 0.28 6.9
TSin
T
T TSin newtons
0 2
0
2
16
16 25 0.28 6.9
TSin
T
T TSin newtons
1 2 13.8T T N Resultant force
T1
T2
160 160
T T
Pressure
• Many other units of pressure
Atmosphere, lb/in.2 (PSI), bar, mbar, hectoPascal
mm Hg
Fluids- pressure (P) is important concept
Definition of pressure
•Force per unit area •P=F/A
SI unit of pressure (Newton per square metre)
1Nm-2 = 1 Pa (Pascal)
1 atmosphere =1.013 x105Pa =1013 hectoPascals
1 hectoPascal (hPa) = 1 mbar
Blaise Pascal (French) (1623-1662)
mathematician, physicist & religious philosopher
Hydraulic fluids
1 atmosphere =760 mm mercury
Conversion
Pressure
Same force — different contact surface area
→ different pressure
Large pressure
Small contact
area
same force
Contact area 1.5 cm radius
Force = 0.5N
P= F/A = 0.5 N
r2
0.5 N (3.14) (0.015m)2
= 7.07 x102 Pa
Contact 0.5mm radius
P= F/A = 0.5N
r2
0.5N
(3.14) (5x10-4m)2
= 6.37 x105 Pa
P =
=
effect of force depends on area of contact
small pressure
large contact
area
force
Pressure
Force per unit area P=F/A
Examples
•Snow shoes
•Golf shoes
•Stiletto heels
Compare the pressure exerted on a piece of food
by the biting force of a molar with that of an incisor.
Assume that the force is the same in each case and
that the food-teeth contact areas are 45mm2 and 5mm2
respectively.
Pmolar = F/Amolar = F/45mm2
Pincisor = F/Aincisor = F/5mm2
Pmolar = F/45mm2
Pincisor = F/5mm2 Ratio = =
1 9
Example
Teeth
•Incisors.→ cutting
•Molars. → crushing
Pincisor is 9 times greater than Pmolar
Pressure
Compare the pressure exerted on a floor by a person
of mass 60 kg wearing stiletto heeled shoes with that of
an elephant. Assume the person is standing on their heels
each of which is 1.0 cm diameter. The elephant has a
mass of 5000 kg and the area of its footprint is 700 cm2.
Person
Heel contact area =
P= F/A = 60 kg. 9.8 ms-2
2 (0.5x10-2 m)2
588 N
(6.28) (2.5x10-5m2)
= 37.5 x105 Pa
=
r2
Elephant
Foot contact area =
P= F/A = 5000kg. 9.81ms-2
4x700x10-4m2 17.5 x104Pa =
700 cm2
Pressure Skeletal system
Pressure on joints is larger
than fluid pressures in the body
Force concerned with a bone or joint divided
by the contact area
Contact area at end of
bone much larger than
cross-sectional area
along length of bone
result: pressure on joint
reduced.
Gripping side flattened. Larger contact area
reduces pressure on soft tissue covering finger.
Gripping side
However Femur
Tibia
Knee Joint
Cross-section of finger bone.
Fluids
Fluid -----either a gas or a liquid
Gases and liquids:
• many common characteristics
•but some notable differences
Example •Liquids nearly incompressible
•Gases easily compressed
•Liquids much greater densities
Gaseous phase --substance usually
at higher temperature than liquid phase
Substance-- liquid or gaseous phase
Pressure Fluids can also exert pressure
Water in container with straight sides
Water has mass of 50kg
so its weight (w) is
W = mg =50kg x 9.8ms-2
= 490N =
force on bottom of container
due to the weight of the water
Let area of container bottom A = 4m2
Therefore pressure due to weight of water is
490N/4m2 = 122.5N/m2 = 122.5 Pa
If area of container bottom =0.5m2
pressure due to same weight of water is
490N/0.5m2 = 980N/m2 = 980 Pa
Force = mg = (rV)g Pressure = (rVg/A) = rgh
h
A
Gas Pressure
Atmospheric pressure results
from weight of air
1.013x 105N/m2 (Pa) at sea level
Column of air: 1m2 “footprint”, weighs 1.013 x 105N
Air exerts considerable pressure on everything
1 standard atmosphere = 1.013 x105Pa
=1013 hectoPascals
Atmospheric pressure depends on;
temperature, altitude, weather changes.
Area 1m2
Pressure
Pressure depends on depth and density Intravenous (IV) solution administration makes
use of the pressure difference generated by
height difference between reservoir and needle
at point of entry.
Ball point pen (open at top (Pa))
Seal hole at top– will not work
h
Prhg
Pressure Measurement
Barometer
Pa = rgh
Long tube, filled with
mercury, closed at one
end, and inverted into
a beaker of mercury.
Barometer measures atmospheric pressure.
1 atmosphere = 1.013 x105Pa (definition)
Height at which the mercury
settles depends on the
atmospheric pressure
exerted on the mercury
in the beaker
h= Pa /(rg)
Atmospheric pressure 760mm of mercury
h Pa
P = 0
5 -2
3 -3 -2
(1.013 X10 Nm )h = 0.76
(13.6x10 kgm )(9.8ms )m
Pressure (Pascal’s principle) Pascal’s principle If an external force causes a pressure change
in a confined incompressible fluid the same
pressure change is experienced at every point
in the fluid.
P = =
Weight =mg =rVg
rVg
A P =
F
A
weight
A
rAh)g
A = rgh =
h
A
A
Density (r) Volume (V)
Pressure due to the column of fluid = rgh
P= rgh
Total pressure at bottom of fluid = Pa + rhg
where Pa = atmospheric pressure
Pa
Pressure Pascal’s principle
The same pressure is transmitted throughout
an incompressible fluid –not necessarily the
same force.
Hydraulic Systems
Force F1 on small piston results in pressure P1
P1 = F1
A1
But P1 = P2 (Pascal principle)
F1
A1
F2
A2
= F2 A1
F1 A2 = If A2 » A1
then F2 » F1
F1 F2
Piston
A1 A2
P1 P2
Pressure
Calculate the force( F1) a dentist must exert on a
small piston to raise a patient and chair of mass
120kg. Small piston diameter is 1.0cm and the
large piston diameter is 5cm.
1 2
1 2
2
1 11 2 2
2 2
22
1 2
1
(0.5 )120 (9.8 )
(2.5 )
47
F F
A A
A rF F mg
A r
cmF kg ms
cm
F N
F1 F2
Piston
A1 A2
P1 P2
P1 = P2
Dentist’s chair operates in a similar manner
Weight of patient & chair =120kg.9.8ms-2 = 1176N
Pressure Measurement
FP
A
4 2
4 2
46.01.88 10
24.5 10
F NP Nm
A m
Example
The maximum force exerted by the blood on an
aneurysm of area 24.5cm2 in 46.0 Newtons.
What is the maximum blood pressure in mm Hg
4 2
3 3 1
1.88 10
13.6 10 9.8
141
P gh
P Nmh
g kgm ms
h mmHgx
r
r
Density of mercury =1.36x104kgm-3
Pressure When diving into a swimming pool you reach a depth
of 6m below the surface of the water. Estimate the net
force on your eardrum at this depth. Assume
approximate area (A) of eardrum is 1cm2.
Air inside your ear is normally at atmospheric
pressure
There is an additional pressure associated with
the depth below the surface of the water given by
rgh Pressure inside eardrum
= atmospheric pressure (Pa)
Pressure outside eardrum
=atmospheric pressure (Pa) + rgh
Therefore net pressure (P) = rgh
P =(1.00x103kgm-3)(9.8ms-2)(6m)
=5.88x104Pa
Net Force (F) = P x A
= (5.88 x104Pa)(1.00 x10-4m2) = 5.88N
Pressure Measurement
Blood pressure measurement
Utilises Pascal’s principle
Fluids used to transmit pressure
Blood pressure measured at upper arm (≈ at heart)
Rubber bulb pressurises inflatable cuff with air.
Attached manometer (reads mm of mercury)
Blood flow through the brachial artery stopped.
Valve opened on bulb. Pressure in cuff reduces;
artery opens momentarily with each heart
beat->>blood flow (turbulent) resumes >>
detected by stethoscope. Maximum heart
pressure (systolic pressure) measured by
manometer.
Cuff pressure lowered further->>> continuous
blood detected->>> minimum heart pressure
(diastolic pressure)