Lecture 7
Solids:
Properties:
Fluids
Properties
•Pressure
•Pascal’s principle
Density
Objects of the same volume are not necessarily
the same mass.
Depends on density.
Example: block of aeroboard & block of metal
Density (r) is defined as mass divided by
volume:
m
r=
SI unit for density is kg.m-3
V
Density of object depends on temperature
Because of thermal expansion,
volume increases with increasing temperature,
Density decreases
with increasing temperature.
Density
Dental Materials
Density is important
Example:
Choosing an alloy with which to construct
components of an upper denture
Heavy alloy would result in large displacing forces
making retention difficult.
To reduce such forces choose
lower density alloy
A rigid (high modulus of elasticity) low
density material is used
equal performance can be achieved
considerable saving in weight.
Density
Density of some common materials
Material
Density (r) kg/m3
Enamel
Dentine
3,000
2,100
Composite
Amalgam
Gold
Silver
Water
≈2,000
11,600
19,300
10,500
1,000
Density value may be significant influence on
thermal characteristics of dental materials
Example see “thermal diffusivity”
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
Solids
Simple model
Atoms of the solid are
assumed to be held
together with
“spring-like” forces
Materials characteristics
Simple Cubic
Crystalline structure
Highly ordered arrangement of atoms
Amalgam, gold, pure ceramic materials
Non-crystalline structure
Amorphous, disordered atom positions
Dental waxes (thermoplastics)
Glasses
Properties
No definite melting temperature
Gradually softens
Solids
Simple model
The elasticity of the springs represent
the resilient nature of the inter-atomic
forces
The elastic properties of solids are usually
referred to in terms of stress and strain
Force F
stress 

area
A
Units: Nm-2
F is the magnitude of the force applied
perpendicular to the cross sectional area A
Compressive
Tensile
Shear
Solids
Force F
Solid: subjected to tensile stress  area  A
Result: strain
A
l0
l
F
F
l
change  in  length l  l0 l
strain 


original  length
l0
l0
Strain is positive and has no units
Solid: subjected to compressive stress
Result: strain
strain 
l  l0
l0

l
l0
l0
F
l
Compressive stress
l
F
A
Solids
Resulting strain depends on the applied stress
Stress is proportional strain
stressstrain
stress   constant )  strain
stress   elastic modulus )  strain
Depends on nature of material
stress
elastic modulus 
strain
Defines the rigidity of the material
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
Solids
Length :
Stress
: Young’s modulus E =
Strain
Thomas Young (1773 – 1829) England
Elastic limit
Greater stress
permanent deformation
Stress
(F/A)
F/A
E
L / L0
fracture
Stress & strain
directly proportional,
known as Hooke’s law
Slope of linear portion is a measure
of the rigidity of the material
Strain
L/L0)
 EA 
F 
 L  kx
 L0 
Solids
Aluminium
Bone (tension)
Bone (compression)
Dentine (compression)
Nylon
Steel
Young’s Modulus
(Nm-2)
7x1010
1.6x1010
9.3x109
6.8x109
7x108
20x1010
Tendon
2x107
Substance
F/A
E
L / L0
 FL0  1
L  

 A E
1
L 
E
Large Young’s modulus :
Large force required to produce small
charge in length
Solids
Tensile (or compressive)
stress
Strength
fracture
strain
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
Forces and stresses
Biting stresses during chewing
Region
Bite force (N)
Molar
400-800
Premolar
220-450
Cuspid
incisor
130-330
90-110
Cusp tip area of a molar ≈ 0.04cm2
Applied biting force = 600N
Compressive stress during chewing?
Force
600 N
600 N
8
2



1.5

10
Nm
area 0.04cm2 4 106 m2
Forces and stresses
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
Stress Nm-2 x106
Compressive stress versus strain
EEnamel ≈33x109 Nm-2
200
enamel
EDentine ≈12x109 Nm-2
100
dentine
J.W. Stanford et. al.
J Am Dent Assoc
60, 746,1960
0.01
0.02
Enamel:
relatively high elastic modulus
Brittle material
strain
Dentine:
More flexible
Tougher
Amalgams, ceramics, composites: Brittle
>>small strain before fracture
Solids
Resilience of a material
stress
fracture
syield
strain
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
Modulus of resilience
s )

 area 
yield
2E
2
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
L / L0  0.01%  110
F/A
E
L / L0
4
E  1.6 1010 Nm2
A   r   (0.02m)
2
2
F  E  L / L0 ) A
F  1.6 1010 Nm2 )1104 )  (0.02m)2
F  2.0 10 N
3
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
Substance-- liquid or gaseous phase
Gaseous phase --substance usually
at higher temperature than liquid phase
Pressure
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)
Blaise Pascal (French) (1623-1662)
mathematician, physicist & religious philosopher
Hydraulic fluids
• Many other units of pressure
Atmosphere, lb/in.2 (PSI), bar, mbar, hectoPascal
mm Hg
Pressure
Same force — different contact surface area
→ different pressure
Contact area 1.5 cm radius
force
Force = 0.5N
0.5 N
small pressure
P= F/A =
Pressure = r2
=
large contact
area
0.5 N
(3.14) (0.015m)2
= 7.07 x102 Pa
same force
Large pressure
Small contact
area
Contact 0.5mm radius
Pressure
P= F/A == 0.5N2
r
P=
0.5N
(3.14) (5x10-4m)2
= 6.37 x105 Pa
effect of force depends on area of contact
Pressure
Force per unit area
Examples
•Snow shoes
P=F/A
Teeth
•Golf shoes
•Incisors.→ cutting
•Stiletto heels
•Molars. → crushing
Example
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
1
=
Ratio =
9
Pincisor = F/5mm2
Pincisor is 9 times greater than Pmolar
Pressure
Compare the pressure exerted on a floor by a person
of mass 60kg wearing stiletto heeled shoes with that of
an elephant. Assume the person is standing on their heels
each of which is 1.0cm diameter. The elephant has a
mass of 5000kg and the area of its footprint is 700cm2.
Person
Heel contact area = r2
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
Elephant
Foot contact area = 700 cm2
P= F/A =
5000kg. 9.81ms-2
4x700x10-4m2
= 17.5 x104Pa
Pressure
Skeletal system
Force concerned with a bone or joint divided
by the contact area
Pressure on joints is larger
than fluid pressures in the body
Contact area increased
Pressure reduced
Gripping side
Cross-section of finger bone.
Gripping side flattened. Larger contact area
reduces pressure on soft tissue covering finger.
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 =
h
A
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
Gas Pressure
Air exerts considerable pressure on everything
1 standard atmosphere = 1.013 x105Pa
=1013 hectoPascals
Atmospheric pressure results
from weight of air
Area 1m2
1.013x 105N/m2 (Pa) at sea level
Column of air: 1m2 “footprint”, weighs 1.013 x 105N
Atmospheric pressure depends on;
temperature, altitude, weather changes.
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.
Prhg
h
Ball point pen (open at top (Pa))
Seal hole at top– will not work
Pressure Measurement
Barometer
Long tube, filled with
mercury, closed at one
end, and inverted into
a beaker of mercury.
P=0
h
Pa
Height at which the mercury
settles depends on the
atmospheric pressure
exerted on the mercury
in the beaker
1 atmosphere = 1.013 x105Pa (definition)
Pa = rhg
h= Pa /(rg)
(1.013 X105 Nm-2 )
h=
 0.76m
3
-3
-2
(13.6x10 kgm )(9.8ms )
Atmospheric pressure
760mm of mercury
Barometer measures atmospheric pressure.
Manometer measures pressure in an
enclosed fluid.
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.
Density (r)
Volume (V)
A
Pa
h
A
P= rhg
Weight =mg =rVg
weight
F
P=
=
A
A
rVg rAh)g rhg
=
=
P=
A
A
Pressure due to the column of fluid = rhg
Total pressure at bottom of fluid = Pa + rhg
where Pa = atmospheric pressure
Pressure
Pascal’s principle
The same pressure is transmitted throughout
an incompressible fluid –not necessarily the
same force.
Hydraulic Systems
F1
F2
Piston
A1
P2
P1
A2
Force F1 on small piston results in pressure P1
F1
P1 =
A1
F1
A1
=
F2
A2
But P1 = P2 (Pascal principle)
F2 = F1
A2
A1
If A2 » A1
then F2 » F1
Pressure
Dentist’s chair operates in a similar manner
F1
F2
Piston
A1
P1
P1 = P2
P2
A2
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.
F1 F2

A1 A2
A1
 r12
F1  F2
 mg 2
A2
 r2
(0.5cm) 2
F1  120kg (9.8ms )
(2.5cm) 2
F1  47 N
2
Weight of patient & chair =120kg.9.8ms-2 = 1176N
Pressure Measurement
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
F
P
A
F
46.0 N
4
2
P 

1.88

10
Nm
A 24.5 104 m2
Density of mercury =1.36x104kgm-3
P  r gh
P
1.88 104 Nm 2
h

r g 13.6 103 kgm 3 9.8ms 1
h  141xmmHg
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
Pressure inside eardrum
= atmospheric pressure (Pa)
rgh
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