Fluids

Lecture 7

Solids

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)