SELECTION OF PLASTICS

The previous lecture has given an indication of the range of plastics available to the design engineer.How do we decide which plastic, if any, is best for a particular application?

FIRST steps in the design process are to define clearly the purpose and function of the proposed product and to identify the service environment.

Then to assess the suitability of a range candidate materials.The most important characteristic for engineering components:

Mechanical properties-strength stiffness, fatigue and toughness Corrosion susceptibility Special properties; permeability, thermal, electrical, optical, damping capacity etc Method of fabrication Cost and manufacturing route

MECHANICAL PROPERTIES

1. Tensile,2. Stress3. Strain4. Young Modulus, E

storage and loss modulus,E and E

3. Shear Modulus, G -storage and loss modulus, G and G

Stress

AFNormal

AFShear

Tension: >0 Compression: <0

F F FA A A

Strain

oo

o

ll

lll

Strain tan

ha

strain Shear

lol lo l h

a

Simple Shear Deformation

= FA

h

x(t)

1 x(t)

h t

Strain =

x(t)h

Strain Rate = . Vh

Shear Modulus G =

V

y

x

Az

Shear Deformation

=

Shear Flow

Shear Stress, Shear Strain, Shear Rate

Stress: Force per unit area.

Symbol: Units: Pa (SI) or dynes/cm² (cgs)

Shear Strain: Relative deformation in shear.Symbol: Units: None (is displacement over height)

Shear Rate: Change of shear strain per unit time.Symbol: Units: [1/s] = s-1

Tensile Test (ASTM D638)

lengthwidth

thickness

Control length (l)

Measure force (F) with load cell

oo

o

ll

lll

Strain

twF

AF

Stress

Reduced section used to limit portion of sample undergoing deformation

Stress-Strain Curve

Str

ess

Strain

Ultimate Tensile Strength

Yield Point

Slope = E (Young’s Modulus)

Percent Elongation(total plastic deformation)

Elastic Limit

Proportionality Limit

Force decreases due to necking

Stress-strain Behavior of Polymers

Brittle

Plastic

Highly elastic

Do not like ceramics and metals, polymer materials exhibit Various stress-strain behaviors, ranging from very brittle to highly

deformableModulus of elasticity ~7MPa to 4GPa.Plastic elongation can be >100%

very sensitive characteristics to strain rate, temperature, chemical environment…

Ductile

Definition and Terminology used

Toughness from the Area of Tensile Tests

1. Tensile Strength and Strain at Break

0A

F

The stress, , and strain, , can be calculated as follows:

A0 is the initial cross-sectional area.

Strain, i is defined as

0

0 )(

l

llii

Tensile Tests

2. Young's Modulus

d

dE 0

0,lim

The modulus, E value, was calculated using a polynomial curve-fit (power of 5) on the stress-strain data over the first 1-2 % strain

3. Toughness

This parameter is the energy for failure or rupture energy.The energy or work of failure can be written as

breakl

lFdlW

0(1)

Tensile Tests

If both sides of equation (1) are divided by A0, which will

bring 0 into the relationship, and realizing that  

0l

dld (2)

000

0

ldA

F

A

W breakl

l

b

dV

W

lA

W

0000

The tensile toughness can be calculated from the integral of the stress-strain curve that is the energy of rupture normalized per unit volume, V0 of the initial sample.

(3)

Tensile response: brittle and rigid•Crosslinked or networked polymers

(e.g. thermoset)

Tensile response: elastomers (rubbery)•Heavily crosslinking => alignment &crystallization difficult

Tensile response: plastic

Upper yield pt

lower yield pt

•Semi-crystalline polymers (linear or branched) (e.g. thermoplastic)

Elastic region Neck formation Elongation

Compare to metals: Deformation is confined within the neck region

Stretching of covalent bond Chains aligned

Neck propagation, localized crystallization -strengthening

Molecular weight↑, TS↑Crystallinity↑, TS↑Drawing↑, TS↑ (introduce deformation during process)Annealing↑, TS↑

Highly Temperature Dependent Near Room Temperature

Polymer (PMMA)

Metal (Iron)

As T increases: • E decreases• TS decreases• Ductility increases

-200oC~25oC

4oC~60oC

W = load at a given point on the load deflection curve, (N) L = Support span, (mm) b = Width of test beam, (mm) d = Depth of tested beam, (mm)

FLEXURE (BENDING) –three point bendingASTM D790

A

W

0L

0AL

W

stress, σ =

strain, ε =

modulus, E =

=

In flexure (bending) situations, these equations do not apply. For the three point bending loading

In tensile analysis:

=

the relevant equations are:

stress, σ = I

My

Where M = bending moment at loading point (=WL/4)y = half depth of beam (= d/2)I = second moment of area (bd3/12)

hence stress, σ = 22

3

bd

WL

deflection δ =EI

WL

48

3

d

L

Ed

L

66

22

=

Hence strain, ε =

2

6

L

d

3

3

4bd

WLModulus, E = =

Material Density(kg/m3)

TensileStrength(MN/m2)

FlexuralModulus(GN/m2)

% Elongation at

break

Price*

ABS 1040 38 2.2 8 2.1

Acrylic 1180 70 2.9 2 2.5

Cellulose acetate 1280 30 1.7 30 3.2

Epoxy 1200 70 3.0 3 8.3

Nylon 66 1140 70 2.8 60 3.9

PEEK 1300 60 3.8 4 42

PET 1360 75 3 70 3

Polycarbonate 1150 65 2.8 100 4.2

Polyethersulfone 1370 84 2.6 60 13.3

Polyimide 1420 72 2.5 8 150

polypropylene 905 33 1.5 150 1

Polysulfone 1240 70 2.6 80 11

PTFE 2100 25 0.5 200 13.3

Polystyrene 1050 40 3 1.5 1.1

LDPE 920 10 0.2 400 0.83

HDPE 950 32 1.2 150 1.1

PVC (rigid) 1400 50 3 80 0.88

PVC (flexible) 1300 14 0.007 300 0.92

DMC (Polyester) 1800 40 9 2 1.5

SMC (Polyester) 1800 70 11 3 1.3

Short-term properties of some important plasticTaken from; RJ Crawford in Plastics Engineering, 3rd Edition,1998, Butterworth Heinemann, p23

* On a weight basis, relative to polypropylene

Sheet moulding compound - SMC •

Dough moulding compound - DMC

Material Selection for StrengthIn service, a material is required to have a certain strength in order to perform its function satisfactorily then a useful way to compare the structural efficiency of a range of materials is to calculate their strengthdesirability factor. Consider a structural member which is essentially a beam subjected to bending (Fig 1). Irrespective of the precise nature of the beam loading the maximum stress, σ, in the beam will be given by:

bd

L

I

dM )2/(max 12/

)2/(2

max

bd

dM

……………………………………. (1.1)

Assuming that we are comparing different materials on the basis that the mean length, width and loading is fixed but the beam depth is variable then equation (1.1) may be written as: σ = β1/d2 (1.2)

where β1 is a constant

But the weight, w, of the beam is given byw = ρbdL …………………. (1.3) (1.3)

So substituting for d from (1.2) into (1.3) w = β2 ρ/ σ1/2 ………………(1.4)

where β2 is the same constant for all materials.Hence, if we adopt loading/weight as a desirability factor, Df , then this will be given by

Where σy and ρ are the strength and density values for the materials being compared.

Similar desirability factors may be derived for other geometries such as struts, columns etc.

2/1y

fD ………………………………………(1.5)

…………………(1.2)

Material Selection for Stiffness If in the service of a component it is the deflection, or stiffness, which is the limiting factor rather than strength, then it is necessary to look for a different desirability factor in the candidate materials. Consider the beam situation described above. This time, irrespective of the loading, the deflection, δ, will be give:

(1.6)

where α1 is a constant and W represent the loading.The stiffness may be expressed as:

δ = α1

EI

WL3

1

1

3L

EI

W

32 Ed

W

(1.7)

where α2 is a constant and again it is assumed that the beam width and length are the same in all cases.

31

/3 Ew

Once again the beam weight will be given by equation (1.3) so substituting for d from equation (1.7)

(1.8) Hence, the desirability factor, Df, expressed as maximum stiffness for

minimum weight will be given by

3/1ED f

Where E is the elastic modulus of material in question and ρ is the density

A ball point pen made from polypropylene has a clip design shown in Fig. 5.1 When the pen is inserted into a pocket, the clip is subjected to a deflection of 2 mm at point A. If the limiting strain in the material is to be 0.5 % calculate (i) a suitable thickness, d, for the clip (ii) the initial stress in the clip when it is first inserted into the pocket and (iii) the stress in the clip when it has been in the pocket for 1 week. The creep curves in Fig. 5 may be used and the short –term modulus of polypropylene is 1.6 GN/m2.

PROBLEM 5.1

2.4 mm

40mm

dWidth6 mm

Figure 5.1

d

w

L

PROB LEM 5.2

A Polypropylene beam is 100 mm long, simply supported at each end and issubjected to a load W at its mid span. If the maximum permissible strain in the material is to be 1.5%, calculate the largest load which may be appliedso that the deflection of the beam does not exceed 5 mm in a service life of 1 year. For the beam I = 28 mm4. and the creep curves in Fig. 5 should be used

L

W

Permeability

The low density of plastics is an advantage in many situations but the relatively loose packing of the molecules means that gases and liquids can permeate through the plastic. This can be important in many applications such as packaging or fuel tanks. It is not possible to generalize about the performance of plastics relative to each other or in respect to the performance of a specific plastic in contact with different liquids and gases. Some plastics are poor at offering resistance to the passage of fluids through them whereas others are excellent. Their relative performance may be quantified in terms of a permeation constant, k, is given by

Atp

Qdk

Where Q = volume of fluid passing through the plasticd = thickness of plasticA = exposed areat = timep = pressure different across surfaces of plastic.

The main fluids of interest with plastics are oxygen and water vapour (for packaging applications) and CO2 (for carbonated drinks applications)

In some cases it is necessary to use multiple layers of plastics because no single plastic offers the combination of price, permeation resistance, printability, etc. required for the application

When multi-layers are used, an overall permeation constant for the composite wall :

Ni

i ik

di

dk 1

11