Chapter 4c

Mechanical Properties

Heat Distortion Temperature• The maximum temperature at which a polymer can be used in rigid

material applications is called the softening or heat distortion temperature (HDT).

• A typical test (plastic sheeting) involves application of a static load, and heating at a rate of 2oC per min. The HDT is defined as the temperature at which the

• elongation becomes 2%.

• A: Rigid poly(vinyl chloride)

• 50 psi load.

• B: Low-density poly(ethylene)

• 50 psi load.

• C: Poly(styrene-co-acrylonitrile)

• 25 psi load.

• D: Cellulose acetate

• (Plasticized) 25 psi load.

Transient Testing: Resilience of Cured Elastomers

• Resilience tests reflect the ability of

• an elastomeric compound to store

• and return energy at a given• frequency and temperature.

• Change of rebound • resilience (h/ho) with • temperature T for: • 1. cis-poly(isoprene); • 2. poly(isobutylene); • 3. poly(chloroprene); • 4. poly(methyl methacrylate).

Polyethylene33%Vinyls16%Polypropylene15%PMMAABSNylonPolycarbonateSaturated PolyesterPEEKPolyurethaneSome are thermosets as well.PVC

Not Cross-Linked90% of market

ThermoplasticsWill reform when melted

EpoxyMelamine FormaldehydePhenolicPolyester (unsaturated)PolyimidePolyurethaneSome are thermoplastic as well.SiliconeUrea Formaldehyde

Cross-linked10% of market

Thermosets/ElastomersWill not reform

Polymer Family Tree

Types of Polymers

Tensile strengths

Polymers: ~ 10 - 100 MPa

Metals: 100’s - 1000’s MPa

Elongation

Polymers: up to 1000 % in some cases

Metals: < 100%

Moduli (Elastic or Young’s)

Polymers: ~ 10 MPa - 4 GPa

Metals: ~ 50 - 400 GPa

Ballpark Comparisons

Amorphous v Crystalline Polymers Thermo-mechanical properties

Thermal ExpansionIf a part is to be produced within a close dimensional tolerance, careful

consideration of thermal expansion/contraction must be made.

Parts are produced in the melt state, and solidify to amorphous or semi-crystalline states.

Changes in density must

be taken into account

when designing the mold.

Thermal Expansion

Stress Strain Studies

Initial slope is the Young’s Modulus (E’ or sometimes G)TS = tensile strengthy = yield strengthToughness = Energy required to break (area under curve)

/strain

Elongation = 100% x

Anatomy of a Stress Strain Graph

Compression and Shear vs. Tensile TestsStress-strain curves are very dependent on the test method. A modulus

determined under compression is generally higher than one derived from a tensile experiment, as shown below for polystyrene.

Tensile testing is most sensitive

to material flaws and microscopic

cracks.

Compression tests tend to

be characteristic of the polymer,

while tension tests are more

characteristic of sample flaws.

Note also that flexural and shear

test modes are commonly

employed.

Elongation by extension of neck

Chains in neck align along elongation direction: strengthening

Stress Strain Graphs

Beyond “B”, the yield strength, deformations are plastic

Thermosets = strong & brittleNot Ductile

Thermolastics = depends on T

Ductility & Elongation (EL)EL < 5% BrittleEL > 5% Ductile

Cold Drawing above the Tg

25

• Compare to responses of other polymers: --brittle response (aligned, cross linked & networked case) --plastic response (semi-crystalline case)

Stress-strain curves adapted from Fig. 15.1, Callister 6e. Inset figures along elastomer curve (green) adapted from Fig. 15.14, Callister 6e. (Fig. 15.14 is from Z.D. Jastrzebski, The Nature and Properties of Engineering Materials, 3rd ed., John Wiley and Sons, 1987.)

TENSILE RESPONSE:initial: amorphous chains are kinked, heavily cross-linked.final: chainsare straight,

stillcross-linked

02040600246( )MPa8xxxelastomer plastic failure brittle failureDeformation !is reversible

Elastomer Molecules

Model of long elastomer molecules, with low degree of cross‑linking: (a) unstretched, and (b) under tensile stress.

High entropy Low entropy

Low energy High energy

13

0.280.61Magnesium,AluminumPlatinumSilver, GoldTantalumZinc, TiSteel, NiMolybdenumGraphiteSi crystalGlass-sodaConcreteSi nitrideAl oxidePCWood( grain)AFRE( fibers)*CFRE*GFRE*Glass fibers onlyCarbon fibers onlyAramid fibers onlyEpoxy only0.40.8246102040608010020060080010001200400TinCu alloysTungsten<100><111>Si carbideDiamondPTFEHDPELDPEPPPolyesterPSPETCFRE( fibers)*GFRE( fibers)*GFRE(|| fibers)*AFRE(|| fibers)*CFRE(|| fibers)*

MetalsAlloys

GraphiteCeramicsSemicond

PolymersComposites

/fibers

E(GPa)

Eceramics> Emetals>> Epolymers

109 Pa

Based on data in Table B2,Callister 6e.Composite data based onreinforced epoxy with 60 vol%of alignedcarbon (CFRE),aramid (AFRE), orglass (GFRE)fibers.

YOUNG’S MODULI: COMPARISON

• Hooke's Law: = E

• Poisson's ratio, :

metals: ~ 0.33 ceramics: ~0.25 polymers: ~0.40

Lε1-ν

FFsimpletensiontest

Linear-elastic1Eε

Units:E: [GPa] or [psi]: dimensionless

Linear Elasticity: Possion Effect

€

=−width strain

axial strain= −

Δw /w

Δl /l= −ε Lε

Why does have minus sign?

Poisson Ratio

€

=−Δw /w

Δl /l= −1

• Poisson Ratio has a range –1 1/2

Look at extremes• No change in aspect ratio:

€

Δw /w = Δl /l

• Volume (V = AL) remains constant: ΔV =0.

Hence, ΔV = (L ΔA+A ΔL) = 0. So,

In terms of width, A = w2, then ΔA/A = 2 w Δw/w2 = 2Δw/w = –ΔL/L.

Hence, €

ΔA /A = −ΔL /L

€

=−Δw /w

Δl /l= −

(−1

2Δl /l )

Δl /l=1/2 Incompressible solid.

Water (almost).

Poisson Ratio: materials specific

Metals: Ir W Ni Cu Al Ag Au 0.26 0.29 0.31 0.34 0.34 0.38

0.42generic value ~

1/3Solid Argon: 0.25

Covalent Solids: Si Ge Al2O3 TiC 0.27 0.28 0.23 0.19 generic

value ~ 1/4

Ionic Solids: MgO 0.19

Silica Glass: 0.20

Polymers: Network (Bakelite) 0.49 Chain (PE) 0.40

Elastomer: Hard Rubber (Ebonite) 0.39 (Natural) 0.49

Tensile stress is applied along cylindrical brass rod (10 mm diameter). Poisson ratio is = 0.34 and E = 97 GPa.

• Determine load needed for 2.5x10–3 mm change in diameter if the deformation is entirely elastic?

FFsimpletensiontest

Example: Poisson Effect

Width strain: (note reduction in diameter)x= Δd/d = –(2.5x10–3 mm)/(10 mm) = –2.5x10–4

Axial strain: Given Poisson ratioz= –x/ = –(–2.5x10–4)/0.34 = +7.35x10–4

Axial Stress: z = Ez = (97x103 MPa)(7.35x10–4) = 71.3 MPa.

Required Load: F = zA0 = (71.3 MPa)(5 mm)2 = 5600 N.

Negtive poisson’s ratio• foams

Lakes, R. S., "No contractile obligations", Nature, 1992, 358, 713-714.

CompressionRadialn = -1.24Axialn = 0

Anisotropic Materials

1) Compaction of UHMWPE powder2) Sintering3) Extrusion

• Mechanical properties are sensitive to temperature

FIGURE 10.9 Effect of temperature on the stress-strain curve for cellulose acetate, a thermoplastic. Note the large drop in strength and increase in ductility with a relatively small increase in temperature. Source: After T.S. Carswell and H.K. Nason. Manufacturing Processes for Engineering Materials, 5th ed. Kalpakjian • Schmid Prentice Hall, 2008.

Poly(methyl methacrylate)

Str

ess

Strain

Polymers

Metals

Ceramics

•Lower elastic modulus, yield and ultimate properties

•Greater post-yield deformability

•Greater failure strain

Polymers: Thermal Properties

• In the liquid/melt state enough thermal energy for random motion (Brownian motion) of chains

• Motions decrease as the melt is cooled

• Motion ceases at “glass transition temperature”

• Polymer hard and glassy below Tg, rubbery above Tg

Polymers: Thermal Properties

linear amorphous

log

(Mo

du

lus)

Temperature

semicrystalline

crosslinked

TmTg

Polymers: Thermal PropertiesS

tres

s

Strain

decreasing temperature or increasing crystallinity

• Properties depend on amount of cross-linking

Figure 8.13 M. P. Groover, “Fundamentals of Modern Manufacturing 3/e” John Wiley, 2007

Increasing cross-linking