The importance of orientation

• For unoriented samples there exists a “universalcurve” plotting Modulus vs. Yield Strength. Highmodulus implies high yield strength.

1 2 3 4

Tensile Modulus (GPa)

Tensile Yield (MPa)

100

50

PMMA

SAN

PET

PVC

NY6/6

PPO

HDPE

PTFE

• Conversely it's difficult tomove off this line. It's'impossible' to find a materialwith a low modulus and ahigh tensile strength.

• Modulus of “cold drawn” material increases as afunction of draw or deformation ratio. Differentrelationship for different polymers. Shown is POM.

2 4 6 8 Deformation Ratio

14

10

6

2

Modulus x10 psi-3

• Many polymers have a"natural" draw ratio.Polymers tend to deform sothat at some temperature,draw ratio is relativelyindependent of drawingconditions.

• To obtain a wide range ofdraw ratios we need to havea very wide range ofdrawing conditions.

• Similar effect of draw ratio on Yield strength.Shown below is the effect on Nylon yield afteruniaxial or biaxial drawing, and rolling.

1 2 4 6

Deformation Ratio

10 5

10 4

Yield Stress, psi• Deformation ratios >1,

sample tested in the drawdirection;<1, sample was testedperpendicular to the drawdirection.

• Both modulus and yield strength increase as drawratio increases. However, draw ratio affectsmodulus and yield/tensile strength differently.

• In particular, for many polymers there appears to bea greater effect of draw ratio on yield or tensilestrength. Modulus doesn't continue to increase asdraw ratio increases but may either plateau out orshow a change in slope.

POM Nylon 66

PP

HDPE

10 6

5 10

Tensile Modulus, psi

10 4 510Tensile strength, psi

• The net result is a complex relationship betweenmodulus and strength in drawn materials.

Orientation

• Polymer chains are oriented during processing.Such orientation may be deliberate, and lead toproperty improvement, or it may not.

• Orientation does not imply crystallinity. Polymerscan be amorphous and yet oriented.

• Herman's orientation function (f) is simply amathematical construction that allows us todescribe the degree or extent of orientation of thechain axis relative to some other axis of interest.

Ø

Axis of Interest (X)

Chainf = (3 {cos2Ø} - 1)/2

• Herman's orientation function 'f' has the propertiesthat, relative to the direction of interest, if 'onaverage' the chain axis is:-

completely aligned f = 1

randomly oriented f = 0

perpendicular f = - 0.5

• Physical properties are assumed to be a linearfunction of 'f' and the intrinsic property, and areadditive based on the amount of each component.

Stretch Direction

Aligned Parallel

(fc = 1)

Intrinsic Birefringence = ∆°c

Aligned Parallel (fam = 1)

Intrinsic Birefringence = ∆°am

Orientation function fam ≠ 1

Birefringence = fam. ∆°am

Orientation function fc 1

Birefringence

= fc. ∆°c

Measured Birefringence (∆t) is

∆t = ß. fc. ∆°c + (1 - ß). fam. ∆°am

ß (1 - ß)

• 'f" helps to define chain orientation relative to somedirection of interest, we pick directions of interestthat are important relative to processing.

• For fiber production the direction of interest is thefiber axis; the fiber is loaded along this axis.

• For more 3 dimensional objects we usually select one of three orthogonal axes:-

Extruder

Die Polymer

Sheet

N (Normal)

MD (Machine)

TD (Transverse)

• Note there may be simplerelationships between 'f' asmeasured relative to onedirection of interest, say TDand some other 'f' measuredrelative to another directionof interest say MD.

• Alternative ways of defining orientation have beenused:-

Draw or Orient Lx°

Lx

• The draw ratio (R) is defined as:- R = Lx/Lx°

• For 'cold drawn' polymers R and f are related as:-

-0.5 0 1.0

1.0 2 4 10

(f)

(R)

f ≈ log R

• 'Cold drawn' implies drawing polymer in the solidstate below Tm; it introduces lots of orientation.

• In melt processing, particularly fiber or filmproduction, draw ratio is defined similarly and drawratios of 50 to 100 are not uncommon.

• However, after being oriented a melt tends to 'relax'losing some of the imposed orientation before itsolidifies.

• In fiber production, additional cold drawing is usedto give better polymer chain orientation.

Review of X-ray diffraction, to determine theorientation of chains in a crystal lattice

a

b c

X

Y

Z

• Unit Cell - the smallest volume the crystal can bedivided into and still retain the basic crystal structure.Analogous to atom/element.

• Unit cell dimensions are a, b and c, measured alongthe X, Y and Z directions respectively.

• Miller Indices - method for identifying planes (orfamilies of planes) in the crystal. Recipe:-

X

Y

Z

Plane cuts X, Yand Z, in units ofa, b or c, at :-

Take reciprocal

1, 0.5,

1, 2, 0

• Poles or Normals -that direction thatcuts a particularplane at 90°.

N

I D

Ø Ø

• For diffraction the incident, normal and diffractedbeams must be in the same plane.

• Bragg equation

nλ = 2d sinØ

I

Diffraction Angle (2Ø)

• Use film as a detector of X-rays

Diffraction

Spot

Main

Beam

Side Front

λ

• A diffraction spot on thefilm implies a set ofplanes inclined at thecorrect Ø to theincident beam.

• With enough planes inthe sample that are:-

a) inclined at the correctØ, and yet

b) rotated about theincident beam

• We see a diffraction circle; this classic powderpattern shows random orientation.

The problem

• Note that in the previous example, diffraction onlyarises from planes that are inclined at the correct Øto the main beam. For diffraction, planes must beinclined at the correct Ø.

• Consider an example where the planes of interestare all lying in some sample plane (TD/N plane).

N

MD TD

Film

• No matter along which major axis you send theincident beam, you never get diffraction.

• The X-ray diffraction experiment only 'samples'those planes that are inclined at the correct Øangle to the main beam.

The solution

• Basically, rotate the sample.

N

2Ø

Detector N

TD MD ß

ß

• Set the diffractometer so incident and detector areat the correct 2Ø for some set of planes.

• Start with the incident beam along one of the majoraxes (MD) and rotate the sample about TD, so thatN sweeps through an angle β.

I

0 ß 180

• Record intensity asmeasured at thedetector as a functionof angle β.

• When the maximum number of planes are inclined Øto the incident beam you'll see a maximum inintensity vs. β. Distribution in β models thedistribution of plane inclinations.

• Determining plane positions is now all geometry:-* suppose the correct Ø for a set of planes is 28°* suppose the maximum in the β plot is at 17°

• This means I had to rotate planes by 17° beforethey were at 28° to the main incident beam.

Ø = 28

β = 17

Incident

Detector

2Ø = 56

N

• Therefore the original position of the planes waseither, 28 - 17 = 11° to the incident beamor 90 - 11 = 79° to the original N position.

• Since the poles to the planes are at 90° to theplanes this means that the poles to these planesare lying ≈11° from the original N direction.

• All the above will work assuming that planes areparallel to TD (in this case). Since there is noguarantee of this orientation, we need to sampleALL possible angles/orientations. Such operationsproduce lots of data that needs presenting in anunderstandable manner.

Data presentation: Pole Figures

• Draw a sphere with the sample in the center andshowing TD, MD and N (or ND)

• Instead of reporting planes that we are interested in,we show their poles. Poles are derived from theseries of X-ray experiments described earlier.

• Poles are projected onto the surface of the sphere.

TD

ND

MD

TD

MD ND

TD

MD N

• In practice, instead of asingle spot we may seea band of high intensityindicating a highconcentration of polesat a particular location.

• As an exercise describe both the distribution ofpoles and the distribution of planes as indicatedby the pole figures shown below.

TD

MD ND

TD

MD ND

TD

MD ND

(90 - ß)

• Conventionally, in X-ray studies, the polymer chainaxis is usually reported as lying along the 'Z' or 'c'axis. So, if we can determine the pole figure for001 planes we will then also know the distributionof 'c' axes, and therefore, the distribution ofpolymer chain axes.

• If we know the chain axis distribution, we can workout Herman's orientation function for polymerchains that are in crystalline regions.

• So far we have focused on a single set of planesand primarily 001 planes, that are used to show theorientation of polymer chain axes.

• To completely describe orientation we need to knoworientation of other crystallographic axes relative tosample process axes.

• We need to find the distribution of two orthogonalcrystallographic axes relative to TD, MD and N.

• Such a requirement implies that we need todetermine pole figures for two of the following setsof planes (see below).

TD

MD ND

TD

MD ND

100

"a"

010

"b"

001

"c"

TD

MD ND

• Based on discussions so far this requirement simplymeans we first set-up the X-ray diffractometer sothat 2Ø for the 001 is satisfied.

• After completing these experiments thediffractometer is reconfigured for the correct 2Ø foreither the 100 or 010 planes.

• These requirements may appear trivial, however,there is a problem. Because of the way chains arepacked within the crystal, it may be that the crystaldoesn't have the required 100, 010 or 001reflections. (The electron density distribution issuch that these planes don't have sufficientcontrast to permit diffraction.)

• In that case it is necessary to deduce thedistribution of the primary crystallographic axesfrom any reflections that do exist.

• For example, the two strongest reflections in PE arefrom 110 and 200 planes. With other weakerreflections one can deduce, using rather complexand messy geometry a, b and c axis distributions.