8 X-Rays Diffraction From Fibres

8/3/2019 8 X-Rays Diffraction From Fibres

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X-rays Diffractionfrom Fibres


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Typical fibre X-ray diffraction pattern


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Intensity curve-Equatorial scan of

A Nylon6 fibre


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Crystallinity by X-rays

X-ray Diffractometer


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Braggs condition for diffraction: 2dsin = n

Crystallinity:

Scattering from the sample = I(s) (s = 2sin/)

Scattering from the crystallinePart = Ic(s)

Xc = 0 Ic(s)dxs / 0

I(s) dxs

0 I(s) dxs = 4 0

s2 I(s) ds

(For spherical symmetry dxs = 4 s2 ds)

Xc = 0 s2Ic(s) ds / 0

s2I(s) ds ------------------(1)

Powered sample measurementEffect of voids no effect.


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Slope = Cc

Ic - Ia

(Iu Ia)

Crystallinity Index:

Method for materials when 100% amorphous not available

Standard Crystalline and amorphous samples:

Cellulose:-c- hydrolysed in HCl,

a- Alcohol extracted.

(Iu Ia) = Cc(Ic Ia) +B


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X-ray intensity curves of the crystalline and amorphous standards

used for the analysis of cotton cellulose.

X-ray intensity curves of the

crystallinity and amorphous standard


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Crystal Sizes

Scherrers Equation:

Size of the crystal: Lnkl =K / cos

crystal defects also give rise to broadening

hkl diff. peak

Io

I

Small angle Xray scattering for crystal sizes:

2dsin = n

Pseudo Lattice with d ~ 100 Ao, = 0o

Long Periodicity:d = C+A, X = C / (C+A)C

A

1 1.5 2

I


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Small angle Xray scattering for crystal sizes:

Diffracted X rays

Sample

Xray sound

Void Sizes Guiniers Law

lnI(s) = lnI(0) 1/3 s2 R2

I lnI(s) Slope = -1/3 R2

s2


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SASX


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Modulus of PE by X-ray diffraction

Highly oriented PE

002 = 37.379o d002 = 0.127nm

Stress of 1GN/m2, 002 = 37.226o

Strain(?)

2 * 0.127(1 +)Sin(37.226) = 0.1542

= 1/285, Modulus = 1/(1/285) =285 Gn/m2

0 0.5 1.0 1.5

I

Xc = 0.60, = 1000 kg/m3, c = 855 kg/m3

= 0.15 nm

Maximum At 2 of 0.5oLong Period is d 2dsin = n n=1

2*d sin(0.5/2) = .15nm

d =~ 20 nm

d= c+a X = c/(c+a) = 0.7

c=(c+a) * 0.7 = 20 * 0.7 = 14 nm


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Schematic Presentation of various small angle patterns

Two point pattern:

Four point pattern:

Equatorial Pattern:


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Typical Results from SAXS in Ao

PE PP Nylon PET

Long Period 90 90 150 180

Crystallite Length 60 60

Amorphous Length 30 30

Diameter 160 120 90 330-160

Nylon 6 Long Period 86, amorp = 28.7; N66 91Ao & 30.3 Ao

Nylon 6 Fibres PET Fibres


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X-ray photographs of nylon fibres


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Crystalline Orientation by X-ray diffraction

Preferred orientation Primary bonds for deformation.

Presence of arcs in diffraction pattern

Hermans Orientation function(Affine deformation):

f = [3(cos2)-1]

fc,z = [3(cos2 c,z)-1]

fa+fb+fc = 0For perfect orientation

= 0,(cos2 ) = 1, f = 1

For random orientation

(cos2) = 1/3, f = 0

For cylindrical symmetry

(cos

2

) = [o/2

I()S

incos

2

d] / [o/2

I()sin d]Experimental Measurement:

Calculate f from the following data:

I() = cos

(cos2 ) = [o /2Sin cos3 d] / [o

/2 cossind]

= o 1 x3 dx / o

1 xdx = x4 *2 / 4* x2 lo1

= [3/2 1] = 1/4


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For uniaxial orientation, a series of (hkl) planes pref. Perpendicular

to fibre direction.

Polyethylene: (110);(200);(020) to determine fa and fbOrthorhombic(a = b = c)

Polyethylene terephthalic lacks (h o o) or (o k o) reflections.

(T, o , 5) off meridian reflection.

Triclinic

cos2 hkl = e2 cos2 n + f2 cos2 + g2 cos2 + 2ef cosa cosb+2fg cosb cos c + 2ge cosa cosb..(1)

e, f, g are geometric constants of the unit cells.

In general cos2 hkl are required to determine fn(1)

Polypropylene: (110) & (040) are used for fc


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Monoclinic 2 = 14.2 2 =16.9

For PP < cos2 c, z > = 1-1.099 < cos2 110, z > -0.901 < cos

2 040, z >

For PET T05 reflection is used for fc determination.

For nylon < cos2 b, z > = 1.0-1.20 < cos2 200, z >-0.795 < cos2 202, z >

Orientation in amorphous regions:-

oriented amorphous halo observed in the case of PET

Biaxial orientation.


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Small-angle diagram of linear polyethylene

drawn to an extension ratio of approx.6:1

at 115oC.Nickel-filtered CuK radiation;

specimen-to-film distance, 400mm fibre

axis,b3

Relationship between the length of themeridian streak and the transverse

dimension of the diffraction source.


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The decomposition of the diffraction

pattern for polypropylene.


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Plot of Iu Ia versus Ic Ia for a typical cotton sample


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Relaxation time distribution controlling the width of loss peak

and length distributions are related to long peak.

8 X-Rays Diffraction From Fibres

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