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Non Metalic Materials Assignment 3

December 2 2003

 

1) Plastics produced from the following monomers.

      a)      Polytetrafluoroethylene (PTFE)

b)      Polyvinyl chloride (PVC)

c)      Polystyrene (PS)

d)     Poly(p-phenyleneterephthalamide) (PPTA)

 

2) True or False

     (a)    High molecular weight increases crystallinity – FALSE

(b)   Density of semi-crystalline polymer increases with crystallinity-TRUE

(c)    Isotatic PP is an amorphous polymer-FALSE

(d)   Cannialistic’ effect occurs only in amorphous polymers-FALSE

(e)    Chain branching increases crystallinity in polymers-FALSE

(f)    Sperulites are generally larger in amorphous polymers-FALSE

(g)   It is the cross-linking in thermosets which prevents thermal softening-TRUE

(h)   In sperulitic polymers their tensile strength at yield always increases with decreasing sperulitic size-TRUE

(i)     Thermosetting polymers release a by-product upon their polymerization-TRUE

(j)     Thermosetting polymers soften over a large temperature region-FALSE

(k)   Thermosetting polymers are frequently recycled for repeated use-FALSE

(l)     The glass transition temperature is relatively high if degree of crystallinity is high-TRUE

(m) Thermoplastic polymer has a viscoelastic modulus that depends on degree of cross-linking or crystallinity-TRUE

(n)   Every crack with atomistically harp tip will propagate catastrophically-FALSE


 

3)      Glass transition temperature is the temperature at which the mechanical properties of the polymer changes from elastic behaviour to brittle behaviour.  The Glass transition temperature is dependent on many factors.  Natural rubber has a glass transition temperature of 200k while that of PET is 338K.  This is due to differences in molecular weight.  The molecular weight of rubber is 68.12 g/mol and that of PET is 102.2 g/mol.  A polymer with a high molecular weight is more densely packed than a polymer with a low molecular weigh.  In a more densely pack polymer, the chains have less freedom to move and requires a higher energy (or higher temperature) for.  Since PET is more densely packed it reaches its glass transition temperature higher than rubber.

 

4)      Table 1 below shows the crystallization rates in various polymers.

 

 

5)      The most flexible chain, that of high-density polyethylene (HDPE), has the higher growth rate because chains can rotate very easily into folded chains forming crystallite lamella.  Nylon and acetal also both have polyethylene-like chains so their growth rates are also high.  Chains hindered by side groups, such as PP and isotactic PS show severely reduced growth rates.  This is because the side chains make folding and rotating of the chain more difficult.  Both polyethylene terephthalate (PET) and polycarbonate are hindered by aromatic groups in the chain itself, so have very slow groth rates.  Thus the more easily a chain can rotates the faster the rate of crystallization.

 

6)      The figure below, demonstrates the relationship between crystallization temperature and the lamellar thickness of the crystallite.

            The thickness of a crystalline develops in two stages:

Stage 1 - The polymer crystallizes directly at a certain constant crystallite thickness which is larger for higher crystallization temperatures.

Stage 2 – Surface energies drive the already formed crystallites to further thicken as they are held at the crystallization temperature for a fixed time,  This second stage is an “annealing phenomenon” which occurs at all crystallization temperature bat at a rate which increases with temperature.

As the temperature increases the lamellar thickness increases.  This is because with increased temperature the rate of nucleation and crystal growth increases.

3)      A copolymer of PP with 10 wt% ethylene is used in many products.  The ethylene repeat units are randomly distributed in the pp chain.

a.       The copolymer of PP will have a lower density.  This is because the molecular weight of ethylene is lower than the molecular weight of the PP mer (c2H4).

 b.      The Tg and Tm of PP copolymer will be is lower than that of PE.  This relates to the fact the density will be lower 

c.       The tensile strength of the PP copolymer will be lower 

d.      The impact strength of the PP copolymer will be lower

  

4)      Calculations of crystal Density

 a.       PE: 736, 492, 254 pm; 2[-CH2-CH2-] units per unit cell

 a= 736E-12      b= 492E-12      c=254E-12

volume = abc = 9.2E-29

molarMass= 2(2*(12.01) + 4*(1.01)) = 56.12 g/mol

cellMass = molarMass/A0 = 56.12/6.02E23 = 9.32E-23 g/ unit cell

Density =  cellMass/volume = 9.32E-23/9.2E-29 = 1.01E6  g/m3

 

b.      Syndiotactic PVC: 1040, 530, 510 pm; 4[-CH2-CHCl-] per cell

 a= 1040E-12    b= 530E-12      c=510E-12

volume = abc = 2.81E-28

molarMass= 62.5 g/mol

cellMass = molarMass/A0 = 1.04E-22 g/ unit cell

Density =  cellMass/volume = 3.6E5 g/m3

  

c.       Polyvinyl fluoride: 857, 495, 252 pm; 2[-CH2-CHF-] per cell

 

a= 857E-12      b= 495E-12      c=252E-12

volume = abc = 1.07E-28

molarMass= 92.1 g/mol

cellMass = molarMass/A0 = 1.53E-22 g/ unit cell

Density =  cellMass/volume = 1.43E6 g/m3

  

d.      Natural rubber: 1246, 886, 810 pm; 8[-CH2-CH=C-CH2-]

 

a= 1246E-12    b= 886E-12      c=810E-12

volume = abc = 8.94E-28

molarMass= 544 g/mol

cellMass = molarMass/A0 = 1.53E-22 g/ unit cell

Density =  cellMass/volume = 1.01E6 g/m3

  

5)      The creep curve for PVC is shown below.

 

 

For 20000 service hours the service the stress is 17.5 Mpa, and a strain of 2%

Thus a suitable wall thickness for a 150 mm diameter and a pressure of 0.7Mpa is given by:

  

6)      The Stress intensity factor is defined as:

 

 

a.       w = 100 mm, 2a = 40 mm, σ = 3.91 MPa K=??

 

 

b.      w = 100, 2a = 14, σ = 9 MPa, K = 1.09  K=??

 

 

 

Since (Ka = 1.09) < (Kb = 1.35) the crack will propagate

 

c.       2a = 3 mm, w = infinite, σ = 10 MPa

 

 

            Using the new equation.

           

           

            Since (Kc=0.6865) < (Ka=1.09) the crack will not propagate

 

7)      From the ‘Deformation and Fracture Mechanics of Engineering Metals  

2a = 60 mm, σ = 10 MPa w = infinite 

a.       Find K

 

 

 

b.      Find: Gc given that elastic modulus E = 3 GPa

 

 

 

c.       Given: 2a = 2 mm, σ = 10 MPa, find if it will fracture

 

 

since () < () the crack will not propergate

  

8)      The major mechanism of plastic deformation in tension for  glassy polymers is crazing and shear banding.  Crazes are narrow zones of highly deformed polymer contiang voids.  The zone are oriented perpendicult to the stress axis as shown below.

 

 

In the crazed zone, the molecular chains get aligned along the stress axis, but they are interspersed with voids.  A craze, however, is capable of supporting some reduced stress relative to that of the uncrazed matrix and hence can neither be classified as a void nor a crack in the material.  Crazing occurs in brittle glassy polymers.

 Shear band form at 45 degrees to the larges principal stress.  The polymeric molecular chains get oriented within the shear bands with any change in volume.  Shear yielding takes two forms diffusd shear yielding and localized shear yielding.

 Semi-Crystalline

The typical stress strain curve for a semi-crystalline is shown below

It is characterized by a liner elastic region; a yielding followed by a drop in stress, a formation of a neck, an increase in stress due to straightening of polymer chain, and finally fracture.. 

  Deformation in semi-crystalline polymers is shown in the figure below

 

 

 

9)      The figure below shows the variation of stress intensity factor with crack velocity in PMMA

 

It shows that the stress intensity factor increases exponentially as the crack velocity increases. 

Most polymer s are viscoelastic, and steady state cracks growth occurs at an applied stress intensity that is less than the critical value.

 

10)  In creep tests the polymers specimen are loaded in tension under a constant load for a certain timer.  The elongation ir strain of the polymer is recorded.  The expected behaviour of polystyrene under different conditions are shown below.

 

 

a.       Crystalline at 70C (assume 100% Crystallainty)

 

 

The figure above shows the strain response for a crystalline at 70C.  It shows no viscoelatic or viscous behaviour.  Because the polymer is crystallised the chains cannot move in-between each other, and its behaviour is time independent.

 

b.      Amorphous at 180°C

 

 

Amorphous polystyrene at 180° will display the above behaviour.   During the initial stages the polymer chains are unentangling and straighengin

 

c.       Cross-linked at 180°C

 

 

The above figure shows the effect of cross-linking.  It is similar to that of a 100% crystalline.  Since the polymer chains are  bonded strongly by crosslink they restrict the sliding of chains, thus resist elongation.

  

d.      Amorphous at 100°C

 

  

11)  To improve fracture toughness micro structural modification can be done.

 

a.       The fracture toughness of a glassy polymer can be enhances wither by suppressing the crazing tendency and replacing it with shear yielding, or by induction of high density by evenly distributed micro-crazing.   

In glassy thermoplastics, the introduction of elastomeric micro constituent can be achieved by copolymerization.

 

b.      In semi-crystalline (crystalline) polymers, some improvement of fracture toughness can be achieved by refinement of spherulitic size. 

 At low strain rates, the fracture follows an interspherulitic path, while at high strain rates the fracture becomes transspherulitic. Since the transspherulitic path is more difficult for crack propagation, it has higher fracture toughness.

 The highest toughness is exhibited by a very fine spherulitic structure, and the worse by course spherulitic structure.

 Fracture depends on two things.  Interspherulitic path or transspherulitic path and Spherulite size.  Thus by modifying the Spherulite size fracture toughness can be improved.

  

12)  Calculation of composition(%) of LDPE and PE in blend

            A = tensile strength of LDPE = 28 MPa

B = tensile strength of PE = 7 MPa

 

Composition of LDPE (%) = (A)/ (A +B))  = 28 / (28 + 7)   = 0.8 = 80 %

      Therefore

      Composition of PP (%) = 1 – 0.8 = 0.2 = 20%

  

13)  Cross-linking is the physical linking between polymer chains as seen in the figure below.

 

 

Crosslink occur in themosets and elastomers.  In themosets crosslinks anchor the polymer chains together to resist vibrational motion of the chains at high temperatures.  It give themoset it high melting points and strong characteristics.

 In elastomers crosslinks enable extremely high strain rates.  In elastomers crosslinks occurs less frequently along the chain than in themosetts.  In elastomers the crosslinks and the main chains are flexible rather than stiff in themosets.  The flexibility exists because og the geometry at the carbon-carbon double bond causes a bend in the chain.  Theses bends have a cumulative effect over a ling chain, such that the chain is coiled between the cross-linked points.  Thus upon loading the coils unwind between the crosslinks attachment points, and after removal of the stress, the coils recover..

 The initial modulus of rubbers without any cross-links is very low as it is only associated with uncoiling of the chains resulting in a value on the order of E = 1 MPa. Vulcanization, the process of forming cross-links long the chains.  It is carried out by adding sulphur compounds to a heated elastomer.  Increasinfg the sulphur conten varies the amount of vulcanizaion, resulting in different degress of crosslinking.  This affect the elastic modulus.

 

 

From the figure above it is evident that increase vulcanisation of increased crosslinking cause the modulus to increase at a given temperature.