tutorial 4 solutions

ENGG103 –Materials in Design

Tutorial 4 – Weeks 9 & 10

Metals Processing (Welding) Polymer Structures Polymer Properties

Upon successfully completing these tutorial exercises, students should be able to: • Demonstrate an understanding of welding as a joining method

including its advantages and disadvantages • Identify and describe the molecular structure of common

polymeric materials • Perform calculations related to the structure of polymeric

materials • Describe and explain structure-property relationships in

polymeric materials

ENGG103 - Tutorial 4 2

Aims

Exercise 4.1 – Welding


The figure below shows the basic cross section profile of a butt weld produced by fusion welding. a) Identify each of the following regions:

i. Parent Metal ii. Heat Affected Zone (HAZ) iii. Partially Melted Zone iv. Weld Metal/Fusion Zone



Weld Metal

Parent Metal

Partially Melted Zone

Heat Affected Zone (original joint profile)



The figure below shows the basic cross section profile of a butt weld produced by fusion welding. b) On the weld cross section presented, identify three regions where the welded joint

may be susceptible to failure. Complete the table below to briefly describe why failure may occur at these locations.



Weld Toe

Parent Metal

Weld Metal

Heat Affected Zone Weld Root



Location Feature Possible Causes of Failure

1 Weld Toe Weld Root

Geometric discontinuities - Stress concentrators - Weld defects (e.g. undercut and misalignment) Hydrogen assisted cracking

2

3

4






2 Weld Metal

Microstructural changes - Segregation of alloying elements, grain size and

morphology, precipitates and inclusions - Localised hardening or softening Residual stress Weld defects (e.g. porosity, inclusions, lack of fusion)

3

4






2 Weld Metal



3 Heat affected zones


morphology, precipitates and inclusions - Localised hardening or softening Residual stress

4






2 Weld Metal



3 Heat affected zones


morphology, precipitates and inclusions - Localised hardening or softening Residual stress

4 Parent Material Residual Stress

Exercise 4.2 – Polymer Structures


Complete the table below by drawing the repeat unit structure and calculating the molar mass of the repeat unit for each polymer.

Polymer (Abbreviation)

Repeat Unit Structure Molar mass

(g/mol)

Polyethylene (PE)

Polytetrafluoroethylene (PTFE)

Poly(vinyl chloride) (PVC)

Polypropylene (PP)

Polystyrene (PS)



i. Polyethylene: • Most basic polymer structure • Repeat unit has two carbon atoms, each with two hydrogen atoms bonded

• Molar mass = 2 × 12.01 + 4 × 1.008 𝑔𝑚𝑚𝑚

= 28.052 𝑔𝑚𝑚𝑚



ii. Polytetrafluoroethylene: • Similar to the polyethylene structure • Hydrogen atoms are replaced by fluorine atoms • (tetra = 4)

• Molar mass = 2 × 12.01 + 4 × 19.00 𝑔𝑚𝑚𝑚

= 100.02 𝑔𝑚𝑚𝑚



iii. Poly(vinyl chloride): • Similar to the polyethylene structure • One hydrogen atom is replaced by chlorine atom

• Molar mass = 2 × 12.01 + 3 × 1.008+35.45 𝑔𝑚𝑚𝑚

= 62.494 𝑔𝑚𝑚𝑚



iv. Polypropylene: • Repeat unit contains 3 carbon atoms • (prop = 3) • One carbon atom sits as a side group with 3 of its own hydrogen atoms

• Molar mass = 3 × 12.01 + 6 × 1.008 𝑔𝑚𝑚𝑚

= 42.078 𝑔𝑚𝑚𝑚



v. Polystyrene: • Similar to the polypropylene structure • Side group is a phenyl group (6 carbon atoms in a ring)

• Molar mass = 8 × 12.01 + 8 × 1.008 𝑔𝑚𝑚𝑚

= 104.144 𝑔𝑚𝑚𝑚

R is the link to the polymer chain



Polymer (Abbreviation)

Repeat Unit Structure

Carbon Atoms

Hydrogen Atoms

Fluorine Atoms

Chlorine Atoms

Molar mass (g/mol)

Polyethylene (PE)

2 4 0 0 28.052

Polytetrafluoroethylene (PTFE)

2 0 4 0 100.02

Poly(vinyl chloride)

(PVC) 2 3 0 1 62.494

Polypropylene (PP)

3 6 0 0 42.078

Polystyrene (PS)

8 8 0 0 104.144



Analysis of a sample of polypropylene showed the molecular weight distribution given in Table 4. From this information determine: a) The number-average molecular mass, 𝑀𝑛 b) The weight-average molecular mass, 𝑀𝑤 c) The number-average degree of polymerisation, DPn d) The weight-average degree of polymerisation, DPw

Molecular Weight (g/mol)

Number Fraction Xi (%)

Weight Fraction Wi (%)

8,000 5 2

16,000 16 10

24,000 24 20

32,000 28 30

40,000 20 27

48,000 7 11



Step 1 – Define a) The number-average molecular mass, 𝑀𝑛 b) The weight-average molecular mass, 𝑀𝑤 c) The number-average degree of polymerisation, DPn d) The weight-average degree of polymerisation, DPw

Step 2 – Data Polypropylene (from previous, molecular mass of repeat unit 𝑚 = 42.078 𝑔/𝑚𝑚𝑚)


Number Fraction Xi (%)

Weight Fraction Wi (%)

8,000 5 2

16,000 16 10

24,000 24 20

32,000 28 30

40,000 20 27

48,000 7 11



Step 3 – Theory

a) 𝑀𝑛 = ∑ 𝑁𝑖𝑀𝑖𝑖𝑁𝑖

b) 𝑀𝑤 =∑ 𝑊𝑖𝑀𝑖𝑖𝑊𝑖

c) 𝐷𝐷𝑛 = 𝑀𝑛𝑚

d) 𝐷𝐷𝑤 = 𝑀𝑤𝑚

Step 4 – Estimate

a) and b) should be similar with weight-average slightly higher (around 30,000 g/mol)

c) and d) should be similar with weight average slightly higher



Step 5 – Solve NEXT: Determine mean molecular weight for each range (Mi)


Number Fraction

xi (%)

Weight Fraction wi (%)

8,000 5 2

16,000 16 10

24,000 24 20

32,000 28 30

40,000 20 27

48,000 7 11

SUM 100 100


Number Fraction

xi (%)


8,000 5 2 16,000 16 10 24,000 24 20 32,000 28 30 40,000 20 27 48,000 7 11

Exercise 4.3

𝑀𝑛 =∑ 𝑁𝑖𝑀𝑖𝑖𝑁𝑖

number-average molecular mass (𝑀𝑛):

𝑀𝑖 - molecular weight of a chain 𝑁𝑖 - is the number of molecules with weight 𝑀𝑖

𝑀𝑛 =8,000 ∗ 5 + 16,0000 ∗ 16 + 24,000 ∗ 24 + 32,000 ∗ 28 + 40,000 ∗ 20 + 48,000 ∗ 7

5 + 16 + 24 + 28 + 20 + 7

29,040 g/mol

a)

• We know the molecular weight and the number faction present in the samples for of the different polymer chains.

• By looking to the formula, we have to multiply the polymer molecular weight by the number fraction present in the polymer for the different polymer chain molecular weight and divide by the total number fraction.

=

b)


Number Fraction

xi (%)


8,000 5 2 16,000 16 10 24,000 24 20 32,000 28 30 40,000 20 27 48,000 7 11

Exercise 4.3

𝑀𝑤 =∑ 𝑊𝑖𝑀𝑖𝑖𝑊𝑖

weight-average molecular mass (𝑀𝑤):

𝑀𝑖 - molecular weight of a chain 𝑁𝑖 - is the number of chains 𝑊𝑖 - weight fraction of a polymer with molecular weight 𝑀𝑖

𝑀𝑤 =8,000 ∗ 2 + 16,0000 ∗ 10 + 24,000 ∗ 20 + 32,000 ∗ 30 + 40,000 ∗ 27 + 48,000 ∗ 11

2 + 10 + 20 + 30 + 27 + 11=

32,240 g/mol

• We know the molecular weight and the weight fraction present in the samples for of the different polymer chains.

• By looking to the formula, we have to multiply the polymer molecular weight by the weight fraction present in the polymer for the different polymer chain molecular weight and divide by the total weight fraction.

=



Step 5 – Solve Then: From our previous question, 𝑚 = 42.078 𝑔/𝑚𝑚𝑚 for polypropylene

c) 𝐷𝐷𝑛 = 𝑀𝑛𝑚

= 29,040 𝑔/𝑚𝑚𝑚42.078 𝑔/𝑚𝑚𝑚

≈ 690.147

d) 𝐷𝐷𝑤 = 𝑀𝑤𝑚

= 32,240 𝑔/𝑚𝑚𝑚42.078 𝑔/𝑚𝑚𝑚

≈ 766.196

Step 6 – Verify

Mass and weight fractions sum to 1.00 or 100% All answers agree with initial estimates and so seem to be reasonable solutions

Exercise 4.4

a) Figure 1 shows that HDPE - high density polyethylene, cools at slower rate, giving origin to a more packed structure or after crystallization, the polymer occupies less volume (figure 1), and consequently should be a linear polymer chain. On the other way, branched polymer chain occupies higher volume (figure 2), it should be a LDPE (figure 1)

V

Vs

T

Vf

𝑇𝑔𝑠 𝑇𝑔𝑓

HDPE

LDPE

Figure 1

Linear Polymer chain

Branched Polymer chain

Remember that we are talking of homopolymers only…..

Exercise 4.4

If the material is amorphous is transparent But Due to the light interaction with the crystalline and amorphous regions, there will be diffraction and the material will appear to be white REMEMBER that material density is different

𝝆𝒇 < 𝝆𝒔

na

nc

𝒏𝒂 ≠ 𝒏𝒄

Interface between the amorphous and crystalline regions

• In semi-crystalline polymers, the refractive index of amorphous and crystalline regions are different, so the light will scatter when travel through the polymer due to the interaction of light between the crystalline region and the amorphous one.

• In amorphous polymers we only have one refractive index, and no interaction between the different regions occurs.

b)

Exercise 4.4

c) Linear Polymer chain Branched Polymer chain

• If we look to the pictures above, the crystalline regions will enhance the mechanical

properties of the polymers because they are more ordered and occupy less volume

when compared to branched polymer ones, so it should be more easy to stretch them

with less effort.

• HDPE – linear chains – high mechanical properties

• LDPE – branched chains – lower mechanical properties

Exercise 4.4

d)

• HDPE – linear chains – high mechanical properties, higher amount of crystalline

regions, high melting point

• LDPE – branched chains – lower mechanical properties, lower crystalline regions, lower

melting point

V

Vs

T

Vf

𝑇𝑔𝑠 𝑇𝑔𝑓 𝑇𝑚𝑠 𝑇𝑚

𝑓

Does a pure amorphous Polymer melt?

𝑇𝑔 Glass transition temperature

𝑇𝑚 Melting temperature

HDPE

LDPE



Calculate the heat generated (in kW.h) during an electrical resistance welding process performed with a current of I=8000 A for t=3 seconds on a steel with resistance R=1.4x10-7

Ω.

Answer:

Heat generated = I 2 Rt

where: I is the welding current (A) R is the resistance at the joint interface (Ω) t is the time (s) that the current is allowed to flow.

Therefore: Heat generated = (8000)2(1.4 x 10-7)(3) = 26.88 W.s = 7.47x10-6 kW.h

Discussion Questions


a) Would you expect the tensile strength of polychlorotrifluoroethylene to be greater than, the same, or less than that of polytetrafluoroethylene for samples with the same molecular weight and degree of crystallinity? Justify your answer

Points to consider: i. What is the origins of tensile strength in polymers? ii. What is the structure of each, particularly in terms of side groups? iii. What is the nature of the side groups? iv. How will the side groups influence the plastic deformation process? v. Consider not only the size of side groups but their ability to influence ponding

(primary and secondary)

Polychlorotrifluoroethylene (PCTFE) Polytetrafluoroethylene (PTFE)

PTFE is a regular Polymer, organized in a Isotatic structure – no side groups present PCTFE have a side group, a vinyl group (Cl), depending on the place occupied in the macromolecule, the Mechanical properties will be influenced. Stength is higher for Isotatic structure, less volume, higher crystallinity, and the other ones (Syndiotactic and Atactic), the Polymer occupies higher volume, and consequently the mechanical properties decrease

Isotactic Polymer

Syndiotactic Polymer

Atactic Polymer

Discussion Questions


b) During the winter months the temperature in some parts of Alaska may be as low as -55 °C. Which of the following elastomers would be most suited to the manufacture of automobile tyres to be used to be used in these conditions? Justify your answer.

Elastomers:

• natural polyisoprene • styrene-butadiene • acrylonitrile-butadiene • chloroprene • polysiloxane

The tyres most be soft at -55 ºC, so the glass transition most be below that temperature • Polyisoprene, Tg = -70 ºC • Styrene-Butadiene, Tg = -65 ºC • Acrylonitrile-Butadiene, Tg = -38 ºC • Choroprene, Tg = -50 ºC • Polysiloxane, Tg = -127ºC Polysiloxane is the polymer with lower glass transition and could be used for the tyres, the other ones and less expensive is Polyisoprene and Styrene-Butadiene.


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tutorial 4 solutions

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