L Manufacturing of Polymer Composites FS 17 Exercise 2. Exercise 2: Solution

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Exercise 2: Solution MATRIX SYSTEMS ETH Zürich Laboratory of Composite Materials and Adaptive Structures Page 1

Task 1: Polymeric materials a) What part of the composite determines its thermal properties? The matrix is the component limiting the thermal range where a composite can be used. b) What is the difference between glass transition temperature and melting temperature? In polymer science, the melting temperature is the temperature at which the melting of the crystalline part of a semi-crystalline polymer takes place. The glass transition temperature is the temperature at which a polymer passes from a glass state (hard and relatively brittle) into a rubber-like state. While the melting represents an effective phase transition (first order transformation, involving a heat exchange with the external environment to break up the inter-chains bonds), the glass transition is second order transformation that involves only the amorphous region of the polymer and entails the passage into a state with higher (rubber-like) or lower (glass-like) mobility of the polymeric chains (regulated by the available thermal energy from the environment). The glass transition temperature is always lower than the possible melting temperature. c) Draw the qualitative behavior of the stiffness as a function of temperature for the following classes of polymers and comment on the curves (The glass transition temperature TG is the same for all three polymers) Partially crystalline thermoplastic Amorphous thermoplastic Thermoset Range I: It is found bellow Tg, and is also known as glass state. The polymer chains remain frozen in space, as they have not enough energy to undergo any type of motion. Range II: It is found between Tg and Tm, and can also be called rubber-elastic state. Thermal energy activates additional molecular degrees of freedom. The polymer chains are now free to rotate, inducing a loss of stiffness in the polymer. This effect is less pronounced for semicrystalline thermoplastics: only the amorphous regions soften at TG. The crystalline regions still contribute substantially to the overall stiffness of a component. ETH Zürich Laboratory of Composite Materials and Adaptive Structures Page 2

Range II: Shortly ahead of the melting temperature stiffness begins to drop drastically. The polymer is in the melting range, where the first crystallites start to melt. Interesting remark: in contrast to metals, which exhibit an almost fixed defined melting point, semicrystalline thermoplastics do not feature a unique melting point, but an interval or a range of melting, d) What is the degree of polymerization and what influence has on the final properties of the resin? The degree of polymerization is the number of times a monomer is repeated on a polymer chain. It is defined as the ratio between the number-average molecular weight (Mn) and the molecular weight of the monomer unit (Mo). DDDD = MM nn MM 0 The higher the degree of polymerization is the better thermal and chemical resistance of the polymer as we have more intermolecular bonds. However, very high viscosities (almost solid state) are also obtained at room temperature, which make the processability of the material more difficult. High temperatures (above the melting points) or solvents are in this case needed. e) Discuss also the influence of the crosslink density on the material properties. Following the same reasoning, the higher the crosslink density is the higher the number of intramolecular bonds on a polymer. The chains cannot slide or rotate because they are fixed into position. There are less available degrees of freedom and hence higher Tg. f) How does the branching of the molecules affect the glass transition temperature? The free volume of a molecule is the space between this molecule and the ones surrounding it. Branching increases the distance between molecules, resulting in a reduction of the intermolecular forces. Therefore, less energy is necessary to release some degrees of freedom, so the Tg reduces. ETH Zürich Laboratory of Composite Materials and Adaptive Structures Page

Consider the reaction, taking place in alkaline medium, between bisphenol-a and epichlorohydrin. g) If the resin has a degree of polymerization of 0 (1 bisphenol-a + 2 epichlorohydrin) and is cured with a diamine hardener, with a molecular weight of 82g/mol and 4 reactive hydrogens. What is the mass of hardener necessary to cure 450g of resin? AAAAAAAAAA HH eeee wwww = MMMM haaaaaaaaaaaaaa nnnn. oooo aaaaaaaaaaaa hyyyyyyyyyyyyyy = 82 4 = 20.5gg/mmmmmm MMMM rrrrrrrrrr = 21 12 + 24 1 + 4 16 = 40gg/mmmmmm EEEE vv = nnnn. oooo eeeeeeeeee gggggggggg 100 MMMM rrrrrrrrrr = 2 100 40 = 0.588 MM haaaaaa = EEEE vv AAAAAAAAAA HH eeee wwww 4.5 = 0.588 20.5 4.5 = 54.2gg ETH Zürich Laboratory of Composite Materials and Adaptive Structures Page 4

Task 2: Thermomechanical properties a) A table with polymer material data is given to you at the end of this exercise. Why for some polymers the maximum service temperature is above its glass transition? Why this is not the case for amorphous polymers? Using semicrystalline polymers above their glass transition temperature permits to improve their ductility without a critical impact on their thermomechanical properties (stiffness for example), as the glass transition takes place only in their amorphous regions. However, amorphous polymers cannot be used above their glass transition temperature as this critically lowers their thermomechanical properties. b) Two well-established techniques in thermomechanical analysis are DSC and TGA. What are the parameters that can be obtain through each of these analyses? DSC: Tg, Tm, enthalpy, degree of crystallinity and heat capacity. TGA: fiber volume content and type of fibers. c) You are given a unidirectional composite having an unknown composition with the following information: 1. A DSC curve of the unknown material; 2. Two TGA curves obtained at different operating conditions: O 2 atmosphere (A complete combustion of the material is observed.) Ar atmosphere (Ar is a noble gas and the atmosphere is considered to be inert.). The properties of the most common composite materials. - What material is the matrix composed of? According to the DSC, T G is 190 C. This is most likely true for PSU. - What material is the fibre? From the TGA we know on the one hand that the fibres can be burnt in oxygen, on the other hand that they stay stable up to 2500 C under inert conditions (Argon atmosphere). This condition can only apply for carbon fibres. - What is the fibre volume content? Composite mass = 10 g, Fibres mass = 7g and Matrix mass = g V V V f m = m f 7 g 4.02 cm ρ = 1.74 g cm = f = mm g 2.45 cm ρ = m 1.24 g cm = V f 4.02 cm = = = 0.621 = 62 % V + V 4.02 cm + 2.45 cm f. cont f m ETH Zürich Laboratory of Composite Materials and Adaptive Structures Page 5

TGA (Ar) TGA (O2) DSC ETH Zürich Laboratory of Composite Materials and Adaptive Structures Page 6

1) Fibres Carbon Glass Density [g/cm ] 1.74 2.55 E-Modulus [GPa] 240 7 Elongation at break [%] 1.5 4.5 Tensile strength [GPa].6.5 2) Thermoplastic polymers PP PET PBT PA 6 PA 66 Structure Semicryst. Semicryst. Semicryst. Semicryst. Semicryst. Density [g/cm ] 0.94 1.2 1.1 1.12 1.12 E-Modulus [GPa] 1.9 2.7 2.5 1.9 2.1 Elongation at break [%] 120 10 120 94 8 Tensile strength [MPa] 6 55 56 7 7 Gass transition temperature [ C] -18 77 60 50-60 50-60 Max. Service temperature [ C] 85 74 70 100 100 Impact strength (IZOD, unnotched) No break No break No break No break No break [J/m] Coefficient of thermal expansion [K -1 ] 150 10-6 70 10-6 10 10-6 70 10-6 70 10-6 Chemical resistance Good High High Good Good Fire behaviour at 45 C PES PSU PEI PPS PEEK Structure Amorphous Amorphous Amorphous Semicryst. Semicryst. Density [g/cm ] 1.4 1.24 1.5 1.4 1. E-Modulus [GPa].7 2.5.7.6 4.5 Elongation at break [%] 0 10-75 42 4 7 Tensile strength [MPa] 99 75 100 87 110 Glass transition temperature [ C] 20 190 220 88 140 Max. Service temperature [ C] 200 170 200 160 260 Impact strength (IZOD, unnotched) No break No break 100 900 No break [J/m] Coefficient of thermal expansion [K -1 ] 55 10-6 54 10-6 62 10-6 49 10-6 47 10-6 Chemical resistance High Good High High High Fire behaviour Flame retardant ) Thermosetting polymers Epoxy Polyester Vinyl ester BMI Structure Amorphous Amorphous Amorphous Amorphous Density [g/cm ] 1.2 1.2 1.1 1. E-Modulus [GPa].2.9.2 4.1 Elongation at break [%] 1.5-8.0 1.0-6.5 2.0-8.0 1.5-. Tensile strength [MPa] 7 6 76 79 Glass transition temperature [ C] 65-175 70-120 70 20-45 Max. Service temperature [ C] 10 150 100 190 Fracture toughness [J/m 2 ] 5-15 10-20 - 24- Coefficient of thermal expansion [K -1 ] 45 10-6 54 10-6 - 1 10-6 Chemical resistance Good Limited Good High Fire behaviour Fire retardant - ETH Zürich Laboratory of Composite Materials and Adaptive Structures Page 7

Task : Composition of the fibres Discuss which type of fibres would be more convenient for each case. 1- Bulletproof vest A Bulletproof vest requires high impact strength, because it must transfer the kinetic energy of the projectile to the vest without failure at the impact point. The suitable material is Kevlar: It is a long molecule. The summation of all Van der Walls forces gives a high material strength. Carbon would be in this case too brittle for impact resistance. Another good point is the protection from high temperatures as it is not thermal conductive. 2- Nose of an airplane. The airplane nose is usually accommodates communication and navigation instruments. Glass fibres: they are radiolucent, which means they permit the radiations to pass through them. - Aerospace structures. Two important parameters on aerospace are stiffness and lightweight design. Carbon fibres: they offer the best specific stiffness and strengths, so they are ideal for high performance lightweight design in an industry that is willing to pay a premium price for a small increase in performance. Sources: (1) and () Airbus, (2) Dupont. ETH Zürich Laboratory of Composite Materials and Adaptive Structures Page 8