Rubber Elasticity. Scope. Sources of Rubber. What is Rubber? MSE 383, Unit 2-6

Transcription

1 Rubber Elasticity MSE 383, Unit 2-6 Joshua U. Otaigbe Iowa State University Materials Science and Engineering Dept. Scope Physical properties of rubber Molecular interpretation of physical properties Relation of rubber to other polymers Sources of Rubber Natural Rubber >> Hevea Braziliensis (Hevea or wild rubber) >> Plantation rubber from Malaysia, Sri- Lanka, Indonesia Synthetic Rubber >> Developed in Germany/World Wars I and II >> 1/2 total global consumption of rubber What is Rubber? >> name from pencil rubber [Priestley (1770)] N.R. obtained from incisions in bark >> milky white in appearance >> water suspension of rubber globules ( µm) >> flows freely 2.6.1

2 Rubber Latex Composition Faraday (1826) >> rubber hydrocarbon 37 wt. % >> other solids (proteins, sugars, waxes, etc.) 3 >> water 60* Chemical Formula and Structure of N.R. Poly (cis-isoprene) (C 5 H 8 ) n Show chemical structure Why unique props. from such a simple molecule? >> physics of rubber elasticity Properties of Rubber Rubber Steel Similar hydrocarbons E (N/mm 2 ) 1 2 x 10 5 Liquids Extension (%)* ~ 700* ~ 1 Unique Properties of N.R. Thermoelastic effects [Gough (1805)] >> heat evolved reversibly on extension >> stretched rubber contracts reversibly on heating Confirmed by Joule (1859) >> Demonstrate rubber strips/wheel 2.6.2

3 Heat of Extension Dart, Anthony and Guth (1942) Up to 10 C Temp. rise at 600% extension Effects related to thermodynamics T C Extension % Molecular Structure of N.R. Recall planar zig-zag, bond rotation versus random chain conformations Length of chain approx. 1 kilometer!! Random conformation is most probable due to molecular motions >> statistical form of chain (cf link chain) Randomly-Jointed Chain r * No valence angle or other restrictions Let n = # of links; λ = length of link r 2 = n λ 2 ( r 2 ) 1/2 = n 1/2 λ R.M.S. length nλ / n 1/2 λ = n 1/2 Extensibility => importance of chain length* 2.6.3

4 Thermodynamics of Rubber Elasticity Entropy >> related to probability of a given state >> stretched state has lower entropy (or lower probability) >> contracted state has higher entropy >> no charge in internal energy (or P.E.) & intermolecular forces in elastic deformation Heat of Extension (Gough-Joule Effect) F l dl F dw = Fdl (work done to extend rubber) du = dq + dw(1st Law of Thermodynamics) du = 0 (assumption of kinetic theory)** (=> all states have the same U) Hence dq = - dw But dw is +ive, dq is -ive => Heat given out on stretching Increase of Tension on Heating (F) Meyer found Tg T( C) (F)λ T abs Compare Gas (P) V T Recall wheel demo Increased tension due to increased thermal energy or energy of molecular bombardment Bombardment by surrounding molecules Brittleness in liquid N 2 and T g ** F F 2.6.4

5 Increase of Tension on Heating (Cont d) Contrast with ordinary solids >> force raises distance b/w atoms >> work done against intermolecular forces raises U of system Rubber vs. Ordinary Solids Rubber >> du = 0 on const. T >> falls in entropy on const. T (cf. perfect gas in compression) Ordinary Solid >> rising U of system on deformation >> no significant change in entropy Stress-Strain Relation (review derivations in class text) F = Ai G [ λ - 1/ λ 2 ] = N κ T N = ν / V λx λy λz = 1 Technologically Important Rubbers Vulcanized N.R. (Goodyear 1839)** >> typically crosslinked with sulfur >> (approx. 1 crosslink per isoprene units) 2.6.5

6 Technologically Important Rubbers N.R. mixed with carbon >> Improved stiffness and strength, liquid absorption, sunlight and ozone defradation PU elastomers >> endlinked networks from condensation of polyols and polyisocyanates Synthetic PIsoP, PBD SBR NBR >> copolymer of styrene and acrylonitrile >> resistant to swelling in oil and petrol >> belong to nitrile rubber family Thermoplastic rubbers (TPR) >> PBD end-tipped with PS blocks >> glassy PS domains in rubbery PBD domains >> processable above T g of PS phase >> easy alloying with PP and processable EP copolymers >> E/P = 40/60 to 60/40 >> completely amorphous in this range Chloroprene rubbers >> PBD with a H-atom replaced by Cl-atom >> heat resistant, gas impermeable, resistant to ozone attack Theory of Rubber Elasticity Chain statistics >> randomly jointed chain Affine deformation of chains >> uniform deformation of specimen Network theory >> network of chains joined together by unbreakable bonds (Details outside scope of course) 2.6.6

7 Summary Rubber elasticity is predominantly entropy-driven and statistical in nature Stiffness increases with increasing T Heat is reversibly generated upon stretching Hooke's law obeyed in shear only but not in tension and compression Elastic constant, G = NkT (= ρ RT/M c ) Crosslinking increases stiffness Stress-strain behavior is predicatable End of Lecture Read Class text, Ch. 2 >> review worked examples Exam #1 on../../.. >> Units 1 and 2 ONLY Practice Problems #2 (Rubber Elasticity) I. Reading Guide: McCrum et at (Class text) and lecture notes II. Problems** 1. Problem 3.8. (Class text, McCrum et al., 1991) Recall that G=NkT (Given 40 C, tetrafunctional xlinks) N=G/kT=0.8*10 6 /1.38*10-23 * =1.8512*10 26 chains per m³ Note: only one xlinks joins 4 chain ends, each chain has 2 chain ends 1 xlink is required per 2 chains # xlinks per m³ = N/2 = 9.26*10 25 m

8 Practice Problems #2 (Rubber Elasticity) 2. Problem 3.10 (Class text, McCrum et al., 1991) By definition E = [dσ/dε]ε 0 = [dσ/dλ]λ 1 (since λ=ε+1) But from Eqn. 3.38: σ = G(λ - λ -2 ) Hence E = [G(1 + 2λ -3 )] = 3G This is a direct result of the assumption of incompressibility. In pure tension, lateral contraction strains = ½ * (axial extension strain) At constant volume. Hence ν = ½. From E = 2G(1+ν) it follows that E = 3G. 3. A spherical rubber balloon is of initial wall thickness 0.5 mm and diameter 100 mm. It is inflated to a final diameter of 500 mm. Calculate the final thickness. Balloon radius = 0.05 m initially, 0.25 m finally. If X, Y axes are Defined in the plane of the balloon wall, the stress state is equal biaxial with: λ X = λ Y = 0.25/0.05 = 5 Volume of spherical shell = 4πr²h; where h = thickness If volume remains constant : r 1 ²h 1 = r 2 ²h 2 Final thickness h 2 = h1(r 1 /r 2 )² = 0.5(0.05/0.25)² = 0.02 mm 4. Problem 3.19 (Class text, McCrum et al., 1991) (See ch. 8, Class text for a discussion of some applications) Rubbers are used in a wide variety of engineering applications, especially where their ability to support high hydrostatic stresses whilst deforming relatively easily under shear stresses make them uniquely suitable. Examples include: Engine mountings, vehicle suspensions, bridge bearings, anti-vibration and earthquake-protection supports for buildings, seals and gaskets. Tensile properties, flexibility, and impermeability are some requirements in tires, hoses, inflatable rafts, diving suits, etc