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### 16 askeland chap

1. 1. 16Composites: Teamwork andSynergy in Materials 16–7 Nickel containing 2 wt% thorium is produced in powder form, consolidated into a part, and sintered in the presence of oxygen, causing all of the thorium to produce ThO2 spheres 80 nm in diameter. Calculate the number of spheres per cm3. The den- sity of ThO2 is 9.86 g/cm3. Solution: In 100 g of material, there are 98 g͞8.902 g/cm3 ϭ 11.0088 cm3 of nickel. From the reaction Th ϩ O2 ϭ ThO2, 2 g Th ր232 g/mol ϭ x g ThO2ր264 g/mol x ϭ 2.2759 g ThO2 The total volume of the oxide is: Voxide ϭ 2.2759 gր9.86 g/cm3 ϭ 0.2308 cm3 The volume fraction of the oxide is 0.2308 foxide ϭ ϭ 0.0205 0.2308 ϩ 11.0088 The volume of each oxide sphere is: Vsphere ϭ 14␲ր32r3 ϭ 14␲ր32140 ϫ 10Ϫ7 cm2 3 ϭ 2.68 ϫ 10Ϫ16 cm3 The total number of oxide particles in 1 cm3 is: particles ϭ 0.0205 cm3 ThO2 ր2.68 ϫ 10Ϫ16 cm3/particle ϭ 7.65 ϫ 1013 particles/cm3 16–8 Spherical aluminum powder (SAP) 0.002 mm in diameter is treated to create a thin oxide layer and is then used to produce a SAP dispersion-strengthened material con- taining 10 vol% Al2O3. Calculate the average thickness of the oxide film prior to compaction and sintering of the powders into the part. 177
2. 2. 178 The Science and Engineering of Materials Instructor’s Solution Manual Solution: The volume of an aluminum powder particle is: VAl ϭ 14␲ ր3210.002 mm ր22 3 ϭ 4.19 ϫ 10Ϫ9 mm3 The volume fraction Al2O3 is Voxide Voxide 0.10 ϭ ϭ Voxide ϩ VAl Voxide ϩ 4.19 ϫ 10Ϫ9 Voxide ϭ 4.654 ϫ 10Ϫ10 mm3 We can then calculate the radius of the particle after oxidation has occurred: Voxide ϭ 14␲ր32r3 Ϫ 4.19 ϫ 10Ϫ9 ϭ 4.65 ϫ 10Ϫ10 r ϭ 1.0358 ϫ 10Ϫ3 mm ϭ 0.0010358 mm The thickness of the oxide layer must therefore be: thickness ϭ 0.0010358 Ϫ 0.001 ϭ 0.0000358 mm ϭ 3.58 ϫ 10Ϫ5 mm 16–9 Yttria (Y2O3) particles 750 Å in diameter are introduced into tungsten by internal oxidation. Measurements using an electron microscope show that there are 5 ϫ 1014 oxide particles per cm3. Calculate the wt% Y originally in the alloy. The density of Y2O3 is 5.01 g/cm3. Solution: The volume of each particle is: Voxide ϭ 14␲ր321750 ϫ 10Ϫ8ր2 cm2 3 ϭ 2.209 ϫ 10Ϫ16 cm3 The total volume of oxide particles per cm3 is given by: Vyttria ϭ 12.209 ϫ 10Ϫ16 cm3 215 ϫ 1014 particles2 ϭ 0.11 cm3 The volume fraction of yttria is therefore foxide ϭ 0.11 The weight percentages of oxide and tungsten are 10.11215.01 g/cm3 2 wt% Y2O3 ϭ ϫ 100 ϭ 3.116% 10.11215.012 ϩ 10.892119.2542 wt% W ϭ 96.884% In 1 g of material, there are 0.03116 g of oxide. From the equation 2Y ϩ 13 ր22O2 ϭ Y2O3 x g of Yր2 188.91 g/mol2 ϭ 0.03116 g of Y2O3 ր225.82 g/mol x ϭ 0.0245 g of Y The weight percent Y in the original alloy was therefore: 0.0245 g Y wt% Y ϭ ϫ 100 ϭ 2.47% 0.0245 g Y ϩ 0.96884 g W 16–10 With no special treatment, aluminum is typically found to have an Al2O3 layer that is 3 nm thick. If spherical aluminum powder prepared with a total diameter of 0.01 mm is used to produce the SAP dispersion-strengthened material, calculate the vol- ume percent Al2O3 in the material and the number of oxide particles per cm3. Assume
3. 3. CHAPTER 16 Composites: Teamwork and Synergy in Materials 179 that the oxide breaks into disk-shaped flakes 3 nm thick and 3 ϫ 10Ϫ4 mm in diameter. Compare the number of oxide particles per cm3 to the number of solid solution atoms per cm3 when 3 at% of an alloying element is added to aluminum. Solution: The total volume of the powder particle is: Vtotal ϭ 14␲ր3210.01ր22 3 ϭ 5.235988 ϫ 10Ϫ7 mm3 The volume of just the oxide layer is: Voxide ϭ 5.235988 ϫ 10Ϫ7 Ϫ 14␲ր3210.005 Ϫ 3 ϫ 10Ϫ6 2 3 ϭ 0.009419 ϫ 10Ϫ7 mm3 The volume fraction of the oxide is: foxide ϭ 0.009419 ϫ 10Ϫ7ր5.235988 ϫ 10Ϫ7 ϭ 0.001799 The volume of one disk-shaped oxide flake is: Vflake ϭ 1␲ ր4213 ϫ 10Ϫ4 mm2 2 13 ϫ 10Ϫ6 mm2 ϭ 2.12 ϫ 10Ϫ13 mm3 ϭ 2.12 ϫ 10Ϫ16 cm3 In one cm3 of SAP, there is 0.001799 cm3 of oxide. The number of oxide particles per cm3 is therefore number ϭ 0.001799 cm3 ր2.12 ϫ 10Ϫ16 cm3/particle ϭ 8.49 ϫ 1012 flakes/cm3 The number of solid solution atoms per cm3 in an Al–3 at% alloying ele- ment alloy is calculated by first determining the volume of the unit cell: Vcell ϭ 14.04958 ϫ 10Ϫ8 cm2 3 ϭ 66.41 ϫ 10Ϫ24 cm3 In 25 unit cells of FCC aluminum, there are 100 atom sites. In the alloy, 3 of these sites are filled with substitutional atoms, the other 97 sites by aluminum atoms. The number of solid solution atoms per cm3 is therefore: number ϭ 3 atoms in 25 cellsր 125 cells2166.41 ϫ 10Ϫ24 cm3/cell2 ϭ 18.1 ϫ 1020 substitutional atoms/cm3 The number of substitutional point defects is eight orders of magnitude larger than the number of oxide flakes.16–13 Calculate the density of a cemented carbide, or cermet, based on a titanium matrix if the composite contains 50 wt% WC, 22 wt% TaC, and 14 wt% TiC. (See Example 16–2 for densities of the carbides.) Solution: We must find the volume fractions from the weight percentages. Using a basis of 100 g of the cemented carbide: 50 g WC ր15.77 g/cm3 fWC ϭ ϭ 0.298 150 ր15.772 ϩ 122 ր14.52 ϩ 114 ր4.942 ϩ 114 ր4.5072 22 g TaC ր14.5 g/cm3 fTaC ϭ ϭ 0.143 150 ր15.772 ϩ 122 ր14.52 ϩ 114 ր4.942 ϩ 114 ր4.5072 14 g TiC ր4.94 g/cm3 fTiC ϭ ϭ 0.267 150 ր15.772 ϩ 122 ր14.52 ϩ 114 ր4.942 ϩ 114 ր4.5072 14 g Ti ր4.507 g/cm3 fTi ϭ ϭ 0.292 150 ր15.772 ϩ 122 ր14.52 ϩ 114 ր4.942 ϩ 114 ր4.5072
4. 4. 180 The Science and Engineering of Materials Instructor’s Solution Manual The density is then found from the rule of mixtures: rc ϭ 10.2982115.772 ϩ 10.1432114.52 ϩ 10.267214.942 ϩ 10.292214.5072 ϭ 9.408 g/cm3 16–14 A typical grinding wheel is 9 in. in diameter, 1 in. thick, and weighs 6 lb. The wheel contains SiC (density of 3.2 g/cm3) bonded by a silica glass (density of 2.5 g/cm3); 5 vol% of the wheel is porous. The SiC is in the form of 0.04 cm cubes. Calculate (a) the volume fraction of SiC particles in the wheel and (b) the number of SiC par- ticles lost from the wheel after it is worn to a diameter of 8 in. Solution: (a) To find the volume fraction of SiC: Vwheel ϭ 1␲ր42D2h ϭ 1␲ր4219 in.2 2 11 in.2 ϭ 63.617 in.3 ϭ 1042.5 cm3 Wwheel ϭ 6 lb ϭ 2724 g rwheel ϭ 2724 g ր1042.5 cm3 ϭ 2.6129 g/cm3 From the rule of mixtures: 2.6129 ϭ fpore rpore ϩ fSiC rSiC ϩ fglass rglass 2.6129 ϭ 10.052 102 ϩ fSiC 13.22 ϩ 11 Ϫ 0.05 Ϫ fSiC 2 12.52 fSiC ϭ 0.34 (b) First we can determine the volume of the wheel that is lost; then we can find the number of particles of SiC per cm3. Vlost ϭ 1␲ր42192 2 112 Ϫ 1␲ր42182 2 112 ϭ 13.352 in.3 ϭ 218.8 cm3 Vparticles ϭ 10.04 cm2 3 ϭ 6.4 ϫ 10Ϫ5 cm3 In 1 cm3 of the wheel, there are (0.34)(1 cm3) ϭ 0.34 cm3 of SiC. The number of SiC particles per cm3 of the wheel is: 0.34 cm3 ր6.4 ϫ 10Ϫ5 cm3/particle ϭ 5312.5 particles/cm3 The number of particles lost during use of the wheel is: particles lost ϭ 15312.5/cm3 21218.8 cm3 2 ϭ 1.16 ϫ 106 particles 16–15 An electrical contact material is produced by infiltrating copper into a porous tungsten carbide (WC) compact. The density of the final composite is 12.3 g/cm3. Assuming that all of the pores are filled with copper, calculate (a) the volume fraction of copper in the composite, (b) the volume fraction of pores in the WC compact prior to infiltration, and (c) the original density of the WC compact before infiltration. Solution: (a) rc ϭ 12.3 g/cm3 ϭ fCu rCu ϩ fWC rWC ϭ fCu 18.932 ϩ 11 Ϫ fCu 2115.772 fCu ϭ 0.507 (b) The copper fills the pores. Therefore the volume fraction of the pores prior to infiltration is equal to that of the copper, or fpores ϭ 0.507. (c) Before infiltration, the composite contains tungsten carbide and pores (which have zero density): rcompact ϭ fWC rWC ϩ fpore rpore ϭ 10.4932115.772 ϩ 10.5072102 ϭ 7.775 g/cm3
5. 5. CHAPTER 16 Composites: Teamwork and Synergy in Materials 18116–16 An electrical contact material is produced by first making a porous tungsten com- pact that weighs 125 g. Liquid silver is introduced into the compact; careful meas- urement indicates that 105 g of silver is infiltrated. The final density of the composite is 13.8 g/cm3. Calculate the volume fraction of the original compact that is interconnected porosity and the volume fraction that is closed porosity (no silver infiltration). Solution: First we can find the volume of the tungsten and silver: VW ϭ 125 g ր19.254 g/cm3 ϭ 6.492 cm3 VAg ϭ 105 g ր10.49 g/cm3 ϭ 10.010 cm3 The volume fractions of each constituent are: 6.492 fW ϭ 6.492 ϩ 10.010 ϩ Vpore 10.010 fAg ϭ 6.492 ϩ 10.010 ϩ Vpore Vpore fpore ϭ 6.494 ϩ 10.010 ϩ Vpore From the rule of mixtures: 13.8 ϭ 36.492 ր 116.502 ϩ Vpore 2 4 119.2542 ϩ 310.010ր 116.502 ϩ Vpore 2 4 110.492 ϩ 0 Vpore ϭ 0.165 cm3 The total volume is 6.492 ϩ 10.010 ϩ 0.165 ϭ 16.667 cm3. The fraction of the contact material that is interconnected porosity prior to silver infil- tration is equal to the volume fraction of silver; the volume fraction of closed porosity is obtained from Vpore. finterconnected ϭ 10.010 ր16.667 ϭ 0.6005 fclosed ϭ 0.165 ր16.667 ϭ 0.009916–17 How much clay must be added to 10 kg of polyethylene to produce a low-cost com- posite having a modulus of elasticity greater than 120,000 psi and a tensile strength greater than 2000 psi? The density of the clay is 2.4 g/cm3 and that of polyethylene is 0.92 g/cm3. Solution: From Figure 16–6, we find that fclay must be greater than 0.3 if the modu- lus is to exceed 120,000 psi; however the fclay must be less than 0.46 to assure that the tensile strength exceeds 2000 psi. Therefore any clay frac- tion between 0.30 and 0.46 should be satisfactory. Wclayր2.4 g/cm3 fclay ϭ 1Wclay ր2.42 ϩ 110,000 gր0.922 If fclay ϭ 0.3, then Wclay ϭ 11,200 g ϭ 11.2 kg If fclay ϭ 0.46, then Wclay ϭ 22,250 g ϭ 22.25 kg The overall cost of the composite will be reduced as the amount of clay added to the composite increases.