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Chemistry

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Feb 20, 2024

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Page 1 of 5 Part A (26 pts, 2 pts each): True or False (T / F). 1. _ T _ Ceramics typically have larger hardnesses than metals. 2. _ T _ Microscopic dimples are evidence of ductile fracture. 3. _ F _ Thermoplastics are usually stiff and brittle above the glass transition temperature (T g ). 4. _ T _ Thermosets are generally stronger but more brittle than thermoplastics. 5. _ T _ During plastic deformation in metals, the number of dislocations usually increases. 6. _ F _ Hot working usually results in excellent dimensional tolerances and surface finishes. 7. _ T _ Permanent mold casting provides both reusability and good surface dimensional accuracy. 8. _ T _ Most metals shrink upon solidification, which can produce cavities in molded metal parts. 9. _ F _ Most metals solidify via homogeneous nucleation. 10. _ T _ A metallic binary alloy freezes over a range of temperatures. 11. _ T _ A liquid, solid, and gas phase of a material can coexist in thermodynamic equilibrium. 12. _ F _ Separate phases must be separate states of matter (e.g., liquid, solid, gas). 13. _ T _ In a solid solution, the strength usually increases but the ductility decreases relative to the pure metal.

Page 2 of 5 Part B (24 pts, 6 pts each): Short Answers. Be concise. 1. (a) Based on the Hume-Rothery rules, which of the elements listed in the table would be expected to exhibit unlimited solid solubility in Au ? Why? (3 pts) • Ag satisfies the H-R rules • Al has the wrong valence • Li has the wrong crystal structure • Pb is too big (>15% difference in radius) (b) Assume that all the elements in the table have at least 1% solubility in Au. Which of the above elements will likely have the highest strength when 1% of the solute is added to Au ? Why? (3 pts) Pb because it has the largest difference in atomic size. 2. (a) Does bending a copper wire make it stronger? Why or why not? (3 pts) Yes, bending a Cu wire makes it stronger due to strain hardening. If we deform the Cu beyond its yield strength and unload, the material’s yield strength increases. Microscopically, this process is due to dislocations being generated and impeding each other’s motion (piling up, entangling, etc.). (b) Will a copper wire break after repeated bending in alternating directions (alternating curvature)? Why or why not? (3 pts) Yes. Each time we do bend beyond yield, the ductility is decreased. Eventually, no ductility remains and the wire will break. At stresses below the yield strength, we will also get fatigue that causes failure. Au: FCC, size = 144 pm, valence = +1 Ag: FCC, size = 145 pm, valence = +1 Al: FCC, size = 143 pm, valence = +3 Li: BCC, size = 152 pm, valence = +1 Pb: FCC, size = 175 pm, valence = +2

Page 3 of 5 3. A Weibull plot of an identical ceramic material sold by two different companies is shown on the right. Which company produces a more consistent product? How do you know? (6 pts) XYZ – slope is larger, i.e., larger Weibull modulus 4. Explain the difference between annealed and tempered glass. (6 pts) Annealed means heated up as to remove residual stresses. Tempered means heat treated to induce residual stress with compression at surface and tension at the center (e.g. Prince Rupert’s drop by rapidly quenching). The compression at the surface increases the resistance to fracture from surface flaws. Part C (50 pts): Numerical Problems 1. The plot below is an engineering stress-strain curve from a tensile test of an aluminum alloy conducted at room temperature. The part is cylindrical in shape with a diameter of 10 mm and a length of 150 mm. Use the data to answer the following questions. (32 pts) (a) Estimate the elastic modulus. (4 pts) Slope of the linear portion of the curve 70 MPa 28 GPa 0.0025 E = = (m/m

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Page 4 of 5 (b) Estimate the load required to plastically deform the material. (4 pts) (c) Estimate the modulus of resilience. (4 pts) (d) Determine the true strain after loading to an engineering stress of 170 MPa. (4 pts) (e) Now consider a specimen made of this same material but with an edge crack of 2 mm. If the fracture toughness of this material is and f = 1.12, will this specimen fail by yielding (i.e., general global yielding) or by fracture (i.e., fast fracture) during tensile testing? Show your work. (8 pts) So, the part will yield first. (f) A (new) specimen made of this same material with an edge crack of 2 mm is subjected to an alternating axial stress between -80 and 100 MPa at a frequency of 200 cycles per second. What is the crack propagation rate (in m/s) at the beginning of the cycling (i.e., when the 2 mm crack is present)? Given f = 1.12, C = 2.0×10 -11 (units of ), and n = 3. Also, will the crack propagation rate change over time? If so, will it increase or decrease and why? (8 pts) ( ) ( ) ( ) 2 130 MPa 130 MPa 0.005 m 10.2 s s p = = = = Y Y P A kN ( ) ( )( )( ) 3 1/ 2 1/ 2 130 MPa .006 0.39 MPa (or 0.39 MJ/m ) s e = = = y y ( ) ( ) ln 1 ln 1 .026 0.0257 e = + = + = e 15 MPa m ( ) 15 MPa 1.12 0.002 m 169 s p s p s s = = = ® = > c c c c y K m f a MPa 3 1/2 MPa m cycle - - ( ) ( )( ) ( ) 3 8 3 2 11 m 1.12 100 MPa 2 3 m 1.40 10 cycle cycle MPa m p - æ ö - é ù = D = - = ´ ç ÷ ë û × è ø n da e C K e dN

Page 5 of 5 Yes, it changes because delta K increases during each cycle since the crack will grow in length. Note: compressive stress does not contribute to delta K. 2. Questions related to polyisoprene (natural rubber). (18 pts) (a) A band of polyisoprene needs to hold together a bundle of steel rods for up to one year (52 weeks). If the tensile stress in the band is less than 2000 psi, it will not hold the rods together tight enough. Determine the initial stress that must be applied to the polyisoprene band when it is wrapped around the bundle. A previous test demonstrated that an initial stress in the band of 1500 psi decreased to 1450 psi after 5 weeks. (8 pts) (b) If the band in (2a) must withstand higher temperatures, would the initial stress calculated in (2a) need to be increased or could we decrease the initial stress and maintain a sufficient level of tension? Justify your answer. (4 pts) We would need to increase the levels of tension because the relaxation time decreases with increasing temperature: . (c) The molecular weight of polyisoprene is 100,000 g/mol. The isoprene monomer has a chemical formula of C 5 H 8 . The molar mass of C is 12.0 g/mol, and the molar mass of H is 1.0 g/mol. Assume all the polymer chains have the same length. Calculate the degree of polymerization. (6 pts) The molecular weight of the isoprene monomer is: 6 m cycles m 1.40 8 200 2.80 10 cycle s s - æ ö æ ö = - = ´ ç ÷ç ÷ è ø è ø da e dt ( ) ( ) ( ) ( ) 0 0 0 0 0 exp ln ln / 5weeks 147weeks ln 1450 /1500 exp / exp / 2000 psi/ exp 52 weeks/147 weeks 2850 psi t t t psi psi t t s s s l l s l s s l s s l s s l æ ö æ ö = - Þ = - Þ = - ç ÷ ç ÷ è ø è ø Þ = - = æ ö = - ç ÷ è ø Þ = - = - = 0 exp Q RT l l æ ö = ç ÷ è ø ( ) ( ) 5 8 5 12 g/mol 8 1 g/mol 68 g/mol C H + = + = = = = 100,000 / 1471 68 / molecular weight of polymer g mol DOP molecular weight of repeat unit g mol