TestsFa2005 - ENSC 3313, Fall 2005 TEST 3 Name_ Answer all...

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ENSC 3313, Fall 2005 TEST 3 Name ________________________________ Answer all 10 questions, 2 points each . 1. Ceramic green bodies formed by pressing, slip casting, tape casting, extrusion, etc., must be sintered. Explain, qualitatively, the process of sintering and the relation between strength and density for ceramics. Sintering is the high temperature process whereby the total surface area (high energy) to volume ratio of the ceramic green body is reduced. At these temperatures, surface diffusion rates are high enough such that particles can coalesce, reshape, and effectively eliminate porosity from the green body. The strength of the ceramic depends on the effectiveness of the sintering operation in removing pores, which act as stress risers and lead to crack initiation. Recall that the strength of the ceramic decreases exponentially with increasing porosity. 2. The fracture strength of a ceramic decays exponentially with increasing porosity. Determine what level of porosity would reduce the strength to 0.75 that for a fully dense ceramic. Note the decay constant is n = 4.5. This requires application of the exponential decay formula as follows: P 5 . 4 o nP o e 75 . 0 e = = σ σ σ = σ Now solve for P () 064 . 0 ~ 5 . 4 75 . 0 ln P = thus, ~ 6.4% porosity results in a 25% reduction in strength.
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3. Chloroprene rubber (trade name Neoprene) is not cross-linked by sulfur because the Cl atom hides the double bond. If it could (say) be cross-linked with S at but 8% efficiency, what proportion of the sites would be cross-linked when 100 kg of S is added to 900 kg of chloroprene? The amu are H = 1, C = 12, Cl = 35.5, S = 32. Chloroprene mer: H Cl H H | | | | C C = C C | | H H For complete cross-linking 1 double bond opens up to provide 2 add-on sites to link 2 chains, hence need 1 S per mer. Complete cross-linking needs 32 g of S per 88.5 g of PC. Complete cross-linking needs 325.4 kg of S per 900 kg of PC. We have, effectively, 8 kg of S, hence proportion linked is 8/325.4 ~ 0.025. 4. Why was this figure shown in class? 5. A Kevlar/epoxy composite was prepared with a density of 1.4 gm/cc. Please determine the volume fraction of Kevlar. If necessary, you make take values at the mid point of a range. ρ = f i ρ I 1.4 = f f (1.45) + f m (1.3) f f + f m = 1 Two equations in two unknowns which can be solved by any of several standard techniques: f f = 0.67 Material Density E σ UTS Fiber E-glass 2.56 76 1.4-2.5 Carbon High-modulus High- strength 1.75 1.95 390 2.50 2.2 2.7 Kevlar 1.45 125 3.2 Matrix Epoxy 1.2-1.4 2.1-5.5 0.063 Polyester 1.1-1.4 1.3-4.5 0.060
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6. A valve in an automatic sprinkler failed after 21 months in service. Why? Galvanic attack of the malleable iron latch in contact with a copper alloy clapper in stagnant water. ASM Hdbk. 8(10 ), p. 183. 7. Deduce what “metal” is likely to have been blacked out in each instance! Add explanatory comments. This is the standard electrode potential series. You should recognize that hydrogen is the reference with a voltage of zero.
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This note was uploaded on 01/10/2012 for the course ENSC 3313 taught by Professor Staff during the Fall '11 term at Oklahoma State.

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TestsFa2005 - ENSC 3313, Fall 2005 TEST 3 Name_ Answer all...

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