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Unformatted text preview: Problem 12.35
J. F. Shackelford, Introduction to Materials Science for Engineers, 7th Edition, Prentice Hall, New Jersey (2009) 12.35 Calculate the density of a fiberreinforced composite composed of 14 vol % Al2O3 whiskers in a matrix of epoxy. The density of Al2O3 is 3.97 Mg/m3 and of epoxy is 1.1 Mg/m3. SOLUTION Assume a basis of 1 m3 for the volume of the composite. On this basis, the volume of the Al2O3 is 0.14 m3, and the volume of the epoxy is 0.86 m3. The mass of each component can then be calculated using the densities given in the problem statement. These are
mAl2 03 = (3.97 Mg/m3 ) (0.14 m3 ) = 0.556 Mg and mepoxy = (1.1 Mg/m3 ) (0.86 m3 ) = 0.946 Mg From these the density of the composite can be calculated,
ρ= m (0.556 + 0.946) Mg = V 1 m3 Solving, ρ = 1.50 Mg/m3. Problem 12.35 Solution Professor R. Gronsky page 1 of 1 Problem 12.38
J. F. Shackelford, Introduction to Materials Science for Engineers, 7th Edition, Prentice Hall, New Jersey (2009) 12.38 Calculate the density of the Kevlar fiberreinforced epoxy composite in Table 12.7. The density of Kevlar is 1.44 Mg/m3 and of epoxy is 1.1 Mg/m3. SOLUTION The solution given to Example 12.6 on page 401 of the text serves as a guide here. First note that the density of the composite is defined by
ρcomposite = mcomposite Vcomposite Begin by assuming a "basis" of 1 m3 of composite, establishing its "volume" to be used in the denominator. Consulting Table 12.7 (p. 412 of the text) as instructed in the problem statement reveals that the "loading" of Kevlar in this epoxy matrix composite is 82 vol %. Therefore on our assumed basis, the Kevlar occupies a volume of 0.82 m3 and the epoxy, 0.18 m3. The mass of the composite in the numerator is calculated next. From conservation of mass, mcomposite = mKevlar + mepoxy Substituting (volume × density) for mass, mcomposite = VKevlar ρKevlar + Vepoxy ρepoxy and inserting values from the given information and volume fraction data, mcomposite = (0.82 m3 × 1.44 Mg/m3) + (0.18 m3 × 1.1 Mg/m3) mcomposite = 1.38 Mg The density of the composite is then calculated,
ρcomposite = 1.38 Mg 1 m3 Solving, ρcomposite = 1.38 Mg/m3. Problem 12.38 Solution Professor R. Gronsky page 1 of 1 Problem 12.56
J. F. Shackelford, Introduction to Materials Science for Engineers, 7th Edition, Prentice Hall, New Jersey (2009) 12.56 Calculate the specific strength of the Kevlar/epoxy composite in table 12.7. (Note Problem 12.38). Reference 12.38 Calculate the density of the Kevlar fiberreinforced epoxy composite in Table 12.7. The density of Kevlar is 1.44 Mg/m3 and of epoxy is 1.1 Mg/m3. SOLUTION The "specific strength" of a material is illustrated in Example 12.11 on pages 414 and 415 of the text, beginning with a definition that normalizes strength against density,
sp. str. = T. S . ρ and expresses the result in units of mm (length). This is a convention used throughout the text, with a caveat about the "cancellation" of units of force by units of mass described in the "Note" to the solution on page 415. The same approach is used here. Table 12.7 lists the tensile strength of the Kevlar/epoxy composite as T.S. = 1,517 MPa, the numerator in the above equation; the denominator is found in the solution to Problem 12.38, reproduced in the indented insert below. From the definition of density,
ρcomposite = mcomposite Vcomposite Begin by assuming a "basis" of 1 m3 of composite, establishing its "volume." Consulting Table 12.7 (p. 412 of the text) as instructed in the problem statement reveals that the "loading" of Kevlar in this epoxy matrix composite is 82 vol %. Therefore on the assumed basis, the Kevlar occupies a volume of 0.82 m3 and the epoxy, 0.18 m3. The mass of the composite in the numerator is calculated next. From conservation of mass, mcomposite = mKevlar + mepoxy Substituting (volume × density) for mass, mcomposite = VKevlar ρKevlar + Vepoxy ρepoxy Problem 12.56 Solution Professor R. Gronsky page 1 of 2 Problem 12.56
J. F. Shackelford, Introduction to Materials Science for Engineers, 7th Edition, Prentice Hall, New Jersey (2009) and inserting values from the given information and volume fraction data, mcomposite = (0.82 m3 × 1.44 Mg/m3) + (0.18 m3 × 1.1 Mg/m3) mcomposite = 1.38 Mg The density of the composite is then calculated,
ρcomposite = 1.38 Mg 1 m3 Solving, ρcomposite = 1.38 Mg/m3. Now substituting into the original equation, and executing the same unit conversions adopted in Example 12.11,
sp. str. = (1517 MPa)([1.02 × 10−1 kg/mm2 ] / MPa) (1.38 Mg/m3 )(103 kg/Mg)(1 m3 /109 mm3 ) and solving, sp. str. = 112 × 106 mm Note: this value is nearly an order of magnitude higher than the specific strength (12.6 × 106 mm) of the aluminareinforced aluminum matrix composite in Example 12.11. Problem 12.56 Solution Professor R. Gronsky page 2 of 2 Problem 12.62
J. F. Shackelford, Introduction to Materials Science for Engineers, 7th Edition, Prentice Hall, New Jersey (2009) 12.62 In Problem 6.9, a competition among various metallic pressurevessel materials was illustrated. We can expand the selection process by including some composites, as listed in the following table: Material 1040 carbon steel 304 stainless steel 3003H14 aluminum Ti5Al2.5Sn Reinforced concrete Fiberglass Cfiber reinforced polymer ρ (Mg/m3) 7.80 7.80 2.73 4.46 2.50 1.80 1.50 Cost ($/kg) 0.63 3.70 3.00 15.00 0.40 3.30 70.00 Y.S. (MPa) 200 200 600 (a) From this expanded list, select the material that will produce the lightest vessel. (b) Select the materials that will produce the minimumcost vessel. Reference: 6.9 Suppose that you were asked to select a material for a spherical pressure vessel to be used in an aerospace application. The stress in the vessel wall is pr σ= 2t where p is the internal pressure, r the outer radius of the sphere, and t the wall thickness. The mass of the vessel is m = 4π r2 tρ where ρ is the material density. The operating stress of the vessel will always be Y . S. σ≤ S where S is a safety factor. (a) Show that the minimum mass of the pressure vessel will be ρ m = 2S π pr3 Y .S. (b) Given Table 6.1 and the following data, select the alloy that will produce the lightest vessel. Alloy ρ (Mg/m3) 1040 carbon steel 7.80 304 stainless steel 7.80 3003H14 aluminum 2.73 Ti5Al2.5Sn 4.46 aApproximate in U.S. dollars Costa ($/kg) 0.63 3.70 3.00 15.00 (c) Given Table 6.2 and the data in the preceding table, select the alloy that will produce the minimum cost vessel. Problem 12.62 Solution Professor R. Gronsky page 1 of 5 Problem 12.62
J. F. Shackelford, Introduction to Materials Science for Engineers, 7th Edition, Prentice Hall, New Jersey (2009) SOLUTION The solution to Problem 6.9 is reproduced in indented text here, offering the best path forward to the solution of the current problem. (a) The solution to part (a) is found be working with the definitions given in the problem statement and seeking simplifications. Begin by equating the general expression for the stress in a spherical pressure vessel to the “design” operating stress with safety factor included,
pr Y .S. ≤ 2t S Note that an expression for the thickness of the vessel is needed. This is obtained by rearranging the mass equation and solving for t.
t= m 4π r2 ρ Substituting,
pr 4π r2 ρ Y . S. 2π r 3 p ρ ≥ = S 2 m m and solving for the mass
m ≥ 2S π r 3 p ρ Y .S. shows from the inequality that the minimum mass is the one given in the problem statement,
m = 2S π pr3 ρ Y .S. completing the proof. (b) The lightest vessel will obviously have the lowest mass. By examination of the expression for mass (m) derived in part (a) above, Problem 12.62 Solution Professor R. Gronsky page 2 of 5 Problem 12.62
J. F. Shackelford, Introduction to Materials Science for Engineers, 7th Edition, Prentice Hall, New Jersey (2009) m = 2S π pr3 ρ Y .S. a path to minimizing mass can be readily found. Note that the safety factor (S) is NOT adjustable, the radius of the spherical pressure vessel (r) is NOT adjustable, and the internal pressure (p) is NOT adjustable, since these are fixed by the engineering design. Consequently the mass of the pressure vessel can be minimized by minimizing the last term, (ρ/Y.S.) on the RHS of this equation, since both density and yield strength vary with the material selected for the application. Densities of all candidate materials are given in the problem statement, but not the yield strengths; these are found in Table 6.1, which is why the problem statement points you there. To simplify comparisons among the four candidate materials, it is easiest to construct a spreadsheet to include the Y.S. data from Table 6.1, and enter a quotient function in the last column to calculate the term of interest. Alloy 1040 carbon steel 304 stainless steel 3003H14 aluminum Ti5Al2.5Sn ρ (Mg/m3) 7.80 7.80 2.73 4.46 Costa ($/kg) 0.63 3.70 3.00 15.00 Y.S. (MPa) 600 205 145 827 ρ/Y.S. (Mg/m3•MPa) 0.0130 0.0380 0.0188 0.0054 Answer: This readily reveals that the Ti5Al2.5Sn alloy will produce the lightest vessel. (c) Another spreadsheet is best here. Information given in the problem statement shows that cost is specified on a “per mass” basis, so the “lowest cost” vessel will have the lowest product of (ρ/Y.S. × Cost), which can be programmed as another column in the spreadsheet. Problem 12.62 Solution Professor R. Gronsky page 3 of 5 Problem 12.62
J. F. Shackelford, Introduction to Materials Science for Engineers, 7th Edition, Prentice Hall, New Jersey (2009) Alloy 1040 carbon steel 304 stainless steel 3003H14 aluminum Ti5Al2.5Sn ρ (Mg/m3) 7.80 7.80 2.73 4.46 Costa ($/kg) 0.63 3.70 3.00 15.00 Y.S. (MPa) 600 205 145 827 ρ/Y.S. (Mg/m3•MPa) 0.0130 0.0380 0.0188 0.0054 ρ/Y.S. × $
0.0082 0.1408 0.0565 0.0809 Answer: By a significant factor, 1040 carbon steel will produce the lowest cost vessel. (a) As in the solution to Problem 6.9, we are looking for the minimum ratio of ρ/Y.S., which can be assessed in the same manner as above, beginning with an expansion of the given table to include the missing values of Y.S. (found in Table 6.1 on page 158 of the text).
Material 1040 carbon steel 304 stainless steel 3003H14 aluminum Ti5Al2.5Sn Reinforced concrete Fiberglass Cfiber reinforced polymer ρ (Mg/m3) 7.80 7.80 2.73 4.46 2.50 1.80 1.50 Y.S. (MPa) 600 205 145 827 200 200 600 ρ/Y.S. (Mg/m3•MPa) 0.0130 0.0380 0.0188 0.00539 0.0125 0.009 0.0025 The lightest vessel is therefore made of a carbon fiber reinforced polymer composite. Problem 12.62 Solution Professor R. Gronsky page 4 of 5 Problem 12.62
J. F. Shackelford, Introduction to Materials Science for Engineers, 7th Edition, Prentice Hall, New Jersey (2009) (b) Again in parallel with the solution to Problem 6.9, the “lowest cost” vessel will have the lowest product of (ρ/Y.S. × Cost), which can be programmed as another column in the spreadsheet.
Material 1040 carbon steel 304 stainless steel 3003H14 aluminum Ti5Al2.5Sn Reinforced concrete Fiberglass Cfiber reinforced polymer ρ/Y.S. (Mg/m3•MPa) 0.0130 0.0380 0.0188 0.00539 0.0125 0.009 0.0025 Cost ($/kg) 0.63 3.70 3.00 15.00 0.40 3.30 70.00 ρ/Y.S. × $
0.0082 0.1406 0.0564 0.0809 0.0050 0.0297 0.1750 The lowest cost vessel is therefore made from reinforced concrete. Problem 12.62 Solution Professor R. Gronsky page 5 of 5 Problem 12.67
J. F. Shackelford, Introduction to Materials Science for Engineers, 7th Edition, Prentice Hall, New Jersey (2009) 12.67 Consider the injection molding of lowcost casings using a polyethyleneclay particle composite system. The modulus of elasticity of the composite increases and the tensile strength of the composite decreases with volume fraction of clay as follows: Volume fraction clay 0.3 0.6 Modulus of Elasticity (MPa) 830 2,070 Tensile strength (MPa) 24.0 3.4 Assuming both modulus and strength change linearly with volume fraction of clay, determine the allowable composition range that ensures a product with a modulus of at least 1,000 MPa and a strength of at least 10 MPa. SOLUTION Begin by invoking the assumption given in the problem statement that the modulus and the tensile strength of the composite vary linearly with volume fraction of the clay component. The operating equation for the elastic modulus is then written in the form y = mx + b, or E = mvclay + b Substituting for the two values given in the problem statement 830 = m(0.3) + b 2,070 = m(0.6) + b eliminating b, 830 – m(0.3) = 2,070 – m(0.6) and solving, m = (2,070 – 830) / (0.6 – 0.3) = 4,133 Substituting, b = 830 – 0.3 (4,133) = –410 Problem 12.67 Solution Professor R. Gronsky page 1 of 2 Problem 12.67
J. F. Shackelford, Introduction to Materials Science for Engineers, 7th Edition, Prentice Hall, New Jersey (2009) Consequently the lower limit of 1,000 MPa on the elastic modulus will occur at a volume fraction of clay given by vclay = (1000 – (–400)) / (4,133) = 0.341 Similarly, TS = m'vclay + b' Substituting for the two values given in the problem statement 24 = m'(0.3) + b' 3.4 = m'(0.6) + b' eliminating b, 24 – m'(0.3) = 3.4 – m'(0.6) and solving, m' = (3.4 – 24) / (0.6 – 0.3) = –68.7 Substituting, b' = 24 – 0.3 (–68.7) = 44.6 Consequently the lower limit of 10 MPa on the tensile strength will occur at a volume fraction of clay given by vclay = (10 – (–44.6)) / (–68.7) = 0.504 The composition range on the clay is therefore 0.341 ≤ vclay ≤ 0.504 Problem 12.67 Solution Professor R. Gronsky page 2 of 2 ...
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This note was uploaded on 01/21/2011 for the course E 45 taught by Professor Gronsky during the Fall '08 term at Berkeley.
 Fall '08
 GRONSKY

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