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Unformatted text preview: Problem 6.7
J. F. Shackelford, Introduction to Materials Science for Engineers, 7th Edition, Prentice Hall, New Jersey (2009) 6.7 In normal motion, the load exerted on the hip joint is 2.5 times body weight. (a) Calculate the corresponding stress (in MPa) on an artificial hip implant with a crosssectional area of 5.64 cm2 in a patient weighing 150 lbf. (b) Calculate the corresponding strain if the implant is made of Ti6Al4V, which has an elastic modulus of 124 GPa. SOLUTION (a) The stress on a hip implant can be calculated using master Equation 6.1 of the text (found on page 153), expressing stress as load divided by the crosssectional area over which it is borne.
σ= P A0 In the case the load (P) is taken as twice the body weight of the patient (given), and the crosssectional area of the implant is also given, making this a very simple problem solved by substitution. The only challenging aspect of the solution is the required unit conversion, which can be executed as follows, using the conversion factors in the inner front cover of the text.
σ= 2.5(150 lbf )(1 N/0.2248 lbf ) 5.64 cm2 (1 m/100 cm)2 Solving,
σ = 2.96 × 106 N/m2 or equivalently (using the same conversion tables), σ = 2.96 MPa. Problem 6.7 Solution Professor R. Gronsky page 1 of 2 Problem 6.7
J. F. Shackelford, Introduction to Materials Science for Engineers, 7th Edition, Prentice Hall, New Jersey (2009) (b) The strain experienced by the hip implant under this load is calculated assuming that the implant is still within its elastic limit and has not deformed plastically. This assumption allows an application of Hooke's Law, Equation 6.3 of the text (p. 154), rearranged here to solve for strain,
= σ E where E is Young's modulus or the modulus of elasticity, given in the problem statement. Substituting,
= 2.96 × 106 MPa 124 × 103 MPa and solving, the strain is found to be ϵ = 2.39 × 10–5. Problem 6.7 Solution Professor R. Gronsky page 2 of 2 Problem 6.8
J. F. Shackelford, Introduction to Materials Science for Engineers, 7th Edition, Prentice Hall, New Jersey (2009) 6.8 Repeat Problem 6.7 for the case of an athlete who undergoes a hip implant. The same alloy is used but, because the athlete weighs 200 lbf, a larger implant is required (with a crosssectional area of 6.90 cm2). Also, consider the situation in which the athlete expends his maximum effort exerting a load of five times his body weight. Reference 6.7 In normal motion, the load exerted on the hip joint is 2.5 times body weight. (a) Calculate the corresponding stress (in MPa) on an artificial hip implant with a crosssectional area of 5.64 cm2 in a patient weighing 150 lbf. (b) Calculate the corresponding strain if the implant is made of Ti6Al4V, which has an elastic modulus of 124 GPa. SOLUTION Reference to Problem 6.7 reveals that there are two questions to answer. The first asks for the stress exerted by a patient on a hip implant, and the second asks for the resulting strain experienced by the implant. (a) The stress on a hip implant can be calculated using master Equation 6.1 of the text (found on page 153), expressing stress as load divided by the crosssectional area over which it is borne.
σ= P A0 In the case the load (P) is taken as five times the body weight (given) of the athlete who has an implant, and the crosssectional area of the implant is also given, making this a problem solved by substitution. Unit conversion is required however, which can be executed as follows, using the conversion factors in the inner front cover of the text.
σ= 5(200 lbf )(1 N/0.2248 lbf ) 6.90 cm2 (1 m/100 cm)2 Solving,
σ = 6.45 × 106 N/m2 or equivalently (using the same conversion tables), σ = 6.45 MPa. Problem 6.8 Solution Professor R. Gronsky page 1 of 2 Problem 6.8
J. F. Shackelford, Introduction to Materials Science for Engineers, 7th Edition, Prentice Hall, New Jersey (2009) (b) The strain experienced by the hip implant under this extreme load is calculated assuming that the implant is still within its elastic limit and has not deformed plastically. This assumption allows an application of Hooke's Law, Equation 6.3 of the text (p. 154), rearranged here to solve for strain,
= σ E where E is Young's modulus or the modulus of elasticity, given in the problem statement. Substituting,
= 6.45 × 106 MPa 124 × 103 MPa and solving, the strain is found to be ϵ = 5.20 × 10–5. Problem 6.8 Solution Professor R. Gronsky page 2 of 2 Problem 6.9
J. F. Shackelford, Introduction to Materials Science for Engineers, 7th Edition, Prentice Hall, New Jersey (2009) 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
σ= pr 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
σ≤ Y . S. S where S is a safety factor. (a) Show that the minimum mass of the pressure vessel will be
m = 2S π pr3 ρ Y .S. (b) Given Table 6.1 and the following data, select the alloy that will produce the lightest vessel. Alloy 1040 carbon steel 304 stainless steel 3003H14 aluminum Ti5Al2.5Sn
aApproximate ρ (Mg/m3) 7.80 7.80 2.73 4.46 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. SOLUTION (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, Problem 6.9 Solution Professor R. Gronsky page 1 of 3 Problem 6.9
J. F. Shackelford, Introduction to Materials Science for Engineers, 7th Edition, Prentice Hall, New Jersey (2009) 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,
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. Problem 6.9 Solution Professor R. Gronsky page 2 of 3 Problem 6.9
J. F. Shackelford, Introduction to Materials Science for Engineers, 7th Edition, Prentice Hall, New Jersey (2009) 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.
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) ρ/Y.S. × $ 0.0130 0.0380 0.0188 0.0054 0.0082 0.1408 0.0565 0.0809 Answer: By a significant factor, 1040 carbon steel will produce the lowest cost vessel. Problem 6.9 Solution Professor R. Gronsky page 3 of 3 Problem 6.39
J. F. Shackelford, Introduction to Materials Science for Engineers, 7th Edition, Prentice Hall, New Jersey (2009) 6.39 You are provided an unknown alloy with a measured Brinell hardness value of 100. Having no other information than the data of Figure 6. 29a, estimate the tensile strength of the alloy (Express your answer in the form x ± y). SOLUTION Engineering solutions often have to be based upon the “best available data” and this problem offers an opportunity to explore how that might be done. The objective is to estimate a tensile strength of an “unknown” alloy from a Brinell hardness value. Figure 6.29a, the only available data set from which to draw your conclusion, presents a scatter plot of Brinell hardness – tensile strength correlations, explicit values for which are found in Table 6.10. The most unnerving aspect of this exercise is that there are NO data points shown for a BHN of 100. It is therefore possible that the unknown alloy is NOT one of those in Table 6.11, or possibly it is one of those alloys, but with a wider variation in properties than the test samples measured. So the only way to proceed is to specify a range of possible tensile strengths, as guided by the instruction to express your answer as “x ± y.” Consider again the plot of Fig 6.29a, reproduced here, noting that all data falls within a broad band. Problem 6.39 Solution Professor R. Gronsky page 1 of 2 Problem 6.39
J. F. Shackelford, Introduction to Materials Science for Engineers, 7th Edition, Prentice Hall, New Jersey (2009) From the scatter across this band, a BHN of 100 can correspond to a tensile strength ranging from approximately 260 to 540 MPa. Consequently the average tensile strength in this interval is [260 + 540] /2 = 400 MPa, and the range of the interval is [540 – 260] = 280 MPa or ± 140 MPa from the average value of 400. This enables you to specify the answer in the usual convention of (avg ± range/2). Therefore the estimated tensile strength of this unknown alloy is 400 ± 140 MPa. Problem 6.39 Solution Professor R. Gronsky page 2 of 2 Problem 6.47
J. F. Shackelford, Introduction to Materials Science for Engineers, 7th Edition, Prentice Hall, New Jersey (2009) 6.47 The Ti6Al4V orthopedic implant material introduced in Problem 6.7 gives a 3.27mmdiameter impression when a 10mmdiameter tungsten carbide sphere is applied to the surface with a 3000kg load. What is the Brinell hardness number of this alloy? SOLUTION The formula for the Brinell Hardness Number (BHN) is given in Table 6.9 (p. 182), relating the diameter of the indenter (D) to the diameter of the impression (d) that it makes in the surface of the sample under load P. Note that the resulting hardness is expressed as a "number" or an index without units. The BHN appears to be "dimensionless" but there are units associated with the variables in the formula. Consequently those units must be standardized, to enable universal comparisons of hardness numbers. The correct units are specified in the original ASTM standard (Designation E1008) for the Brinell hardness test, and are also found in the text, at the bottom of page 184, in the sample problem labeled "Example 6.9." For all BHN calculations, the load (P) must be expressed in kg and all dimensions (D and d) must be expressed in mm. Beginning with the formula,
BHN = 2P √ π D D − D 2 − d2 substituting for the variables in appropriate units,
BHN = 2(3000) π (10) (10) − (10)2 − (3.27)2 and solving, the Brinell Hardness Number is BHN = 347. Problem 6.47 Solution Professor R. Gronsky page 1 of 1 ...
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