Chapter 10 Why are metals ductile and ceramics brittle? Metals are generally ductile because the structure consists of close-packed layers of atoms that allow for low energy dislocation movement. Slip is then the preferred method of releasing stress, resulting in observable ductility. Ceramics are generally regarded as brittle because dislocation movement is hampered, partly due to the stronger bonding but also because of electrostatic repulsion between similarly charged ions. Crack propagation is often the preferred mechanism of releasing stress. The reason why ceramics tend to be brittle and metals ductile is therefore not due to the presence of dislocations in metals and their absence in ceramics, but because of the greater difficultly of slip in ceramics at normal temperatures. However, temperature is important and ceramics show ductility at high temperatures while metals become brittle and loose ductility at low temperatures. What is the elastic modulus of a solid? The simplest description of stress and strain is in terms of isotropic linear elastic materials. For this group of solids, the ratio of the applied stress to the resulting strain is the elastic modulus of the material. There is no single elastic modulus, but a number of moduli, depending upon the nature of the stress although the elastic properties of isotropic elastic materials are uniquely determined by any two of them. However, the term the elastic modulus used alone usually refers to just one of these, Young s modulus, that relates tensile stress and strain. Elastic moduli can also be specified for anisotropic materials, but take a more complicated form. What are solid lubricants?
Solid lubricants are materials that possess low shear strength in at least one dimension so that they are soft and feel greasy to the touch. They fall into three main classes, inorganic solids with lamellar crystal structures, solids that suffer plastic deformation easily and polymers in which the constituent chains can slip past each other in an unrestricted way. The two categories of most importance are layer structures and soft inorganic compounds. Typical layer structures used in this way are graphite derivatives such as fluorinated graphite and molybdenum disulphide. A soft inorganic compound often used at high temperatures is a eutectic mixture of calcium fluoride and barium fluoride. Solid lubricants have advantages over the normal liquid lubricants in certain conditions such as at high temperatures, low temperatures and in a vacuum. Quick quiz 1. a; 2. b; 3. a; 4. c; 5. a; 6. b; 7. c; 8. b; 9. a; 10. c; 11. a; 12. b; 13. a; 14. c; 15. b; 16. a; 17. b; 18. c; 19. c; 20. b; 21. a; 22. b; 23. c; 24. a; 25. c; 26. b; 27. b; 28. c; 29. a; 30. a; 31. c; 32. c; 33. b. Calculations and questions 1. 15.6 MPa. 2. 194.2 MPa. 3. 85.4 cm. 4. 156.0 cm. 5. 60.14 cm. 6. 100.36 cm. 7. 152.2 GPa. 8. 108.4 GPa.
9. 0.5. 10. 102 x 9.93 x 9.93 mm. 11. 12.46 mm. 12. 21.2 kn. 13. (a) 0.29 mm, (b) -0.0055 mm. 14. (a) 111.1 kpa; (b) 0.32 mm; (c) 0.0011, (0.11 %); (d) 6.3 x 10-6 m. 15. (a) 733.4 kpa; (b) 0.63 mm; (c) 0.0018, (0.18 %); (d) 0.0063 mm. 16. (a) 0.126, (b) 12.6 %. 17. (a) 0.210, (b) 21 %. 18. (b) ~70 GPa; (c) ~1.2 GPa; (d) ~6500 N; (e) 6.4 %. 19. (b) ~138 GPa; (c) ~1.3 GPa; (d) ~973 MPa; (e) 4.1 %. 20. (a) 1.2 GPa; (b) ~350 MPa; (c) ~55.8 kn; 37.5 mm. 21. (a) ~1.8 GPa; (b) ~1.1 GPa; (c) ~0.7 GPa; (d) ~0.9 GPa; (e) ~0.6 GPa; (f) 1.4 %. 22. (a) 45.5 kn; (b) 100.27 mm. 23. 117 MPa. 24. 1.63 m. 25. 0.45 N m -2. 26. (a) 94.4 GPa; (b) 58.5 GPa. 27. (a) 202.7 GPa; (b) 125.4 GPa. 28. (a) Long, 39.5 GPa; (b) Trans, 58.5 GPa. 29. (a) Long. 155.2 GPa; (b) Trans. 6.72 GPa. 30. (a) V, 287.7 GPa; (b) R, 167.1 GPa. 31. (a) Parallel, 114.2 GPa; (b) Perpendicular, 90.4 GPa. 32. 167.2 GPa. 33. 18.6 %. Solutions
1 A weight of 500 kg is hung from a 2 cm diameter rod of brass. What is the engineering stress? Engineering stress, = force / original area = F / A 0 F = mass x acceleration = m g = 500 x 9.8067 = 4903 N A 0 = r 2 = x 0.01 2 = 4903 / x 0.01 2 = 15.6 MPa 2 A weight of 3500 kg is hung from a 1.5 cm diameter rod of nickel. What is the engineering stress? Engineering stress, = force / original area = F / A 0 F = mass x acceleration = m g = 3500 x 9.8067 = 34323 N A 0 = r 2 = x 0.0075 2 = 4903 / x 0.0075 2 = 194.2 MPa 3 A steel wire 75 cm long and 1 mm diameter is subjected to a load of 22 kn. Young s modulus of the steel 201.9 GPa. Calculate the new length. Stress and strain are related by = E = l / l 0 = (l - l 0 ) / l 0 l = l 0 / E = load / area = 22000 / x 0.0005 2
l = (0.75 x 22000) / x 0.0005 2 x 201.9 x 10 9 = 0.104 m New length = 75 + 10.4 = 85.4 cm 4 A rod of bronze 150 cm long and 3 mm diameter is subjected to a load of 30 kn. Young s modulus of the bronze is 105.3 GPa. Calculate the new length. Stress and strain are related by = E = l / l 0 = (l - l 0 ) / l 0 l = l 0 / E = load / area = 30000 / x 0.0015 2 l = (1.5 x 30000) / x 0.0015 2 x 105.3 x 10 9 = 0.060 m New length = 1.5 + 0.06 = 1.56 m 5 A rod of copper 60 cm long is subjected to a tensile stress of 300MPa. Young s modulus of copper is 129.8 GPa. Calculate the new length. l = l 0 / E l = (0.6 x 300 x 10 6 ) / 129.8 x 10 9 = 0.0014 m New length = 60 + 0.14 = 60.14 cm 6 A rod of aluminium 100 cm long is subjected to a tensile stress of 250 MPa. Young s modulus of aluminium is 70.3 GPa. Calculate the new length. l = l 0 / E l = (1 x 250 x 10 6 ) / 70.3 x 10 9 = 0.0036 m
New length = 1 + 0.0036 = 1.0036 m 7 A cast iron rod of length 200 mm and dimensions 10 mm x 20 mm is subjected to a load of 70 kn. An extension of 0.46 mm is observed. Calculate Young s modulus of the cast iron. / = E = l / l 0 = (l - l 0 ) / l 0 l = l 0 / E = load / area = 70,000 / (0.01 x 0.02) = l / l 0 = 0.00046 / 0.2 E = 70,000 x 0.2 / (0.01 x 0.02) x 0.00046 = 152.2 GPa 8 A zinc bar of length 125 mm and dimensions 5 mm x 7.5 mm is subjected to a load of 40 kn. An extension of 1.23 mm is observed. Calculate Young s modulus of zinc. / = E = l / l 0 = (l - l 0 ) / l 0 l = l 0 / E = load / area = 40,000 / (0.005 x 0.0075) = l / l 0 = 0.00123 / 0.125 E = 40,000 x 0.125 / (0.005 x 0.0075) x 0.00123 = 108.4 GPa 9 Calculate Poisson s ratio for a bar of metal originally 10 mm x 10 mm x 100 mm, which is extended to 101 mm, if there is no change in the overall volume of the sample.
Assuming that the bar is isotropic V(initial) = V(final) = 100 x 10 x 10 = 10 4 mm 3 V(final) = 101 x w x w = 101w 2 mm 3 where the cross section of the bar is now w x w. Hence w = (10 4 / 101) = 9.95 mm z (longitudinal strain) = l / l 0 = (101 100) / 100 = 0.01 x (lateral strain) = w / w 0 = (9.95 10) / 10 = -0.005 = - x (lateral strain) / z (longitudinal strain) = -(-0.005 / 0.01) = 0.5 10 A copper bar of 10 mm square section is subjected to a tensile load that increases its length from 100 mm to 102 mm. The value of Poisson s ratio for copper is 0.343. Calculate the new dimensions of the bar. = - x (lateral strain) / z (longitudinal strain) z = l / l 0 = 2 / 100 = 0.01 x = w / w 0 w / w 0 = -0.343 x 2 / 100 = -0.0069 As w 0 = 10 mm, w = 10 x 0.0069 = 0.069 mm New cross-sectional dimensions are 9.93 x 9.93 mm The new dimensions are 102 x 9.93 x 9.93 mm 11 A brass rod of 12.5 mm diameter is subjected to a tensile load that increases its length from 150 mm to 151.5 mm. The value of Poisson s ratio for brass is 0.350. Calculate the new diameter.
= - x (lateral strain) / z (longitudinal strain) z = l / l 0 = 1.5 / 150 = 0.01 x = - 0.350 x 0.01 = -0.00350 The diameter of the rod is 12.5 mm and the change in diameter is d - x = - d / 12.5 = -0.00350 - d = -0.04375 New diameter = 12.5-0.044 = 12.46 mm The new dimensions are 151.5 mm x 12.46 mm diameter 12 A cylindrical titanium rod of diameter 15 mm is subjected to a tensile load applied along the long axis. Young s modulus of the metal is 115.7 GPa and Poisson s ratio is 0.321. Determine the magnitude of the load needed to produce a contraction in diameter of 5 x 10-3 mm if the deformation is elastic. = - x (lateral strain) / z (longitudinal strain) z = l / l 0 - x = - d / d 0 x = -5 x 10-3 / 15 = -3.33 x 10-4 z = - x / = 3.33 x 10-4 / 0.321 = 1.038 x 10-2 = z E = 1.038 x 10-2 x 115.7 x 10 9 = 120.15 x 10 6 N m -2 Load = F = A 0 = 120.15 x 10 6 x (7.5 x 10-3 ) 2 = 21.2 kn
13 A steel rod of diameter 16.2 mm and length 25 cm is subjected to a force of 50,000 N in tension along the long axis. Young s modulus is 210 GPa and Poisson s ratio is 0.293. Determine (a) the amount that the specimen will elongate in the direction of the applied force and (b) the change in diameter of the rod. (a) = F / A 0 = 50,000 / (8.1 x 10-3 ) 2 z = l / l 0 = l / 0.25 = z E i.e. = 50,000 / (8.1 x 10-3 ) 2 = 210 x 10 9 x l / 0.25 l = 0.25 x 50,000 / (8.1 x 10-3 ) 2 x 210 x 10 9 = 0.00029 m =.29 mm The bar will elongate by 0.29 mm (b) = - x (lateral strain) / z (longitudinal strain) - x = - d / d 0 = (- d / d 0 ) / ( l / l 0 ) d = - x ( l / l 0 ) x d 0 = -0.293 x (0.29 / 250) x 16.2 = 0.0055 mm The diameter will decrease by 0.0055 mm 14 A niobium bar of dimensions 15 mm square and of length 300 mm is subjected to a tensile force of 25,000 N. Young s modulus of niobium is 104.9 GPa NS Poisson s ratio is 0.397. Determine (a), the engineering stress; (b), the elongation; (c), the engineering strain; (d), the change in the cross section of the bar. (a) = F / A 0 = 25,000 / (0.015 x 0.015) = 111.1 MPa (b) z = l / l 0 = l / 0.3 = z E
Hence 111.1 x 10 6 = 104.9 x 10 9 x l / 0.3 l = 0.3 x 111.1 x 10 6 / 104.9 x 10 9 = 0.00032 m = 0.32 mm (c) z = 0.00032 /.3 = 0.00106 = 0.106 % (d) = - x / z = ( w / w 0 ) / z i.e. ( w / 0.015 ) / 0.00106 = 0.397 w = 0.397 x 0.00106 x 0.015 = 6.3 x 10-6 m 15 A tungsten rod of 12.5 mm diameter and of length 350 mm is subjected to a tensile force of 90,000 N. Young s modulus of tungsten is 411.0 GPa and Poisson s ratio is 0.280. Determine (a), the engineering stress; (b), the elongation; (c), the engineering strain; (d), the change in the diameter of the rod. (a) = F / A 0 = 90,000 / (0.00625) 2 = 733.4 MPa (b) z = l / l 0 = l / 0.350 = z E Hence 733.4 x 10 6 = 411 x 10 9 x l / 0.35 l = 0.35 x 733.4 x 10 6 / 411 x 10 9 = 0.000625 m = 0.625 mm (c) z = 0.000625 /.35 = 0.00179 = 0.179 % (d) = - x / z = ( d / d 0 ) / z i.e. ( d / 0.0125 ) / 0.00179 = 0.280 d = 0.280 x 0.00179 x 0.0125 = 6.265 x 10-6 m = 0.006265 mm
16 A tensile test specimen of magnesium has a gauge length of 5 cm. The metal is subjected to a tensile loading until the gauge markings are 5.63 cm apart. Calculate (a) the engineering stress and (b) the percentage elongation. (a) = l / l 0 = (5.63 5) / 5 = 0.126 (b) 12.6% 17 A tensile test specimen of brass has a gauge length of 5 cm. The metal is subjected to a tensile loading until the gauge markings are 6.05 cm apart. Calculate (a) the engineering stress and (b) the percentage elongation. (a) = l / l 0 = (6.05 5) / 5 = 0.210 (b) 21% 18 An aluminium alloy specimen 3 mm diameter with 50 mm gauge length was tested to destruction in a tensile test. The results are given in the Table. The maximum load applied was 810 N and the final length between the gauge marks was 54 mm. (a) Plot an engineering stress versus engineering strain curve. (b) Determine Young s modulus of the alloy. (c) The tensile strength. (d) The 0.2% offset yield strength of the alloy. (e) The percentage elongation at fracture. Load / N Extension / mm 1000 0.10 2000 0.20 3000 0.290
4000 0.402 5000 0.504 6000 0.697 7000 0.900 7500 1.297 8000 2.204 7150, Fracture 3.200 (a) Data for the graph: Load / Stress / Extension / Strain N MPa mm 1000 141.4 0.10 0.0020 2000 282.8 0.20 0.0040 3000 424.3 0.290 0.0058 4000 565.8 0.402 0.00804 5000 707.2 0.504 0.01008 6000 848.7 0.697 0.01394 7000 990.1 0.900 0.01800 7500 1060.8 1.297 0.02594 8000 1131.5 2.204 0.04408 7150 1011.3 3.200 0.06400
Note: values derived from the graph are only approximate. (b) E is given by the initial slope 707.2 x 10 6 / 0.01008 = 70.2 GPa (c) The tensile strength is the maximum value of : 1.2 GPa (d) The 0.2 % offset line intersects the curve at 920 MPa (e) The % elongation at fracture = ( l / l 0 ) x 100 = (3.2 / 50) x 100 = 6.4 % 19. A steel specimen 12 mm diameter with 50 mm gauge length was tested to destruction in a tensile test. The results are given in the Table. The maximum load applied was 152 kn. (a) Plot an engineering stress versus engineering strain curve. (b) Determine Young s modulus of the alloy. (c) The tensile strength. (d) The 0.1% offset yield stress of the alloy. (e) The percentage elongation at fracture. Load / kn Extension / mm 10 0.030 20 0.064 30 0.098 40 0.130 50 0.170 60 0.195
70 0.218 80 0.256 90 0.294 100 0.335 110 0.400 120 0.505 130 0.660 140 0.898 150 1.300 152 (max) 1.500 150 1.700 140 1.960 133, 2.070 Fracture (a) Data for the graph: Load / kn Stress / MPa Extension / mm Strain 10 88.4 0.030 0.00060 20 176.8 0.064 0.00128 30 265.3 0.098 0.00196 40 353.7 0.130 0.00260 50 442.1 0.170 0.00340 60 530.5 0.195 0.00390 70 618.9 0.218 0.00436
80 707.4 0.256 0.00512 90 795.8 0.294 0.00588 100 884.2 0.335 0.00666 110 972.6 0.400 0.00800 120 1081.0 0.505 0.01010 130 1149.5 0.660 0.01320 140 1237.9 0.898 0.01796 150 1326.3 1.300 0.02600 152 (max) 1346.0 1.500 0.03000 150 1326.3 1.700 0.03400 140 1257.9 1.960 0.03920 133 1176.0 2.070 0.04140 Note: values derived from the graph are only approximate. (b) E is given by the initial slope 707.4 x 10 6 / 0.00512 = 138 GPa (c) The tensile strength is the maximum value of : 1.34 GPa (d) The 0.1 % offset line intersects the curve at 972.6 MPa (e) The % elongation at fracture = ( l / l 0 ) x 100 = (2.070 / 50) x 100 = 4.1 %
20. Figure 10.41 shows the engineering stress engineering strain behaviour of a carbon steel. Determine: (a) Young s modulus. (b) The stress at 0.2% offset strain, (proof stress). (c) The maximum load that can be sustained by a rod of diameter 12.5 mm. (d) The change in length of a rod originally 250 mm long subjected to an axial stress of 400 MPa. Working from an enlarged photocopy of Fig 10.41: (a) Young s modulus is given by the initial slope of the graph: E 300 / 0.025 1.2 x 10 4 Mpa = 12 GPa (b) The 0.02 % offset line intersects the curve at 350 MPa0.15 (c) The maximum load corresponds to the tensile strength, 455 MPa. = F / A 0 so F max 455 x A 0 = 455 x (0.00625) 2 F max 45508 kn (d) At = 400 MPa, 0.15 = l / l 0 l / 250 0.15 l 37.5 mm
21 A tensile test carried out on a sample of polypropylene of dimensions 12.5 mm width, 3.5 mm thick, gauge length 50 mm, gave the data in the Table. The maximum load applied was 625 N and the length between the gauge marks at fracture was 53.8 mm. Estimate: (a) The initial modulus. (b) The secant modulus at 0.2% strain. (c) The tangent modulus at 0.2% strain. (d) The secant modulus at 0.4% strain. (e) The tangent modulus at 0.4% strain. (f) The percentage elongation at break. Force / N Extension / mm 25 0.018 50 0.042 75 0.071 100 0.115 125 0.145 150 0.187 175 0.230 200 0.285 225 0.345 250 0.387 275 0.460 300 0.543 286, Break 0.720 Plot extension versus load:
(a) The initial slope of the graph 300 / 0.188 N mm -1 = (F / l) = F / A 0 ; = l / l 0 ; = E E = / = (F / A 0 ) x (l 0 / l) E (300 x 50) / (0.0125 x 0.0035 x 0.188) 1.82 GPa (b) 0.2 % strain = 0.2 % of 50 = 1 mm The slope for the secant modulus, from the graph, 96 / 0.1 N mm -1 E = / = (F / A 0 ) x (l 0 / l) E (96 x 50) / (0.0125 x 0.0035 x 0.1) 1.10 GPa (c) Taking the slope for the tangent modulus at 0.2 % strain, from the graph: 128 / 0.2 N mm -1 [Note: not very accurate.] E = / = (F / A 0 ) x (l 0 / l) E (128 x 50) / (0.0125 x 0.0035 x 0.2) 0.73 GPa (d) 0.4 % strain = 0.4 % of 50 = 0.2 mm The slope for the secant modulus, from the graph, 159 / 0.2 N mm -1 E = / = (F / A 0 ) x (l 0 / l) E (159 x 50) / (0.0125 x 0.0035 x 0.2) 0.91 GPa (e) Taking the slope for the tangent modulus at 0.4 % strain, from the graph:
111 / 0.2 N mm -1 [Note: not very accurate.] E = / = (F / A 0 ) x (l 0 / l) E (111 x 50) / (0.0125 x 0.0035 x 0.2) 0.63 GPa (f) The % elongation at break (0.720 / 50) x 100 = 1.4 % 22 A copper nickel alloy has a 0.1% offset yield strength of 350 MPa and Young s modulus of 130 GPa. (a) Determine the maximum load that may be applied to a specimen of cross section 10 x 13 mm, without significant plastic deformation occurring. (b) If the original specimen length is 100 mm, what is the maximum length that it can be stretched to elastically? (a) Significant deformation occurs at loads higher than the yield strength, hence: yield strength = 350 x 10 6 Pa = (F / A 0 ) = F / (0.01 x 0.013) F = 350 x 10 6 x 0.01 x 0.013 = 45.5 kn (b) The maximum extension is equal to that at the yield point, hence: = l / l 0 = l / 0.1 = / E = 350 x 10 6 / 130 x 10 9 l = (350 x 10 6 x 0.01 / 130 x 10 9 = 0.269 mm Maximum length is l 0 + 0.269 = 100.269 mm 23 Using the Griffith criterion, [ c = (2 E / a) ½ ], estimate the stress at which a glass plate containing a surface crack of 1.2 m deep will fracture due to a force applied perpendicular to the length of the crack. Young s modulus of the glass is 71.3 GPa and the surface energy of the glass is 0.360 J m -2. c = (2 E / a) ½
= [( 2 x 0.360 x 71.3 x 10 9 ) / ( x 1.2 x 10-6 )] ½ = 116.7 x 10 6 N m -2 = 116.7 MPa 24 A glass plate has to withstand a stress of 10 8 N m -2. Using the data in the previous question, what will be the critical crack size for this to be achieved? c = (2 E / a) ½ 1 x 10 8 = [( 2 x 0.360 x 71.3 x 10 9 ) / ( x a)] ½ 1 x 10 16 = ( 2 x 0.360 x 71.3 x 10 9 ) / ( x a) a = 1.63 x 10-6 m = 1.63 m 25 A plate of high-density polyethylene has a surface crack 7.5 m in one face. The plate fractures in a brittle fashion when a force of 6 x 10 6 N m -2 is applied in a direction perpendicular to the crack. Young s modulus of the polyethylene is 0.95 GPa. Estimate the surface energy of the material. c = (2 E / a) ½ 6 x 10 6 = [( 2 x x 0.95 x 10 9 ) / ( x 7.5 x 10-6 )] ½ 3.6 x 10 13 = (2 x x 0.95 x 10 9 ) / ( x 7.5 x 10-6 ) = (3.6 x 10 13 x x 7.5 x 10-6 ) / (2 x 0.95 x 10 9 ) = 0.446 Jm -2
26 Determine (a) the upper bound and (b) the lower bound Young s modulus of an ingot of magnesium metal containing 30 volume % magnesia (MgO) particles. Young s modulus of magnesium is 44.7 GPa and that of magnesia is 210.3 GPa. (a) The upper bound limit, E c = E m V m + E p V p = 44.7 x 0.7 + 210.3 x 0.3 = 94.4 GPa (b) The lower bound limit, E c = E m E p / (E m V p + E p V m ) = (44.7 x 210.3) / (44.7 x 0.3 + 210.3 x 0.7) = 58.5 GPa 27 An aluminium alloy is to be strengthened by the incorporation of beryllium oxide (BeO) particles. Calculate (a) the upper and (b) the lower bound elastic moduli of a composite consisting of 40 weight % alloy and 60 weight percent BeO. Young s modulus of the alloy is 70.3 GPa and its density is 2698 kg m -3. Young s modulus of BeO is 301.3 GPa and its density is 3010 kg m -3. The composite contains 40 wt % alloy and 60 wt % BeO, i.e. we have 40 kg alloy + 60 kg BeO volume of alloy = 40 / 2696 = 0.01484 m 3 volume of BeO = 60 / 3010 = 0.01993 m 3 volume fraction of alloy = 0.01484 / (0.01484 + 0.01993) = 0.4267 volume fraction of BeO = 0.01993 / (0.01484 + 0.01993) = 0.5733 (a) The upper bound limit, E c = E m V m + E p V p
= 70.3 x 0.4267 + 301.3 x 0.5733 = 202.7 GPa (b) The lower bound limit, E c = E m E p / (E m V p + E p V m ) = (70.3 x 301.3) / (70.3 x 0.5733 + 301.3 x 0.4267) = 125.4 GPa 28 Compute Young s modulus of a composite consisting of continuous and aligned glass fibres of 50 % volume fraction in an epoxy resin matrix under (a) longitudinal and (b) transverse loading. Young s modulus of the glass fibres is 76 GPa and that of the resin is 3 GPa. (a) Under longitudinal loading, E long = E m V m + E f V f = 3 x 0.5 + 76 x 0.5 = 39.5 GPa (b) Under transverse loading, E trans = E m E f / (E m V f + E f V m ) = (3 x 76) / (76 x 0.5 + 3 x 0.5) = 5.77 GPa 29 Compute Young s modulus of a composite consisting of continuous and aligned carbon fibres of 60 % weight fraction in an epoxy resin matrix under (a) longitudinal and (b) transverse loading. Young s modulus of the carbon fibres is 290 GPa and the density is 1785 kg m -3. Young s modulus of the resin is 3.2 GPa and its density is 1350 kg m -3. The composite contains 60 wt % C fibre and 40 wt % epoxy, i.e. we have 60 kg C fibre + 40 kg epoxy volume of C fibre = 60 / 1785 = 0.0336 m 3 volume of epoxy = 40 / 1350 = 0.0296 m 3
volume fraction of C fibre = 0.0336 / (0.0336 + 0.0296) = 0.53 volume fraction of epoxy = 0.0296 / (0.0336 + 0.0296) = 0.47 (a) Under longitudinal loading, E long = E m V m + E f V f = 3.2 x 0.47 + 290 x 0.53 = 155.2 GPa (b) Under transverse loading, E trans = E m E f / (E m V f + E f V m ) = (3.2 x 290) / (290 x 0.47 + 3.2 x 0.53) = 6.72 GPa 30 Determine (a) the Voigt and (b) the Reuss bounds to Young s modulus of a ceramic material consisting of layers of alumina and a high silica glass. Young s modulus of the alumina is 380 GPa and that of the glass is 72.4 GPa, and the glass comprises 30 volume % of the solid. (a) For the Voigt bound, E(Voigt) = E m V m + E p V p = 380 x 0.7 + 72.4 x 0.3 = 287.7 GPa (b) For the Reuss bound, E(Reuss) = E m E p / (E m V p + E p V m ) = (380 x 72.4) / (380 x 0.3 + 72.4 x 0.7) = 167.1 GPa 31 A mineral with an approximate formula Mg 7 Si 8 O 23 has a structure that is made up of alternating layers with compositions of 7MgO and 8SiO 2. Estimate Young s modulus of the material when stressed (a) parallel and (b) perpendicular to the layers. Young s modulus
and density of MgO are 210.3 GPa and 3580 kg m -3, and for silica are 72.4 GPa and 2650 kg m -3. The mass of an MgO layer is 7 x molar mass MgO = 7 x 40.31 = 282.17 g The mass of an SiO 2 layer is 8 x molar mass an SiO 2 = 8 x 60.09 = 480.72 g To simplify, take these as kg so: volume of MgO fraction = 6282.17 / 3580 = 0.0788 m 3 volume of SiO 2 fraction = 480.72 / 2650 = 0.1814 m 3 volume fraction of MgO = 0.0788 / (0.0788 + 0.1814) = 0.303 volume fraction of SiO 2 = 0.1814 / (0.0788 + 0.1814) = 0.697 As SiO 2 is the greater volume fraction, take this as the matrix. (a) E parallel to layers is equal to the Voigt bound, E(Voigt) = E m V m + E p V p = 72.4 x 0.697 + 210.3 x 0.303 = 114.2 GPa (b) E perpendicular to the layers is equal to the Reuss bound, E(Reuss) = E m E p / (E m V p + E p V m ) = (72.4 x 210.3) / (72.4 x 0.303 + 210.3 x 0.697) = 90.4 GPa
32 Young s modulus of sintered calcia stabilised zirconia with a porosity of 5 % is 151.7 GPa. Estimate Young s modulus of completely pore free material. Young s modulus is approximately given by: E c = E 0 [1-1.9 V p + 0.9 V p 2 ] where V p is the volume fraction of pores. Hence 151.7 = E 0 [1-1.9 (0.05) + 0.9 x 0.05 2 ] = 0.9072 E 0 E 0 = 167.2 GPa 33 Young s modulus of sintered silicon carbide with 5 % porosity is 468.9 GPa. What is the porosity of a specimen with a Young s modulus of 350 GPa? Young s modulus is approximately given by: E c = E 0 [1-1.9 V p + 0.9 V p 2 ] where V p is the volume fraction of pores. Hence 468.9 = E 0 [1-1.9 (0.05) + 0.9 x 0.05 2 ] 350 = E 0 [1-1.9x + 0.9 x 2 ] 468.9 / [1-1.9 (0.05) + 0.9 x 0.05 2 ] = 350 / [1-1.9x + 0.9 x 2 ] 468.9 / 0.90725 = 350 / [1-1.9x + 0.9 x 2 ] [1-1.9x + 0.9 x 2 ] = 0.6772 Solving the quadratic equation gives x = 0.186 as the only reasonable answer, thus porosity = 0.186 = 18.6 %