official ielts practice materials vol 2 free download
Engineering Materials vol 2 Part 1 doc
... 100 (140) 7.9 21 1 50 20 0 Mild steel 20 0 23 0 (26 0–300) 7.9 21 0 22 0 430 High-carbon steel 150 (20 0) 7.8 21 0 350–1600 650 20 00 Low-alloy steels 180 25 0 (23 0–330) 7.8 20 3 29 0–1600 420 20 00 High-alloy ... 30–100 1 120 85 19 0.4 >100 1 728 450 89 13 0.5 >100 1600 420 22 14 0 .2 >100 1550 450 11 12 0.1–0.5 45 933 917 24 0 24 0.1–0.45 45 915 24 0.1–0 .25 10–50 860 180 24 0.1–0.35 30–40 890 130 22 0.1–0.17 ... 12 0 .21 140 1765 4 82 60 12 0.1–0 .2 20–50 1570 460 40 12 0.1–0 .2 50–170 1750 460 40 12 0.1–0.5 50–170 1680 500 12 30 10–18 0–0.18 6 20 1403 0.5–0.9 >100 1356 385 397 17 0.5 30–100 1190 121 20 0.5...
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Engineering Materials vol 2 Part 2 doc
... High-strength; high-temperature. (free- flowing) + 16 Zn + 25 Cd Silver; general-purpose 38 Ag + 20 Cu 605–650 High-strength; high-temperature. (pasty) + 22 Zn + 20 Cd Case studies in phase diagrams ... hot-water layout, you will already have had some direct experience of this system. 22 Engineering Materials 2 Fig. 2. 7. Many metals are made up of two phases. This figure shows some of the shapes ... (Fig. 2. 2c). But if A atoms prefer to have A neigh- bours, or B atoms prefer B neighbours, the solution can cluster (Fig. 2. 2d); and when A atoms prefer B neighbours the solution can order (Fig. 2. 2e). Many...
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Engineering Materials vol 2 Part 3 pot
... ∆H = −334 kJ kg −1 . For water at 27 2 K, with T m − T = 1 K, we find that W f = 1 .22 kJ kg −1 (or 22 J mol −1 ). 1 kg of water at 27 2 K thus has 1 .22 kJ of free work avail- able to make it turn ... Conservation of volume gives 4 3 4 3 4 3 3 3 1 3 2 3 πππ rrr =+ . (5.30) Combining eqns (5 .29 ) and (5.30) gives ∆A = 4 1 3 2 323 1 2 2 2 πγ [( ) ( )]. / rr rr+−+ (5.31) For r 1 /r 2 in the range ... coarsening process? If we put r 1 = r 2 /2 in eqn. (5.31) we get ∆A = −4 πγ (−0.17 r 2 2 ). If γ = 0.5 J m 2 and r 2 = 10 −7 m our two precipitates give us a free work of 10 −14 J, or about...
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Engineering Materials vol 2 Part 4 doc
... the vol- ume of the nucleus. For 0 ഛ θ ഛ 90° these are: solid–liquid area = 2 π r 2 (1 − cos θ ); (7.4) catalyst–solid area = π r 2 (1 − cos 2 θ ); (7.5) nucleus volume = 2 3 1 3 2 1 2 3 3 πr ... (7.1) we can write W f = 2 π r 2 (1 − cos θ ) γ SL + π r 2 (1 − cos 2 θ ) γ CS − π r 2 (1 − cos 2 θ ) γ CL −−+ − cos cos ( ) . 2 3 1 3 2 1 2 3 3 πr H TT T m m θθ ∆ (7.7) Note ... the interfacial energy terms to reduce to 2 π r 2 (1 − cos θ ) γ SL + π r 2 (1 − cos 2 θ )(– γ SL cos θ ), or 2 π r 2 {1 − 3 2 cos θ + 1 2 cos 3 θ } γ SL . (7.9) The complete result...
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Engineering Materials vol 2 Part 5 pps
... 2. 1 41–160 150 25 0 Ti alloys 4.5 120 170– 128 0 27 2. 4 1.1 38 28 0 400–600 (Steels) (7.9) (21 0) (22 0–1600) 27 1.8 0.75 28 20 0 (400–600) * See Chapter 25 and Fig. 25 .7 for more information about these ... strength E / r*E 1 /2 / r*E 1/3 / r* s y / r* Creep r (Mg m − 3 ) modulus s y (MPa) temperature E (GPa) (°C) Al alloys 2. 7 71 25 –600 26 3.1 1.5 9 22 0 150 25 0 Mg alloys 1.7 45 70 27 0 25 4.0 2. 1 41–160 150 25 0 Ti ... reading M. F. Ashby and D. R. H. Jones, Engineering Materials I, 2nd edition, Butterworth-Heinemann, 1996, Chapters 7 (Case study 2) , 10, 12 (Case study 2) , 27 . Further reading I. J. Polmear, Light Alloys,...
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Engineering Materials vol 2 Part 6 pot
... 0 .20 0.55 B 0.40 0.60 1 .20 0.30 1.50 C 0.36 0.70 1.50 0 .25 1.50 D 0.40 0.60 1 .20 0.15 1.50 E 0.41 0.85 0.50 0 .25 0.55 F 0.40 0.65 0.75 0 .25 0.85 G 0.40 0.60 0.65 0.55 2. 55 116 Engineering Materials ... Jones, Engineering Materials I, 2nd edition, Butterworth-Heinemann, 1996, Chapters 21 , 22 , 23 and 24 . Further reading K. J. Pascoe, An Introduction to the Properties of Engineering Materials, Van ... started the crack which led to the Alexander Keilland failure.) Example 2: Pressure vessel steel to A 387 grade 22 class 2 A 387 22 (2) is a creep-resistant steel which can be used at about 450°C. It...
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Engineering Materials vol 2 Part 7 pps
... 1000 10 3–5 23 23 (1470) 795 25 .6 8.5 150 40 – 3110 – 1 422 84 4.3 300 40 4 21 73 – 627 17 3 .2 500 10 4– 12 2843 – 670 1.5 8 500 10 5 –– 710 20 25 3 .2 510 40 0 .2 ––1.8 10–14 # <50 40 0 .2 – 21 0–14 $ – ... 20 0 20 00 20 0–500 10 21 Sialons (490–1400) 3 .2 300 20 00 500–830 15 Cement, etc . Cement 52 (73) 2. 4 2. 5 20 –30 50 7 12 Concrete 26 (36) 2. 4 30–50 50 7 12 Rocks and ice Limestone Cost of mining 2. 7 ... − b) 2 + w 2 = r 2 . (14.11) Provided b Ӷ 2r this can be expanded to give wrb .= 2 (14. 12) Thus w d rb d r d b d / / == 22 12 12 . (14.13) 148 Engineering Materials...
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Engineering Materials vol 2 Part 9 pps
... concretes 20 9 Fig. 20 .2. (a) The hardening of Portland cement. The setting reaction (eqn. 20 .5) is followed by the hardening reactions (eqns 20 .6 and 20 .7). Each is associated with the evolution ... 20 .7). Each is associated with the evolution of heat (b). 2C 2 S + 4H → C 3 S 2 H 3 + CH + heat (20 .6) 2C 3 S + 6H → C 3 S 2 H 3 + 3CH + heat. (20 .7) d Tobomorite gel. Portland cement is stronger than ... any cement are, in this nomenclature Lime CaO = C Alumina Al 2 O 3 = A Silica SiO 2 = S Water H 2 O = H. 20 0 Engineering Materials 2 Fig. 19.9. Forming methods for glass: pressing, rolling, float-moulding...
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