Engineering Materials vol 2 Part 1 doc

... 397 17 0.5 30 10 0 11 90 12 1 20 0.5 30 10 0 1 12 0 85 19 0.4 > ;10 0 1 728 450 89 13 0.5 > ;10 0 16 00 420 22 14 0 .2 > ;10 0 15 50 450 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 ... coefficient (MK − 1 ) 0.3 80 18 09 456 78 12 0 . 21 14 0 17 65 4 82 60 12 0 .1 0 .2 20–50 15 70 460 40 12 0 .1 0 .2 50 17 0 17 50 460...

Ngày tải lên: 11/08/2014, 02:22

... at annealing temperature (°C) (minutes) 600 18 0 0 16 0 10 13 5 20 11 5 30 11 5 60 620 18 0 0 16 0 4 13 5 9 11 5 13 11 5 26 645 18 0 0 16 0 1. 5 13 5 3.5 11 5 5 11 5 10 Estimate the time that it takes for recrystallisation ... analogy with eqn. (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 )...

Ngày tải lên: 11/08/2014, 02:22

... 0 .1 0.3 4 0 .2 23 12 0.05 1 .2 Mahogany 0.53 13 .5 0.8 90 46 0 .25 6.3 Douglas fir 0.55 16 .4 1. 1 70 42 0.34 6 .2 Scots pine 0.55 16 .3 0.8 89 47 0.35 6 .1 Birch 0. 62 16 .3 0.9 – – 0.56 – Ash 0.67 15 .8 1. 1 ... cheaper. Table 26 .4 Specific strength of structural materials Material E r s r y K IC r Woods 20 –30 12 0 17 0 1 12 Al-alloy 25 17 9 8 16 Mild steel...

Ngày tải lên: 11/08/2014, 02:22

... E w|| = 11 .6 GPa, E w⊥ = 0. 71 GPa and ρ w = 0.39 Mg m −3 . Equations (28 .22 ) and (28 .23 ) then give us the following data. f || (s 1 ) ≈ 59 23 6 5 31 944 14 75 21 24 28 91 etc. ! 21 7 868 19 53 34 72 etc. f ⊥ (s 1 ) ... eqn. (28 .22 ) that this requires n l Ebd bd n l Ebd bd ww ww cc cc 2 2 3 12 2 2 3 12 2 12 2 12 π ρ π ρ || || || || || || || ||    ...

Ngày tải lên: 11/08/2014, 02:22

... alloy. (a) 2 wt% lead + 98 wt% tin, (b) one phase (just), (c) 2 wt% lead, (d) 10 0 wt% (Sn). 16 Engineering Materials 2 back to b.c.c. at 13 91 C; and titanium changes from c.p.h. to b.c.c. at 8 82 C. ... 29 6–3 01 Higher temperatures. (free flowing) + 93.5 Pb Silver; eutectic 42 Ag + 19 Cu 610 – 620 High-strength; high-temperature. (free-flowing) + 16 Zn + 25 Cd Silver; gene...

Ngày tải lên: 11/08/2014, 02:22

... otherwise. Product Number: 1 023 1A Part Number: 06 623 Released: 09 /20 10 i. “Student Content” means the learning materials accompanying these license terms that are ... PROHIBITED Designing a Microsoft® SharePoint® 20 10 Infrastructure iii Microsoft Confidential. © 20 10 Microsoft Corporation. All rights reserved. These materials are confiden...

Ngày tải lên: 04/07/2014, 13:20

... improvement in load capacity. 10 5 0 .1 Sliding speed [m/s] Load (length × viscosity × speed) 0 .2 0.5 1 2 5 10 20 50 10 0 2 × 10 5 3 × 10 5 4 × 10 5 5 × 10 5 6 × 10 5 7 × 10 5 [dimensionless] Adiabatic Isothermal 0.05 ... 36°. 1 1. 3 0 0 .1 Groove len g th/axial bearin g len g th Dimensionless load 1 .2 0 .2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0 4...

Ngày tải lên: 05/08/2014, 09:19

... scientist, Fred Griffith (18 81 19 41) in the late 1 920 s working with Streptococcus pneumoniae and the process was later explained in the 19 30s by Oswald Avery (18 77 19 55) and his associates at ... number of individual organisms in the particular cat- egory): algae ( 12 0 ), bacteria (14 400), fungi (20 200), yeast (4300), protozoa (10 90), animal viruses (13 50), plant viruses (59...

Ngày tải lên: 10/08/2014, 03:20

... 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 0 to 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 into ice. The reverse is tr...

Ngày tải lên: 11/08/2014, 02:22

... alloys 2. 7 71 25 –600 26 3 .1 1.5 9 22 0 15 0 25 0 Mg alloys 1. 7 45 70 27 0 25 4.0 2. 1 41 16 0 15 0 25 0 Ti alloys 4.5 12 0 17 0– 12 8 0 27 2. 4 1. 1 38 28 0 400–600 (Steels) (7.9) ( 21 0) (22 0 16 00) 27 1. 8 0.75 28 20 0 ... following data: Ageing time (h) 0 10 10 0 20 0 10 00 Hardness (MPa) 650 950 12 0 0 11 50 10 00 Account briefly for this behaviour. Peak hardn...

Ngày tải lên: 11/08/2014, 02:22