Mechanics of Materials 1 Part 3 ppsx

Mechanics of Materials 1 Part 3 ppsx

Mechanics of Materials 1 Part 3 ppsx

... 0 1 2 2 3 4 5 6 0 10 0~ 6= 600 10 0x12 =12 00 12 00 13 20 14 40 16 80 19 56 - 72 69 69 68 66 .3 61. 6 54.5 - 10 0 10 0 12 12 12 12 12 - 3. 2 6.2 51. 3 55.6 59 .1 64 .1 66.0 ... = 10 0 x 37 .5 x 31 . 25 x = 11 0.25 x 10 -6m3 - 6.75 x b‘ = 2Rcosn/6 = 2 x 25 x x J3/2 = 43. 3 x m b = (10 0- 43. 3 )10 -3 = 56....

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Mechanics of Materials 1 Part 7 ppsx

Mechanics of Materials 1 Part 7 ppsx

... criterion, taking account of the safety factor, - OY = 01- 03= 7-(-7) n = 27 = 2 x 21. 8~ x 10 3 225 x lo6 2.25 10 0 x 10 6 T= = 2.3x 10 3~ m 2 x 21. 8 x 10 3 The torque which can ... limit of the steel in tension is 225 MN/m2 and Poisson’s ratio v is 0 .3. 422 Mechanics of Materials 11 32 x lo6 (67.5 x 10 6)2 d4 = 33 .6...

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High Cycle Fatigue: A Mechanics of Materials Perspective part 3 docx

High Cycle Fatigue: A Mechanics of Materials Perspective part 3 docx

... causes of individual HCF failures. The combined findings of these teams centered around many aspects of the design, field usage, and materials and their characterization. In particular, the use of ... history. Introduction 15 the very high engine speeds. In a series of these types of failures, there were no failed parts recovered from which to deduce the failure scenario in te...

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High Cycle Fatigue: A Mechanics of Materials Perspective part 7 ppsx

High Cycle Fatigue: A Mechanics of Materials Perspective part 7 ppsx

... 12 420E +27 −0 33 3 3 01 9E +11 0 1 −5956E +14 05 −4 91 0E +12 08 −7 51 9E +11 0 10 20 30 40 50 010 2 030 4050 59 Hz HCF data Alternating stress (ksi) Mean stress (ksi) R = 1 R = –0 .33 R = 0.8 R ... subset Number of tests k m Walker exponent, w A R ≤−0 1 @ 59 Hz 24 5.83E16 −7 17 0 16 5 C R ≤−0 33 3 @ 59HzR= 0 1 @ 37 0–400 Hz 20 7.63E16 −698 0 3 817 52 Introdu...

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High Cycle Fatigue: A Mechanics of Materials Perspective part 8 ppsx

High Cycle Fatigue: A Mechanics of Materials Perspective part 8 ppsx

... Introduction and Background 0 0.5 1 1.5 2 1 –0.5 0 0.5 1 Haigh equation k 1 = 0.0, k 2 = 1. 0 k 1 = 0.25, k 2 = 0.75 k 1 = 0.5, k 2 = 0.5 k 1 = 0.75, k 2 = 0.25 k 1 = 1. 0, k 2 = 0.0 σ mean /σ ult σ alt ... 10 6 6 × 10 6 8 × 10 6 1 × 10 7 –4 3 –2 1 0 1 ML data ASE data Last block cycles Stress ratio, R Ti-6Al-4V plate Smooth bar step tests 10 7 cycle b...

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High Cycle Fatigue: A Mechanics of Materials Perspective part 15 ppsx

High Cycle Fatigue: A Mechanics of Materials Perspective part 15 ppsx

... steels: 0.86% carbon, 0 .11 % carbon, rolled, Armco iron, Copper Aluminium a b 0.86% carbon Armco iron FREQUENCY-CYCLES PER SEC. 10 ,000 3, 000 1, 000 30 0 14 16 18 20 22 24 26 28 30 32 34 4 5 6 FATIGUE LIMIT-TONS ... SEC. 10 ,000 3, 000 1, 000 30 0 14 16 18 20 22 24 26 28 30 32 34 4 5 6 FATIGUE LIMIT-TONS PER SQ. IN. 0 .11 % carbon, rolled 0 .11 % carbon, normalized F...

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High Cycle Fatigue: A Mechanics of Materials Perspective part 18 ppsx

High Cycle Fatigue: A Mechanics of Materials Perspective part 18 ppsx

... (4 .17 ) K B =  a 0 x m B  x a a c  dx (4 .18 ) m A x a = 2  2a −x  1+ M 1A  1 x a  1/ 2 +M 2A  1 x a  +M 3A  1 x a  3/ 2  (4 .19 ) m B x a = 2  x  1+ M 1B  x a  1/ 2 +M 2B  x a  +M 3B  x a  3/ 2  (4.20) In ... transients # of tests  LCF N LCF % of life  f / e 3 855 50000 25 0.99 5 925 25000 25 0.92 3 925 15 000 15 0.94 1 925...

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High Cycle Fatigue: A Mechanics of Materials Perspective part 21 ppsx

High Cycle Fatigue: A Mechanics of Materials Perspective part 21 ppsx

... total (MPa m) 10 –5 10 –4 10 3 10 –2 10 1 10 0 1 10 10 0 Test 1 Test 2 Linear sum, worst case Linear sum, average da /dBlock (mm/block) Ti- 533 1S 550¡C n = 10 ,000 Figure 4 .38 . Combined LCF–HCF ... point of intersection from Figure 4 .37 . The Ti-6-4 alloy LCF–HCF Interactions 18 9 σ th /σ end 0.2 0.4 0.6 0.8 1 3 0 .1 1 10 R = 3 R = 3, SRA R = 1 R = 0...

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High Cycle Fatigue: A Mechanics of Materials Perspective part 54 ppsx

High Cycle Fatigue: A Mechanics of Materials Perspective part 54 ppsx

... is Table D .1. Non-linear correction constants based on number of specimens Number of specimens AB m 8 1. 30 1. 2 1. 72 10 1. 08 1. 2 1. 10 12 1. 04 1. 2 0.78 15 0.97 1. 2 0.55 20 1. 00 1. 2 0.45 30 1. 00 1. 2 0.22 50 ... to Appendix D 5 21 39 8.5 39 9.0 39 9.5 400.0 400.5 4 01. 0 4 01. 5 0 50 10 0 15 0 200 0 50 10 0 15 0 200 Number of specimens Point estimate...

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Mechanics of Materials 1 Part 1 potx

Mechanics of Materials 1 Part 1 potx

... 17 3 17 6 17 6 17 7 17 9 18 0 18 1 18 2 18 2 18 2 18 2 18 3 18 4 18 4 18 6 18 6 18 7 18 7 18 8 18 9 18 9 19 0 19 5 Simple Stress and Strain 21 Fig. 1. 21. Solution From the proportions of Fig. ... 14 .8 17 .27 19 .14 22.2 24.7 Extension (m x 5.6 11 .9 18 .2 24.5 31 . 5 38 .5 45.5 52.5 59.5 66.5 Load (kN) 27 . 13 29.6 32 .1 33 .3 31 . 2 32...

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