Mechanics of Materials 1 Part 15 pps

Mechanics of Materials 1 Part 15 pps

Mechanics of Materials 1 Part 15 pps

... del 2 (11 t dsl + (12 t dS2 ~ ) = pdsl .dS2 2 2 Fatigue, Creep and Fracture 50 1 a- da = i 1 (1 12~&iii)~.* 0.005 = 14 .8( -11 .88 + 11 78) = 11 56 cycles. Example 11 .13 In ... + C) 10 23 (1, 3000 + C) = 10 73 (1, 500 + C) 10 23(8.0064 + C) = 10 73(6. 214 6 + C) 819 0.5 + 10 23C = 6668.27 + 10 73C 15 22.23 = 50C C = 30.44. 518 Mech...

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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

... behavior of materials. Dealing ∗ See Chapter 2, Section 2, for a discussion of gigacycle fatigue and testing at 20 kHz. 12 8 Introduction and Background 5 5.5 6 6.5 7 7.5 8 . 01 .1 1 5 10 20 30 50 70 80 90 95 99 99.9 99.99 Log ... the purpose of translating strain rates into equivalent frequencies, consider that nominal maximum strains in the HCF regime are of the order of les...

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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 ... centre of the web. 1. 3 x 503 25 x 1. 33 IN.A. = [ 12 + 2( 12 + 25 x 1. 3 x 252) = [1. 354+ 2(0.00046+2.03)+2(0. 011 +0.52) ]10 -' = 6.48 x 10 -...

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Mechanics of Materials 1 Part 5 pps

Mechanics of Materials 1 Part 5 pps

... (Fig. 11 .7) ( :) ;; 25’ lo6 volume of bar = in x ~ x 2.6 = 12 .76 x O’ x 12 .76 x Then 10 x 10 1. 30 O2 30+- = 10 9 3i3x 10 12 1. 30 30 x 313 x 10 ” f- x 313 x 10 ” ... aM My, = 0 .1 W-Hs, and ~ aH 0.25 .I 0 0.25 1 E1 = - 1 (-0 .1 Ws, + Hs:)ds, 0 1 0. 015 6258 +3 (- 3 .12 5 W+ 5.2088) 1 - EI x 10 3 278 M...

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Mechanics of Materials 1 Part 6 pps

Mechanics of Materials 1 Part 6 pps

... 0,)' + 4~2 ~1 336 Mechanics of Materials $13 .7 re I up =ux uq =uy =o r,, :O 1 1 1 I 1 I I O0 45" 90" 13 5' 18 0" 225' 270" 315 ' ? ... circle (Fig. 14 .13 c), (E1 + 62) OA' x strain scale = ___ 2 But and 1 E 1 E1 + E2 = - [(a, + a,) - v(o1+ a,)] = E (a1 +a,) (1 - v)...

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Mechanics of Materials 1 Part 7 ppsx

Mechanics of Materials 1 Part 7 ppsx

... of Fracture Mechanics (Butterworths, London), 19 73. 416 Mechanics of Materials 415 .13 lead to the adoption of higher values. Since this could well result in uneconomic utilisation of ... the points of intersection of the loci and the load line with a slope of 10 0/ - 10 0 = - 1. 400k2 Distortion energy Max principal stress Fig. 15 .18 . Thus fro...

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Mechanics of Materials 1 Part 8 pps

Mechanics of Materials 1 Part 8 pps

... Introduction 11 .1. 1 The SIN curve 1 1. 1.2 PISIN curves 1 1 .1. 3 Effect of mean stress 1 1 .1. 4 Effect of stress concentration 11 .1. 5 Cumulative damage 1 1. 1.6 Cyclic stress-strain 1 1 .1. 7 ... eqns. ( 1 .12 ) and (1 .13 ) gives (1. 14) and combining eqns. (1 .lo) and (1. 1 1) gives I, + I,, = I, + I, (1. 15) Substitution into eqns. (1. 1...

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High Cycle Fatigue: A Mechanics of Materials Perspective part 2 pps

High Cycle Fatigue: A Mechanics of Materials Perspective part 2 pps

... University Press, New York, 19 98. Hertzberg, R.W., Deformation and Fracture Mechanics of Engineering Materials, 3rd ed., Wiley, New York, 19 89. x Chapter 1 Introduction 1. 1. HISTORICAL BACKGROUND Fatigue ... consideration in the disciplines of solid mechanics or strength of materials. In fact, one of the earliest researchers to address the topic was Wöhler [1] , who, in...

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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

... −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 20304050 59 Hz HCF data Alternating stress (ksi) Mean stress (ksi) R = 1 R = –0.33 R = 0.8 R = 0.5 R = 0 .1 46 ... <0 01& gt; oriented specimens with the exception of eight spec- imens oriented at < 0 01+ 15& gt; that were tested at R = 08. The <0 01 +15 > oriented data fa...

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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 ... aid of Figure 2.40, where the conditions for initiation compared to propagation are shown schematically. The initiation of a crack, if associated with the 0 ×...

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