Mechanics of Materials 2010 Part 5 pptx

Mechanics of Materials 2010 Part 5 pptx

Mechanics of Materials 2010 Part 5 pptx

... Saouma Mechanics of Materials II Draft 6 FUNDAMENTAL LAWS of CONTINUUM MECHANICS 6.3.2 †Moment of Momentum Principle 25 The moment of momentum principle states that the time rate of change of the ... P 2 (5. 5-a)  C A·dr = 0 along a closed contour line (5. 5-b) 5. 3 Integration by Parts 5 The integration by part formula is  b a u(x)v  (x)dx = u(x)v(x)| b a −  b a...

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High Cycle Fatigue: A Mechanics of Materials Perspective part 5 doc

High Cycle Fatigue: A Mechanics of Materials Perspective part 5 doc

... field of “gigacycle fatigue,” indicating lives of the order of 10 9 cycles or higher, data have been generated indicating that some materials do not have a fatigue limit within the range of cycles ... to understand the very long-life of materials and the apparent lack of a fatigue limit using ultrasonic test machines [5] . They show that in higher strength materials such as...

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High Cycle Fatigue: A Mechanics of Materials Perspective part 17 pptx

High Cycle Fatigue: A Mechanics of Materials Perspective part 17 pptx

... = 0 .5 R = 0.8 HCF LCF R = 0.1 R = 0 .5 R = 0 .5 700 790 900 53 4 639 910 (A) (B) (C) Figure 4.9. Schematic showing amplitudes of LCF and HCF fatigue limit stresses. LCF–HCF Interactions 151 10 100 1000 1 ... 151 10 100 1000 1 10 100 1000 10,000 c = 5 μm, short crack c = 5 μm, long crack c = 50 0 μm, short crack c = 50 0 μm, long crack Stress (MPa) Crack length, a (μm) K th...

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High Cycle Fatigue: A Mechanics of Materials Perspective part 24 pptx

High Cycle Fatigue: A Mechanics of Materials Perspective part 24 pptx

... +a 0  (5. 10) Notch Fatigue 2 25 Evaluation of Figures 5. 6 and 5. 7 points out some of the difficulties in making generaliza- tions about the transition from crack initiation to crack propagation, particularly ... form k f =1+ k t −1 1+  a m r (5. 3) The equation of Peterson [4] is k f =1+ k t −1 1+ a m r (5. 4) The equation of Heywood [5] is k f = k t 1+2  a m r (5. 5) whe...

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High Cycle Fatigue: A Mechanics of Materials Perspective part 26 pptx

High Cycle Fatigue: A Mechanics of Materials Perspective part 26 pptx

... Equation (5. 11). Numerical results showing the stress distributions ahead of a circular notch of radius r, and of a crack of length a, are shown in Figure 5. 23, for three different values of a/a 0 ; ... parameter has the form for an S–N type curve, FIN = 65 739N −06779 +4335N −004 15 (5. 25) To apply the parameter to notch data, the average value of the parameter over a di...

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High Cycle Fatigue: A Mechanics of Materials Perspective part 27 pptx

High Cycle Fatigue: A Mechanics of Materials Perspective part 27 pptx

... strength of a steel is easily calculated from Equations (5. 48) and (5. 49). The results, in terms of the parameter √ area, are presented in Figure 5. 29 which shows that for all but the softest of steels, ... as a D = a 0  2 a N = a ∗  2 = k 2 t a 0  2 (5. 39) 246 Effects of Damage on HCF Properties 0 100 200 300 400 50 0 10 4 10 5 10 6 10 7 R = –1 R = 0.1 R = 0 .5 Predicted...

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High Cycle Fatigue: A Mechanics of Materials Perspective part 38 pptx

High Cycle Fatigue: A Mechanics of Materials Perspective part 38 pptx

... as-FODed Bending S, with SR Figure 7.34. Prediction of the HCF capability of specimens with FOD using the unnotched stresses. 0 0 .5 1 1 .5 2 2 .5 3 3 .5 4 0 5 10 15 20 25 30 Peak local stress predictions Axial ... there is no model for values of k f . 362 Effects of Damage on HCF Properties 0 0 .5 1 1 .5 2 2 .5 3 3 .5 4 0 5 10 15 20 25 30 Un-notched predictions S un-n...

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Mechanics of Materials 1 Part 5 pps

Mechanics of Materials 1 Part 5 pps

... 9.3 758 ) - (1. 25 W - 6. 258 ) 1 - EI x 103 1 =- (1 .56 W+ 15. 6 258 ) EI x 103 Now the total horizontal deflection of A = 0 -3.1 25 W+ 5. 2088 + 1 .56 W+ 15. 6 258 = 0 - 1 .56 5 ... = - [ + 0.06 258 ~~ El 2 1 = -!- {["':" x 0.02 25+ 0.06 258 x 0. 15 El x 0.01 +0.06 258 (-0.1) 1 =- { (1. 25 x 2. 25 W+6. 25 x...

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Mechanics of Materials 1 Part 9 pptx

Mechanics of Materials 1 Part 9 pptx

... integral part of this expression is the first moment of area of the shaded parts of Fig. 3.23(a) about the vertical axis and evaluation of this integral allows the determination of M for ... distribution being given by eqns. (10 .5) and (10.6)t as: (3.42) (3.43) E.J. Hem, Mechanics of Materials I, Butterworth-Heinernann, 1997. 66 Mechanics of Materials...

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