Non rotating shaft. Problem: static failure, a very large deformation will occur on the structure or machine members..

Transcription

1 Chapter 8 Fatigue

2 Introduction

3 F Non rotating shaft Under static failure, the stress on the member is constant Problem: static failure, a very large deformation will occur on the structure or machine members..

4 F F Rotating shaft - F to F F -F Non rotating shaft Alternating or fluctuating stresses on member will cause the member subjected to fatigue failure.

5 R. R Moore Rotating Beam: Figure 8-3 Testing specimen Subject to pure bending (no traverse shear) Very well machined and polished No circumferential scratches

6 Fatigue Strength and Cycle Graph (S-N Graph): Figure 8-5

7 Terminology N: Cycle : one rotation of the specimen = 1 cycle Low Cycle 10 3 High Cycle Cycle of alternating stress Finite Life S e Infinite Life Life Sf: fatigue strength is the limit of strength where failure occurs when the alternating stress above the fatigue strength. However, when cycle is larger than 10 6, the fatigue strength is constant to a value Se. This value is called Endurance Limit Fatigue Strength Sf A Cycle, N NA Endurance limit S e Cycle: Cycle is related to the cycle of the specimen: boundary is 10 3 Life: The life is related to the stress of the specimen: boundary is Se. Significance to design Designers have a choice to set the design life whether finite life or infinite life. Designers can predict the failure to occur (finite life): i.e. when the stress of the member is A the predicted life will be NA cycles.

8 Question: Do all the material will have the same behavior? NO (Refer to Figure 8.8: Endurance limit of various aluminum 10 7 Refer to Figure 8.10 for Magnesium Alloy) You must refer to specific handbook if different material. However, for this class, it focuses on Carbon Steel material (Figure 8.5)

9 Endurance limit (Carbon Steel Material) Sn : actual endurance limit Sn : experimental endurance limit (inside controlled environment) CL : load factor CG : size factor CS : surface factor CT : strength of material decreases when temperature increases CR : the reliability factor (50% : CR =1)

10 Experimental endurance limit (Sn) Since S 0. 5 ' n S ut S ut 3. 45Bhn for MPa. ' Therefore S n Bhn 0. for ksi. S ut 5Bhn ' Therefore S n Bhn Note: Bhn: Brinell Hardness Number These formulas only valid when Bhn <= 400 Bhn

11 Load Factor (C L ) Note: ignoring possible yielding means fatigue failure occurs below the yield strength

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13 Example Calculate the endurance limit Se, if Material AISI 1020 HR, surface is machined until D = 30 mm, d = 20mm and r (fillet radius) = 2 mm is achieved. d D r The part is subject to M = 20 Nm. Determine whether the part is failed subjected to fatigue load

14 Endurance limit is the fatique properties of the material. This properties must withstand the fatique stresses induced by the internal bending.

15 Kf: fatique stress concentration factor Kt: static stress concentration factor q: notch sensitivity Internal stress raiser may also happen with the presence of other material inside the material such as graphite in the gray cast iron insensitive to the notches Kf = 1 While uniform material will be sensitive to the presence of notches Kt = Kf In equation 8.2, the value of q is from 0 to 1, when q = 0, Kf = 1 (insensitive to the notches) q = 1, Kf = Kt (uniform material structure) Note: Stress concentration factor will result in the stress increases.

16 STRESS CONCENTRATION FACTOR (Kt : Section 4.13) Refer to Figure 4.35 shaft with fillet, Figure 4.36 grooved shaft, Figure 4.37 shaft with radius hole, Figure 4.38 Bar with shoulder fillet, Figure 4.39 Notched flat bar

17 Notch Sensitivity

18 Example 2

19 Midrange Stress Alternating Stress a 2 m max min 2 max min

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21 How to determine the mean or alternating stress? Moment on rotating shaft Stress Time M 4 I A 4 4 When the shaft rotates in ccw direction, element A will rotate from position 1, 2, 3 and 4 for one full cycle. At the same time, the stresses of A will fluctuate from 0, max, 0 and min. Therefore, moment on rotating shaft will generate a = Mc/I and m = 0.

22 Torsion on rotating shaft 3 2 Stress Time A 4 I A a = 0 and m = Tr/J.

23 Determine the alternating and midrange stresses for the following cases? Rotating shaft i. Axial load (P) ii. Combination of T and M Non rotating shaft Moment (M) between -10Nm and 10Nm 20 Nm and 50 Nm -30 Nm and 10Nm Torsion (T) between 10Nm and 10Nm 20 Nm and 50 Nm -30 Nm and 10Nm

24 Mean stress vs Alternating stress graph

25 Notes: Refer to Table 8-2

26 FATIGUE SAFETY FACTOR: GRAPHICAL METHOD Safety factyor against infinite life S a Sm a m

27 Discuss the situation of the parts if the load point is either at A, B, E.

28 FATIGUE SAFETY FACTOR: CALCULATION METHOD a m Basic Equation 1 S S n S S ut Since S a S m a Sa a m & m m S a m Therefore 1 S n S ut a m 1 S S n ut

29 Determine whether the component is yielding first before fatigue. Point A will yield first before failure against fatigue happens. Therefore, it will face static failure first. To check which the part is yielding first a m 1 S y L n L : SF against yielding a m 1 n : SF against fatigue S S n ut Since it yields first n L <= 1 and n > 1 Or n L (yield) < n (fatigue) it will yield first.

30 When the stress is above the yield strength, the residual stress will result in increase of the yield stress due to strain hardening of the plastic zone. However, the increment of yield stress is limited when the part faces the fatigue failure. Therefore, point B can be considered as the maximum possible alternating stress can be achieved.