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CHAPTER 7 Fatigue Failure Resulting from Variable Loading

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1 CHAPTER 7 Fatigue Failure Resulting from Variable Loading
6/25/2007 CHAPTER 7 Fatigue Failure Resulting from Variable Loading Dr. A. Aziz Bazoune King Fahd University of Petroleum & Minerals Mechanical Engineering Department Dr. Abdelaziz Bazoune

2 Lecture 23

3 7-10 Stress Concentration Factor and Notch Sensitivity
In fatigue: Stress concentration should always be taken into account.

4 Some materials are not fully sensitive to notches and a reduced value of Kt is used and the maximum stress is calculated as follows: (7-29) Kf is the fatigue stress concentration factor, for simple loading: (Ex 7.7) or

5 Notch sensitivity q index is defined by
(7-30) To find use Fig for steel and Al alloys , for reversed bending or reversed axial load. For reversed torsion use Fig For cast iron use to be conservative. For , then and the material has no sensitivity at notch at all. For , then and the material has full notch sensitivity.

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7 In analysis or design work
Find first from the geometry of the part Specify the material Find Solve for from the following Equation (7-31) Figure 7-20 has its basis the Neuber equation, which is given by (7-32) Where is defined as the Neuber constant and is a material constant.

8 Equating Eqs. (7-31) and (7-32) gives the notch sensitivity equation
(7-33) For steel, with in kpsi, the Neuber equation can be approximated by a third polynomial fit of data as (7-34) Where is defined as the Neuber constant and is a material constant.

9 Table 7-8 Heywood’s Parameter for steels
A distinction in the configuration of the notch is accounted for in the modified Neuber equation (after Heywood), where the fatigue stress-concentration factor is given as (7-35) Table 7-8 Heywood’s Parameter for steels where Table 7-8 gives values of for steels for transverse holes, shoulders and grooves. Feature (a)1/2 (in)1/2 Sut in kpsi (a)1/2 (mm)1/2 Sut in MPa Transverse hole 5/Sut 174/Sut Shoulder 4/Sut 139/Sut Groove 3/Sut 104/Sut

10 Example 7-3 (Textbook) A steel has a minimum strength of 520 MPa and a machined Surface. Estimate ka . SOLUTION From Table 7-4, a=4.51 and b=-0.265 From Eq.(7-18) therefore

11 Remark From this chart , it is seen that the surface factor Ka for machined or cold drawn is always less than 0.8.

12 Example 7-4 (Textbook) SOLUTION From Eq. (7-19) From Table (7-5)

13 Example 7-5 (Textbook) SOLUTION From Eq. (7-26)
Interpolating from Table (7-6) gives From Eq. (7-8)

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15 Example 7-6 (Textbook) SOLUTION See next page Figure 7-20
Equations (7-32) and (7-34) Equations (7-35) SOLUTION From Figure A-15-9 From Figure (7-20), for Sut = 690 MPa and r =3 mm From Eq. (7-31) See next page

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18 From Eq. (7-34) with Sut = 690 Mpa=100kpsi, and Substitute into Eq. (7-32) with r =3 mm From Table 7-8 From Eq. (7-35) With 2.5% lower than a) and b)


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