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School of Engineering – EN334 M. Daniel K. Wood DESIGN FOR FATIGUE STRENGTH In real life situation, mechanical elements are not only loaded statically,

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Presentation on theme: "School of Engineering – EN334 M. Daniel K. Wood DESIGN FOR FATIGUE STRENGTH In real life situation, mechanical elements are not only loaded statically,"— Presentation transcript:

1 School of Engineering – EN334 M. Daniel K. Wood DESIGN FOR FATIGUE STRENGTH In real life situation, mechanical elements are not only loaded statically, but they are also loaded in such a manner that the stresses in the members vary from a maximum value to a minimum value during infinite number of cycles. A shock absorber in a car is a typical example where the springs are loaded cyclically as the car is driven through a rough road that has several holes. The springs are repeatedly loaded by forces that are at one time at a maximum value and at the other time at a minimum value. The same can be said of a rotating shaft that experiences bending moments. The effect of this is that at one time some fibres on the surface experience compressive stress and at other times they experience tensile stress. This variation between compressive and tensile stresses may be repeated several times within one minute, depending on the speed of rotation.

2 School of Engineering – EN334 M. Daniel K. Wood Stresses of this nature are referred to as fluctuating stresses and they result in mechanical members failing under fatigue failure mode. In fatigue failure, ten million or 10 7 cycles are referred to as an infinite life. What this means is that if a shaft rotates ten million times, then it is assumed that it has attained its design life. Fatigue failure mode is very dangerous to mechanical parts because the stress required to cause it, is normally below the ultimate strength and the yield strength of the material. The designer should be familiar with this kind of failure mode so that care should be taken to design a machine part that is resistant to the failure mode. In fact, stress concentration factor is linked with fatigue failure. A small crack developed on a turbine blade is dangerous and it can indeed cause a major failure. This is because a small crack can propagate very easily under fluctuating stresses and can very easily lead to a catastrophic failure of the engine.

3 School of Engineering – EN334 M. Daniel K. Wood The concept of fluctuating stresses is shown in Figure 4.1. For fluctuating stresses the stress at a point, is given by a range stress,  r, and a mean stress,  m. These stresses, are functions of the maximum stress and minimum stress respectively. Figure 4.1 Range and mean stresses

4 School of Engineering – EN334 M. Daniel K. Wood The maximum stress is given as: (4.1) The stress range is given as (4.2)

5 School of Engineering – EN334 M. Daniel K. Wood Endurance Limit Engineering parts, which fail under fatigue loading, experience extreme stresses, σ max and σ min. Such parts are more likely to fail than those parts that experience only maximum stress, σ max. The stress amplitude, at which the machine member will fail after a given number of stress-cycle is known as the fatigue strength, designated as S n. As n approaches ∞ the fatigue strength approaches the endurance limit, designated as S’ n, as shown in Figure 4.2. Figure 4.2

6 School of Engineering – EN334 M. Daniel K. Wood However, surface imperfections due to manufacturing processes and environmental affects negatively impacts the endurance limit, and results in lower values compared to those published. Therefore a modified endurance limit is given as –where k sf = surface finish factor k r = reliability factor k s = size factor k t = temperature factor k m = stress concentration modifying factor = inverse of fatigue stress concentration factor (k m = 1/k f ) k sf = surface finish factor = half of the ultimate strength of the material (S u /2)

7 School of Engineering – EN334 M. Daniel K. Wood k sf : The surface finish factor depends on the quality of the finish and the tensile strength of the material. Graphs of these are normally available. k r : The reliability factor can be obtained from the Table 4.1. k s : The size factor is obtained from the following relations: Table 4.1 Reliability factors ReliabilityReliability factor, k r 50%1 90%0.897 95%0.868 99%0.814

8 School of Engineering – EN334 M. Daniel K. Wood kt : The temperature factor is obtained from the relation km : The stress concentration modifying factor is given as where the fatigue stress concentration factor kf, is given as

9 School of Engineering – EN334 M. Daniel K. Wood and q is the notch sensitivity usually given graphically and is a function of the material’s ultimate strength and notch radius of the part. The stress concentration factor, k st, is also a function of the part’s geometry. When q and k st have been established, k f can then be obtained. For no notch round shaft, k t = 1, which implies k f = 1, and hence k m = 1. By using these modified factors in equation (4.3), the modified endurance limit can then be obtained.

10 School of Engineering – EN334 M. Daniel K. Wood Fatigue Strength When the mean stress and the stress range are varied, the fatigue resistance of parts subjected to these fluctuating stresses can be studied using Goodman diagram shown in Figure 4.3. Figure 4.3. Goodman diagram Alternating Stress Component σ a Midrange stress σ m

11 School of Engineering – EN334 M. Daniel K. Wood The diagram is drawn by marking out the yield strength of a material in both the x- and y-axes, S y. The ultimate strength of the material is marked out in the x-axis, S ult, and this is usually greater than the yield strength. The endurance limit is marked out in the y-axis, S e. The points corresponding to the yield points on the x-axis and y-axis are connected by a straight line, so are the points corresponding to the ultimate strength along the x-axis and the endurance limit along the y-axis. A region is defined by the intersection of the endurance limit-ultimate strength line and the yield strength-yield strength lines, within which design is feasible (shaded area). The stress range is associated with the y-axis and the mean stress is associated with the x-axis. By plotting a line defined by the gradient obtained from the loads related to the stress range and mean stress respectively, it is possible to locate the stress range and mean stress values.

12 School of Engineering – EN334 M. Daniel K. Wood As the mean stress increases, the semi-range reduces. Within the region bounded by the yield strength-yield strength line and ultimate strength-endurance limit line, the sum of the mean stress and the stress range is equal to the yield strength. The stress range and the mean stress in a machine member can be estimated if we superimpose a line whose gradient is given by the ratio of the stress range to mean load as shown in Figure 4.4. Alternating Stress Component σ a Midrange stress σ m Figure 4.4 Deriving mean stress and stress range from Goodman diagram

13 School of Engineering – EN334 M. Daniel K. Wood The maximum stress is given as (4.8) The mean stress is given as (4.9) The stress range is given as (4.10)

14 School of Engineering – EN334 M. Daniel K. Wood The ratio of the stress range to the mean stress is given as follows: (4.11) The ratio of the stress range to the mean stress given in equation (4.11) is a much quicker way of obtaining the stress range and mean stress from the Goodman diagram if the associated loads are known.

15 School of Engineering – EN334 M. Daniel K. Wood Example 4.1 Problem: An AISI 1020 CR steel bar is to withstand a tensile pre-load of 356kN and a fluctuating tensile load varying from 0 to 71.2kN. The steel bar has a geometric stress-concentration factor of 1.995 corresponding to a fillet radius of 5mm. Assume the notch sensitivity is 0.70, and the surface factor k sf = 0.78. Determine the bar diameter for a factor of safety of 1.5 and an infinite life. Solution: From the appendix (material properties), the yield and ultimate stresses for AISI 1020 CR steel are S y = 414 x10 6 N/m 2 and S u = 552 x10 6 N/m 2 respectively.

16 School of Engineering – EN334 M. Daniel K. Wood S’ n = half of the ultimate strength of the material (S u /2) = 0.5×552 ×10 6 N/m 2 = 276 ×10 6 N/m 2. k sf = surface finish factor = 0.78 (given) k r = reliability factor = 1 (assume since no information is given) k s = size factor = 0.85 (mid-value assumed since initial diameter is not known) k t = temperature factor = 1 (assume since no information is given) = fatigue stress concentration factor = 1 + 0.7×(1.995 – 1) = 1.6965 k m = 1 / k f = the stress concentration modifying factor = 1/1.6956 = 0.5894 Therefore, = 0.78×1×0.85×1×0.5894×276 ×10 6 = 107.85 ×10 6 N/m 2 (Pa).

17 School of Engineering – EN334 M. Daniel K. Wood The stress induced in the steel bar is: The stress range in the steel bar is:

18 School of Engineering – EN334 M. Daniel K. Wood The stress amplitude in the steel bar is: The mean stress in the steel bar is: We now use the Goodman diagram to find the stress range as follows:

19 School of Engineering – EN334 M. Daniel K. Wood The load line can now be drawn with a gradient of. Using basic knowledge of intersection of straight line, the points of intersection can be found from the ultimate strength-endurance limit line, and the superimposed design line-equations as follows: Ultimate strength-endurance limit line: Design line equation: Where Y is the alternating stress and X is the midrange stress

20 School of Engineering – EN334 M. Daniel K. Wood At the point of intersection, Y 1 = Y 2. Solving the simultaneous equation leads to: We find the value of X = 155.179Pa. The value of Y 1 is found as follows: Using the relation

21 School of Engineering – EN334 M. Daniel K. Wood Solving for the diameter, Voila!


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