Problem 1:

Describe the main disadvantages of using the fatigue life (S/N) approach.  For what types of applications would this not be an appropriate method?

Solution:

The main disadvantages are:

•    the large amount of uncertainty due to scatter in the data. At a given stress level, lifetimes can differ by orders of magnitude due to the highly unpredictable nature of crack                 initiation.

•    the empirical nature of the relations used (such as Goodman, Gerber, etc.) to account for different loadings.

Accordingly, there is always a possibility of failure and the S/N approach should not be used in highly safety critical applications.

Problem 2:

Some fatigue data for a steel alloy are given below.

a)    Make an S-N plot using the above data and comment on the amount of scatter present.

b)    What is the fatigue limit for this alloy?

c)    From the plot, estimate the fatigue lifetimes for stress amplitudes of 415 MPa and 275 MPa.

d)    From the plot, estimate the fatigue strengths at 2 × 104  cycles and 6 × 105  cycles.

Part a:

The fatigue data for this alloy are plotted below.

These are clearly idealized data. In a real stress-life fatigue data set, we should expect much more scatter in cycles to failure.

Part b:

The fatigue limit is the stress level at which the curve becomes horizontal, which is 290 MPa.

Part c:

From the plot, the fatigue lifetimes at a stress amplitude of 415 MPa is ~50,000 cycles (log N = 4.7). At 275 MPa the fatigue lifetime is essentially an infinite number of cycles since this stress amplitude is below the fatigue limit.

Part d:

Also from the plot, the fatigue strengths at 2 x 104  cycles (log N = 4.30) and 6 x 105  cycles (log N =5.78) are ~440 MPa and ~325 MPa , respectively.

Problem 3:

One of the barriers to wider adoption of additive manufacturing (3D printing) is that additive     processes tend to introduce defects into the material that reduce the fatigue life. These defects can take the form of additional porosity, residual stresses, excessive surface roughness, lack of

adhesion between layers, and/or metallurgical defects. Depending on the chosen process and material, the resulting parts often need to be post processed in order to achieve reasonable   tensile and fatigue properties.

Your task is to evaluate shot peening as a post-processing method to improve the fatigue life of components made with a relatively new method for 3D printing aluminium. Three SN data sets are posted on Moodle: 3D printed material (heat treated and machined), 3D printed material    (heat treated, machined, and shot peened) as well as data for A356 cast aluminium, which has  similar tensile properties to the 3D printed material.

a)   Generate an SN plot of these three data sets including Basquin fit curves.

b)   What is the allowable stress amplitude for a fatigue lifetime of 105  cycles?

c)    Comment on the amount of scatter present in each data set. Which data set is likely to have the highest and lowest Weibull modulus?

d)   Is either condition of the 3D printed material suitable as a direct replacement for an A356 casting? Why or why not? What design changes should be considered when moving from casting to 3D printing?

e)   Shot peening is primarily a surface treatment. In this case, it only affected the outer ~0.5 mm of a 6.35 mm sample. These data are from fully reversed rotating beam (bending)     fatigue tests, where only the surface of the sample experiences the full stress amplitude. If the loading were uniaxial instead of bending, do you expect the magnitude of the shot peening effect to be increased or decreased? Why?

Part a:

A simple plot with all three data sets and Basquin fit parameters:

To extract Basquin parameters from our fitting, we need to select a fit equation matching the        form of the Basquin law. In excel, this is the ‘power’ option. Be careful with definitions: X = 2Nf     and y = Ga  =  . The Basquin parameters are just the fit parameters that Excel gives you. For the cast material:

 = Ga  = Gf′  (2Nf )b  = 1758 (2Nf ) −0.208

Part b:

This is just evaluating the fit with 2Nf  = 105:

Ga,Nf=105,3D  = 993(105) −0.198  = 101.6 MPa

Ga,Nf=105,3D peened  = 1219(105) −0.188  = 140 MPa

Ga,Nf=105,cast  = 1758(105) −0.208  = 160 MPa

Part c:

The data set with the most scatter is the shot peened 3D printed material. At the 117 MPa stress level, the fatigue life varied by more than an order of magnitude. In comparison, the 3D printed  material that was not shot peened varied by at most a factor of 2.2 at a particular stress level.

It is difficult to compare these directly to the cast material curve, since it was generated with a   single sample at each stress level. However, the cast material appears to cluster tightly to the fit line, suggesting that it has the least scatter.

Weibull modulus is a measure of how tightly clustered a data set is. There’s not enough data here to calculate a Weibull modulus, but we should expect that the shot peened material has the          lowest Weibull modulus due to its high scatter, and that the cast material has the highest Weibull modulus (tightest distribution).

Part d:

In general, no. The fatigue life of the 3D printed material without shot peening is ~10x shorter at a given stress level. If we were to substitute it without any other changes, we should expect the        component to fail in fatigue much faster than a cast part.

Shot peening does appear to have a beneficial effect. The best fit Basquin curve for the shot          peened material is shifted to the right compared to the non-peened material. In fact, the fit curve appears to be quite close to the cast material curve.

While there is an apparent shot peening effect, we should be cautious about relying on the         improvement because of the increased scatter in the shot peening data. If we wanted to apply   Weibull statistics to calculate allowable stress amplitude for a high level of reliability (e.g. stress for 99% survival at a certain lifetime) we would find that the higher Weibull modulus of the shot peened material would push the allowable stress amplitude down much more than the other     conditions, reducing the comparative advantage of shot peening.

Of course, if the part is not highly stressed, or not subject to a large number of cycles, it may be  possible to directly substitute the 3D printed material. Alternatively, the part can be redesigned  with more material in high stress areas to achieve the same overall lifetime. One of the                 advantages of additive manufacturing is additional geometric freedom. Perhaps we can come up with a clever design with lower stresses that wouldn’t be possible with a cast part?

Part e:

In bending, the maximum stress amplitude is only achieved at the surface of the sample, the          region most affected by the shot peening process. In uniaxial tension-compression, the full stress  amplitude is applied through the entire sample thickness. Since the full stress amplitude is applied to unmodified material in addition to shot peening affected material, we should expect that the    effect of shot peening would be much smaller in uniaxial tension.

(We may even find that shot peening reduces the lifetime in uniaxial tension. The surface               compressive residual stresses must be balanced by tensile residual stress in the interior of the      part. This would add a positive mean stress to the full stress amplitude in uniaxial tension,             possibly reducing the fatigue lifetime, depending on the type and spatial distribution of defects in the sample.)

Problem 4:

The fatigue limit for 1045 carbon steel is 300 MPa at zero mean stress. The ultimate tensile         strength is 750 MPa.  Assuming a Goodman relation, what is the fatigue limit at a mean stress of

σm  = 250 MPa?

Solution:

Ga = Ga,Gm=0 ( 1 − ) = 300 MPa ( 1 − ) = 200 MPa

Note that this is the fatigue limit where we predict infinite life. The mean stress relations all         modify the allowable stress amplitude in order to give you the same lifetime you started with. In order to calculate anything about a different lifetime, you need the Basquin relation.

Problem 5:

A steel alloy was found to have an ultimate tensile strength of 1200 MPa and an endurance limit   of 550 MPa for fully reversed fatigue cycling (R = -1).  You have an application that will require the material to withstand a static load of 600 MPa, while dynamic loading will cause the load to           oscillate a total range of 700 MPa around that static value.  Would you expect this steel alloy to    fail by fatigue in this application?  Plot a diagram to aid in presenting your answer.

Solution:

Notice that the problem states ‘total range’ . This means the stress amplitude is 350 MPa and not

700 MPa.

Gendurance limit,  Gm=0 = 550 MPa

GUTS = 1200 MPa

Gm = 600 MPa

ΔG = 700 MPa    →     Ga = 350 MPa

600

500

400

300

200

100

0

0          200        400        600        800       1000      1200

σm [MPa]

Using the Goodman relation, the endurance limit should be:

Ga = Gendurance limit,am=0 ( 1 − ) = 550 MPa ( 1 − ) = 275 MPa

Because 275 MPa < 350 MPa, fatigue failure is a possibility.

Using the Gerber relation?

Ga = Gendurance limit,am=0 ( 1 − (GUTS) ) = 550 MPa ( 1 − ( 1200 MPa) ) = 412.5 MPa

Since 412.5 MPa > 350 MPa, the Gerber relation says failure won’t occur.

We can’t apply Solderman relation without knowing σy .

Based on the Goodman relation we should assume that fatigue failure is a possibility.  However, the Gerber relation doesn’t give the same prediction; such uncertainties are one of the many       problems with the S/N approach to fatigue.

Problem 6:

A multipurpose bridge has been in service for three years and each day has carried a large number of cars, trucks, and trains.  Each time one vehicle passes over the bridge, it experiences one stress cycle. For simplicity assume only one vehicle passes over at a time.  The stress cycle for cars is the  smallest, with higher stresses for the trucks and trains.  Based on the original stress analysis that    was conducted when the bridge was built, you know the lifetime for only car traffic would be 108     cycles, for only truck traffic should be 2*106  cycles, and for only train traffic should be 105  cycles.

Over that time 5000 cars, 100 trucks and 30 trains have passed each day. Based on this information and using Miner’s rule, answer the following:

a)   What fraction of the useful life of the bridge has been used so far in the first three years?

b)   Since trains stress the bridge the most and use the life quickest, it is decided that trains will no longer be able to use the bridge.  How many years of useful life will remain if only cars and       trucks use the bridge from now on?

Part a:

Use Miner’s rule; failure occurs when

3

x  = 1

i=1

After 3 years, 1095 days have passed so Miner’s rule tells us:

5000  × 1095 days       100  × 1095 days       30  × 1095 days

+                                                            +

So 43.8% of the life has been used up after 3 years!  Those trains are a big problem.

Part b:

56.2% of the life remains.  If only cars and trucks use the bridge, they will use this remaining life up more slowly. Applying Miner’s rule again:

5000  × “ days       100  × “ days

1 =                                                      +                                                   + already used life

→         “ = 5620 days = 15.4 years

Problem 7:

The fatigue life for a certain materials at stress amplitudes of σ1, σ2, and σ3, are 10,000, 50,000, and 500,000 cycles, respectively.  If a component made of this materials is subjected to 2,500    cycles at a stress amplitude of σ1  and 10,000 cycles at a stress amplitude of σ2, estimate how     many cycles the component can withstand at a stress amplitude of σ3 .

Solution:

Again, this is a simple Miner’s rule problem.