You have just been told that some copper water-cooling plates have begun failing

Question

You have just been told that some copper water-cooling plates have begun failing in a furnace. The suspected cause is low-cycle fatigue, caused by thermal expansion movements between the cooling plates and the furnace shell. To make a preliminary assessment, you urgently need the strain-life plot for annealed deoxidized copper-the material of the plates. But where do you start? Remember that Figure 17.4 shows the typical form of the strain-life plot. Looking at the left side of the plot, you can see that it should at least be possible to find standard tensile testing data for copper, which would give you values for the true fracture stress and strain. There are plenty of handbooks and/or websites that can give you tensile test data, but of course this is always listed as nominal stress and strain, so needs to be converted into true stress and strain. This is OK, because you already know about the equations that do this conversion. You also need a value for Young's modulus, but again that is easily available (and will be the same whatever the grade of copper). You easily locate the necessary data for annealed, copper: sigma_TS = 216 MN m^-2, epsilon_f = 48% rightarrow 0.48, E = 130 GN m^-2. Then: epsilon'_f = ln(1 + epsilon_f) = ln(1 + 0.48) = ln(1.48) = 0.39 sigma'_f = sigma_TS (1 + epsilon_f) = 216 (1 + 0.48) = 320 MN m^-2, sigma'_f/E = 320 MN m^-2/130 GN m^-2 = 0.0025 So you already have two critical points to put on your plot-we have marked them on the diagram. We could now do with something at the right side of the plot, to define the elas­tic fatigue line. There are lots of data available for HCF-in fact, much more than there are for LCF. But almost always, handbooks and/or websites list just one value-the stress amplitude for failure after typically 10^8 cycles. In the case of annealed copper, a quick handbook search comes up with a stress of plusminus 76 MN m^-2 after 2 times 10^7 cycles (4 times 10^7) reversals. The strain amplitude corresponding to the stress amplitude of 76 MN m^-2 is: Delta epsilon^el/2 = Delta sigma/2E = 76 MN m^-2/130 GN m^-2 = 0.00059 Marking this point on the diagram fixes the elastic fatigue line. Finally, we need some points in the middle of the plot, to define the plastic line Then we can just add the plastic line and the elastic line to get the overall strain-life plot (but remember that the vertical axis is not a linear scale!). Fortunately, a quick trawl through some standard textbooks comes up with a drawing of the cyclic stress-strain loop for annealed copper under a fixed strain range of 0.0084 (Hertzberg). The stabilized stress range is 252 MN m^-2, and the number of reversals to failure is 8060. The total strain amplitude is then 0.0042. The elastic part of the total strain amplitude is: Delta epsilon^el/2 = Delta sigma/2 E = 252 MN m^-2/2 times 130 GN m^-2 = 0.001 The plastic part of the total strain amplitude is then 0.0042 - 0.001 = 0.0032. The three strain amplitudes are marked on the preceding diagram, and they allow the overall strain-life plot to be drawn. This is sufficient for your prelim­inary assessment. However, predictive design should always rely on actual fatigue test data-and plenty of it, so that the variability of the data (mean and standard deviation) can be established. The copper water-cooling plates in the Worked Example failed at welded joints which connected the plates to the inlet and outlet water pipes. During thermal cycling of the furnace, the pipes were bent back and forth, which resulted in cyclic straining of the welds. The strain amplitude was estimated to have been 0.01. Using the strain-life plot in the diagram, estimate the number of cycles required to cause the welded joints to fail. Your estimate of the strain amplitude could be in error by a factor of 2. What would be the failure life if the strain amplitude were as large as 0.02?

Explanation / Answer

ef = ln(1+0.01)= 0.00995

sigmaf= 216 *(1+0.01) = 218.16

sigmaf/E = 218.16/130 = 1.6782

delta sigma =0.01*E = 1.3

So from table for anealed copper and plot its come out to be 1000 reversals or 500 cycles.

for 0.02 it comes out to be 150 cycles or reversals.

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