Learning Forum

[momidor]

 

Good afternoon, 

I've got simple issue with crack. The load is pure tensile ~ 70% plastic limits. Material is low carbon steel. 

The stress life ( SL ) with scale factor gave me results as hereunder, i.e. isolines of SL. 

How to interpret it ?

Having specific combimation of crack geometry & mat ref & loads, I would like to obtain the maximum numers of cycles to failure.

How to do this in Ansys ?

Greetings  

 

 

[peteroznewman]

Hello,

What do you mean 70% plastic limits?  Do you mean the peak stress in the model is only 70% of the yield strength?

If the stress is below the yield strength, then the Stress-Life fatigue analysis shows that at the tip of the crack, there is a 50% probability of fracture after 3264 cycles. This is the minimum number of cycles to fracture.

Regards,
Peter

[momidor]

Hello Peter 

The only load case is the horizontal tensile 180 MPa which is roughly 70 % of yield of material. 

At the tip of the crack is the pick stress 247 MPa equal to yield which is correspond to the G.R Irwin theory.

At the tip of the carck is the 11 100 cycles to failure. 

Fatigue sensitivity chart is as hereunder.

"there is a 50% probability of fracture after 3264 cycles." - why 50% probability ? 

Regards 

Jurek 

 

 

 

[peteroznewman]

Hello Jurek,

Why 50% Probability

Below is a typical high-cycle stress-life experimental fatigue data. Figure 15 is copied from this reference. The line refers to a failure probability of 50% in experiments at 20 kHz. Look at the spread of the points about the line along the life axis. In some cases, there is a factor of 10 on the cycles to failure when one sample failed vs. another sample tested at the same stress amplitude. That is the nature of fatigue data.

The stress-life curves are almost always published at the best fit where 50% of the samples failed. That is why engineers apply large factors of safety on their designs to stay away from the 50% probability of failure.  Most engineers don't want any failures during the design life of the product. Unfortunately those curves don't often come with the variation of life around the mean-life curve, so how is an engineer supposed to know how far away to stay from the mean-life curve?

Strain-Life vs Stress-Life

The plot above is for stress levels that are below yield. When the stress is above yield, a different curve is fitted to the data and that is the Strain-Life curve. The equation that represents the 50% probability failure curve is shown below.

Since your crack tip is yielding with each cycle, this is the appropriate fatigue model, not the stress-life data you used. The ANSYS Fatigue Tool has both stress-life and strain-life calculations of predicted 50% probability of failure.

Stress Singularity at Crack Tip

I expect you have a stress singularity in this model. On the Stress plot, the maximum value of stress is 376 MPa. I don't know where that is, a Max flag will show, but I assume it is at the crack tip. Similarly, on the Life plot (set aside the point above that it's not using strain-life), the minimum life is shown as 3265, and again, I assume that is at the crack tip, but you take a probe and read 11100 cycles.

If you have only linear elastic materials in the model (no plasticity) and were to refine your mesh, I expect the stress will go up and the life will go down. If you refine the mesh some more, the stress will increase some more and the life will continue to fall. That is an indication of a stress singularity in the model. That occurs whenever the geometry has a sharp interior corner under tension. If you add plasticity to the model, then the stress can't increase without limit.

It is possible to eliminate the stress singularity from a linear elastic model by editing the geometry and inserting a radius (blend) at the sharp interior corner. If mesh defeaturing is turned off, and mesh controls put several elements around this radius, a new stress and life result can be calculated. If the elements around this radius are doubled (half the size), the stress and life results will change very little.

The stress vs radius plot you show illustrates that for a linear material, the stress approaches infinity as the radius approaches zero. But for a material with plasticity, the yield stress is the limit out to some small radius.

When you have a sharp interior corner in a linear elastic model, the exact answer is infinity, but because we have finite sized elements, a finite value of stress is computed. However, as the elements get smaller, the stress keeps increasing without limit. If a radius equal to the plastic zone is created in the geometry, the exact answer is finite and the elements can converge on that number.

Fracture Mechanics

You mention the G. R. Irwin theory, which built on Griffith's research and came to be known as Fracture Mechanics. Irwin developed the following expression for the idealized radius of the zone of plastic deformation at the crack tip, which you could use to calculate a radius for your geometry.

Fracture Mechanics is different than Fatigue for predicting when a part will fail due to fracture.

Fatigue analysis uses the geometry and stress to predict how many cycles it will take, on average, to fracture the part.

Fracture Mechanics takes geometry with a crack in it, and asks at what stress will the crack propagate across the part and cause a fracture failure.

ANSYS has a Fracture Tool that can be used to predict whether a specific crack modeled in the geometry will fail at the applied stress.

Regards,
Peter

[momidor]

Hi Peter, 

Fully agree and two more things. 

Singularity

The tip radius is 0,05 mm and gives some sort of singularity.  I have model with plasticity, i.e. bilinear kinematic hardering, large deflection is on,  ect.. As long as I have singularity, the computing of the strain life doesn't make sense does it ? 

 Fracture Mech. 

For plane stress ( PSN ) and for plane strain ( PSO ) I used a bit different formulas. 

Regards

Jurek

[peteroznewman]

Hello Jurek,

In the graph you show above, if the horizontal axis is element size in mm and the vertical axis is stress in MPa then extrapolating gives the zero element size stress as 817 MPa. That is not a singularity because it is a linear graph with a finite stress intercept. A singularity would show some curvature that gets steeper as the elements get smaller.

A model with plasticity can be used to compute the Strain-Life fatigue estimate because it uses strain in the formulation of the estimate.

Regards,
Peter

[momidor]

Hi Peter,

Your hunch were right, i.e. ~817 MPa.

One more thing.. Even if it is not singularity why the simulation converged to such enormous value when ultimate stress is 360 MPa only ?

Regards

Jurek

[peteroznewman]

Hello Jurek,

The Tensile Ultimate Strength is not used by the solver during the solution, the solver doesn't stop at that value, it just keeps on following that hardening slope for as long as it can. Either it reaches the full load or an element distortion might stop the solver.

Tensile Ultimate Strength is only used in post-processing by the Stress Tool to plot the Factor of Safety contour plot by dividing that value by the Equivalent Stress.

Regards,
Peter

You can show your appreciation by clicking Like below the posts that are helpful.

 

[momidor]

Hello Peter, 

if so, can I rely on the value strain life (14300 cycles ) calculated inside the plastic zone ? 

Area of  plasticity: 225 MPa / 210000 MPa = 0,00107 ( ~0,1 %), i.e. from the black line to the tip of the crack. 

Regards 

Jurek 

[peteroznewman]

Hello Jurek,

In the Life plot on the right, the probe shows 1,450 cycles but the minimum value of life shown on the legend is 528 cycles.

In the Equivalent Elastic Strain plot on the left, the probe shows a strain of 0.001 but the maximum value shown on the legend is a strain of 0.00483.

It's the Max and Min values that determine when materials will fail, the probe can pickup values far from those extremes.

Regards,
Peter

[momidor]

Hello Peter, 

I really regret but in 18.1 Academic fracture tool is disabled. Do you know how to turn it on, if any ?

Regards 

Jurek 

[peteroznewman]

Did you insert a Fracture branch in your model and insert a Crack?

 

[momidor]

Thanks !