**Moderator:** bernd

You mean the gap? I think that’s a visual issue of perspective. The plate thins more where you see the gap and the surface facing us is therefore further away from the view point than at the top, causing visual shrinkage. I will check and post a view with perspective switched off.

I'm guessing the same after you described the BCs.

Edit: in the other pictures, the left face stays flat, so for sure it's a visualization issue.

Edit: in the other pictures, the left face stays flat, so for sure it's a visualization issue.

And here is the final proof with perspective switched off:

This got the academic juices flowing again.

The applied stress is 2000 MPa and yield stress 100MPa. With the hole halving the cross section, the load multiplier at failure is therefore 100/2/2000=0.025, as I mention above and predicted by the analysis.

However, what does theory say about the angle of the shear band? Well, if we take a shear band at angle alpha, then with virtual work you can work out that

P-ult = 0.5*s_y/sin(2*alpha)

The minimum value for P_ult is attained at alpha = 45 degrees and is equal to 0.5 s_y.

Both the shear band angle and the ultimate load are accurately predicted.

The applied stress is 2000 MPa and yield stress 100MPa. With the hole halving the cross section, the load multiplier at failure is therefore 100/2/2000=0.025, as I mention above and predicted by the analysis.

However, what does theory say about the angle of the shear band? Well, if we take a shear band at angle alpha, then with virtual work you can work out that

P-ult = 0.5*s_y/sin(2*alpha)

The minimum value for P_ult is attained at alpha = 45 degrees and is equal to 0.5 s_y.

Both the shear band angle and the ultimate load are accurately predicted.

A soil mechanics problem this time. We want to dig a 4m deep trench in very soft clay. Can we do this without support?

Undrained shear strength (Su): 10kPa

Shear modulus (G): 500kPa

Wet weight (Gamma): 20kN/m3

For Tresca material we can translate undrained shear strength into yield strength by simply doubling it. For von Mises material doing so will introduce some error, as I will explain later.

I created the model by extruding a simple sketch. The bottom and left boundaries are fully fixed and the right boundary represents a symmetry plane, so is allowed to slide. The facing boundary and its opposite are fixed out of plane to create the plane strain conditions of a long trench.

The weight load is incremented until failure occurs. The load factor is therefore the proportion of the weight load applied.

As can be seen from the load-deflection curve, the trench will collapse at a load factor of 0.58 and is therefore not safe!

The plastic deformation plot shows a soil wedge sliding into the trench at collapse.

A hand calculation assuming a straight slip surface at 45 degrees shows that collapse occurs at a load factor of 0.5.

So why this 16% error in the analysis? Well part is always related to lack of mesh refinement. However previous analyses showed that even for coarse meshes rather accurate collapse loads can be predicted.

The answer is in the translation of undrained shear strength to yield stress. The von Mises stress (S_vm) for pure shear (Tau) equals S_vm = Sqrt(3) * Tau. So a von Mises material with yield stress Sy has a shear strength of of Sqrt(3) * Sy and not 2 * Sy as is the case with Tresca (= Mohr Coulomb with Phi=0) material. So by simply doubling the undrained shear strength to get a yield stress, we introduce as much as a 2/Sqrt(3) ~ 1.155 (or 15.5%) error in the collapse load.

This simple case study shows that the macro accurately predicts collapse, but that care should be taken with translation of measured soil parameters into model parameters

Undrained shear strength (Su): 10kPa

Shear modulus (G): 500kPa

Wet weight (Gamma): 20kN/m3

For Tresca material we can translate undrained shear strength into yield strength by simply doubling it. For von Mises material doing so will introduce some error, as I will explain later.

I created the model by extruding a simple sketch. The bottom and left boundaries are fully fixed and the right boundary represents a symmetry plane, so is allowed to slide. The facing boundary and its opposite are fixed out of plane to create the plane strain conditions of a long trench.

The weight load is incremented until failure occurs. The load factor is therefore the proportion of the weight load applied.

As can be seen from the load-deflection curve, the trench will collapse at a load factor of 0.58 and is therefore not safe!

The plastic deformation plot shows a soil wedge sliding into the trench at collapse.

A hand calculation assuming a straight slip surface at 45 degrees shows that collapse occurs at a load factor of 0.5.

So why this 16% error in the analysis? Well part is always related to lack of mesh refinement. However previous analyses showed that even for coarse meshes rather accurate collapse loads can be predicted.

The answer is in the translation of undrained shear strength to yield stress. The von Mises stress (S_vm) for pure shear (Tau) equals S_vm = Sqrt(3) * Tau. So a von Mises material with yield stress Sy has a shear strength of of Sqrt(3) * Sy and not 2 * Sy as is the case with Tresca (= Mohr Coulomb with Phi=0) material. So by simply doubling the undrained shear strength to get a yield stress, we introduce as much as a 2/Sqrt(3) ~ 1.155 (or 15.5%) error in the collapse load.

This simple case study shows that the macro accurately predicts collapse, but that care should be taken with translation of measured soil parameters into model parameters

Last edited by HarryvL on Sun Mar 24, 2019 5:46 am, edited 3 times in total.

very cool. I would like to do this too as well as your concrete analysis

Please note that I made a small edit to the explanation of undrained shear strength for von Mises plasticity to make it a bit clearer.

I will now focus on cleaning my macro up so I can share it.

This means that some exciting stuff, like interface plasticity, beams, plates and shells, buckling, maximizing the power of BLAS and LAPACK to increase speed all need to wait.

Once available, I could do with some help with input and output facilities and getting some OOP where it helps extensibility, clarity and speed. I need to get rid of tons of FOR loops.

This means that some exciting stuff, like interface plasticity, beams, plates and shells, buckling, maximizing the power of BLAS and LAPACK to increase speed all need to wait.

Once available, I could do with some help with input and output facilities and getting some OOP where it helps extensibility, clarity and speed. I need to get rid of tons of FOR loops.

sure! We will find a smart way to integrate this into FEM!

Of course I couldn't resist ... what about a square pit in the same material and of the same dimensions?

well, it is slightly stronger (about 0.7/0.58 = 20%), but not enough to stand up.

well, it is slightly stronger (about 0.7/0.58 = 20%), but not enough to stand up.