Practical application of smoothing

Practical application of smoothing

Example of the Br wheel in Handbook 61, title II

Finite Elements Working Group

Editor: D. Guth, with the support of V. Bruno and F. Dubois.

Introduction

1.1 Purpose 

The purpose of this document is, on the basis of an extremely simple example, to show the differences that can be obtained according to the modeling which is chosen, depending on a certain number of parameters such as: mesh size, modeling of the load, extent of smoothing, and on several software.

It is used as an example in §D.5.2.1 of the Wiki (https://wiki-gtef.frama.wiki/accueil-gtef:partie-2:chapitre-d:d5 ).

It does not claim to exhaustively represent the scenarios that the modeler may encounter, however it is easily replicable by everyone on their favorite software and can be used as a benchmark.

It aims at illustrating the common problems that engineers encounter in their daily work: how to model loads (point or distributed?), over which length to smooth (2h, 4h, other?), where to read the values (at the nodes, at the mesh centers?), what shape for the elements? (triangular? quadrilateral?), is the smoothing (or "averaging") performed by the software correct? etc.

If you want to share your results coming from other software than Robot (R), Sofistik (S) or Pythagoras (P), the right address is elements.finis@afgc.asso.fr .

1.2 Study configuration

We are going to calculate the forces resulting from the impact of a Br wheel (Handbook 61 title II), centered on a one-way joist slab bridge.

The calculations were carried out by SETRA with a Poisson's ratio of 0.15 for the concrete.

The guide does not specify whether the guide values have been smoothed or not.

→ We calculate the effect of a Br wheel, with a diffusion on E=10 cm.

The impact of 60cm (transverse) x 30 cm (direction of traffic) is, therefore, to be modeled as diffused on an impact of 80 cm x 50 cm².

The impact force is 10 tons, which we take equal to 100 kN, i.e. a pressure of 250 kPa.

The modeled slab has dimensions of b=40 m (b≈∞) and a=6 m:

The size of the meshes is 25x25 cm², where the thickness h of the slab is modeled as squared.

In addition, a calculation with 100x100 cm² meshes is also performed to evaluate the impact of increasing the mesh size on the results.

1.3 Studied cases 

We study, with 3 software (R, S and P), in whole or in a part, the following configurations:

• 25x25cm² mesh / distributed load / point load - quadrilateral elements (potentially triangular);

• 100x100cm² mesh / distributed load / point load - quadrilateral elements.

Synthesis and conclusion

All graphic outputs are provided in the Appendix.

We compare the calculated values to the BT1 values that serve as a reference point.

Moments in kN.m/m

Conclusion:

From these calculations, we can already extract the following conclusions:

• It is important to understand what the software does in terms of display and smoothing. Of course, we can always size with the worst case scenario, but even if we could justify it, this approach would not be economical. You should not hesitate to insist to the editor to ensure that the software does not remain a black box for the modeler;

• Avoid point forces, they are too unfavorable and, moreover, have no physical meaning. If describing impact surfaces for rolling loads is complex, for example, the single force must at least be transformed into several point forces;

• Given the differences observed, it is useless to provide results with 8 decimal places, 1 is enough;

• This example illustrates why, during a verification, an external or outside control of the studies, it is necessary to specify in the hypothesis/modeling note the mesh size, the type of element and the chosen smoothing logic. It is also possible to set the tolerance which will be accepted to consider the results as "correct". It is indeed unlikely that two software will provide the same results; the software must be able to converge with a maximum difference of 5% (given it is agreed on at the beginning of the study and in the case of a linear calculation).

• As indicated in the Wiki, the smoothing logic remains subject to the engineer's judgment, however the thickness of the slab and the load impact area must guide the choice.

Once the force is distributed over an impact area defined at the mean slab layer and the mesh size is in a reasonable ratio relative to the slab’s thickness and the dimensions of the load impact rectangle, then the peak value for reinforcement may be selected, where the value smoothed over 2 hours does not lead to too important deviations - which one can see in the table above.

It should be remembered that rolling loads will not stress the entire cross-sectional area at the same time, because they move by nature. The choice between a 2h or 4h smoothing must also be made on the basis of a reflection on the transversal redistributive capacity and/or the incidence of increased constraints by a few %; the consequences are not necessarily the same between a calculation in SLS characteristics and a calculation of crack opening or fatigue;

• If we consider that the SETRA values are correct values (although we have no indication about a possible smoothing), we realize that the approach consisting in smoothing for 2 hours seems to be the most valid;

• The triangular elements give worse values for the same mesh size, which confirms what the Wiki says in Part 2 - A2;

In any case, it is the responsibility of the engineer to get an idea of the impact of the mesh size or of how to model the loads, through tests on simplified parts of models and evaluating the sensibility of such tests.

In summary:

3) Appendix 1 - Model R1 - manual smoothing

This model has a node under the center of the impact rectangle as opposed to Model 2 which was built to have the center of a 25x25 mesh under the center of the impact rectangle - See Appendix 3.

This § calculates the smoothed forces, manually, as opposed to Appendix 2. We will see that the differences between the two approaches are minimal.

• Load impact area with mesh size

• Moment My (transversal)

Values taken every ≈12.5 cm at the central axis of the one-way joist slab:

• Average value on 2 meshes: 28.04 kN.m/m

• Average value on 4 meshes: 26.7 kN.m/m

• Moment Mx (longitudinal)

• Values taken every ≈12.5 cm at the transverse axis:

• Average value on 2 meshes: 23.3 kN.m/m

• Average value on 4 meshes: 21.9 kN.m/m

• Smoothing detail (from the integral):

We read with 2 meshes = 0.50 cm=2h and with 4 meshes=1 m=4h.

Reminder: effect of a load shift of 0.125 m in X (longitudinal): changes practically nothing, see below.

My, smoothed over 1.00 m -> My=26. kN.m/m

My, smoothed over 0.50 m ->My=28.0 kN.m/m

These values are almost identical to the case without shift.

4) Appendix 2 - Using the smoothing functions of a software - Model R1

Many tests are carried out to judge the sensibility of the mesh size.

A calculation with a point force is also computed.

4.1 Use of automated smoothing - distributed load - mesh size 25x25 cm².

On 0.50 m: My=14.03/0.50=28.06 kN.m/m

With a mesh refinement: My=13.96/0.50m=27.92≈28.06 kN.m/m

• For 1 m, My=26.73 kN.m/m

• With a mesh refinement: My=26.57 kN.m/m ≈26.73 kN.m/m

• Mx, smoothed on 0.50 m=11.71/0.50=23.42 kN.m/m

• Mx, smoothed on 1.00 m =21.95 kN.m/m

4.2 Use of automated smoothing - distributed load - mesh size 100x100 cm².

• My=13.97/0.5=27.94 kN.m/m

• My=26.57 kN.m/m

• Mx=10.95/0.5=21.90 kN.m/m

• Mx=20.53 kN.m/m

4.3 Smoothing with point force

Whether you smooth on 1.00 m or 50 cm, the smoothed value remains excessive.

• My=20.9/0.50 m= 41.80 kN.m/m

• My=34.85 kN.m/m, on 1 m.

• Mx=18.56/0.5=37.12 kN.m/m

• Mx=28.09 kN.m/m

4.3 Reminder: load distribution over several point loads

5) Appendix 3 - Model R2 - Diffused load centered mesh

(With centered mesh under the impact i.e. the center of a mesh is under the center of a load)

• Moment My

• Moment Mx

6) Appendix 4 - FE calculation data - models R1 and R2

Materials v=0.15

Sum of reactions:

7 Appendix 5 - Sofistik Model

7.1 Distributed load - mesh ≈25 x 25 cm².

(Curve generated from nodal values - method recommended by the editor).

• My=28.6 kN.m/m

Displayed on a grid with a mesh size of 25x25 cm:

Smoothing over 1 m would lead to (28.6+27.0)/2=27.8 kN.m/m

The same calculation with triangular elements: My=26.9 kN.m/m

• Mx=23.6 kN.m/m

Values displayed in 25x25 cm² meshes

Smoothing over 1 m would lead to (23.6+21.9)/2=22.8 kN.m/m

Same calculation with triangular elements: Mx=21 kN.m/m

Model view with triangular meshes

7.2 Point force- mesh ≈25 x 25 cm².

• My=52.6 kN.m/m

• Mx=45.7 kN.m/m

Data:

Loading (wheel impact):

Impact rectangle 80x50 with 250 Kpa load

• Materials (Poisson’s ratio for concrete)

8) Appendix 6 – Pythagore software

Material and load data:

• Poisson’s ratio for slab

• Distributed load

• Point load

Note on the smoothing performed by the software:

The node smoothing performed by the Pythagore software consists of an average, at a given node, of the results obtained for this node in the four finite elements having this node in common.

As a result, for the node located exactly in the middle of the plate (and load), these four values are symmetrically identical, so the smoothed and unsmoothed maximum values coincide, as shown in the following maps.

8.1 Distributed load - mesh size = 25 x 25 cm².

8.1.1 Without smoothing

My = 27.41 kN.m/m

Mx = 21.55 kN.m/m

8.1.2 With smoothing at the nodes

My = 27.41 kN.m /m (no change in peak value)

Mx = 21.55 kN.m/m (no change in peak value due to load symmetry)

8.2 Concentrated load - mesh size = 25 x 25 cm².

8.2.1 Without smoothing

My = 34.83 kN.m/m

Mx = 28.11 kN.m/m

8.2.2 With smoothing

Same peaks as without smoothing

8.3 Distributed load - mesh size = 100 x 100 cm2

8.3.1 Without smoothing

My = 19.69 kN.m/m

Mx = 13.48 kN.m/m

8.3.2 With smoothing

My = 19.69 kN.m/m (no change in peak value due to load symmetry)

Mx = 13.48 kN.m/m (no change in peak value)

8.4 Concentrated load - mesh size = 100 x 100 cm2

8.4.1 Without smoothing

Same peaks as with smoothing (see below).

8.4.2 With smoothing at the nodes

My = 21.71 kN.m/m

Mx = 15.21 kN.m/m

8.5 Distributed load - automatic meshing ≈ 25 x 25 cm2

8.5.1 Without smoothing

My = 27.54 kN.m/m (almost the same result as with the square mesh)

Mx = 22.45 kN.m/m

8.5.2 With smoothing at the nodes

Same peak values as without smoothing.

9) Appendix 7 – SETRA abacus (operating society for transport and automobile repairs) - Technical Bulletin n°1

Ma (transverse)=2870 t.m.

Mb=2350 t.m

10) Appendix 8 – Pücher abacus

M=Average value read/8/π*100 kN - this method includes a "certain" margin of error given how each person reads the contours.

Mxx=Mx+νMy et Myy=My+νMx

Numerical application:

We have Mx≈24.3 kN.m/m My≈19.9 kN.m/m

• Mxx=24.3+0.15x19.9=27.3 kN.m/m

• Myy=19.9+0.15x24.3=23.5 kN.m/m

Transverse moment (according to x)

Longitudinal moment (according to y)