Does anyone know how to model both the longitudinal and perpendicular values of "2nd moment of area" in a finite element program. I am familiar with the LEAP5 finite element suite (and i assume others are similar) but there does not appear to be a facility to input both Ixx and Iyy.
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Stiffness of concrete deck slabs 1
- Thread starter ljr
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It depends whether you are looking to model 1 dimensional finite elements (i.e. beams) or plates/shells. In the FE formulation of plates/shells Ixx/Iyy is not an issue as these values do not appear anywhere in the equations to be solved - so in order to model a deck/floor with different stiffness in two directions you need to be able to use an element with Orthotropic Material capability. It is my guess that Leap5 does not have this capability as its FE section is quite limited however Robot Millennium does. When you set up a surface to represent a floor in this package you can choose Isotropic or Orthotropic material types, within the orthotropic material section they hav a database of floor types such as grillages, concrete floor on trapezoidal plate etc, you fill in the criteria and the stiffness matrix for this material type is automatically formulated for you.
I'm not about to try and flog you a copy of this software, it works for me but maybe you should find out more for yourself. I am in the Uk so I can give you the UK email address technical@issrobot.co.uk and maybe they can forward you to the correct people
Good luck.
A
I'm not about to try and flog you a copy of this software, it works for me but maybe you should find out more for yourself. I am in the Uk so I can give you the UK email address technical@issrobot.co.uk and maybe they can forward you to the correct people
Good luck.
A
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- #3
Thanks for the reply arnie_c. I was beginning to think the question had beaten the experts in this forum.
I agree; the values of Ixx and Iyy do not appear in the formulation of the FE equations. However using elements with orthotropic material capabilities does not solve the problem. For a concrete deck slab (which forgot to mention in my original message) the values of Youngs modulus is the same in both longitudinal and transverse directions. However the amount of reinforcement in the longitudinal and transverse directions may be very significantly different. Thus to correctly model the structure one needs to input values of Ixx/Iyy.
you cannot imagine the number of finite element models I have seen that appear to model concrete deck slabs with the same value of stiffness in both longitudinal and transverse directions. This must be wrong!!!!!!!
Further comments appreciated.
I agree; the values of Ixx and Iyy do not appear in the formulation of the FE equations. However using elements with orthotropic material capabilities does not solve the problem. For a concrete deck slab (which forgot to mention in my original message) the values of Youngs modulus is the same in both longitudinal and transverse directions. However the amount of reinforcement in the longitudinal and transverse directions may be very significantly different. Thus to correctly model the structure one needs to input values of Ixx/Iyy.
you cannot imagine the number of finite element models I have seen that appear to model concrete deck slabs with the same value of stiffness in both longitudinal and transverse directions. This must be wrong!!!!!!!
Further comments appreciated.
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- #5
Sorry ludvik but i would refer you to arnie_c statement (which i agree with) that the values of Ixx/Iyy do not appear in the formulation of FE equations.
Consequently if they do not appear in the equation; how does one model the different values of Ixx /Iyy. If you cannot model these values, can you be confident of the accuracy of the output.
2D/3D capability is not the problem
Further comments appreciated.
Consequently if they do not appear in the equation; how does one model the different values of Ixx /Iyy. If you cannot model these values, can you be confident of the accuracy of the output.
2D/3D capability is not the problem
Further comments appreciated.
Sorry, I thought you were talking about I values for beam elements. You only specify thickness for shells.
Concrete elements, including slabs, are *normally* modeled with their gross concrete section properties, and assumed to act elastically. The contribution of the reinforcing to stiffness is normally ignored.
If there is a major difference in reinforcing transversely as there is longitudinally, you may have a difference in bending stiffness of around 10% in the two directions (the difference in strength would obviously be much higher).
I have only heard of orthotropic shell modeling in automotive and aero engineering, though I would be interested to hear if anyone has used it in structural eng.
Note also that concrete material non-linearity and cracking introduce inaccuracy to your model too.
Concrete elements, including slabs, are *normally* modeled with their gross concrete section properties, and assumed to act elastically. The contribution of the reinforcing to stiffness is normally ignored.
If there is a major difference in reinforcing transversely as there is longitudinally, you may have a difference in bending stiffness of around 10% in the two directions (the difference in strength would obviously be much higher).
I have only heard of orthotropic shell modeling in automotive and aero engineering, though I would be interested to hear if anyone has used it in structural eng.
Note also that concrete material non-linearity and cracking introduce inaccuracy to your model too.
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1
- #8
I must preface the following statement by saying I do not know much at all about structures-specific packages; however I do enough about general-purpose FE codes that I think I now understand the details behind the question.
Effectively what you are asking is 'how do I augment the bending stiffness of the concrete due to the strengthening of the rerods'.
Some general purpose codes actually have reinforcing rod capabilities. ABAQUS, for one, does have this capability and it is used primarily for reinforced concrete purposes. The concrete is modeled (in your case, with shells), and then reinforcement rods are essentially modeled within those shells, resulting in the strengthening which you are aiming for. If a general-purpose FEA code can do this, I guess I would expect applications-specific codes (such as this one) to be able to have some similar effect.
Another possibility, if that is not available, would be layered-composite shells. This would certainly require orthotropic behavior in the layers which are intended to model the reinforcement rods.
Although values of Ixx/Iyy do not discretely appear in any formulation for shells, there is effectively the concept of Ixx/Iyy in shell formulation (this is, just as in beams, what resists bending in the shell elements). I would think augmenting this bending stiffness with moment of inertia values is a somewhat phony way to accomplish this stiffening effect (but, as I said, I don't know structures-specifics FEA work, so maybe that is a standard of which I am not familiar).
Brad
Effectively what you are asking is 'how do I augment the bending stiffness of the concrete due to the strengthening of the rerods'.
Some general purpose codes actually have reinforcing rod capabilities. ABAQUS, for one, does have this capability and it is used primarily for reinforced concrete purposes. The concrete is modeled (in your case, with shells), and then reinforcement rods are essentially modeled within those shells, resulting in the strengthening which you are aiming for. If a general-purpose FEA code can do this, I guess I would expect applications-specific codes (such as this one) to be able to have some similar effect.
Another possibility, if that is not available, would be layered-composite shells. This would certainly require orthotropic behavior in the layers which are intended to model the reinforcement rods.
Although values of Ixx/Iyy do not discretely appear in any formulation for shells, there is effectively the concept of Ixx/Iyy in shell formulation (this is, just as in beams, what resists bending in the shell elements). I would think augmenting this bending stiffness with moment of inertia values is a somewhat phony way to accomplish this stiffening effect (but, as I said, I don't know structures-specifics FEA work, so maybe that is a standard of which I am not familiar).
Brad
Guest
ludvik,
You seemed to have missed/misunderstood my comments.
Skanska FEM allows orthotropic analysis and design of plates. You input your reinforcement meshes and it does a full cracked section analysis to several codes. Most good finite element packages have this capability but Skanska has applied it to a specialist beam/slab design package.
See
You seemed to have missed/misunderstood my comments.
Skanska FEM allows orthotropic analysis and design of plates. You input your reinforcement meshes and it does a full cracked section analysis to several codes. Most good finite element packages have this capability but Skanska has applied it to a specialist beam/slab design package.
See
- Thread starter
- #10
All
There seems to be a lot of misunderstandings regarding my original query. Either that or a lot of misunderstanding of finite elements.
My query had nothing to do with design codes. FE is purely an analysis tool. It does not matter whether the model is concrete, brickwork, steelwork or a combination of the three. One has to model the existing/proposed structure realistically. In concrete this means modelling values of ExIx and EyIy. As there is no facility to input values of Ix or Iy the computer has to be manipulated slightly.
The user inputs the element thickness from which the computer calculates a value of Ix.
The user has to input Ex
The user has to input kEy where k= the ratio of Iy/Ix
Thus k*Ey*Ix
Iy*Ey*Ix/Ix
Just because the computer doesn't ask for the number does not mean that it is not required.
Regards
There seems to be a lot of misunderstandings regarding my original query. Either that or a lot of misunderstanding of finite elements.
My query had nothing to do with design codes. FE is purely an analysis tool. It does not matter whether the model is concrete, brickwork, steelwork or a combination of the three. One has to model the existing/proposed structure realistically. In concrete this means modelling values of ExIx and EyIy. As there is no facility to input values of Ix or Iy the computer has to be manipulated slightly.
The user inputs the element thickness from which the computer calculates a value of Ix.
The user has to input Ex
The user has to input kEy where k= the ratio of Iy/Ix
Thus k*Ey*Ix
Iy*Ey*Ix/Ix
Just because the computer doesn't ask for the number does not mean that it is not required.
Regards
Guest
Skanska FEM requires the input of ratio E1 to E2 as previously mentioned. It performs a full cracked section analysis which does require input from a code of practice. It also calculates the values of I1 and I2 for the input reinforcement. Extracts from the help file below.
The program approximates the effect of the crack at reinforced concrete plates with iteration technique.
Since the crack analysis is non-linear the method of superposition is not valid, so it can be used only for combinations. The iteration must be used separately for all combinations. The steps of crack analysis for each combination are the following:
1. Loading of the structure with full load, determination of internal forces with linear method and (if needed) calculation of the required reinforcement.
2. Determination of the load-level causing the first crack from the internal forces.
3. Calculation of the number of the load steps between the full load and the load-level causing crack from the defined load step value. The default value is 10% of the original load, but the user can change it.
4. Increasing of the load by steps and decreasing of the stiffness of finite elements by using the current codes.
5. Converging to the final result with final iteration at the full load (final load-step). The final iteration ends if the relative error of the squared sum of the displacements becomes less than the allowed one (see ch. 3.4.1). The default value of the maximum allowed relative error is 1%.
At every step the load value is increased with the actual step size after modifying the cracked elements. At the first step the applied load value is equal with the cracked-force + one load-step value, at final step it is equal with the full load.
At crack solution, there is an opportunity for using the reinforcement pre-defined by the user or determined from the linear analysis. Note: At the latter case the user can get false results or the reinforcement may not resist for the modified initial forces, because of the change of the reinforcement and the cracked elements strength in every point or the reinforcement quantity calculated from the linear analysis.
A finite element can be considered cracked if the tensile stress in an outer part becomes larger than the tensile strength of concrete. At plate elements it means, that one of the main moments in the middle of the element exceeds the moment that causes crack. At beam elements it means, that the tensile stress in an outer part becomes larger than the tensile strength at one of the end points of the element.
The program calculates the stiffness of the cracked elements from the stiffness determined by the second stress condition and the crack-distribution factor defined in the codes. (If the current code does not contain such factor, the program uses the factor value defined in EC2).
The stiffness of cracked elements, which is parallel with the crack, is not modified, so the plate element will behave as an orthotropic one. The crack direction at an element is calculated by the method called less reserved moments (see. Appendix ch. 7.2): the direction of the crack is perpendicular to the direction, in which the plate losses its load-capacity. If a plate element can be cracked in both directions at the same plane (lower or upper), the stiffness will be decreased in those directions. If an element can be cracked at both sides (lower and upper), the directions of the cracks will be (about) perpendicular to each other.
Because of the limitations and features of the finite element method the shear- and moment diagrams determined for cracked elements become unequal. The moment values at the border of two orthotropic elements can be different from the expected value of that point.
The program approximates the effect of the crack at reinforced concrete plates with iteration technique.
Since the crack analysis is non-linear the method of superposition is not valid, so it can be used only for combinations. The iteration must be used separately for all combinations. The steps of crack analysis for each combination are the following:
1. Loading of the structure with full load, determination of internal forces with linear method and (if needed) calculation of the required reinforcement.
2. Determination of the load-level causing the first crack from the internal forces.
3. Calculation of the number of the load steps between the full load and the load-level causing crack from the defined load step value. The default value is 10% of the original load, but the user can change it.
4. Increasing of the load by steps and decreasing of the stiffness of finite elements by using the current codes.
5. Converging to the final result with final iteration at the full load (final load-step). The final iteration ends if the relative error of the squared sum of the displacements becomes less than the allowed one (see ch. 3.4.1). The default value of the maximum allowed relative error is 1%.
At every step the load value is increased with the actual step size after modifying the cracked elements. At the first step the applied load value is equal with the cracked-force + one load-step value, at final step it is equal with the full load.
At crack solution, there is an opportunity for using the reinforcement pre-defined by the user or determined from the linear analysis. Note: At the latter case the user can get false results or the reinforcement may not resist for the modified initial forces, because of the change of the reinforcement and the cracked elements strength in every point or the reinforcement quantity calculated from the linear analysis.
A finite element can be considered cracked if the tensile stress in an outer part becomes larger than the tensile strength of concrete. At plate elements it means, that one of the main moments in the middle of the element exceeds the moment that causes crack. At beam elements it means, that the tensile stress in an outer part becomes larger than the tensile strength at one of the end points of the element.
The program calculates the stiffness of the cracked elements from the stiffness determined by the second stress condition and the crack-distribution factor defined in the codes. (If the current code does not contain such factor, the program uses the factor value defined in EC2).
The stiffness of cracked elements, which is parallel with the crack, is not modified, so the plate element will behave as an orthotropic one. The crack direction at an element is calculated by the method called less reserved moments (see. Appendix ch. 7.2): the direction of the crack is perpendicular to the direction, in which the plate losses its load-capacity. If a plate element can be cracked in both directions at the same plane (lower or upper), the stiffness will be decreased in those directions. If an element can be cracked at both sides (lower and upper), the directions of the cracks will be (about) perpendicular to each other.
Because of the limitations and features of the finite element method the shear- and moment diagrams determined for cracked elements become unequal. The moment values at the border of two orthotropic elements can be different from the expected value of that point.
discombobulation: You outline the procedure for a non-linear cracked analysis. I know about these, but I have never done one. There are obviously more accurate than linear elastic analysis... how much more accurate are they? Do you have problems in the case of a moving load on a bridge? For a large structure with moving loads, what is the processing time like? Can the same model be used for seismic/wind/dynamic analysis?
ljr: You also have the options of modeling orthotropic material properties, or embedding reinforcing as ABAQUS apparently has the capacity for.
I was taught at school, and my experience has been, that FE models of concrete slabs for normal structures can be sucessfully completed with isotropic shells. This is of course a good step up from the grillages and 2-d beam element approximations of yesteryear.
My feeling is that the more sophisticated techniques can be saved for special structures. Does your structure have any special requirements ljr?
ljr: You also have the options of modeling orthotropic material properties, or embedding reinforcing as ABAQUS apparently has the capacity for.
I was taught at school, and my experience has been, that FE models of concrete slabs for normal structures can be sucessfully completed with isotropic shells. This is of course a good step up from the grillages and 2-d beam element approximations of yesteryear.
My feeling is that the more sophisticated techniques can be saved for special structures. Does your structure have any special requirements ljr?
ljr,
Based on your last posting, it sounds to me that all you are doing is describing an orthotropic elastic behavior. You are using the terms Ixx/Iyy to describe the ratio in bending behavior between the two directions, but in reality the Ixx/Iyy concept is (in my view) getting in the way of the underlying discussion.
For a unitary-dimension slab modeled with shells, the stiffness effects due to geometry (the 'Ixx' and 'Iyy', for lack of better terms) are the same in both directions. The difference in bending stiffness is not geometric, but rather material. (In your case, it is due to a difference in material orientation in the orthogonal directions--a difference in 'E'). The bending stiffness in both directions is dependent on these geometric and material values (what we are calling Ixx,Iyy, Ex, and Ey). So the bending stiffnesses are Kx=Ixx*Ex and Ky=Iyy*Ey
However, Ixx=Iyy (on a per-unit basis). Therefore, what is changing is not the Ixx and Iyy values, but rather the Ex and Ey values. The Ixx and Iyy issues are superfluous to the whole issue, as it boils down to merely being an issue of orthotropic elasticity.
One final note (so as not to further confuse the issue)--my above description is a simplification of what is happening, but is correct. The stiffnesses are dependent on the values of E integrated over the section, so that the precise calculation of this stiffness for reinforced concrete involves integrating the various material (concrete and steel) moduli over the cross-section of the element. However, based on the postings, it sounds as if the general approach is to describe this effect with an augmented Young's modulus. Within that assumption, the above directly describes what is happening. My earlier posting regarding discrete modeling of the reinforcement rods would take into account this integrated effect directly.
I hope this clarifies things (rather than further confusing)
Brad
Based on your last posting, it sounds to me that all you are doing is describing an orthotropic elastic behavior. You are using the terms Ixx/Iyy to describe the ratio in bending behavior between the two directions, but in reality the Ixx/Iyy concept is (in my view) getting in the way of the underlying discussion.
For a unitary-dimension slab modeled with shells, the stiffness effects due to geometry (the 'Ixx' and 'Iyy', for lack of better terms) are the same in both directions. The difference in bending stiffness is not geometric, but rather material. (In your case, it is due to a difference in material orientation in the orthogonal directions--a difference in 'E'). The bending stiffness in both directions is dependent on these geometric and material values (what we are calling Ixx,Iyy, Ex, and Ey). So the bending stiffnesses are Kx=Ixx*Ex and Ky=Iyy*Ey
However, Ixx=Iyy (on a per-unit basis). Therefore, what is changing is not the Ixx and Iyy values, but rather the Ex and Ey values. The Ixx and Iyy issues are superfluous to the whole issue, as it boils down to merely being an issue of orthotropic elasticity.
One final note (so as not to further confuse the issue)--my above description is a simplification of what is happening, but is correct. The stiffnesses are dependent on the values of E integrated over the section, so that the precise calculation of this stiffness for reinforced concrete involves integrating the various material (concrete and steel) moduli over the cross-section of the element. However, based on the postings, it sounds as if the general approach is to describe this effect with an augmented Young's modulus. Within that assumption, the above directly describes what is happening. My earlier posting regarding discrete modeling of the reinforcement rods would take into account this integrated effect directly.
I hope this clarifies things (rather than further confusing)
Brad
Allow my 2-cents in this. Ixx and Iyy are geometrically appropriate for beams, but for plates, the equivalent would be (width x thickness^3/12) for a slab. Plate theory expresses dimensions and loads in terms of per unit width (i.e., in-lb/in line moment), so the appropriate equivalent unit moment of inertias for an orthotropic plate would be the same in both directions, T^3/12, which is combined with the EFFECTIVE BENDING MODULI, Ex & Ey, to develop the effective orthotropic bending stiffness. If the code only accepts a single E but different "Ix &Iy", then one may effect the orthotropicity by calculating effective T's in x and y such that Ty = Tx * cuberoot(Ex/Ey).
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