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question on FEA Basics...Element Decision 3
- Thread starter nodalDOF
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If you set up the boundary conditions correctly you should get the same displacements (PL^3/3EI) and bending stresses (Mc/I) using either a bar/beam element (1d), shells or plates (2d) or a solid (3d). The solid will use a general stress state, the shell will be most likely plain stress. For shells and solids, as you add more elements you will start to get edge effects which will disappear when you get about one beam depth away from the wall. At that point all 3 models should agree. Another factor is how long is this beam compared to the other dimensions. For short, stubby beams the shear displacement may become significant (VL/AsG).
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- #3
Well, it depends on what you may define as the "correct stresses and deflections". You can make a very good model of a beam with either, but a good model will take more work than you might think.
First, it depends on the cross sectional shape of your beam. If you have an I-shaped member, you are going to have to construct an I-shape from the elements you want to use. This is decidedly more complicated than computing the moment of inertia by hand, which is one step in modeling the beam using rod elements.
Second, it depends on the sophistication of your element to model the variation in stress thru the depth of the beam. With elements (bricks or plane stress plates) that can model a linear variation in strain, it will not be too much work because that is the assumption used in bending theory. However, if you were to use "constant stress triangular elements" (CSTs), then you would need more small elements to capture a variation in strain.
Once you have your element, you decide on your mesh density. The poorer quality of your element, the finer your mesh must be to compensate. After you get your element and mesh decided, you then have to create boundary conditions for your cantilever beam. The fixed end is pretty obvious, but the free end (where I assume you are applying the load) is not as obvious.
Consider a FE cantilever beam made from linearly varying rectangular elements. Also say the cross section of the beam is rectangular in shape and we proceed by creating a model in the X-Y plane with X being the axis of the beam and Y being the depth. The model requires a fixed support which we will place on the left hand side, and a point of 30 pounds load acting in th -Y direction on the right hand end.
The length of the beam is ten feet, and the depth is ten inches. With these proportions, we decide to use five elements deep by thirty elements long (this creates a series of rectangles with a 2:1 aspect ratio, a good rule of thumb). Each element is two inches deep and four inches long.
Some may just apply the point load to one of the six nodes at the end of the beam. In this model you would get very bad results for about the first foot of the beam as the load spread out into the beam. After about three elements (horizontally) the load will automatically redistribute itself through the cross section and begin to agree with Sophmore level mechanics of beams. In a ten foot long beam, overall this would be a small error.
The proper application of the load would be to distribute the load to the six end nodes in proportion to emulate the parabolic distribution of shear stress thru the depth of the beam. If you were to compute these nodal loads, you would eliminate most of the error in the first foot of the beam. This is of little importance in a ten foot long member described, but could become important for short members (just don't ask me how short is short).
In the end, you will get deflections in the X and Y directions, which may be interpreted to rotation by taking the difference between the top and bottom nodes and dividing by the depth. However, you will also get influence from shear deformation which is not found in a Sophmore level analysis. The shear deformation influence would account for most of the difference between any hand calculations you are likely to undertake and your model. This is why I made the early statement that "it depends on what you define as the correct stresses and displacements."
Finally, always remember the wisdom, stiffness models are always too stiff, and flexibility models are too flexible. Translated, this means that most significant errors in stiffness solutions result from the mathmatical consequence of the assumed displaced shape being too stiff. Unfortunately, while I can say with confidence the displacements will be larger in labratory tests, I can not tell you that this means the stresses will be high or low.
I suspect this is more than you wanted to know, but some may find a bit of helpful knowledge in the discussion. Good Luck
First, it depends on the cross sectional shape of your beam. If you have an I-shaped member, you are going to have to construct an I-shape from the elements you want to use. This is decidedly more complicated than computing the moment of inertia by hand, which is one step in modeling the beam using rod elements.
Second, it depends on the sophistication of your element to model the variation in stress thru the depth of the beam. With elements (bricks or plane stress plates) that can model a linear variation in strain, it will not be too much work because that is the assumption used in bending theory. However, if you were to use "constant stress triangular elements" (CSTs), then you would need more small elements to capture a variation in strain.
Once you have your element, you decide on your mesh density. The poorer quality of your element, the finer your mesh must be to compensate. After you get your element and mesh decided, you then have to create boundary conditions for your cantilever beam. The fixed end is pretty obvious, but the free end (where I assume you are applying the load) is not as obvious.
Consider a FE cantilever beam made from linearly varying rectangular elements. Also say the cross section of the beam is rectangular in shape and we proceed by creating a model in the X-Y plane with X being the axis of the beam and Y being the depth. The model requires a fixed support which we will place on the left hand side, and a point of 30 pounds load acting in th -Y direction on the right hand end.
The length of the beam is ten feet, and the depth is ten inches. With these proportions, we decide to use five elements deep by thirty elements long (this creates a series of rectangles with a 2:1 aspect ratio, a good rule of thumb). Each element is two inches deep and four inches long.
Some may just apply the point load to one of the six nodes at the end of the beam. In this model you would get very bad results for about the first foot of the beam as the load spread out into the beam. After about three elements (horizontally) the load will automatically redistribute itself through the cross section and begin to agree with Sophmore level mechanics of beams. In a ten foot long beam, overall this would be a small error.
The proper application of the load would be to distribute the load to the six end nodes in proportion to emulate the parabolic distribution of shear stress thru the depth of the beam. If you were to compute these nodal loads, you would eliminate most of the error in the first foot of the beam. This is of little importance in a ten foot long member described, but could become important for short members (just don't ask me how short is short).
In the end, you will get deflections in the X and Y directions, which may be interpreted to rotation by taking the difference between the top and bottom nodes and dividing by the depth. However, you will also get influence from shear deformation which is not found in a Sophmore level analysis. The shear deformation influence would account for most of the difference between any hand calculations you are likely to undertake and your model. This is why I made the early statement that "it depends on what you define as the correct stresses and displacements."
Finally, always remember the wisdom, stiffness models are always too stiff, and flexibility models are too flexible. Translated, this means that most significant errors in stiffness solutions result from the mathmatical consequence of the assumed displaced shape being too stiff. Unfortunately, while I can say with confidence the displacements will be larger in labratory tests, I can not tell you that this means the stresses will be high or low.
I suspect this is more than you wanted to know, but some may find a bit of helpful knowledge in the discussion. Good Luck
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- #4
not a bad question, nodalDOF, assuming you're new to FEA. i apologise to dinosaur 'cause i didn't read your long post, and maybe you've said what i'm about to add.
nodal, most elements today assume a displacement field (linear or parabolic) and calculate the resulting stresses based on energy principles. thus if the applied strain field is linear, then the FEA results should be "right", and in more complex cases they're approximate; and this is before you consider non-linearities.
i'd suggest you try it for yourself. it's called a patch test, put one element under some loads (tension, bending), and see how it behaves, compared to theory. there's nothing like experience to learn from !
good luck
nodal, most elements today assume a displacement field (linear or parabolic) and calculate the resulting stresses based on energy principles. thus if the applied strain field is linear, then the FEA results should be "right", and in more complex cases they're approximate; and this is before you consider non-linearities.
i'd suggest you try it for yourself. it's called a patch test, put one element under some loads (tension, bending), and see how it behaves, compared to theory. there's nothing like experience to learn from !
good luck
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