Guess I would add that "no Yielding or Plastic Deformation" would be very difficult to justify any yielding anywhere if it ever came to court, etc.....
I would be very careful with this one....we might know (or think we do) what is safe, etc. but I'm sure any explanations would quickly become...
Without test data I guess I would start with the properties of the fibers (steel?) and rubber and estimate the composite properties based on the ratios of each (see a book or article on how to do this)
Ed.R.
Just curious?? Could part of the problem also be that a combination of things (such as decreased snow loads, more "precise" structures solutions, LFRD.....etc.) has decreased the overall factor of safety that used to built into designs to a point where we are getting these failures.....
I...
An integration point is the point within an element at which integrals are evaluated numerically. These points are chosen in such a way that the results for a particular numerical integration scheme are the most accurate. Depending on the integration scheme used the location of these points will...
I guess I'm no longer sure what you are asking....As both rb & esp have indicated, and indeed what you said in your last post, if the stress in the corner exceeds the yield stress then plastic deformation will occur and the loads will be redistributed. If you can accept this inelastic behavior...
Could be what RB describes.....I really can't tell from the picture but if my eyes are seeing correctly it appears that the area may be an area of stress concentration.....I would suggest you take the structural dimensions and a text (Roark maybe) and investigate what kind of stresses you get...
Without more information there is no way anyone can answer your question.....It could be real if it is in an area of stress concentration and is modeled properly but more likely it is some kind of mistake in your model or just a poorly modeled structure. There is no way for us to know......
Ed.R.
It seems clear to me that you do not understand what you are doing for the following reasons:
1. Stress is NOT a vector but rather is a tensor. Stresses cannot be manipulated using the rules of vector math....
2. Maximum principal stresses occur at points of zero shear so trying to combine...
Agree with 40818.....
Also disagree with statement that:
"The accuracy of the solution depends upon the shape function"
Assuming that the shape function has no errors in it you can obtain just as accurate a solution using very different shape functions i.e. linear, parabolic, cubic .... etc...
rb:
You are correct....I just assumed he would have information about the stresses at the top, bottom, and maybe mid-plane..
By the way the old saw he mentioned about only getting good results for less than 1/2 the thickness depends on the type of element being used as well as what the model...
rb:
If its linear the load (stress) ratio (axial/total) should be the same (or very close) as the membrane/total stress ratio....If it is not linear (material) then would need to give it more thought....and get an estimate of how much the linear range is exceeded by......
Ed.R.
The stress distribution across the plate thickness is a linear combination of bending and axial (membrane) stresses so if you have the distribution across the thickness of the plate the axial component is simply the average of stresses on the top and bottom and the bending stresses is the total...
The product of inertia, just like the moment of inertia, is a purely mathematically defined quantity i.e. integral( x y dA).
Just as with stresses when the product of inertia goes to zero the principal axes of a cross section are obtained....
Ed.R.
I think JStephen has it right....
Use circular plate of bolt circle diameter, fixed edges, concentrated load in center..
Resulting thickness computed for maximum specified deflection at center of plate should give a reasonable value for thickness.
Ed.R.
Going back to the basic definition of Von Mises stress you find that it related by a constant to the maximum shearing stress on the octahedral plane (sigvm = (3/sqrt(2))tauoct)) so basically a Von Mises criterion is a shearing stress criterion while a principal stress criterion is a normal...
It is impossible to provide an answer to the question you pose as the results obtained using (any) elements are dependent on the modeling not on the specific element....
Good results can be obtained with linear elements, parabolic elements, and even higher order elements than parabolic...
Deflection at free end is 4 P L^3/(3 pi E r r0^3) where
r=radius at free end and
r0= radius at fixed end....
obtained by integration of y''=M(x)/(E I(x))
Ed.R.
GLAD221:
You have the dimensions of the cylinder therefore you can write the equation of the cylinder and get the slope equation (dy/dx)....The perpendicular slope is then just the inverse of the slope of the cylinder.....
No idea how to do it in the software....
Ed.R.
dsd133:
Maybe I don't understand the structure but if you are modeling with frame (I assume this means truss-axial stiffness only) members then I think you should probably model using the area of the weakest part of the cross section...Both stresses and displacements will be correct if you use...
