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Problem physics- shallow truss element-Non linear FEM
I have got a shallow truss element-same as in Crisfield’s book.
I have a truss configuration as in the attached file (simple 1 member as in Crisfield as Bathe) . I have a load W = 42 units applied (the force causes compression in the member).I have used incremental approach wherein the laod is applied in 6 increments of 7 units each.
My load displacement plot is also shown in the attached file. The figure is similar to shown in Crisfield’s book for a similar problem. However, my question is that at the end of each increment the axial forces in the member were -651, -1507, --776, 483, 1106, 1629 units respectively.
What surprises me is how is that after 3rd increment, the force is turning axial tensile-this is not physically possible? IS there something wrong?
Also attaching the derivation of equations for a shallow truss element from Crisfield's book
Please help.
| Attachment | Size |
|---|---|
 problem physics.pdf | 18.82 KB |
| chapter 2 derivation.pdf | 110.58 KB |
| rough work.pdf | 32.96 KB |
 My solution-kajal.xls | 1.21 MB |
| for peymann and i mechanica - clarifying.pdf | 52.46 KB |
 My solution-peyman's answers.xls | 54 KB |
the bar does go into tension
As the load increases the bar compresses and the right node moves downwards until the bar is horizontal. At that point the compressive force in the bar is maximum. If the vertical displacement downwards is increased further, ***incorrect "then snap through occurs and" *** the right node is below the horizontal position and the bar starts to relax and decompress. Loading continues as the bar is angled downwards and the bar goes into tension. Thus the rapid increase in resistance to vertical load(at the end of your plot) is due to tension stiffening (ie, the bar going into tension).
You need to plot the deflected shape of the bar at each load increment and it should be obvious then that the bar does start in compression, but at some point will change to tension.
(I edited this per Peyman's comments below, he is right, so I have indicated the incorrect words in quotes.)
Dear Sir, Thank you for
Dear Sir,
Thank you for the response--So , can we say my answers are realistic because:
I have a total load 42 N.
I am performing 6 iterations and applying -7 N in each iteration.The bar is 25 mm above the horizontal position (initial position).
At end of every iteration, I get following (total) displacements and corresponding interal forces (tension +ve and comp -ve):
Iteration 1: -3.5 mm ; Bar force: -651 N
Iteration 2: -9.243 mm ; Bar force: -1507 N
Iteration 3: -45.758 mm ; Bar force: -776 N
Iteration 4: -52.311 mm ; Bar force: 483.465 N
Iteration 5: -55.024 mm ; Bar force: 1106 N
Iteration 6: -57.128 mm ; Bar force: 1629 N
As you see once the bar goes below 25 mm (25 mm downward indicates horizontal position) compression keeps on decreasing and from iteration 4 onwards it has tension.
The graph is there in the pdf attached above.
So, are the results realistic?
snap through
Julie
snaps through
Also, sir, what is the meaning of the word 'snaps-through'?
Snap-through and shallow truss
Hi Kajal and Louie;
Snap-through means that the structure reaches to a point that changes the shape suddenly because a small increase in load corresponds to a very different shape. in this structure which we can find an analytical answer for it too, snap-through does not happen when the bar becomes horizintal. Infact it happens before that due to the comression force in the bar that has a negative effect on the stiffness and makes the total stiffness zero (and then we have snap-through). See the Bathe's book for analytical result and solve a problem with this properties:
bar length=20, E=2e6, A=0.5, initial angle of the bar= 15 degrees
for this problem snap-throgh happens when the bar moves down a little bit more that 2 cm which is way less than becomming horizontal. However as Louie said, after the snap-through (jumping down to become a down slope bar) the bar takes the applied load as tension in it.
My suggestion for you is to try to solve a problem that you have the answer for it and compare your result with it (anlytical formula is in Bathe's book). the curve you are showing seems similar to the answer but doesn't show snap-through. check if the equilibrium is satisfied in every step, this is a simple check that you have to do and it should completely be satisfied. Honestly your answers seem wrong to me. if you are using the same code as before, the bar should be shallow.
Personally I do not prefer the methods which are not general and are good for only special cases like this shallow truss. That is why I suggested from the beginning to study corotational. For the case of 2D trusses it is not that difficult and you will learn it very fast, and it will answer many of your questions. Download this file from Louie's website and read it carefully:
http://people.wallawalla.edu/~louie.yaw/Co-rotational_docs/2Dcorot_truss.pdf
Cheers
Peyman
thanks, glad you caught the error
Hi Peyman,
Thanks for correcting my error. I see you are right, snap through does occur before the bar is horizontal. I'm glad you caught that.
I edited my statement above to indicate which words need to be removed so it is correct.
regards,
Louie
Analytical hand calculations attached
Dear Sir's,
Thanks a million for the input.
I've worked the hand calculations as per page 6 and 7 of Crisfield's book (attached pdf in the first post of this thread "rough work.pdf") and get the results exactly as same as my MathCAD code, i.e. those I mentioned above (again pasted below):-
Iteration 1: -3.5 mm ; Bar force: -651 N
Iteration 2: -9.243 mm ; Bar force: -1507 N
Iteration 3: -45.758 mm ; Bar force: -776 N
Iteration 4: -52.311 mm ; Bar force: 483.465 N
Iteration 5: -55.024 mm ; Bar force: 1106 N
Iteration 6: -57.128 mm ; Bar force: 1629 N
The hand calculations pdf is also attached.
Do you still say results are wrong?
Plz help
Yes I think the result are
Yes I think the result are wrong. I ill email you my solution since I don't know how to upload it here.
Peyman
Julie
Julie
Hi Sirs, Please see my
Hi Sirs,
Please see my attachment, Mysolution-kajal in the first post of this thread above.I have exactly coded as per expression in Crisfield (displacement control) and find that snap through acually is occuring at 1cm .
Is it that expressions derived in crisfield cannot handle snap through properly?plz help
Julie
Julie
Please see my attached file
Please see my attached file for peymann and i mechanica - clarifying.pdf- in the first post of this thread.
Clearly, using Crisfield's expression, whrn the bar is horizontal i.e. w = -25 ; axial force N is not equal to 0, and hence stiffness is not equal to zero.
So, no snap through is reflected in the analytical solution.
IS there something wrong / limitation in Crisfield's expressions/derivation?
Please please help anyone?
Peymans reply to the above
Peymans reply to the above post is as follows:
Ok Kajal, I just finished working on it and here it is. read my answer carefully and put it in imechanica too since I don't know how to upload files.
first of all, I did calculation with Crisfield's and I got what you see in sheet 2 of the attached file in comparison to Bathe's (for ks=0). Note that the positive load and deflection in the curve are downward. The difference in the beginning and the end of the curves are due to the non-iterative nature of Crisfield's method and is natural. However as you see Crisfield's method is a load control method and is not capable to capture the part of the curve which has negative stiffness. When the load increases beyond the limit load (i.e. the first maximum point in Bathe's curve) it jumps to the next point corresponding to the new increased load (snaps through).
Now answer to your next question. You asked why at w=-25 where the bar becomes horizontal and Kt is not zero, load is zero. It seems to me that you say so based on load=stiffness*deflection and you expect that for zero stiffness, we should have zero load. However there is no such thing in nonlinear analysis because in nonlinear we have:
Kt*(small change in deflection)=(small change in load). When the bar is horizontal, it is compressed and its Kt is negative. You can see it from Crisfield's equation too. If the horizontal bar was in tension, it had positive Kt. It is obvious that if you move a horizontal bar, which is in compression, just a little bit, you need a load in the reverse direction to keep it there (negative Kt). Vice versa, if the horizontal bar is in tension and you move it you need a bigger load in the same direction to keep it there (Kt is positive). Remember that LOAD IS ALWAYS FOUND FROM THE EQUILIBRIUM OF THE SYSTEM. When you apply equilibrium at the node where the load is applied, you get zero load for a horizontal bar anyway.
The file is attached in the first post of this thread-refer- My solution-peyman's answers.xls