CSHEAR Element Output

[Kristina Sornikova]

Dear All,

I looking at the nastran output of shear panel element CSHEAR and I confused with the Shear-IJ output, where I=1..4, J=1..4

Example: So Shear-IJ, what units it has? lb/in? Also what is kick output? Thanks. Kristina.

Sorry about how it looks, when I paste it it shows good format but preview does not show good format.

F O R C E S A C T I N G O N S H E A R P A N E L E L E M E N T S (CSHEAR)

====== POINT 1 ====== ====== POINT 2 ====== ====== POINT 3 ====== ====== POINT 4 ======
ELEMENT F-FROM-4 F-FROM-2 F-FROM-1 F-FROM-3 F-FROM-2 F-FROM-4 F-FROM-3 F-FROM-1
ID KICK-1 SHEAR-12 KICK-2 SHEAR-23 KICK-3 SHEAR-34 KICK-4 SHEAR-41
0 40069 2.50181E+02 4.25804E+01 -1.47293E+02 2.94144E+00 1.22708E+02 2.62269E+01 -1.63647E+02 1.30414E+02
0.0 2.92114E+01 0.0 2.92114E+01 0.0 2.92114E+01 0.0 2.92114E+01

 

[IRstuff]

ISn't that something you find in the manual?

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[rb1957]

i pretty sure they are lbs/in (thk) ... forces from NASTRAN tend to be that.

to be sure, either run a test case, or it should be in the manual

another day in paradise, or is paradise one day closer ?

 

[Kristina Sornikova]

Thank you all for respond. I check NXNastran user guide, pages 897 and 913.

Page 897, section 7.2 say "All items are element forces (or moments) unless otherwise indicated".

Page 913 shows CSHEAR O/P, only list all items, and Shear-IJ is real number magnitude (RM). That's why I confuse.

 

[Kristina Sornikova]

rb1957 Sir, can you give me example of test case? One element and how to set up and what to check?

 

[rb1957]

there are so many ways to create a shear test ... a simple shear panel, a tube in torsion, a web in bending ...

an example ..
take 1 CSHEAR element, say 2" square and 0.05" thk
surround it with RODs,
constrain 1 edge (2 nodes) to restrain rigid body motion,
load the opposite edge, in-plane, some load
then verify shear stress = element force/thk.

another day in paradise, or is paradise one day closer ?

 

[Kristina Sornikova]

Thank you rb1957 Sir, I try what you say, 2.0" element bottom edge pinned, top edge total load 1Lbf to right, and I check the f06 file.

I see this now:
F O R C E S A C T I N G O N S H E A R P A N E L E L E M E N T S (CSHEAR)

====== POINT 1 ====== ====== POINT 2 ====== ====== POINT 3 ====== ====== POINT 4 ======
ELEMENT F-FROM-4 F-FROM-2 F-FROM-1 F-FROM-3 F-FROM-2 F-FROM-4 F-FROM-3 F-FROM-1
ID KICK-1 SHEAR-12 KICK-2 SHEAR-23 KICK-3 SHEAR-34 KICK-4 SHEAR-41
0 5 -5.00000E-01 -5.00000E-01 5.00000E-01 5.00000E-01 -5.00000E-01 -5.00000E-01 5.00000E-01 5.00000E-01
0.0 -5.00000E-01 0.0 -5.00000E-01 0.0 -5.00000E-01 0.0 -5.00000E-01

Element is like below (corner nodes 1234, middle rod elements 1234, and CSHEAR 5):
---------3
4 ======= 3 --> 1 lbf

4[-------5-----]2

1 ======= 2
---------1
In first try I make 1.0" square element, then Shear-12 was 1.0.
Next I increase size to 2.0" same load and constraint, now Shear-12 is 0.5, so I think it is lb/in.

Shear-12 = (End Load 1 + End Load 2)/L where L=2.0" now.
Shear-12 = (0.5 + 0.5)/2 = 0.5 lb/in

I also check element max shear stress:

S T R E S S E S I N S H E A R P A N E L S ( C S H E A R )
ELEMENT MAX AVG SAFETY ELEMENT MAX AVG SAFETY
ID. SHEAR SHEAR MARGIN ID. SHEAR SHEAR MARGIN
5 3.125000E+01 -3.125000E+01

So shear flow using max shear = 31.25 psi*t, where t = 0.016"

q = 31.25*0.016 = 0.5 lb/in

But I don't understand the signs, can you help?
Also, I apply 1lbf at top right, but below is rod forces:

F O R C E S I N R O D E L E M E N T S ( C R O D )
ELEMENT AXIAL ELEMENT AXIAL
ID. FORCE TORQUE ID. FORCE TORQUE
1 0.0 0.0 2 -5.000000E-01 0.0
3 0.0 0.0 4 5.000000E-01 0.0


Why only 0.5lbf in right and left rods and no load in top rod? I understand no load in bottom rod pinned.

 

[rb1957]

well you've found an error in my original response ... element force = load/length (which follows from stress = element force/thk).

carefully read sign conventions in the NASTRAN manual, sometimes the +ve sense is determined by the node numbers, sometime by the node sequence, sometimes by the co-ord system.

rod forces ... draw a FBD ... how are you applying the load ? (1/2 to each corner ? then the rod is doing nothing)
the element between the constraints probably also (particularly if over-constrained)
the other two sides need to react load to preserve moment balance, considering the moment about the constraints from the load (yes?)

another day in paradise, or is paradise one day closer ?

 

[Kristina Sornikova]

OK I clear now.

I apply 1 Lbf at top edge. OK that is also clear now, nastran apply 0.5 lbf at each top node so no load in rod.

Yes, moment balance also I get, bottom left node 1.0lbf Down reaction, and bottom right node Up reaction.

Thank you very much.