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#20 - 01/26/00 10:22 AM Restraint load at a snubber
Ohliger Offline
Member

Registered: 12/16/99
Posts: 246
Loc: Mannheim,Germany
How calculate CAESAR 2 Restraint load in dyn. Case ? In a static case the restraint load do CAESAR calculate exact with the globel element force with the forces before and after the node (with looking for the sign).
In the dyn. output (global forces and restraint) are different. The sum of global force is not the restraint load ! At the most restraint points the different are smal. But in my snubber point the different a big. Here is the restraint load much lower as the forces from globel load output. When i take the globel force values for the calculation the restraint load then this value is near the value from a dyn.calculation with another program!

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#21 - 01/26/00 04:58 PM Re: Restraint load at a snubber
Richard Ay Offline
Member

Registered: 12/13/99
Posts: 5995
Loc: Houston, Texas, USA
The best way to answer this question is to relate things in the <em>static</em> sense first. Consider the case (shown in the figure below) where two elements share a support, and a vertical load is applied.





For this model, assume the load on the restraint from each of the two beams is also 100. Viewing the static restraint report will show a reaction of 200, not 300! Why? Because the applied load of 100 goes straight out the restraint (it doesn't load the beams).

Now, in the case of dynamics, each element's mass is determined, then this value is split equally between the two nodes (of the element). So any given node will have a mass equal to 1/2 of each element framing into it. This mass is then accelerated (F=Ma), to determine the system load(s).

The system of equations is solved and the nodal movements are determined. From these movements, elemental forces and stresses are determined. The restraint report may differ from the sum of the element forces based on whether or not the sum of these forces is equal to (Ma) - it doesn't have to be. This is the same idea as in the static example above.


------------------
Regards,
Richard Ay (COADE, Inc.)

[This message has been edited by rich_ay (edited January 26, 2000).]

[This message has been edited by rich_ay (edited January 26, 2000).]
_________________________
Regards,
Richard Ay
Hexagon PPM (CAS)
[img] [/img]

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#22 - 02/01/00 03:07 AM Re: Restraint load at a snubber
Ohliger Offline
Member

Registered: 12/16/99
Posts: 246
Loc: Mannheim,Germany
Thanks for you quick answer!
If the lokal forces an moments for the dyn.load case and combination load case (dyn+static)not equal the restraint forces and moments(lokal values will be come from a mode and the restraint values will be come from missing mass),then the user must allways selve calculate the local restraint values with coordinate transformation!
I think that is not a good service.
The best way is Caesar gives local restraint
values.German pipe stress programs (KWU ROHR
and ROHR 2)do it. I hope Caesar can give the
restraint values and the displacements in the future also in the local coordinate system.

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#23 - 02/01/00 07:44 AM Re: Restraint load at a snubber
Richard Ay Offline
Member

Registered: 12/13/99
Posts: 5995
Loc: Houston, Texas, USA
Local restraints really don't make sense, except in the case of a skewed nozzle.

Consider a restraint at the mid-point of an elbow. What is the local coordinate system for this restraint; is it aligned with the incoming tangent, the out going tangent, or is it aligned with a tangent to the elbow at the mid-point?

The only time a "local restraint coordinate" system makes sense is when a single (skewed) pipe runs into the restraint, such as a nozzle connection. In this case, you can use the local force report to obtain the same information. Just be sure to flip the signs on all the values.


------------------
Regards,
Richard Ay (COADE, Inc.)
_________________________
Regards,
Richard Ay
Hexagon PPM (CAS)
[img] [/img]

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#24 - 02/02/00 04:59 AM Re: Restraint load at a snubber
Ohliger Offline
Member

Registered: 12/16/99
Posts: 246
Loc: Mannheim,Germany
I think you dont understand the problem!
I give you a output example:

A snubber at node 350 is skewed -27,3 degrees about globel Z.
The dyn.case has following results:

Restraint Forces (global system)

Node FX FY FZ

350 1096 0 566
527 272
1 X/3M 1 X/3M

With this forces the snubber load can be calculated
Fres.= SQRT ( FX*FX + FZ*FZ).
The force at the snubber is Fres.= 1223,5.

Now look in the output local element forces:

Node Fa Fb Fc

350 1247 310 510 Element 340 to 350
840 73 241
2 X(2) 2 X(2) 5 Y(6)

350 1233 831 476 Element 350 to 360
829 221 178
2 X(2) 7 X(2) 2 X(2)

When i assume the sum of Fb is the snubber load then i get 310+831 = 1141 !
This value is not Fres. = 1223,5.
Resume:The local forces can not be a base for calculation restraint forces in dynamic case or dyn+static combination case. Look the restraint forces have the biggest part from
1 X/3M (missing mass)and the local forces come from 2 X(2) and 7 X(2) !!
The calculation for the local restraint load at the snubber with the globel force is
very simple. But if you have restraint in the space with 3 skewed axis then it is not fast and simple to calculate from hand.

In the staic load case the local forces are correct for looking the restraint load !
I think in dynamic load case not !!

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#25 - 02/02/00 09:37 AM Re: Restraint load at a snubber
Richard Ay Offline
Member

Registered: 12/13/99
Posts: 5995
Loc: Houston, Texas, USA
For a dynamics load case, it is unlikely that the local forces of the adjacent elements will add up to the restraint load. This is due to the following reasons:

1) All loads are reported as positive. Therefore, if one element load is positive, and the other is negative, the sum of the two reported (positive) element loads will of course not equal the restraint load.

Example: Element load #1 = 500; Element load #2 = -300; Restraint load = 200. Reported values are 500, 300, and 200 respectively. Of course, 500 + 300 does not equal 200!!

2) Reported results are the sum of a large number of modal, pseudostatic, and missing mass results, which are summed either using the SRSS or absolute methods. Therefore, in the above example, the user may be able to determine what happened, but any relationship is lost as the summation becomes more intricate.
<table>
<tr>
<th>Example:</th> <th>Element Load #1 </th><th>Element Load #2</th><th>Restraint Load</th>
</tr>
<tr>
<td>Mode #1</td> <td align="center">500</td><td align="center">-300</td><td align="center"> 200</td>
</tr>
<tr>
<td>Mode #2</td><td align="center"> 400</td> <td align="center"> 200</td> <td align="center">600</td>
</tr>
<tr>
<td>Missing Mass</td><td align="center">200</td><td align="center">-200 </td><td align="center">0</td>
</tr>
<tr><td> </td><td> </td><td> </td><td> </td>
</tr>

<tr>
<td>SRSS Totals</td><td align="center">671</td><td align="center"> 412 </td><td align="center">632</td>
</tr>
<tr>
<td>ABS Totals</td> <td align="center">1100</td><td align="center">700</td> <td align="center"> 800</td>
</tr>
</table>

Obviously, 671 + 412 does not equal 632, and 1100 + 700 does not equal 800!! (And we can't see any other apparent relationship.)

3) Dynamic models are constructed of discrete mass points, not as continuous elements. Therefore there are no body loads in dynamic analysis, only point loads. Therefore, the free-body diagram for the restraint includes not just the loads from each adjacent element, but also the force due to the acceleration times the mass of the restrained node (this force usually goes directly into the restraint, not into the adjacent elements). (The contribution of this mass point can be decreased if the user creates a finer mesh of mass points, i.e., so less mass goes directly into the restraints.)
<table>
<tr>
<th>Example:</th> <th>(Element Load #1)</th> <th>(Element Load #2)</th> <th>(Mass Point x Acceleration)</th> <th> (Restraint Load)</th>
</tr>

<tr>
<td>Mode #1</td> <td align="center">500</td> <td align="center">-300</td> <td align="center">50</td> <td align="center">250</td>
</tr>

<tr>
<td>Mode #2</td> <td align="center">400</td> <td align="center">200</td> <td align="center"> -20</td> <td align="center">180</td>
</tr>

<tr>
<td>Missing Mass</td> <td align="center">200</td> <td align="center">-200</td> <td align="center">1000</td> <td align="center">1000</td>
</tr>

<tr>
<td>SRSS Totals</td> <td align="center">671</td> <td align="center">412</td> <td align="center"> (not reported)</td> <td align="center">1046</td>
</tr>

<tr>
<td>ABS Totals</td> <td align="center">1100</td> <td align="center">700</td> <td align="center">(not reported)</td> <td align="center">1430</td>
</tr>
</table>

Obviously, 671 + 412 does not equal 1046, and 1100 + 700 does not equal 1430!! (And again there isn't any apparent relationship.) Note that the bulk of the Mass Point contribution will often show up (on the restraint) as the missing mass contribution, since it is rigidly restrained. (That explains why the greatest contributor to your restraint was missing mass, but the greatest contributor to your element was due to a mode.)

Due to summation methods as demonstrated above, the Response Spectrum method will never give "exact" results. Rather, the intent of the Response Spectrum method is to provide probabilistic results: i.e., probabilistically the largest element load, probabilistically the largest restraint load, probabilistically the highest stress, probabilistically the largest displacement, etc.

As far as you particular problem (calculating the load on the skewed snubber) is concerned, I don't see a problem. Your component loads are FX=1096, FZ=566. Therefore, adding these vectorially, your resultant load can be found as SQRT(1096x1096+566x566)=1234. The line of action of the snubber is confirmed by calculating ARCSIN(566/1234)=27.3 degrees.


------------------
Regards,
Richard Ay (COADE, Inc.)



[This message has been edited by rich_ay (edited February 04, 2000).]
_________________________
Regards,
Richard Ay
Hexagon PPM (CAS)
[img] [/img]

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