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#71948 - 07/25/18 05:36 AM Support Displacements in Skewed line
sharu Offline
Member

Registered: 04/09/13
Posts: 21
Loc: india
Hi All,

Does anybody know any way to extract displacements (axial, lateral, circumferential) on support locations for a skewed line.

For loads, we are referring to local restraint report. But, Displacements are not displayed in the report.

One way to find out displacements is to rotate the line to any of the orthogonal axis and to find out from Global restraint summary.

Ours is a pipeline with several turns and each skewed line having different orientation. Several times rotating the model and extracting displacements makes it a tedious job.
Is there any time saving way to do this activity.

Kindly share your thoughts..

Regards,
Sharun

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#71949 - 07/25/18 09:26 AM Re: Support Displacements in Skewed line [Re: sharu]
Michael_Fletcher Offline
Member

Registered: 01/29/10
Posts: 1025
Loc: Louisiana, US
There's no way out of some amount of tedium, here. Instead, I will offer what I would do.

Step 1: Copy and paste info from the elements tab into a spreadsheet, such as Excel.

Step 2: Use left() right() len() formulas to convert the verbal text into numerical values.

Step 3: Calculate axial and lateral unit vectors. Use IF() function to ignore vertical pipes. Assuming Y = vertical, then the provided X and Z values are axial. Lateral vectors are Z-> X and X->-Z OR Z->-X and X->Z. (Take your pick.) Unit vectors are these values divided by the Square Root Sum of the Squares of these values.

Step 4: Copy and paste info from support tab into Excel.

Step 5: Use index() & match() or hlookup()/vlookup() formulas to extract unit vectors to only nodes with supports.

Step 6: Export displacements to Excel and bring into this spreadsheet.

Step 7: Marry global displacements with the support nodes with index() & match().

Step 8: Axial movement is axial unit vector dot product with displacement vector. Lateral movement is lateral unit vector dot product with displacement.

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#71953 - 07/26/18 12:15 AM Re: Support Displacements in Skewed line [Re: sharu]
sharu Offline
Member

Registered: 04/09/13
Posts: 21
Loc: india
Hi Michael,

Thanks for the reply.

I would like to appreciate your efforts to find a solution for this.

Could you please explain Step 2 and Step 3 in more detail.

Using left() right() len() formulas which values are we converting from the elements data, and how are they used in further steps.

Also Step 3 if you could elaborate with an example it be we more clear.


Thank You
_________________________
sharun

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#71954 - 07/26/18 12:31 AM Re: Support Displacements in Skewed line [Re: sharu]
Vannella Offline
Member

Registered: 10/24/10
Posts: 39
Loc: IT
I normally use an Excel files that combines data coming from "Elements" (used to detect the element direction) and a customized displacement summary report.
Global displacements are converted by a formula that makes a vlookup in "Elements", detect the element direction and apply basic trigoniometry to get the local displacements.

Corrado Vannella
www.vannella.com

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#71959 - 07/26/18 09:44 AM Re: Support Displacements in Skewed line [Re: sharu]
Michael_Fletcher Offline
Member

Registered: 01/29/10
Posts: 1025
Loc: Louisiana, US
I left out a few steps. Here's some additional help.

LEN([reference]) returns the total number of characters in a cell.

If A1 is "10 ft. 5.125 in"
LEN(A1) will return the number 15.

FIND("X",[reference]) returns a number that represents the first character of a text reference.

If A1 is "10 ft. 5.125 in"
FIND("ft."A1) will return 4. Note that spaces count.

LEFT([reference],[number]) will return a string of text characters.

If A1 is "10 ft. 5.125 in"
LEFT(A1, 4) will return "10 f".
LEFT(A1, 2) will return "10".
VALUE(LEFT(A1,2) will return the number 10.

ISERROR([reference]) will return TRUE if an error message occurs, and FALSE if not.

ISERROR(FIND("CAESAR","10 ft. 5.125 in")) will return TRUE.

ISNUMBER("10 ft. 5.125 in") will return FALSE.

IF([reference],[operation 1],[operation 2]) if [reference] resolves to TRUE, then perform operation 1, else it'll perform operation 2.

If our formula for B1 were:
IF(A1="",0,A1)

And A1 is blank, B1 will be 0. If A1 isn't blank, then B1 will be A1.

So combining the above information, let's say we don't know for sure what's in A1, but it's one of the following values:

1. Nothing
2. Some number plus "in" (e.g. 5.125 in)
3. Some number plus "ft." (e.g. 10 ft.)
4. Some number plus "ft." with some number plus "in" (e.g. 10 ft 5.125 in)

Step 1: B1 will check to see if it's blank. If it's not blank, then push it on to C1. If it is blank, resolve it to the value 0.
B1 = IF(A1="",0,A1)

Step 2: If it's 0, push it forward, else, check to see if it's case 2 by determining that it's not case 3 or 4.

C1 = IF(B1=0,0,IF(ISERROR(FIND("ft.",B1),VALUE(LEFT(B1,FIND("in",B1)-2)/12))

Step 3: If C1 is a number, push it forward, else, determine if it's Case 3 by omitting Case 4 as a possibility. We divide by 12 to convert to feet.

D1 = IF(ISNUMBER(C1),C1,IF(ISERROR(FIND("in",C1),VALUE(LEFT(C1,FIND("ft.",C1)-2))))

Step 4: If D1 is a number, push it forward, else you need to convert 10 ft. 5.125 in. into a number that represents calculated feet.

I'll let you figure that one out, but I recommend that you use as many columns as you need, as well as to use RIGHT() coupled with LEN() at some point.

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#71960 - 07/26/18 10:21 AM Re: Support Displacements in Skewed line [Re: sharu]
Richard Ay Offline
Member

Registered: 12/13/99
Posts: 6226
Loc: Houston, Texas, USA
You can also try this utility program, available here:
"Global to Local"

This should be self explanatory.
_________________________
Regards,
Richard Ay - Consultant

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#71961 - 07/26/18 11:59 AM Re: Support Displacements in Skewed line [Re: sharu]
Michael_Fletcher Offline
Member

Registered: 01/29/10
Posts: 1025
Loc: Louisiana, US
Richard,

Do you know where to find the DFORMD.dll required to run the program?

Thanks

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#71962 - 07/27/18 03:44 AM Re: Support Displacements in Skewed line [Re: sharu]
sharu Offline
Member

Registered: 04/09/13
Posts: 21
Loc: india
Hi Richard,

Thanks for the program. It is really a wonderful tool.

I hope this would be helpful for all.

Michael,

Appreciate your efforts to provide more explanation.

Regards,
_________________________
sharun

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#71975 - 07/31/18 06:44 AM Re: Support Displacements in Skewed line [Re: sharu]
Richard Ay Offline
Member

Registered: 12/13/99
Posts: 6226
Loc: Houston, Texas, USA
Michael,

Ouch - that is one of the Fortran DLLs you need. I'll have to dig around to see if I can find that specific version (that matches the EXE).

Try this one, put it next to the EXE:
https://intergraph.box.com/s/xszf340977rjcj7glpz7an56jsjg9nh0

_________________________
Regards,
Richard Ay - Consultant

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