wind load case

Posted by: raja

wind load case - 03/30/08 07:33 AM

dear Sir,

When we assigned a wind shape factor in CII, the load case shown four wind cases (Win-1 , to Win-4). Why this four case? How to apply this four case? Is it require to make four operating case like

W+P1+T1+Win1+D1+H
W+P1+T1+Win2+D1+H
W+P1+T1+Win3+D1+H
W+P1+T1+Win4+D1+H

and four sus case like

W+P1+Win1+H
W+P1+Win2+H
W+P1+Win3+H
W+P1+Win4+H

Why four case? please read the above and clear my doubts?

regards
Posted by: Shahid Rafiq

Re: wind load case - 03/30/08 01:24 PM

These are the loadcases with directions of wind in East, west, north and south direction, apply it to X, -X, Z, -Z.

Posted by: Richard Ay

Re: wind load case - 03/30/08 09:33 PM

You have the option (possibility) of defining up to 4 different "wind vectors". The differences could be in direction, or speed, or some other significant parameter. You don't have to use all 4, perhaps one wind vector is sufficient in some cases.
Posted by: Sam Manik

Re: wind load case - 04/02/08 02:43 AM

Dear Raja,

Hope this will help you. Assume we only has simple load W, P, T, D, Wind and using CODE ASME B 31.3.

L1 WW+HP (HYD)
L2 W+P+T+D (OPE)
L3 W+P+T+D+Win1 (OPE)
L4 W+P+T+D+Win2 (OPE)
L5 W+P+T+D+Win3 (OPE)
L6 W+P+T+D+Win4 (OPE)
--------------------------
L7 W+P (SUS)
L8 T+D (EXP); just created to generate HOT SUS
L9 L2-L7 (EXP)*
L10 L2-L8 (HOT SUS)*
--------------------------
L11 L2-L3 (OCC)*
L12 L2-L4 (OCC)*
L13 L2-L5 (OCC)*
L14 L2-L6 (OCC)*
L15 L11+L12+L13+L14 (OCC)****
----------------------------
L16 L7+L15 (OCC)***; combined of SUS+Max wind to have OCC

Note:
* Use algebraic combination method under the Load Case Options tab.
** Use SRSS combination method under the Load Case Options tab.
*** Use ABS or Scalar combination method under the Load Case Options tab.
**** Use Maximum combination method under the Load Case Options tab.

Now, you can see we set up L3, L4, L5, and L6
L3 W+P+T+D+Win1 (OPE)
L4 W+P+T+D+Win2 (OPE)
L5 W+P+T+D+Win3 (OPE)
L6 W+P+T+D+Win4 (OPE)

to have L11, L12, L13, and L14:
L11 L2-L3 (OCC)*
L12 L2-L4 (OCC)*
L13 L2-L5 (OCC)*
L14 L2-L6 (OCC)*

in order to have the maximum efect of 4 direction of wind to piping system using ***max combination methode in L15:
L15 L11+L12+L13+L14 (OCC)****
Why? wind assume will not come in 4 direction in the same time. We just take the bad effect from which direction (corretc me if i'm wrong).

In my opinion, the combination you mentioned for SUS are not correct:
W+P1+Win1+H
W+P1+Win2+H
W+P1+Win3+H
W+P1+Win4+H

why? SUS will not include WIN which are occasional.



Posted by: Omdo

Re: wind load case - 04/02/08 11:11 PM

Radja,

Since you put displacement in your operating case, it has to be followed non linear as syamsul showed you how to extract the maximum wind.

second, Sustain loads is function of [Pressure, Weight, WNC, WW). Definitely the cases you create are Occasional Loads.


Regards
Posted by: raja

Re: wind load case - 04/03/08 03:00 AM



Mr.Samsul P manik,

But i think when we create the OCC load case, it has to subject with OPE. case and SUS. case like below
L11 L3-L7 (OCC)*
L12 L4-L7 (OCC)*
L13 L5-L7 (OCC)*
L14 L6-L7 (OCC)*

Please check ans confirm.

regards,
raja


Posted by: Sam Manik

Re: wind load case - 04/03/08 04:06 AM

Dear Raja,

Assume for ASME B31.3 code, the correct combination for (OCC)*** are:
SUS+(max wind)
SUS+(U = earthquake)
SUS+(F = concetric force due to PSV discharge)


L?? W+P+T+D+U1 (OPE)
L?? W+P+T+D-U1 (OPE)
L?? W+P+T+D+U2 (OPE)
L?? W+P+T+D-U2 (OPE)
L?? W+P+T+D+U3 (OPE)
L?? W+P+T+D-U3 (OPE)
---------------------
L?? W+P+T+D+F (OPE)
---------------------
to have U1,-U1, U2, -U2, U3, -U3, and F use the same like L11~L14. The case number drop starting from L7 if we include U and F. Note U1, U2, and U3 could occur in the same time which will generate many load cases if you want in complete combination.
Posted by: manu

Re: wind load case - 06/07/09 11:22 PM

I have one doubt

suppose the "wind 2= -wind 1" (say in x and -x direction)
and all my supports are bi-directional

can i still have different stresses in load case 11 and 12

load case as coppied from above:


L1 WW+HP (HYD)
L2 W+P+T+D (OPE)
L3 W+P+T+D+Win1 (OPE)
L4 W+P+T+D+Win2 (OPE)
L5 W+P+T+D+Win3 (OPE)
L6 W+P+T+D+Win4 (OPE)
--------------------------
L7 W+P (SUS)
L8 T+D (EXP); just created to generate HOT SUS
L9 L2-L7 (EXP)*
L10 L2-L8 (HOT SUS)*
--------------------------
L11 L2-L3 (OCC)*
L12 L2-L4 (OCC)*
L13 L2-L5 (OCC)*
L14 L2-L6 (OCC)*
L15 L11+L12+L13+L14 (OCC)****
----------------------------
Posted by: AAD

Re: wind load case - 06/08/09 02:51 AM

Mr. Manu

for you case as below :

L15 L11+L12+L13+L14 (OCC)****

in this case you taken OCC case as L15 : -(win1+win2+win3+win4).

this is not possible. at a time wind from all direction ( north,east,west & south ) not possible. So your last case L15 is not correct to be considered as OCC case.
anyway this is my opinion & exeprts( in this forum) opinion awaited.

Regards,
ajay damle

Posted by: manu

Re: wind load case - 06/08/09 03:42 AM

hi ajay

i had just copied the load cases from a post above.

anyway, my query was that, can we have different value of stresses due to wind acting in +x and wind acting in -X direction.(or for that matter due to any occasional loads)

and if so why?

I have seen that it is usualy the case, but can't think why?


Posted by: shr

Re: wind load case - 06/08/09 04:31 AM

Hi ajay damle

You might have overlook load case L15 as mention by Sam manik is the maximum value among all possible (only wind effect) load case that is suppose to add with sustain stress the get OCC stress. It should be OK.
You can add only wind effect coming from all each possible direction with sustain stress get get 4 different Occ stress or just pick the highest wind effect then add with sustain to get worse OCC stress.

Hi manu

load on support & nozzle can definitely be different from different direction wind load. so we can not avoid those cases.

Regarding stress since we may have non-linearity like single directional support, gap in support, support with friction so effect on wind direction also be different , Hence we need to follow those load cases.

Regards

Habib
Posted by: mariantrifan

Re: wind load case - 06/08/09 04:52 AM

MANU

1. You can make the maximum from wind, in this case L15 = L11, L12, L13, L14 (MAX) if you want.
Or :

L1 WW+HP (HYD)
L2 W+P+T+D (OPE)
L3 W+P+T+D+Win1 (OPE)
L4 W+P+T+D+Win2 (OPE)
L5 W+P+T+D+Win3 (OPE)
L6 W+P+T+D+Win4 (OPE)
--------------------------
L7 W+P (SUS)
L8 T+D (EXP); just created to generate HOT SUS
L9 L2-L7 (EXP)*
L10 L2-L8 (HOT SUS)*
--------------------------
L11 L2-L3 (OCC)*
L12 L2-L4 (OCC)*
L13 L2-L5 (OCC)*
L14 L2-L6 (OCC)*

L15 L11+L7 (occ)
............
L18 L14+L7 (occ)

L19 L15,...,L18 (max)


2. You can have small diferent between wind case +X and -X, how much depends by system.

3. Why because you not have a symmetric system and a small difference in your pipe configuration have effect in your calculation results ( for see this influence just make a small and simple system and you see the difference).
Posted by: AAD

Re: wind load case - 06/08/09 05:02 AM

hi manu,

sorry for misinterpreting your question.

anyway reply given by Marian is correct.

regards,
ajay damle
Posted by: manu

Re: wind load case - 06/08/09 06:48 AM

Thanks Habib
I had't thought of friction.
Suppose it's the one causing difference, i will check it without friction.

Thanks marian!
i can't think why a symmetry should cause a difference, though i am not sure.I will check and return

Thanks Ajay smile