I need to create 3 TB capacity shared folder for the customer.

I need to create  3 TB capacity shared folder for the customer.
but when I checked for the storage, we have three 1TB HDDS to increase the space i the file server.
but user requirement is, user needs a shared folder with size of 3TB.
in this case, can I add the new 1TB x 3 HDDS to the server, and can I create the spanned volume across it?
if I do so, customer will get the 3TB single shared folder.

Please suggest.
satheesh kumarAsked:
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rindiConnect With a Mentor Commented:
You can setup your RAID controller to use JBOD for those 3 disks, if it has that option. Personally I don't recommend that though. Either get 2 more disks, then setup RAID 6, or get larger disks (for example 2x3TB disks, and set them up in a RAID 1 array. That way you get redundancy, which you don't have with JBOD or spanned disks.
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Handy HolderSaggar maker's bottom knockerCommented:
What's the server make/model, what RAID controller if any and what disks and arrays do they currently have?
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Lee W, MVPConnect With a Mentor Technology and Business Process AdvisorCommented:
You want a MINIMUM of 4 disks and ideally 5 to 7.  While you can create a RAID 0, if the data is AT ALL important, it's stupidity to put i on a RAID 0.  If ANY of the three drives fail, you lose EVERYTHING, on ALL the drives.  A RAID 5 allows you to lose one drive and a RAID 6 allows you to lose 2 drives... and a RAID 10/0+1 allows you to POTENTIALLY lose 2-3 disks but requires 6 drives.  And the additional drive is to be a hot spare.
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satheesh kumarAuthor Commented:
Nice information
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Handy HolderSaggar maker's bottom knockerCommented:
It may be wrong though, if you currently have a RAID controller with several 1TB disks added to it you can expand the array and maintain parity; that's why I asked you for the current configuration.
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DavidPresidentCommented:
You have one other constraint ... not all access methods & client software. will work with a share > 2TB (or more specifically  512 * fffffffe hex blocks.     The issue is that 2TB worth of 512-byte blocks is more than 2^32 bits worth of address information.

So I would just make sure that you already know their software will work OK, before you go to a lot of trouble
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rindiCommented:
Satheesh-kumar, if you need more info, please come back and ask. If you don't, close the Question.
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satheesh kumarAuthor Commented:
Hi Rindi,

Thanks alot, because you are asking me to come back and ask the question, you are ready to give the solution even it may be small or very bit question thanks a ton.
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satheesh kumarAuthor Commented:
Thanks Rindi and Lee, for your help, some time I will ask more questions to clarify deeply, thanks for your timely help.
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