SQL XML Insert multiple elements - sql-server

Question

Please help! I have a field of xml within a database. I am trying to insert a new element for each element in the field. E.g:
<some-element>
<a> </a>
<b> </b>
</some-element>
<some-element>
<a> </a>
<b> </b>
</some-element>
would insert element c and go to:
<some-element>
<a> </a>
<b> </b>
<c> </c>
</some-element>
<some-element>
<a> </a>
<b> </b>
<c> </c>
</some-element>
I know I can use #fieldname.modify('insert #CElement into (/some-element)[1]') but this only changes the first element! I want to repeat this for all elements.
Any help would be very much appreciated! (I am using SQL server 2008)
Moved from answer:
Thank you for your reply! Perhaps it would help if I was more specific... I am describing a simple table of data within the xml. I am trying to add a new column. I can add the column within the column descriptions but I need to add the column element to all the rows. (It would not be a simple task to change the structure and I would like to avoid this!) E.g:
<Table>
<Columns>
<Column ID="0">
<Column-Name>0NAME</Column-Name>
</Column>
<Column ID="1">
<Column-Name>1NAME</Column-Name>
</Column>
<Column ID="2">
<Column-Name>2NAME</Column-Name>
</Column>
<Column ID="3">
<Column-Name>!!!! THIS COLUMN IS BEING ADDED !!!!!</Column-Name>
</Column>
</Columns>
<Rows>
<Row ID="0">
<C ID="0">0 contents here</C>
<C ID="1">0 contents here</C>
<C ID="2">0 contents here</C>
<!-- NEW COLUMN NEEDS TO BE CREATED HERE -->
</Row>
<Row ID="1">
<C ID="0">1 contents here</C>
<C ID="1">1 contents here</C>
<C ID="2">1 contents here</C>
<!-- NEW COLUMN NEEDS TO BE CREATED HERE -->
</Row>
<Row ID="2">
<C ID="0">2 contents here</C>
<C ID="1">2 contents here</C>
<C ID="2">2 contents here</C>
<!-- NEW COLUMN NEEDS TO BE CREATED HERE -->
</Row>
</Rows>
</Table>

Answer 1

You can recreate your XML, e.g.:
declare #x xml = '<some-element>
<a>1</a>
<b>2</b>
</some-element>
<some-element>
<a>3</a>
<b>4</b>
</some-element>'
select #x.query
('
for $e in some-element
return
<some-element>
{ $e/a }
{ $e/b }
<c/>
</some-element>
')

Answer 2

Adapting the answer provided by Kirill Polishchuk to the new XML structure:
select #XML.query
('
for $c in Table/Columns
return
<Table>
<Columns>
{ $c/* }
</Columns>
<Rows>
{
for $r in Table/Rows/Row
return
<Row>
{ $r/* }
<C ID="3"/>
</Row>
}
</Rows>
</Table>
')