String.Trim() only trims from the beginning and end of strings, so it has no effect in your command, because the . only occurs inside your input string.
If you truly want to remove just the . and keep the post-decimal-point digits, use the -replace operator:
$MyDouble2 -replace '\.' # -> '55'
Note:
* -replace takes a regex (regular expression) as the search operand, hence the need to escape regex metacharacter . as \.
* The above is short for $MyDouble2 -replace '\.', ''. Since the replacement string is the empty string in this case, it can be omitted.
If you only want to extract the integer portion, use either 4c74356b41's .Split()-based answer, or adapt the regex passed to -replace to match everything from the . through the end of the string.
$MyDouble2 -replace '\..*' # -> '5'
@Matt mentions the following alternatives:
For removing the . only: Using String.Replace() to perform literal substring replacement (note how . therefore does not need \-escaping, as it did with -replace, and that specifying the replacement string is mandatory):
$MyDouble2.Replace('.', '') # -> '55'
For removing the fractional part of the number (extracting the integer part only), using a numerical operation directly on $MyDouble (as opposed to via the string representation stored in $MyDouble2), via Math.Floor():
[math]::Floor($MyDouble) # -> 5 (still a [double])