When you launch
shutdown -i to shutdown a computer from the command prompt (CMD), does it damage the computer that is being turned off (either local or remote)?
shutdown command in Windows performs a normal shutdown/reboot/logoff on the computer (local or remote). If you did close all your running programs and saved your files, there's no difference between running any
shutdown command (with any parameter) or using the windows user interface from the Start menu, and the command doesn't damage the computer or filesystem in any way.
More about running programs: if some programs preventing the shutdown are running, Windows will automatically try to terminate them, and, if that isn't possible, it will prompt you about it. The only problem you can have here arises for programs performing unsaved changes to files: if they provide an automatic recovery of not properly closed sessions (e.g. like Microsoft Office) they will save their state to recover it on the next start, otherwise you'll lose the unsaved data.
No it won't.
When you use the start menu to shutdown the computer, it will also use the same shutdown command, just with different parameters.
shutdown -i option was made to expose features that the normal windows startmenu does not offer without requiring to use commandline. For example,
shutdown -i can be used to send a shutdown command to another PC.
In a similar fashion that
shutdown -i won't damage your computer, neither will
shutdown -s -t 0.
There is also the
-f option, which is to force a shutdown. This will not damage your windows, but it may cause any unsaved data (such as an open word document that was not saved) to be lost.
Normally windows will show you a dialog with: There's a program preventing a shutdown. Do you want to cancel or continue anyway? A timeout will eventually abort the shutdown. With the
-f option, instead of showing the message, it will assume the "continue anyway" and close any remaining programs.