Similar to Kaspar's answer but with the g flag to replace all the occurrences on a line.

find ./ -type f -exec sed -i 's/string1/string2/g' {} \;

For global case insensitive:

find ./ -type f -exec sed -i 's/string1/string2/gI' {} \;

@kev's answer is good, but only affects files in the immediate directory.The example below uses grep to recursively find files. It works for me everytime.

grep -rli 'old-word' * | xargs -i@ sed -i 's/old-word/new-word/g' @

Command breakdown

grep -r: --recursive, recursively read all files under each directory.
grep -l: --print-with-matches, prints the name of each file that has a match, instead of printing matching lines.
grep -i: --ignore-case.

xargs: transform the STDIN to arguments, follow this answer.
xargs -i@ ~command contains @~: a placeholder for the argument to be used in a specific position in the ~command~, the @ sign is a placeholder which could replaced by any string.

sed -i: edit files in place, without backups.
sed s/regexp/replacement/: substitute string matching regexp with replacement.
sed s/regexp/replacement/g: global, make the substitution for each match instead of only the first match.

There are a few standard answers to this already listed. Generally, you can use find to recursively list the files and then do the operations with sed or perl.

rpl

For most quick uses, you may find the command rpl is much easier to remember.

Replace foo with bar on all .txt files:

rpl -v foo bar '*.txt' 

Simulate replacing the regex foo.* with bar in all .txt files recursively:

rpl --dry-run 'foo.*' bar '**/*.txt'

You'll probably need to install it (apt-get install rpl or similar).