The Tampa Bay Buccaneers and Shaquil Barrett have agreed to a four-year deal worth up $72 million that keeps the star pass-rusher with the reigning Super Bowl champions, agent Drew Rosenhaus told ESPN's Adam Schefter.
The deal includes $36 million fully guaranteed, according to Rosenhaus. The contract also includes escalators -- related to sacks and playoff appearances -- that could increase Barrett's annual earnings, a source told ESPN's Jenna Laine.
Barrett played last season under the $15.8 million franchise tag and recorded eight sacks in 15 games, and he led the NFL with 74 pressures, according to ESPN Stats & Information. The former Pro Bowler was recently rated as the ninth-best available free agent by ESPN.
Barrett was outstanding during Tampa Bay's championship run, recording three sacks in the NFC Championship Game and another in Super Bowl LV to go along with his four hits on Chiefs quarterback Patrick Mahomes.
Barrett, 28, has 41.5 sacks, 15 forced fumbles and 266 tackles in 92 regular-season games over seven seasons. He led the NFL with 19.5 sacks in 2019.
ESPN's Jenna Laine contributed to this report.
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