Buccaneers place franchise tag on Shaquil Barrett

The Buccaneers have tagged Shaquil Barrett, keeping him in Tampa Bay for one more year.

The Tampa Bay Buccaneers aren’t going to risk losing the NFL’s 2019 sack leader in free agency. According to ESPN’s Adam Schefter, the Bucs are placing the franchise tag on Shaquil Barrett, preventing him from becoming an unrestricted free agent.

Tampa Bay had until 11:59 a.m. ET on Monday to use the franchise tag.

Barrett led the league with 19.5 sacks last season, breaking out in a big way in his first season with the Buccaneers. His previous career-high was 5.5 sacks with the Broncos in 2015, and his 37 QB hits in 2019 were more than he had in the previous five years combined.

With Barrett tagged, Jameis Winston is set to hit the free-agent market on Wednesday afternoon. The legal tampering period begins on Monday at noon, which is when Winston’s agent can begin contacting teams and discussing terms of a contract.

Assuming Barrett is tagged with an outside linebacker designation instead of defensive end, the franchise tag is expected to be worth $15.8 million for the 2020 season. Last offseason, Barrett signed a one-year, $4 million deal with Tampa Bay.

The Buccaneers now have until July 15 to sign Barrett to a long-term extension, otherwise he’ll play the season out on the tag – which he’s said he’s willing to do.