This will not win the Raiders’ weekend news cycle, but Johnathan Hankins agreed to terms to stay in the Bay Area. Hankins will stay with the Raiders on a two-year deal, Adam Schefter of ESPN.com tweets.
Hankins was scheduled to be a free agent for the third time in three offseasons, and his most recent stay on the market was extensive. Following a release from the Colts after one season, Hankins did not agree to terms elsewhere until September.
In 2017, he lingered in free agency after the first wave but signed a $9MM-AAV Colts deal. He will eschew another stay on the market.
Oakland initially inked Hankins to a one-year pact after Week 1 of last season. The parties will now work together again in 2019, with Hankins likely in line to stay a first-unit player under Paul Guenther. The second-year Oakland DC wanted the mammoth lineman back, per Vic Tafur of The Athletic (on Twitter).
Hankins started 14 games for the Raiders last season, doing so following his 2017 work as a Colts starter and three-year run as a Giants first-stringer. Despite going into his seventh season, Hankins is only set to turn 27 this offseason. He graded out as one of the league’s best interior defenders as a Colt, and although Pro Football Focus did not view his 2018 work with the Raiders in the same light (No. 76 out of 112 full-time interior defenders), the Raiders will use some of their extensive cap space to retain him.
The former second-round pick will now have a full offseason to work in Guenther’s system, a 4-3 look like he played with the Giants, this year.