Former Alabama defensive lineman and Super Bowl LVI champion A’Shawn Robinson signed with the New York Giants on Monday afternoon. The two sides came to terms on a one-year deal worth up to $8-million.
Robinson began his professional career with the Detroit Lions in 2016. In four seasons with the team, he tallied 172 tackles, 20 tackles for loss, and seven sacks.
Following the 2019 season, Robinson signed a two-year contract with the Los Angeles Rams. There, he spent the next three seasons. Despite having multiple injuries, Robinson still found himself to be an intrical piece of the Rams’ defensive line unit.
He recorded 121 tackles, four tackles for loss, and two sacks in his Rams career.
The Giants have been unable to reach a contract extension with defensive tackle Dexter Lawrence. Therefore, the addition of Robinson is huge at this point in time. He is a great run-stopper and has proven that he can rush the passer effectively as well. Robinson will benefit from being a part of a unit that was in desperate need a presence along the interior of the defensive line.
The #Giants have agreed to terms with #Rams FA DT A’Shawn Robinson, continuing to beef up their defensive front right before the Draft, sources tell me and @MikeGarafolo. He gets a 1-year deal worth up to $8M max value with the opportunity to find a home with NYG. pic.twitter.com/K53jVNi77Y
— Ian Rapoport (@RapSheet) April 24, 2023
Roll Tide Wire will continue to follow Alabama players in the NFL.
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