Defensive Tackle A’Shawn Robinson Signs With LA Rams
The tea has been spilled and according to a reliable source former Detroit Lions defensive tackle A’Shawn Robinson is joining the Los Angeles Rams on a two-year contract worth $17 million. Read on for more on A’Shawn Robinson signing with LA…
Hmmmm. Gonna need a second source. … this wins the day. https://t.co/hMeheZDA7w
— Ian Rapoport (@RapSheet) March 19, 2020
CelebNSports247.com reports that A’Shawn Robinson just announced that he will be signing with Los Angeles Rams on a two-year contract worth $17 million.
Following the Los Angeles Rams posted that A’Shawn Robinson had signed a 2-year deal with the Rams on Wednesday night
Tom Pelissero with NFL.com tweeted that Robinson would be signing a two-year, $17 million deal with the Rams:
The #Rams are signing former #Lions DT A’Shawn Robinson to a two-year, $17 million deal, his agent, Sean Kiernan of @SelectSports, tells me and @RapSheet.
— Tom Pelissero (@TomPelissero) March 19, 2020
Robinson played in all 16 regular-season games during his first two years in the league, which included starting all 16 games and recording a career-high 53 tackles in 2017. He was limited to 13 games (seven starts) both in 2018 and 2019.
Overall, Robinson recorded 5.0 sacks, three forced fumbles, 16 passes defended and 172 tackles (120 solo) as a Lion.
The Rams defensive line will look different in 2020 with outside linebacker Dante Fowler Jr. departing for the Atlanta Falcons and defensive end Michael Brockers set to sign with the Baltimore Ravens. However three-time AP NFL Defensive Player of the Year Aaron Donald will still anchor the unit in the trenches. – Bleacher Report
