March 19 (UPI) -- The Los Angeles Rams have agreed to a contract with free agent defensive tackle A'Shawn Robinson.
Robinson announced the move with a video on social media. The Rams' deal with Robinson is for two years and $17 million, according to his agent at the Select Sports Group.
"Breaking news: I heard from a credible source that A'Shawn Robinson is going to sign with the Los Angeles Rams," Robinson tweeted. "Details to follow."
Robinson, 24, was a second-round pick by the Detroit Lions in the 2016 NFL Draft. The Alabama product had 40 total tackles, three tackles for a loss, three passes defensed, two quarterback hits, two fumble recoveries, 1.5 sacks and a forced fumble in 13 games last season for Detroit.
Robinson has missed six games in the last two seasons after appearing in all 16 games in each of his first two seasons.
The 6-foot-4, 330-pound defensive lineman has 172 tackles, 16 tackles for a loss, 16 passes defensed, 14 quarterback hits, five sacks, three forced fumbles, three fumble recoveries, an interception and a touchdown in 58 career games.