Former Cleveland Browns safety Juan Thornhill reached an agreement with the Pittsburgh Steelers on a one-year contract Monday night.
The deal was initially reported by Ian Rapoport of NFL Network on social media. Thornhill had been deciding between the Steelers and the San Francisco 49ers, whom he visited last Thursday. Ultimately, Thornhill felt the Steelers offered him the best opportunity.
Thornhill was drafted by the Kansas City Chiefs in the second round of the 2019 NFL Draft out of Virginia. He played four seasons with the Chiefs before becoming an unrestricted free agent.
He then signed a three-year, $21 million deal with the Browns. However, he was released by Cleveland in February, heading into the final year of his contract, which was worth $7 million. It’s believed he was let go due to injury concerns, as he missed six games in each of the last two seasons. Additionally, his high cap hit for a player with that injury history may have been a factor.
Thornhill played in 11 games for the Browns in 2024, recording 49 tackles and three passes defended. Over his career, he’s accumulated eight interceptions, including a touchdown return in his rookie season. He has appeared in 87 games and started 74 of them.
Thornhill is expected to back up Minkah Fitzpatrick at free safety and contribute in nickel packages. DeShon Elliott is likely to remain the starter at strong safety after a strong performance last season. By joining the Steelers, Thornhill will have two opportunities each year to show the Browns they made a mistake in releasing him, as the Browns and Steelers, both AFC North teams, play twice annually.
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