The Pittsburgh Steelers signed defensive tackle Larry Ogunjobi to a one-year contract last week, and the terms related to that deal are now finally known thanks to them hitting the NFLPA.
According to Joel Corry of CBS Sports, the base value of the Ogunjobi contract reportedly totals out at $8 million. The base salary is listed as $1.535 million, and Ogunjobi also received a $6.465 signing bonus as part of the one-year deal, according to Corry. That makes Ogunjobi’s 2022 salary cap charge an even $8 million.
Previous reports indicated that Ogunjobi’s deal had an “up to” amount, thus signifying that incentives were likely part of it. According to Corry, no incentives, either likely to be earned or not likely to be earned, are listed. We will keep monitoring this to see if any changes are made.
Ogunjobi was originally selected by the Cleveland Browns in the third round of the 2017 NFL Draft out of Charlotte, and he played his first four years in the NFL for them. Last season, Ogunjobi played for the Cinninciniti Bengals after signing a one-year, $6.2 million contract last March.
In 2021 with the Bengals, Ogunjobi played in 16 regular season games on his way to registering 49 total tackles, of which 12 resulted in lost yardage. He was credited with seven sacks last season and 16 total quarterback hits in 724 defensive snaps played.
In the Bengals playoff win against the Las Vegas Raiders, Ogunjobi suffered a foot injury that landed him on the team’s Reserve/Injured list. That injury required surgery.
This past March, Ogunjobi agreed to a three-year $40.5 million contract with the Chicago Bears, with $26.35 million reportedly guaranteed. That agreement was voided, however, as Ogunjobi reportedly failed his physical.