Steelers outside linebacker Bud Dupree was named AFC Defensive Player of the Week after his dominating week 8 performance against the Colts.
In the Steelers 26-24 home win, Dupree filled the stat sheet with two sacks, a forced fumble, a fumble recovery, three quarterback hits and three solo tackles.
1️⃣ forced fumble
1️⃣ fumble recovery
2️⃣.0️⃣ sacks@Bud_Dupree has been named the AFC Defensive Player of the Week for his performance vs. the Colts!
— Pittsburgh Steelers (@steelers) November 6, 2019
Dupree is having his most productive year of his career in 2019. Through his first eight games, Dupree has racked up eight tackles for loss, six sacks, two forced fumbles, and nine quarterback hits.
The Steelers management gave Dupree another chance by picking up his fifth-year option a year ago after his productive 2017 campaign. This year is the final year before Dupree will be looking for a new contract.
After having only 5.5 sacks all of last season, the talented but inconsistent linebacker is clearly playing for a new contract when free agency comes around. Being healthy for the first time in a few years has been beneficial to the energized play that Dupree has showcased during the first half of this 2019 season.
Dupree has done well causing havoc for opposing quarterbacks all season long and has also stepped up his tackling in the running game and that played a role in him being named AFC DPOY. His tackle on 3rd-down forced the Colts to attempt a field goal instead of running the clock down, a field goal that Vinatieri went on to miss.
Dupree and T.J. Watt have one of the best current outside linebacking tandems in the league and have been key in the Steelers getting back to .500 on the season–and will be even more important when it comes down to the team’s playoff push.