PITTSBURGH (KDKA) — After a major performance Sunday against the Colts, Bud Dupree has been named AFC Defensive Player of the Week.
The team and league made the announcement Wednesday morning.
“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!”
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!MORE: https://t.co/Ic7VB2W78v pic.twitter.com/GUreu7WYvR
— Pittsburgh Steelers (@steelers) November 6, 2019
Dupree stepped up his game against the Colts, forcing and recovering a fumble and accruing two sacks.
The outside linebacker is in a contract year and playing for a big payday in the offseason. He has played in all eight games this year, and has 32 tackles, six sacks, nine QB hits, two forced fumbles and two pass deflections.