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Stephon Tuitt named AFC Defensive Player of the Week


(Photo by Joe Sargent/Getty Images)


Stephon Tuitt had a dominating performance against the Ravens on Sunday with nine total tackles (eight solos), two sacks, three tackles for a loss and three quarterback hits. It was another outstanding game by Tuitt as he continues to wreak havoc against opposing teams this year. The league took notice of his efforts in Baltimore, as well, as he was named AFC Defensive Player of the Week for Week 8 of the NFL season.


It’s the second time Tuitt has received a Defensive Player of the Week award in his career. The first time came in Week 11 of the 2016 season against the Browns when he had five total tackles (four solos), two sacks, two tackles for a loss and four quarterback hits.


Tuitt is the fourth player from the Steelers to be honored for a league award this season. Chase Claypool was named AFC Offensive Player of the Week for his four-touchdown performance against the Eagles in Week 5, T.J. Watt won AFC Defensive Player of the Week for his Week 2 performance against the Broncos and was named AFC Defensive Player of the Month for September.

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