Steelers' Stephon Tuitt named AFC defensive player of week

On the heels of a two-sack game that helped the Pittsburgh Steelers remain undefeated, veteran lineman Stephon Tuitt was named AFC defensive player of the week by the NFL on Wednesday.

Tuitt had nine tackles — eight solo, three for loss — during the Steelers’ 28-24 victory at the Baltimore Ravens that allowed the Steelers to open up a two-game lead on the reigning AFC North champions.

In seven games this season, Tuitt has six sacks, a half-sack off his career high. The seven-year veteran has six tackles for loss and 17 QB hits. Pro Football Focus rates Tuitt as the NFL’s 10th-best interior defensive lineman this season.

“Stephon is having a great year,” Steelers defensive lineman Cameron Heyward said. “He came in with the right mindset, not only in good physical condition, but his mental aspect, I thought he grew a lot. And he understands that the plays were going to come to him. He was always around the quarterbacks and always about to make plays, but now you’re seeing him finish plays, and I’m happy for him.

“But he’s got to continue to do it. We understand he’s a great talent for our defense, and I think T.J. (Watt), myself, Bud (Dupree) all benefit from him being out there.”

Defensive coordinator Keith Butler said the rotation of the defensive line has helped the Steelers work to shield Cameron Heyward and Stephon Tuitt from too much wear-and-tear early.https://t.co/SM6aekFUS8

— Tribune-ReviewSports (@TribSports) September 25, 2020

This is the third time among the seven games the Steelers have played that they have had a conference player of the week honor. Watt was defensive player of Week 2 and receiver Chase Claypool was honored as the AFC’s offensive player of the week after he had four touchdowns during a win against the Philadelphia Eagles on Oct. 11.

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