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Aaron Donald Repeats as NFL Defensive Player of Year

Aaron Donald Repeats as NFL Defensive Player of Year

Former Pitt and Penn Hills defensive tackle Aaron Donald has been named NFL Defensive Player of the Year for the second consecutive season by the Associated Press.

Donald, whose Los Angeles Rams will play in Super Bowl LIII against the New England Patriots on Sunday, finished the 2018 season with an NFL-best 20.5 sacks, four forced fumbles, two fumble recoveries, 263 tackles and 97 tackles for loss.

He becomes just the third player in the history of the award to repeat as winner, joining Lawrence Taylor (1981-82) and J.J. Watt (2014-15).

“Any time you put a body of work into it and then it pays off and you’re rewarded for it, it means a lot,” Donald said to the AP. “It’s a blessing.”

Donald was the only unanimous member fo the 2018 AP All-Pro team and had 45 votes for defensive player of the year compared to five for second-place Khalil Mack of the Chicago Bears.

In total, Donald has been selected to five Pro Bowl, has been named first-team All-Pro four times and was the NFL’s defensive rookie of the year in 2014.

In 2013 with Pitt, Donald won the Bronko Nagurski Trophy, the Chuck Bednarik Award, the Lombardi Award and the Outland Trophy and was a unanimous selection as All-American.

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