The accolades continue to pile up for T.J. Watt and his outstanding 2021 season, this time culminating with the most notable award of the NFL’s Defensive Player of the Year. On Thursday night at the NFL awards ceremony, it was announced Watt was selected the choice for AP Defensive Player of the Year in his third-straight year as a finalist.
T.J. WATT IS THE AP DEFENSIVE PLAYER OF THE YEAR!#NFLHonors | : https://t.co/xuborBgBLd pic.twitter.com/qVlrFW91FZ— Pittsburgh Steelers (@steelers) February 11, 2022
T.J. Watt brings home the award in 2021 after tying the NFL single-season sack record of 22.5. Watt also led the NFL with 21 tackles for loss and three fumble recoveries and also had five force fumbles and seven passes defensed on the season.
With Watt bringing home the Defensive Player of the Year, it marks the eighth time it was won by a member of the Pittsburgh Steelers by seven different players. Awarded every year since 1971, the first recipient by the Steelers was Joe Greene in 1972. Greene won the award again in 1974 and is the only member of the Steelers to win the ward twice. The other Steelers who have been selected are Mel Blount (1975), Jack Lambert (1976), Rod Woodson (1993), James Harrison (2008), and Troy Polamalu (2010).
T.J. Watt had previously been selected First-Team All-Pro for 2021 along with the 101 Awards AFC Defensive Player of the Year as well as Defensive Player of the Year by the Pro Football Writers of America. Watt was also selected the Steelers team Most Valuable Player for the third-straight season.
On behalf of BTSC, congratulations to T.J. Watt on his selection as the NFL’s Defensive Player of the Year for 2021.