南昌工程学院本科毕业设计
上游坡1/3坝高水深H=26m ③α2=11 , α1=27m1:11:m2图5-4
③由图5-4可知：?2?11?,?1?27?, m 1 = 1.963 ,m 2 = 5.145.
W1?1?(19.33?96.5?4?52)??1036.67? 21W2??19.33?138??1333.77?
2?C?0,??3630?. 又已知：
土块BCDE的平衡式为：
P1?W1sin?1?1W1cos?1tan?1?0 K土体ADE的平衡式为：
11W2cos?2tan?2?P1sin(?1??2)tan?2?W2sin?2?P1cos(?1??2)?0 KKtan?因为 ?1??2??，令 ?f ，而
Kf?则：
A?BA?B2B?()?(?C) 22m21?m2m21?1.96325.145A?????3.998
m2?m1m15.145?1.9631.963B?W21333.77?A??3.998?5.143 W11.36.67?-40-
南昌工程学院本科毕业设计
C?1?m1m21?1.963?5.145??1.777
m1(m2?m1)1.963?(5.145?1.963) f?A?BA?B2B?()?(?C) 22m23.998?5.1433.998?5.14325.143?()?(?1.777) 225.145?0.315?tan?36?30???2.349 所以 K?f0.315
5.2.2上游坡死水位2796m
上游坡死水位2796m ①α2=12.6 ,α1=44.1m1:11:m2图5-5
①由图5-5可知：?2?12.6?,?1?44.1?, m 1 = 1.05 ,m 2 = 4.478.
W1?1?(22?28.6?4?31)??376.6? 21W2??22?206??2266?
2又已知： C?0,??36?30?. 土块BCDE的平衡式为：
P1?W1sin?1?1W1cos?1tan?1?0 K土体ADE的平衡式为：
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南昌工程学院本科毕业设计
11W2cos?2tan?2?P1sin(?1??2)tan?2?W2sin?2?P1cos(?1??2)?0 KKtan?因为 ?1??2??，令 ?f ，而
Kf?则：
A?BA?B2B?()?(?C) 22m21?m2m21?1.0524.478A?????2.616
m2?m1m14.478?1.051.05B?W22266?A??2.616?15.739 W1376.6?C?1?m1m21?1.05?4.478??1.584
m1(m2?m1)1.05?(4.478?1.05) f?A?BA?B2B?()?(?C) 22m22.616?15.7392.616?15.739215.739?()?(?1.584) 224.478?0.282?tan?36?30???2.62 所以 K?f0.282 上游坡死水位2796m 1: ②α2=12.6 ,α1=47.91:m2图5-6
②由图5-6可知：?2?12.6?,?1?47.9?, m 1 = 0.923 ,m 2 = 4.478.
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m1南昌工程学院本科毕业设计
W1?W2?1?22?28.6??314.6? 21?(22?65.4?65.4?46)??2223.6? 2?C?0,??3630?. 又已知：
土块BCDE的平衡式为：
P1?W1sin?1?1W1cos?1tan?1?0 K土体ADE的平衡式为：
11W2cos?2tan?2?P1sin(?1??2)tan?2?W2sin?2?P1cos(?1??2)?0 KKtan?因为 ?1??2??，令 ?f ，而
Kf?则：
A?BA?B2B?()?(?C) 22m21?m2m21?0.92324.478A?????2.527
m2?m1m14.478?0.9230.923B?W22223.6?A??2.527?17.863 W1314.6?C?1?m1m21?0.923?4.478??1.564
m1(m2?m1)0.923?(4.478?0.923) f?A?BA?B2B?()?(?C) 22m22.527?17.8632.527?17.863217.863?()?(?1.564) 224.478?0.276?tan?36?30???2.68 所以 K?f0.276-43-
南昌工程学院本科毕业设计
上游坡死水位2796m ③α2=13.5 ,α1=34.11:m11:m2图5-7
③由图5-7可知：?2?13.5?,?1?34.1?, m 1 = 1.523 ,m 2 = 4.174.
W1?1?(16.8?43.2?4?31)??424.88? 21W2??16.8?192??1612.8?
2又已知： C?0,??36?30?. 土块BCDE的平衡式为：
P1?W1sin?1?1W1cos?1tan?1?0 K土体ADE的平衡式为：
11W2cos?2tan?2?P1sin(?1??2)tan?2?W2sin?2?P1cos(?1??2)?0 KKtan?因为 ?1??2??，令 ?f ，而
Kf?则：
A?BA?B2B?()?(?C) 22m21?m2m21?1.52324.174A?????3.432
m2?m1m14.174?1.5231.523B?W21612.8?A??3.432?13.027 W1424.88?-44-
虞江心墙土石坝水利枢纽计算00002 - 图文
