毕业设计(8)

2019-04-01 16:42

当x=T时 b?'''?y??yT?12?MT3??10.57?Kpa?

当x=0时 ?y?a?bx?1370.78?Kpa?

1当x?T时 ?y?a?bx?1232.31?Kpa?

32当x?T时 ?y?a?bx?1093.85?Kpa?

3当x=T时 ?y?a?bx?955.38?Kpa? 2、坝内剪应力?x

?x?a1?b1x?c1x2

a1=?''=1006.46(Kpa)

?6?P1?'''?b1???2??4?????135.98?Kpa? T?T?C1??6?P1?'''???2.81?Kpa??3??3?2? ?T?T?当x=0时 ?x?a1?b1x?c1x2?1006.46?Kpa?

1当x?T时 ?x?a1?b1x?c1x2??292.66?Kpa?

32当x?T时 ?x?a1?b1x?c1x2??627.32?Kpa?

3当x?T时 ?x?a1?b1x?c1x2?2.47?Kpa? 3、坝内铅直面上的正应力?x

?x?a2?b2x?C2x2?d2x2

对于特定的水平截面a2、b2、c2、d2均为常数,可由边界条件和平衡条件求得,但计算较为复杂。实际上,?x的三次曲线分布与直线相当接近。 所以:?x?a3?b3x

''a3??y?917.87?Kpa?

36

b3?'''?x??xB??4.41

当x=0时 ?x?a3?b3x?917.87?Kpa?

1当x?T时 ?x?a3?b3x?860.1?Kpa?

32当x?T时 ?x?a3?b3x?802.33?Kpa?

3当x?T时 ?x?a3?b3x?744.56?Kpa? 4、坝体内主应力?1、?2

坝内主应力简图

1?2??x??y2??????x??y?222?????

将坝底四等分,分别计算各等分点的主应力

(1)A点的应力X=0m时

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?1??x??y2??x??y???2???x??y???2??2????2175.95?Kpa? ???2?????112.71?Kpa? ?22?2??x??y2(2)B点的应力X=13.1m时

?1??x??y2??x??y???2???x??y???2???x??y???2???x??y???2???x??y???2???x??y???2??2????1393.03?Kpa? ???2????699.39?Kpa? ???2????1194.26?Kpa? ???2????701.92?Kpa? ???2????955.41?Kpa? ???2????744.53?Kpa? ??22?2??x??y22(3)C点的应力X=26.2m时

?1??x??y22?2??x??y22(4)D点的应力X=39.3m时

?1??x??y22?2??x??y2所以,经过计算,在正常蓄水位时坝体应力没有出现拉应力,满足应力要求。

(二)校核洪水位时 ?M??14311.27?KN?m?T=39.3m

?W?38577.03?KN?

?P?12492.02?KN?

1、坝内水平截面上的正应力

?y?a?bx

''?1537.57?Kpa? 当x=0时 a=?y当X=T时 b?'''?y??yT?12?MT3??28.29?Kpa?

38

当X=0时 ?y?a?bx?1537.57?Kpa?

1当X?T时 ?y?a?bx?1166.97?Kpa?

32当X?T时 ?y?a?bx?796.37?Kpa?

3当X?T时 ?y?a?bx?425.77?Kpa? 2、坝内剪应力?x

?x?a1?b1x?c1x2

a1=?''=1133.62(Kpa)

b1??6P1????2?'?4?''???163.91?Kpa? ?T??T??6?P1?'''?C1?2??3??3????3.44?Kpa? T?T?当X=0时 ?x?a1?b1x?c1x2?1133.62?Kpa?

1当X?T时 ?x?a1?b1x?c1x2??423.26?Kpa?

32当X?T时 ?x?a1?b1x?c1x2??799.47?Kpa?

3当X?T时 ?x?a1?b1x?c1x2?5.01?Kpa? 3、坝体内铅直面上的正应力?x

?x?a2?b2x?C2x2?d2x2

对于特定的水平截面a2、b2、c2、d2均为常数,可由边界条件和平衡条件求得,但计算较为复杂。实际上,?x的三次曲线分布与直线相当接近。 所以:?x?a3?b3x

''a3??y?1537.57?Kpa?

b3?'''?x??xB??5.00?Kpa?

当X=0时 ?x?a3?b3x?1537.57?Kpa?

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1当X?T时 ?x?a3?b3x?1472.07?Kpa?

32当X?T时 ?x?a3?b3x?1406.57?Kpa?

3当X?T时 ?x?a3?b3x?1341.07?Kpa?

4、坝体内主应力?1、?2

坝内主应力简图

1?2?

??x??y???2??2?????2?x??y2

将坝底四等分,分别计算各等分点的主应力 (1)A点的应力X=0m时

?1??2??x??y2??????x??y?222?????2444.47?Kpa? ??x??y2??x??y???2??2????120.55?Kpa? ??2(2)B点的应力X=13.1m时

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