tg2(45o??2)tan2(450?3402)K???0.602 o00cos(45??2)cos(45?342)K1?1?2h02?0.29?1??1.242 H12.411E1??H12KK1??18?2.42?0.602?1.242?38.76KN
22E1X?E1cos(?i??)?38.76?cos(280?340)?18.20KN
E1y?E1sin(?i??)?38.76?sin(280?340)?34.22KN
对上墙O1的力臂为
Z1x?H1h02.40.29????0.88m 33K133?1.2423. 下墙土压力计算 （1）求破裂角?2 假定破裂面交于
荷载外，采用表27-6中第五类公式计算。破裂角计算示意图如图：
则有：
?????2??2?340?170?14002'?65002'tan??tan6502'?2.148
0
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11A0?(H1?H2)2??(2.4?3.6)2?18
2211/H2(H2?2H1)tg?2?H12tg?i??b?d?H(tg??tg?i)?h0011??2211??3.6?(3.6?2?2.4)?(?0.25)??2.42?0.532?[5.5?0?2.4?(0.725?0.532)]?0.2922??2.97B0? R1?E1x18.20??38.6KN
cos(???i)cos(340?280)tg?2??tg??(tg??ctg?)(tg??0B0Rsin(???i))?1A0A0?sin?cos??2.9738.76?sin(340?280) ??2.148?(2.148?cot34)(2.148?)?001818?18?sin34?cos6502'?0.450?2?24015'现验算破裂面位置如下：
破裂面顶端至墙顶内缘的距离为：
Htg?2?H2tg?2?H1tg?1/?6?0.45?3.6?0.25?2.4?0.725?3.54m
比较b0?5.5m?3.54m，故分析与原假设不相符，即破裂面交于荷载内。
计算图示如下：
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?????2??2?340?170?14002'?36058'
tan??0.753A??tan?2?0.25
tg?2??tg??(tg??ctg?)(tg??A)??0.753?(0.753?cot340)(0.753?0.25)?0.744
?2?36040'现验算破裂面位置如下：
破裂面顶端至墙顶内缘的距离为：
Htg?2?H2tg?2?H1tg?1/?6?0.744?3.6?0.25?2.4?0.725?5.30m
比较b0?5.5m?5.30m，故破裂面交于荷载内。 （2） 计算土压力E2
cos(?2??)cos(36040'?340)K?(tan?2?tan?2)??(0.744?0.25)?0.170 00sin(?2??)sin(3640'?3658')K1?1?2(H1?h0)2?(2.4?0.29)?1??2.494 H23.6112E2??H2KK1??18?3.62?0.170?2.494?49.45KN
22E2x?E2cos(?2??2)?49.45?cos(?14002'?170)?49.38KN
E2y?E2sin(?2??2)?49.45?sin(?14002'?170)?2.56KN
Z2x?H2H1?h03.62.4?0.29????1.56m 33K133?2.4944. 墙身截面计算
因上墙顶宽b1?0.5m，则上墙底宽b2?1.46m，下墙底宽B?1.64m。 （1） 计算墙的重力及力臂
①上墙墙身自重G1
11G1??KH1(b1?b2)??22?2.4?(0.5?1.46)?51.74KN
22
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对墙趾的力臂
2b12?b1b2?b2?(2b1?b2)nH1ZG1?nH2?3(b1?b2)0.52?0.5?1.46?1.462?(2?0.5?1.46)?0.05?2.4?0.05?3.6?
3?(0.5?1.46)?0.76mn取0.05
②下墙墙身自重G2
11G2??KH2?b2?d1?B???22?3.6?(1.46?0.9?1.64)?116.98KN
22对墙趾的力臂
ZG2[B2?B?b2?d1???b2?d1?]??2?b2?d1??B?nH2???3??B??b2?d1???2[1.642?1.64?(1.64?0.9)?(1.46?0.9)2]?[2?(1.46?0.9)?1.64]?0.05?3.6 ?3?[1.64?(1.46?0.9)]?1.11m③第二破裂面与墙背之间的土楔重G3
11G3??H1?d1?d2???18?2.4?(0.9?0.46)?29.38KN
22g/1??tig?(0.7?25其中 d2?H1?t???2.40?.532) m0.46对墙趾的力臂
ZG3?n?H1?H2??b1?H1tg?1?2d2?d1d2?d12??2d2?d1?H1tg?13?d2?d1?0.462?0.9?0.46?0.92?(2?0.46?0.9)?2.4?0.35?0.05?(2.4?3.6)?0.5?2.4?0.35?3?(0.46?0.9)?1.99m ④土楔上荷载重G4
G4??h0d2?18?0.29?0.46?2.40KN
对墙趾的力臂
11ZG4?n?H1?H2??b1?d1?0.05?(2.4?3.6)?0.5??0.9?1.25m
22（2） 抗滑稳定性验算
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KC???G1?G2?G3?G4?E1Y?E2Y?fE1X?E2X(51.74?116.98?29.38?2.40?34.22?2.56)?0.6237.28?0.6??2.1118.20?49.3867.58
Kc?2.11?[Kc]?1.3，故挡土墙抗滑稳定性满足要求。
（3） 抗倾覆稳定性验算
Z1Y?nH2?b2?d1?Z1Xtg?i?0.05?3.6?1.46?0.9?0.88?0.532?2.07m Z2Y?B?Z2Xtg?2?1.64?1.56?(?0.25)?1.25m
K0???M?MYX?G1ZG1?G2ZG2?G3ZG3?G4ZG4?E1YZ1Y?E2YZ2YE1X?Z1X?H2??E2XZ2X51.74?0.76?116.98?1.11?29.38?1.99?2.40?1.25?34.22?2.07?2.56?1.2518.20?(0.88?3.6)?49.38?1.56277.58??1.75158.57 K0?1.75?[K0]?1.5，因此该挡土墙的抗倾覆稳定性满足要求。 （4） 基底应力计算
ZNe?M??M???(G?E)yyx?277.58?158.57?0.50m237.28B1.64?ZN??0.50?0.32m22
e0?0.75?B1.64?0.75??0.21m?e?0.32m 66即偏心距大于容许偏心距，基底出现拉应力，由于土地基不能承受拉力，则基底发生应力重分布。于是：
?max?2?N3ZN2??G?EY?2237.28?????316.37Kpa 3ZN30.50则?max?316.37Kpa?[?0]?800Kpa，故地基承载力满足要求。
（5） 截面应力验算
①求上墙实际墙背上的土压力E1'
E1'x?E1x?18.20KNE?Etan?1?18.20?0.35?6.37KN
'1y'1x
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