土木工程道路专业毕业设计---三级公路两阶段施工图设计

毕业设计

5.41:0.254°14.01.00.38.01:51:0.25

图6.1 挡土墙结构图

6.5.2 墙后土设计计算

1、主动土压力计算: 假设破裂面交于路基宽度内: 破裂角?:

ta?n??ta?n?(ta?n?co?t)(ta?n?A) (6.4) 故A??A0ab?2h0(b?d)?H(H?2a?2h0)tan? (6.5) ?B0(H?a)(H?a?2h0)不计车辆荷载,h0?0

ab?2h0(b?d)?H(H?2a)tan?5.4?8.1?9.3?(9.3?2?5.4)tan(?14.04?)A???0.42 22(H?a)(9.3?5.4)????????35??14.04??23??43.96??90?

故:

tan???tan??(tan??cot?)(tan??A)

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毕业设计

??tan43.96??(tan43.96??cot35?)(tan43.96??0.419)?0.855 即 tan??0.855 , ??40.52? 则 B'?(H?a)tan??Htan??b

?14.7?tan40.52??8?tan(?14.04?)?8.1?2.469m?8.5m

与假设符合。

2、等代均布土层厚换算:

(1)汽车-超20级一辆重车的扩散长度L'为

L'?l?(h?2a)tan30??13?(8?10.8)?tan30??23.85m?20m 取 L'?20m

又,在B'范围内布置车轮的宽度B0为 B0?1.652?0.75?(0.5?0.3)?0.702m 只可布置一侧车轮

故在B'?L'面积内布置的轮重为半辆重车,即 ?G?550?275kN 2得等代土层厚h0: h0??B'L'?G?275?0.315m

17.658?2.469?20(2)查《公路路基设计规范》H?10m,q?10kN/m3,??17.658KN/m3 得等代土层厚h0: h0?q??10?0.566m

17.658取两者之间较大值,故h0?0.566m 根据求得的h0重新计算破裂角?: A??A0ab?2h0(b?d)?H(H?2a?2h0)tan?? B0(H?a)(H?a?2h0) 57

毕业设计

?5.4?8.?1?2?0.5?66?(8.1?0.7?5)9?.3?(9.?3?2a5n.41?4.2040.566) (9.?35.4)?(9.3?5?.420.566)t?0.442????????35??14.04??23??43.96??90?

故:

tan???tan??(tan??cot?)(tan??A)

??tan43.96??(tan43.96??cot35?)(tan43.96??0.442)?0.87 故 tan??0.870 , ??41.02? 3、计算土压力系数:

cos(???)cos(41.02??35?)???K?(tan??tan?)??tan41.02?tan(?14.04)??0.150 ????sin(???)sin(41.02?43.96)b?atan?8.1?5.4?tan41.02?又 h3???5.487m ??tan??tan?tan41.02?tan(?14.04)

h4?H?h3?h1?8?5.487?0.806?1.707m K1?1?

K1?1?得墙背土压力为

1 Ea??H2KK1

2h2hh2a(1?3)?024 H2HH2?5.45.4872?0.566?1.707(1?)??1.841 29.32?9.39.31 Ea??17.658?9.32?0.150?1.707?195.52kN

2 则 Eax?Eacos(???)?195.52?cos(?14.04??23?)?193.14kN Eay?Easin(???)?195.52?sin(?14.04??23?)?30.45kN 对O点的力臂为

E的作用点距墙角的竖直距离:

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毕业设计

Ha(H?h3)2?h0h4(3h4?2H) Zax?? 233HK19.35.?4? Zax?3(?9.325.?487)?0.5?66?1.7?07?(31.70729.3)?3.24m8 23?9.?31.707E的作用点距墙角的水平距离:

Zay?B?Zaxtan??3.675?2.776?tan(?14.04?)?4.487m 4、 墙重计算及对O点的力臂计算 W1?2.5?8.0?21.582?431.64kN Z1?1.5?2.58??0.25?3.75m 22 W2?2.175?1.3?21.582?61.023kN Z2?1.5?2.175?1.588m 21 W3??0.325?1.3?21.582?4.559kN

2 Z3?1.5?2.175?0.325?3.783m 31 W4??0.3?1.5?21.582?4.856kN

22 Z4??1.5?1m

3 W5?1?1.5?21.582?32.373kN Z5?0.75m

6.5.3验算

1、滑动稳定性验算

基底合力的法向分力N和切向分力T为 N?Eay?W1?W2?W3?W4?W5

?564.901kN

T?Eax?193.14kN

得沿基底的抗滑稳定系数为: 取 f?0.3

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毕业设计

Kc?Nf564.901?0.3??0.877 T193.14因 Kc?1.3,不满足要求, 需改善地基,

地基处理:在粘性地基内夯打碎石以增大地基摩擦系数,取f?0.5 Kc?Nf564.901?0.5??1.46 T193.14因 Kc?1.3,满足要求 2、

倾覆稳定性验算

挡土墙上作用的竖直力和水平力对墙趾O点的力矩分别为

?My?Eay?Zay?W1Z1?W2Z2?W3Z3?W4Z4?W5Z5

?1959.59kN?Mx?Eax?Zax?627.32kN

得抗倾覆稳定性系数 K0M???Myx?1959.59?3.12?1.5

627.32 符合要求 3、

基底应力验算,由式子

e?BMy??Mx?ZN?, ZN? 2N

得基底合力的合力偏心距:

B?My??Mx3.6751959.59?627.32B???0.521m?(?0.613m) e??2N2564.9016又 e?

B

,得基底最大压应力为: 6

B6e(1?) NB ?max?564.9016?0.521?(1?) 3.6753.675

?284.47kPa?因?max?284.47kPa???0??245kPa

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