衡重式挡土墙课程设计
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衡重式挡土墙课程设计
1、设计资料
(1)地形
中浅切割褶皱剥蚀中低山地区,地势起伏较大。
(2)工程地质条件
自上而下土层依次如下:
①号土层:耕植土,厚约0.9m ,黄褐色,含大量有机质。
②号土层,中风化砾岩,厚度未揭露,地基容许承载力R=800kPa 。
(3k j (4(521:0.273(1'1tan =0
.3133.00.3tan 1111 H d H =0.67 则'1 = 82.33
假想破裂面交于荷载内,按表5.6第一类公式计算,得
验核破裂面位置如下:第一破裂面距墙顶内缘距离为
)tan (tan '11 i H =3.0×(0.521+0.67)
=3.573m <0b =5.5m
破裂面交于荷载内,与假设相符,而且也不可能出现其他情况,故采用此类计算公式。因为i <'1 ,故出现第二破裂面。
(2)计算第二破裂面上的主动土压力1E
有表5.6第一类公式,有 K=)2
45cos()245(tan 2
=)
5.1745cos()5.1745(tan 2 =0.587 1K =1+1
02H h =1+0.355.02 =1.367 1E =122
1KK H =0.5×18×3.0×3.0×0.587×1.367=64.98kN
(34(122 =235061435' ='3666
tan 2 =-tan +))(tan tan (cot A
=-2.311+)27.0311.2)(311.2428.1(
=0.795
2 = 48.38
验核破裂面位置:堤顶破裂面距墙顶内缘的距离为
=3×(0.67+0.521)+5.1×(0.795-0.25)
=3.573+2.779
=6.353m >0b =5.5m
则破裂面位于荷载外,与假设不符。应采用5.7表中第四类公式,
0'0b b -)tan (tan '11i H =5.5-3.573=1.93m A=-21220'0tan )
2(2 H H H h b =-25.0)321.5(1.555.093.12 =0.21 22 ='3666
tan 2 =-tan +))(tan tan (cot A =-2.311+)21.0311.2)(311.2428.1(
=0.759
2 ='1237 验核破裂面位置:堤顶破裂面距墙顶内缘的距离为
=3×(0.67+0.521)+5.1×(0.759-0.25)
=3.573+2.596
=6.169>0b =5.5m
则破裂面位于荷载外,与假设相符。
(2)计算土压力2E
采用5.7表中第四类公式,得
土压力系数:
K=)tan (tan )
sin()cos(2222 =)25.048.38(tan 08
.105sin 48.73cos
=0.16
25
.0759.093.1tan tan 22'02 b h =3.54m 22
20211221H h h H H K 01.2654.355.021.561 2.326 主动土压力:
E =21KK H =0.5×18×5.1×5.1×0.16×2.326=87.12kN x E 22E cos (22 )=87.12×cos ('30171.15 )=87.04kN
y E 22E sin (22 )=87.12×sin (''30170615 )=3.65kN
压力作用点:
=326
.21.51.53)1.5454.33(54.355.0326.23331.5 =2.23m
222tan y x Z B Z =1.53+2.23×0.25=2.09m
5、墙身截面计算
墙身截面计算如下图所示:
(1)墙身重及力臂
①上墙墙身重1G 为
1G =)(2
1211b b H k =0.5×22×3×(0.5+1.65)=70.95kN 对墙趾的力臂1G Z 为
1G Z =)
(3)2(211212
221212b b nH b b b b b b nH =305.0)65.11(65.165.15.025.01.505.02 G ④下墙墙身重2G 为
2G =)(2
1122B d b H k =0.5×22×5.1×(1.65+1+1.53) =234.5kN
对墙趾的力臂为
2G Z =)]
([3])(2[])()([21212212122b d B nH B d b d b d b B B =)
65.253.1(31.505.0]53.165.22[]65.2)165.1(53.153.1[22
=1.21m
(2)滑动稳定验算 =04
.8703.306.0)65.364.5743.41.395.23495.70( =2.06>1.3
所以满足抗滑稳定性要求。
(3)倾覆稳定性验算 =09
.204.87)1.5134.1(03.3023.265.306.264.5713.143.481.11.3921.15.23491.095.70 =1.5≥1.5
(4( =
10
95.70103.3032.110)1.505.091.0(95.70265.1 =0.46m
1e <232.0b (=0.53),且1e >6
2b (=0.28)。 =65
.146.061(65.11095.70 =-32.87kN <l =100a kP