衡重式挡土墙算例下载

1、本算例来自于:书 名 特种结构作 者 黄太华 袁健 成洁筠 出 版 社 中国电力出版社 书 号 5083-8990-5丛 书 普通高等教育“十一五”规划教材衡重式挡土墙算例某衡重式挡土墙,截面尺寸如图2.1所示。墙后填土水平即0b =o ,填土面均布荷载标准值210/k q kN m =,填土内摩擦角标准值30k j =o ,墙后填土及基底土重度均为318/kN m g =,无地下水,墙背很粗糙,排水良好。基底摩擦系数0.30m =,地基承载力特征值180ak f kPa =,地基承载力宽度修正系数为0.3b h =,深度修正系数为1.6d h =。该挡土墙处于非抗震区,采用M7.5水泥砂浆、

2、MU30毛石砌筑,墙身容重322/s kN m g =,挡土墙安全等级为二级。验算该挡土墙的稳定性、地基承载力及墙身承载力是否满足规范要求。i414133221121006000240036009009009003009001800300600300图2.1 衡重式挡土墙示意解:取每延米长度挡土墙作为研究对象。 1挡土墙自重及土压力计算0.9 1.8 1.8 2.7122(0.30.9 2.4 3.60.3 2.1248.5222s G V kN g +=´´+´+´-´´=; 重力G 作用线到墙趾的距离为m x 959.030.11

3、221.0888.3888.362.162.1041.01.23.0216.327.28.14.228.19.09.03.0/(31.21.23.02139.08.13.0(6.39.02128.13.0(6.38.12139.09.03.0(4.29.02129.03.0(4.29.023.09.03.00=-+=´´-´+´+´´´´-+´´´+´´´+´´´+´´+´´= 下墙按延

4、长墙背法的仰斜式挡土墙计算主动土压力。墙背很粗糙,排水良好,取2203k d j =o ,°-=-=04.146.39.0arctan a 。上述a 为墙背与竖直线的夹角,仅用于主动土压力系数的计算。当a 角用作其它用途时,取墙背与水平线的夹角,此时,9014.04104.04a =+=o o o 。作用于墙背的主动土压力系数为:22 2222cos (cos cos(1cos (3014.04 cos (14.04cos(14.042010.5170.2082.487a k j a a a d -=é+êë+=é-+êê&

5、#235;=o o o o o上墙假想墙背的倾角为:1 1.8arctan 36.872.4a =o ,若出现第二破裂面,第二破裂面与竖直线的夹角为45302ki j a =-=o o 。由于1i a a <,故上墙将出现第二破裂面,按第二破裂面理论计算土压力。第二破裂面与地面交点离墙顶的距离为:1.82.4tan 1.8 2.4tan 300.414i m a -=-=o 。第二破裂面与墙背间土体的重量为:0.4140.918 2.428.42s G kN +=´´=;该重力作用线到墙趾的水平距离为:mx s 05.2577.1324.0457.1206.02.14

6、.29.0414.0(2139.04.29.0213414.08.1414.0(4.2414.08.1(212414.04.2414.09.03.00=-+=´+´´´´-+´´-´+´´+= 上墙的主动土压力系数为: 2230tan (45tan (45220.66730cos(45cos(4522a k j j -=+o ooo o o; 上墙的土压力合力为: 2211( 1.1(18 2.410 2.40.66755.6222ak c a E h qh k kN y g =+=´

7、´´+´´=;上墙土压力合力作用点到墙踵的竖向距离为:221 2.4 2.418 2.410 2.42323.6 3.60.927 4.527118 2.410 2.42z m ´´´+´´=+=+=´´+´ 上墙土压力合力作用线与水平线的夹角为:303060i a j +=+=o o o 。 下墙土压力合力为:2211( 1.118 3.6(18 2.410 3.60.20870.5122ak c a E h qh k kN y g =+=´´´

8、;+´+´´=;下墙土压力合力作用点到墙踵的竖向距离为:221 3.6 3.618 3.6(18 2.410 3.6232 1.57118 3.6(18 2.410 3.62z m ´´´+´+´´=´´+´+´ 下墙土压力合力作用线与水平线的夹角为:14.0420 5.96a d +=-+=o o o 。 2抗倾覆稳定验算 上墙:sin(55.62sin 6048.17az a i E E kN a j =+=o ; cos(55.62cos6027.81ax a

9、 i E E kN a j =+=o ;3.00.927tan 30 2.465f x m =-=o ;4.5270.3 4.227f z m =-=;下墙:sin(70.51sin(104.042070.13ax a E E kN a d =-=-=o o ;cos(70.51cos(104.04207.32az a E E kN a d =-=-=o o ;cot 2.1 1.57cot104.04 2.493f x b z m a =-=-=o ; 0tan 1.570.3 1.27f z z b m a =-=-=;248.50.95928.4 2.0548.17 2.4657.32

10、2.493433.522.1 1.627.81 4.22770.13 1.27206.62t K ´+´+´+´=>´+´,抗倾覆稳定满足规范要求。 3抗滑移稳定验算00.3arctan 8.132.1a =o ;上墙:0cos(9055.62cos(9030308.1351.62an a i E E kN a d a =-=-=o o o o o ;0sin(9055.62sin(9030308.1320.72at a i E E kN a d a =-=-=o o o o o 。下墙:0cos(70.51cos(104.04

11、208.1317.17an a E E kN a d a =-=-=o o o ;0sin(70.51sin(104.04208.1368.39at a E E kN a d a =-=-=o o o ; 0(cos (248.528.4cos8.13274.12n s G G G kN a =+=+=o ; 0(sin (248.528.4sin8.1339.16t s G G G kN a =+=+=o ;(51.6217.17274.120.32.06 1.320.7268.3939.16s K +´=>+-,抗滑移满足规范要求。4地基承载力验算2.1433.52206.

12、62 1.050.820.232248.528.4e m -=-=-=+;0.070.35306b bm e m =<<=; ,max 6248.528.460.23(1(1218.52.1 2.1k k N e p kPa b b +´=+=´+=;( 3.0(0.51800 1.618(1.20.5200.2a ak b d f f b d kPa h g h g =+-+-=+´´-=;,max 218.5 1.2240.2k a p kPa f kPa =<=,地基承载力满足规范要求。5墙身承载力验算查砌体结构设计规范(GB50

13、003-2001得,0.69c f MPa =,0.07t f MPa =,0.19f MPa t =,墙体内部摩擦系数0.60f =。按规定应对错误!未找到引用源。中的1-1、2-2及3-3剖面进行墙身承载力验算,1-1剖面的墙身承载力验算方法参见例1或例2,3-3剖面的墙身承载力验算参考相关文献。本例选择2-2剖面进行墙身承载力验算。第二破裂面上的土压力水平分量为:1cos(55.62cos6027.81x ak i E E kN a j =+=o ;竖向分量为:kN E E i ak z 17.4860sin 62.55sin(1=°=+=j a10.927z m =;m z

14、b x f 452.14.29.0927.08.14.29.0111=´-=´-= 10.9 1.822 2.471.282G kN +=´´=; 010.910.90.9 2.40.9 2.4(0.92230.71(0.9 1.8 2.42x m ´´+´´´+=´+´ 10111111111 1.871.280.748.17 1.45227.810.9270.90.7870.1132271.2848.17z f x z G x E x E z b e mG E +-´+´-´=-=-=-=+;1110.060.30306b bm e m =<<=,说明墙背不会出现拉应力。 111max 11671.2848.1760.1131.35(1 1.35(1123.30.1230.691.8 1.8z c G E e kPa M