重力式挡土墙算例wwww
某二级公路,路基宽8.5m,拟
一段路堤挡土墙,进行稳定性验算。
一、 计算资料
1(墙身构造:拟采用浆砌片石重力式路堤墙,见下图。墙高H=6m,填土高a=2m,填土边坡1:1.5(),墙背俯,,33:41,
,,18:26,斜,倾角(1:0.33),墙身分段长度10m,初拟墙顶宽b=0.94m,墙底宽B=2.92m。 1
2(车辆荷载:二级荷载
33(填料:砂土,容重,计算内摩擦角,填料,,35:,,18KN/m
,,与墙背的摩擦角。 ,2
4(地基情况:中密砾石土,地基承载力抗力,基底f,500KPa摩擦系数。 ,,0.5
35(墙身材料:10#砌浆片石,砌体容重,容许压应,,22KN/ma
力[,],1250KP,容许剪应力[],175KP ,aaa
β
α
θ
二、计算公式
采用路堤墙,破裂面交于荷载内的主动土压力计算公式
tg,,,tg,,(ctg,,tg,)(tg,,A),,,,,,,,
ab,2h(b,d),H(H,2a,2h)tg,00A, (H,a)(H,a,2h)0
12EHKK,,E,Esin(,,,)E,Ecos(,,,) ,, 1yx2
,,2ah2hhcos(,)103K,1,(1,), K,(tg,,tg,), 12H2HHsin(,),,
,b,atgdh,,h,,h,H,h,h 12312,,,,tg,tgtg,tg
2a(H,h),hh(3h,2H)H1033Z,,,Z,B,Ztg, xyx233HK1
三、计算与验算
1(车辆荷载换算
,,2mH,10m 当时,;当时, q,20.0KPq,10KPaa
6,2,, 由直线内插法得:H=6m时, ,,q,,20,10,10,15KP,,a10,2,,
q15 换算均布土层厚度: h,,,0.83m0r18
2(主动土压力计算(假设破裂面交于荷载中部)
, (1)破裂角
,由,18:26,,,35:,,,17:30, ,,,2
得: ,,,,,,,,35:,18:26,,17:30,,70:56,
ab,2h(b,d),H(H,2a,2h)tg,00A,(H,a)(H,a,2h)0
2,2,1.5,2,0.83(3,0.5),(66,4,2,0.83)tg18:26,, (6,2)(6,2,2,0.83)
11.81,23.318,,,0.14977.28
tg,,tg,(ctg,tg)(tg,A),,,,,
,,tg70:56,,(ctg35:,tg70:56,)(tg70:56,,0.149)
,,2.893,(1.428,2.893)(2.893,0.149),,2.893,3.443
,0.55
,,28.81:,28:49,
验核破裂面位置:
路堤破裂面距路基内侧水平距离:
(H,a)tg,,Htg,,b,(6,2),0.55,6,0.33,3,3.4m 荷载外边缘距路基内侧水平距离: 5.5+0.5=6m 因为:3.4〈6,所以破裂面交于荷载内,假设成立
, (2)主动土压力系数K和 1
,b,atg3,2,0.55h,,,2.1521tg,,tg,0.55,tg18:26,
d0.5 h,,,0.5662,,tg,tg0.55,tg18:26,
h,H,h,h,6,2.152,0.566,3.282312
,,cos(,)cos(28:49,,35:)K,(tg,tg),(0.55,tg18:26),, sin(,)sin(28:49,,70:56,),,
,0.395
2hhh2a42.1522,0.83,3.282031K,1,(1,),,1,(1,),122 H2H612H6
,1,0.547,0.151,1.698
(3)求主动土压力及土压力的作用点 ,a
1122,,,,KK,,18,6,0.395,1.698,217.31KN a122
E,,cos(,,,),217.31,cos(18:26,,17:30,),175.96KNxa
E,,sin(,,,),217.31,sin(18:26,,17:30,),127.53KN ya
2a(H,h),hh(3h,2H)H1033Z,,x233HK1
262(6,2.152),0.83,3.282(3,3.282,2,6),, 233,6,1.698
,2,0.129,2.13m
Z,B,Ztg,,2.92,2.13,tg18:26,,2.22m yx
Z因基底倾斜,土压力对墙趾的力臂改为: x
,Zx,Zx,B,tan,,2.13,2.92,0.2,1.55m 0
3(稳定性验算
一般情况下,挡土墙的抗倾覆稳定性较容易满足,墙身
断面尺寸主要由抗滑稳定性和基底承载力来控制,故选择
基底倾斜1:5() ,,11:18,0
(1)计算墙身重G及力臂(取墙长1m计) ,,
11G,A,[(b,B)H,B,Btg],,1,,,a10a22
1 ,[3,(0.94,2.92),,2.92,2.92,tg11:18,],222
,(11.58,0.85),22,236KN
由力矩平衡原理:
,bBbB11,,,[,,,(,),,],, GAAbA G1213a233
11A,Bh (其中:,A,H(B,b),) A,bH21311122
则:
236,,,G
0.9412.92,0.94[0.94,6,,,6,(2.92,0.94),(0.94,) 223
12.92,,2.92,2.92,tg11:18,,],2223
,,1.05mG
(2)抗滑稳定性验算
, [1.1G,,(,,Etan,],,(1.1G,,E)tan,,,,Q1yx0Q0Q1x1y,,1.4,,,0.5() Q1
,,,,,[1.1G,(,,Etan)],(1.1G,E)tanQ1yx0y0Q1
,[1.1,236,1.4,(127.53,175.96,0.2],0.5
,(1.1,236,1.4,127.53),tg11:18,
,331.34
,,,1.4,175.96,246.34 Q1X
331.34 〉246.34,故满足抗滑稳定性的方程。
,,(N,Etan)x0抗滑稳定性系数Kc= E,Ntan,x0
GEE(,,tan),0.5,yx0 Kc=〉1.3 ,1.9EGE,(,),0.2xy
故满足抗滑稳定性的要求
(3)抗倾覆稳定性验算
0.8G,,,(,,,,,),0GyyxxQ1
G,0.8,,(,,,,,)GyyxxQ1
,0.8,236,1.05,1.4,(127.53,2.22,175.96,1.55),212.77,0
故满足抗倾覆稳定性的方程
GZ,EZGyy抗倾覆稳定性系数K= 0EZxx
236,1.05,127.53,2.22,1.94 K=〉1.5 0175,1.55
故满足抗倾覆稳定性的要求 4(基底应力与偏心距验算
Be,,, 1) 偏心荷载作用时,作用于基底的合力偏心距: 0N2
,,,GZ,EZ,EZ,,0yGyyxx,,, N,G,E,y
236,1.05,127.53,2.22,175.96,1.55,,0.71 236,127.53
2.92B,,,,,0.71,0.75e 0,22
B2.92B,,0.487e,因为,故,不满足要求。 0666
Be,由于,将墙趾加宽成宽0.8m,高0.8m的台阶后再06
进行验算。
'',GZ,EZ,EZGyXxy,,N2G,E,0.8,22y
2(236,0.8,22),(1.05,0.8),127.53,(2.22,0.8),175.96,1.55, 2236,127.53,0.8,22,1.50m
2.92,0.8Bm ,,,,,1.50,0.36e0,22
BB3.72因为m,故, ,,0.620e,666
故满足基底合力偏心距的要求。 2) 基底应力验算
NNee66dd,,,,1.0,,(1,),,(1,) ,,其中 maxminABAB当基底有倾斜时:
,,G,E dy
,,250.0,127.53,377.53KNd
377.536,0.36,,(1,),160.41Kp,[,],500KP maxa0a3.723.72
377.536,0.36,,(1,),42.56Kp,[,],500KP mina0a3.723.72
验算通过,所拟尺寸合理,可以使用。