扶壁式挡土墙计算实例.doc
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”规划教材扶壁式挡土墙算例某工程要求挡土高度为 8.3m，墙后地面均布荷载标准值按 qk =10 kN / m2 考虑，墙后填土为砂类土，填土的内摩擦角标准值 jk = 35 o，填土重度 g m =18 kN / m3 ，墙后填土水平，无地下水。地基为粘性土，孔隙比 e =0.786 ，液性指数 IL =0.245 ，地基承载力特征值 fak =230 kPa ，地基土重度 g=18.5kN / m3 。根据挡土墙所处的地理位置及墙高等因素综合考虑，选择采用扶壁式挡土墙，挡土墙安全等级为二级，试设计该挡土墙。解: IL = 0.245 <0.25 属坚硬粘性土，土对挡土墙基底的摩擦系数 m.(0.35,0.45) ，取m=0.35 。查规范得 hb =0.3 、hd =1.6 。1）主要尺寸的拟定为保证基础埋深大于 0.5m，取 d=0.7m，挡土墙总高 H=8.3m+d=9m。两扶壁净距ln 取挡墙高度的 1/31/4，可取 ln=3.00 2.25 m，取 ln=3.00m。用墙踵的竖直面作为假想墙背，计算得主动土压力系数2 jk 2 35 ° Ka = tan (45 °-) = tan (45 °-) = 0.271 22 根据抗滑移稳定要求，按式（3-6）计算得： 22 kaB2 + B3 3 1.3( qH + 0.5g H )K = 1.3(10 9 + 0.5189) 0.271 = 4.79 ，取 m(qk +g H) 0.35(10 +189) B2 + B3 =4.80m，其中 B2=0.30m，B3=4.50m。 22E = (qH + 0.5 g H )K = (10 9 + 0.5 18 9) 0.271 = 221.95 kN ax k a 91 910 9 +18 9 9 22 3z = 3.165 m 110 9 +18 9 9 2 1.6Ez 1B 3 -(B + B )12323()2kBBqHg+( ) ax1.6 221.95 3.165 1 = - 4.8 =-1.04m < 0 4.8(10 +189) 2 计算结果为负数说明若仅为了保证稳定性的要求不需设置墙趾板，但为了减少墙踵板配筋及使地基反力趋于均匀，取 B1=0.60m。 B = B1 + B2 + B3 = 0.6 + 0.3 + 4.5 =5.4m 。 B C图 3. 1挡土墙基本尺寸2）土压力计算由于填土表面水平，第一破裂面与铅垂面夹角 qi = 45°-jk = 45°-35°= 27.5 °；22 第二破裂面与铅垂面的夹角 ai = 45 °-jk = 45 °-35 °= 27.5° 。22 4.5墙顶 A与墙踵 C连线AC与铅垂面的夹角 a= arctan 8.7 = 27.35 °因为 a= 27.35°< ai = 27.5°，因此，不会在土体中出现第二破裂面，AC连线为实际破裂面。按库伦理论计算土压力。 a= 27.35 °，d=jk = 35°，b= 0°。 2cos ( jk -a)Ka = 2é sin( d +j )sin( j -b ) ù2 kkcos a×cos( d +a)1 ê+ úcos( d +a)cos( a -b ). . 2 °cos (35 °-27.35 ) = ù2 2 é °-0) sin(35 °+ 35 )sin(35 °° cos 27.35°× cos(35 °+ 27.35 ) 1 °ê+ ú°-0) cos(35 °+ 27.35 )cos(27.35 °° . . = 0.584 12 12E = q HK +g HK =1090.584 +189 0.584 = 478.3 kN aka a22 ax Ea ×cos( d +a) = 478.3cos(35 °+ °= 220.0kN E = 27.35 ) H 12 H 91 29qH × +g H × 109+189 k 223 223z = K =0.584 = 3.165mfEa a 478.3 az Ea ×sin( d +a) = 478.3sin(35 °+ °= 423.7kN E = 27.35 ) xf = 5.4 -3.165 tan 27.35 °=3.76 m 3）自重与填土重力a、立板+底板自重钢筋混凝土标准重度 gc = 25 kN / m3 ，其自重为： G1K = (0.3 5.4 + 0.3 8.7) 25 = 105.8 kN 5.4 0.3 0.3 5.4 + 0.3 8.7 (0.6 + )22x1 = 1.50 m 0.3 5.4 + 0.3 8.7 b填土重及地面均布荷载总量G2K = 1 BH 31g= 1 4.5 8.7 18 = 352.4 kN 22 4.5 x2 = 0.6 + 0.3 += 2.4 m 3 墙身自重计算时，扶壁自重按填土计算，另墙趾上少量填土重量略去不计导致各项验算稍偏安全。4）抗倾覆稳定验算稳定力矩：= Gx x + ExM zk 1k 1 + G2k 2 f = 105.8 1.5 + 352.4 2(az) .4+423.7 3.76 = 2597.6kN m × 倾覆力矩：M qk = Eax z f = 220.0 3.165 = 702.6kN m × Mzk 2597.6 Kl = = 3.7 >> 1.6 M 702.6 qk 满足抗倾覆稳定性要求。5）抗滑移稳定性竖向力之和： .N = G1k + G2k + Eaz =105.8 + 352.4 +423.7 = 881.9kN ×m 抗滑力： m.N = 881.9 0.35 = 308.7 kN 滑移力： Eax= 220.0kN m.N 308.7 Ks = =1.39 <1.3 Eax 220.0 满足抗滑移稳定性要求。6）地基承载力验算按正常使用极限状态下荷载效应的标准组合，各分项系数取 1.0。欲求基底偏心距e，先计算总竖向力到墙趾的距离：Mzk -Mqk 2597.6 -702.6 a = = 2.15 m. N 881.9 BB 5.4 B = 0.18 m < e =-a =-2.15 = 0.55 m <= 0.90 m 30 0226 地基承载力修正值计算：a = fak +h g (b -3) +h g dm (d -0.5) fb = 230 + 0.3 18.5 (5.4 -3)+1.6 18 (0.7 -0.5 ) = 249.1 kPa 地基压力计算：.N . 6e . pk . + 0max1=÷B è B . 881.9 . 6 0.55 . e > b由/ 30 ，于则 = 5.4 .è 1+ 5.4 ÷. = 263.1 kPa < 1.2 fa = 1.2 249.1 = 298.9 kPa pk max <1.2 fa 时必然满足 pk < fa 。 7）结构设计采用 C25混凝土和 HRB335级钢筋， fc =11.9N / mm2 ， ft =1.27 N / mm2 ， fy =300 N / mm 2 。立板设计 a土压力计算立板荷载均考虑墙后静止土压力。由于土体不存在滑动趋势，且墙面光滑，则静止土压力为水平方向。 s= K = 10 (1 -sin 35 ° ) = 10 0 .4 2 6 = 4 .2 6 kN / m 2 0 a qk 0 d 3.1m3.6mec b s 0 b = ( qk +g H 1) K 0 = (1 0 + 18 8.7 ) 0 .4 2 6 =7 0 .9 7 kN / m 2 图 3. 2立板荷载图 b. 将立板划分为上、下两部分，在离底板顶面 1.5Ln=1.5×3.00=4.50m高度以下的立板，可视为三边固定一边自由的双向板；离底板顶面 4500mm高度以上的部分视为沿高度方向划分为单位高的水平板带，以扶壁为支撑，按水平单向连续板计算。现以距墙底 4.55.5m（即距墙顶 4.23.2m）区间的立板为例进行计算。距离墙顶 03.2m区间的立板计算可参照进行。对于砂性土 K0 = 1-sin jk =1-sin 35 °= 0.426 s0 = (qk +g )0 = (10 +18 3.7) 0.426 = kN / m2zK 32.63 M = 0 n = 14.68 kN m 中 s 20 l 2 32.63 20 3.0 2 ×1.35 M中1.35 14.68 10 6 2As中 = 0.9 fh = 0.9 300 (300 -40) = 282.4 mm y 0 A 282.4 r =-中中bh s0= 1000 (300 -40) = 0.11% < 0.79% ，简化公式适用 ft 1.27 r= 0.0019 0.002 min max0.45 f = 0.45 300 = ,= 0.2% >r中 y 取 r中=rmin =0.2% ， As=r中bh0 = 0.002 1000 260 =520 mm 2 选用 12200 As= 565 mm2 s0ln 2 32.63 3.0 2 ×M支 =-12 =-12 =-24.47 kN m As =支支1.35 M = 1.35 24.47 10 6 = 470.6 mm 2 < Asmin = 520 mm 2 ，取 0.9 fh 0.9 300 (300 -40) y 0 As = Asmin =520 mm2 由于计算的配筋低于最小配筋，故简化公式适用。选用 12200 As= 565 mm2 s0ln 32.63 3.0 V = = 48.9 kN 22 1.35 V = 1.35 48.9 = 66.1 kN 0.7 f bh = 0.7 1.27 1000 260 = 231.1 10 3 N = 231.1 kN >1.35 Vt0 混凝土的抗剪能力满足要求，不需配置腹筋。 U型筋计算1.35 V 66.1 10 3 As = =220.3 mm2 fy 300 上式计算出的 As为每延米单肢 U型筋的截面积。配 10250， As=314 mm2 的 U型筋，开口方向朝向墙背，水平放置。距墙底 04.5m（即距墙顶 8.74.2m）区间按三边固定，上边缘自由的矩形板计算。 (qk +g )0 = (10 +18 4.2) 0.426 = 36.5 kN /2zK m b 2 1 0(qHg+) (10 18 8.7) 0.426 71.0 /kKkNm=+=图 3. 3立板较下部分荷载图可将荷载简化为一三角形荷载和一矩形荷载的叠加，查表计算。Mx = 0.0384 36.5 3.0 2 + 0.0184 (71.0 -36.5) 3.0 2 = 18.3 kN m × 1.35 Mx 1.35 18.3 10 62 2A = = 352.5 mm < A = 520 mm sx smin 0.9 fh 0.9 300 260 y 0 取 Asx = Asmin =520 mm2 ，由于计算的配筋低于最小配筋，故简化公式适用。选用 12200 Asx=565 mm2 My = 0.0113 36.5 3.0 2 + 0.0071 (71.0 -36.5) 3.0 2 = 5.9 kN m × 1.35 My 1.35 5.9 10 62 2A = = 113.8 mm < A = 520 mm sy smin 0.9 fh 0.9 300 260 y 0 取 Asy = Asmin =520 mm 2 ，由于计算的配筋低于最小配筋，故简化公式适用。选用 12200 Asy=565 mm 2 02 2Mx =-0.0794 36.5 3.0 -0.0387 (71.0 -36.5) 3.0 38.1 kN m × 6 0 =-1.35 Mx 0 1.35 38.1 10 2 2A = = 732.7 mm > A = 520 mm sx smin 0.9 fh 0.9 300 260 y 0 rx 0 = Asx 0 = 732.7 = 0.28% < 0.79% ，简化公式适用bh0 1000 (300 -40) 选用 12150 Asx0 =754 mm2 。 同理，可计算 My 0 、 Mxz 0 、 M0x 及其对应的 Asy0 、 A0 、 A0，配筋最大值应为 sxzsxAsx0 =732.7 mm2 ，其他配筋均小于最小配筋率。底板设计881.9 60.55 3p2pkp=4pmin =(1 -)5.4 5.4 p = p = 263.1 kN / m2 = 63.5 / 2kN m 1 k max 4.2图 3. 4底板基底应力分布图 p1 -p2 263.1 -63.5 2p3 = p1 - B1 = 263.1 - 0.6 = 240.9 kN / m B 5.4 p1 -p2 263.1 -63.5 2p = p - (B + B ) = 263.1 - 0.9 = 229.8 kN / m41 12B 5.4 a.墙趾板端部弯矩与剪力按式（ 3-15）：M = 1 B 211311233(cpphdhgg+- )6 = 1 0.6 2 2 263.1+ 240.9 -3 25 0.400 -318(0.700 -0.400) 6 = 43.3kN m × 1.35 M 1.35 43.3 10 6 2As = = 609.2 mm 0.9 fh 0 0.9 300 (400 -45) y r= As = 609.2 = 0.17% < 0.79% ，简化公式适用。bh0 1000 (400 -45) 选用 12180 As=628 mm2 。按式（ 3-13）： é p + p ùV =ê 13 -gch1 -g(d-h1)ú×B1. 2 . é263.1 +240.9 ù =-25 0.400 -18 (0.700 -0.400) 0.6 êú. 2 . =142.0 kN 1.35 V = 1.35 142.0 = 191.6 kN 0.7 f bh = 0.7 1.27 1000 355 = 315.6 10 3 N = 315.6 kN t0 1.35 <0.7 f bh 0 ，板厚满足抗剪要求。 Vt b墙踵板计算B 4.2 ln3 = 3.0 = 1.4 < 1.5 ，按三边固定，一边自由的双向板计算。 q +g h +g H = 10 + 25 0.3 + 18 8.7 = 174.1 kN / m2 kc 21 2 63.5 p=kN / m2 p4 = 229.8 kN / m2 174.1 -63.5 = 110.6kN / m2 174.1 -229.8 =-55.7 kN / m 2 图 3. 5墙踵板荷载计算图可简化为一个矩形荷载和一个三角形荷载，计算方法类似于立板的距墙底 04.5m（即距墙顶 8.74.2m）区间。本例仅选用 Mx 、Mx 0 及 Asx、Asx0 进行计算，其它内力及配筋计算读者可参照进行。图 3. 6中的荷载可简化为一个三角形荷载 q =110.6kN / m2 和一个矩形荷载 q =-55.7kN / m2 。 Mx = 0.0384 ( 55.7) 32 + 0.0200 110.6 32 = 0.65 kN m -× 0 22Mx =-0.0794 ( 55.7) 3 -0.0407 110.6 3 =-0.71 kN m -×由于弯矩计算值均太小，无配筋计算的必要，均按构造配筋。扶壁设计立板在扶壁与墙踵板交接处的铰线应不为 45o，该铰线与水平线的夹角应略小于 45o，但为设计方便，将其按 45o计算，这样做的计算结果将稍偏小。土压力的作用线临底板处应力为外凸的曲线，亦为计算的方便将其取为虚线所示的直线，亦使计算结果稍微偏小。 l × K = 10 3.0 (1 -sin35 ) = 12.8 kN /q × ° mkn 0 é ln ùlq +g (H -) Kn ê k 1 ú 0. 2 . 3.0 (10 +18 7.2) (1 -sin35 ) =° =178.7 kN / m 1.5m图 3. 7扶壁土压应力分布图 7.2 1 1.5 1 7.2 M = 12.8 7.2 (1.5 + ) +178.4 1.5 + (178.4 -12.8) 7.2 (1.5 + )2232 3 = 2861.9 kN m × 11V =178.4 1.5 + (178.4 +12.8) 7.2 = 822.1 kN 2 2 4.2 oq= arctan = 25.8 8.7 1.35 M 1.35 2861.9 10 6 As = o = 3339 mm 2 0.9 fh cos q 0.9 300 (4800 -40) cos 25.8 y 0 As3339 0.79% r= bh0 = 400 (4800 -40) = 0.18% < cos 25.8 o= 0.87% ，简化公式适用 0.2% r= = 0.22% min ocos 25.8 2r<rmin ，故按最小配筋率配筋，取 As= 0.22% 400 (4800 -40) = 4189 mm 0.7 f bh = 0.7 1.27 400 (4800 -40) = 1692.7 10 3 N = 1692.7 kN > V =822.1 kN ，故扶壁混t0 凝土的抗剪能力即能满足抗剪要求，不必配置抗剪腹筋。 图3. 8 扶壁式挡墙配筋图 layout development cities in developed countries show that developing rail transit Guide and an important means of achieving sustainable urban development. More than the bulk of the urban rail transit vehicles, rail traffic in the urban space layout and guide the evacuation, optimizing the structure and other aspects also play an important role in its role as the scale becomes more important. Rail transit on land development and stimulation is achieved through its good accessibility, under the dual role of mechanisms in land and planning, transport accessibility high land along the rail transit development and high strength, promotion of urban morphology and land use patterns to adjust accordingly. 1, and Copenhagen 1947, Copenhagen refers to shaped planning introduced, requirements land development along narrow of finger corridor concentrated, refers to shaped planning guide big Copenhagen area along track traffic line clear defined of corridor for Metro development, through track traffic and city development of integration, will along development concentrated in track station around a km around, around track site land implementation integrated intensive development, and provides good of non-mobile of way received connection facilities, Ensure that a larger proportion of the region's residents commute using mass transit. Figure 5.1-3 Copenhagen Metro City Development 2, Tokyo 5.1-4 Tokyo Metro to guide regional development (left: 1950, right: 2000) formation is the mass transit system of the city of Tokyo, which have been developed under the guidance of. Eastern suburbs . Established to track traffic based on the network-city, using rail-oriented intensive development mode has been recognized in many cities. Created the urban structure of the node, forming the urban rail transit-dependent and lifestyle, as well as mass transit systems provide adequate passenger protection. 5.1.3 to guide the healthy development of
