桥梁-沉降量计算

项目空心板铰缝桥面铺装防撞墙人行道支座垫石体积(m3)17.32.285.744.76重力密度(KN/m3)262626262626应力(KN)449.859.28149.24123.7600合计P上(KN)782.08项目人行道人群荷载标准值S人(KN/m2)3人行道面积a(m2)0合计P人(KN)0P人=S人×a1）台后无汽车荷载时桥梁宽B(m)基底至填土顶H(m)回填土按γ=17(KN/m)72.1填土内摩擦角：0.246Ea1=(B×μ×γ×H2)/2=64.550.70其水平力：Eax1=Eacos(δ+α)=61.56其竖直力：Eay1=Easin(δ+α)=19.412）台后有汽车荷载时破坏棱体长度l0l0=H×tanθ1.22h=ΣG/Bl0γ=3.84Ea2={B×μ×γ×H×(H+h)}/2=182.700.97其水平力：Eax2=Eacos(δ+α)=174.24其竖直力：Eay2=Easin(δ+α)=54.94项目台帽台身基础支撑梁耳墙搭板体积(m3)6.025.6315.7812.800重力密度(KN/m3)262626262626应力(KN)156.52146.38410.28332.800合计P下(KN)1045.98总计P总(KN)基底总面积A(m2)基底压应力p'(KPa)桥梁沉降量计算P上=P空+P铰+P铺+P防+P人行道+P支座=三、台后主动土压力车辆荷载等代均布土层厚度h:(tanθ公式见《公路桥涵设计通用规范》JTGD60-2004中4.2.3-7)ψ=35°δ=ψ/2=17.5°β=0土与基底的摩擦系数μ=桥墩基底至填土顶距离为水底至基底距离四、下构恒载1883.0020.58(μ公式见《公路桥涵设计通用规范》JTGD60-2004中4.2.3-5)主动土压力着力点C=H2/3=91.50P下=P台帽+P台身+P基础+P支撑梁+P耳墙+P搭板=一、上构恒载二、人群荷载主动土压力着力点C=（H/3）×{（H+3h）/(H+2h)}=α=0（台背竖直）公式见《公路桥涵设计通用规范》JTGD60-2004中4.3.4-1荷载长期效应组合永久作用（SGik）人群荷载（SQjk）台后汽车荷载（SQlk）设计安全等级荷载值(KN)1883.00054.942结构重要性系数γ01永久作用效应分项系数γGi1.2可变荷载效应分项系数γQl1.41.4可变荷载效应组合系数γQj0.8合计Sud(KN)2336.510675基底压应力p(KPa)p=Sud／A113.53基础尺寸(m)b×l=2.1×22.11查表4.3.6得Δz=0.6m计算深度zn=b×(2.5-0.4×lnb)（b为基础宽度2.7m）基底附加压应力p0(KPa)p0=p-γh'p0=59.53fa0=70(KPa)计算表编号分层厚度z(m)z/bl/bEsiψsαXi=ψsp0/EsiAi=zi×αi-zi-1×αi-1sn=Xi×AiAi/Esi10.60.2857142869.523809524350.60.9931.0205669930.59580.6080538150.0170228621.20.5714285719.52380952450.60.9157.1439689530.50223.5877012080.1004431.80.8571428579.52380952450.60.847.1439689530.4142.9576031470.082842.41.1428571439.52380952450.60.7687.1439689530.33122.3660825170.06624531.4285714299.52380952450.60.77.1439689530.25681.8345712270.0513663.61.7142857149.52380952450.60.6517.1439689530.24361.7402708370.0487274.229.52380952450.60.67.1439689530.17641.2601961230.0352884.82.2857142869.52380952450.60.5547.1439689530.13920.9944404780.02784910∑Ai=2.6592∑(Ai/Esi)=0.4297028576.188462459附注：本公式根据《公路桥涵地基与基础设计规范》JTGD63-2007中4.3编成。zn=4.627mEs=∑Ai/∑(Ai/Esi)=s=ψs×p0×∑(zi×αi-zi-1×αi-1)/Esis=∑sn=∑(Xi×Ai)=15.34891935(h'为基底埋置深度；γ为土重度，取18)五、沉降量计算（公路桥涵地基与基础设计规范JTGD63-2007）Sud=γ0×(SGik×γGi+SQjk×γGl×γQj+SQlk×γQl)=