1 工程概述瓦洪公路(随塘河路~平庄公路)新建工程中的南横河桥为三跨简支预制板梁桥(8m+13m+8m),本计算书为桥墩单桩承载力的验算。
2 基本设计资料2.1 主要设计规范及标准《公路桥涵设计通用规范》(JTG D60 2004)《公路钢筋混凝土及预应力混凝土桥涵设计规范》(JTG D62 2004)《公路工程抗震设计规范》(JTJ004 89)《公路桥涵地基与基础设计规范》(JTJ024-85)《地基基础设计规范》(上海)(DGJ08-11-1999)《公路设计手册-路基》2.2 荷载标准车道荷载:公路-Ⅱ级设计车道数n车:3人群荷载q人群: 3.0kN/m22.3 桥宽0.3m(栏杆)+3m(人行道)+3.5m(非机动车道)+1m(机非分隔带)+12m(机动车道)+1m(防撞墙)=20.3m2.4 跨径布置8m(边跨)+13m(中跨)+8m(边跨)2.5 斜角顺交7度2.6 材料容重钢筋混凝土γ1:26.0kN/m3沥青混凝土γ2:23.0kN/m32.7 铺装边跨每延米铺装重:q铺边=26×19.8×0.07+23×15.5×0.09=68.121kN/m中跨每延米铺装重:q铺中=26×19.8×0.07+23×15.5×0.09=68.121kN/m2.8 边跨和中跨板梁布置边跨主梁每延米自重:q边=26×(0.396868×2+0.30093×17)=153.648196kN/m中跨主梁每延米自重:q中=26×(0.49818×2+0.39993×17)=202.67442kN/m2.9 桥墩和支座布置桥墩形状见右图B盖梁=21.3/cos(7)=21.4599592822735mB墩柱=0.8mB承台=21.3/cos(7)=21.4599592822735mn墩柱=0mh1=0.1mh2=1.06mh3=0mh4=0mb1=0.65mb2=0.6mb3=0.37mb4=0.32mb5=1.35mb6=0mb7=1.35mb8=0mb9=0.639079422382671mb10=0.675mb11=0.675m盖梁面积A盖梁=1.385m2墩柱面积A墩柱0m2承台面积A承台=0m2盖梁形心距墩底中心e1=-0.0359205776173286m边跨支座距距墩底中心e2=0.255m中跨支点距距墩底中心e3=-0.305m盖梁形心距桩群中心e4=-0.0359205776173286m墩身形心距桩群中心e5=0m边跨支座距桩群中心e6=0.255m中跨支点距桩群中心e7=-0.305m3 作用效应计算3.1 永久作用效应3.1.1 边跨和中跨主梁自重P边=153.648196×8/2=614.592784kN P中=202.67442×13/2=1317.38373kN 3.1.2 铺装P边铺=68.121×8/2=272.484M边铺纵=272.484×(-0.788545382481173)=-214.866P中铺=68.121×13/2=442.7865M中铺纵=442.7865×(-0.788545382481173)=-349.157253.1.3 人行道板P边人行=4.8×3×8/2=57.6M边人行纵=57.6×(8.35)=480.96P中人行=4.8×3×13/2=93.6M中人行纵=93.6×(8.35)=781.563.1.4 栏杆P边栏杆=3.38×8/2=13.52kN M边栏杆纵=13.52×(10)=135.2kN-m(4)-1单孔加载边跨:2车道Pq边1=2×1×(1+0.45)×(7.875×8/2+172.8)=592.47kN Mq边1纵=592.47×(3.9)=2310.633kN-m 3车道Pq边2=3×0.78×(1+0.45)×(7.875×8/2+172.8)=693.1899kN Mq边2纵=693.1899×(2.35)=1628.996265kN-m 4车道Pq边3=4×0.67×(1+0.45)×(7.875×8/2+172.8)=793.9098kN Mq边3纵=793.9098×(0.800000000000001)=635.12784kN-m 5车道Pq边4=5×0.6×(1+0.45)×(7.875×8/2+172.8)=888.705kN Mq边4纵=888.705×(-0.75)=-666.52875kN-m 中跨:2车道Pq中1=2×1×(1+0.325352902776602)×(7.875×13/2+190.8)=641.437671121306kN Mq中1纵=641.437671121306×(3.9)=2501.60691737309kN-m 3车道Pq中2=3×0.78×(1+0.325352902776602)×(7.875×13/2+190.8)=750.482075211928kN Mq中2纵=750.482075211928×(2.35)=1763.63287674803kN-m 4车道Pq中3=4×0.67×(1+0.325352902776602)×(7.875×13/2+190.8)=859.52647930255k N Mq中3纵=859.52647930255×(0.800000000000001)=687.621183442041kN-m 5车道Pq中4=5×0.6×(1+0.325352902776602)×(7.875×13/2+190.8)=962.156506681959k N Mq中4纵=962.156506681959×(-0.75)=-721.617380011469kN-m (4)-2双孔加载a、 2车道Pq2边1=2×1×(1+0.45)×(7.875×8/2+172.8)=91.35kN Mq2边1纵=91.35×(3.9)=356.265kN-m Pq2中1=2×1×(1+0.325352902776602)×(7.875×13/2)=641.437671121306kN Mq2中1纵=641.437671121306×(3.9)=2501.60691737309kN-m b、3车道Pq2边2=3×0.78×(1+0.45)×(7.875×8/2+172.8)=106.8795kN Mq2边2纵=106.8795×(2.35)=251.166825kN-m Pq2中2=3×0.78×(1+0.325352902776602)×(7.875×13/2)=750.482075211928kN Mq2中2纵=750.482075211928×(2.35)=1763.63287674803kN-m C、4车道Pq2边3=4×0.67×(1+0.45)×(7.875×8/2+172.8)=122.409kN Mq2边3纵=122.409×(0.800000000000001)=97.9272000000001kN-m Pq2中3=4×0.67×(1+0.325352902776602)×(7.875×13/2)=859.52647930255kN Mq2中3纵=859.52647930255×(0.800000000000001)=687.621183442041kN-mD、5车道Pq2边4=5×0.6×(1+0.45)×(7.875×8/2+172.8)=137.025kN Mq2边4纵=137.025×(-0.75)=-102.76875kN-m Pq2中4=5×0.6×(1+0.325352902776602)×(7.875×13/2)=962.156506681959kN Mq2中4纵=962.156506681959×(-0.75)=-721.617380011469kN-m3.2.2 汽车制动力一个车道汽车制动力:Tq制动=max{(7.875×(2×8+2×8+13)+172.8)×10%,90}=90kN 按2个桥墩均摊计算Tq2=±(3×0.78×90/2)=±(105.3)kN M制动纵=±(105.3×(2.35))=±(247.455)kN-m 3.2.3 人群荷载P人群=3×3×(8+13)/2=94.5kN M人群纵=94.5×8.35=789.075kN-m 3.3 作用效应汇总3.3.1 永久作用作用在边跨支座上竖向力:Pg1=614.592784+272.484+57.6+13.52+40+19.2+0=1017.396784kN 作用在边跨支座上纵向弯矩:Mgz1=-214.866+480.96+135.2+(-396)+54.72+0=60.0139999999999kN-m作用在中跨支座上竖向力:Pg2=1317.38373+442.7865+93.6+21.97+65+31.2+0=1971.94023kN 作用在中跨支座上纵向弯矩:Mgz2=-349.15725+781.56+219.7+(-643.5)+88.92+0=97.52275kN-m桥墩自重:Pg3=772.773133754668kN 3.3.2 可变作用车道荷载∑xi 2=#NAME?m 2∑yi 2=#NAME?m 2ximax=0m yimax=9.9m桩基根数n桩11根基本组合1Pmax 1#NAME?2#NAME?3#NAME?4#NAME?5#NAME?6#NAME?7#NAME?8#NAME?9#NAME?10#NAME?11#NAME?12#NAME?最大值#NAME?基本组合2Pmax 1#NAME?4.2 桩基承载力计算桩基根数n桩11根桩直径d 0.6m 桩长l 30m 桩周长U 1.88m 桩截面面积A 0.28m 2Σli×гi 813.9kN/m 极限承载力[σR]1300kPa [P]=0.5×(U×Σli×гi+A×σR)-γ×A×l 容许承载力[P]841kN 桩最不利受力Pmx #NAME?kNPmax/[P]#NAME?判定:#NAME?∑∑∑++=2max 2max max i i y i i x x x M y y M n P P。