预览加载中,请您耐心等待几秒...
1/4
2/4
3/4
4/4

在线预览结束,喜欢就下载吧,查找使用更方便

如果您无法下载资料,请参考说明:

1、部分资料下载需要金币,请确保您的账户上有足够的金币

2、已购买过的文档,再次下载不重复扣费

3、资料包下载后请先用软件解压,在使用对应软件打开

裂缝宽度验算:按《公路桥涵设计通用规范》(JTGD60—2004)第1.0.7条规定:该箱涵所处环境类别为:Ⅰ类按《公路桥涵设计通用规范》(JTGD60—2004)第6.4.2条规定:钢筋砼构件其计算裂缝宽度不应超过:0.20mm1、顶板(1)跨中l0=4.35m,h=0.34m,a=0.04m,AS=805.54mm2ys=0.12m,h0=0.30m,b=1.00m,d=20mmMd=74.16kN·m,Nd=9.95kN,Vd=45.18kNe0=Md/Nd=7.453mNl=46.90kN,NS=36.95kN按《公路桥涵设计通用规范》(JTGD60—2004)第6.4.4条规定:偏心受压构件:σss=Ns(es-z)/Aszηs=1+h0/4000/e0(l0/h)2=1.00es=ηse0+ys=7.57rf'=(bf'-b)hf'/b/h0=0Z=(0.87-0.12(1-rf')(h0/es)2)h0=0.26σss=1.29Mpa按《公路桥涵设计通用规范》(JTGD60—2004)第6.4.3条规定:Wtk=C1C2C3σss/Es((30+d)/(0.28+10ρ))C1=1.0C2=1+0.5Nl/Ns=1.63C3=0.9Es=2.0E+05Mpaρ=(As+Ap)/(bh0+(bf-b)hf=0.60%Wtk=0.0014mm满足(2)角点l0=4.35m,h=0.44m,a=0.04m,AS=1570.80mm2ys=0.17m,h0=0.40m,b=1.00m,d=20mmMd=206.56kN·m,Nd=9.95kN,Vd=365.26kNe0=Md/Nd=20.758mNl=99.25kN·m,NS=107.31kN·m按《公路桥涵设计通用规范》(JTGD60—2004)第6.4.4条规定:偏心受压构件:σss=Ns(es-z)/Aszηs=1+h0/4000/e0(l0/h)2=1.00es=ηse0+ys=20.93rf'=(bf'-b)hf'/b/h0=0Z=(0.87-0.12(1-rf')(h0/es)2)h0=0.35σss=4.04Mpa按《公路桥涵设计通用规范》(JTGD60—2004)第6.4.3条规定:Wtk=C1C2C3σss/Es((30+d)/(0.28+10ρ))C1=1.0C2=1+0.5Nl/Ns=1.46C3=0.9Es=2.0E+05Mpaρ=(As+Ap)/(bh0+(bf-b)hf=0.60%Wtk=0.0039mm满足2、底板(1)跨中l0=4.35m,h=0.34m,a=0.04m,AS=1365.91mm2ys=0.12m,h0=0.30m,b=1.00m,d=20mmMd=132.18kN·m,Nd=116.83kN,Vd=10.62kNe0=Md/Nd=1.131mNl=65.50kN,NS=51.33kN按《公路桥涵设计通用规范》(JTGD60—2004)第6.4.4条规定:偏心受压构件:σss=Ns(es-z)/Aszηs=1+h0/4000/e0(l0/h)2=1.00es=ηse0+ys=1.25rf'=(bf'-b)hf'/b/h0=0Z=(0.87-0.12(1-rf')(h0/es)2)h0=0.26σss=0.14Mpa按《公路桥涵设计通用规范》(JTGD60—2004)第6.4.3条规定:Wtk=C1C2C3σss/Es((30+d)/(0.28+10ρ))C1=1.0C2=1+0.5Nl/Ns=1.64C3=0.9Es=2.0E+05Mpaρ=(As+Ap)/(bh0+(bf-b)hf=0.60%Wtk=0.0002mm满足(2)角点l0=4.35m,h=0.44m,a=0.04m,AS=1570.80mm2ys=0.17m,h0=0.40m,b=1.00m,d=20mmMd=203.76kN·m,Nd=116.83kN,Vd=365.26kNe0=Md/Nd=1.744mNl=99.92kN·m,NS=103.84kN·m按《公路桥涵设计通用规范》(JTGD60—2004)第6.4.4条规定:偏心受压构件:σss=Ns(es-z)/Aszηs=1+h0/4000/e0(l0/h)2=1.00es=ηse0+ys=1.91rf'=(bf'-b)hf'/b/h0=0Z=(0.87-0.12(1-rf')(h0/es)2)h0=0.35σss=0.30Mpa按《公路桥涵设计通用规范》(JTGD60—2004)第6.4.3条规定:Wtk=C1C2C3σss/Es((30+d)/(0.28+10ρ))C1=1.0C2=1+0.5Nl/Ns=1.48C3=0.9Es=2.0E+05Mpaρ=(As+Ap)/(bh0+(bf-b)hf=