双孔箱涵结构设计

钢筋混凝土箱涵结构设计,,,,,,,,,,,,,

一、设计资料,,,,,,,,,,,,,

,1、孔径及净空,,,,,,,,,,,,

,净跨径,,,L0=,4,m,,,,,,,

,净高,,,h0=,3,m,,,,,,,

,孔数,,,m=,2,,,,,,,,

,2、设计安全等级,,,,一级,,,,,,,,

,结构重要性系数,,,r0=,1.1,,,,,,,,

,3、汽车荷载,,,,,,,,,,,,

,荷载等级,,,公路—,Ⅰ级,,,,,,,,

,4、填土情况,,,,,,,,,,,,

,涵顶填土高度,,,H=,1.02,m,,,,,,,

,土的内摩擦角,,,Φ=,30,°,,,,,,,

,填土容重,,,γ1=,18,kN/m3,,,,,,,

,地基容许承载力,,,[σ0]=,150,kPa,,,,,,,

,5、建筑材料,,,,,,,,,,,,

,普通钢筋种类,,,,HRB400,,,,,,,,

,主钢筋直径,,,,22,mm,,,,,,,

,钢筋抗拉强度设计值,,,fsd=,330,MPa,,,,,,,

,涵身混凝土强度等级,,,C,30,,,,,,,,

,涵身混凝土抗压强度设计值,,,fcd=,13.8,MPa,,,,,,,

,涵身混凝土抗拉强度设计值,,,ftd=,1.39,MPa,,,,,,,

,钢筋混凝土重力密度,,,γ2=,25,kN/m3,,,,,,,

,基础混凝土强度等级,,,C,15,,,,,,,,

,混凝土重力密度,,,γ3=,24,kN/m3,,,,,,,

二、设计计算,,,,,,,,,,,,,

,(一)截面尺寸拟定(见图L-01),,,,,,,,,,,,

,顶板、底板厚度,,,δ=,0.5,m,,,,,,,

,,,,C1=,0.2,m,,,,,,,

,侧墙厚度,,,t=,0.5,m,,,,,,,

,,,,C2=,0.2,m,,,,,,,

,横梁计算跨径,,,LP=L0+t=,4.5,m,,,,,,,

,,,,L=2L0+3t=,9.5,m,,,,,,,

,侧墙计算高度,,,hP=h0+δ=,3.5,m,,,,,,,

,,,,h=h0+2δ=,4,m,,,,,,,

,基础襟边,,,c=,0.2,m,,,,,,,

,基础高度,,,d=,0.1,m,,,,,,,

,基础宽度,,,B=,9.9,m,,,,,,,

,(二)荷载计算,,,,,,,,,,,,

,1、恒载,,,,,,,,,,,,

,恒载竖向压力,,,p恒=γ1H+γ2δ=,30.86,kN/m2,,,,,,,

,恒载水平压力,,,,,,,,,,,,

,顶板处,,,eP1=γ1Htan2(45°-φ/2)=,6.12,kN/m2,,,图L-01,,,,

,底板处,,,eP2=γ1(H+h)tan2(45°-φ/3)=,30.12,kN/m2,,,,,,,

,2、活载,,,,,,,,,,,,

,汽车后轮着地宽度0.6m，由《公路桥涵设计通用规范》(JTGD60—2015)第4.3.4条规定，按30°角向下分布。,,,,,,,,,,,,

,一个汽车后轮横向分布宽,,,,,,,,,,,,

,,,0.6/2+Htan30°=,,0.89m,>1.3/2m,,,,,,,

,,,,,,<1.8/2m,,,,,,,

,故横向分布宽度,,,,,,,,,,,,

,,,,a=(0.6/2+Htan30°)×2+1.3=,3.078,m,,,,,,,

,同理，纵向，汽车后轮着地长度0.2m,,,,,,,,,,,,

,,,,0.2/2+Htan30°=,0.689m,<1.4/2m,,,,,,,

,故,,,b=(0.2/2+Htan30°)×2=,1.378,m,,,,,,,

,,,,∑G=,140,kN,,,,,,,

,车辆荷载垂直压力,,,q车=∑G/(a×b)=,33.01,kN/m2,,,,,,,

,车辆荷载水平压力,,,e车=q车tan2(45°-φ/2)=,11.00,kN/m2,,,,,,,

,(三)内力计算,,,,,,,,,,,,

,1、构件刚度比,,,,,,,,,,,,

,,,,K=(I1/I2)×(hP/LP)=,0.78,,,,,,,,

,,,,u=2K+1=,2.56,,,,,,,,

,2、节点弯矩和轴向力计算,,,,,,,,,,,,

,(1)a种荷载作用下(图L-02),,,,,,,,,,,,

,涵洞四角节点弯矩,,,MaA=MaC=MaE=MaF=,-1/u·pLP2/12,,,,,,,,

,,,,MBA=MBE=MDC=MDF=,-(3K+1)/u·pLP2/12,,,,,,,,

,,,,MBD=MDB=,0,,,,,,,,

,横梁内法向力,,,Na1=Na2=Na1'=Na2'=,0,,,,,,,,

,侧墙内法向力,,,Na3=Na4=,(MBA-MaA+pLp2/2)/Lp,,,,,,,,

,,,,Na5=,-(Na3+Na4),,,,,,,,

,恒载,,,p=p恒=,30.86,kN/m2,,,,,,,

,,,,MaA=MaC=MaE=MaF=,-20.38,kN·m,,,,,,,

,,,,MBA=MBE=MDC=MDF=,-67.93,kN·m,,,,,,,

,,,,Na3=Na4=,58.87,kN,,,,,,,

,,,,Na5=,-117.74,kN,,,,,,,

,车辆荷载,,,p=q车=,33.01,kN/m2,,,,,,,

,,,,MaA=MaC=MaE=MaF=,-21.80,kN·m,,图L-02,,,,,

,,,,MBA=MBE=MDC=MDF=,-72.67,kN·m,,,,,,,

,,,,Na3=Na4=,62.98,kN,,,,,,,

,,,,Na5=,-125.96,kN,,,,,,,

,(2)b种荷载作用下(图L-03),,,,,,,,,,,,

,,,,MbA=MbC=MbE=MbF=,-K·phP2/6u,,,,,,,,

,,,,MBA=MBE=MDC=MDF=,K·phP2/12u,,,,,,,,

,,,,MBD=MDB=,0,,,,,,,,

,,,,Nb1=Nb2=Nb1'=Nb2'=,phP/2,,,,,,,,

,,,,Nb3=Nb4=,(MBA-MbA)/Lp,,,,,,,,

,,,,Nb5=,-(Nb3+Nb4),,,,,,,,

,恒载,,,p=eP1=,6.12,kN/m2,,,,,,,

,,,,MbA=MbC=MbE=MbF=,-3.80,kN·m,,,,,,,

,,,,MBA=MBE=MDC=MDF=,1.90,kN·m,,图L-03,,,,,

,,,,Nb1=Nb2=Nb1'=Nb2'=,10.71,kN,,,,,,,

,,,,Nb3=Nb4=,-1.27,kN,,,,,,,

,,,,Nb5=,2.54,kN,,,,,,,

,(3)c种荷载作用下(图L-04),,,,,,,,,,,,

,,,,Φ=20u(K+6)/K=,445.40,,,,,,,,

,,,,McA=McE=,-(8K+59)·phP2/6Φ,,,,,,,,

,,,,McC=McF=,-(12K+61)·phP2/6Φ,,,,,,,,

,,,,MBA=MBE=,(7K+31)·phP2/6Φ,,,,,,,,

,,,,MDC=MDF=,(3K+29)·phP2/6Φ,,,,,,,,

,,,,MBD=MDB=,0,,,,,,,,

,,,,Nc1=Nc1'=,phP/6+(McC-McA)/hP,,,,,,,,

,,,,Nc2=Nc2'=,phP/3-(McC-McA)/hP,,,,,,,,

,,,,Nc3=Nc4=,(MBA-McA)/Lp,,,,,,,,

,,,,Nc5=,-(Nc3+Nc4),,,,,,,,

,恒载,,,p=eP2-eP1=,24.00,kN/m2,,,,,,,

,,,,McA=McE=,-7.18,kN·m,,,,,,,

,,,,McC=McF=,-7.74,kN·m,,,,,,,

,,,,MBA=MBE=,4.01,kN·m,,图L-04,,,,,

,,,,MDC=MDF=,3.45,kN·m,,,,,,,

,,,,Nc1=Nc1'=,13.84,kN,,,,,,,

,,,,Nc2=Nc2'=,28.16,kN,,,,,,,

,,,,Nc3=Nc4=,2.49,kN,,,,,,,

,,,,Nc5=,-4.97,kN,,,,,,,

,(4)d种荷载作用下(图L-05),,,,,,,,,,,,

,,,,Φ1=20(K+2)(6K2+6K+1)=,516.46,,,,,,,,

,,,,Φ2=u/K=,3.29,,,,,,,,

,,,,Φ3=120K3+278K2+335K+63=,548.19,,,,,,,,

,,,,Φ4=120K3+529K2+382K+63=,736.58,,,,,,,,

,,,,Φ5=360K3+742K2+285K+27=,866.91,,,,,,,,

,,,,Φ6=120K3+611K2+558K+87=,947.08,,,,,,,,

,,,,MdA=,(-2/Φ2+Φ3/Φ1)·phP2/4,,,,,,,,

,,,,MdE=,(-2/Φ2-Φ3/Φ1)·phP2/4,,,,,,,,

,,,,MdC=,-(2/Φ2+Φ5/Φ1)·phP2/24,,,,,,,,

,,,,MdF=,-(2/Φ2-Φ5/Φ1)·phP2/24,,,,,,,,

,,,,MBA=,-(-2/Φ2+Φ4/Φ1)·phP2/24,,,,,,,,

,,,,MBE=,-(-2/Φ2-Φ4/Φ1)·phP2/24,,,,,,,,

,,,,MDC=,(1/Φ2+Φ6/Φ1)·phP2/24,,,,,,,,

,,,,MDF=,(1/Φ2-Φ6/Φ1)·phP2/24,,,,,,,,

,,,,MBD=,-Φ4·phP2/12Φ1,,,,,,,,

,,,,MDB=,Φ6·phP2/12Φ1,,,,,,,,

,,,,Nd1=,(MdC+phP2/2-MdA)/hP,,,图L-05,,,,,

,,,,Nd2=,php-Nd1,,,,,,,,

,,,,Nd1'=,(MdF-MdE)/hP,,,,,,,,

,,,,Nd2'=,php-Nd1',,,,,,,,

,,,,Nd3=,(MBA+MBD-MdA)/LP,,,,,,,,

,,,,Nd4=,(MBE+MBD-MdE)/LP,,,,,,,,

,,,,Nd5=,-(Nd3+Nd4),,,,,,,,

,车辆荷载,,,p=e车=,11.00,kN/m2,,,,,,,

,,,,MdA=,15.26,kN·m,,,,,,,

,,,,MdE=,-56.29,kN·m,,,,,,,

,,,,MdC=,-12.85,kN·m,,,,,,,

,,,,MdF=,6.01,kN·m,,,,,,,

,,,,MBA=,-11.43,kN·m,,,,,,,

,,,,MBE=,4.59,kN·m,,,,,,,

,,,,MDC=,12.01,kN·m,,,,,,,

,,,,MDF=,-8.59,kN·m,,,,,,,

,,,,MBD=,-16.02,kN·m,,,,,,,

,,,,MDB=,20.60,kN·m,,,,,,,

,,,,Nd1=,-27.29,kN,,,,,,,

,,,,Nd2=,65.81,kN,,,,,,,

,,,,Nd1'=,0.40,kN,,,,,,,

,,,,Nd2'=,38.11,kN,,,,,,,

,,,,Nd3=,-9.49,kN,,,,,,,

,,,,Nd4=,9.97,kN,,,,,,,

,,,,Nd5=,-0.48,kN,,,,,,,

,(5)节点弯矩、轴力计算及荷载效应组合汇总表,,,,,,,,,,,,

,按《公路桥涵设计通用规范》(JTGD60—2015)第4.1.6条进行承载能力极限状态效应组合,,,,,,,,,,,,

,,,,,,,,,,,,,,

,,,,,,,,,,,,,,

,荷载种类,,,M(kN·m),,,,,,,,,

,,,,MA,ME,MC,MF,MBA,MBE,MDC,MDF,MBD,MDB

,恒载,a,,-20.38,-20.38,-20.38,-20.38,-67.93,-67.93,-67.93,-67.93,0.00,0.00

,,1.2×∑结构、土的重力,,-24.45,-24.45,-24.45,-24.45,-81.51,-81.51,-81.51,-81.51,0.00,0.00

,,b,,-3.80,-3.80,-3.80,-3.80,1.90,1.90,1.90,1.90,0.00,0.00

,,c,,-7.18,-7.18,-7.74,-7.74,4.01,4.01,3.45,3.45,0.00,0.00

,,1.4×∑土侧压力,,-15.37,-15.37,-16.16,-16.16,8.28,8.28,7.49,7.49,0.00,0.00

,车辆荷载,a,,-21.80,-21.80,-21.80,-21.80,-72.67,-72.67,-72.67,-72.67,0.00,0.00

,,d,,15.26,-56.29,-12.85,6.01,-11.43,4.59,12.01,-8.59,-16.02,20.60

,,1.4×∑汽车,,-9.16,-109.32,-48.51,-22.11,-117.74,-95.31,-84.92,-113.76,-22.43,28.84

,荷载效应组合,,,-48.98,-149.15,-89.12,-62.72,-190.97,-168.54,-158.94,-187.78,-22.43,28.84

,,,,,,,,,,,,,,

,荷载种类,,,N(kN),,,,,,,,,

,,,,N1,N2,N1',N2',N3,N4,N5,,,

,恒载,a,,0.00,0.00,0.00,0.00,58.87,58.87,-117.74,,,

,,1.2×∑结构、土的重力,,0.00,0.00,0.00,0.00,70.64,70.64,-141.29,,,

,,b,,10.71,10.71,10.71,10.71,-1.27,-1.27,2.54,,,

,,c,,13.84,28.16,13.84,28.16,2.49,2.49,-4.97,,,

,,1.4×∑土侧压力,,34.37,54.42,34.37,54.42,1.71,1.71,-3.41,,,

,车辆荷载,a,,0.00,0.00,0.00,0.00,62.98,62.98,-125.96,,,

,,d,,-27.29,65.81,0.40,38.11,-9.49,9.97,-0.48,,,

,,1.4×∑汽车,,-38.20,92.13,0.57,53.36,74.88,102.13,-177.01,,,

,荷载效应组合,,,-3.84,146.55,34.94,107.8,147.2,174.5,-321.70,,,

,,,,,,,,,,,,,,

,3、构件内力计算(跨中截面内力),,,,,,,,,,,,

,(1)顶板1(图L-06),,,,,,,,,,,,

,,,,x=,LP/2,,,,,,,,

,,,,P=1.2p恒+1.4q车=,83.25,kN,,,,,,,

,,,,Nx=N1=,-3.84,kN,,,,,,,

,,,,Mx=MA+N3x-Px2/2=,71.55,kN·m,,,,,,,

,,,,Vx=Px-N3=,40.09,kN,,,,,,,

,顶板1',,,x=,LP/2,,,,,,,,

,,,,P=1.2p恒+1.4q车=,83.25,kN,,,,图L-06,,,

,,,,Nx=N1'=,34.94,kN,,,,,,,

,,,,Mx=ME+N4x-Px2/2=,32.69,kN·m,,,,,,,

,,,,Vx=Px-N4=,12.84,kN,,,,,,,

,(2)底板2(图L-07),,,,,,,,,,,,

,,,,ω1=,1.2p恒+1.4(q车+3e车HP2/4LP2),,,,,,,,

,,,,=,90.24,kN/m2,,,,,,,

,,,,ω2=,1.2p恒+1.4q车,,,,,,,,

,,,,=,83.25,kN/m2,,,,,,,

,,,,x=,LP/2,,,,,,,,

,,,,Nx=N2=,146.55,kN,,,,,,,

,,,,Mx=,MC+N3x-ω2·x2/2-5x3(ω1-ω2)/12LP,,,,,,,,

,,,,=,24.05,kN·m,,,,,,,

,,,,Vx=,ω2x+3x2(ω1-ω2)/2LP-N3,,,,,图L-07,,,

,,,,=,51.88,kN,,,,,,,

,底板2',,,ω1=,1.2p恒+1.4q车,,,,,,,,

,,,,=,83.25,kN/m2,,,,,,,

,,,,ω2=,1.2p恒+1.4(q车-3e车HP2/4LP2),,,,,,,,

,,,,=,76.26,kN/m2,,,,,,,

,,,,x=,LP/2,,,,,,,,

,,,,Nx=N2'=,107.78,kN,,,,,,,

,,,,Mx=,MF+N4x-ω2·x2/2-x3(ω1-ω2)/6LP,,,,,,,,

,,,,=,133.86,kN·m,,,,,,,

,,,,Vx=,ω2x+x2(ω1-ω2)/2LP-N4,,,,,,,,

,,,,=,4.98,kN,,,,,,,

,(3)左侧墙(图L-08),,,,,,,,,,,,

,,,,ω1=,1.4eP1+1.4e车,,,,,,,,

,,,,=,23.97,kN/m2,,,,,,,

,,,,ω2=,1.4eP2+1.4e车,,,,,,,,

,,,,,57.57,kN/m2,,,,,,,

,,,,x=,hP/2,,,,,图L-08,,,

,,,,Nx=N3=,147.23,kN,,,,,,,

,,,,Mx=,MA+N1x-ω1·x2/2-x3(ω2-ω1)/6hP,,,,,,,,

,,,,=,-100.98,kN·m,,,,,,,

,,,,Vx=,ω1x+x2(ω2-ω1)/2hP-N1,,,,,,,,

,,,,=,60.49,kN,,,,,,,

,(4)右侧墙(图L-09),,,,,,,,,,,,

,,,,ω1=1.4eP1=,8.57,kN/m2,,,,,,,

,,,,ω2=1.4eP2=,42.17,kN/m2,,,,,,,

,,,,x=,hP/2,,,,,,,,

,,,,Nx=N4=,174.47,kN,,,,图L-09,,,

,,,,Mx=,ME+N1'x-ω1·x2/2-x3(ω2-ω1)/6hP,,,,,,,,

,,,,=,-109.70,kN·m,,,,,,,

,,,,Vx=,ω1x+x2(ω2-ω1)/2hP-N1',,,,,,,,

,,,,=,-5.24,kN,,,,,,,

,(5)中间墙(图L-10),,,,,,,,,,,,

,,,,x=,hP/2,,,,,,,,

,,,,Nx=N5=,-321.70,kN,,,,,,,

,,,,Mx=,MBD+(N1+N1')x,,,,,,,,

,,,,=,32.00,kN·m,,,,,,,

,,,,Vx=,-(N1+N1'),,,,,图L-10,,,

,,,,=,-31.10,kN,,,,,,,

,,,,,,,,,,,,,,

,,,,,,,,,,,,,,

,,,,,,,,,,,,,,

,,,,,,,,,,,,,,

,(5)构件内力汇总表,,,,,,,,,,,,

,,,,,,,,,,,,,,

,构件,Md,Nd,Vd,Md,Nd,Vd,Md,Nd,Vd,,,

,A-B,A,,,A-B,,,B,,,,,

,,-48.98,-3.84,147.23,71.55,-3.84,40.09,-190.97,-3.84,-321.70,,,

,B-E,B,,,B-E,,,E,,,,,

,,-168.54,34.94,-321.70,32.69,34.94,12.84,-149.15,34.94,174.47,,,

,C-D,C,,,C-D,,,D,,,,,

,,-89.12,146.55,147.23,24.05,146.55,51.88,-158.94,146.55,-321.70,,,

,D-F,D,,,D-F,,,F,,,,,

,,-187.78,107.78,-321.70,133.86,107.78,4.98,-62.72,107.78,174.47,,,

,A-C,A,,,A-C,,,C,,,,,

,,-48.98,147.23,-3.84,-100.98,147.23,60.49,-89.12,147.23,146.55,,,

,E-F,E,,,E-F,,,F,,,,,

,,-149.15,174.47,34.94,-109.70,174.47,-5.24,-62.72,174.47,107.78,,,

,B-D,B,,,B-D,,,D,,,,,

,,-22.43,-321.70,31.10,32.00,-321.70,-31.10,28.84,-321.70,254.32,,,

,,,,,,,,,,,,,,

,(四)截面设计,,,,,,,,,,,,

,1、顶板(A-B\B-E),,,,,,,,,,,,

,钢筋按左、右对称，用最不利荷载计算。,,,,,,,,,,,,

,(1)跨中,,,,,,,,,,,,

,l0=,4.50m，,h=,0.50m，,a=,0.05m，,h0=,0.45m，,b=,1.00m，,,,

,,Md=,71.55kN,·m，,Nd=,-3.84kN,，Vd=,40.09kN,,,,,

,,,,e0=Md/Nd=,-18.658,m,,,,,,,

,,,,i=h/121/2=,0.144,m,,,,,,,

,长细比,,,l0/i=,31.18,>17.5,,,,,,,

,由《公路钢筋混凝土及预应力混凝土桥涵设计规范》(JTG3362-2018)第5.3.9条,,,,,,,,,,,,

,,,,ξ1=0.2+2.7e0/h0=,-111.746,≤1.0，,取ξ1=,-111.75,,,,,

,,,,ξ2=1.15-0.01l0/h=,1.060,>1.0，,取ξ2=,1.00,,,,,

,,,,η=,1+(l0/h)2ξ1ξ2h0/1300e0,,,,,,,,

,,,,η=,1.168,,,,,,,,

,由《公路钢筋混凝土及预应力混凝土桥涵设计规范》(JTG3362—2018)第5.3.4条,,,,,,,,,,,,

,,,,e=ηe0+h/2-a=,-21.591,m,,,,,,,

,,,,r0Nde=,fcdbx(h0-x/2),,,,,,,,

,,,,91.08=,13800x(,0.45-x/2),,,,,,,

,解得,,,x=,0.015m,≤,ξbh0=,0.53×,0.45=,0.239m,,,

,故为大偏心受压构件。,,,,,,,,,,,,

,,,,As=(fcdbx-r0Nd)/fsd=,0.000636473,m2,=636.5,mm2,,,,,

,,,,μ=100As/(bh0)=,0.14%,<,0.2%,,,,,,

,应按最小配筋率配置受拉钢筋。,,,,,,,,,,,,

,选用φ22@,420mm,，,实际As=,905.1,mm2,,,,,,,

,,,,"0.51×10-3fcu,k1/2bh0=",1257.0kN,>,r0Vd=,44.1kN,,,,,

,故抗剪截面符合《公路钢筋混凝土及预应力混凝土桥涵设计规范》(JTG3362—2018)第5.2.11条的要求。,,,,,,,,,,,,

,由《公路钢筋混凝土及预应力混凝土桥涵设计规范》(JTG3362—2018)第5.2.12条,,,,,,,,,,,,

,,,,0.50×10-3α2ftdbh0=,312.8kN,>,r0Vd=,44.1kN,,,,,

,故可不进行斜截面抗剪承载力的验算，仅需按(JTG3362—2018)第9.3.12条构造要求配置箍筋。,,,,,,,,,,,,

,(2)结点(A\E),,,,,,,,,,,,

,l0=,4.50m，,,h=δ+C1=,0.70m，,a=,0.05m，,h0=,0.65m，,b=,1.00m，,,

,,Md=,190.97kN,·m，,Nd=,-3.84kN,，Vd=,321.70kN,,,,,

,,,,e0=Md/Nd=,-49.796,m,,,,,,,

,,,,i=h/121/2=,0.202,m,,,,,,,

,长细比,,,l0/i=,22.27,>17.5,,,,,,,

,由《公路钢筋混凝土及预应力混凝土桥涵设计规范》(JTG3362—2018)第5.3.9条,,,,,,,,,,,,

,,,,ξ1=0.2+2.7e0/h0=,-206.645,≤1.0，,取ξ1=,-206.65,,,,,

,,,,ξ2=1.15-0.01l0/h=,1.086,>1.0，,取ξ2=,1.00,,,,,

,,,,η=,1+(l0/h)2ξ1ξ2h0/1300e0,,,,,,,,

,,,,η=,1.086,,,,,,,,

,由《公路钢筋混凝土及预应力混凝土桥涵设计规范》(JTG3362—2018)第5.3.4条,,,,,,,,,,,,

,,,,e=ηe0+h/2-a=,-53.766,m,,,,,,,

,,,,r0Nde=,fcdbx(h0-x/2),,,,,,,,

,,,,226.82=,13800x(,0.65-x/2),,,,,,,

,解得,,,x=,0.026m,≤,ξbh0=,0.53×,0.65=,0.345m,,,

,故为大偏心受压构件。,,,,,,,,,,,,

,,,,As=(fcdbx-r0Nd)/fsd=,0.001091617,m2,=1091.6,mm2,,,,,

,,,,μ=100As/(bh0)=,0.17%,<,0.2%,,,,,,

,应按最小配筋率配置受拉钢筋。,,,,,,,,,,,,

,选用φ22@,290mm,，,实际As=,1310.8,mm2,,,,,,,

,,,,"0.51×10-3fcu,k1/2bh0=",1815.7kN,>,r0Vd=,353.9kN,,,,,

,故抗剪截面符合《公路钢筋混凝土及预应力混凝土桥涵设计规范》(JTG3362—2018)第5.2.11条的要求。,,,,,,,,,,,,

,由《公路钢筋混凝土及预应力混凝土桥涵设计规范》(JTG3362—2018)第5.2.12条,,,,,,,,,,,,

,,,,0.50×10-3α2ftdbh0=,451.8kN,>,r0Vd=,353.9kN,,,,,

,故可不进行斜截面抗剪承载力的验算，仅需按(JTG3362—2018)第9.3.12条构造要求配置箍筋。,,,,,,,,,,,,

,由《公路钢筋混凝土及预应力混凝土桥涵设计规范》(JTG3362—2018)第5.2.9条,,,,,,,,,,,,

,"VCS=a1a2a30.45*10-3bh0((2+0.6ρ)fcu,k0.5ρsvfsv)0.5=",,,,909.96kN,>,r0Vd=,353.9kN,,,,,

,故斜截面内混凝土与箍筋共同的抗剪承载力已满足要求。,,,,,,,,,,,,

,2、底板(C-D\D-F),,,,,,,,,,,,

,钢筋按左、右对称，用最不利荷载计算。,,,,,,,,,,,,

,(1)跨中,,,,,,,,,,,,

,l0=,4.50m，,h=,0.50m，,a=,0.05m，,h0=,0.45m，,b=,1.00m，,,,

,,Md=,133.86kN,·m，,Nd=,107.78kN,，Vd=,4.98kN,,,,,

,,,,e0=Md/Nd=,1.242,m,,,,,,,

,,,,i=h/121/2=,0.144,m,,,,,,,

,长细比,,,l0/i=,31.18,>17.5,,,,,,,

,由《公路钢筋混凝土及预应力混凝土桥涵设计规范》(JTG3362—2018)第5.3.9条,,,,,,,,,,,,

,,,,ξ1=0.2+2.7e0/h0=,7.652,>1.0，,取ξ1=,1.00,,,,,

,,,,ξ2=1.15-0.01l0/h=,1.060,>1.0，,取ξ2=,1.00,,,,,

,,,,η=,1+(l0/h)2ξ1ξ2h0/1300e0,,,,,,,,

,,,,η=,1.023,,,,,,,,

,由《公路钢筋混凝土及预应力混凝土桥涵设计规范》(JTG3362—2018)第5.3.4条,,,,,,,,,,,,

,,,,e=ηe0+h/2-a=,1.470,m,,,,,,,

,,,,r0Nde=,fcdbx(h0-x/2),,,,,,,,

,,,,174.28=,13800x(,0.45-x/2),,,,,,,

,解得,,,x=,0.029m,≤,ξbh0=,0.53×,0.45=,0.239m,,,

,故为大偏心受压构件。,,,,,,,,,,,,

,,,,As=(fcdbx-r0Nd)/fsd=,0.000853431,m2,=853.4,mm2,,,,,

,,,,μ=100As/(bh0)=,0.19%,<,0.2%,,,,,,

,应按最小配筋率配置受拉钢筋。,,,,,,,,,,,,

,选用φ22@,420mm,，,实际As=,905.1,mm2,,,,,,,

,,,,"0.51×10-3fcu,k1/2bh0=",1257.0kN,>,r0Vd=,5.5kN,,,,,

,故抗剪截面符合《公路钢筋混凝土及预应力混凝土桥涵设计规范》(JTG3362—2018)第5.2.11条的要求。,,,,,,,,,,,,

,由《公路钢筋混凝土及预应力混凝土桥涵设计规范》(JTG3362—2018)第5.2.12条,,,,,,,,,,,,

,,,,0.50×10-3α2ftdbh0=,312.8kN,>,r0Vd=,5.5kN,,,,,

,故可不进行斜截面抗剪承载力的验算，仅需按(JTG3362—2018)第9.3.12条构造要求配置箍筋。,,,,,,,,,,,,

,(2)结点,,,,,,,,,,,,

,l0=,4.50m，,,h=δ+C1=,0.70m，,a=,0.05m，,h0=,0.65m，,b=,1.00m，,,

,,Md=,187.78kN,·m，,Nd=,107.78kN,，Vd=,321.70kN,,,,,

,,,,e0=Md/Nd=,1.742,m,,,,,,,

,,,,i=h/121/2=,0.202,m,,,,,,,

,长细比,,,l0/i=,22.27,>17.5,,,,,,,

,由《公路钢筋混凝土及预应力混凝土桥涵设计规范》(JTG3362—2018)第5.3.9条,,,,,,,,,,,,

,,,,ξ1=0.2+2.7e0/h0=,7.438,>1.0，,取ξ1=,1.00,,,,,

,,,,ξ2=1.15-0.01l0/h=,1.086,>1.0，,取ξ2=,1.00,,,,,

,,,,η=,1+(l0/h)2ξ1ξ2h0/1300e0,,,,,,,,

,,,,η=,1.012,,,,,,,,

,由《公路钢筋混凝土及预应力混凝土桥涵设计规范》(JTG3362—2018)第5.3.4条,,,,,,,,,,,,

,,,,e=ηe0+h/2-a=,2.063,m,,,,,,,

,,,,r0Nde=,fcdbx(h0-x/2),,,,,,,,

,,,,244.58=,13800x(,0.65-x/2),,,,,,,

,解得,,,x=,0.028m,≤,ξbh0=,0.53×,0.65=,0.345m,,,

,故为大偏心受压构件。,,,,,,,,,,,,

,,,,As=(fcdbx-r0Nd)/fsd=,0.00080595,m2,=806.0,mm2,,,,,

,,,,μ=100As/(bh0)=,0.12%,<,0.2%,,,,,,

,应按最小配筋率配置受拉钢筋。,,,,,,,,,,,,

,选用φ22@,290mm,，,实际As=,1310.8,mm2,,,,,,,

,,,,"0.51×10-3fcu,k1/2bh0=",1815.7kN,>,r0Vd=,353.9kN,,,,,

,故抗剪截面符合《公路钢筋混凝土及预应力混凝土桥涵设计规范》(JTG3362—2018)第5.2.11条的要求。,,,,,,,,,,,,

,由《公路钢筋混凝土及预应力混凝土桥涵设计规范》(JTG3362—2018)第5.2.12条,,,,,,,,,,,,

,,,,0.50×10-3α2ftdbh0=,451.8kN,>,r0Vd=,353.9kN,,,,,

,故可不进行斜截面抗剪承载力的验算，仅需按(JTG3362—2018)第9.3.12条构造要求配置箍筋。,,,,,,,,,,,,

,由《公路钢筋混凝土及预应力混凝土桥涵设计规范》(JTG3362—2018)第5.2.9条,,,,,,,,,,,,

,"VCS=a1a2a30.45*10-3bh0((2+0.6ρ)fcu,k0.5ρsvfsv=",,,,804.51kN,>,r0Vd=,353.9kN,,,,,

,故斜截面内混凝土与箍筋共同的抗剪承载力已满足要求。,,,,,,,,,,,,

,3、左、右侧板(A-C，E-F),,,,,,,,,,,,

,(1)板中,,,,,,,,,,,,

,l0=,3.50m，,h=,0.50m，,a=,0.05m，,h0=,0.45m，,b=,1.00m，,,,

,,Md=,109.70kN,·m，,Nd=,174.47kN,，Vd=,5.24kN,,,,,

,,,,e0=Md/Nd=,0.629,m,,,,,,,

,,,,i=h/121/2=,0.144,m,,,,,,,

,长细比,,,l0/i=,24.25,>17.5,,,,,,,

,由《公路钢筋混凝土及预应力混凝土桥涵设计规范》(JTG3362—2018)第5.3.9条,,,,,,,,,,,,

,,,,ξ1=0.2+2.7e0/h0=,3.989,>1.0，,取ξ1=,1.00,,,,,

,,,,ξ2=1.15-0.01l0/h=,1.080,>1.0，,取ξ2=,1.00,,,,,

,,,,η=,1+(l0/h)2ξ1ξ2h0/1300e0,,,,,,,,

,,,,η=,1.027,,,,,,,,

,由《公路钢筋混凝土及预应力混凝土桥涵设计规范》(JTG3362—2018)第5.3.4条,,,,,,,,,,,,

,,,,e=ηe0+h/2-a=,0.844,m,,,,,,,

,,,,r0Nde=,fcdbx(h0-x/2),,,,,,,,

,,,,161.91=,13800x(,0.45-x/2),,,,,,,

,解得,,,x=,0.027m,≤,ξbh0=,0.53×,0.45=,0.237m,,,

,故为大偏心受压构件。,,,,,,,,,,,,

,,,,As=(fcdbx-r0Nd)/fsd=,0.000547645,m2,=547.6,mm2,,,,,

,,,,μ=100As/(bh0)=,0.12%,<,0.2%,,,,,,

,应按最小配筋率配置受拉钢筋。,,,,,,,,,,,,

,选用φ22@,420mm,，,实际As=,905.1,mm2,,,,,,,

,,,,"0.51×10-3fcu,k1/2bh0=",1251.4kN,>,r0Vd=,5.8kN,,,,,

,故抗剪截面符合《公路钢筋混凝土及预应力混凝土桥涵设计规范》(JTG3362—2018)第5.2.11条的要求。,,,,,,,,,,,,

,由《公路钢筋混凝土及预应力混凝土桥涵设计规范》(JTG3362—2018)第5.2.12条,,,,,,,,,,,,

,,,,0.50×10-3α2ftdbh0=,311.4kN,>,r0Vd=,5.8kN,,,,,

,故可不进行斜截面抗剪承载力的验算，仅需按(JTG3362—2018)第9.3.12条构造要求配置箍筋。,,,,,,,,,,,,

,(2)结点,,,,,,,,,,,,

,l0=,3.50m，,,h=t+C2=,0.70m，,a=,0.05m，,h0=,0.65m，,b=,1.00m，,,

,,Md=,149.15kN,·m，,Nd=,174.47kN,，Vd=,34.94kN,,,,,

,,,,e0=Md/Nd=,0.855,m,,,,,,,

,,,,i=h/121/2=,0.202,m,,,,,,,

,长细比,,,l0/i=,17.32,<17.5,,,,,,,

,由《公路钢筋混凝土及预应力混凝土桥涵设计规范》(JTG3362—2018)第5.3.9条，不考虑偏心距增大系数。,,,,,,,,,,,,

,,,,ξ1=0.2+2.7e0/h0=,3.762,>1.0，,取ξ1=,1.00,,,,,

,,,,ξ2=1.15-0.01l0/h=,1.100,>1.0，,取ξ2=,1.00,,,,,

,,,,η=,1+(l0/h)2ξ1ξ2h0/1300e0,,,,,,,,

,,,,η=,1.000,,,,,,,,

,由《公路钢筋混凝土及预应力混凝土桥涵设计规范》(JTG3362—2018)第5.3.4条,,,,,,,,,,,,

,,,,e=ηe0+h/2-a=,1.153,m,,,,,,,

,,,,r0Nde=,fcdbx(h0-x/2),,,,,,,,

,,,,221.25=,13800x(,0.65-x/2),,,,,,,

,解得,,,x=,0.025m,≤,ξbh0=,0.53×,0.65=,0.343m,,,

,故为大偏心受压构件。,,,,,,,,,,,,

,,,,As=(fcdbx-r0Nd)/fsd=,0.000473628,m2,=473.6,mm2,,,,,

,,,,μ=100As/(bh0)=,0.07%,<,0.2%,,,,,,

,应按最小配筋率配置受拉钢筋。,,,,,,,,,,,,

,选用φ22@,290mm,，,实际As=,1310.8,mm2,,,,,,,

,,,,"0.51×10-3fcu,k1/2bh0=",1810.1kN,>,r0Vd=,38.4kN,,,,,

,故抗剪截面符合《公路钢筋混凝土及预应力混凝土桥涵设计规范》(JTG3362—2018)第5.2.11条的要求。,,,,,,,,,,,,

,由《公路钢筋混凝土及预应力混凝土桥涵设计规范》(JTG3362—2018)第5.2.12条,,,,,,,,,,,,

,,,,0.50×10-3α2ftdbh0=,450.4kN,>,r0Vd=,38.4kN,,,,,

,故可不进行斜截面抗剪承载力的验算，仅需按(JTG3362—2018)第9.3.12条构造要求配置箍筋。,,,,,,,,,,,,

,3、中间板(B-D),,,,,,,,,,,,

,(1)板中,,,,,,,,,,,,

,l0=,3.50m，,h=,0.50m，,a=,0.05m，,h0=,0.45m，,b=,1.00m，,,,

,,Md=,32.00kN,·m，,Nd=,321.70kN,，Vd=,31.10kN,,,,,

,,,,e0=Md/Nd=,0.099,m,,,,,,,

,,,,i=h/121/2=,0.144,m,,,,,,,

,长细比,,,l0/i=,24.25,>17.5,,,,,,,

,由《公路钢筋混凝土及预应力混凝土桥涵设计规范》(JTG3362—2018)第5.3.9条,,,,,,,,,,,,

,,,,ξ1=0.2+2.7e0/h0=,0.797,≤1.0，,取ξ1=,0.80,,,,,

,,,,ξ2=1.15-0.01l0/h=,1.080,>1.0，,取ξ2=,1.00,,,,,

,,,,η=,1+(l0/h)2ξ1ξ2h0/1300e0,,,,,,,,

,,,,η=,1.136,,,,,,,,

,由《公路钢筋混凝土及预应力混凝土桥涵设计规范》(JTG3362—2018)第5.3.4条,,,,,,,,,,,,

,,,,e=ηe0+h/2-a=,0.313,m,,,,,,,

,,,,r0Nde=,fcdbx(h0-x/2),,,,,,,,

,,,,110.75=,13800x(,0.45-x/2),,,,,,,

,解得,,,x=,0.018m,≤,ξbh0=,0.53×,0.45=,0.239m,,,

,故为大偏心受压构件。,,,,,,,,,,,,

,,,,As=(fcdbx-r0Nd)/fsd=,-0.000311143,m2,-=311.1,mm2,,,,,

,,,,μ=100As/(bh0)=,-0.07%,<,0.2%,,,,,,

,应按最小配筋率配置受拉钢筋。,,,,,,,,,,,,

,选用φ22@,420mm,，,实际As=,905.1,mm2,,,,,,,

,,,,"0.51×10-3fcu,k1/2bh0=",1257.0kN,>,r0Vd=,34.2kN,,,,,

,故抗剪截面符合《公路钢筋混凝土及预应力混凝土桥涵设计规范》(JTG3362—2018)第5.2.11条的要求。,,,,,,,,,,,,

,由《公路钢筋混凝土及预应力混凝土桥涵设计规范》(JTG3362—2018)第5.2.12条,,,,,,,,,,,,

,,,,0.50×10-3α2ftdbh0=,312.8kN,>,r0Vd=,79.5kN,,,,,

,故可不进行斜截面抗剪承载力的验算，仅需按(JTG3362—2018)第9.3.12条构造要求配置箍筋。,,,,,,,,,,,,

,(2)结点,,,,,,,,,,,,

,l0=,3.50m，,,h=t+C2=,0.90m，,a=,0.05m，,h0=,0.85m，,b=,1.00m，,,

,,