1、孔径及净空净跨径L 0 =4m 净高h 0 = 2.5m孔数m=22、设计安全等级一级结构重要性系数r 0 = 1.13、汽车荷载荷载等级公路 —Ⅰ级4、填土情况涵顶填土高度H =0.55m 土的内摩擦角Φ =30°填土容重γ1 =18kN/m 3地基容许承载力[σ0] =150kPa5、建筑材料普通钢筋种类HRB335主钢筋直径20mm 钢筋抗拉强度设计值f sd =280MPa涵身混凝土强度等级C 25涵身混凝土抗压强度设计值f cd =11.5MPa 涵身混凝土抗拉强度设计值f td = 1.23MPa 钢筋混凝土重力密度γ2 =25kN/m 3基础混凝土强度等级C 15混凝土重力密度γ3 =24kN/m 3(一)截面尺寸拟定 (见图L-01)顶板、底板厚度δ =0.35m C 1 =0.25m 侧墙厚度t =0.35m C 2 =0.25m 横梁计算跨径L P = L 0+t = 4.35m L = 2L 0+3t =9.05m 侧墙计算高度h P = h 0+δ = 2.85m h = h 0+2δ =3.2m 基础襟边 c =0.1m 基础高度 d =0.1m 基础宽度 B =9.25m(二)荷载计算1、恒载恒载竖向压力p 恒 = γ1H+γ2δ =18.65kN/m 2恒载水平压力顶板处e P1 = γ1Htan 2(45°-φ/2) = 3.30kN/m 2图 L-01底板处e P2 = γ1(H+h)tan 2(45°-φ/3) =22.50kN/m 22、活载汽车后轮着地宽度0.6m ，由《公路桥涵设计通用规范》(JTG D60—2004)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<1.3/2 m <1.8/2m故横向分布宽度a = (0.6/2+Htan30°)×2+1.3 =2.535m 同理，纵向，汽车后轮着地长度0.2m0.2/2+Htan30°=0.418 m <1.4/2m 故b = (0.2/2+Htan30°)×2 =0.835m ∑G =140kN 车辆荷载垂直压力q 车 = ∑G/(a×b) =66.13kN/m 2车辆荷载水平压力e 车 = q 车tan 2(45°-φ/2) =22.04kN/m2(三)内力计算1、构件刚度比K = (I 1/I 2)×(h P /L P ) =0.66钢筋混凝土双孔箱涵结构设计0.62 m0.6/2+Htan30°=一 、 设 计 资 料二 、 设 计 计算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M aA = M aC = M aE = M aF =-1/u·pL P2/12M BA = M BE = M DC = M DF =-(3K+1)/u·pL P2/12M BD = M DB =0横梁内法向力N a1 = N a2 = Na1' = Na2'=0侧墙内法向力N a3 = N a4 =(M BA-M aA+pL p2/2)/Lp图 L-02Na5=-(N a3+N a4)恒载p = p恒 =18.65kN/m2M aA = M aC = M aE = M aF =-12.73kN·mM BA = M BE = M DC = M DF =-37.75kN·mN a3 = N a4 =34.81kNNa5=-69.62kN车辆荷载p = q车 =66.13kN/m2M aA = M aC = M aE = M aF =-45.14kN·mM BA = M BE = M DC = M DF =-133.85kN·mN a3 = N a4 =123.44kNNa5=-246.88kN(2)b种荷载作用下 (图L-03)M bA = M bC = M bE = M bF =-K·ph P2/6uM BA = M BE = M DC = M DF =K·ph P2/12uM BD = M DB =0N b1 = N b2 = Nb1' = Nb2'=ph P/2N b3 = N b4 =(M BA-M bA)/L pN b5=-(N b3+N b4)恒载p = e P1 = 3.30kN/m2M bA = M bC = M bE = M bF =-1.27kN·mM BA = M BE = M DC = M DF =0.63kN·m图 L-03N b1 = N b2 = N b1' = N b2'= 4.70kNN b3 = N b4 =-0.44kNN b5=0.87kN(3)c种荷载作用下 (图L-04)Φ=20u(K+6)/K=469.36M cA = M cE =-(8K+59)·ph P2/6ΦM cC = M cF =-(12K+61)·ph P2/6ΦM BA = M BE =(7K+31)·ph P2/6ΦM DC = M DF =(3K+29)·ph P2/6ΦM BD = M DB =0N c1 = N c1'=ph P/6+(M cC-M cA)/h PN c2 = N c2'=ph P/3-(M cC-M cA)/h PN c3 = N c4 =(M BA-M cA)/L pN c5 =-(N c3+N c4)恒载p = e P2-e P1 =19.20kN/m2M cA = M cE =-3.56kN·mM cC = M cF =-3.81kN·mM BA = M BE = 1.97kN·m图 L-04M DC = M DF = 1.71kN·mN c1 = N c1'=9.03kNN c2 = N c2'=18.33kNN c3 = N c4 = 1.27kNN c5 =-2.54kN(4)d种荷载作用下 (图L-05)Φ1=20(K+2)(6K2+6K+1)=398.62Φ2=u/K= 3.53Φ3=120K3+278K2+335K+63=435.56Φ4=120K3+529K2+382K+63=574.10Φ5=360K3+742K2+285K+27=633.47Φ6=120K3+611K2+558K+87=748.61M dA =(-2/Φ2+Φ3/Φ1)·ph P2/4M dE =(-2/Φ2-Φ3/Φ1)·ph P2/4M dC =-(2/Φ2+Φ5/Φ1)·ph P2/24M dF =-(2/Φ2-Φ5/Φ1)·ph P2/24M BE =-(-2/Φ2-Φ4/Φ1)·ph P 2/24M DC =(1/Φ2+Φ6/Φ1)·ph P 2/24M DF =(1/Φ2-Φ6/Φ1)·ph P 2/24M BD =-Φ4·ph P 2/12Φ1M DB =Φ6·ph P 2/12Φ1N d1 =(M dC +ph P 2/2-M dA )/h P 图 L-05N d2 =ph p -N d1N d1' =(M dF -M dE )/h P N d2' =ph p -N d1'N d3 =(M BA +M BD -M dA )/L P N d4 =(M BE +M BD -M dE )/L P N d5 =-(N d3+N d4)车辆荷载p = e 车 =22.04kN/m 2M dA =23.52kN ·m M dE =-74.30kN ·m M dC =-16.09kN ·m M dF =7.62kN ·m M BA =-14.98kN ·m M BE = 6.51kN ·m M DC =16.13kN ·m M DF =-11.89kN ·m M BD =-21.49kN ·m M DB =28.02kN ·m N d1 =-45.31kN N d2 =108.13kN N d1' =0.39kN N d2' =62.43kN N d3 =-13.79kN N d4 =13.64kN N d5 =0.15kN(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》(JTG D60—2004)第4.1.6条进行承载能力极限状态效应组合3、构件内力计算(跨中截面内力)(1)顶板1 (图L-06)x =L P /2P = 1.2p 恒+1.4q 车 =114.96kN N x = N 1 =-44.21kNV x = Px-N 3 =53.59kN 顶板1'x =L P /2P = 1.2p 恒+1.4q 车 =114.96kN N x = N 1' =19.77kN M x = M E +N 4x-Px 2/2 =49.64kN ·m V x = Px-N 4 =15.19kN(2)底板2 (图L-07)ω1 =1.2p 恒+1.4(q 车+3e 车H P 2/4L P 2)=124.90kN/m 2ω2 =1.2p 恒+1.4q 车=114.96kN/m 2x =L P /2N x = N 2 =183.63kNM x =M C +N 3x-ω2·x 2/2-5x 3(ω1-ω2)/12L P =37.47kN ·m V x =ω2x+3x 2(ω1-ω2)/2L P -N 3=69.80kN底板2'ω1 =1.2p 恒+1.4q 车=114.96kN/m 2ω2 =1.2p 恒+1.4(q 车-3e 车H P 2/4L P 2)=105.03kN/m 2x =L P /2N x = N 2' =119.65kNM x =M F +N 4x-ω2·x 2/2-x 3(ω1-ω2)/6L P=183.56kN ·m V x =ω2x+x 2(ω1-ω2)/2L P -N 4=4.39kN(3)左侧墙 (图L-08)ω1 =1.4e P1+1.4e 车=35.48kN/m 2ω2 =1.4e P2+1.4e 车62.36kN/m 2x =h P /2N x = N 3 =196.45kNM x =M A +N 1x-ω1·x 2/2-x 3(ω2-ω1)/6h P =-155.86kN ·m V x =ω1x+x 2(ω2-ω1)/2h P -N 1=104.35kN(4)右侧墙 (图L-09)ω1 = 1.4e P1 = 4.62kN/m 2ω2 = 1.4e P2 =31.50kN/m 2x =h P /2N x = N 4=234.85kNM x =M E +N 1'x-ω1·x 2/2-x 3(ω2-ω1)/6h P =-170.30kN ·m V x =ω1x+x 2(ω2-ω1)/2h P -N 1'=-3.61kN(5)中间墙 (图L-10)x =h P /2N x = N 5=-431.30kNM x =M BD +(N1+N 1')x =-64.91kN ·m V x =-(N 1+N 1')=24.44kN(5)构件内力汇总表图 L-09图 L-10图 L-06图 L-07图L-08(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。