1、孔径及净空净跨径L 0 = 4.00m 净高h 0 =4.00m2、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 =7.2m 土的内摩擦角Φ =35°填土容重γ1 =19kN/m 3地基容许承载力[σ0] =260kPa5、建筑材料普通钢筋种类HRB335主钢筋直径25mm 钢筋抗拉强度设计值f sd =280MPa涵身混凝土强度等级C40涵身混凝土抗压强度设计值f cd =18.4MPa 涵身混凝土抗拉强度设计值f td = 1.65MPa 钢筋混凝土重力密度γ2 =25kN/m 3基础混凝土强度等级C10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 =0.4m C 1 =0.05m 侧墙厚度t =0.4m C 2 =0.05m 横梁计算跨径L P = L 0+t = 4.4m L = L 0+2t = 4.8m 侧墙计算高度h P = h 0+δ =4.4m钢 筋 混 凝 土 箱 涵 结 构 设一 、 设 计 资 料二 、 设 计 计 算h = h 0+2δ =4.8m 基础襟边 c =0.2m 基础高度 d =0.2m 基础宽度 B =5.2m(二)荷载计算1、恒载恒载竖向压力p 恒 = γ1H+γ2δ =146.80kN/m 2恒载水平压力顶板处e P1 = γ1Htan 2(45°-φ/2) =37.07kN/m 2底板处e P2 = γ1(H+h)tan 2(45°-φ/2) =61.79kN/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°角向一个汽车后轮横向分布宽> 1.3/2 m > 1.8/2 m故车轮压力扩散线相重叠，应按如下计算横向分布宽度a = (0.6/2+Htan30°)×2+1.3 =3.100m同理，纵向，汽车后轮着地长度0.2m0.2/2+Htan30°= 4.257 m > 1.4/2 m故b = (0.2/2+Htan30°)×2 =1.400m ∑G =140kN 车辆荷载垂直压力q 车 = ∑G /(a×b) =32.26kN/m 2车辆荷载水平压力e 车 = q 车tan 2(45°-φ/2) =8.74kN/m 2(三)内力计算1、构件刚度比K = (I 1/I 2)×(h P /L P ) =1.002、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 aB = M aC = M aD =-1/(K+1)·pL P 2/12横梁内法向力N a1 = N a2 =0侧墙内法向力N a3 = N a4 =pL P /2恒载p = p 恒 =146.80kN/m 2M aA = M aB = M aC = M aD =-118.42kN ·m N a3 = N a4 =322.96kN 车辆荷载p = q 车 =32.26kN/m 20.6/2+Htan30°=4.46 mM aA = M aB = M aC = M aD =-26.02kN·mN a3 = N a4 =70.97kN(2)b种荷载作用下 (图L-03)M bA = M bB = M bC = M bD =-K/(K+1)·ph P2/12N b1 = N b2 =ph P/2N b3 = N b4 =0恒载p = e P1 =37.07kN/m2M bA = M bB = M bC = M bD =-29.90kN·mN b1 = N b2 =81.56kN(3)c种荷载作用下 (图L-04)M cA = M cD =-K(3K+8)/[(K+1)(K+3)]·ph P2/60M cB = M cC =-K(2K+7)/[(K+1)(K+3)]·ph P2/60N c1 =ph P/6+(M cA-M cB)/h PN c2 =ph P/3-(M cA-M cB)/h PN c3 = N c4 =0恒载p = e P2-e P1 =24.71kN/m2M cA = M cD =-10.96kN·mM cB = M cC =-8.97kN·mN c1 =17.67kNN c2 =36.70kN(4)d种荷载作用下 (图L-05)M dA =-[K(K+3)/[6(K2+4K+3)]+(10K+2)/(15K+5)]·ph P2/4M dB =-[K(K+3)/[6(K2+4K+3)]-(5K+3)/(15K+5)]·ph P2/4M dC =-[K(K+3)/[6(K2+4K+3)]+(5K+3)/(15K+5)]·ph P2/4M dD =-[K(K+3)/[6(K2+4K+3)]-(10K+2)/(15K+5)]·ph P2/4N d1 =(M dD-M dC)/h PN d2 =ph P-(M dD-M dC)/h PN d3 = N d4 =-(M dB-M dC)/L P车辆荷载p = e车 =8.74kN/m2M dA =-28.91kN·mM dB =13.40kN·mM dC =-20.45kN·mM dD =21.86kN·mN d1 =9.62kNN d2 =28.85kNN d3 = N d4 =-7.69kN(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3、构件内力计算(跨中截面内力)(1)顶板 (图L-06)x =L P/2P = p恒+0.7q车 =169.38kNN x = N1 =105.96kNM x = M B+N3x-Px2/2 =231.92kN·mV x = Px-N3 = 5.38kN(2)底板 (图L-07)ω1 =p恒+0.7(q车-3e车H P2/L P2)=151.02kN/m2ω2 =p恒+0.7(q车+3e车H P2/L P2)=187.74kN/m2x =L P/2N x = N2 =138.45kNM x =M A+N3x-ω1·x2/2-x3(ω2-ω1)/6L P=229.93kN·mV x =ω1x+x2(ω2-ω1)/2L P-N3=-14.81kN(3)左侧墙 (图L-08)ω1 =0.7e P1+0.7e车=32.07kN/m2ω2 =0.7e P2+0.7e车49.37kN/m2x =h P/2N x = N3 =367.25kNM x =M B+N1x-ω1·x2/2-x3(ω2-ω1)/6h P=-17.61kN·mV x =ω1x+x2(ω2-ω1)/2h P-N1=-25.89kN(4)右侧墙 (图L-09)ω1 = 0.7e P1 =25.95kN/m2ω2 = 0.7e P2 =43.25kN/m2x =h P/2N x = N4 =367.25kNM x =M C+N1x-ω1·x2/2-x3(ω2-ω1)/6h P=-26.49kN·mV x =ω1x+x2(ω2-ω1)/2h P-N1=-39.35kN(5)短期组合下构件内力汇总表(四)截面设计1、顶板 (B-C)钢筋按左、右对称，用最不利荷载计算。

(1)跨中l0 =4.40 m ，h =0.40 m ， a =0.05 m ，h0 = M d =231.92 kN·m ，N d =105.96 kN， V d =e0 = M d/N d = 2.189i =h/121/2 =0.115m长细比l0/i =38.11> 17.5由《公路钢筋混凝土及预应力混凝土桥涵设计规范》(JTG D62—2004)第5.3.10条ξ1 = 0.2+2.7e0/h0 =17.085> 1.0 ，取ξ1 =ξ2 = 1.15-0.01l0/h = 1.040> 1.0 ，取ξ2 =η =1+(l0/h)2ξ1ξ2h0/1400e0η = 1.014由《公路钢筋混凝土及预应力混凝土桥涵设计规范》(JTG D62—2004)第5.3.5条e = ηe0+h/2-a = 2.369mr0N d e =f cd bx(h0-x/2)276.13 =18400x(0.35-x/2)解得x =0.046 m≤ξb h0 =故为大偏心受压构件。

A s = (f cd bx-r0N d)/f sd =0.002599m2= 2599.0μ = 100A s/(bh0) =0.74 %>0.2 %符合《公路钢筋混凝土及预应力混凝土桥涵设计规范》(JTG D62—2004)第9.1.12条的选用 φ25 @180 mm，实际 A s =2945.2mm20.51×10-3f cu,k1/2bh0 =1128.9 kN>r0V d =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》(JTG D62—2004)第由《公路钢筋混凝土及预应力混凝土桥涵设计规范》(JTG D62—2004)第5.2.10条0.50×10-3α2f td bh0 =288.8 kN>r0V d =故可不进行斜截面抗剪承载力的验算，仅需按(JTG D62—2004)第9.3.13条构造要求配(2)结点l0 =4.40 m ，h = δ+C1 =0.45 m ， a =0.05 m ，M d =189.82 kN·m ，N d =105.96 kN， V d =e0 = M d/N d = 1.791mi =h/121/2 =0.130m长细比l0/i =33.87> 17.5由《公路钢筋混凝土及预应力混凝土桥涵设计规范》(JTG D62—2004)第5.3.10条ξ1 = 0.2+2.7e0/h0 =12.293> 1.0 ，取ξ1 =ξ2 = 1.15-0.01l0/h = 1.052> 1.0 ，取ξ2 =η =1+(l0/h)2ξ1ξ2h0/1400e0η = 1.015由《公路钢筋混凝土及预应力混凝土桥涵设计规范》(JTG D62—2004)第5.3.5条e = ηe0+h/2-a = 1.994mr0N d e =f cd bx(h0-x/2)232.39 =18400x(0.40-x/2)解得x =0.033 m≤ξb h0 =故为大偏心受压构件。