1、孔径及净空净跨径L 0 =  1.9m 净高h 0 =2m
孔数
m=
12、设计安全等级二级结构重要性系数r 0 =  1.0
3、汽车荷载荷载等级公路 —
Ⅱ级
4、填土情况涵顶填土高度H =7.3m 土的内摩擦角Φ =30°填土容重γ1 =19kN/m 3地基容许承载力[σ0] =
300
kPa
5、建筑材料普通钢筋种类HRB400主钢筋直径25mm 钢筋抗拉强度设计值f sd  =
330MPa
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 20混凝土重力密度γ3 =
23.5
kN/m 3(一)截面尺寸拟定 (见图L-01)顶板、底板厚度
δ =0.5m C 1 =
0.5
m
钢 筋 混 凝 土 箱 涵 结 构 设 计
一 、 设  计  资  料
二 、 设  计  计  算
侧墙厚度t =0.6m C 2 =
0.6m 横梁计算跨径L P  = L 0+t =  2.5m L = 3L 0+4t =8.1m 侧墙计算高度h P  = h 0+δ =  2.5m h = h 0+2δ =
3m 基础襟边  c =0m 基础高度  d =0m 基础宽度  B =
8.1
m
(二)荷载计算1、恒载恒载竖向压力p 恒 = γ1H+γ2δ =151.20
kN/m 2恒载水平压力顶板处e P1 = γ1Htan 2(45°-φ/2) =46.23kN/m 2图 L-01
底板处e P2 = γ1(H+h)tan 2(45°-φ/3) =
65.23
kN/m 2
2、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
>1.3/2m >1.8/2m
故横向分布宽度
a = 1.8+1.3 =
3.100
m
同理，纵向，汽车后轮着地长度0.2m
0.2/2+Htan30°=  4.315 m >1.4/2m
故
b =    1.400m ∑G =
140kN 车辆荷载垂直压力
q 车 = ∑G/(a³b) =
32.26kN/m 2车辆荷载水平压力e 车 = q 车tan 2(45°-φ/2) =10.75
kN/m 2
(三)内力计算1、构件刚度比
4.51 m
0.6/2+Htan30°=
K = (I1/I2)³(h P/L P) =0.58
u=2K+1=  2.16
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/u²pL P2/12
M BA = M BE = M DC = M DF =-(3K+1)/u²pL P2/12
M 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
Na5=-(N a3+N a4)
恒载p = p恒 =151.20kN/m2
M aA = M aC = M aE = M aF =-36.50kN²m
M BA = M BE = M DC = M DF =-99.87kN²m
N a3 = N a4 =163.65kN
Na5=-327.30kN
车辆荷载p = q车 =32.26kN/m2
M aA = M aC = M aE = M aF =-7.79kN²m图 L-02
M BA = M BE = M DC = M DF =-21.31kN²m
N a3 = N a4 =34.91kN
Na5=-69.83kN
(2)b种荷载作用下 (图L-03)
M bA = M bC = M bE = M bF =-K²ph P2/6u
M BA = M BE = M DC = M DF =K²ph P2/12u
M BD = M DB =0
N b1 = N b2 = Nb1' = Nb2'=ph P/2
N b3 = N b4 =(M BA-M bA)/L p
N b5=-(N b3+N b4)
恒载p = e P1 =46.23kN/m2
M bA = M bC = M bE = M bF =-12.92kN²m
M BA = M BE = M DC = M DF =  6.46kN²m图 L-03
N b1 = N b2 = N b1' = N b2'=57.79kN
N b3 = N b4 =-7.75kN
N b5=15.50kN
(3)c种荷载作用下 (图L-04)
Φ=20u(K+6)/K=490.51
M cA = M cE =-(8K+59)²ph P2/6Φ
M cC = M cF =-(12K+61)²ph P2/6Φ
M BA = M BE =(7K+31)²ph P2/6Φ
M DC = M DF =(3K+29)²ph P2/6Φ
M BD = M DB =0
N c1 = N c1'=ph P/6+(M cC-M cA)/h P
N c2 = N c2'=ph P/3-(M cC-M cA)/h P
N c3 = N c4 =(M BA-M cA)/L p
N c5 =-(N c3+N c4)
恒载p = e P2-e P1 =19.00kN/m2
M cA = M cE =-2.57kN²m
M cC = M cF =-2.74kN²m
M BA = M BE =  1.41kN²m图 L-04
M DC = M DF =  1.24kN²m
N c1 = N c1'=7.85kN
N c2 = N c2'=15.90kN
N c3 = N c4 =  1.59kN
N c5 =-3.19kNphp实例计算
(4)d种荷载作用下 (图L-05)
Φ1=20(K+2)(6K2+6K+1)=334.28
Φ2=u/K=  3.73
Φ3=120K3+278K2+335K+63=373.22
Φ4=120K3+529K2+382K+63=484.48
Φ5=360K3+742K2+285K+27=510.20
Φ6=120K3+611K2+558K+87=637.80
M dA =(-2/Φ2+Φ3/Φ1)²ph P2/4
M dE =(-2/Φ2-Φ3/Φ1)²ph P2/4
M dC =-(2/Φ2+Φ5/Φ1)²ph P2/24
M dF =-(2/Φ2-Φ5/Φ1)²ph P2/24
M BA =-(-2/Φ2+Φ4/Φ1)²ph P2/24
M BE =-(-2/Φ2-Φ4/Φ1)²ph P2/24
M DC =(1/Φ2+Φ6/Φ1)²ph P2/24
M DF =(1/Φ2-Φ6/Φ1)²ph P2/24
M BD =-Φ4²ph P2/12Φ1
M DB =Φ6²ph P2/12Φ1
N d1 =(M dC+ph P2/2-M dA)/h P图 L-05
N d2 =ph p-N d1
N 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 =10.75kN/m2
M dA =9.74kN²m
M dE =-27.77kN²m
M dC =-5.78kN²m
M dF =  2.77kN²m
M BA =-5.56kN²m
M BE =  2.56kN²m
M DC =  6.09kN²m
M DF =-4.59kN²m
M BD =-8.12kN²m
M DB =10.69kN²m
N d1 =-19.65kN
N d2 =46.53kN
N d1' =0.09kN
N d2' =26.80kN
N d3 =-9.37kN
N d4 =8.88kN
N d5 =0.48kN
(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

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