箱涵结构和配筋计算

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 =-3.19kN

(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