发布网友 发布时间:2022-04-26 18:20
共1个回答
热心网友 时间:2023-10-20 16:58
涵洞与路线右交角为120°(α=90°-120°=-30°),路基边坡m0=1.5(即1:1.5),冀墙正截面背侧坡比n0=4(即4:1),正截面顶宽c0=40cm,洞口截面高H=479cm,冀尾截面高h=70cm,正侧面线转到涵洞轴线转角β=-20°(绕O点逆时针取负),涵洞轴线流水坡度i=2%。相关计算如下:1.墙身计算考虑流水坡度i合成:m=m0/(1±m0i/cosα),在上游取正,在下游取负, m=1.5/(1+1.5*2%/cos30°)=1.4498;2.涵洞轴线冀长:L=(H-h)m/cosα=(4.79-0.7)*1.4498/cos30°=6.847m;3.洞口截面墙顶宽:c= c0/cos(β-α)=0.40/cos(-20°+30°)=0.406m; 4.洞口截面墙背侧坡比: n=[n0+signβsin(β-α)/m]cos(β-α),sign为取符号函数signβ=sign(-20°)=-1,n=[4-sin(-20°+30°)/1.4498]cos(-20°+30°)=3.8213忽略流水坡度i影响, n’=[4-sin(-20°+30°)/1.5]cos(-20°+30°)=3.82525.洞口截面底宽:a=c+H/n=0.406+4.79/3.8213=1.660m,(n’→1.658m);6.冀尾截面底宽:b=c+h/n=0.406+0.70/3.8213=0.589m,(n’→0.588m);7.墙身体积计算(任取与洞口截面平行的一超薄dz段分析,如立面图阴影部分),其体积为: dV≈[(c+x) y /2]dz,其中x=c+y/n,dz =mdy代入得:dV≈ [y2/2/n+ cy]mdy,对y从h~H积分并整理得:V=0.5[(H3-h3)/3/n+c(H2-h2)]mV=0.5*[(4.793-0.73)/3/3.8213+0.406*(4.792-0.72)]*1.4498 =13.536m3 (n’→V=13.529m3) 令(H-h)m=Lcosα代入得:V=0.5[(H2+Hh+h2)/3/n+c(H+h)]Lcosα