土木工程施工练习题与答案

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1、土木工程施工练习题与答案1-1某场地如图1-58所示，方格边长为40m，①试按挖、填平衡原则确定场地的计划标高H0，然后据以算出方格角点的施工高度、绘出零线，计算挖方量和填方量。②当ix=2‰，iy=0，试确定方格角点计划标高。③当ix=2‰，iy=3‰时试确定方格角点计划标高。解：①计算场地的计划标高H0∑H1=45.8+43.0+42.2+44.1=175.12∑H2=2×(45.2+44.0+42.8+42.9+44.2+44.8)=527.83∑H3=04∑H4=4×(44.6+43.7)=

2、353.2åH1+2åH2+3åH3+4åH4175.1+527.8+0+353.2H===44.00(m)04N4´6计算方格各角点施工高度h=H0-Hixh1=44.00－45.8=－1.80(m)h2=44.00－45.2=－1.20(m)1－1.82－1.2304+1.0iyh3=44.00－44.0=0(m)45.844.045.244.044.044.043.044.0h4=44.00－43.0=1.00(m)1-11-21-3h5=44.00－44.8=－0.80(m)5－0.86－0

3、.67+0.38+1.2h6=44.00－44.6=－0.60(m)44.844.044.644.043.744.042.844.0h7=44.00－43.7=0.30(m)2-12-22-3h8=44.00－42.8=1.20(m)9－0.110－0.211+1.112+1.8h9=44.00－44.1=－0.10(m)44.144.044.244.042.944.042.244.0h10=44.00－44.2=－0.20(m)h11=44.00－42.9=1.10(m)角点编号施工高度h12=4

4、4.00－42.2=1.80(m)图例地面标高设计标高确定零线角点3的施工高度h3=0即为零点角点6施工高度位（－）、角点7施工高度（+），距角点7零点的距离www.docin.com/mydoc-638096-1.htmlah740´0.30x===13.33(m)7-6h+h0.3+0.676角点10施工高度位（－）、角点11施工高度（+），距角点11零点的距离ah1140´1.10x===33.85(m)11-10h+h1.10+0.201110零线为以上三点的连线计算挖方量22挖a403V=

5、(h+h+h+h)=(-)(1.8+1.2+0.8+0.6)=(-)1760.00(m)1-1125644更多试题查看www.docin.com/p-64828948.html22挖a403V=(h+h+h+h)=(-)(0.8+0.6+0.1+0.2)=(-)680.00(m)2-15691044222222挖ah10h6400.20.63V=(+)=(-)(+)=(-)172.31(m)2-24h+hh+h40.2+1.10.6+0.310116722挖a40V=(2h+h+2h-h)+V=(-

6、)(2´0.6+1.2+2´0-0.3)+26.67=(-)586.671-262377663V挖=（－）1760.00+680.00+172.31+586.67=（－）3198.98(m)计算填方量22填a403V=(h+h+h+h)=(+)(0+1.0+0.3+1.2)=(+)1000.00(m)1-334784422填a403V=(h+h+h+h)=(+)(0.3+1.2+1.1+1.8)=(+)1760.00(m)2-378111244222222填ah7h11400.31.13V=(+)=

7、(+)(+)=(+)412.31(m)2-24h+hh+h40.6+0.30.2+1.16710112323填ah7400.33V==(+)=（+）26.67(m)1-26(h+h)(h+h)6(0.6+0.3)(0+0.3)67373V填=(+)1000.00+1760.00+412.31+26.67=(+)3198.98(m)3V挖=V填=3198.98(m)②当ix=2‰，iy=0，计算方格角点计划标高Hn=H0±l•iH1=44.00－(40+20)×2‰=43.88(m)H2=44.00－

8、20×2‰=43.96(m)H3=44.00+20×2‰=44.04(m)H4=44.00+(40+20)×2‰=44.12(m)H5=44.00－(40+20)×2‰=43.88(m)H6=44.00－20×2‰=43.96(m)H7=44.00www.docin.com/mydoc-638096-1.html+20×2‰=44.04(m)H8=44.00+(40+20)×2‰=44.12(m)H9=44.00－(40+20)×2‰=43.88(m)H10=44.0