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卫星桥应力空心板梁钢绞线张拉理论伸长量计算书FDNMECBA钢绞线示意图如下:M、N为张拉端计算钢绞线长度LL=LMA(直线)+LAB(直线)+LBc(曲线)+LCD(直线)+LDE(曲线)+LEF(直线)+LNF(直线)其中:LMA、LFN为锚垫板,工具锚及千斤顶总长度,按60cm计。则钢铰线由三个直线段及两个曲线段构成。L=LMB(直线)+LBC(曲线)+LCD(直线)+LDE(曲线)+LEN(直线)1、边板N1束钢绞线计算:各段长度为:LMB=84cm+60.6cm=144.6cm=1.446m;LBC=1.309m、θ=2×=0.0349rad;LCD=11.771m、θ=0;LDE=1.309m、θ=0.0349rad;LEN=1.446m、θ=0。张拉理论伸长量:△L1=2(△LMB+△LBC)+△LCD分段计算理论伸长量取:u=0.20k=0.0015Ay=140mm2×5=700mm2Ey=1.95×105Mpab=1.0σk=0.75RybMB段:张拉端控制力:P=σk·Ay·N××b=1860×0.75×700××1.0=1302×0.75=976.5KN应力损失:PSMB=P(1-e-(μθ+kx))=976.5×(1-e-(0.20×0+0.0015×1.446))=976.5×(1-e-0.00217)=976.5×(1-0.9978)=2.148KN终端力PB=976.5-2.148=974.35KN平均张拉力:BM=(P+PB)/2=975.4KN采用简算法:△LBM=BM·L/(Ay·Ey)=(975.4×103×1.446)/(700×1.95×105)=0.01033m(2)BC段:θ=0.0349rad应力损失:PSBC=PB(1-e-(μθ+kx)=974.35×(1-e-(0.2×0.0349+0.0015×1.309))=974.35×(1-0.9911)=8.672KN终端力PC=PB-PSBC=974.35-8.672=965.68KN平均张拉力:BC=(PB+PC)/2=(974.35+965.68)/2=970.0KN伸长值:△LBC=BC·LBC/(Ay·Ey)=(970.0×1.309×103)/(700×1.95×105)=0.0093mCD段:θ=0应力损失:PSCD=PC(1-e-kx)=965.68×(1-e-0.0015×11.771)=965.68×(1-e-0.0177)=965.68×(1-0.9825)=16.90KN终端力PD=PC-PSCD=965.68-16.90=948.78KN平均张拉力:CD=(965.68+948.78)/2=957.2KN计算伸长量:△LCD=(957.2×103×11.771)/(700×1.95×105)=0.0825m各计算数据见下表:N1束钢铰线理论伸长量计算表:线段θ(rad)L(m)KL+μθe-(kx+μθ)应力损失KN终点力KN平均张力KN△L1cmMB01.4460.002170.99782.148974.35975.41.033BC0.03491.3090.008940.99118.672965.68970.00.930CD011.7710.01770.982516.90948.78957.28.250DE0.03491.3090.008940.99118.672965.68970.00.930EN01.4460.002170.99782.148974.35975.41.033总理论伸长量△L1=2×(△LMB+△LBC)+LCD=2×(1.033+0.930)+8.25=3.926+8.25=12.176(cm)N2束伸长量计算:LMB=1.522(m)θ=0LBC=2.618(m)θ=12×π/180=0.209(rad)LCD=9.068(m)θ=0LDE=2.618(m)θ=0.209(rad)LEN=1.522(m)θ=0取μ=0.2k=0.0015Ay=140×5=700mm2Ey=1.95×105MPσK=0.75Ryb=1395MPaP=976.5KN。计算结果见下表:N2束钢铰线理论伸长量计算表:线段L(m)θKL+μθe-(kx+μθ)应力损失KN终点力KN平均张力KN△L1cmMB1.52200.002280.99772.2460974.25975.41.088BC2.6180.2090.04570.955343.549930.70952.51.827CD9.06800.01360.986512.564918.14924.46.14