(整理)温度场和温度应力计算

1、精品文档附计算书3：温度场和温度应力计算一、温度场计算计算以本工程1.2m厚底板为例，用差分法计算底板28d水化热温升曲线。计算中各参数的取值如下：W每m3胶凝材料用量，440kg/m3；Q胶凝材料水化热总量(kJ/kg)；,本例采用实测值260kJ/kg；C混凝土的比热，取1.0kJ/(kg.。C)；P混凝土的质量密度，取2400kg/m3；导温系数，取0.0035m2/h；m，取0.5。混凝土的入模温度取10。C，地基温度为18。C，大气温度为18。C。温度场计算差分公式如下：T+TAtAt、T=k十2a-T(2a-1)+AT(B.4.2-1)n,k+12Ax2n,kAx2n,k试算At、

2、Ax，确定u。Ax2取At=0.5天=12小时，Ax=0.4m，即分3层贝yu竺=0.0035X旦=0.2625沁1，可行。Ax20.424代入该值得出相应的差分法公式为T+TT二0.525-n,k+12n,kk画出相应的计算示意图，并进行计算。底板厚1.2m，分3层，每层0.4m，相应的计算示意如下图。从上至下各层混凝土的温度分别用T、T、T表示，相应k时刻各层的温度即为T、1231kT2,k、T3,k。混凝土与大气接触的上表面边界温度用T表示，与地基接触的下表面边界温度用T0表示。k=0，即第k-At二0-0.5二0天，上表面边界T，取大气温度，T=18。C00各层混凝土温度取入模温度，即

3、T1,0=T2,0=T30=10。C下表面边界T0，取地基温度，T0=18。C；k=1，即第k-At二1-0.5二0.5天，温升AT=T1max(e一m-(k-1)-Ate一m-k-At)二440-260-(e-0，(1-1).0.5-e-0，1.0.5)二1.0-240010.544。C上表面边界温度T，散热温升为0,始终保持不变，T=18。C00第一层混凝土温度，见计算图示中方框1，的边界为T0和，在的基础上考虑温升AT,即T+TT二0.525-0辺+0.475-T+AT二22.644。C1,121,01第二层混凝土温度，见计算图示中方框2,的边界为一,0和一,0，在的基础上考虑温升AT，

4、即1T+TT二0.525-丄辺+0.475-T+AT二20.544。C2,122,01第三层混凝土温度T3,1，见计算图示中方框3,T3,1的边界为T2,0和丁，在T3,0的基础上考虑温升AT,即1T+TT=0.525t+0.475-T+AT=22.644C23,01下表面边界温度T0，需要考虑散热温升AT/2，所以需每一步都需进行修正。见计算01图示中方框6，T0的边界为T30和地基温度18。C，在T0的基础上考虑温升AT/2，即03,001,T+18,ATT=0.525亠+0.475-T+1=26.444C0202以上即完成了一遍k=1时，各温度计算。同理k=2，即第k-At二2-0.5二

5、1天，温升AT=8.212。C，2上表面T0=18oC，T+TT=0.525亠21+0.475-T+AT=29.085C21,12T+TT=0.525+31+0.475-T+AT=29.858C22,12T+TT=0.525亠0+0.475-T+AT=31.302C23,12T+18ATT=0.525亠+0.475T+2=31.441C0202同理可计算k=356，即第1.528天的各层温度值，本算例中不再进行详细计算，最终计算结果如表1。表1不同混凝土龄期下各层混凝土温度值（单位：。C）混凝土龄期上表面第一层第二层第三层底界边层018101010180.51822.64420.54422.6

6、4426.44411829.08529.85831.30231.4411.51832.77336.42937.35434.27121834.83640.69341.28335.7902.51835.83343.18943.56536.44131836.10344.37844.61736.4913.51835.87644.62144.77436.12341835.31144.19844.29535.469混凝土龄期上表面第一层第二层第三层底界边层4.51834.52743.31843.38034.62751833.60742.13842.17733.6715.51832.61540.77440.

7、79932.65661831.59439.31339.32931.6206.51830.57737.81637.82630.59471829.58536.32736.33429.5957.51828.63234.87834.88228.63981827.72933.48733.49027.7338.51826.88032.16932.17126.88291826.08830.93230.93326.0899.51825.35329.77729.77825.354101824.67628.70828.70824.67610.51824.05327.72027.72124.053111823.48

8、226.81326.81323.48211.51822.96025.98225.98222.961121822.48525.22225.22222.48512.51822.05224.53024.53022.052131821.65923.90023.90021.65913.51821.30323.32823.32821.303141820.98022.80822.80820.98014.51820.68722.33822.33820.687151820.42321.91221.91220.42315.51820.18321.52721.52720.183161819.96821.17921.

9、17919.96816.51819.77320.86420.86419.773171819.59720.58120.58119.59717.51819.43820.32420.32419.438181819.29520.09320.09319.29518.51819.16619.88519.88519.166191819.05019.69719.69719.05019.51818.94519.52819.52818.945201818.85119.37619.37618.85120.51818.76619.23819.23818.766211818.68919.11519.11518.6892

10、1.51818.62019.00319.00318.620221818.55818.90318.90318.55822.51818.50218.81318.81318.502231818.45218.73118.73118.452混凝土龄期上表面第一层第二层第三层底界边层23.51818.40718.65818.65818.407241818.36618.59218.59218.36624.51818.32918.53318.53318.329251818.29618.48018.48018.29625.51818.26718.43118.43118.267261818.24018.38818

11、.38818.24026.51818.21618.34918.34918.216271818.19418.31418.31418.19427.51818.17518.28318.28318.175281818.15718.25418.25418.157二、温度应力计算示例以下计算示例按照步长为1天进行。1、里表温差4T1计算公式如下AT(t)二T(t)-T(t)(B.5.1)1mb计算结果见表2各龄期混凝土收缩当量温差计算根据公式8(t)=8o(1e-o.oit)-M-M-MM(B.2.1)yy12311T(t)=8(t)/a(B.2.2)yy其中80取4.0X10-4y石家庄市内年平均气温值

12、取14.5C计算结果见表3综合降温差t2计算公式如下AT(t)=-4T(tT(细T(t卡T)T他/)26mbmdmyw计算结果见表2表2综合温差计算结果混凝土龄期第一层(Tb)第三层(Tm)第三层(Td)T1(Tm-Tb)T2*T1*T2混凝土龄期第一层(Tb)第三层(Tm)第三层(Td)T1(Tm-Tb)T2*T1*T201010100.0000.000/-8.6610.522.64422.64422.6440.0008.6612.217-8.803129.08531.30231.3022.21717.4652.365-6.1711.532.77337.35437.3544.58123.63

13、61.866-4.127234.83641.28341.2836.44727.7631.285-2.5522.535.83343.56543.5657.73230.3150.781-1.422336.10344.61744.6178.51431.7370.384-0.2693.535.87644.77444.7748.89832.0060.0860.043435.31144.29544.2958.98431.963-0.1310.9394.534.52743.38043.3808.85331.024-0.2830.763533.60742.17742.1778.57030.261-0.3850

14、.9235.532.61540.79940.7998.18429.337-0.4501.007631.59439.32939.3297.73528.330-0.4851.0366.530.57737.82637.8267.24927.295-0.5001.720729.58536.33436.3346.74925.575-0.5001.0347.528.63234.88234.8826.25024.541-0.4880.977827.72933.49033.4905.76123.563-0.4700.9098.526.88032.17132.1715.29122.654-0.4460.8349

15、26.08830.93330.9334.84521.820-0.4200.7569.525.35329.77829.7784.42521.064-0.393-1.0121024.67628.70828.7084.03222.075-0.3640.52210.524.05327.72127.7213.66821.554-0.3370.4481123.48226.81326.8133.33121.105-0.3100.37911.522.96025.98225.9823.02120.726-0.2840.3131222.48525.22225.2222.73720.413-0.2600.25212

16、.522.05224.53024.5302.47820.161-0.2370.1961321.65923.90023.9002.24119.965-0.2160.14313.521.30323.32823.3282.02519.822-0.1960.4621420.98022.80822.8081.82919.360-0.1780.06414.520.68722.33822.3381.65119.296-0.1610.0241520.42321.91221.9121.49019.273-0.146-0.01215.520.18321.52721.5271.34319.285-0.132-0.0

17、451619.96821.17921.1791.21119.330-0.119-0.07516.519.77320.86420.8641.09219.405-0.108-0.1021719.59720.58120.5810.98419.507-0.097-0.12617.519.43820.32420.3240.88619.633-0.088-0.1481819.29520.09320.0930.79919.780-0.079-0.16718.519.16619.88519.8850.71919.947-0.072-0.1841919.05019.69719.6970.64820.132-0.

18、064-0.20019.518.94519.52819.5280.58320.332-0.058-0.2142018.85119.37619.3760.52520.546-0.052-0.226混凝土龄期第一层(Tb)第三层(Tm)第三层(Td)T1(Tm-Tb)T2*T1*T220.518.76619.23819.2380.47320.772-0.047-0.2372118.68919.11519.1150.42521.010-0.042-0.24721.518.62019.00319.0030.38321.257-0.038-0.2562218.55818.90318.9030.34521

19、.512-0.034-0.26322.518.50218.81318.8130.31021.776-0.031-0.2702318.45218.73118.7310.27922.046-0.028-0.27623.518.40718.65818.6580.25122.322-0.025-0.2812418.36618.59218.5920.22622.603-0.023-0.28624.518.32918.53318.5330.20322.888-0.020-0.2892518.29618.48018.4800.18323.178-0.018-0.29325.518.26718.43118.4

20、310.16523.471-0.017-0.2962618.24018.38818.3880.14823.766-0.015-0.29826.518.21618.34918.3490.13324.064-0.013-0.3002718.19418.31418.3140.12024.364-0.012-0.30127.518.17518.28318.2830.10824.665-0.011-0.3032818.15718.25418.2540.09724.968-0.097/注：表中*4片和*4T2列中括号内数值为按照以下公式计算结果，其值在计算自约束应力和外约束应力时使用。AT(t)=AT(t

21、)-AT(t-j)(B.6.1-2)1i11AT(t)二AT(t-j)-AT(t)(B.6.2-2)2i22表3各龄期混凝土收缩当量温差计算结果混凝土龄期y(t)(10=4)Ty(t)Ty(t)00.0000.0000.5170.50.0520.5170.51510.1031.0320.5121.50.1541.5450.51020.2052.0540.4842.50.2542.5390.50030.3043.0390.1763.50.3213.2150.45040.3673.665-0.0464.50.3623.6190.39250.4014.0120.3905.50.4404.4020.3

22、8860.4794.7900.386混凝土龄期8y(t)(10-4)Ty(t)Ty(t)6.50.5185.177-0.31170.4874.8660.3357.50.5205.2010.33380.5535.5340.3318.50.5875.8650.33090.6196.1950.3289.50.6526.5232.017100.8548.5390.40510.50.8948.9440.403110.9359.3470.40111.50.9759.7480.399121.01510.1470.39712.51.05410.5440.395131.09410.9390.39313.51.1

23、3311.3320.025141.13611.3570.37714.51.17311.7340.375151.21112.1090.37315.51.24812.4820.371161.28512.8530.36916.51.32213.2230.368171.35913.5900.36617.51.39613.9560.364181.43214.3200.36218.51.46814.6820.360191.50415.0420.35919.51.54015.4010.357201.57615.7580.35520.51.61116.1130.353211.64716.4660.35121.

24、51.68216.8170.350221.71717.1670.34822.51.75117.5150.346231.78617.8610.34423.51.82118.2060.343241.85518.5480.34124.51.88918.8890.339251.92319.2290.33825.51.95719.5670.336261.99019.9020.33426.52.02420.2370.333272.05720.5690.331混凝土龄期&y(t)(10-4)Ty(t)Ty(t)27.52.09020.9000.329282.12321.2301、各龄期的混凝土弹性模量基础混

25、凝土浇灌初期，处于升温阶段，呈塑性状态，混凝土的弹性模量很小，由变形变化引起的应力也很小，温度应力一般可忽略不计。经过数日，弹性模量随时间迅速上升，此时由变形变化引起的应力状态(即混凝土降温引起拉应力)随着弹性模量的上升显著增加，因此必需考虑弹性模量的变化规律，一般按下列公式计算：E二E(1一e-o.09t)(t)(0)式中E任意龄期的弹性模量；(t)Eo最终的弹性模量，一般取成龄的弹性模量，取3.15x104N/mm2；t混凝土浇灌后到计算时的天数。各阶段弹性模量：E=0.271x104N/mm2(1)E=0.519x104N/mm2(2)E=0.745x104N/mm2(3)E=1.314

26、x104N/mm2(6)E=1.749x104N/mm2(9)E=2.080 x104N/mm2(12)E=2.333x104N/mm2(15)E=2.527x104N/mm2(18)E=2.674x104N/mm2(21)E=2.787x104N/mm2(24)E=2.873x104N/mm2(27)E=2.938x104N/mm2(30)2、自约束应力计算根据公式(XnB.6.1-1)o(t)=AT(t)E(t)H(t,t)z21iiii=1o(1)=1.0X10-42X4.581X2710X1=0.62Mpao(2)=1.0X10-42X(4.581X2710X1+3.151x5190

27、x1)=1.44Mpao=1.0X10-4x(4.581x2710 x0.278+3.151x5190 x0.278+1.166x7450 x1)=0.83Mpa2根据表2中计算结果，3.5天后AT(t)变为负值，同时由于徐变效应，自约束应力计算值将减小，本算例中不再计算。3、外约束应力的计算公式如下Xno(t)=AT(t)E(t)H(t)xR(t)(B.6.2-1)x1H2iii1ii=1式中L底板长度，考虑单组底板最长为27000mm；|CHE其中H基础底板厚度，本例为1200mm；C总阻力系数(地基水平剪切刚度)，N/mm3，本工程采用钢制模版加支撑x钢结构，考虑成型混凝土结构要满足验收标准，近似将混凝土外侧结构看做高强度混凝土阻力系数取0.01N/mm3。前期升温外约束应力为压应力，由于早期弹性模量较小且徐变时间较长，其值不大，实际应用中可将这部分压应力作为安全储备。外约束应力计算时从降温阶段开始。1)9天(第1台阶降温)自第9天至第30天的徐变应力：0=001=2.183x10-51200 x1.749xl04pL=2.183x10-5x27000=0.294722chp查表得（9）=1749x104x