




版权说明:本文档由用户提供并上传,收益归属内容提供方,若内容存在侵权,请进行举报或认领
文档简介
1、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: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;混凝土的比热,取1.0kJ/ (kg); 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,地基温度为18,大气温度为18。温度场计算差分公式如下: (B.4.2-1)试算、,确定。取 = 0.5天 = 12小时, = 0.4m,即分3层则,可行。代入该值得出相应的差分法公式为画出相
2、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、表示,相应时刻各层的温度即为、。混凝土与大气接触的上表面边界温度用表示,与地基接触的下表面边界温度用表示。 = 0,即第天,上表面边界,取大气温度, = 18各层混凝土温度取入模温度,即= = = 10下表面边界,取地基温度,= 18; = 1,即第天,温升10.544上表面边界温度,散热温升为0,始终保持不变, = 18 第一层混凝土温度,见计算图示中方框1,的边界为和,在的基础上考虑温升,即第二层混凝土温度,见计算图示中方框2,的边界为和,在的基础上考虑温升,即第三层混凝
3、土温度,见计算图示中方框3,的边界为和,在的基础上考虑温升,即下表面边界温度,需要考虑散热温升,所以需每一步都需进行修正。见计算图示中方框6,的边界为和地基温度18,在的基础上考虑温升,即以上即完成了一遍 = 1时,各温度计算。同理 = 2,即第天,温升 = 8.212,上表面 = 18,同理可计算 = 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:)混凝土龄期上表面第一层第二层第三层底界边层018101010180.51822.64420.54422.64426.44411829.08529.85831.30
4、231.4411.51832.77336.42937.35434.27121834.83640.69341.28335.7902.51835.83343.18943.56536.44131836.10344.37844.61736.4913.51835.87644.62144.77436.12341835.31144.19844.29535.4694.51834.52743.31843.38034.62751833.60742.13842.17733.6715.51832.61540.77440.79932.65661831.59439.31339.32931.6206.51830.57737
5、.81637.82630.59471829.58536.32736.33429.5957.51828.63234.87834.88228.63981827.72933.48733.49027.7338.51826.88032.16932.17126.88291826.08830.93230.93326.0899.51825.35329.77729.77825.354101824.67628.70828.70824.67610.51824.05327.72027.72124.053111823.48226.81326.81323.48211.51822.96025.98225.98222.961
6、121822.48525.22225.22222.48512.51822.05224.53024.53022.052131821.65923.90023.90021.65913.51821.30323.32823.32821.303141820.98022.80822.80820.98014.51820.68722.33822.33820.687151820.42321.91221.91220.42315.51820.18321.52721.52720.183161819.96821.17921.17919.96816.51819.77320.86420.86419.773171819.597
7、20.58120.58119.59717.51819.43820.32420.32419.438181819.29520.09320.09319.29518.51819.16619.88519.88519.166191819.05019.69719.69719.05019.51818.94519.52819.52818.945201818.85119.37619.37618.85120.51818.76619.23819.23818.766211818.68919.11519.11518.68921.51818.62019.00319.00318.620221818.55818.90318.9
8、0318.55822.51818.50218.81318.81318.502231818.45218.73118.73118.45223.51818.40718.65818.65818.407241818.36618.59218.59218.36624.51818.32918.53318.53318.329251818.29618.48018.48018.29625.51818.26718.43118.43118.267261818.24018.38818.38818.24026.51818.21618.34918.34918.216271818.19418.31418.31418.19427
9、.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T1计算公式如下 (B.5.1)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 (B.2.1) (B.2.2)其中 取4.0×10-4石家庄市内年平均气温值取14.5计算结果见表3 综合降温差T2计算公式如下 (B.5.2)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*T201010100.0000.000/-8.6610.522.
10、64422.64422.6440.0008.6612.217 -8.803129.08531.30231.3022.21717.4652.365 -6.1711.532.77337.35437.3544.58123.6361.866 -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 -0.2693.535.87644.77444.7748.89832.0060.086 0.043435.31
11、144.29544.2958.98431.963-0.131 0.9394.534.52743.38043.3808.85331.024-0.283 0.763533.60742.17742.1778.57030.261-0.385 0.9235.532.61540.79940.7998.18429.337-0.450 1.007631.59439.32939.3297.73528.330-0.485 1.0366.530.57737.82637.8267.24927.295-0.500 1.720729.58536.33436.3346.74925.575-0.500 1.0347.528.
12、63234.88234.8826.25024.541-0.488 0.977827.72933.49033.4905.76123.563-0.470 0.9098.526.88032.17132.1715.29122.654-0.446 0.834926.08830.93330.9334.84521.820-0.420 0.7569.525.35329.77829.7784.42521.064-0.393 -1.0121024.67628.70828.7084.03222.075-0.364 0.52210.524.05327.72127.7213.66821.554-0.337 0.4481
13、123.48226.81326.8133.33121.105-0.310 0.37911.522.96025.98225.9823.02120.726-0.284 0.3131222.48525.22225.2222.73720.413-0.260 0.25212.522.05224.53024.5302.47820.161-0.237 0.1961321.65923.90023.9002.24119.965-0.216 0.14313.521.30323.32823.3282.02519.822-0.196 0.4621420.98022.80822.8081.82919.360-0.178
14、 0.06414.520.68722.33822.3381.65119.296-0.161 0.0241520.42321.91221.9121.49019.273-0.146 -0.01215.520.18321.52721.5271.34319.285-0.132 -0.0451619.96821.17921.1791.21119.330-0.119 -0.07516.519.77320.86420.8641.09219.405-0.108 -0.1021719.59720.58120.5810.98419.507-0.097 -0.12617.519.43820.32420.3240.8
15、8619.633-0.088 -0.1481819.29520.09320.0930.79919.780-0.079 -0.16718.519.16619.88519.8850.71919.947-0.072 -0.1841919.05019.69719.6970.64820.132-0.064 -0.20019.518.94519.52819.5280.58320.332-0.058 -0.2142018.85119.37619.3760.52520.546-0.052 -0.22620.518.76619.23819.2380.47320.772-0.047 -0.2372118.6891
16、9.11519.1150.42521.010-0.042 -0.24721.518.62019.00319.0030.38321.257-0.038 -0.2562218.55818.90318.9030.34521.512-0.034 -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 -
17、0.28624.518.32918.53318.5330.20322.888-0.020 -0.2892518.29618.48018.4800.18323.178-0.018 -0.29325.518.26718.43118.4310.16523.471-0.017 -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 -0.30127.518.17518.28318.2830.1
18、0824.665-0.011 -0.3032818.15718.25418.2540.09724.968-0.097 /注:表中*T1和*T2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。 (B.6.1-2) (B.6.2-2) 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)00.000 0.000 0.517 0.50.052 0.517 0.515 10.103 1.032 0.512 1.50.154 1.545 0.510 20.205 2.054 0.484 2.50.254 2.539 0.500 30.304
19、 3.039 0.176 3.50.321 3.215 0.450 40.367 3.665 -0.046 4.50.362 3.619 0.392 50.401 4.012 0.390 5.50.440 4.402 0.388 60.479 4.790 0.386 6.50.518 5.177 -0.311 70.487 4.866 0.335 7.50.520 5.201 0.333 80.553 5.534 0.331 8.50.587 5.865 0.330 90.619 6.195 0.328 9.50.652 6.523 2.017 100.854 8.539 0.405 10.5
20、0.894 8.944 0.403 110.935 9.347 0.401 11.50.975 9.748 0.399 121.015 10.147 0.397 12.51.054 10.544 0.395 131.094 10.939 0.393 13.51.133 11.332 0.025 141.136 11.357 0.377 14.51.173 11.734 0.375 151.211 12.109 0.373 15.51.248 12.482 0.371 161.285 12.853 0.369 16.51.322 13.223 0.368 171.359 13.590 0.366
21、 17.51.396 13.956 0.364 181.432 14.320 0.362 18.51.468 14.682 0.360 191.504 15.042 0.359 19.51.540 15.401 0.357 201.576 15.758 0.355 20.51.611 16.113 0.353 211.647 16.466 0.351 21.51.682 16.817 0.350 221.717 17.167 0.348 22.51.751 17.515 0.346 231.786 17.861 0.344 23.51.821 18.206 0.343 241.855 18.5
22、48 0.341 24.51.889 18.889 0.339 251.923 19.229 0.338 25.51.957 19.567 0.336 261.990 19.902 0.334 26.52.024 20.237 0.333 272.057 20.569 0.331 27.52.090 20.900 0.329 282.123 21.230 1、 各龄期的混凝土弹性模量基础混凝土浇灌初期,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
温馨提示
- 1. 本站所有资源如无特殊说明,都需要本地电脑安装OFFICE2007和PDF阅读器。图纸软件为CAD,CAXA,PROE,UG,SolidWorks等.压缩文件请下载最新的WinRAR软件解压。
- 2. 本站的文档不包含任何第三方提供的附件图纸等,如果需要附件,请联系上传者。文件的所有权益归上传用户所有。
- 3. 本站RAR压缩包中若带图纸,网页内容里面会有图纸预览,若没有图纸预览就没有图纸。
- 4. 未经权益所有人同意不得将文件中的内容挪作商业或盈利用途。
- 5. 人人文库网仅提供信息存储空间,仅对用户上传内容的表现方式做保护处理,对用户上传分享的文档内容本身不做任何修改或编辑,并不能对任何下载内容负责。
- 6. 下载文件中如有侵权或不适当内容,请与我们联系,我们立即纠正。
- 7. 本站不保证下载资源的准确性、安全性和完整性, 同时也不承担用户因使用这些下载资源对自己和他人造成任何形式的伤害或损失。
最新文档
- 解聘协议书和离职合同
- 船运装卸服务合同协议
- 花艺合作协议月结合同
- 蔬菜水果配送合同协议
- 茶餐厅加盟合同协议
- 面试题及答案三道
- 菜鸟驿站托管合同协议
- 颜色理论与搭配纺织品设计师试题及答案
- 花园种植协议书范本
- 蛋奶煎饼采购合同协议
- 六年级家长会心理教育专题课件
- 四川宜宾环球集团有限公司招聘笔试题库2025
- 如何设计高质量的课件:制作教学指南
- 2025年农村商业银行人员招聘考试笔试试题(含答案)
- 2025年江苏省新高考高三联考数学试卷试题(含答案详解)
- 2025年江苏省苏州市昆山八校联考中考零模英语试题(原卷版+解析版)
- 生物技术与生物医药产业发展趋势分析
- 工程第一次监理例会会议纪要
- DBJ-T13-200-2025 福建省桩基础与地下结构防腐蚀技术标准
- 2022长大桥梁养护指南 第 2 部分:机电系统维护管理指南
- 外研版(三起)(2024)三年级下册英语Unit 1 单元测试卷(含答案)
评论
0/150
提交评论