锅炉温度应力计算

1、.、 基本资料锅炉基础承台最大承台为JK1-2长宽14.8m5.8m2.8m。采用冀东水泥厂出的普通硅酸盐水泥。混凝土强度等级C30强度等级水灰比W/C塌落度mm水32.5R水泥砂石子外加剂SDB-800C300.4215030mm15036076511253施工配合比见下表：2、最大绝热温升： h= mCQ/Ch : 最大绝热温升mC : 水泥用量360KgMQ0 : 水泥水化热总量289C ：混凝土比热0.96KJ/Kg : 密度2400 h=mCQ/C=(360289)(0.962400)=45.23.浇筑温度:T =T浇=28 4.不同龄期砼内部温度: Tmax=Tj+T(h).Tma

2、x：混凝土内部温度Tj：浇注温度： 降温系数查施工手册列表计算如下：龄期d03691215182124273000.670.650.610.530.420.330.270.230.190.17T(h).030.28429.38027.57223.95618.98414.91612.20410.3968.5887.684Tmax2858.28457.38055.57251.95646.98442.91640.20438.39636.58835.684时段划分及各时段温差ABCDEFGHIJd033669912121515181821212424272732T30.284-0.904-1.808-

3、3.616-4.972-4.068-2.712-1.808-1.808-0.9045.各时段弹性模量计算砼浇筑后强度逐步提高，弹性模量也在不断上升。引起应力增加。E（）=E（h）(1-e-0 .09)C30砼 E（h） =3.0104 N/mm2各时段砼的弹性模量如下：E0E3dE6dE9 dE12 dE15 dE18 dE21 dE24 dE27 dE30 d00.70991.25181.66541.98122.22222.40632.54682.65402.73592.79846.各龄期收缩当量温差y() =y0(1-e-0.01)M1M2M10y0=3.24104（砼最终弹性模量）Mi各

4、修正值，查施工手册为：普通硅酸盐水泥水泥细度5000孔骨料石灰石水灰比0.42水泥浆量0.21M1M2M3M4M51. 01.351.01.01.0 自然养护30天环境相对湿度71%水里半径倒数0.48机械振捣配筋率7.7%M6M7M8M9M100.930.771.131.01.0M1M2 M10=1.092 y()=3.2410-4(1-e-0.01) 1.092 T y()=- y()/ :砼线膨胀系数,取1.010-5T y() :各龄期当量温差(d)36912151821242730y1040.1050.2060.3050.4000.4900.5830.6700.7540.8380.9

5、18Ty-1.05-2.06-3.05-4.00-4.90-5.83-6.70-7.54-8.38-9.18T-1.01 -0.99 -0.95 -0.90 -0.93 -0.87 -0.84 -0.84 -0.80 7.各龄期砼的应力松弛系数 查施工手册：H30.186 H60.208 H90.214 H12 0.215 H150.233 H180.2521 H210.301 H240.524H270.570 H301.08.砼温度应力计算：=-(/1-)EiT1-1/(chL/2)H ()=Cx/(hEi) Cx=地基刚度系数Cx=Cx1Cx2Cx1查手册10-81 取为0.01N/mm3

6、Cx2桩的阻力系数 Cx2=Q/FF=24=8106mm2 Q=2EJ(KhD/4EJ)Kh-侧向压缩刚度系数110-2N/ mm2E-灌注桩的弹性模量2.8104D-灌注桩直径取900J-灌注桩的惯性矩J=1/643.14(900)4=322108 Q=2EJ(KhD/4EJ)=22.810432210890010-2/（42.8104322108） =2.013104 N/mm2 CX2=Q/F =0.25210-2 N/mm3 CX= CX1CX2=(1.010-2) (0.25210-2) =1.25210-2 N/mm3=CX/HE(t)6= 1.24810-2/28001.2518

7、104 =1.8910-5 =0.13999= 1.24810-2/28001.6654104 =1.6410-5 =0.1214同样，有：12=1.5010-5 =0.111015=1.4210-5 =0.105118=1.3610-5 =0.100621=1.3310-5 =0.098424=1.3010-5 =0.096227=1.2810-5 =0.094730=1.2610-5 =0.0932T=各龄期综合温差T=2/3T（1）- T（2）+ Ty（1）- Ty（2）T6=2/3(-0.904)-1.01 = -1.613T9=2/3(-1.808)-0.99 = -2.195T12

8、=2/3(-3.616)-0.95 = -3.361T15=2/3(-4.972)-0.90 = -4.215T18=2/3(-4.068)-0.93 = -3.642T21=2/3(-2.712)-0.87 = -2.678T24=2/3(-1.808)-0.84= -2.045T27=2/3(-1.808)-0.84 = -2.045T30=2/3(-0.904)-0.80 = -1.252拉应力计算公式：(t)= (1-1/chiL/2)E(i)T(i)H(i)/1-(t)-各龄期砼基础所承受的温度应力E(i)- 各龄期砼的弹性模量-砼线膨胀系数1.010-5 T(i)-各龄期的总和温差

9、H(i)-各龄期砼的松弛系数-泊松比0.15列表计算：(d)6912151821242730Ei1041.25181.66541.98122.22222.40632.54682.65402.73592.7984ch(.L/2)1.0101.0071.0061.0061.0051.0051.0051.0041.004T-1.613-2.195-3.361-4.215-3.342-2.678-2.045-2.045-1.252Hi0.2080.2140.2150.2330.2520.3010.5240.5701.0001-0.850.850.850.850.850.850.850.850.8510-10.0050.0060.010.0150.0130.0120.0170.0150.016最大拉应