锅炉温度应力计算

1、1、 基本资料锅炉基础承台最大承台为 JK1-2长X宽x 14.8mx 5.8mX 2.8m采用冀东水泥厂出的普通硅酸盐水泥。混凝土强度等级C30施工配合比见下表：强度 等级水灰比W/C塌落度mm水32.5R 水泥砂后于外加剂SDB-800C300.42150± 30mm150360765112532、最大绝热温升:工=me Q/CpTh :最大绝热温升me :水泥用量360Kg/M3Qo :水泥水化热总量289C :混凝土比热 0.96KJ/Kg 0GP :密度2400Th=me Q/C - p =(3>289)也.96 2400)=45.2C3 .浇筑温度：丁入=T 浇=2

2、8 C4 .不同龄期硅内部温度：Tmax=Tj+T(h). §Tmax：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d036912151821242730§00.670.650.610.530.420.330.270.230.190.17T（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ABCDEFG

3、HIJd033 66991212 1515 1818 2121 2424 272732ZT30.284-0.904-1.808-3.616-4.972-4.068-2.712-1.808-1.808-0.9045 .各时段弹性模量计算硅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 丫C30硅 E (h) =3.0 X04 N/mm2各时段硅的弹性模量如下：E0E3dE6dE9 dE12 dE15 dE18 dE21 dE24 dE27 dE30 d00.70991.25181.66541.98122.22222.40632.54682.65402

4、.73592.79846 .各龄期收缩当量温差y j0(1-e-0.01 )M1 - M2 - M10*3.24 104 (硅最终弹性模量)Mi各修正值，查施工手册为:普通硅酸盐水 泥水泥细度5000 孔石灰石水灰比 0.42水泥殳笈量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%M6M7M8M9M100.930.771.131.01.0M1 M2 M10.092y 了3.24 X0-4(1-e-0-01 ) X092T y（石-q（ J） %硅线膨月系数，取1.0 X0-5Ty（1 :各龄期当量温差a

5、 (d)36912151821242730a X1040.1050.2060.3050.4000.4900.5830.6700.7540.8380.918Ty-1.05-2.06-3.05-4.00-4.90-5.83-6.70-7.54-8.38-9.18 T-1.01-0.99 -0.95 -0.90 -0.93 -0.87-0.84-0.84-0.807 .各龄期硅的应力松弛系数查施工手册：H3 = 0.186 H6 = 0.208 H9= 0.214 H12 = 0.215H15= 0.233H18= 0.2521H21 = 0.301 H24 = 0.524H27= 0.570H30

6、 =1.08 .硅温度应力计算：>(a/V )通 T1-1/(ch BLH T)(3= Cx/(h . Ei)Cx=地基刚度系数Cx=Cxi+Cx2Cxi 查手册 10-81 取为 0.01N/mm3CX2桩的阻力系数Cx2=Q/FF=2 X4=8xi06mm k 1.248 10-2/2800 1.6654 104 = 1.64 X0-5Q=2EJ(KhD/4EJ)?Kh-侧向压缩刚度系数1 M0-2N/ mm2E-灌注桩的弹性模量2.8 X04D-灌注桩直径取900J-灌注桩的惯性矩 J=1/64 3.14 (900)4=32208Q=2EJ(KhD/4EJ)?=2X2.8 X04刈

7、22 X 08 900X10-2/ (4X2.8 X04>322X108) ?=2.013 X04 N/mm2CX2=Q/F =0.252 10-2 N/mm3Cx= Cx1 + Cx2=(1.0 10-2) +(0.252 10-2)=1.252 X0-2 N/mm3:L =0.13992Cx/HE(t) =0.12142妒卜 1.248 10-2/2800 1.2518 104 = 1.89 X0-5同样，有：觞=1.50 10-5加=1.42 10-5觞=1.36 10-5角= 1.33 10-5物=1.30 10-5囱=1.28 10-5觊=1.26 10-5T寿龄期综合温差I

8、=0.11102 =0.10512上=0.10062L =0.09842 =0.09622 =0.09472 =0.09322 T=2/3T ( rj - T (- + Ty ( r ) - Ty E 16=2/3 卜0.904)-1.01 = -1.613 19=2/3 卜1.808)-0.99 = -2.195 T12=2/3 )(-3.616)-0.95 = -3.361 T15=2/3 )(-4.972)-0.90 = -4.215 %=2/3 )(-4.068)-0.93 = -3.642 121=2/3 )(-2.712)-0.87 = -2.678 124=2/3 )(-1.80

9、8)-0.84= -2.045 127=2/3 )(-1.808)-0.84 = -2.045 130=2/3 )(-0.904)-0.80 = -1.252拉应力计算公式： 吁 0c -1/ch iLE(QT(i)H(i)/1”qt）_各龄期硅基础所承受的温度应力E（i） 各龄期硅的弹性模量e-硅线膨胀系数1.010-5 T（i）各龄期的总和温差H（i）各龄期硅的松弛系数-泊松比 0.15列表计算:a (d)69121518212427304Ei X01.25181.66541.98122.22222.40632.54682.65402.73592.7984ch( B .L/2)1.0101.0071.0061.0061.0051.0051.0051.0041.004AT-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-仙0.850.850.850.850.850.850.850.850.85-1 aX1010.0050.0060.010.0150.0130.0120.0170.0150.016最大拉应力：*=(0.005+0.006