﻿ 基于实测沉降的浮顶储罐变形分析
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 浙江大学学报(工学版)  2017, Vol. 51 Issue (9): 1824-1833  DOI:10.3785/j.issn.1008-973X.2017.09.017 0

### 引用本文 [复制中英文]

dx.doi.org/10.3785/j.issn.1008-973X.2017.09.017
[复制中文]
FAN Hai-gui, CHEN Zhi-ping, XU Feng, TANG Xiao-yu, SU Wen-qiang. Floating-roof tanks' distortion analysis based on measured settlement[J]. Journal of Zhejiang University(Engineering Science), 2017, 51(9): 1824-1833.
dx.doi.org/10.3785/j.issn.1008-973X.2017.09.017
[复制英文]

### 通信联系人

orcid.org/0000-0002-5016-2919.
Email: zhiping@zju.edu.cn

### 文章历史

Floating-roof tanks' distortion analysis based on measured settlement
FAN Hai-gui , CHEN Zhi-ping , XU Feng , TANG Xiao-yu , SU Wen-qiang
Institute of Process Equipment and Control Engineering, Zhejiang University, Hangzhou 310058, China
Abstract: The distortion behavior of floating-roof tanks under differential settlement was investigated based on measured settlement. The radial displacement formula of tanks under uneven settlement was derived by applying the classical elasticity theory. Using finite element method, distortion response of tanks under uneven settlement was simulated by solid element and shell element, respectively. Results of theoretical formula and finite element method were compared and analyzed. Based on that, double exponential attenuation function fitting method was applied to obtain the radial displacement formula of tanks with eave wind girder under uneven settlement. The finite element simulation method was applied to compare with the formula and verify it. Results show that for tanks with eave wind girders which have different forms and sizes, the formula can be used to calculate tanks distortion under uneven settlement accurately, thus to predict the potential failure modes of in-service tanks.
Key words: tanks    measured settlement    deformation response    radial displacement    eave wind girder    finite element simulation

1 不均匀沉降下储罐径向位移公式

 $\left. \begin{array}{l} {N_\varphi } - {p_r}r = 0,\\ \frac{{\partial {N_{x\varphi }}}}{{\partial x}} + \frac{1}{r}\frac{{\partial {N_\varphi }}}{{\partial \varphi }} + {p_\varphi } = 0,\\ \frac{{\partial {N_x}}}{{\partial x}} + \frac{1}{r}\frac{{\partial {N_{x\varphi }}}}{{\partial \varphi }}{p_x} = 0 \end{array} \right\}$ (1)

 $u = {u_n}\cos \left( {n\varphi } \right).$

 ${N_x} = {N_n}\cos \left( {n\varphi } \right).$

 $\left. \begin{array}{l} {N_\varphi } = 0,\\ {N_{x\varphi }} = {f_1}\left( \varphi \right),\\ {N_x} = - \frac{x}{r}\frac{{\partial {f_1}}}{{\partial \varphi }} + {f_2}\left( \varphi \right). \end{array} \right\}$ (2)

 ${N_x} = \left\{ {\begin{array}{*{20}{c}} {\begin{array}{*{20}{c}} {{N_n}\cos \left( {n\varphi } \right),}\\ {0,} \end{array}}&\begin{array}{l} x = 0;\\ x = l. \end{array} \end{array}} \right.$ (3)

 $\left. \begin{array}{l} {N_\varphi } = 0,\\ {N_{x\varphi }} = \frac{{r{N_n}}}{{nl}}\sin \left( {n\varphi } \right),\\ {N_x} = {N_n}\left( {1 - \frac{x}{l}} \right)\cos \left( {n\varphi } \right). \end{array} \right\}$ (4)

 $\left. \begin{array}{l} \frac{{\partial u}}{{\partial x}} = \frac{1}{{E{t_c}}}\left( {{N_x} - v{N_\varphi }} \right),\\ \frac{{\partial v}}{{\partial x}} + \frac{1}{r}\frac{{\partial u}}{{\partial \varphi }} = \frac{{2\left( {1 + v} \right)}}{{E{t_c}}}{N_{x\varphi }},\\ \frac{w}{r} + \frac{1}{r}\frac{{\partial v}}{{\partial \varphi }} = \frac{1}{{E{t_c}}}\left( {{N_\varphi } - v{N_x}} \right). \end{array} \right\}$ (5)

 $\left. \begin{array}{l} u = \frac{{{N_n}}}{{E{t_{\rm{c}}}}}\left( {x - \frac{{{x^2}}}{{2l}}} \right)\cos \left( {n\varphi } \right) + {u_n}\cos \left( {n\varphi } \right),\\ v = \frac{n}{r}\left[ {\frac{{{N_n}}}{{E{t_{\rm{c}}}}}\left( {\frac{{{x^2}}}{2} - \frac{{{x^3}}}{{6l}}} \right) + {u_n}x} \right]\sin \left( {n\varphi } \right) + \\ \;\;\;\;\;\;\;\frac{{2\left( {1 + v} \right)r{N_n}x}}{{nE{t_{\rm{c}}}l}}\sin \left( {n\varphi } \right) - \frac{{rv{N_n}}}{{nE{t_{\rm{c}}}}}\sin \left( {n\varphi } \right)\\ w = - \frac{{{n^2}}}{r}\left[ {\frac{{{N_n}}}{{E{t_{\rm{c}}}}}\left( {\frac{{{x^2}}}{2} - \frac{{{x^3}}}{{6l}}} \right) + {u_n}x} \right]\cos \left( {n\varphi } \right) - \\ \;\;\;\;\;\;\;\frac{{\left( {2 + v} \right)r{N_n}x}}{{E{t_{\rm{c}}}l}}\cos \left( {n\varphi } \right) \end{array} \right\}$ (6)

 $w = - \frac{{{n^2}}}{r}{u_n}x\cos \left( {n\varphi } \right).$ (7)

 $w=-\frac{{{n}^{2}}}{r}u_{n}^{'}x\sin \left( n\varphi \right)$ (8)

 $w=-\sum\limits_{n=2}^{{{n}_{\max }}}{\left\{ \frac{{{n}^{2}}}{r}x\left[ {{u}_{n}}\cos \left( n\varphi \right)+u_{n}^{'}\sin \left( n\varphi \right) \right] \right\}.}$ (9)

2 有限元分析与比较

 图 1 储罐底部的实测沉降数据 Fig. 1 Measured settlement data of tank bottom

 $\begin{array}{l} u = 4.25{\rm{cos}}(2\varphi ) + 1.01{\rm{sin}}(2\varphi ) - 0.667{\rm{cos}}(3\varphi ) + \\ 3.167{\rm{sin}}(3\varphi ) + 1.417{\rm{cos}}(4\varphi ) - 2.454{\rm{sin}}(4\varphi ) + \\ 0.37{\rm{cos}}(5\varphi ) - 1.883{\rm{sin}}(5\varphi ) - 0.75{\rm{cos}}(6\varphi ). \end{array}$ (10)
 $\begin{array}{l} u = 0.088{\rm{cos}}(2\varphi ) - 0.045{\rm{sin}}(2\varphi ) + 0.944{\rm{cos}}(3\varphi ) - \\ 0.872{\rm{sin}}(3\varphi ) + {\rm{cos}}(4\varphi ) - 2.375{\rm{sin}}(4\varphi ) + \\ 1.746{\rm{cos}}(5\varphi ) - 0.706{\rm{sin}}(5\varphi ) - 0.088{\rm{cos}}(6\varphi ) - \\ 1.545{\rm{sin}}(6\varphi ) + 1.011{\rm{cos}}(7\varphi ) - 0.601{\rm{sin}}(7\varphi ) - \\ 0188{\rm{cos}}(8\varphi ). \end{array}$ (11)

 $\begin{array}{l} w = [ - 0.739{\rm{cos}}(2\varphi ) - 0.176{\rm{sin}}(2\varphi ) + \\ 0.261{\rm{cos}}(3\varphi ) - 1.239{\rm{sin}}(3\varphi ) - 0.986{\rm{cos}}(4\varphi ) + \\ 1.707{\rm{sin}}(4\varphi ) - 0.402{\rm{cos}}(5\varphi ) + 2.047{\rm{sin}}(5\varphi ) + \\ 1.174{\rm{cos}}(6\varphi )]x{\rm{ }}. \end{array}$ (12)

 $\begin{array}{l} w = [ - 0.0088{\rm{cos}}(2\varphi ) + 0.0045{\rm{sin}}(2\varphi ) - \\ 0.212{\rm{cos}}(3\varphi ) + 0.196{\rm{sin}}(3\varphi ) - 0.4{\rm{cos}}(4\varphi ) + \\ 0.95{\rm{sin}}(4\varphi ) - 1.091{\rm{cos}}(5\varphi ) + 0.441{\rm{sin}}(5\varphi ) + \\ 0.0792{\rm{cos}}(6\varphi ) + 1.391{\rm{sin}}(6\varphi ) - 1.238{\rm{cos}}(7\varphi ) + \\ 0.736{\rm{sin}}(7\varphi ) + 0.301{\rm{cos}}(8\varphi )]x. \end{array}$ (13)
2.1 实体单元、壳单元和理论公式的比较

 图 2 采用壳单元和实体单元模拟得到的1号储罐径向位移云图 Fig. 2 Radial displacement nephograms of tank 1 using shell element and solid element
 图 3 由3种不同方法计算得到的1号储罐顶部径向位移沿圆周方向的分布 Fig. 3 Radial displacement distributions of tank 1 top in circumferential direction obtained from three different methods

2.2 壁厚的影响

 图 4 不同壁厚储罐顶部径向位移沿圆周方向的分布 Fig. 4 Radial displacement distributions of tank with different thicknesses in circumferential direction

2.3 储液的影响

 图 5 1号储罐在式(10) 作用下的最大径向位移与储液高度的关系 Fig. 5 Relation of maximum radial displacement and liquid height of tank 1 under Formula Ten

3 含抗风圈储罐的变形研究

3.1 理论公式的求解

 $w=-\sum\limits_{n=2}^{{{n}_{\max }}}{\left\{ {{A}_{n}}\frac{{{n}^{2}}}{r}x\left[ {{u}_{n}}\cos \left( n\varphi \right)+u_{n}^{'}\sin \left( n\varphi \right) \right] \right\}}.$ (14)

 图 6 当抗风圈宽度为700 mm、厚度为15 mm时, 储罐在谐波沉降下的径向位移云图 Fig. 6 Radial displacement nephogram of tank under harmonic settlement with eave wind girder's width of 700 mm and its thickness of 15 mm

IG为横坐标, K为纵坐标, 将对应不同抗风圈的计算结果绘制于图 7中.根据图中数据的分布规律, 采用双指数衰减函数对图中数据进行曲线拟合, 拟合结果如图 7所示.拟合曲线的相关系数平方R2=0.998 21, 表明曲线的拟合度非常高, 能够准确地反映储罐径向位移随抗风圈截面惯性矩的变化情况.该拟合曲线即为1号储罐在3阶谐波沉降作用下的衰减系数A3.采用同样的方法, 可以求解1号储罐在各阶谐波沉降作用下的衰减系数An.

 图 7 1号储罐在3阶谐波沉降下的径向位移模拟值和拟合曲线 Fig. 7 Radial displacement simulation result and fitting curve of tank 1 under harmonic settlement with 3rd order

3.2 验证与比较

 图 8 1号储罐在式(10) 作用下的径向位移云图(抗风圈截面模量为853.33×106 mm4) Fig. 8 Radial displacement nephogram of tank 1 under Formula Ten (Section modulu of eave wind girder is 853.33×106 mm4)
 图 9 式(10) 作用下1号储罐罐顶径向位移沿圆周方向的分布 Fig. 9 Radial displacement distribution of tank 1 top under Formula Ten in circumferential direction

 图 10 2号储罐在式(11) 作用下的径向位移云图(抗风圈截面模量为1 215×106 mm4) Fig. 10 Radial displacement nephogram of tank 2 under Formula Eleven (Section modulus of eave wind girder is 1 215×106 mm4)
 图 11 式(11) 作用下2号储罐罐顶径向位移沿圆周方向的分布 Fig. 11 Radial displacement distribution of top of tank 2 under Formula Eleven in circumferentia direction

3.3 抗风圈结构形式的影响

 图 12 角钢抗风圈储罐顶部径向位移沿圆周方向的分布 Fig. 12 Radial displacement of tank with angle steel eave wind girder in circumferential direction

4 结论

(1) 储罐在不均匀沉降作用下的径向位移可以应用理论公式(式(9))计算得到, 理论公式结果与有限元模拟结果几乎相同.

(2) 作为一种典型的薄壁结构, 大型储罐在沉降作用下的变形与其罐壁厚度无关.

(3) 罐内储液可以提高储罐抵抗沉降变形的能力, 随着储液的增多, 储罐在沉降作用下的变形逐渐减小.

(4) 应用理论公式(式(14)), 可以准确地计算具有不同尺寸、不同结构形式抗风圈的储罐在不均匀沉降作用下的径向位移.

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