38 5 216 1 1),2) 163318) 163318). API. TE256 A doi 1.652/1-879-15-298 MODE OF CASING EXTERNA EXTRUSION BASED ON THE PRINCIPE OF VIRTUA WORK 1) ZHAO Wanchun,2) ZENG Jia WANG Tingting FENG Xiaohan School of Petroleum Engineering, Northeast Petroleum University, Daqing 163318, Heilongjiang, China) School of Electrical Engineering & Information, Northeast Petroleum University, Daqing 163318, Heilongjiang, China) Abstract In order to accurately describe casing collapse under lateral load, we assume that the shape of the stress deformation zone of the casing is two arcs in plastic deformation. We establish casing damage mechanics balance relationship based on theprinciple of virtual work under lateral loads, then derive the casing damage mechanics model by the Fresnel integral. With the engineering practice of Daqing oilfield,weconduct casing collapse test. The new formula leads to a far smaller average relative error than that obtained from the API formula. From the viewpoint of energy, the paper provides a new research method for the study of casing collapse deformation under external load. Key words virtual work principle, lateral load, plastic deformation, fresnel integral. [1-2].. 215 11 13 1 216 2 5. 1) 5173, 515788) UNPYSCT-2168) 216T9268) 21M5518) BH- TZ-53) 12519) YJSCX216-1NEPU). 2). E-mail zhaowanchun52@163.com,.., 216, 385): 55-55 Zhao Wanchun, Zeng Jia, Wang Tingting, et al. Model of casing external extrusion based on the principle of virtual work. Mechanics in Engineering, 216, 385): 55-55
5 551 [3-].. [5]. [6]. [7].. API API. 1 [8-9]. [1]. W = W 1) 2. 1. 2 y). x [11-12]. 1 2 3. D 2 2πR = 2z E + 2R arcsin D δ R z E = πr 1 + πr ) 2 arcsin [R δ)/r] + π 2π 2 3 ) R δ α = arccos R R = πr π α ) R sin α 2) 3) ) R sin α = Rα 5) R cos α = R δ R R) 6) α = arctan α R 2R R δ 7)
力 552 学 与 实 践 216 年 第 38 卷 a) 金属圆管塑性变形区 y) 方向的投影 b) III-III 方向剖面图 图3 xe = xg = 8) ye = y G = D δ 9) E 点 G 点以及 A 点的坐标为 Ã! µ 2 arcsin [R δ)/r] + π E, D δ, πr 1 2π G, D δ, ) A, D, ) 易求得空间曲线 EA 的方程组表达式为 b) I-I 方向剖面图 δ 2 2δ x x+d 2 Rα 2 2Rα z= 2 x + x y= 图2 则空间曲线 EA 的长度为 s µ 2 µ 2 Z dz dy EA = + dx = 1+ dx dx p 2 δ 2 + R2 α2 ) + 2 + p δ 2 + R2 α2 p p δ 2 + R2 α2 ) + 2 + 2 δ 2 + R2 α2 ln 上式代入数据后 可用软件求出结果. 曲线 GA 的方程为 a) II-II 方向剖面图 y= 2δ 2 δ x x+d 2 1)
5 553 GA = GA 2δ 1 + ) 2 dy dx = dx δ 2 + 2 + 2 δ ln 2 + δ 2 + 2δ δ2 + 2 2 11). h σ q M N [13-1] N = σ h 12) M = σ h 2 / 13) 1) W 1 3b) ds F E W s = 2 s dw s = 2M αds 1) 2M αds = 2 αmax 2M Rαdα = 2M Rα 2 max 15) 3b) y = δ 2 x2 2δ x + D w F E w = D y = 2δ x δ 2 x2 16) x α max α α max x) = 2α x α ) 2x 2 x2 = x2 2 α 17) α max x) = 2α x α 2 x2 = ) 2x x2 2 α 18) 15) W 1 = 2 W s dx = 32 15 M Rα 2 19) 2) W 2 W 2 W 1 W 2 = 32 15 M R π α α α 3) W 3 ) ε = 2 GA 2) 21) W 3 = ε N dε = 2N GA ) ) W W = EA M α 2 M 2x x2 EA 2 ) α dx = 22) EA 2 EA 3 23) 3 AF F B W = W = W = W BE 5) EA W = W 2) W = qwdxdy = qv V = 2 [ 2 πr 2 αmax 2π R sin α max R cos α ) ] max dx = 2 3 R2 α + R2 π 2 α ) 2 α cos 2α FresenlS π ) ) 2 α FresenlC sin 2α π 25) Mathematica.
55 216 38 1) W = W 1 + W 2 + W 3 + W 26) [ 32 q = 15 M Rα 2 + 32 15 M R π α α α + ) 2N GA + M α 2 EA 2 EA 3 ]/ 3 [ 3 R2 α + R2 π 2 α ) 2 α cos 2α FresenlS π ) ) ] 2 α FresenlC sin 2α 27) π 3 5 1 / 2 J-55 h 6.2 mm 6.99 mm 7.72 mm. 27) 1. 1 / / API API mm MPa /MPa /MPa /% /% 6.2 16. 1.97 21.51 6.67 3.1 6.99 23.63 22.27 27.86 5.76 17.9 7.72 31.2 29.68 33.85.32 9.12 1 API 5.58% API 2.37% API API. 6.99 mm 5. 5.. 1),. 2) API API. 3). ). 1,.. :, 1992 2,,.., 28, 272): 377-382 3,.., 22, 221): 8-51 566 )
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