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[分享] CosmosWorks算法验证文件 10 (Static, Plastic Analysis)

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发表于 2007-11-9 15:22:36 | 显示全部楼层 |阅读模式 来自: 中国上海

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# C8 R4 B  c$ C5 d" ]' ?Title      Plastic Compression of a Pipe Assembly
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Reference:S. H. Crandall, N. C. Dahl, An Introduction to the Mechanics of Solids, McGraw-Hill Book Co., Inc., New York, NY, 1959, pg. 180, ex. 5.1.
Analysis Type(s):Static, Plastic Analysis

: h8 e" @, R6 a! kTest Case! e0 `; L& r) D$ G. G) S" b# v: U
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Two coaxial tubes, the inner one of 1020 CR steel and cross-sectional area As, and the outer one of 2024-T4 aluminum alloy and of area Aa, are compressed between heavy, flat end plates, as shown below. Determine the load-deflection curve of the assembly as it is compressed into the plastic region by an axial displacement. Assume that the end plates are so stiff that both tubes are shortened by exactly the same amount.' J: Y) H7 E. j# t

7 P2 V! f- A& Z8 f! }& M0 `Figure 7.1  Pipe Assembly Problem Sketch; B0 d+ g( Q/ m5 c1 ~7 V$ q  V+ F$ c
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1 g+ I" t5 t0 MFigure 7.2  Pipe Assembly Finite Element Models; K# b$ b$ N. l& W9 c1 [- W) K
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Material Properties
Es = 26,875,000 psi
σ(yp)s = 86,000 psi
Ea = 11,000,000 psi
σ(yp)a = 55,000 psi
υ = 0.3
Geometric Properties
L= 10 in
As = 7 in2
Aa = 12 in2
Loading
1st Load Step: δ = 0.032 in
2nd Load Step: δ = 0.05 in
3rd Load Step: δ = 0.10 in
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Analysis Assumptions and Modeling NotesThe following tube dimensions, which provide the desired cross-sectional areas, are arbitrarily chosen. Inner (steel) tube: inside radius = 1.9781692 in., wall thickness = 0.5 in. Outer (aluminum) tube: inside radius = 3.5697185 in., wall thickness = 0.5 in.2 I/ ~: A& f5 }5 j1 E. V+ C
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The problem can be solved in one of three ways: 0 N( s3 H" T- w' i9 L% u. w! N9 y

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  • using PIPE20 - the plastic straight pipe element
  • using SOLID45 - the 3-D structural solid element
  • using SHELL43 - the 2-D plastic large strain shell element

3 T+ Z' [/ P3 n- c) u' F, I9 OIn the SOLID45 and SHELL43 cases, since the problem is axisymmetric, only a one element  Θ-sector is modeled. A small angle Θ = 6° is arbitrarily chosen to reasonably approximate the circular boundary with straight sided elements. The nodes at the boundaries have the UX (radial) degree of freedom coupled. In the SHELL43 model, the nodes at the boundaries additionally have the ROTY degree of freedom coupled.
( |) d; ?* i1 R6 MAn ANSYS warning message is issued stating that element 1, (PIPE20 pipe element) has a radius to thickness ratio less than 5. Because the model involves only axial loading, this does not affect the accuracy of the results. 9 M# W3 ~7 F; u+ \3 p8 r

' x1 o( Y3 l! o# Z: B# eResults Comparison
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TargetANSYS[1]Ratio
PIPE20Load, lb for Deflection = 0.032 in1,024,4001,024,4001.000
Load, lb for Deflection = 0.05 in1,262,0001,262,0001.000
Load, lb for Deflection = 0.1 in1,262,0001,262,0001.000
SOLID45Load, lb for Deflection = 0.032 in1,024,4001,022,5290.998
Load, lb for Deflection = 0.05 in1,262,0001,259,6950.998
Load, lb for Deflection = 0.1 in1,262,0001,259,6950.998

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  • From POST1 FSUM of bottom nodal forces (ΣFZ).
TargetANSYS[1]Ratio
SHELL43Load, lb for Deflection = 0.032 in1,024,4001,023,9321.000
Load, lb for Deflection = 0.05 in1,262,0001,261,4231.000
Load, lb for Deflection = 0.1 in1,262,0001,261,4231.000
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  • From POST1 FSUM of bottom nodal forces (ΣFZ) X 360°/6°(Identified as parameter "LOAD").
发表于 2007-11-14 20:55:10 | 显示全部楼层 来自: 中国河北邢台
看来楼主是有限元的高手,能不能给我这个初学者介绍几本书好入门,谢谢!!!
 楼主| 发表于 2007-11-15 00:44:25 | 显示全部楼层 来自: 中国上海
Cosmos帮助文件很好,ANSYS清华有一体资料我觉得比较好,字体清楚,写得也有条理,错误少,在ANSYS可以下到,PDF格式,因为没有封面,我不知道名字是什么。
发表于 2008-1-16 12:27:06 | 显示全部楼层 来自: 中国上海
楼上可以告诉你那本清华书的下载连接地址么?谢谢
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