二辊矫直过程中棒材中性层偏置的研究与验证外文文献翻译、中英文翻译、外文翻译.doc
二辊矫直过程中棒材中性层偏置的研究与验证外文文献翻译、中英文翻译、外文翻译
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二辊矫直过程中棒材中性层偏置的研究与验证摘 要： 金属材料变形过程中必然存在中性层。传统的二辊矫直理论忽略了中性层的迁移现象，特别是大截面棒材矫直过程中。中性层对矫直回弹影响较大，进而影响辊形设计、工艺参数和棒材直线度精度。基于三点弯曲和弹塑性压力，建立了棒材矫直过程中的中性层偏移模型。通过模型研究棒材矫直过程中的中性层迁移现象，并结合室温拉伸试验和弯曲试验。得到了中性层偏移量与反向弯曲半径和金属塑性变形能力的关系。中性层偏移模型为棒材矫直机理和变形的进一步研究提供了参考。关键词： 中性层偏移； 棒材矫直； 回弹； 三点弯曲； 实验分析1 引言高强度合金钢在石油、汽车、造船、工程机械等领域的应用日益广泛。矫直作为最后的精整工序，是保证生产棒材质量的关键工序 1 。由于棒材的原始弯曲很可能存在于任何方向，特别适用于横辊矫直机。两辊矫直和多辊矫直是横辊矫直机的两种主要形式。然而，现有的多辊矫直机存在矫直盲区大、矫直全程和全程连续缺陷、矫直后需要去除头端等问题。而且多辊矫直机结构复杂， 体积大， 造价高。多辊矫直机由于矫直精度低，无法实现棒材矫直的高精度。与多辊矫直机相比， 双辊矫直机不仅能消除棒材盲区， 提高表面粗糙度， 还能提高椭圆度， 解决矫直后的缩口问题。更重要的是，其矫直后棒材的残余挠度可达到 0.1-0.5mm/m， 满足当前高精度棒材的精度要求。国产小型二辊矫直装置已经制造出来， 但未能掌握辊型设计和自动矫直过程的核心技术，大截面棒材二辊矫直机完全依赖进口。因此，在设备国产化的基础上，我们课题组与中国河北省某公司合作开发了高精度双辊棒材矫直机。中性层偏移模型的偏差和建立是问题2 , 3 ， 直接关系到辊形设计的精度和矫直精度。二辊矫直是一个复杂的弹塑性变形过程。中性层偏移会改变棒材截面的应力分布，影响弯矩比和计算的反弯曲率，最终影响辊型设计的精度。遗憾的是，现有的棒材矫直理论分析忽略了中性层偏移现象;因此， 设计了轧辊形状和工艺参数误差较大。Guan 等人的研究表明， 在一定的相对拐角半径下， 中性层弯曲回弹的相对误差可达 70% 以上 4 。目前， 对于中性层偏移问题， 许多学者做了大量的研究 5 - 8 。他们的研究主要集中在两辊矫直过程中用板和管代替中性层偏移。因此，根据其矫直过程的特点，结合三点弯曲理论和弹塑性理论，建立了矫直过程中的中性层偏移模型。该工作为进一步研究大截面高强钢筋二辊矫直机理和研制装置提供了重要参考。2 建立两辊矫直过程中中性层偏移的理论模型2.1 矫直变形分析及基本假设1. 钢筋变形前后截面保持平坦， 垂直于变形钢筋的轴线 9 。两个相邻截面之间没有反向和倾斜。2. 材料是连续的、均匀的、各向同性的， 中性层的应力和应变被认为是一致的 1 0 。3.忽略棒材矫直过程中直径的变化。4. 弯曲塑性变形过程符合等体积原理。5. 等效应力s和等效应变e具有s= Ben 的硬化指数关系，当 B 为塑性材料系数，n 为材料硬化指数 1 1 。二辊矫直机矫直棒材时，弯曲挠度由凹辊角和凹辊间隙决定。通过调整辊缝， 实际弯曲挠度可以从 0 变化到最大弯曲挠度 1 2 。更合理的矫直状态如图 1 所示; 在试验过程中， 矫直变形近似于三点弯曲。从棒材矫直变形区取一个 ABCD 单元， 如图 2 所示。图中棒材弯曲中心为柱面坐标系的原点。根据假设(3)， r 方向的变形可以忽略不计er = 0 ， Y 方向应力要小得多， 可以忽略不计， 即sY = 0 。因此， 棒材矫直变形区单元应力状态可简化为平面应变问题tqY =tYr = 0 和gqY = gYr = 0 ， 同时， 假设(1)表示tqr = gqr = 0 。2.2 矫直过程中的应变关系在进入矫直机前，由加工或热处理引起的棒材弯曲的原始形式在长度范围内有单弯、S 弯、多峰弯和空间弯1 3 ，如图 3 所示。在矫直辊中心段，将所有的弯曲方式统一成单一的弯曲方式。一般将原始弯曲形式简化为单一弯曲形式，便于理论分析。可以假设单元 ABCD 初始处于拉伸状态， 反向弯曲14 后处于压缩状态。进入矫直机棒材弯曲程度小。考虑中性层与棒材几何中心轴线重合，则棒材单位初始长度 l0 为:l0 = R0 q0式中， R0 为单位原始弯曲半径， q0 为单位原始弯曲角。单元 ABCD l0 D 原纤维长度：l0 D = r0 q0式中 r0 为单位的原始弯曲半径。单元弯曲后， 中性层长度 lw为： lw = rqw（ 1）（ 2）（ 3）式中， 为反弯后中性层的半径， qw为反弯角。此时， 单元 ABCD lw的光纤长度：lwD = r qw（ 4）因此单位在切向上的真应变为：eq = ln(lwD / l0D ) = ln(r qw / r0 q0 )（ 5）矫直前后棒材中性层长度为常数， 即：R0 q0 = rqw（ 6）最后，qe = ln (r R0)（ 7）R0 + Rw - r r其中 Rw为杆的反向弯曲半径。2.3 塑性变形区应力与应变关系棒材的塑性变形面积符合相应的塑性变形规律。根据假设(4)， eq = -eY 。 在平面塑性变形中， r 方向无变形， 即 der = 0 。根据增量理论15：sr = (1/ 2)sq（ 8）23因此， 等效应变ep 等效应力sp 销塑性变形区分别为：23(e -e ) + (2qre -e ) + (2qYe -e )2Yr32ep =eq（ 9）12(s -s ) + (2qrs -s ) + (2qYs -s )2Yrsp =sq（ 10）2.4 外力与塑性变形区应力的关系半空间如图 4 所示， C(x,h)在加载区域 S 的内表面， 而 A(x, y, z)是固体中的一点16。所以距离是：r0 = (x-x)2 + (h-y)2 +1z 2 2（ 11）势函数定义为：H1 = S p(x,h) W dxdh（ 12）当 W = z ln(r0 + z)- r0势函数定义为：H = H1 =z Sp(x,h)ln(r0 + z)dxdh（ 13）定义为：y = H1 ，y = H =p(x,h) 10 dxdh1zz Srz 方向位移函数为：u =1(1- 2uy) - z y（ 14）z4pG z z 方向的应力函数为：u = 1z2 y - z2y2（ 15）p zz 边界条件是：=s- p(x,h) 在s的地区z0不在s的地区10xyz当集中的力 P 垂直作用于原点表面时，r = ( 2 + 2 + 2 )2 ， Sp(x,h) dxdh= P 将定义的 Boussinesq 势能函数简化为：y = H1 = P ln(r + z)1z0y = H = Pzr0（ 16）z 方向的应力分量表示为：r30s = s = - 3P zzr2p5（ 17）2.5 中性层偏移的数学模型棒材在矫直过程中，除塑性变形区外，中性层附近还存在弹性变形区。在它们的边界面上画出s= sS 的关系， 即1r - R +d 2w23 - 3P 2p= sS（ 18）根据假设(5)得出结论nB 2 e = s（ 19）3q S即n 2R rB ln 0 = sS3(R0 + Rw - r ) r（ 20）用联立方程消去 r+3 3P 2psS2R e A R - d0 wR + d -3 3P022psSr=（ 21）理论推导中采用的半空间假设与杆件与凸辊的接触条件不同，因此采用了修正系数R e A R - d + T 0r=hw2R + d - T02（ 22）3 3P2psS 1 这里 A =3 sS n ，T =两者都是与材料相关的系数,d 是棒材直径,s 是指S2 B 棒材的屈服强度， B 为塑性硬化系数， n 为硬化指数， 而 是与材料和弯曲程度有关的修正系数， 由模拟和实验的数据拟合确定。因此， 中性层偏移值的数学模型为R e A R - d + T d= R0-h w2wR + d - T02（ 23）3 两辊矫直过程中中性层偏移的模拟3.1 弯曲的三维有限元模型建立的有限元分析模型如图 5 所示。棒子为疟原虫， 压头为刚体。棒材长度为340 mm， 原始最大挠度为 10 mm/m。材料模型为双线性运动硬化模型， 屈服强度、弹性模量和泊松比见表 2。棒材和压头采用六面体单元 solid164 和扫描17 进行网格划分。在变形过程中，所有的接触都定义为地对地自动接触，静摩擦系数为 0.25， 动摩擦系数为 0.15。3.2 仿真结果图 6 为相同压下量下 40Cr 和 42CrMo 的塑性变形深度图。可以看出，两种材料在相同还原度下的塑性变形层深度是不同的。40Cr 和 42CrMo 的塑性变形深度和半径比值分别约为 0.91 和 0.74,40Cr 的塑性变形深度大于 42CrMo，拉伸侧的塑性变形深度大于压力侧的塑性变形深度。40Cr 和 42CrMo 的拉伸边塑性变形深度比为 1.077， 压力边塑性变形深度比为 1.094， 间接证明中性层会向压力边移动。杆在轴向上分为两半。取杆内不同单元，得到它们在切线方向(X 方向)的应力曲线， 从图 7a 所示的应力曲线中可以看出， 第 7 号单元在切线方向(X 方向) 上的应力曲线。104824 在 0.0402 s 内受拉应力拉伸，单元号为 104824。10471 0 被压应力压缩。同时，应力中性层必须在两者之间。元素没有。104710 在 0.0804s 受拉应力作用，104596 受压应力;应力中性层在它们之间转移。在 0.1474 s 时， 应力中性层向单元号之间偏移。104596 和元素编号 104482. 同理，如图 7b 所示， 应力中性层首先位于 104824 104938 号单元之间。随着弯曲程度的逐渐增加， 它在 104824 104710、104710 104596、104596 104824 之间移动。因此，根据表 1 所示的两种材料的单元应力曲线和模拟数据， 可以确定应力中性层相对于几何中心轴的偏移量。4 .两辊矫直过程中的中性层偏移实验4.1 实验路线及目的实验路线及目的如图 8 所示。4.2 室温单轴拉伸试验拉伸试验是测量材料力学性能的最基本的实验之一。通过拉伸试验，可以得到材料的屈服强度、伸长率、抗拉强度等参数，这些参数是影响棒材弯曲变形的重要因素 1 8 。本文采用 40Cr 和 42CrMo 用于拉伸试验。对拉伸试验数据进行了分析， 为中性层偏移模型的构建提供了参数支持。拉伸试验是在 WAW 1000 万能试验机上进行的。对得到的数据进行处理， 得到如图 9 所示的应力应变曲线。应用 Origin 软件对拉伸实验数据进行拟合。数据是从屈服点到抗拉强度。拟合结果表明， 40Cr 和 42CrMo 进入后， 幂函数拟合的测定校正系数分别为 0.9539 和 0.9647 塑性变形。幂函数调整后的决定系数接近于 1。由于拟合结果很好，40Cr 和 42CrMo 的应力应变关系可以用s= Ben n 的幂函数硬化关系有效描述。两种材料的幂函数拟合结果见表 2。4.3 弯曲测试弯曲试验用于确定材料在弯曲载荷下的力学性能。它是力学性能测试的基本方法之一。钢筋两端简支，中间施加集中荷载。负载由 100 千牛顿的液压负载测试机器。液压试验机由电液伺服压力试验机测控系统控制。负载值可由电液伺服试验机压力测控系统控制。4.3.1 压力过程荷载-位移曲线压力过程的荷载-位移曲线如图 10 所示。4.3.2 弯杆的后处理首先， 使用尼康 S4150 相机对横截面进行拍照， 示意图如图 11 所示。使用计算机辅助设计(CAD) 软件导入照片，然后放大 10 倍，进行精确的尺寸测量，弯曲钢筋截面网格如图 12 所示。以压头附近的哈希网格为重点进行应变分析。在本研究中， 假设平面应变， 并且钢筋弯曲变形主要是纤维的纵向变形。因此，本文仅对网格的纵向应变状态进行分析。如图 12 所示，选择网格 A、B、C 三列，在 CAD 中绘制;对比变形前后网格形状;测量前、后三列网格的纵向网格线的长度后弯曲;并对弯曲前后网格线的长度比取自然对数，求其值为杆体纵向应变。设置杆纤维拉伸应变为正，压力应变为负19。两杆三列网格应变数据如表 3 所示。根 据 位于杆直径方向的网格线位置分布，从下边缘到上边缘，在杆直径方向上的位置应变曲线如图 13 所示。图 13a 为 40Cr 在半径方向弯曲后的应变分布图。原始中立层坐标为零(加载前几何中立层与应变中立层重合)。弯曲后， 从中性层附近的局部放大图可以看出， 在原始应变中性层位置产生了正应变。根据连续变形原理，钢筋下缘的应变值为负。在 t 之间应该有一个应变为 0 的纤维层原始中性层(0 mm)和棒材下边缘(14 mm)的位置， 该纤维层为弯曲后的应变中性层位置。众所周知， 在的过程中塑性弯曲时， 应变中性层会向压力侧偏移。三柱网格应变应变中性层偏移量分别为0.6912、1.0496 和 0.8741 mm。图 13b 为 42CrMo 弯曲后在半径方向上的应变分布， 其中原始中性层坐标为零。可以看出，A、B、C 三列网格中性层的偏移量分别为 0.8746、0.9937、0.9310 mm。通过比较两种材料杆的偏置值，结合不同的加载情况可以看出，应变中性层偏置由杆直径、弯曲量和荷载等多种因素决定。反向半径下棒材中性层偏移实验值如表 4 所示。5 讨论5.1 反向弯曲半径 Rw 对中性层偏移的影响三点弯曲的压下量的变化是由外力来实现的。因此， 反向弯曲半径 Rw 与外力 P有关， 采用原始弯曲半径 R0 =12,505 mm 的 40Cr 和 42CrMo 可以说明相关趋势。图 14 为 40Cr 和 42CrMo 的中性层偏移值随反向弯曲半径的变化。从图 14 可以看出， 无论是 40Cr 还是 42CrMo ， 其中性层的偏移量都随着弯曲半径的减小而增大;分析结果与相关文献相似5-8。理论值与仿真值有较大偏差;究其原因， 是仿真中使用的材料模型是双线性运动硬化模型， 而不是理论推导中使用的幂指数模型。实验值与理论值之间也存在一定的误差，这主要是由于测试方法和测量过程中存在的数据误差造成的。但误差在 0.1 mm 以内， 变化趋势相似。因此可以得出计算式(23)近似可以接受的结论。5.2 材料对中性层棒的影响对比图 15 中两种材料的中性层偏移值， 不同材料的中性层偏移值是不同的， 它们之间的差异比较大。这意味着偏移值与材料20的力学性能有关。塑料材质越好， 中性层偏移的次数就越多。从图 15 中还可以看出， 当反向弯曲半径值较大时， 两种材料的理论数据与实验数据差异较大。当计算最大弯曲程度较小的棒材中性层的偏移量，说明公式(23)的精度不够好。弯曲程度越小，中性层的偏移量越小2 1 。公式计算误差在计算结果中所占比例较大， 影响计算精度。由以上分析可知， 式(23) 仅适用于一般规格和最大弯曲挠度大于 15mm / m 的大截面杆件;不适用于弯曲程度小或直径小的棒材矫直。6 结论1.建立了棒材矫直过程中中性层的迁移模型。对弹塑性弯曲中性层进行了理论估计和分析。实验结果表明，该模型基本适用，为今后棒材矫直及变形机理的持续研究提供了理论参考。2.通过建立棒材矫直过程中性层迁移模型，得出了一些结论。中性层偏移量的大小不仅与反向弯曲半径、材料力学性能、棒材规格有关，还与弯曲力和矫直力的初始程度有关。这在文献中没有得到反映5-8。3.棒材矫直时中性层偏移不仅与工艺结构参数 Rw 有关， 还与棒材规格及材料力学性能有关。中性层偏移量随反向弯曲半径的减小而增大，随塑性变形的增大而增大。4.对于弯曲程度小、直径小的棒材，矫直过程中中性层偏移量小，通常可以忽略不计。但对于一般规格， 尺寸较大， 最大反向弯曲挠度大于 15mm /m 时， 中性层偏移量较大， 这就意味着必须考虑中性层偏移量。参考文献1. 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Jin X, Li Q (2007) Theoretical analysis of thin-walled pipe distortion.Mech Res Appl 20(5):3031ORIGINAL ARTICLEInt J Adv Manuf Technol (2015) 79:15191529 DOI 10.1007/s00170-015-6899-3Research and verification on neutral layer offset of bar in two-roll straightening processLifeng Ma & Ziyong Ma & Weitao Jia & Yangyang Lv &Yaping Jiang & Haijie Xu & Pengtao LiuReceived: 3 August 2014 / Accepted: 9 February 2015 / Published online: 6 March 2015# Springer-Verlag London 2015Abstract A neutral layer must be existed in the metal material deformation. Traditional two-roll straightening theory ignores the migration phenomenon of neutral layer, especially for large cross-sectional bar in the straightening process. Neutral layer has a large impact on straightening springback and then affects the roller shape design, process parameters, and straightness accuracy of bar. In the present work, a neutral layer offset model was established in the bar straightening process based on the three-point bending and elastic-plastic pressure. The neutral layer migration phenomenon had been studied through the model in the bar straightening process, combined with room-temperature tensile test and bending test. The relationship of neutral layer offset values and reverse bending radius and plastic deformation capacity of the metal has been obtained. The neutral layer offset model provides a reference to further study of bar straightening mechanism and deformation.Keywords Neutral layer offset . Bar straightening . Springback . Three-point bending . Experimental analysis1 IntroductionHigh-strength alloy steel bars are increasingly used in the areas of petroleum, automobile, shipbuilding, and construc- tion machinery. As the final finishing step, straightening is the key process to ensure the quality of bar produced 1.L. Ma: Z. Ma (*): W. Jia : Y. Lv: Y. Jiang: H. Xu: P. Liu Heavy Machinery Engineering Research Center of Ministry Education, Taiyuan University of Science and Technology,Taiyuan 030024, Chinae-mail: Z. Mae-mail: 13kelaBecause original bending of bar is likely to exist in any direc- tion, it is particularly suitable for cross-roll straightener. Two- roll straightening and multi-roll straightening are two main forms of cross-roll straightener. However, the existing multi- roll straightening machine has many problems, such as big blind area of straightening, disability of full-length and full- continuous straightening, and need of removing the head and end after straightening. Moreover, the complex structure and large volume of multi-roll straightener generally cause high cost. Because of low precision of straightening, multi-roll straightener cannot realize the high precision of bar straight- ening. Compared to the multi-roll straightener, two-roll straightening machine cannot only eliminate the blind area of bar and improve surface roughness, but also improve oval- ity and solve necking after straightening. More importantly, its residual deflection of bar can achieve 0.10.5 mm/m after straightening, which meets the current accuracy requirements of high precision bar. Domestic small two-roll straightening device has been manufactured, but it has been failed to master the roller design, and core technology of automatic straight- ening process, and the two-roll straightening machine of large cross-sectional bar entirely depends on import. Therefore, based on the localization of device, our research group con- ducted a collaborative work with a company in Hebei Province, China, to develop the high precision two-roll bar straightening machine. The deviation and establishment of neutral layer offset model are problems 2, 3, which are di- rectly related to the precision of the roll shape design and straightening accuracy.Two-roll straightening is a complex process of elastic- plastic deformation. Neutral layer offset changes stress distri- bution of cross section of bar, affects the bending moment ratio and the calculated reverse bend curvature, finally affects the accuracy of roll shape design. Unfortunately, the existing theoretical analysis on bar straightening has ignored the phe- nomenon of neutral layer offset; therefore, the designed roll1520Int J Adv Manuf Technol (2015) 79:15191529shape and process parameters have large error. The study of Guan et al. showed that under certain relative corner radius, the relative error in the neutral layer bending springback could be up to 70 % or more 4. At present, for the issue on the neutral layer offset, many scholars have done lots of research 58. Their research has focused on plate and pipe instead of the neutral layer offset in the two-roll straightening process. So based on the characteristics of its straightening process, combined with three-point bending theory and elastic-plastic theory, a neutral layer offset model is established in the straightening process. This work provides an important refer- ence for further study on the two-roll straightening mechanism of large cross section and high-strength steel bars and device development.2 Establishment of the theoretical model of neutral layer offset in two-roll straightening2.1 Straightening deformation analysis and basic assumptions1. Cross section of bar remains flat before and after defor- mation is perpendicular to the axis of the deformed bar 9. There are no reverse and tilt between two adjacent cross sections.2. The material is continuous, homogeneous, and isotropic, and the stress and strain of neutral layers are considered to be coincident 10.3. The diameter change during bar straightening process is ignored.4. The bending plastic deformation process is consistent with the principle of constant volume.5. Equivalent stress and equivalent strain have a harden-Fig. 1 Diagram of actual straightening state2.2 Strain relation in straightening processBefore entering straightener, the original forms of bar bending caused by machining or heat treatment have single, S bending, multi-peak, and space bending types 13 in the length range, as shown in Fig. 3. All of them will be unified into a single bending type in the central section of straightening roller. Generally, the original bending form is simplified as single bending type to facilitate the theoretical analysis.It can be assumed that the unit ABCD is initially in tensile state and in the compression state after reverse bending 14. The entering straightener bar has small degree of bending. The neu- tral layer and the geometric center axis of the bar are considered to be coincident, so the original length of the bar unit l0 is:l0 R0 01where R0 is the original bending radius and 0 is the original bending angle of unit.The original fiber length of unit ABCD l0D is:l0D r0 02ing exponent relationship of Bn, where B is the co-where r0 is the original bending radius of unit.efficient of plastic material and n is the material hardeningexponent 11.When two-roll straightener straightens bars, bending de- flection is determined by concave roll angle and roll gap. By adjusting the roll gap, the actual bending deflection can vary from 0 to the maximum bending deflection 12. A more rea- sonable straightening state is shown in Fig. 1; at this point, straightening deformation is approximate to three-point bending.A unit of ABCD is taken from the bar straightening defor-After unit being bent, the length of neutral layer lw is:lw w3where is the radius of neutral layer after reverse bending and w is the reverse bending angle. At this point, the fiber length of the unit ABCD lwD is:lwD r w4So the true strain of the unit in the tangential direction is:mation zone, as shown in Fig. 2. In the figure, bar bending center is the original point of cylindrical coordinate system. According to hypothesis (3), deformation in r direction is lnlwD lnl0Dr wr0 05negligible, i.e., r=0. direction stress is much smaller than others and can be ignored, namely = 0. Therefore, the units stress state of bar straightening deformation zone can be sim- plified to plane strain problem, i.e., = r = 0 and = r= 0. Meanwhile, the hypothesis (1) means r=r=0.The length of bar neutral layer is a constant before and after straightening that means:R0 0 w6Int J Adv Manuf Technol (2015) 79:151915291521Fig. 2 Diagram of bar straightening stress and strain in deformation zoneFinally, 1 q222r R0 ln7p p2r3 r100wR R r pj j2where Rw is the reverse bending radius of bar.2.3 Stress and strain relationship of plastic deformation zoneThe area of plastic deformation of the bar meets the relevant laws of plastic deformation. According to hypothesis (4), =.2.4 The relationship between external force and plastic deformation zone stressHalf-space as shown in Fig. 4, C(, ) is taken in the inner surface of loading area S, and A(x, y, z) is a point in the solid 16. So the distance is:hi1222 2In the plane plastic deformation, r direction has no defor-mation, i.e., dr=0.According to the incremental theory 15:0 x y z111r 2 8The potential function is defined as:H p dd12 So, the equivalent strain p and the equivalent stress p in plastic deformation zone are:1 S ; where =zln(0+z)0The potential function is defined as: p2q222 1p3rr2 9H H p; ln zdd13 p3 j jzS0Fig. 3 Original bending types of bar1522Int J Adv Manuf Technol (2015) 79:15191529Fig. 4 Stress and displacement diagram of half-space concentrationDefinition as:p3 . 3P .1 s . 2218 H 1 H 1 2r Rw d .1 z ; z Sp ;d d0.A conclusion is drawn by hypothesis (5)u j jB Displacement function in z direction is:. 2n 1 z 4G12 14p3 s19z z 1 .2Stress function in z direction is:z 2z z z215namely lnB p3 . R r . s20. 2 . R0 r.n0 RwEliminate r by simultaneous equations. z Boundary condition is:p ; In the sregion0Outside the sregionR0 eA 0Rw d s3p3P1A2S2 21R d 1s3p3PWhen concentrated force P acts on the surface of originvertically, 0 x2 y2 z22, Sp(,)d d=P0 2 2SSimplify defined Boussinesq potential function to:The half-space assumptions applied in the theoretical deri-vation differs from the contact condition of bar and the convex H 1 Pln zroll, so a correction factor is applied1z016 H P A . dz0 R0 e Rw 2 Td22Stress component in z direction is expressed as:R0 2 T 3Pzr 2z30 517where A p3. 1, T q3p3P, both of which are coeffi-2SBn2S2.5 Mathematical model of the neutral layer offsetAn elastic deformation zone exists near neutral layer, in addi- tion to the plastic deformation zone, when bar is in the straightening process. Such a relationship s is drawn in their boundary surface that meanscients related to the material, d is the bar diameter, s is the yield strength of the bar, B is the plastic hardening coefficient, n is the hardening exponent, and is the correction factor related with material and bending de- gree and determined by the data fitting of simulation and experiment.Int J Adv Manuf Technol (2015) 79:151915291523Therefore, the mathematical model of neutral layer offset value isR eA .R dT 0 Rww 2 Rd0 2 T23Fig. 5 Finite element modelWith the degree of bending increases gradually, it shifts be- tween element nos. 104824 and 104710, element nos. 104710and 104596, and element nos. 104596 and 104824. Therefore,3 Simulation of neutral layer offset in two-roll straightening process3.1 3D finite element model of bendingFinite element analysis model developed is shown in Fig. 5. Bar is plasmodium, and indenter is assumed as rigid body. The bar length is 340 mm, and the original maximum deflection is 10 mm/m. Material model is the bilinear kinematic hardening model, and the yield strength, elastic modulus, and Poissons ratio are shown in Table 2. Bar and indenter are meshed by using hexahedral elements solid164 and sweep 17. All con- tacts are defined as surface-to-surface automatic contact in the deformation process, the static friction coefficient is 0.25, and dynamic friction coefficient is Simulation resultsFigure 6 is the plastic deformation depth diagrams of 40Cr and 42CrMo with the same reductions. It can be seen that the depth of plastic deformation layer of two materials in the same reductions is different. The ratios of plastic deformation depth and radius of 40Cr and 42CrMo are approximately 0.91 and 0.74, respectively, and so, plastic deformation depth of 40Cr is greater than that of 42CrMo, and the plastic deformation depth of tensile side is greater than the pressure side. The depth ratios of tensile side and pressure side plastic deformation of 40Cr and 42CrMo are 1.077 and 1.094, respectively, which indirectly proved that the neutral layer will move to the pres- sure side.The bar is divided into two halves in the axial direction. Taking different units inside the bar and obtaining their stress curves in the tangent direction (X direction), it can be seen from the stress curves shown in Fig. 7a that element no. 104824 is stretched by tensile stress in 0.0402 s and element no. 104710 is compressed by compressive stress. At the same time, the stress neutral layer must be between them. Element no. 104710 subjects to tensile stress in 0.0804 s, and element no. 104596 subjects to compressive stress; stress neutral layer shifts between them. In 0.1474 s, the stress neutral layer devi- ates to the place between element no. 104596 and element no. 104482. Similarly, as shown in Fig. 7b, the stress neutral layer is between element nos. 104824 and 104938 in the first place.it is possible to determine the value of stress neutral layer offset relative to the geometric center axis according to the element stress curves and simulation data of two materials as shown in Table 1.4 Neutral layer offset experiments in the process of two rollers straightening4.1 Experimental route and purposeThe experimental route and purpose are shown in Fig. 8.4.2 Uniaxial tensile tests at room temperatureTensile test is one of the most basic experiments to measure the mechanical properties of materials. By a tensile test, the yield strength, elongation, tensile strength, and other parame- ters of a material can be obtained, which are important factors of bar bending deformation 18. In this paper, 40Cr and(a)(b)Fig. 6 Plastic deformation depth diagrams of materials with the same reductions: a 40Cr and b 42CrMo1524Int J Adv Manuf Technol (2015) 79:15191529Fig. 7 Stress change curves of materials in the X direction: a 40Cr and b 42CrMo(a)(b)42CrMo are used for tensile test. Experimental data of tensile test is analyzed, which provides parameters support for the construction of a neutral layer offset model.Table 1 Simulated data of neutral layer offset under different reverse bending radiusTensile tests were conducted on a WAW 1000 universal testing machine. The resulting data had been processed to get the stress and strain curves as shown in Fig. 9.The Origin Software is applied to fitting the tensile exper- imental data. The data is taken from the yield point to tensile strength. The fitting results show that the correction coeffi- cients of determination fitted by the power function are re- spectively 0.9539 and 0.9647 after 40Cr and 42CrMo enterMaterialDiameterTimeReductionBending radiusOffset valued (mm)T (s) (mm)RW (mm) (mm)40Cr280.04025.022875.8680.19530.0678.3751725.8180.42330.080410.051438.2640.48840.093811.7251232.8840.61860.120615.075959.0770.78140.147418.425784.8060.911642CrMo250.04025.022875.9020.13020.0678.3751725.1590.22790.080410.051437.9090.32560.093811.7251231.0330.39070.120615.075959.0770.65120.147418.425784.6310.7814 Fig. 8 Study route and purposeInt J Adv Manuf Technol (2015) 79:151915291525(a)8007006003040Cr25Load /KN2040Cr42CrMoStress / MPa500154003001020010005e t 5e t0(b)0.000.020.040.060.040.16Strain051015202530Displacement /mmFig. 10 The load-displacement curve12001000Stress / MPa80060040020042CrMotesting machine. Hydraulic pressure testing machine is con- trolled by electro-hydraulic servo pressure testing machine measurement and control system. The load value can be con- trolled by electro-hydraulic servo testing machine pressure measurement and control system.4.3.1 Load-displacement curve of pressure processThe load-displacement curve of pressure process is shown in Fig. 10.00.000.020.040.060.080.10strainFig. 9 Relationship curve of stress-strain: a 40Cr and (b) 42CrMothe plastic deformation. The power function adjusted determi- nation coefficients are close to 1. Since the fitting result is very good, the stress and strain relationship of 40Cr and 42CrMo can be effectively described by the power function hardening relationship of Bn. The power function fitting results of the two kinds of materials are shown in Table 2.4.3 Bending testsBending test is applied to determine the mechanical properties when material subjected to bending load. It is one of the basic methods of mechanical property tests. The bar is simply sup- ported at both ends with a concentrated load applied in the middle. The load is loaded by 100 kN of hydraulic pressureTable 2 Mechanical properties of different material parameterss (MPa)E (GPa)B (MPa)n4.3.2 The post-processing of bending barFirstly, using Nikon S4150 camera was used to take pictures of the bar cross section, as schematically shown in Fig. 11. Using computer-aided design (CAD) software to import photos and then amplifying ten times for precise size measure- ment, the grid of bending bar section is shown in Fig. 12.Taking the hash grid near pressure head the earnest for strain analysis. In this study, the plane strain is assumed, and40Cr4102060.311820.2242CrMo9302100.31119.90.058 Fig. 11 Diagram of taking photos1526Int J Adv Manuf Technol (2015) 79:15191529Fig. 12 Grids of bar bending: a(a)(b)40Cr and b 42CrMobar bending deformation is mainly longitudinal deformation of fiber. Therefore, only the longitudinal strain state of grid is analyzed. As shown in Fig. 12, choosing three column of the grids, A, B, and C, to paint in CAD; comparing grid shapes before and back deformation; measuring the length of longi- tudinal grid lines of the three columns of the grids before andafter bending; and taking the natural logarithm of the grid lines length ratio after bending and before bending, the value as the bar longitudinal strain is worked out. Setting the bar fiber tensile strain as positive, and pressure strain as negative 19. The three-column grid strain data of two bars is shown in Table 3.Table 3Strain data of bar grid linesRW=912.6112345678940CrA0.1370.1220.1060.0900.0740.0580.0420.0250.009B0.1250.1090.0940.0800.0650.0490.0350.0190.004C0.1100.0980.0850.0720.0580.0450.0330.0200.007101112131415161718A0.01050.02710.04460.06210.07940.09700.11360.13080.1471B0.01390.02910.04690.06120.07690.09190.10690.12270.1370C0.01260.02620.04100.05540.06990.08390.09780.11150.1251RW=800.9412345678942CrMoA0.0810.0710.0590.0480.0390.0290.0180.0070.0059B0.0950.0820.0700.0570.0430.0310.0170.0060.0064C0.0730.0660.0540.0440.0340.0230.0140.0040.00481011121314151617A0.01620.02660.03760.04780.05830.06810.07870.0900B0.01810.03050.04220.05450.06630.07810.09080.1041C0.01510.02520.03540.04630.05620.06550.07580.0861Int J Adv Manuf Technol (2015) 79:151915291527(a)5(a)Theoretical data Experimental data Simulation data1.2Offset value /mm1.0Linear strain0.000.8-0.050.6AB C-15-10-5051015-0.100.4(b)-0.15AB C-15-12-9-6-3036912150.120.08Coordinate of grid lines(b)0.21.0750100012501500175020002250Reverse bending radius Rw/mmTheoretical data Experimental data Simulation data0.040.8Linear strainOffset value /mm0.000.6-0.040.4-0.080.2-0.12Coordinate of grid lines0.0500100015002000250030003500Fig. 13 Strain relation of radial direction for bars: a 40Cr and b 42CrMoAccording to the location distribution of the grid lines lo- cated in the direction of bar diameter, from lower edge to upper edge of the bar, the position strain curve on the direction of bar diameter is shown in Fig. 13.Figure 13a is the strain distribution of 40Cr after bending on the direction of radius. The coordinates of original neutral layer is zero (before loading, the geometric neutral layer and the strain neutral layer are coincident). After bending, it can be seen from the local amplification figure near the neutral layer that the positive strain is generated on the position of original strain neutral layer. The strain value of lower edge of the bar is negative according to the deformation principle of continuity. There should be a fiber layer with a strain of 0 between theTable 4Experimental data of neutral layer offset under different reverse bending radiusMaterialDiameter (mm)Reverse bending radius (RW/mm)40Cr282136.811282.30912.61713.160.42070.67721.04961.196842CrMo253322.171615.171076.53800.940.04140.47570.73210.9937Reverse bending radius Rw/mmFig. 14 Curves of neutral layer offset and bending radius: a 40Cr and b42CrMoposition of original neutral layer (0 mm) and the lower edge of the bar (14 mm), and this fibrous layer is strain neutral layer position after bending. It is known that during the process ofTheoretical data of 40Cr Experimental data of 40CrTheoretical data of 42CrMo Experimental data of 42CrMo1.21.0Offset value /mm0.20.0500100015002000250030003500Reverse bending radius Rw /mmFig. 15 Curve of neutral layer offset and bending radius for different materials1528Int J Adv Manuf Technol (2015) 79:15191529plastic bending, strain neutral layer will be offset to the pres- sure side. The offset of strain neutral layer of three-column grid strain is 0.6912, 1.0496, and 0.8741 mm, respectively.Figure 13b shows strain distribution of 42CrMo after bend- ing on the direction of radius, where the coordinates of orig- inal neutral layer is zero. It can be seen that the offset of neutral layer belonging to the three columns of the grids called A, B, and C is 0.8746, 0.9937, and 0.9310 mm, respectively. By comparing the offset values of the two material bars, combin- ing with the different loading conditions, it can be seen that the strain neutral layer offset is determined by many factors, such as bar diameter, amount of bending and load. Neutral layer offset experimental values of bars under the reverse radius are shown in Table 4.5 Discussions5.1 The impact of reverse bending radius Rw on neutral layer offsetThe change of reduction in three-point bending is achieved by an external force. Therefore, reverse bending radius Rw is associated with the external force P. Using 40Cr and 42CrMo with original bending radius of R0 =12,505 mm, it illustrates relevant trends. Figure 14 shows the changes of neutral layer offset value of 40Cr and 42CrMo with variation of the reverse bending radius.It can be seen from Fig. 14 that either 40Cr or 42CrMo, the offset value of their neutral layer increases with the decreasing of bending radius; the analysis results are similar with the related literatures 58. Theoretical and simulation values have large deviation; the reason for which is that the material model used in simulation is bilinear kinematic hardening model, rather than power index model used in theoretical der- ivation. There is also certain error between experimental and theoretical values, which are mainly caused by the data errors existed in the test methods and measurement process. But the error is in 0.1 mm or less, and the trends of change are similar. Therefore, a conclusion can be drawn that the calculation for- mula (23) is approximately acceptable.5.2 The impact of material on the neutral layer barComparing the neutral layer offset values of two materials in Fig. 15, the neutral layer offset values of different materials are different, and the difference between them is relatively large. This means that the offset value is related to the mechanical properties of the material 20. The better plastic the material is, the more the neutral layer offset.In Fig. 15, it can also be seen that the difference between theoretical and experimental data of two materials is large when the values of reverse bending radius are large. Whencalculating the offset value of bar neutral layer, which has less degree of maximum bending, it shows that the accuracy of formula (23) is not good enough. The smaller the degree of inflection is, the smaller the offset value of neutral layer is 21. The higher proportion of formula calculation error in calculation results affects the calculation accuracy.It can be seen from the analysis above that formula (23) is only applied to general specification and large cross section bar whose biggest bending deflection is greater than 15 mm/ m; it is not applied to straightening bar with a small degree of bending or a small diameter.6 Conclusions1. The migration model of neutral layer of bar in straighten- ing process is established in this study. Theoretical esti- mation and analysis can be performed on the elastic- plastic bending neutral layer. Experiments show that the model is approximately applicable, which provides a the- oretical reference for a future continuous research on bar straightening and deformation mechanism.2. Through the establishment of the migration model of neu- tral layer in bar straightening process, some conclusions can be foun
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