二輥棒材矯直機(jī)的設(shè)計(jì)含5張CAD圖
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論文題目: 二輥棒材矯直機(jī)的設(shè)計(jì)
姓
名:
蔣少偉
學(xué) 號(hào):
2017008126
系
部:
機(jī)電工程系
專
業(yè):
機(jī)械設(shè)計(jì)制造及其自動(dòng)化
班 級(jí):
機(jī)械 1704
指導(dǎo)教師:
郭曉娥
職
稱:
完成日期:
2021
年
月
日
二輥矯直過程中棒材中性層偏置的研究與驗(yàn)證
摘 要: 金屬材料變形過程中必然存在中性層。傳統(tǒng)的二輥矯直理論忽略了中性層的遷移現(xiàn)象,特別是大截面棒材矯直過程中。中性層對(duì)矯直回彈影響較大,進(jìn)而影響輥形設(shè)計(jì)、工藝參數(shù)和棒材直線度精度?;谌c(diǎn)彎曲和彈塑性壓力,建立了棒材矯直過程中的中性層偏移模型。通過模型研究棒材矯直過程中的中性層遷移現(xiàn)象,并結(jié)合室溫拉伸試驗(yàn)和彎曲試驗(yàn)。得到了中性層偏移量與反向彎曲半徑和金屬塑性變形能力的關(guān)系。中性層偏移模型為棒材矯直機(jī)理和變形的進(jìn)一步研究提供了參考。
關(guān)鍵詞: 中性層偏移; 棒材矯直; 回彈; 三點(diǎn)彎曲; 實(shí)驗(yàn)分析
1 引言
高強(qiáng)度合金鋼在石油、汽車、造船、工程機(jī)械等領(lǐng)域的應(yīng)用日益廣泛。矯直作為最后的精整工序,是保證生產(chǎn)棒材質(zhì)量的關(guān)鍵工序[ 1 ] 。由于棒材的原始彎曲很可能存在于任何方向,特別適用于橫輥矯直機(jī)。兩輥矯直和多輥矯直是橫輥矯直機(jī)的兩種主要形式。然而,現(xiàn)有的多輥矯直機(jī)存在矯直盲區(qū)大、矯直全程和全程連續(xù)缺陷、矯直后需要去除頭端等問題。而且多輥矯直機(jī)結(jié)構(gòu)復(fù)雜, 體積大, 造價(jià)高。多輥矯直機(jī)由于矯直精度低,無法實(shí)現(xiàn)棒材矯直的高精度。與多輥矯直機(jī)相比, 雙輥矯直機(jī)不僅能消除棒材盲區(qū), 提高表面粗糙度, 還能提高橢圓度, 解決矯直后的縮口問題。更重要的是,其矯直后棒材的殘余撓度可達(dá)到 0.1-0.5
mm/m, 滿足當(dāng)前高精度棒材的精度要求。國產(chǎn)小型二輥矯直裝置已經(jīng)制造出來, 但未能掌握輥型設(shè)計(jì)和自動(dòng)矯直過程的核心技術(shù),大截面棒材二輥矯直機(jī)完全依賴進(jìn)口。因此,在設(shè)備國產(chǎn)化的基礎(chǔ)上,我們課題組與中國河北省某公司合作開發(fā)了高精度雙輥棒材矯直機(jī)。中性層偏移模型的偏差和建立是問題[2 , 3 ] , 直接關(guān)系到輥形設(shè)計(jì)的精度和矯直精度。
二輥矯直是一個(gè)復(fù)雜的彈塑性變形過程。中性層偏移會(huì)改變棒材截面的應(yīng)力分布,影響彎矩比和計(jì)算的反彎曲率,最終影響輥型設(shè)計(jì)的精度。遺憾的是,現(xiàn)有的棒材矯直理論分析忽略了中性層偏移現(xiàn)象;因此, 設(shè)計(jì)了軋輥形狀和工藝參數(shù)誤差較大。Guan 等人的研究表明, 在一定的相對(duì)拐角半徑下, 中性層彎曲回彈的相對(duì)誤差可達(dá) 70% 以上[ 4 ] 。目前, 對(duì)于中性層偏移問題, 許多學(xué)者做了大量的研究[ 5 - 8 ] 。他們的研究主要集中在兩輥矯直過程中用板和管代替中性層偏移。因此,根據(jù)其矯直過程的特點(diǎn),結(jié)合三點(diǎn)彎曲理論和彈塑性理論,建立了矯直過程中的中性層偏移模型。該工作為進(jìn)一步研究大截面高強(qiáng)鋼筋二輥矯直機(jī)理和研制裝置提供了重要參考。
2 建立兩輥矯直過程中中性層偏移的理論模型
2.1 矯直變形分析及基本假設(shè)
1. 鋼筋變形前后截面保持平坦, 垂直于變形鋼筋的軸線[ 9 ] 。兩個(gè)相鄰截面之間沒有反向和傾斜。
2. 材料是連續(xù)的、均勻的、各向同性的, 中性層的應(yīng)力和應(yīng)變被認(rèn)為是一致的
[ 1 0 ] 。
3.忽略棒材矯直過程中直徑的變化。
4. 彎曲塑性變形過程符合等體積原理。
5. 等效應(yīng)力s和等效應(yīng)變e具有s= Ben 的硬化指數(shù)關(guān)系,當(dāng) B 為塑性材料系數(shù),
n 為材料硬化指數(shù)[ 1 1 ] 。
二輥矯直機(jī)矯直棒材時(shí),彎曲撓度由凹輥角和凹輥間隙決定。通過調(diào)整輥縫, 實(shí)際彎曲撓度可以從 0 變化到最大彎曲撓度[ 1 2 ] 。更合理的矯直狀態(tài)如圖 1 所示; 在試驗(yàn)過程中, 矯直變形近似于三點(diǎn)彎曲。
從棒材矯直變形區(qū)取一個(gè) ABCD 單元, 如圖 2 所示。圖中棒材彎曲中心為柱面坐標(biāo)系的原點(diǎn)。根據(jù)假設(shè)(3), r 方向的變形可以忽略不計(jì)er = 0 , Y 方向應(yīng)力要小得多, 可以忽略不計(jì), 即sY = 0 。因此, 棒材矯直變形區(qū)單元應(yīng)力狀態(tài)可簡(jiǎn)化為平面應(yīng)變問題tqY =tYr = 0 和gqY = gYr = 0 , 同時(shí), 假設(shè)(1)表示tqr = gqr = 0 。
2.2 矯直過程中的應(yīng)變關(guān)系
在進(jìn)入矯直機(jī)前,由加工或熱處理引起的棒材彎曲的原始形式在長度范圍內(nèi)有單彎、S 彎、多峰彎和空間彎[1 3 ] ,如圖 3 所示。在矯直輥中心段,將所有的彎曲方式統(tǒng)一成單一的彎曲方式。一般將原始彎曲形式簡(jiǎn)化為單一彎曲形式,便于理論分析。
可以假設(shè)單元 ABCD 初始處于拉伸狀態(tài), 反向彎曲[14] 后處于壓縮狀態(tài)。進(jìn)入矯直機(jī)棒材彎曲程度小??紤]中性層與棒材幾何中心軸線重合,則棒材單位初始長度 l0 為:
l0 = R0 ×q0
式中, R0 為單位原始彎曲半徑, q0 為單位原始彎曲角。單元 ABCD l0 D 原纖維長度:
l0 D = r0 ×q0
式中 r0 為單位的原始彎曲半徑。
單元彎曲后, 中性層長度 lw為: lw = r×qw
( 1)
( 2)
( 3)
式中, ρ 為反彎后中性層的半徑, qw為反彎角。此時(shí), 單元 ABCD lw的光纖長度:
lwD = r ×qw
( 4)
因此單位在切向上的真應(yīng)變?yōu)椋?
eq = ln(lwD / l0D ) = ln(r ×qw / r0 ×q0 )
( 5)
矯直前后棒材中性層長度為常數(shù), 即:
R0 ×q0 = r×qw
( 6)
最后,
q
e = ln (
r × R0
)
( 7)
R0 + Rw - r × r
其中 Rw為桿的反向彎曲半徑。
2.3 塑性變形區(qū)應(yīng)力與應(yīng)變關(guān)系
棒材的塑性變形面積符合相應(yīng)的塑性變形規(guī)律。根據(jù)假設(shè)(4), eq = -eY 。 在平面塑性變形中, r 方向無變形, 即 der = 0 。
根據(jù)增量理論[15]:
sr = (1/ 2)sq
( 8)
2
3
因此, 等效應(yīng)變ep 等效應(yīng)力sp 銷塑性變形區(qū)分別為:
2
3
(
e -e ) + (
2
q
r
e -e ) + (
2
q
Y
e -e )
2
Y
r
3
2
ep =
= eq
( 9)
1
2
(
s -s ) + (
2
q r
s -s ) + (
2
q Y
s -s )
2
Y r
sp =
= sq
( 10)
2.4 外力與塑性變形區(qū)應(yīng)力的關(guān)系
半空間如圖 4 所示, C(x,h)在加載區(qū)域 S 的內(nèi)表面, 而 A(x, y, z)是固體中的一點(diǎn)[16]。所以距離是:
r0 = [(x-
x)2 + (h-
y)2 +
1
]
z 2 2
( 11)
勢(shì)函數(shù)定義為:
H1 = òò S p(x,h)× W × dxdh
( 12)
當(dāng) W = z ln(r0 + z)- r0
勢(shì)函數(shù)定義為:
H = ?H1 =
?z
òò S
p(x,h)ln(r0 + z)dxdh
( 13)
定義為:
y = ?H1 ,y = ?H =
?
p(x,h)× 1
0
× dxdh
1 ?z
?z òò S r
z 方向位移函數(shù)為:
u = 1 é(1- 2uy) - z ?yù
?
( 14)
z 4pG ê? ?z ú?
z 方向的應(yīng)力函數(shù)為:
u = 1
z 2
? ?y - z
?2y?
÷
2
( 15)
?
pè ?z ?z ?
邊界條件是:
{=
s
- p(x,h) 在s的地區(qū)
z 0 不在s的地區(qū)
1
0 x y z
當(dāng)集中的力 P 垂直作用于原點(diǎn)表面時(shí),r = ( 2 + 2 + 2 )2 ,
òò S
p(x,h)× dxdh= P 將定
義的 Boussinesq 勢(shì)能函數(shù)簡(jiǎn)化為:
y = ?H1 = P ln(r + z)
1 ?z 0
y = ?H = P
?z r0
( 16)
z 方向的應(yīng)力分量表示為:
r
3
0
s = s = - 3P × z
z r 2p 5
( 17)
2.5 中性層偏移的數(shù)學(xué)模型
棒材在矯直過程中,除塑性變形區(qū)外,中性層附近還存在彈性變形區(qū)。在它們的邊界面上畫出s= sS 的關(guān)系, 即
1
?
?
è
r - R +
d ?2
w
2
÷
?
3 × - 3P ×
2p
= sS
( 18)
根據(jù)假設(shè)(5)得出結(jié)論
n
B × ? 2 × e ? = s
( 19)
3
? q ÷ S
è ?
即
n
? 2 R × r ?
B × ? ?× ln 0 ÷ = sS
?
3
è
? (R0 + Rw - r )× r÷
( 20)
用聯(lián)立方程消去 r
+
3 3P ?
÷
2
ps
S
÷
?
?
2
R × e A × ? R - d
0 ? w
R + d -
3 3P
0
2 2ps
S
r= è
( 21)
理論推導(dǎo)中采用的半空間假設(shè)與桿件與凸輥的接觸條件不同,因此采用了修正系數(shù)η
R × e A × ? R - d + T ?
è
?
0
r=h×
? w 2 ÷
R + d - T
0 2
( 22)
3 3P
2psS
1
這里 A =
3 ?sS ? n ,T = 兩者都是與材料相關(guān)的系數(shù),d 是棒材直徑,s 是指
? ÷ S
2 è B ?
棒材的屈服強(qiáng)度, B 為塑性硬化系數(shù), n 為硬化指數(shù), 而η 是與材料和彎曲程度有關(guān)的修正系數(shù), 由模擬和實(shí)驗(yàn)的數(shù)據(jù)擬合確定。
因此, 中性層偏移值的數(shù)學(xué)模型為
è
?
R × e A ? R - d + T ?
d= R
0
-h×
? w 2 ÷
w R + d - T
0 2
( 23)
3 兩輥矯直過程中中性層偏移的模擬
3.1 彎曲的三維有限元模型
建立的有限元分析模型如圖 5 所示。棒子為瘧原蟲, 壓頭為剛體。棒材長度為340 mm, 原始最大撓度為 10 mm/m。材料模型為雙線性運(yùn)動(dòng)硬化模型, 屈服強(qiáng)度、
彈性模量和泊松比見表 2。棒材和壓頭采用六面體單元 solid164 和掃描[17] 進(jìn)行網(wǎng)格劃分。在變形過程中,所有的接觸都定義為地對(duì)地自動(dòng)接觸,靜摩擦系數(shù)為 0.25, 動(dòng)摩擦系數(shù)為 0.15。
3.2 仿真結(jié)果
圖 6 為相同壓下量下 40Cr 和 42CrMo 的塑性變形深度圖。可以看出,兩種材料在相同還原度下的塑性變形層深度是不同的。40Cr 和 42CrMo 的塑性變形深度和半徑比值分別約為 0.91 和 0.74,40Cr 的塑性變形深度大于 42CrMo,拉伸側(cè)的塑性變形深度大于壓力側(cè)的塑性變形深度。40Cr 和 42CrMo 的拉伸邊塑性變形深度比為 1.077, 壓力邊塑性變形深度比為 1.094, 間接證明中性層會(huì)向壓力邊移動(dòng)。
桿在軸向上分為兩半。取桿內(nèi)不同單元,得到它們?cè)谇芯€方向(X 方向)的應(yīng)力曲線, 從圖 7a 所示的應(yīng)力曲線中可以看出, 第 7 號(hào)單元在切線方向(X 方向) 上的應(yīng)力曲線。104824 在 0.0402 s 內(nèi)受拉應(yīng)力拉伸,單元號(hào)為 104824。10471 0 被壓應(yīng)力壓縮。同時(shí),應(yīng)力中性層必須在兩者之間。元素沒有。104710 在 0.0804
s 受拉應(yīng)力作用,104596 受壓應(yīng)力;應(yīng)力中性層在它們之間轉(zhuǎn)移。在 0.1474 s 時(shí), 應(yīng)力中性層向單元號(hào)之間偏移。104596 和元素編號(hào) 104482. 同理,如圖 7b 所示, 應(yīng)力中性層首先位于 104824 ~ 104938 號(hào)單元之間。
隨著彎曲程度的逐漸增加, 它在 104824 ~ 104710、104710 ~ 104596、104596 ~
104824 之間移動(dòng)。因此,根據(jù)表 1 所示的兩種材料的單元應(yīng)力曲線和模擬數(shù)據(jù), 可以確定應(yīng)力中性層相對(duì)于幾何中心軸的偏移量。
4 .兩輥矯直過程中的中性層偏移實(shí)驗(yàn)
4.1 實(shí)驗(yàn)路線及目的
實(shí)驗(yàn)路線及目的如圖 8 所示。
4.2 室溫單軸拉伸試驗(yàn)
拉伸試驗(yàn)是測(cè)量材料力學(xué)性能的最基本的實(shí)驗(yàn)之一。通過拉伸試驗(yàn),可以得到材料的屈服強(qiáng)度、伸長率、抗拉強(qiáng)度等參數(shù),這些參數(shù)是影響棒材彎曲變形的重要因素[ 1 8 ] 。本文采用 40Cr 和 42CrMo 用于拉伸試驗(yàn)。對(duì)拉伸試驗(yàn)數(shù)據(jù)進(jìn)行了分析, 為中性層偏移模型的構(gòu)建提供了參數(shù)支持。
拉伸試驗(yàn)是在 WAW 1000 萬能試驗(yàn)機(jī)上進(jìn)行的。對(duì)得到的數(shù)據(jù)進(jìn)行處理, 得到如圖 9 所示的應(yīng)力應(yīng)變曲線。
應(yīng)用 Origin 軟件對(duì)拉伸實(shí)驗(yàn)數(shù)據(jù)進(jìn)行擬合。數(shù)據(jù)是從屈服點(diǎn)到抗拉強(qiáng)度。擬合結(jié)果表明, 40Cr 和 42CrMo 進(jìn)入后, 冪函數(shù)擬合的測(cè)定校正系數(shù)分別為 0.9539 和 0.9647 塑性變形。冪函數(shù)調(diào)整后的決定系數(shù)接近于 1。由于擬合結(jié)果很好,
40Cr 和 42CrMo 的應(yīng)力應(yīng)變關(guān)系可以用s= Ben n 的冪函數(shù)硬化關(guān)系有效描述。兩種材料的冪函數(shù)擬合結(jié)果見表 2。
4.3 彎曲測(cè)試
彎曲試驗(yàn)用于確定材料在彎曲載荷下的力學(xué)性能。它是力學(xué)性能測(cè)試的基本方法之一。鋼筋兩端簡(jiǎn)支,中間施加集中荷載。負(fù)載由 100 千牛頓的液壓負(fù)載測(cè)試機(jī)器。液壓試驗(yàn)機(jī)由電液伺服壓力試驗(yàn)機(jī)測(cè)控系統(tǒng)控制。負(fù)載值可由電液伺服試驗(yàn)機(jī)壓力測(cè)控系統(tǒng)控制。
4.3.1 壓力過程荷載-位移曲線
壓力過程的荷載-位移曲線如圖 10 所示。
4.3.2 彎桿的后處理
首先, 使用尼康 S4150 相機(jī)對(duì)橫截面進(jìn)行拍照, 示意圖如圖 11 所示。使用計(jì)算機(jī)輔助設(shè)計(jì)(CAD) 軟件導(dǎo)入照片,然后放大 10 倍,進(jìn)行精確的尺寸測(cè)量,彎曲鋼筋截面網(wǎng)格如圖 12 所示。
以壓頭附近的哈希網(wǎng)格為重點(diǎn)進(jìn)行應(yīng)變分析。在本研究中, 假設(shè)平面應(yīng)變, 并且鋼筋彎曲變形主要是纖維的縱向變形。因此,本文僅對(duì)網(wǎng)格的縱向應(yīng)變狀態(tài)進(jìn)行分析。如圖 12 所示,選擇網(wǎng)格 A、B、C 三列,在 CAD 中繪制;對(duì)比變形前后網(wǎng)格形狀;測(cè)量前、后三列網(wǎng)格的縱向網(wǎng)格線的長度后彎曲;并對(duì)彎曲前后網(wǎng)格線的長度比取自然對(duì)數(shù),求其值為桿體縱向應(yīng)變。設(shè)置桿纖維拉伸應(yīng)變?yōu)檎瑝毫?yīng)變?yōu)樨?fù)[19]。兩桿三列網(wǎng)格應(yīng)變數(shù)據(jù)如表 3 所示。
根 據(jù) 位于桿直徑方向的網(wǎng)格線位置分布,從下邊緣到上邊緣,在桿直徑方向上的位置應(yīng)變曲線如圖 13 所示。
圖 13a 為 40Cr 在半徑方向彎曲后的應(yīng)變分布圖。原始中立層坐標(biāo)為零(加載前幾何中立層與應(yīng)變中立層重合)。彎曲后, 從中性層附近的局部放大圖可以看出, 在原始應(yīng)變中性層位置產(chǎn)生了正應(yīng)變。根據(jù)連續(xù)變形原理,鋼筋下緣的應(yīng)變值為負(fù)。在 t 之間應(yīng)該有一個(gè)應(yīng)變?yōu)?0 的纖維層原始中性層(0 mm)和棒材下邊緣(?14 mm)的位置, 該纖維層為彎曲后的應(yīng)變中性層位置。眾所周知, 在的過程中塑性彎曲時(shí), 應(yīng)變中性層會(huì)向壓力側(cè)偏移。三柱網(wǎng)格應(yīng)變應(yīng)變中性層偏移量分別為0.6912、1.0496 和 0.8741 mm。
圖 13b 為 42CrMo 彎曲后在半徑方向上的應(yīng)變分布, 其中原始中性層坐標(biāo)為零。可以看出,A、B、C 三列網(wǎng)格中性層的偏移量分別為 0.8746、0.9937、0.9310 mm。通過比較兩種材料桿的偏置值,結(jié)合不同的加載情況可以看出,應(yīng)變中性層偏置由桿直徑、彎曲量和荷載等多種因素決定。反向半徑下棒材中性層偏移實(shí)驗(yàn)值如表 4 所示。
5 討論
5.1 反向彎曲半徑 Rw 對(duì)中性層偏移的影響
三點(diǎn)彎曲的壓下量的變化是由外力來實(shí)現(xiàn)的。因此, 反向彎曲半徑 Rw 與外力 P
有關(guān), 采用原始彎曲半徑 R0 =12,505 mm 的 40Cr 和 42CrMo 可以說明相關(guān)趨勢(shì)。圖 14 為 40Cr 和 42CrMo 的中性層偏移值隨反向彎曲半徑的變化。
從圖 14 可以看出, 無論是 40Cr 還是 42CrMo , 其中性層的偏移量都隨著彎曲半徑的減小而增大;分析結(jié)果與相關(guān)文獻(xiàn)相似[5-8]。理論值與仿真值有較大偏差;究其原因, 是仿真中使用的材料模型是雙線性運(yùn)動(dòng)硬化模型, 而不是理論推導(dǎo)中使用的冪指數(shù)模型。實(shí)驗(yàn)值與理論值之間也存在一定的誤差,這主要是由于測(cè)試方法和測(cè)量過程中存在的數(shù)據(jù)誤差造成的。但誤差在 0.1 mm 以內(nèi), 變化趨勢(shì)相似。因此可以得出計(jì)算式(23)近似可以接受的結(jié)論。
5.2 材料對(duì)中性層棒的影響
對(duì)比圖 15 中兩種材料的中性層偏移值, 不同材料的中性層偏移值是不同的, 它們之間的差異比較大。這意味著偏移值與材料[20]的力學(xué)性能有關(guān)。塑料材質(zhì)越好, 中性層偏移的次數(shù)就越多。
從圖 15 中還可以看出, 當(dāng)反向彎曲半徑值較大時(shí), 兩種材料的理論數(shù)據(jù)與實(shí)驗(yàn)數(shù)據(jù)差異較大。當(dāng)計(jì)算最大彎曲程度較小的棒材中性層的偏移量,說明公式
(23)的精度不夠好。彎曲程度越小,中性層的偏移量越小[2 1 ] 。公式計(jì)算誤差在計(jì)算結(jié)果中所占比例較大, 影響計(jì)算精度。
由以上分析可知, 式(23) 僅適用于一般規(guī)格和最大彎曲撓度大于 15mm / m 的大截面桿件;不適用于彎曲程度小或直徑小的棒材矯直。
6 結(jié)論
1.建立了棒材矯直過程中中性層的遷移模型。對(duì)彈塑性彎曲中性層進(jìn)行了理論估計(jì)和分析。實(shí)驗(yàn)結(jié)果表明,該模型基本適用,為今后棒材矯直及變形機(jī)理的持續(xù)研究提供了理論參考。
2.通過建立棒材矯直過程中性層遷移模型,得出了一些結(jié)論。中性層偏移量的大小不僅與反向彎曲半徑、材料力學(xué)性能、棒材規(guī)格有關(guān),還與彎曲力和矯直力的初始程度有關(guān)。這在文獻(xiàn)中沒有得到反映[5-8]。
3.棒材矯直時(shí)中性層偏移不僅與工藝結(jié)構(gòu)參數(shù) Rw 有關(guān), 還與棒材規(guī)格及材料力學(xué)性能有關(guān)。中性層偏移量隨反向彎曲半徑的減小而增大,隨塑性變形的增大而增大。
4.對(duì)于彎曲程度小、直徑小的棒材,矯直過程中中性層偏移量小,通常可以忽略不計(jì)。但對(duì)于一般規(guī)格, 尺寸較大, 最大反向彎曲撓度大于 15mm /m 時(shí), 中性層偏移量較大, 這就意味著必須考慮中性層偏移量。
參考文獻(xiàn)
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8. Huang DY (1993) Discussion of sheet metal bending neutral layerdisplacement coefficient. Mech Res Appl 4:8–11
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ORIGINAL ARTICLE
Int J Adv Manuf Technol (2015) 79:1519–1529 DOI 10.1007/s00170-015-6899-3
Research and verification on neutral layer offset of bar in two-roll straightening process
Lifeng Ma & Ziyong Ma & Weitao Jia & Yangyang Lv &
Yaping Jiang & Haijie Xu & Pengtao Liu
Received: 3 August 2014 / Accepted: 9 February 2015 / Published online: 6 March 2015
# Springer-Verlag London 2015
Abstract 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 analysis
1 Introduction
High-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, China
e-mail: malifengfqh@163.com
Z. Ma
e-mail: 13kela@sina.com
Because 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 0.1–0.5 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 roll
1520 Int J Adv Manuf Technol (2015) 79:1519–1529
shape 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 [5–8]. 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 straightening
2.1 Straightening deformation analysis and basic assumptions
1. 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 σ and equivalent strain ε have a harden-
Fig. 1 Diagram of actual straightening state
2.2 Strain relation in straightening process
Before 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 ? θ0 e1T
where R0 is the original bending radius and θ0 is the original bending angle of unit.
The original fiber length of unit ABCD l0D is:
l0D ? r0 ? θ0 e2T
ing exponent relationship of σ ? Bεn, where B is the co-
where r0 is the original bending radius of unit.
efficient of plastic material and n is the material hardening
exponent [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 ? ρ ? θw e3T
where ρ is the radius of neutral layer after reverse bending and θw is the reverse bending angle. At this point, the fiber length of the unit ABCD lwD is:
lwD ? r ? θw e4T
So 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
ε lnlwD ln
θ
? ?
l0D
r ? θw
r0 ? θ0
e5T
negligible, i.e., εr=0. Ψ direction stress is much smaller than others and can be ignored, namely σΨ = 0. Therefore, the units’ stress state of bar straightening deformation zone can be sim- plified to plane strain problem, i.e., τθΨ = τΨr = 0 and γθΨ = γΨr
= 0. Meanwhile, the hypothesis (1) means τθr=γθr=0.
The length of bar neutral layer is a constant before and after straightening that means:
R0 ? θ0 ? ρ ? θw e6T
Int J Adv Manuf Technol (2015) 79:1519–1529 1521
Fig. 2 Diagram of bar straightening stress and strain in deformation zone
Finally,
σ ? 1 qe??σ??????—?????σ????T??2???t?????e?σ??????—?????σ?????T??2???t?????e??σ??????—?????σ????T??2??
θ
r ? R0
ε ? ln
e7T
p p?2?? θ r θ Ψ
3
θ
Ψ r
e10T
0
w
eR t R — rT?ρ
? p???jσ j
2
where Rw is the
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