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 哈尔滨工程大学学报  2019, Vol. 40 Issue (6): 1182-1186  DOI: 10.11990/jheu.201804001 0

### 引用本文

LIU Yingwei, ZHANG Yang, DONG Xuefeng. Tensile ability of liquid film in the later stage of casting solidification[J]. Journal of Harbin Engineering University, 2019, 40(6), 1182-1186. DOI: 10.11990/jheu.201804001.

### 文章历史

1. 哈尔滨工程大学 材料科学与化学工程学院, 黑龙江 哈尔滨 150001;
2. 伍伦贡大学, 机械、材料、机电一体化和生物医学工程系, 新南威尔士 2522

Tensile ability of liquid film in the later stage of casting solidification
LIU Yingwei 1, ZHANG Yang 1, DONG Xuefeng 2
1. School of Materials Science and Chemical Engineering, Harbin Engineering University, Harbin 150001, China;
2. School of Mechanical, Materials, Mechatronics and Biomedical Engineering, University of Wollongong, New South Wales 2522, Australia
Abstract: The later stage of metal solidification is a dangerous period for hot crack generation. Investigating the tensile strength of liquid films is important to improve casting quality given that the liquid film between crystal grains has a certain preventative effect on thermal cracking. The analytic formula is obtained through a calculus-based method, and the relationship between the radius of the curvature of the liquid film and grain spacing is first quantified based on mass conservation. Moreover, the indirect relationship between the tensile strength of the liquid film and grain spacing is obtained. Analytical results show that the radius of the curvature of the liquid film gradually decreases as the grain spacing is increased in the absence of liquid metal supplementation, and an extremum of the radius of curvature exists. However, the extreme value obtained in this study is different from that obtained in a previous study. The maximum of the tensile strength of the liquid film corresponds to the extremum of the radius of curvature. Crystal grains will not produce hot cracks if the shrinkage tension does not exceed this maximum value during solidification. This study provides an important reference for the establishment of the solidification process.
Keywords: solidification    surface tension    additional pressure    hot cracking    grain    grain boundary    curvature radius    tensile strength

1 液膜抗拉能力的计算 1.1 弯曲液面的附加压力

 Download: 图 1 液膜曲率半径与晶粒边界间距的关系 Fig. 1 Relationship between the radius of curvature of liquid film and grain spacing

 Download: 图 2 液膜抗裂拉力的形成 Fig. 2 Formation of tensile strength of liquid film
1.2 液膜抗拉能力计算

 ${\rm{d}}f = \Delta P{\rm{d}}s$ (1)

 ${\rm{d}}{f_x} = {\rm{d}}f\sin \theta$ (2)

 ${\rm{d}}{f_x} = \frac{\sigma }{r}\sin \theta r{\rm{d}}\theta$ (3)

 $\begin{array}{l} {F_x} = \int\limits_B^c {\rm{d}} {f_x} = \int\limits_0^{{\theta _1}} {\frac{\sigma }{r}} \sin \theta r{\rm{d}}\theta = \\ - \sigma \cos \left. \theta \right|_{{\theta _1}}^0 = \sigma \left( { - 1 + \cos {\theta _1}} \right) \end{array}$ (4)

 ${\sigma _x} = {F_x}/\Delta y$ (5)

 $\Delta y = \int\limits_B^c {\rm{d}} y$ (6)

 $\Delta y = \left| {\int\limits_{{\theta _1}}^0 r \sin \theta {\rm{d}}\theta } \right| = | - r\cos \theta |_{{\theta _1}}^0| = r\left( {1 - \cos {\theta _1}} \right)$ (7)

 ${\sigma _x} = - (\sigma /r)$ (8)

2 液膜曲率半径与晶粒间隙的关系 2.1 质量守恒

 ${x^2} + {\left[ {y - \left( {r + \frac{H}{2}} \right)} \right]^2} = {r^2}$ (9)
 ${x^2} + {\left[ {y - \left( { - r - \frac{H}{2}} \right)} \right]^2} = {r^2}$ (10)

 ${y_1} = \left( {r + \frac{H}{2}} \right) - \sqrt {{r^2} - {x^2}}$ (11)
 ${y_2} = - \left( {r + \frac{H}{2}} \right) + \sqrt {{r^2} - {x^2}}$ (12)

 $S = \int_{ - \frac{T}{2}}^{\frac{T}{2}} {\left( {{y_1} - {y_2}} \right)} {\rm{d}}x$ (13)

 $S = \int_\limits{ - \frac{T}{2}}^{\frac{T}{2}} {\left[ {2\left( {r + \frac{H}{2}} \right) - 2\sqrt {{r^2} - {x^2}} } \right]} {\rm{d}}x$ (14)

 $S = 2\left( {r + \frac{H}{2}} \right)T - 2\left[ {\frac{T}{2}\sqrt {{r^2} - \frac{{{T^2}}}{4}} + {r^2}\arcsin \left( {\frac{T}{{2r}}} \right)} \right]$ (15)

 $\overline {{O_1}{K^2}} + \overline {K{C^2}} = \overline {{O_1}{C^2}}$

 ${(r - \Delta H)^2} + {\left( {\frac{T}{2}} \right)^2} = {r^2}$ (16)

 $\frac{H}{2} = \frac{{{H_0}}}{2} - r + \sqrt {{r^2} - {{\left( {\frac{T}{2}} \right)}^2}}$

 $S = T{H_0} + T\sqrt {{r^2} - \frac{{{T^2}}}{4}} - 2{r^2}\arcsin \left( {\frac{T}{{2r}}} \right)\$ (17)

 $T{H_0} + T\sqrt {{r^2} - \frac{{{T^2}}}{4}} - 2{r^2}\arcsin \left( {\frac{T}{{2r}}} \right) = {T_0}{H_0}$ (18)
2.2 曲率半径与晶粒边界间距的关系

2.3 曲率半径与晶粒尺寸的关系

2.4 rmin的确定

 $\begin{array}{l} {H_0} + \sqrt {{r^2} - \frac{{{T^2}}}{4}} + T\frac{{2r\frac{{\partial r}}{{\partial T}} - \frac{T}{2}}}{{2\sqrt {{r^2} - \frac{{{T^2}}}{4}} }} - \left( {\frac{{\partial r}}{{\partial T}}} \right) \cdot \\ 4{\mathop{\rm rarcsin}\nolimits} \left( {\frac{T}{{2r}}} \right) - \frac{{\frac{1}{{2r}} - \frac{T}{{2\sqrt r }}\frac{{\partial r}}{{\partial T}}}}{{\sqrt {1 - {{\left( {\frac{T}{{2r}}} \right)}^2}} }}2{r^2} = 0 \end{array}$ (19)

 ${H_0}\sqrt {{r^2} - \frac{{{T^2}}}{4}} - \frac{{{T^2}}}{2} = 0$

 ${r_{\min }} = \sqrt {\frac{{{T^2}}}{4} + \frac{{{T^4}}}{{4H_0^2}}}$ (20)
2.5 分析与讨论