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考慮端部摩擦的中心直裂紋巴西圓盤斷裂參數解析解

劉建 喬蘭 李慶文 趙國彥

劉建, 喬蘭, 李慶文, 趙國彥. 考慮端部摩擦的中心直裂紋巴西圓盤斷裂參數解析解[J]. 工程科學學報, 2022, 44(12): 2040-2047. doi: 10.13374/j.issn2095-9389.2021.06.07.006
引用本文: 劉建, 喬蘭, 李慶文, 趙國彥. 考慮端部摩擦的中心直裂紋巴西圓盤斷裂參數解析解[J]. 工程科學學報, 2022, 44(12): 2040-2047. doi: 10.13374/j.issn2095-9389.2021.06.07.006
LIU Jian, QIAO Lan, LI Qing-wen, ZHAO Guo-yan. Analytical solutions of fracture parameters for a centrally cracked Brazilian disk considering the loading friction[J]. Chinese Journal of Engineering, 2022, 44(12): 2040-2047. doi: 10.13374/j.issn2095-9389.2021.06.07.006
Citation: LIU Jian, QIAO Lan, LI Qing-wen, ZHAO Guo-yan. Analytical solutions of fracture parameters for a centrally cracked Brazilian disk considering the loading friction[J]. Chinese Journal of Engineering, 2022, 44(12): 2040-2047. doi: 10.13374/j.issn2095-9389.2021.06.07.006

考慮端部摩擦的中心直裂紋巴西圓盤斷裂參數解析解

doi: 10.13374/j.issn2095-9389.2021.06.07.006
基金項目: 國家自然科學基金與山東聯合基金重點資助項目(U1806209);中央高校基本科研業務費專項資金資助項目(FRF-TP-19-021A3);北京科技大學青年教師學科交叉研究資助項目(FRF-IDRY-19-002)
詳細信息
    通訊作者:

    E-mail: lanqiao@ustb.edu.cn

  • 中圖分類號: O346.1

Analytical solutions of fracture parameters for a centrally cracked Brazilian disk considering the loading friction

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  • 摘要: 運用權函數法推導出考慮加載端摩擦的四種形式分布載荷加載下,中心直裂紋巴西圓盤試樣在任意I/II復合型斷裂模式下I、II型應力強度因子及T應力的解析解,并探究了端部摩擦及載荷分布角度對斷裂參數的影響。研究結果表明:(1)當中心裂紋相對長度β較小時,純I型、純II型斷裂的YIYIIT*(分別是量綱為一的I型、II型應力強度因子及T應力)均隨摩擦系數及載荷分布角度增大而減小;但是,當β較大時,摩擦系數增大可使純I型YI增大,而載荷分布角度增大可使純II型T*增大。(2)接觸載荷分布形式為常數函數時,載荷分布角度對斷裂參數的影響最顯著,而四次函數下其對斷裂參數的影響相對最小。(3)當β較小時,純II型加載角度隨載荷分布角度增大而減小;當β較大時,其隨載荷分布角度增大而增大;摩擦系數增大可使純II型加載角度增大。

     

  • 圖  1  CCBD試樣與BD試樣示意圖。(a) CCBD試樣;(b) BD試樣

    Figure  1.  Schematic of (a) centrally cracked Brazilian disk and (b) Brazilian disk specimens

    圖  2  圓盤邊界上(r=R)式(13)計算結果與q(θ)的對比

    Figure  2.  Comparison between the calculated results based on Eq. (13) and the applied normal pressures q(θ)

    圖  3  均布載荷加載下本文結果與Dong等[4]及李一凡等[21]結果的對比。(a)YI;(b) YII;(c) T*

    Figure  3.  Comparison between the results of this study and the results of Dong et al.[4] and Li et al.[21] under uniformly distributed pressure: (a) YI; (b) YII; and (c) T*

    圖  4  純I型及純II型斷裂的幾何參數隨摩擦系數的變化特征。(a) β=0.2:純I型YI;(b)β=0.2:純I型T*;(c)β=0.2:純II型YII;(d) β=0.2:純II型T*;(e)β=0.8:純I型YI;(f)β=0.8:純I型T*;(g) β=0.8:純II型YII;(h)β=0.8:純II型T*

    Figure  4.  Variations in the YI, YII and T* of pure mode I and II fractures versus friction coefficient μ: (a) β=0.2: pure mode-I YI; (b) β=0.2: pure mode-I T*; (c) β=0.2: pure mode-II YII; (d) β=0.2: pure mode-II T*; (e) β=0.8: pure mode-I YI; (f) β=0.8: pure mode-I T*; (g) β=0.8: pure mode-II YII; (h) β=0.8: pure mode-II T*

    圖  5  純I型及純II型斷裂的幾何參數隨載荷分布角度的變化特征。(a) β=0.2:純I型YI;(b)β=0.2:純I型T*;(c)β=0.2:純II型YII;(d) β=0.2:純II型T*;(e)β=0.8: 純I型YI;(f)β=0.8:純I型T*;(g) β=0.8:純II型YII;(h)β=0.8:純II型T*

    Figure  5.  Variations in the YI, YII, and T* of pure mode I and II fractures versus the load distribution angle α: (a) β = 0.2: pure mode-I YI; (b) β = 0.2: pure mode-I T*; (c) β = 0.2: pure mode-II YII; (d) β = 0.2: pure mode-II T*; (e) β = 0.8: pure mode-I YI; (f) β = 0.8: pure mode-I T*; (g) β = 0.8: pure mode-II YII; (h) β = 0.8: pure mode-II T*

    圖  6  載荷分布角度對純II型加載角度的影響。(a)β=0.2;(b) β=0.8

    Figure  6.  Effect of the load distribution angle on the critical loading angle for pure mode II fractures: (a) β = 0.2; (b) β = 0.8

    表  1  分布載荷加載下巴西圓盤應力解析解的系數

    Table  1.   Series coefficients Cn of the stress analytical solutions for the Brazilian disk subjected to distributed pressures

    Distribution formf(θ)/qmaxqmaxC0Cn (n=±1, ±2, ···)  
    Uniform1$ {[2(\sin \alpha + \mu (1 - \cos \alpha ))]^{ - 1}} $$ \dfrac{{2\alpha }}{{\text{π}} }{q_{\max }} $$ {q_{\max }}\dfrac{{\sin 2n\alpha - 2\mu {{(\sin n\alpha )}^2}}}{{n{\text{π}} }} $
    Elliptical$ {\left(1 - {\left(\dfrac{\theta }{\alpha }\right)^2}\right)^{1/2}} $$ {[{\text{π}} ({J_1}(\alpha ) + \mu {H_1}(\alpha ))]^{ - 1}} $$ \dfrac{\alpha }{2}{q_{\max }} $$ {q_{\max }}\dfrac{{{J_1}(2n\alpha ) - \mu {H_1}(2n\alpha )}}{{2n}} $
    Parabolic$ 1 - {\left(\dfrac{\theta }{\alpha }\right)^2} $$ \begin{gathered} {\alpha ^2}[4\mu (1 + {\alpha ^2}/2 - \alpha \sin \alpha - \\ \cos \alpha ) + 4\sin \alpha - 4\alpha \cos \alpha {]^{ - 1}} \\ \end{gathered} $$ \dfrac{{4\alpha }}{{3{\text{π}} }}{q_{\max }} $$\begin{gathered} {q_{\max } }[\mu (2\alpha n\sin 2n\alpha + \cos 2n\alpha - 2{\alpha ^2}{n^2} - 1)+ \\ \sin 2n\alpha - 2\alpha n\cos 2n\alpha ]/(2{\text{π} } {\alpha ^2}{n^3}) \\ \end{gathered}$
    Quartic polynomial$ {\left(1 - {\left(\dfrac{\theta }{\alpha }\right)^2}\right)^2} $$\begin{gathered} {\alpha ^4}[16(3 - {\alpha ^2})\sin \alpha - 48\alpha \cos \alpha + \\ 2\mu ({\alpha ^4} + 4{\alpha ^2} + 24 + \\ 8({\alpha ^2} - 3)\cos \alpha - 24\alpha \sin \alpha ){]^{ - 1} } \\ \end{gathered}$$ \dfrac{{16\alpha }}{{15{\text{π}} }}{q_{\max }} $$ \begin{gathered} {q_{\max }}[(3\mu - 4{\alpha ^2}{n^2}\mu - 6\alpha n)\cos 2n\alpha + \\ (6\alpha n\mu + 3 - 4{\alpha ^2}{n^2})\sin 2n\alpha - \\ 2\mu ({\alpha ^4}{n^4} + {\alpha ^2}{n^2} + 3/2)]/(2{\text{π}} {n^5}{\alpha ^4}) \\ \end{gathered} $
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  • 收稿日期:  2021-06-07
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