﻿ 基于应力分布模型的随机疲劳加速试验设计
 文章快速检索 高级检索

1. 东北大学 机械工程与自动化学院, 沈阳 110819;
2. 航空动力装备振动及控制教育部重点实验室, 沈阳 110819;
3. 上海航空材料结构检测股份有限公司, 上海 201210

Accelerated random fatigue test design based on stress distribution model
MU Tong1,2, MENG Ge1,2, XIE Liyang1,2, ZHANG Jianbo3, SHI Chaocheng3
1. School of Mechanical Engineering and Automation, Northeastern University, Shenyang 110819;
2. Key Laboratory of Vibration and Control of Aero-Propulsion System, Ministry of Education, Shenyang 110819;
3. Shanghai Aeronautical Material and Structures Testing Co., LTD, Shanghai 201210, China
Abstract: The two-stage S-N curve is applied to the Dirlik random vibration fatigue life prediction model. The fatigue damage of each stress interval is calculated in segments and superimposed together. Based on the stress amplitude distribution estimated by the Dirlik model, the equivalent relationship between the magnification of excitation and fatigue life under the two-stage S-N curve is deduced. A life conversion method of accelerated life test based on the equivalent relationship is proposed, which provides a reference for the determination of stress level for accelerated life tests. The method can be generalized to the multi-stage S-N curve, like the very high cycle fatigue. For the large ratio of small stress amplitude in the random vibration fatigue, the damage computing method of this part is optimized, and the more realistic equivalent relationship between theoretical fatigue life and accelerated life is obtained. The analysis of an example shows that under the two-stage S-N curve, the relationship between stress and life also shows a logarithmic linear relationship of two-stage, which is verified by using the experimental data in the literature.
Keywords: random vibration    frequency domain    fatigue life    accelerated test    equivalent relation

1 随机振动疲劳理论 1.1 随机过程

 ${R_x}\left( {{t_1},{t_1} + \tau } \right) = {R_x}(\tau ) = E\left[ {x(t)x(t + \tau )} \right]$ （1）

 ${S_x}(\omega ) = \int\limits_{ - \infty }^{ + \infty } {{R_x}(\tau ){{\rm{e}}^{ - {\rm{i}}\omega \tau }}{\rm{d}}\tau }$ （2）
 ${R_x}(\tau ) = \frac{1}{{2{\rm{ \mathsf{ π} }}}}\int\limits_{ - \infty }^{ + \infty } {{R_x}(\omega ){{\rm{e}}^{{\rm{i}}\omega \tau }}{\rm{d}}\omega }$ （3）

 ${G_x}(\omega ) = \left\{ {\begin{array}{*{20}{l}} {2{S_x}(\omega )}&{\omega \ge 0}\\ 0&{\omega < 0} \end{array}} \right.$ （4）

 ${G_{滤波法}}(f) = \mathop {\lim }\limits_{t \to \infty } \mathop {\lim }\limits_{b \to 0} \frac{1}{{bt}}\int\limits_0^t {{y^2}(f,t,b){\rm{d}}t}$ （5）
 ${G_{傅里叶}}(f) = \mathop {\lim }\limits_{t \to \infty } \frac{2}{t}E\left[ {{{\left| {{\rm{fft}}\left( {f,t} \right)} \right|}^2}} \right]$ （6）

 ${m_i} = \int\limits_0^{ + \infty } {{f^i}G\left( f \right){\rm{d}}f} \;\;\;i = 0,1,2, \cdots$ （7）

 ${m_0} = E\left[ {{x^2}(t)} \right]$ （8）
 ${m_2} = E\left[ {{{\dot x}^2}(t)} \right]$ （9）
 ${m_4} = E\left[ {{{\ddot x}^2}(t)} \right]$ （10）

 ${\sigma _{{\rm{RMS}}}} = m_0^{\frac{1}{2}} = \sqrt {\int\limits_0^{ + \infty } {G\left( \omega \right){\rm{d}}\omega } }$ （11）

 $\gamma = \frac{{{m_2}}}{{\sqrt {{m_0}{m_4}} }}$ （12）

 $\varepsilon = \sqrt {1 - \frac{{m_2^2}}{{{m_0}{m_4}}}} = \sqrt {1 - {\gamma ^2}}$ （13）

 ${V_{\rm{p}}} = \sqrt {\frac{{{m_4}}}{{{m_2}}}}$ （14）

 ${V_ + } = \sqrt {\frac{{{m_2}}}{{{m_0}}}}$ （15）
1.2 随机疲劳

 $D = \frac{{n(S)}}{{N(S)}}$ （16）

 ${S^m}N = C$ （17）

 $n(S) = vtp(S)\Delta S$ （18）

 $D = vt\int {\frac{{p(S)}}{{N(S)}}{\rm{d}}S}$ （19）

 ${t_{\rm{L}}} = \frac{1}{{v\int {\frac{{p(S)}}{{N(S)}}{\rm{d}}S} }} = \frac{C}{{v\int {{S^m}} p(S){\rm{d}}S}}$ （20）

Dirlik使用蒙特卡洛方法在计算机中建立了时域模拟信号，对大量不同不规则因子的PSD曲线进行了模拟仿真，总结出了一套经验公式。Dirlik将幅值范围概率密度函数假设为了一个指数分布与2个瑞利分布的和，公式为

 $p(S) = \frac{{\frac{{{D_1}}}{Q}{{\rm{e}}^{\frac{{ - Z}}{Q}}} + \frac{{{D_2}Z}}{{{R^2}}}{{\rm{e}}^{\frac{{ - {Z^2}}}{{2{R^2}}}}} + {D_3}Z{{\rm{e}}^{\frac{{ - {Z^2}}}{2}}}}}{{2\sqrt {{m_0}} }}$ （21）

 $Z = \frac{S}{{2\sqrt {{m_0}} }};{D_1} = \frac{{2\left( {{X_m} - {\gamma ^2}} \right)}}{{1 + {\gamma ^2}}}$
 ${D_2} = \frac{{1 - \gamma - {D_1} + D_1^2}}{{1 - R}};{D_3} = 1 - {D_1} - {D_2}$
 $R = \frac{{\gamma - {X_m} - D_1^2}}{{1 - \gamma - {D_1} + D_1^2}};{X_m} = \frac{{{m_1}}}{{{m_0}}}\sqrt {\frac{{{m_2}}}{{{m_4}}}}$
 $Q = \frac{{1.25\left( {\gamma - {D_3} - {D_2}R} \right)}}{{{D_1}}}$

Dirlik法是频域疲劳寿命预测研究中使用最广泛的方法，对于大部分宽带随机过程都有很好的适应性，因此工程中一般采用该模型进行随机振动疲劳寿命预测。虽然其只是经验公式，没有理论支撑，但在绝大多数情况下，都有比其他模型更好的计算精度[12-14, 16-20]

 $\left\{ {\begin{array}{*{20}{l}} {{S^{{m_{\rm{a}}}}}N = {C_{\rm{a}}}}&{0 < S \le {S_0}}\\ {{S^{{m_{\rm{b}}}}}N = {C_{\rm{b}}}}&{S > {S_0}} \end{array}} \right.$ （22）

 $D = vt\left( {\int\limits_0^{{S_0}} {\frac{{{S^{{m_{\rm{a}}}}}p(S)}}{{{C_{\rm{a}}}}}{\rm{d}}S} + \int\limits_{{S_0}}^\infty {\frac{{{S^{{m_{\rm{b}}}}}p(S)}}{{{C_{\rm{b}}}}}{\rm{d}}S} } \right)$ （23）

 ${t_{\rm{L}}} = \frac{1}{{v\left( {\int\limits_0^{{S_0}} {\frac{{{S^{{m_{\rm{a}}}}}p(S)}}{{{C_{\rm{a}}}}}{\rm{d}}S} + \int\limits_{{S_0}}^\infty {\frac{{{S^{{m_{\rm{b}}}}}p(S)}}{{{C_{\rm{b}}}}}{\rm{d}}S} } \right)}}$ （24）
2 加速试验

2.1 基于Dirlik模型的加速等效关系

 ${G_{\rm{b}}}(f) = {W_{\rm{a}}}(f)H_{{\rm{ba}}}^2(f)$ （25）

 $m_i^\prime = \int\limits_0^{ + \infty } {{f^i}{G^\prime }(f){\rm{d}}f = km_i^\prime }$ （26）

 $\sigma _{{\rm{RMS}}}^\prime = {\left( {k{m_0}} \right)^{\frac{1}{2}}} = \sqrt k {\sigma _{{\rm{RMS}}}}$ （27）
 ${\gamma ^\prime } = \frac{{k{m_2}}}{{\sqrt {k{m_0}k{m_4}} }} = \gamma$ （28）
 ${\varepsilon ^\prime } = \sqrt {1 - \frac{{{k^2}m_2^2}}{{k{m_0}k{m_4}}}} = \varepsilon$ （29）
 $V_{\rm{P}}^\prime = \sqrt {\frac{{k{m_4}}}{{k{m_2}}}} = {V_{\rm{P}}}$ （30）
 $V_ + ^\prime = \sqrt {\frac{{k{m_2}}}{{k{m_0}}}} = {V_ + }$ （31）

 ${Z^\prime } = \frac{S}{{2\sqrt {k{m_0}} }} = \frac{Z}{{\sqrt k }};D_1^\prime = \frac{{2\left( {{X_m} - {\gamma ^2}} \right)}}{{1 + {\gamma ^2}}} = {D_1}$
 $D_2^\prime = \frac{{1 - \gamma - {D_1} + D_1^2}}{{1 - R}} = {D_2}$
 $D_3^\prime = 1 - {D_1} - {D_2} = {D_3}$
 ${Q^\prime } = \frac{{1.25\left( {\gamma - {D_3} - {D_2}R} \right)}}{{{D_1}}} = Q$
 ${R^\prime } = \frac{{\gamma - {X_m} - D_1^2}}{{1 - \gamma - {D_1} + D_1^2}} = R;X_m^\prime = \frac{{{m_1}}}{{{m_0}}}\sqrt {\frac{{{m_2}}}{{{m_4}}}} = {X_m}$

 ${p^\prime }(S) = \frac{{\frac{{{D_1}}}{Q}{{\rm{e}}^{\frac{{ - Z}}{{\sqrt {kQ} }}}} + \frac{{{D_2}Z}}{{{R^2}\sqrt k }}{{\rm{e}}^{\frac{{ - {Z^2}}}{{2k{R^2}}}}} + \frac{{{D_3}Z}}{{\sqrt k }}{{\rm{e}}^{\frac{{ - {Z^2}}}{{2k}}}}}}{{2\sqrt {k{m_0}} }}$ （32）

 $\begin{array}{*{20}{c}} {{D^\prime } = \frac{{{V_{\rm{P}}}}}{{{C_{\rm{a}}}}}\int\limits_0^{{S_0}} {{S^{{m_{\rm{a}}}}}\frac{{\frac{{{D_1}}}{Q}{{\rm{e}}^{\frac{{ - Z}}{{\sqrt {kQ} }}}} + \frac{{{D_2}Z}}{{{R^2}\sqrt k }}{{\rm{e}}^{\frac{{ - {Z^2}}}{{2k{R^2}}}}} + \frac{{{D_3}Z}}{{\sqrt k }}{{\rm{e}}^{\frac{{ - {Z^2}}}{{2k}}}}}}{{2\sqrt {km} }}{\rm{d}}S} + }\\ {\frac{{{V_{\rm{P}}}}}{{{C_{\rm{b}}}}}\int\limits_{{S_0}}^0 {{S^{{m_{\rm{b}}}}}\frac{{\frac{{{D_1}}}{Q}{{\rm{e}}^{\frac{{ - Z}}{{\sqrt {kQ} }}}} + \frac{{{D_2}Z}}{{{R^2}\sqrt k }}{{\rm{e}}^{\frac{{ - {Z^2}}}{{2k{R^2}}}}} + \frac{{{D_3}Z}}{{\sqrt k }}{{\rm{e}}^{\frac{{ - {Z^2}}}{{2k}}}}}}{{2\sqrt {k{m_0}} }}{\rm{d}}S} } \end{array}$ （33）

 ${j_{损伤}} = \frac{{{D^\prime }}}{D} = \frac{1}{{\sqrt k }} \cdot \frac{{{A_1} + {A_2} + {A_3} + {B_1} + {B_2} + {B_3}}}{{{G_1} + {G_2} + {G_3} + {H_1} + {H_2} + {H_3}}}$ （34）

 ${j_{寿命}} = \sqrt k \frac{{{G_1} + {G_2} + {G_3} + {H_1} + {H_2} + {H_3}}}{{{A_1} + {A_2} + {A_3} + {B_1} + {B_2} + {B_3}}}$ （35）

 ${A_1} = \frac{{{D_1}{{\left( {4{m_0}k} \right)}^{\frac{{{m_{\rm{a}}} + 1}}{2}}}{Q^{{m_{\rm{a}}}}}}}{{{C_{\rm{a}}}}}\gamma \left( {{m_{\rm{a}}} + 1,\frac{{{S_0}}}{{2Q\sqrt {{m_0}k} }}} \right)$
 ${A_2} = \frac{{{8^{\frac{{{m_{\rm{a}}}}}{2} + 1}}{D_2}{{\left( {{m_0}k} \right)}^{\frac{{{m_{\rm{a}}} + 1}}{2}}}{R^{{m_{\rm{a}}}}}}}{{{C_{\rm{a}}}}}\gamma \left( {\frac{{{m_{\rm{a}}}}}{2} + 1,\frac{{S_0^2}}{{8k{R^2}{m_0}}}} \right)$
 ${A_3} = \frac{{{8^{\frac{{{m_{\rm{a}}}}}{2} + 1}}{D_3}{{\left( {{m_0}k} \right)}^{\frac{{{m_{\rm{a}}} + 1}}{2}}}}}{{{C_{\rm{a}}}}}\gamma \left( {\frac{{{m_{\rm{a}}}}}{2} + 1,\frac{{S_0^2}}{{8k{m_0}}}} \right)$
 ${B_1} = \frac{{{D_1}{{\left( {4{m_0}k} \right)}^{\frac{{{m_{\rm{b}}} + 1}}{2}}}{Q^{{m_{\rm{b}}}}}}}{{{C_{\rm{b}}}}}\mathit{\Gamma }\left( {{m_{\rm{b}}} + 1,\frac{{{S_0}}}{{2Q\sqrt {{m_0}k} }}} \right)$
 $\begin{array}{*{20}{c}} {{B_2} = \frac{{{8^{\frac{{{m_{\rm{b}}}}}{2} + 1}}{D_2}{{\left( {{m_{\rm{0}}}k} \right)}^{\frac{{{m_{\rm{b}}} + 1}}{2}}}{R^{{m_{\rm{b}}}}}}}{{{C_{\rm{b}}}}} \cdot }\\ {\mathit{\Gamma }\left( {\frac{{{m_{\rm{b}}}}}{2} + 1,\frac{{S_0^2}}{{8k{R^2}{m_0}}}} \right)} \end{array}$
 ${B_3} = \frac{{{8^{\frac{{{m_{\rm{b}}}}}{2} + 1}}{D_3}{{\left( {{m_{\rm{0}}}k} \right)}^{\frac{{{m_{\rm{b}}} + 1}}{2}}}}}{{{C_{\rm{b}}}}}\mathit{\Gamma }\left( {\frac{{{m_{\rm{b}}}}}{2} + 1,\frac{{S_0^2}}{{8k{m_0}}}} \right)$
 ${G_1} = \frac{{{D_1}{{\left( {4{m_0}} \right)}^{\frac{{{m_{\rm{a}}} + 1}}{2}}}{Q^{{m_{\rm{a}}}}}}}{{{C_{\rm{a}}}}}\gamma \left( {{m_{\rm{a}}} + 1,\frac{{{S_0}}}{{2Q\sqrt {{m_0}} }}} \right)$
 ${G_2} = \frac{{{8^{\frac{{{m_{\rm{a}}}}}{2} + 1}}{D_2}{{\left( {{m_0}} \right)}^{\frac{{{m_{\rm{a}}} + 1}}{2}}}{R^{{m_{\rm{a}}}}}}}{{{C_{\rm{a}}}}}\gamma \left( {\frac{{{m_{\rm{a}}}}}{2} + 1,\frac{{S_0^2}}{{8{R^2}{m_0}}}} \right)$
 ${G_3} = \frac{{{8^{\frac{{{m_{\rm{a}}}}}{2} + 1}}{D_3}{{\left( {{m_0}} \right)}^{\frac{{{m_{\rm{a}}} + 1}}{2}}}}}{{{C_{\rm{a}}}}}\gamma \left( {\frac{{{m_{\rm{a}}}}}{2} + 1,\frac{{S_0^2}}{{8{m_0}}}} \right)$
 ${H_1} = \frac{{{D_1}{{\left( {4{m_0}} \right)}^{\frac{{{m_{\rm{b}}} + 1}}{2}}}{Q^{{m_{\rm{b}}}}}}}{{{C_{\rm{b}}}}}\mathit{\Gamma }\left( {{m_{\rm{b}}} + 1,\frac{{{S_0}}}{{2Q\sqrt {{m_0}} }}} \right)$
 ${H_2} = \frac{{{8^{\frac{{{m_{\rm{b}}}}}{2} + 1}}{D_2}{{\left( {{m_0}} \right)}^{\frac{{{m_{\rm{b}}} + 1}}{2}}}{R^{{m_{\rm{b}}}}}}}{{{C_{\rm{b}}}}}\mathit{\Gamma }\left( {\frac{{{m_{\rm{b}}}}}{2} + 1,\frac{{S_0^2}}{{8{R^2}{m_0}}}} \right)$
 ${H_3} = \frac{{{8^{\frac{{{m_{\rm{b}}}}}{2} + 1}}{D_3}{{\left( {{m_0}} \right)}^{\frac{{{m_{\rm{b}}} + 1}}{2}}}}}{{{C_{\rm{b}}}}}\mathit{\Gamma }\left( {\frac{{{m_{\rm{b}}}}}{2} + 1,\frac{{S_0^2}}{{8{m_0}}}} \right)$

2.2 算例

 图 1 某吊挂结构局部模型 Fig. 1 Local model for a suspension structure

 应力/MPa 寿命/次 490 1.023×105 430 2.800×105 410 4.403×105 390 5.946×105 370 8.927×105 331 1×107
 图 2 S-N曲线拟合结果 Fig. 2 Result of S-N curve fitting

 $\left\{ {\begin{array}{*{20}{l}} {{S^{7.7118}}N = 5.701 \times {{10}^{25}}}&{S > 370.0076}\\ {{S^{21.6914}}N = 4.5564 \times {{10}^{61}}}&{0 < S \le 370.0076} \end{array}} \right.$

 图 3 输入加速度功率谱密度 Fig. 3 Inputted acceleration power spectral density
 图 4 危险点响应应力功率谱密度 Fig. 4 Power spectral density of stress response at dangerous point

 图 5 激励与损伤的等效关系 Fig. 5 Equivalent relationship between excitation and damage

 图 6 加速试验结果 Fig. 6 Results of accelerated test

 载荷均方根1/MPa 寿命1/s 载荷均方根2/MPa 寿命2/s 24.5 93 360 24.6 65 580 32.7 61 860, 32 340 30.8 51 660 44.2 11 400, 5 490, 9 660 42.2 8 850 63.2 1 440, 1 200, 1 410 54.4 1 332, 1 320, 1 530 93.1 259.8, 225, 198 77.2 334.8, 319.8, 295.8, 183.6 124.8 64.98 110.5 44.7, 42, 40.02 129.2 30.54 152.6 11.58, 6.48 151.7 90 158.0 7.98, 6.4

3 结论

1) 针对随机振动中小应力循环占比较大的情况，将Dirlik雨流幅值分布模型应用到了双对数坐标系下的2段直线所表征的S-N曲线上，该求解思路使得频域方法更加适用于随机振动疲劳问题，且该方法还可以拓展到多段直线所表征的S-N曲线的情况中。

2) 基于所提出的多段式S-N曲线的求解思路，以2段的情况为例，推导了使用Dirlik模型为基础的振动疲劳频域求解方法支撑下的加速应力等效关系式，总结出了加速寿命换算方法，为振动疲劳的加速试验提供了理论支持。

3) 算例表明，在使用双对数坐标系下的2条直线对S-N曲线进行表达时，所推导出的应力与损伤放大倍数的等效关系近似为2段对数直线关系，并借助相关文献中的试验数据对方法进行了验证。

http://dx.doi.org/10.7527/S1000-6893.2019.23229

0

#### 文章信息

MU Tong, MENG Ge, XIE Liyang, ZHANG Jianbo, SHI Chaocheng

Accelerated random fatigue test design based on stress distribution model

Acta Aeronautica et Astronautica Sinica, 2020, 41(2): 223229.
http://dx.doi.org/10.7527/S1000-6893.2019.23229