# 点焊疲劳分析

## 薄片位置（1 或 2）

$\sigma \left(\theta \right)=-{\sigma }_{\mathrm{max}}\left({f}_{y}\right)\mathrm{cos}\theta -{\sigma }_{\mathrm{max}}\left({f}_{z}\right)\mathrm{sin}\theta +\sigma \left({f}_{x}\right)+{\sigma }_{\mathrm{max}}\left({m}_{y}\right)\mathrm{sin}\theta -{\sigma }_{\mathrm{max}}\left({m}_{z}\right)\mathrm{cos}\theta$

${\sigma }_{\mathrm{max}}\left({f}_{y}\right)=\frac{{f}_{y}}{\pi DT}\text{\hspace{0.17em}}×\text{\hspace{0.17em}}{C}_{fyz}\text{\hspace{0.17em}}×\text{\hspace{0.17em}}{D}^{defyz}\text{\hspace{0.17em}}×\text{\hspace{0.17em}}{T}^{tefyz}$
(3)
${\sigma }_{\mathrm{max}}\left({f}_{z}\right)=\frac{{f}_{z}}{\pi DT}\text{\hspace{0.17em}}×\text{\hspace{0.17em}}{C}_{fyz}\text{\hspace{0.17em}}×\text{\hspace{0.17em}}{D}^{defyz}\text{\hspace{0.17em}}×\text{\hspace{0.17em}}{T}^{tefyz}$
(4)
$\sigma \left({f}_{x}\right)=\left(\frac{1.744{f}_{x}}{{T}^{2}}\right)\text{\hspace{0.17em}}×\text{\hspace{0.17em}}{C}_{fx}\text{\hspace{0.17em}}×\text{\hspace{0.17em}}{D}^{defx}\text{\hspace{0.17em}}×\text{\hspace{0.17em}}{T}^{tefx}\text{ }\text{for}\text{ }{f}_{x}\text{ }>\text{ }0.0$
(5)
(6)
${\sigma }_{\mathrm{max}}\left({m}_{y}\right)=\left(\frac{1.872{m}_{y}}{D{T}^{2}}\right)\text{\hspace{0.17em}}×\text{\hspace{0.17em}}{C}_{myz}\text{\hspace{0.17em}}×\text{\hspace{0.17em}}{D}^{demyz}\text{\hspace{0.17em}}×\text{\hspace{0.17em}}{T}^{temyz}$
(7)
${\sigma }_{\mathrm{max}}\left({m}_{z}\right)=\left(\frac{1.872{m}_{z}}{D{T}^{2}}\right)\text{\hspace{0.17em}}×\text{\hspace{0.17em}}{C}_{myz}\text{\hspace{0.17em}}×\text{\hspace{0.17em}}{D}^{demyz}\text{\hspace{0.17em}}×\text{\hspace{0.17em}}{T}^{temyz}$
$D$

$T$

${C}_{fyz}$ , ${C}_{myz}$ , ${C}_{fx}$

$defyz$ , $demyz$ , $defx$

$tefyz$ , $temyz$ , $tefx$

${C}_{fyz}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}1,\text{\hspace{0.17em}}\text{ }defyz\text{\hspace{0.17em}}=\text{\hspace{0.17em}}0,\text{ }tefyz\text{\hspace{0.17em}}=\text{\hspace{0.17em}}0$
${C}_{myz}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}0.6,\text{\hspace{0.17em}}\text{ }demyz\text{\hspace{0.17em}}=\text{\hspace{0.17em}}0,\text{ }temyz\text{\hspace{0.17em}}=\text{\hspace{0.17em}}0.5$
${C}_{fx}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}0.6,\text{\hspace{0.17em}}\text{ }defx\text{\hspace{0.17em}}=\text{\hspace{0.17em}}0,\text{ }tefx\text{\hspace{0.17em}}=\text{\hspace{0.17em}}0.5$

## 焊点熔核位置

$\tau \left(\theta \right)={\tau }_{\mathrm{max}}\left({f}_{y}\right)\mathrm{sin}\theta +{\tau }_{\mathrm{max}}\left({f}_{z}\right)\mathrm{cos}\theta$
(9)
$\sigma \left(\theta \right)=\sigma \left({f}_{x}\right)+{\sigma }_{\mathrm{max}}\left({m}_{y}\right)\mathrm{sin}\theta -{\sigma }_{\mathrm{max}}\left({m}_{z}\right)\mathrm{cos}\theta$

${\tau }_{\mathrm{max}}\left({f}_{y}\right)=\frac{16{f}_{y}}{3\pi {D}^{2}}$
(11)
${\tau }_{\mathrm{max}}\left({f}_{z}\right)=\frac{16{f}_{z}}{3\pi {D}^{2}}$
(12)
$\sigma \left({f}_{x}\right)\text{ }=\text{ }\frac{4{f}_{x}}{\pi {D}^{2}}\text{ }\text{for}\text{ }{f}_{x}\text{\hspace{0.17em}}>\text{\hspace{0.17em}}0.0$
(13)
$\sigma \left({f}_{x}\right)\text{ }=\text{ }0.0\text{ }\text{for}\text{ }{f}_{x}\text{\hspace{0.17em}}\le \text{\hspace{0.17em}}0.0$
(14)
${\sigma }_{\mathrm{max}}\left({m}_{y}\right)=\frac{32{m}_{y}}{\pi {D}^{3}}$
(15)
${\sigma }_{\mathrm{max}}\left({m}_{z}\right)=\frac{32{m}_{z}}{\pi {D}^{3}}$
$D$

$T$

$\tau \left(\theta \right)$ $\sigma \left(\theta \right)$ 计算每个 $\theta$ 的等效最大绝对主应力。这些应力被用于后续的疲劳分析。采用雨流循环计数法计算各角度下的疲劳寿命和损伤 ( $\theta$ )。然后选择最严重的损伤值进行输出。