螺栓和螺母预紧力背景信息

可以将螺栓力施加于 SimSolid 中的螺栓和螺母几何体。.

螺栓和螺母

SimSolid 中,螺栓会由其几何属性自动标识。螺栓必须有圆柱形的主体和基于六面体的头部。六角形可以位于螺栓头部的外径或内径上。可以使用类似的基于六角的几何特征来识别螺母。


Figure 1.


Figure 2.
SimSolid 中,预紧力载荷可以施加于以下各种几何体,包括:
  • 盲螺栓
  • 螺栓和螺母
  • 螺纹杆上的螺栓
  • 通用杆或手柄上的螺母

扭矩 M 与轴向力 F 的关系

M 是预紧结束时实现的最大力矩,它由螺母与结构之间的摩擦力所产生的力矩平衡。

为简单起见,假设接触中的法向力均匀分布,因此接触压力如下:(1)
P= F ContactArea MathType@MTEF@5@5@+= feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiuaiabg2 da9maalaaabaGaamOraaqaaiaadoeacaWGVbGaamOBaiaadshacaWG HbGaam4yaiaadshacaWGbbGaamOCaiaadwgacaWGHbaaaaaa@42A5@
(2)
P= F π(R 1 2 R 0 2 ) MathType@MTEF@5@5@+= feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiuaiabg2 da9maalaaabaGaamOraaqaaiabec8aWjaacIcacaWGsbGaaGymamaa CaaaleqabaGaaGOmaaaakiabgkHiTiaadkfacaaIWaWaaWbaaSqabe aacaaIYaaaaOGaaiykaaaaaaa@41B5@

R0 和 R1是接触载荷点的内半径和外半径。摩擦力分布力将是 T=fP MathType@MTEF@5@5@+= feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamivaiabg2 da9iaadAgacqGHxiIkcaWGqbaaaa@3A81@ ,其中 f 是摩擦系数。

在极坐标系中,摩擦力相对于螺栓轴的基本力矩为:
dM=T r 2 dRdTet MathType@MTEF@5@5@+= feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamizaiaad2 eacqGH9aqpcaWGubGaey4fIOIaamOCamaaCaaaleqabaGaaGOmaaaa kiabgEHiQiaadsgacaWGsbGaey4fIOIaamizaiaadsfacaWGLbGaam iDaaaa@43A9@
其中,r 是到轴的距离,而 dR 和 dTet 分别是半径和角度差。
对接触区域上的基本力矩进行积分,以获得以下结果:
M= 2Ff(R 1 3 R 0 3 ) 3(R 1 2 R 0 2 ) MathType@MTEF@5@5@+= feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamytaiabg2 da9maalaaabaGaaGOmaiabgEHiQiaadAeacqGHxiIkcaWGMbGaey4f IOIaaiikaiaadkfacaaIXaWaaWbaaSqabeaacaaIZaaaaOGaeyOeI0 IaamOuaiaaicdadaahaaWcbeqaaiaaiodaaaGccaGGPaaabaGaaG4m aiabgEHiQiaacIcacaWGsbGaaGymamaaCaaaleqabaGaaGOmaaaaki abgkHiTiaadkfacaaIWaWaaWbaaSqabeaacaaIYaaaaOGaaiykaaaa aaa@4D66@
该方程式将施加的扭矩、M 和轴向力联系了起来。

轴向力

轴向力取决于结构和螺栓刚度,以及螺母相对于螺栓的位置:(3)
F = K D MathType@MTEF@5@5@+= feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOraiabg2 da9iaadUeacqGHxiIkcaWGebaaaa@3A4C@
K 为结构刚度系数,D 为相对位移。
相对位移可以用以下表达式表示:
D = N H MathType@MTEF@5@5@+= feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiraiabg2 da9iaad6eacqGHxiIkcaWGibaaaa@3A51@
这里,N 是螺母圈数,H 是螺距。因此,
F = K H N MathType@MTEF@5@5@+= feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOraiabg2 da9iaadUeacqGHxiIkcaWGibGaey4fIOIaamOtaaaa@3C12@
(方程式 A)
假设在第一次分析过程中描述了一个螺母转动 (N(1)=1),并从分析中找到了对应的轴向力 F(1)。这种情况下的结构刚度系数可以定义为:(4)
F(1)=KH1 MathType@MTEF@5@5@+= feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOraiaacI cacaaIXaGaaiykaiabg2da9iaadUeacqGHxiIkcaWGibGaey4fIOIa aGymaaaa@3E0E@

这意味着: F=F(1)N MathType@MTEF@5@5@+= feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOraiabg2 da9iaadAeacaGGOaGaaGymaiaacMcacqGHxiIkcaWGobaaaa@3C65@

现在,您可以将扭矩与转数相关联:(5)
M= 2NF(1)f(R 1 3 R 0 3 ) 3(R 1 2 R 0 2 ) MathType@MTEF@5@5@+= feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamytaiabg2 da9maalaaabaGaaGOmaiabgEHiQiaad6eacqGHxiIkcaWGgbGaaiik aiaaigdacaGGPaGaey4fIOIaamOzaiabgEHiQiaacIcacaWGsbGaaG ymamaaCaaaleqabaGaaG4maaaakiabgkHiTiaadkfacaaIWaWaaWba aSqabeaacaaIZaaaaOGaaiykaaqaaiaaiodacqGHxiIkcaGGOaGaam OuaiaaigdadaahaaWcbeqaaiaaikdaaaGccqGHsislcaWGsbGaaGim amaaCaaaleqabaGaaGOmaaaakiaacMcaaaaaaa@513C@
因此,为了实现规定扭矩 M,在 N=1 的情况下进行第一次分析之后,必须使用以下公式进行第二次分析(第二次收敛迭代): (6)
N(2)=M/ 2NF(1)f(R 1 3 R 0 3 ) 3(R 1 2 R 0 2 ) MathType@MTEF@5@5@+= feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOtaiaacI cacaaIYaGaaiykaiabg2da9maalyaabaGaamytaaqaamaaemaabaWa aSaaaeaacaaIYaGaey4fIOIaamOtaiabgEHiQiaadAeacaGGOaGaaG ymaiaacMcacqGHxiIkcaWGMbGaey4fIOIaaiikaiaadkfacaaIXaWa aWbaaSqabeaacaaIZaaaaOGaeyOeI0IaamOuaiaaicdadaahaaWcbe qaaiaaiodaaaGccaGGPaaabaGaaG4maiabgEHiQiaacIcacaWGsbGa aGymamaaCaaaleqabaGaaGOmaaaakiabgkHiTiaadkfacaaIWaWaaW baaSqabeaacaaIYaaaaOGaaiykaaaaaiaawEa7caGLiWoaaaaaaa@575C@
一般来说,在迭代 (i+1) 时,施加的转数如下: (7)
N ( i + 1 ) = M / 2 N ( i ) F ( i ) f ( R 1 3 R 0 3 ) 3 ( R 1 2 R 0 2 ) MathType@MTEF@5@5@+= feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOtaiaacI cacaWGPbGaey4kaSIaaGymaiaacMcacqGH9aqpdaWcgaqaaiaad2ea aeaadaabdaqaamaalaaabaGaaGOmaiabgEHiQiaad6eacaGGOaGaam yAaiaacMcacqGHxiIkcaWGgbGaaiikaiaadMgacaGGPaGaey4fIOIa amOzaiabgEHiQiaacIcacaWGsbGaaGymamaaCaaaleqabaGaaG4maa aakiabgkHiTiaadkfacaaIWaWaaWbaaSqabeaacaaIZaaaaOGaaiyk aaqaaiaaiodacqGHxiIkcaGGOaGaamOuaiaaigdadaahaaWcbeqaai aaikdaaaGccqGHsislcaWGsbGaaGimamaaCaaaleqabaGaaGOmaaaa kiaacMcaaaaacaGLhWUaayjcSdaaaaaa@5BA5@
这里,N(i) 是在前一次迭代时施加的转数,F (i) 是在前一次迭代时评估的结果轴向力。这些对所施加的转数的修正很重要,因为在迭代过程中,求解得到了细化,这改变了上面公式 A 中的结构刚度因子 K。所以,K 不是常数,而是依赖于传递 K(i)。