Bolt and Nut Tightening Background Information

Bolt forces can be applied to bolt and nut geometries in SimSolid.

Bolts and Nuts

In SimSolid, bolts are automatically identified by their geometric attributes. Bolts are required to have cylindrical bodies and a head with a hexahedral based shape. The hex shape can be on an outer or inner diameter in the bolt head. Nuts are identified in a similar manner using this hex based geometric signature.
Figure 1.


Figure 2.


In SimSolid, tightening loads can be applied to a variety of geometries, including the following:
  • Blind bolts
  • Bolts with nuts
  • Nuts on threaded rods
  • Nuts on a generic post or handle

Relationship Between Torque M and Axial Force F

M is the maximum moment realized at the end of the tightening and it is equilibrated by moment from friction forces in contact between nut and the structure.

Assume for simplicity that normal forces in contact are distributed evenly, so the contact pressure is as follows:

P= F ContactArea MathType@MTEF@5@5@+= feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiuaiabg2 da9maalaaabaGaamOraaqaaiaadoeacaWGVbGaamOBaiaadshacaWG HbGaam4yaiaadshacaWGbbGaamOCaiaadwgacaWGHbaaaaaa@42A5@
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 and R1 are inner and outer radii of the contact spot. Friction distributed force will be T=fP MathType@MTEF@5@5@+= feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamivaiabg2 da9iaadAgacqGHxiIkcaWGqbaaaa@3A81@ where f is a friction coefficient.

In a polar coordinate system, the elementary moment of the friction force with respect to the bolt axis is:

dM=T r 2 dRdTet MathType@MTEF@5@5@+= feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamizaiaad2 eacqGH9aqpcaWGubGaey4fIOIaamOCamaaCaaaleqabaGaaGOmaaaa kiabgEHiQiaadsgacaWGsbGaey4fIOIaamizaiaadsfacaWGLbGaam iDaaaa@43A9@
Where r is the distance to axis while dR and dTet are radius and angle differentials respectively.

Integrate the elementary moment over the contact area to obtain the following:

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@
This equation relates applied torque, M, and axial force.

Axial Force

Axial force depends on the structure and bolt stiffness, and on nut placement relative to the bolt:

F = K D MathType@MTEF@5@5@+= feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOraiabg2 da9iaadUeacqGHxiIkcaWGebaaaa@3A4C@
K is structure stiffness factor, and D is relative displacement.

Relative displacement can be expressed by the following:

D = N H MathType@MTEF@5@5@+= feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiraiabg2 da9iaad6eacqGHxiIkcaWGibaaaa@3A51@
Here, N is number of nut turns and H is thread pitch. Therefore,
F = K H N MathType@MTEF@5@5@+= feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOraiabg2 da9iaadUeacqGHxiIkcaWGibGaey4fIOIaamOtaaaa@3C12@
(equation A)

Assume that at first analysis pass one nut turn is described (N(1)=1), and corresponded axial force F(1) is found from the analysis. The structure stiffness factor in this case can be defined as the following:

F(1)=KH1 MathType@MTEF@5@5@+= feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOraiaacI cacaaIXaGaaiykaiabg2da9iaadUeacqGHxiIkcaWGibGaey4fIOIa aGymaaaa@3E0E@

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

Now you can relate torque to the number of turns:

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@

Therefore, in order to realize prescribed torque M, after the first analysis is done with N=1, a second analysis (second convergence pass) must be performed using the following equation:

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@

In general, at pass (i+1) the number of turns applied is as follows:

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@
Here, N(i) is the number of turns applied at previous passes, and F (i) is result axial force evaluated at previous pass. These corrections for number of turns applied are important because in the course of passes solution is refined, which changes structure stiffness factor K in equation A above. So, K is not constant, but depends on pass K(i).