Calculation of torque

(1)

Calculation of torque

Forces on metric v-thread:

Characteristics: d = Nominal thread  (mm) p = Pitch (mm) d2 = Flank  = d-0.64952p (mm)  VSM  = Pitch angle = _       2 arctan d p = _      0.64952 ) ( arctan p d p (°)  = Half flank angle = 30°  VSM

 = Coefficient of friction of steel/ground steel 0.15

 = Friction angle on v-thread (°)

 =

_

_

_

_

_         cos tan arctan cos arctan   is a relatively small angle simplified:  = _        cos arctan =        30 cos 15 . 0 arctan = 9.826°

The following also applies:

Ft = FLtan



 



 with '+' for tightening and '–' for loosening

Calculation of the torque MA/L and axial force FL on the thread:

MA/L = F r = 2 2 d F_t = 0.5d₂F_Ltan



 



Therefore: FL M =F r

M = torque that is exerted on the adjusting nut (Nmm) F = Leverage (N) r = Lever arm (mm) FL = Axial force (N) FN = Normal force (N) FR = Frictional force (N) Ft = Tangential force (N) M =A = tightening torque (Nmm) M =L = loosening torque (Nmm)

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FL =





     tan 5 . 0 ₂ / d M_A _L =                 tan arctan 9.826 5 . 0 2 2 /  d p d M_A _L =





_                 9.826 ) 64952 . 0 ( arctan tan 64952 . 0 5 . 0 /  p d p p d M_A _L

  is a relatively small angle

We assume an average value of 0.9° (please see table below)

M6x0.5 M8x0.75 M11x1 M25x1 M50x1.5 M55x2 M150x2 M155x2 M200x3 1.6° 1.82° 1.76° 0.749° 0.558° 0.679° 0.245° 0.357° 0.276° FL =









    0.64952 tan0.9 9.826 5 . 0 / p d M_A _L

 for further simplification we use only d instead of (d-0.64952p)

FL =





     tan 0.9 9.826 5 . 0 / d M_A _L Tighten: FL = d M_A  0947 . 0  MA 0.0947FLd Loosen: FL = d M_L  0.0785  ML 0.0785FLd

Calculation of the frictional torque adjusting nut – bearing surface M =R: M L M L M L R F d d F d F M       0.075  2 15 . 0 2  Simplified with dM  d: d F M_R 0.075 _L Note:

MR only comes into effect if the adjusting nut is pressing on the bearing surface.

Torque being exerted on the adjusting nut: Tighten: d F d F d F M M M  _A  _R 0.0947 _L  0.075 _L 0.17 _L Loosen: 2 15 . 0 2 075 . 0 0785 . 0 F d F d n F d F d n F d M M M M  _L  _R _G   _L   _L    _R   _L   _R Note: Derivation of MG please see below

Values still to be taken into account:

Surface pressure on the thread flanks  max. axial load

MR = Frictional torque adjusting nut –

bearing surface (Nmm)

 = Coefficient of friction  0.15

dM = Average surface  of adjusting

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zul L L _Pa H d p m F A F Pa       1 2 



d p



p p m Pa F_L  _zul  0.64952 0.54126 Simplified: p d p m Pa F_L  _zul  0.54126 And shear stress on thread

zul L m d F _       2



d p



m F_L _zul  0.64952  Simplified: m d F_L _zul  

Braking effect of gib MRR/MRR-A (radial)      _G cos60 R F F

With  = 3.23°  average value of thread M4-M12 And                 13 30 cos 2 . 0 arctan cos arctan    

_

_

_

_

          13 23 . 3 tan 5 . 0 tan 5 . 0 _St AI St AI G d M d M F 

 with the simplification d2StdSt

Tighten:





St AI St AI G d M d M F         1455 . 0 13 23 . 3 tan 5 . 0 Therefore: St AI G R d M F F cos600.2 0.687

Tightening torque MAI (Nmm) of the grub screw as per Bossard T.037 for M4-M16:

M4 M5 M6 M8 M10 M12 M14 M16

1000 3000 5000 10000 20000 45000 45000 90000

Therefore for FR (N) for one gib:

M4 M5 M6 M8 M10 M12 M14 M16

171.75 412.2 572.5 858.75 1374 2576.25 2208.2 3864.37 Therefore the result for frictional torque MG:

2 687 . 0 2 d d M n d F n M St AI R G        d d M n M St AI G 0.343   Pazul = 80-150 N/mm2  Gieck Z18

m/p = Number of thread pitches = thread length/pitch d2 = Flank  = d0.64952p (mm)

F L = Axial force (N)

H1 = Thread load-bearing depth (VSM) = 0.54126p (mm)

m = Adjusting nut height (mm) zul = 560 N/mm2  Mechanical

engineering p.39

MAI = Tightening torque of grub screw (Nmm)

FG = Axial force of grub screw (N)

FR = Frictional force of gib on adjusting nut thread. (N)

 = Coefficient of friction grub screw – adjusting nut0.2

dSt =  of grub screw (mm)

MG = Braking torque of gib on

adjusting nut thread (Nmm) d =  of adjusting nut thread (mm) dSt =  of grub screw (mm)

n = Number of gibs

MAI = Tightening torque of grub screw

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Braking effect of gib MRA/MRA-A (axial) t E F F sin with E AI t r M F  E AI E r M F sin Therefore:      _E cos60 R F

F (please see above)

1 . 0 sin 2 . 0 5 . 0 sin 60 cos           E AI E AI E R r M r M F F   

Therefore the result for frictional torque MG:

2 d F n M_G   _R d r M n M E AI G 0.05 sin  Note:

The angle  is dependent on which position the gib has been ground. The tolerances when manufacturing the eccentric, as well as the gib, also have a great influence on the angle .

Sample calculation:

Assuming: An adjusting nut MRR 50x1.5 (2 gibs with M6) is used to clamp a spindle bearing.

The maximum axial load on the bearing must not exceed 10,000 N. What are the tightening and loosening torques?

Tightening and loosening torque of the adjusting nut: Tighten:

FE = Force of eccentric on gib (N)

Ft = Tangential force on eccentric (N)

rE = Radius on outer surface of eccentric (mm)

MAI = Tightening torque on eccentric (Nmm)

 = Angle of eccentric to gib (°)

FR = Frictional force of gib on adjusting nut thread.

(N)

 = Coefficient of friction gib – adjusting nut  0.2

Ft



FE

rE

MAI

MG = Braking torque of gib on

adjusting nut thread (Nmm) d =  of adjusting nut thread (mm) n = Number of gibs

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Nmm d F M M M  _A  _R 0.17 _L  0.17100005085000 Loosen: 2 15 . 0 2 075 . 0 0785 . 0 F d F d n F d F d n F d M M M M  _L _R _G   _L   _L   _R   _L   _R Nmm M 0.1510000502572.525103625

Check surface pressure/shear stress  max. axial load:

N p d p m Pa F_L _zul 50 0.54 1.5 178128 5 . 1 14 150 54 . 0               OK N m d F_L _zul   56050141231504 OK

Assuming: An adjusting nut MRR 100x2 (2 gibs with M8) is used to clamp a spindle bearing.

The maximum axial load on the bearing must not exceed 25,000 N. What are the tightening and loosening torques?

Tightening and loosening torque of the adjusting nut: Tighten: Nmm d F M M M  _A  _R 0.17 _L 0.1725000100425000 Loosen: 2 15 . 0 2 075 . 0 0785 . 0 F d F d n F d F d n F d M M M M  _L _R _G   _L   _L   _R   _L   _R Nmm M 0.15250001002858.7550460875

Check surface pressure/shear stress  max. axial load:

N p d p m Pa F_L _zul 100 0.54 2 508938 2 20 150 54 . 0               OK N m d F_L _zul   560100203518583 OK