conversion of torque to force 4

[edison123]

Can someone tell how much is the force applied by a 1.5 inch dia bolt torqued to 420 N.m (Newton.meter) ?

 

[arto]

F = 5T/d

 

[codeeng]

I use the following formula for torque:

T = F/1000 [p/2pi + u(Rm+r)]

where
T = Torque Nm
F = Bolt Load N
p = thread pitch mm
u = coeff friction (0.1 to 0.2 typically)
Rm = mean thread radius mm
r = mean nut bearing radius mm

u will vary considerably depending on type of lubricant, finish and fit of threads.

 

[jte]

Looks roughly the same as the other formulas. I'd use the formula T=kdf where T is torque, k is a friction factor (commonly 0.18 for well lubed nuts and studs), and f is force you will have 61 kN or 14 kips on your bolt. Well within the allowable for most steels under ordinary temperatures.

jt

 

[edison123]

Thank you all.

I used some on-line calculators and arrived at 5.6 Tons. Is this right ?

 

[jte]

Well, since (depending on your definition) a ton is either 2 kips (2000 lb) or 1000 kg (roughly 10 kN) then your result of 5.6 tons is roughly 56 kN or 11 kips. Pretty close to the values I came up with above. I'd guess your on line calculator used a slightly different friction coefficient, but the results are certainly reasonable.

Keep in mind that translating torque on nuts to force on a bolt is a very rough calculation. There are too many details involved (friction/lube, tolerances, temperature, phase of the moon) to pretend that the correlation is extremely accurate. Close enough usually is. If not, use stud tensioners, direct tension indicators, etc.

jt

 

[edison123]

Thx jte

 

[edison123]

jte,

phase of the moon ???

Yeah, it get the picture ... it is a very rough calculation.

 

[shopper1732]

T = .2 * d * W

where:

T = in lb
d = nominal size of a UNC bolt (e.g. .75 for a 3/4-10 bolt)
W = lb of induced load