I'm trying to calculate the clamping force resulting from torquing a nut and bolt to a particular level.

I have found this formula in various forms in a lot of places.

$$T = KDP$$

  • $T$ = Torque (in-lb)
  • $K$ = Constant to account for friction (0.15 - 0.2 for these units)
  • $D$ = Bolt diameter (inches)
  • $P$ = Clamping Force (lb)

I applied this to my problem

  • $T = 0.6\text{ N-m} = 5.3\text{ in-lb}$
  • $D = 3\text{ mm} = 0.12\text{ in}$
  • $K = 0.2$

This gives $P = \dfrac{T}{KD} = 220\text{ lb} = 100\text{ kg}$.

So, I have two questions.

  • The result seems way too high. I'm using a tiny M3 bolt, and not much torque. I can't see how this would result in 100 kg force. Can anyone see the error?
  • The formula doesn't take account of thread pitch. I would expect a fine thread to give more clamping force for the same torque. Is there a formula that accounts for thread pitch?
  • 1
    $\begingroup$ You'd be amaze how much mechanical advantage can do. $\endgroup$ Apr 7, 2016 at 9:35
  • $\begingroup$ As a point of comparison, structural bolts can be pre-tensioned to tens of thousands of pounds just using a spud wrench. Granted, these types of bolts are much larger than your M3 bolt, but 220 lbs is nothing. $\endgroup$
    – grfrazee
    Apr 7, 2016 at 12:54
  • 2
    $\begingroup$ Note that the relationship between torque and clamping force is not very reliable in practical situations, and where it really matters other methods are often used to determine the clamping force. $\endgroup$
    – Ethan48
    Apr 7, 2016 at 13:03
  • $\begingroup$ Thanks @ttonon - That answer makes sense to me. It's really the frictional coefficient that determines the relation between torque and loading. The ramp effect of the thread is small compared to this. $\endgroup$ Aug 24, 2020 at 12:39
  • 1
    $\begingroup$ @CameronAnderson For sure. In the structural steel world, that's called the 'turn of the nut' method. $\endgroup$
    – Ethan48
    Aug 26, 2020 at 14:01

4 Answers 4


The required torque is calculated basically in the way you would calculate how much force you need to push a triangular shaped door stop between the bottom of the door and the floor. This operation necessarily involves friction that for accurate calculations needs to be estimated. All in all the calculated results are maybe only + or - 25% accurate.

There are simple equations, like the one the questioner provides and there are more accurate ones (below). The questioner's formula is erroneous because it doesn't include the important effect of the screw thread. The "K" in that equation should include friction as well as the helical angle of the screw. I believe this simple form of the equation started with the accompaniment of a figure or chart to look up a suitable value for K, and then it got more simplified, but with knowledge of the basic physics lost.

We can start with that equation, but then write K further as

K = {[(0.5 dp)(tan l + mt sec b)/(1 – mt tan l sec b)] + [0.625 mc D]}/D


K = {[0.5 p/p]   +   [0.5 mt (D – 0.75 p sin a)/sin a]   +   [0.625 mc D]}/D  

where D = bolt nominal shank diameter. p = thread pitch (bolt longitudinal distance per thread). a = thread profile angle = 60° (for M, MJ, UN, UNR, and UNJ thread profiles). b = thread profile half angle = 60°/2 = 30°. tan l = thread helix angle tan = p/(p dp). dp = bolt pitch diameter. mt = thread coefficient of friction. mc = collar coefficient of friction.

These expressions contain both the effects of friction and of screw thread. They can be found in the reputable texts, Shigley, Mechanical Engineering Design, 5 ed., McGraw-Hill, 1989, p. 346, Eq. 8-19, and MIL-HDBK-60, 1990, Sect. 100.5.1, p. 26, Eq. 100.5.1, respectively. They may be too much for some people and we can understand the desire to simplify.

I don't have practical experience in comparing these calculations to the real world. It's possible the more complicated expressions are judged not to be worth the effort in comparison to their accuracy. However, in an "Engineering" forum, I think it's important to not lose sight of the fundamental physics.

  • $\begingroup$ This answers my original question about thread pitch - Since for any normal bolt D is much greater than "0.75 p sin(a)", it is safe to leave that second term out (given the other variability in the calculations). $\endgroup$ Dec 1, 2020 at 10:03

That figure is about right for a low tensile bolt. See also this calculator and this table

As a reality check if we approximate to a cross sectional area of 7 mm2 and a load of 1000 N that gives a tensile stress of 140 MPa which is below yield even for low tensile steels.

In this particular context, where torque is known, the thread pitch doesn't come into it as you are calculating based on the relationship between torque, friction and tension.

A fine thread will (all else being equal) be stronger than a coarse one. Some methods involve calculating clamping force by tightening the bolt by a predetermined angle and here pitch does matter.

A screw thread is essentially a variation of a wedge or inclined plane and can provide very high mechanical advantage before you even consider the leverage of the wrench/driver used.

  • $\begingroup$ Thanks Chris, I used the calculator - It came out at 960n which is close enough to my answer to give me confidence, but wow. That's a lot of force for what doesn't feel like much tightening. We use drivers with calibrated torque click at 0.6 nm, and it doesn't' take much turning effort to torque the screw. $\endgroup$ Apr 7, 2016 at 11:35
  • $\begingroup$ "In this particular context, where torque is known, the thread pitch doesn't come into it as you are calculating based on the relationship between torque, friction and tension." This statement is incorrect. Screw pitch always comes into it, and it's the quantity that accounts for the mechanical advantage of a screw. $\endgroup$
    – ttonon
    Aug 24, 2020 at 16:03
  • $\begingroup$ As elaboration and proof, from your claim, different threads will require the same torque, but you will have to make more turns with a finer thread. Since energy is torque times angle your statement violates conservation of energy because, in the frictionless case, it claims you can put different amounts of energy in yet get the same amount of energy stretching the bolt. Where does the extra energy go? $\endgroup$
    – ttonon
    Aug 24, 2020 at 16:11

Poor method to get a known clamping force; frictions are large unknowns. In the real world (when clamping force is important) ,a hydraulic tensioner pulls the stud/bolt and then the nut is tightened. For ordinary applications like car wheel lugs or headbolts , the manufacturer has the experience to know the torque levels to apply.

  • $\begingroup$ Good for a school test. $\endgroup$ Jan 17, 2018 at 15:56

The correct answer is 100 N-m (force) not 100 KG (mass).

The error was in failing to keep all the units consistent throughout.

Off the top of my head 1 N force is about the force exerted by gravity for one "standard apple." That equates to about 0.25 pounds.

  • $\begingroup$ Hi, I did check, and the formula stated the units, which I used throughout. The final answer was 220 Pounds, which I converted to Kg. $\endgroup$ Jun 7, 2023 at 3:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.