I am trying to figure out the math on how much pressure per square inch my hose clamp is applying. The circumference of my hose clamp is 36 inches and I will be applying approximately 36 inch pounds of torque to the hose clamp. The mechanical advantage of the hose clamp appears to be about 30 to 1. Or, 10 revolutions per inch. I come up with about 30 pounds per square inch.\nOr inch of hose clamp? Am I doing this correctly?
The hose clamp is one half inch wide and forgot to add that.
I’m assuming that the torque you are applying is to a fastener that tensions the hose clamp around whatever it is you are clamping. The tension you are creating is tangential to the surface of what you are clamping. So if what you are asking is what pressure you are creating radial to what you are clamping, then no.
I don’t know how to work it out though.
Surface area would be 18 sq inches. ( rounded off, using 3X Diam. instead of Pi. ) so 60 PSI per linear inch of clamp??
Hose clamps will deform the hose as they are tightened. So maximum pressure is the compressive strength of the hose.
I am way below that, I am more worried about not enough pressure. The large diameter spreads the pressure out over a very wide area. I am just curious as to what the pressure actually is.
36" circumference
1/2" wide
That is a massive hose clamp (and hose). What is it used for? Or is this a theoretical exercise?
As @Princhester mentioned, if you are looking to calculate the radial pressure the hose clamp exerts on whatever it is clamping, this value will only be indirectly related to the force you are applying to tighten the clamp. If you imagine that your hose clamp is applied to a thick rod made of rubber, tightening the clamp will compress a disk shaped section of the rod by reducing the radius. You can look up “bulk elastic compression” for details, but calculating even an approximate value for the applied pressure is going to be complicated because you will need to find values for the elastic modulus of the materials that you are clamping. Also note that elastic deformation is the simple case, if the applied pressure causes inelastic deformation then you’re in a much more complicated situation.
Treat the hose clamp as a thin band applying uniform external pressure p around a cylindrical hose. For a thin cylindrical shell (like your hose) under internal pressure, the hoop force in the wall is T=prb, where p is the average radial pressure on the hose, r is the outer radius of the hose under the clamp, and b is the axial width of the clamp band. so p ≈ T/rb. I say approx because finding T is the tricky part, since that hoop force comes from the tension in the clamp band, which comes from the screw torque of the hose clamp. If you do some digging, you might luck out and find the manufacturer’s torque spec, otherwise, you will have to calculate an approximation.
I am clamping a thin PVC coated fabric to a Hard plastic end cap for a bellow. There won’t be much compression because there is nothing to compress. I am sure the pressure is low, I am just concerned about it being too low. I have a working model and it has slipped off a couple of times. Going to add a bead at the bottom of the flange so the clamp can’t slide by it. So I really just want to know if I am using the right formula, ignoring compression. Torque spec is 36” pounds like I showed in the op.
I know for rubber pipe couplings (Fernco) they sell a pre-loaded torque wrench that applies the correct torque of 60 in-lbs.
The spec I am working with is 36 inch pounds,
Here is how I figured it. The worm drive has a 10:1 advantage
I am using inch pounds so each revolution is 3.14 inches X 10 = 31.4 mechanical advantage.
31.4 X 36 inch pounds = 1,130.4 total force / 18 sq inches of clamp = 62.8# per linear inch of clamp.
Does this look correct?
(Deleted)
No, you’re all over the place. Pounds-force per linear inch-clamp is funny, though. I suspect there some pounds inch for torque and pounds inch pressures are getting mixed up.
Thought experiment: visualize the clamp band cut opposite the clamp, vertically straightned out and with artibitrary and unknown weights hung on it. The clamping screw now slightly lowers and raises the weights. What is being measured when you capture the screw adjust torque?
That is exactly how I did it, it came out to 1,130#. That sounds awful high.
Looking at it another way. If I measure out 1” from the worm scrw center I will travel a total of 30” for the clamp to move 1” so I come up with a 30 to one advantage from 1” from center.
That wasn’t a yes or no, it was a demand for clarification. What do you think is being measured? Say you wanted to know the weight of the load*, the tension in the band, but all you could measure is the screw torque and height raised/lowered, the displacement.
*I think that’s what you’re trying for.
Ok, I see what you mean, tension on the band would best describe what I am looking for. If the band is 3” long or 36” long, the tension will be the same but spread out over a wider area, reducing pressure significantly.
Yes but that’s not too useful unless something goes wrong like your band is breaking or the need to spread the load, etc.
You and I know something that the band in tension on the pipe cannot: it isn’t really suspending weights in a line but is bent around a cylinder with half the total load balanced on each side. Our visualization is finished and we now see that there’s no weights lifting at all, only the band’s own ends pulling each other against … something. Since there’s no lifting, what is increasing the tension & associated torque as the clamping screw turns?
Probably because you appear to be ignoring friction.
When torquing a typical bolt only ~15% of the torque goes into actually stretching the bolt (increasing the clamp force/ preload) - the rest of the torque energy goes into friction on the threads and under the bolt head. A hose clamp bolt isn’t going to behave much differently than a standard bolt.