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Tension vs torque 1

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patprimmer

New member
Nov 1, 2002
13,816
I am out of my area of expertise, so the question I am asking is simplistic and only requires a simplistic or rule of thumb answer.

Also, nothing will fall down if I get it wrong, it will simply mean things will wobble around a bit and bolts will need to be tightened more regularly.

I an calculating the creep in plastic spacer plates for pallet racking.

I have no problem with the normal load plus safety factor per pad, but I am having trouble deciding on a figure to use for preload from the retaining bolts.

They are 10 mm by 1.5 mm "dynabolts"

They are supposed to be pulled down snug, then 3 turns to spread the ferrule or sleeve.

The recommended torque is 20 Ft.Lbs or 27.1Nm.

I need the initial clamping force.

I tried calculating the total mechanical advantage times the tightening force by both the specified torque and by estimating probable force and spanner size (I estimated these as 20 Kg force at 300mm to give a mechanical advantage of 628:1 or 12.5 tonnes clamp).

This is OK to a point, but I have no idea of the loss factor for friction, except that it must be quite high for a dry thread on a rough and uneven pressed steel plate , electro galvanised surface in the presence of concrete dust generated while drilling.

My instinct tells me that the real force is likely to be more like 3 tonnes max, and probably realistically 2 tonnes.

Logic also tells me that movement from subsidence will be much greater than the creep in a 30% Glass Filled nylon part at the thickness, bearing area and probable load encountered, and so long as the bolt is tightened annually, some preload will be maintained.

What I really need is the voice of experience as to what load pallet rack installers are really likely to put on the spanner, and what compression that will put on the spacer.

Regards

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The simple equation for torque vs. tension is:

T = k D F

where

T = tightening torque (N m)
k = torque coefficient (m/mm)
D = bolt diameter (mm)
F = clamping force (kN)

Friction can vary enormously. For a typical joint with smooth surfaces and nominal lubrication, k ~ 0.2. I think worst case you may have k ~ 0.4. That would mean a 5 kN force developed from a 27 Nm torque.

Regards,

Cory

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.
 
Thanks Cory.

I knew it would be you who helped

Regards

Pat

Regards

eng-tips, by professional engineers for professional engineers
Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.
 
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