Bolt Tension Brain Teaser! 1

[TomosSmith]

This is a hypothetical question which my colleagues and I have been discussing all afternoon - please help!

A bracket is fixed to the underside of a steel beam with a nut and bolt. The bolt is torqued up to provide a tension in the bolt of 10 Tonnes. An additional weight of 5 tonnes is then suspended from the bracket.


Question: What is the tension in the bolt?

 

[israelkk]

Small addition (much smaller than 5 ton) depends on the bolt connection stiffness. Assuming the connection stiffness is 10 times the bolt itself stiffness, the addition is 10% of the 5 tons. i.e. 500 kgf. Therefore, the tension in the bolt will be 10500 kgf.

 

[PeteDB]

http://www.boltscience.com/

Check out the site above. I have found its information usefull.

 

[DLite30]

The 10 ton tension in the bolt would be equivalent to hanging 10 tons of weight from the bolt. Then you had 5 more tons of weight, for a total of 15 tons.

Pretty straight forward, unless you're mis-stating the scenario.

 

[GregTirevold]

Fb = Fi + ((Kb*Fa)/(Kb+Kp))

Where:

Fb = Force on bolt
Fi = Preload on bolt
Kb = Spring Constant for bolt
Fa = Applied Force on bolt
Kp = Spring Constant for clamped parts

 

[Tmoose]

Unless the bracket positions the "additional weight" to be in line with the fastener the resulting loading on the single bolt may be a function of some leverage relationship. The designers of brackets for bathroom partitions and plant hangers apparently use handbooks that have guidelines to ensure a load that exceeds the component weight by 10% will pull the key fastener out of the wall as easily as a 4 foot crowbar extracts a 10 penny from a 2X4

 

[chicopee]

Based on the original post,what I think that your colleagues and you were thinking is at what load would contact stress between clamp and beam be zero and that would be not less than a 10 ton load and consequently the bolt would be under a 20 ton tensile load.

 

[ivymike]

assuming you're after just the tensile effects, not the prying mentioned by tmoose, Isrealkk and GregTirevold have it right. In the case where the abutment (your i-beam) is infinitely stiff, the bolt tension does not change until the force on the abutment hits zero (when you hang a 10 tonne weight). In reality, both the bolt and the abutment are flexible so the load does change. Greg has a formula above which looks about right - I haven't checked it in detail.

 

[drawoh]

TomosSmith,

For starters, a tonne (as opposed to the various tons) is a unit of mass. You cannot have a tension force or a weight of 10_tonnes or 5_tonnes or 3.14159265897_tonnes. The SI unit for force is the Newton.

If you clamp an infinitely rigid bracket with a force of 100kN (g=10m/s²approx), you generate a contact force between the bracket and the steel beam of 100kN.

If you hang a 50kN weight from the bracket, you subtract 50kN from the contact force. The bolt tension is not affected, unless the weight exceeds the original bolt tension.

Remember Hooke's Law. If your model does not change the strain on the bolt, it does not change the stress.

JHG

 

[EngineerTex]

Since some people decide their belief in science based on "a majority of scientists agreeing" on some subject, I'll cast my vote on israelkk, GregTirevold and Tmoose. I will also say that ivymike, chicopee and drawoh are extremely close, but missed the mark just slightly and that DLiteE30 is wrong.

Israelkk has the best explanation. It is because of the elasticity of the beam and bracket sandwiched between the nut face and bolt head. The bracket and beam are not infinitely stiff. If they were infinitely stiff, then additional load would be added directly to the preload, where 5T on top of 10T would be 15T. However, since as israelkk stated, you are dealing with a system that is NOT theoretical, it is real-world where the clamped members are elastic and about 5 - 10 times stiffer than the bolt, you have to consider that as you add a load to the bracket, and the bolt is elongating a small amount, the very stiff clamped members are losing their preload until there is a gap between the two. At the point where there is a gap between the two sandwiched members, they are no longer applying the 10T load that was found in the preload. So when you put a few percent over 10T on the bolt, the clamped members are no longer able to apply any load, so the bolt is only experiencing a bit over 10T.

And drawoh is correct about the units.

Engineering is not the science behind building. It is the science behind not building.

 

[ivymike]

um, what of what I said was wrong? (I didn't say in my example that the bolt was inf. stiff, as I'm guessing you supposed I did, just the beam)

 

[EngineerTex]

Well, ivymike, if the beam is infinitely stiff, then there would be... dangit -- you're right. My apologies. I confused myself.

But I think that we all agree that in the real world, it still wouldn't be simply additive.

Engineering is not the science behind building. It is the science behind not building.

 

[TomosSmith]

Thanks for your input everyone. Well it seems that we have a consensus that in the real world where beam and bracked are not infinitely stiff, the loads are not simply additive, the bolt tension does change, and we must consider the stiffness of the materials.

Thanks for the formula GregTirevold, that will be useful in solving the actual problem that started our debate!

Tmoose - no we weren't considering any leverage, just a simple case where the load is suspended directly below the bolt.

Thanks again,
Tom

 

[Tmoose]

So, Tom, did you win the bet?

Something I think that is worth considering is that for a full metallic joint of generous proportions the fastener stiffness is relatively quite low, so the variation in fastener load up to the point preload is overcome is a pretty small. Hence even Holokrome has made statements like this -
"Suppose a joint has been tightened to a preload Pi and additional load, Pe, tending to separate the members is applied. In general in rigid assemblies, as long as the external load is less then Pt it primarily decompresses the joint and has little effect on the tension in the screw.
Thus even if such a load is repeatedly applied, the fastener will not fail in fatigue."

Being able to ignore fatigue reduces having to calculate, test and worry about sundries like rounded thread root profiles, proof load vs yield strength, bolt thread engagement and stripping, and exotic materials, and instead waste time on the internet instead.

Dan T

 

[Canadieng]

I concur that under an external load, the bolt takes only a portion of the extra load until joint separation (depending on relative stiffnesses of the members and bolt)

Here is a good summary of the phenomena:

http://www.mece.ualberta.ca/courses/mec403/2009-W/Bolted Joints in Tension-09.pdf

 

[zekeman]

In proof of the majority, since this is a democracy, consider a bolt squeezing a plate .
If the stiffness of the plate is k1 and the stiffness of the bolt is k2, then any external force separating the bolt would cause an equal increment on the plate separation as well as the bolt. So
If the external force is F, then
F/k1=delta bolt tension/k2
and
delta bolt tension=F*k2/k1
valid as long as the plate remains under compression

confirming Isaelkk

 

[TomosSmith]

My colleague demonstrated that the bolt tenstion does not change with the inital load with the following experiment:

Put an elastic band around a ruler lengthways. The elastic band is the nut and bolt, the beam/bracket is the ruler. When adding an initial load to the elastic band at the end of the ruler, the length and therefore the tension the elastic band does not change until the initial preload is exceeded.

But from the discussion above, we have realised that the tension in the elastic band is actually changing with the added load, although it is much too small to see, and it is essentially the case of an infinitely stiff beam/bracket.

 

[zekeman]

Correction

The answer should be

x=increment tension=K2*F/(K1+K2)

The reasoning is

if you call the decrease in plate force against bolt head, y
and the increase in tension x for an external force F
you get 2 equations
x=y=F
y/k1=x/k2

solution:

x=k2*F/(k1+k2)

k1 plate stiffness
k2 bolt stiffness

Tomosmith,

Your colleague demonstrated for

k1>>k2

x goes toward 0

 

[zekeman]

Another one, a typo this time

eq should read

x+y=F
y/k1=x/k2

 

[plasgears]

10 T, the amount of preload.