Understanding Bolt Capacity 7

[jaydigs]

I need to hang an overhead fixture that weighs approxiamtely 500lbs. My company currently uses 4 Ø1/2-13 UNC grade 8 bolts for this purpose. I came up with a torque spec from the Machinery's handbook of 128ft-lbs to torque the bolts down. I am wondering how much capacity is left for the bolt to support a load when you consider the preload you're applying. The preload is 15325lbs for the 128ft-lbs torque. The proof strength is 120000psi so for a nominal Ø of 1/2" would it be 120000 X .19625in^2(cross sectional area of .5in bolt) = 23550 (bolt capacity) - 15325 (preload) = 8225lbs(capacity of bolt remaining)? So then that bolt, torqued to recommended spec of 128ft-lbs can then support a load of 8225lbs? If anyone can explain I would be greatly appreciative.

Thanks

 

[CoryPad]

Calculate the relative stiffnesses, then calculate the additional bolt force, then calculate the new total bolt force and compare to the bolt allowable value, and compare the alternating bolt force/stress to the bolt's allowable fatigue stress.

The rationale for keeping the design load less than the preload is because it is a simple concept, not that it is highly accurate.

 

[drawoh]

CoryPad,

The rationale for keeping the design load less than the preload is because it is a simple concept, not that it is highly accurate.

Lots of concepts are simplied. This one is pretty close to reality, as long as the joint is rigid.

I will have to re-run my model above for bolts that are large, relative to the joint. The above model is a good representation of a housing. It does not model a pipe connection very well.

JHG

 

[jaydigs]

CoryPad;

Thanks for your help in explaining this. I'm sure a lot of people will benefit from this thread since bolted fasteners is not given enough time in a conventional ME curriculum. I know a lot of engineers use fasteners everyday without understanding completely the analysis that goes into specifying them.

 

[Bribyk]

Really?! We covered this in a few different classes including a 3 hour lab using a tension tester and a drilled bolt with a strain gauge inside. We even got to see how a soft gasket material affected the bolt loading. Very informative... but MechEng2005 summed it up more eloquently than I've heard before, in one sentence.

Welds were not covered very well in the curriculum I took but these real life topics (such as bolts) rarely are.

 

[MechEng2005]

Thanks Bribyk!

However, I have re-confused myself on this matter...

The logic from my OP still holds as one argument. The bolt is preloaded, causing the clamped members to compress slightly. Then, a force is applied. This force will increase the stress in the bolt, causing it to elongate. However, as it elongates, clamped memebers are allowed to decompress. This reduces the stress caused by the preloading. For the notation I will use:

P = Preload force
F = Applied force
A = Effective cross-sectional area of bolt
S = Stress in the bolt
1 = State of bolt with only preload
2 = State of bolt with preload and applied load

It makes sense then that:
S1 = P/A
S2 < (P+F)/A

However, in trying to determine a value for S2 I came across the following difficulty. I believe the amount of applied force required to elongate the member to a point where the effective preload is zero should be the same as the preload. The energy used to preload the bolt is effectively stored in the clamped members. The amount of energy required to release this energy should be the same as was required to compress them. Therefore, if F = P then the clamped members are no longer compressed. This would mean that basically there is no preload and the only relevant force is F. However, if there is no preload at that point then P2 = 0, and from F = P we find that F = 0. So the force required to elongate the bolt and release the preload is nothing! How is a force of zero elongating the bolt?! The argument contradicts itself!

Can anybody see where I went wrong?

I think the error must be somewhere in the logic that the force required to elongate the bolt and release the compression in the clamped members is the same as the preload. If this were true, assuming both the elongation of the bolt and compression of the clamped members are linear, with change in length/thickness proportional to the force, then the stress in the bolt at anytime until F>P would be constant, since any increase in F would reduce P by the same amount... However, I don't see how the logic in the storage of potential energy in the clamped memebers is incorrect...

Or perhaps this is just a case of slight variations in assumptions I have made (and are commonly made in theories) such as linear stress/strain, negligible reduction of area, etc, cause the model to fail to produce accurate results. Or maybe there are other effects that occur experimentally that I am not accounting for in this argument?

Anybody who has thoughts or could point me to some other information or flaws, it would be appreciated!

-- MechEng2005

 

[hydtools]

Read again the information in the boltscience link.

http://www.boltscience.com/pages/basics5.htm

I think you may be confusing preload with clamp load. Preload is the force carried by the bolt. Clamp load is the load carried by the joint materials. With no external force clamp load and preload are equal. The external load is transfered to the bolt through the joint material. As the external load is increased the clamp load is decreased, but the bolt load is increased and the bolt elongates the same as the joint materials decompress/expand. The clamp load can go to zero but the bolt load continues to increase; ulitmately to failure.

Ted

 

[MechEng2005]

Thanks hydtools!

I'm still not sure I have a firm grasp on the qualitative results, but I'll keep trying to wrap my head around this and let you know if I come up with anything...

-- MechEng2005

 

[CanSteelConnDsgnr]

Here's how I visualize it:

Picture two U-shaped plates held together by a single bolt. The upper plate is connected to rigid support and the lower plate is connected to a load. The bolt connects the flat portion of both U-shapes to each other and is pretensioned.

The pretension load in the bolt acts on the plates holding them together (clamping force). In order to put more load in the bolt, you need to separate the plates by overcoming the force holding them together (i.e. the pretension load / clamping force).

The only way to add more load directly to a pretensioned bolt would be to weld a bar directly to the bolt head and pull on it. This way the load goes directly into the bolt without having to over come the clamping force between the plates.

I hope this is helpful

 

[trainguy]

Guys,

In general terms, and ignoring the effect of stiffness of the clamped parts:

If you apply an external load that is less than the pre-load, the tension in the bolt is equal to the pre-load.

If you apply an external load that is greater than the preload, the tension in the bolt is equal to the applied load and you have joint separation.

In other words, you have to overcome the pre-load in the bolt for the tension to change.

tg

 

[MechEng2005]

Trainguy -
That has been discussed previously. However, if you have a preload of "P" and add an applied load, "F", the bolt elongates (reducing the clamped load). For the bolt to elongate you MUST have increased the stress/force acting on the bolt. So the rule the if F<P the bolt doesn't seem right. Maybe it is accurate as a rule of thumb, but not exact.

I have been looking at this and done some calculations. I have simplified the problem to be a bolt going through a round tube. This way, when the bolt is preloaded, I can determine the compression of the tube (since I know the cross-sectional area, where-as with clamped members the behavior would be different depending on bolt head size and how the load is transfered through the plates at the connection). Then, I looked at what happens when a force is applied and set the length of the bolt equal to the length of the tube and solve for the clamping force. If everything is correct in my calculation, the force acting on the bolt when a force is applied is ALWAYS greater than just the preload, even when the force is much smaller than the preload.

As an example, I used a 1/2"-13NC bolt going through a tube with OD 1", ID 17/32", a free length of 2", Preload of 7,344 lb, Applied Force of 1,000 lb. Both the bolt and tube have a modulus of elasticity of 29E6 psi.

The clamp load ends up at 6,547 lb. So the total force on the bolt (clamp load + applied force) is 7,547 lb.

Increasing the applied load to 7,000 lb gives a total load of 8,767 lb. So the effect of the applied load is definately less, but if it were assumed that it had no effect and the load was just equal to the preload, there is a difference of almost 20%. Not too bad for a "rule of thumb", but not really good approximation either.

So clearly, the total load is greater than the preload, but less than the preload plus the applied load.

Again, this assumes a bolt running through a tube which is basically useless in practical applications (nevermind that stress concentrations, sticking of materials, etc were ignored). I do not intend to imply that what I have done is quanitatively accurate, but I do think it is qualitatively reasonable.

-- MechEng2005

 

[hydtools]

Look again at the referenced link(s) and the bolted joint diagrams. The portion of the applied load that is added to the bolt preload depends on the ratio of bolt stiffness to joint stiffness. For example if the joint is 5 times as stiff as the bolt, 1/5 of the applied load is added to the bolt preload. This compares with MechEng2005's example.

Ted

 

[trainguy]

Mecheng2005,

We are both right. (How's that for diplomacy...) My point is a practical rule of thumb, based on the fact that in most bolted joints, the stiffness of clamped parts is much greater than the stifness of the bolt itself. Of course you increase load in the pre-loaded bolt whan adding external tension, but not by much, until the joint has separated.

I believe your tube/bolt example is not in agreement with the literature (and my comment) because there is not enough difference between the bolt and tube stiffnesses, as compared to a conventional joint.

I suggest you read Shigley Mechanical Engineering Design, Section 8-4 thru 8-7 (in the 8th ed.), esp. the computation of bolt and member stiffness.

What I've done is to read it and understand it once, then I typically follow the simpler rule of thumb mentioned above...

tg

 

[MechEng2005]

Trainguy -

Ok, we agree. I happened to be sitting next to Shigley (the book, not the person). I am glad that I went through my simplified example, as I now have a greater appreciation and understanding of what is being discussed in the book. Thanks for the info.

-- MechEng2005