Preloaded bolt 10

[NedGan76]

Howdy!

This is another tricky problem to check your knowledge and common engineering sense

Given a steel bolt in an infinitely stiff package. First, we tighten the nut until we prestress the bolt with a force P. Then, we pull the bolt with a force F. What is the final axial force inside the bolt (between the head and the nut):
A) P + F
B) F
C) P
D) other (please explain)

Cheers!
Ned Ganchovski

A better software for your calculation notes, free and open source:

https://github.com/Proektsoftbg/Calcpad

 

[GBTorpenhow]

I'm game.

Spoiler

The answer is D).

The bolt is a spring. If F<P then the nut doesn't move, and the bolt tension is P. If F>P then the nut lifts off of the infinitely stiff block and the tension in the bolt is F.

 

[SWComposites]

please tell us where we can buy a material with E = infinity

 

[GregLocock]

Initially P-F and then F. If you don't like that answer consider what happens when P=F. This is not rocket science. It's FBDs!

Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376

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[BAretired]

Is it really necessary to specify that E = ∞? Would the answer be different if E = 29,000,000psi?

 

[3DDave]

Since the assembly has no offsetting force it accelerates per a=F/m

 

[Sam_G]

I think it is the greater of P or F. There is initially prestress P in the bolt which only increases if F exceeds P.

 

[NedGan76]

Is it really necessary to specify that E = ∞? Would the answer be different if E = 29,000,000psi?

It is a good question. What is your opinion about that?

Cheers!
Ned Ganchovski

A better software for your calculation notes, free and open source:

https://github.com/Proektsoftbg/Calcpad

 

[Stress_Eng]

Once preloaded, the joint is seen as two springs in parallel. If one (plate) is infinitely stiff, the other (bolt) becomes negligible. Therefore the bolt will see P if F < P and F when F > P.

 

[HTURKAK]

If you want to pull the block , the block shall be supported so the rules of Statics equilibrium satisfied..
The answer to your question is (D).

- If F < P , the final axial force at the bolt shank (between the head and the nut) will be P
- If F= P , the final axial force at the bolt shank will be still P
- If F> P , the nut will separate and the final axial force at the bolt shank will be F

My question is;
What will be the final axial force at the bolt shank if F is applied to bolt head ??




Use it up, wear it out;
Make it do, or do without.

NEW ENGLAND MAXIM

 

[GBTorpenhow]

I hand my neighbor the end of a bungee cord, thread it through a hole in the fence between our yards, stretch it a little, then tie a big knot in it and let the knot catch on the hole in the fence. I let go of the cord. The neighbor still feels tension P in the cord because the knot can't slip through the hole in the fence and this keeps the cord taught. The neighbor can't see what I am doing on the other side of the fence. Now I pick up the end of the cord and pull on it with force F. How hard do I have to pull on the cord before he can tell if I am pulling on it or not? Is there any way I can possibly reduce the tension in the cord by pulling on my end?

 

[rb1957]

Man, what a torturous thread. The load in the bolt is never, ever, P-F.

Look at a bolt loading diagram.

People who say "it's simple, FBD" ... are neglecting the companion compression field in the plates, that balances the preload in the bolt.

The external load, F, is typically not directly applied to the bolt, but rather to the bolt through one (or both ?) of the plates.

Imagine there was no preload. Then the external load F would be carried by the bolt form one plate to the other.
Preload allows the joint to carry load without using the bolt, directly, by using the compression between the plates.

Look at a bolt loading diagram.
The minimum load in the bolt, with zero external load, is P.
The load in the bolt increase, as P is divided between the compression field and the bolt according to relative stiffness.
At some external load the joint gaps, preload is lost, and the load in the bolt is F.

"Hoffen wir mal, dass alles gut geht !"
General Paulus, Nov 1942, outside Stalingrad after the launch of Operation Uranus.

 

[BAretired]

It is a good question. What is your opinion about that?

My opinion is that the value of E makes no difference to the final result. Before force F is applied, the bolt is in equilibrium with the block, bolt with tension P and block with compression P. As F is applied, P in bolt and P in block diminish until F = P when bolt tension is P and block compression is zero. When F exceeds P, bolt tension = F and block compression remains at zero.

The bolt head bears against the back of block with a reaction equal to the larger of P or F, and is resisted by bending of the block between unspecified supports.

 

[Stress_Eng]

The slope angle for the two lines in a bolt loading diagram, for the bolt and the compressed plates, are a function of the moduli. You will find that, as the modulus for the plate goes up, the slope for the bolt tends to a straight horizontal line, indicating that the bolt takes less of the applied load.

 

[NedGan76]

BAretired

From my experience, some people tend to make the things much more complicated than necessary. So, by introducing this premise, I avoid turning this otherwise simple problem into a postdoc research. ))

For example, imagine that the block is concrete C15/20. We may loose prestress with time, due to plastic behavior, creep and shrinkage. So, the final force P_t will be less than the initial force P_0 after some time t. Also, the block itself will deform elastically under the compressive strain. So, after we apply the force F, it will slightly increase P. Due to the relaxation of the compressive strain in the block, the bolt will elongate a little. So, making the block infinitely stiff simplifies the problem.

Cheers!
Ned Ganchovski

A better software for your calculation notes, free and open source:

https://github.com/Proektsoftbg/Calcpad

 

[HTURKAK]

Sometimes it is necessary to look reference books. The following figure is useful to remind the prestressing concept..
Copy and paste from Advanced Stress and Stability Analysis ( V. I. Feodosiev )

Consider a spring balance at the figure and the upper ring placed on a nail and the lower hook after tensioning located to a
rigid heavy table edge as shown in Fig. The balance with this set up indicates 4.5 kg /wt . Now let us hang weights to the lower hook of the strained balance. Until the weight of the load remains less than the specified tension force, balance will permanently indicate 4.5 kilograms. If a weight is placed to pan hanged on the hook is greater than 4.5 ( say 6 kg )kilograms then the indicator will move from its place and will indicate the corresponding weight ( 6 kg )..



Use it up, wear it out;
Make it do, or do without.

NEW ENGLAND MAXIM

 

[rb1957]

that example would give the simplistic bolt loading ... bolt load (indicated on the scale) = P until P=F then scale indicates F.
this is I think an infinitely stiff plate (or a zero stiffness bolt ... or both ??).

but the OP notes "For example, imagine that the block is concrete C15/20. We may loose prestress with time ...". This is talking about pre-stressed concrete, and not preloaded bolts ... or am I wrong ?

"Hoffen wir mal, dass alles gut geht !"
General Paulus, Nov 1942, outside Stalingrad after the launch of Operation Uranus.

 

[NedGan76]

rb1957,

Yes, these problems are common for prestressed reinforcement. But this would also apply to our problem, if the block was concrete. It was just an example that the material does matter and specifying Е = infinity can simplify the problem.

In practice, we can have preloaded anchor bolts in concrete. Also, we use sometimes full-penetration steel studs in RC slabs/walls, anchored on the other side with a backing plate. You can also preload them a bit to minimize the deformability of the connection. Bu need to be careful not to crush the concrete.

Cheers!
Ned Ganchovski

A better software for your calculation notes, free and open source:

https://github.com/Proektsoftbg/Calcpad

 

[rb1957]

ok, so your preloaded bolt looks like the original post ? Then previous comments apply ... the bolt preload is reacted by compression in the joint.

But when you say "we lose pre-stress" when talking about concrete, are you talking about losing the prestress in the stress rods in the concrete, which is I imagine a creep behaviour ?

Now it's a whole different thing if the bolt is an eye-bolt and the external load is applied directly to the bolt ... it may be like that "fish scale" analogy (or it may not, have to think about it).

"Hoffen wir mal, dass alles gut geht !"
General Paulus, Nov 1942, outside Stalingrad after the launch of Operation Uranus.

 

[NedGan76]

rb1957

Sorry, but this is exactly the kind of discussion, I tried to avoid by setting E = infinity.
Let's keep the things as simple.

Cheers!
Ned Ganchovski

A better software for your calculation notes, free and open source:

https://github.com/Proektsoftbg/Calcpad