Help me understand how bolts take less load than members

[davidwinth]

TL;DR Summary

A bolt holds two members in compression, and is said to take less of the external load even when the numbers say different.

Hello,

I am studying through a machine design book and ran across this confusing statement which summarizes the results of an example problem. The problem has a bolt and nut that together hold two members in compression. The stiffness ratio is given as $$K_m = 6K_b$$. The bolt is preloaded with $$P_i = 1100 lbs$$ and the members have an external load (that tries to pull them apart) of $$F_e = 1200 lbs$$. The example then finds the resultant compression in the members and the tension in the bolts. The tension in the bolt is 1271.42 lbs and the compression in the members is 71.42 lbs. Then the book gives this statement: "In this problem, it can be seen that the proportion of the load shared by the bolt is very small because the stiffness of the bolt is low when compared to the members."

How is 1271.42 less than 71.42? Thanks.

 

[Baluncore]

I suspect you are missing something, and have also left it out of your question.

I assume the bolt is elastic, and provides a clamping force that is multiplied by static friction between the members.

A diagram would help.
Please provide a reference to the book title, author, edition, and page number.

 

[davidwinth]

The book is called, "The Machine Design Problem Solver" page 215, and there is no consideration of friction mentioned at all. Of course both the bolt and the members are considered as elastic (hence the reference to their stiffnesses). I have not left out anything that I can see. The entire example is only two equations which give the results I showed above. As for a picture, it looks much like the one below except the two members are of equal thickness and there are external forces pulling the members apart. I typed an exact quote of the summarizing the findings. Thank you.

 

[Baluncore]

I believe it is all to do with understanding the implied direction of the applied forces. Here is the problem...

Problem 4-5
Consider a bolted connection shown in Figure 1, in which the stiffness coefficients are given as Km ≈ 6 Kb. If the connection is preloaded with Pi = 1100 lbs. and the members are externally loaded with Fe = 1200 lbs., find the resultant compression of the machine parts and the tension of the bolt.

In this problem, it can be seen that the proportion of the load shared by the bolt is very small because the stiffness of the bolt is low when compared to the members. The result is that the machine parts are still in compression even after applying an external load greater than the preload. Therefore, there is no separation of the parts.

 

[jack action]

"In this problem, it can be seen that the proportion of the load shared by the bolt is very small because the stiffness of the bolt is low when compared to the members."

The load shared by the bolt is:
$$\frac{K_b f_e}{K_b+K_m} = 171.43\text{ lb}$$
The load shared by the members:
$$\frac{K_m f_e}{K_b+K_m} = 1028.57\text{ lb}$$
Thus:
$$171.43\text{ lb} < 1028.57\text{ lb}$$

 

[Juanda]

I was also confused by this. Here is a related thread.
Post #6 from that thread contains the insight that made it click for me.

I think I finally got an intuition about why that happens.
When the external separation force is applied, the tension force on the bolt must increase but simultaneously, the force the clamped parts were doing on the bolt diminishes. The net effect on the bolt depends on the stiffness ratio of the joint.
In the case of the clamped parts being WAY stiffer than the bolt, it is possible to increase the external force until it surpasses the preload so it causes gapping without almost affecting the bolt. At the instant where gapping occurs, all the external force travels exclusively through the bolt but before that, the bolt did not see a significant increment in tension.