Tensile Load Distribution in Fastener Group 4

[happypap]

Hi all,

I am familiar with the classic methodology presented in Bruhn and Niu to distribute a shear load among a group of fasteners, but I have never found a reference which explains what to do to distribute a tensile (out-of-plane) load. Can someone please point me to a good reference?

Thanks
Pap

 

[Sparweb]

Two ways to approach it:

1) assumptions that allow you to analyze it with statics. (Be conservative)
2) finite element analysis either using a matrix of deflections or a FEM software package.



Steven Fahey, CET
"Simplicate, and add more lightness" - Bill Stout

 

[desertfox]

Hi happypap

Take a look at this thread I have just outlined a method of
tensile loading in bolts for an offset load :-

THREAD 404-101797

regards desertfox

 

[happypap]

Thanks for the replies.

Imagine a circular plate with many bolts attached around the perimeter. Clearly if I apply a tensile load to the center of the plate all the bolts will see the approximately same load (unless of course the stiffness of the structure the plate is attached to is not uniform). As I move the load away from the center the bolts closer to the application of the load will see a larger percentage of the force since the system is redundant. What I am looking for is the approximate relation between the distance from the force and the percentage of load, i.e. in a similar fashion to the classic distribution of inplane shear and moments described in bruhn and niu.

Thanks
Pap

 

[Bmtaylor]

Happypap,

I would treat this as a rigid body and use a static to resolve the reactions.

Brian

 

[Lcubed]

Hi Desertfox, How do I get to the thread you have posted?
Lcubed

 

[Lcubed]

Happypap,
Label fasteners A, B, C…...
Calculate distance from load to each fastener
Let the distance to A be a, to B be b, etc.
Square the distances a, b, etc., and sum the squares; call the sum of the squared distances K

Divide K by each of the distances, and sum the results; call the sum M

The load at a is Load P x (K/a) / M, the load at b is Load P x (K/b) / M, etc.
or PK/aM, PK/bM, etc.
Regards,
L. L. Lawson

 

[aviat]

Lcubed:
You can use the search at top of page (change to find a thread) or
thread404-101797

 

[desertfox]

Hi Lcubed

Looks like aviat showed you how to get to that thread for which I thank him.

Lcubed I tried your method on my example on the above thread
and I don't get the same answers as I did following my method. I thought at first that your method was the same as mine but it clearly isn't, my method comes from a book called "mechanical engineering design" by G.D. Redford and
could be applied to happypaps problem.
All that is required is that the square of the distances to
each bolt line from the point at which the lid will pivot due to the load offset needs to be calculated and if there are 2 fixings on the same line then 2 goes in front of the unit load (u) and the square of the distance (la)etc.
Lcubed if you apply your method to the problem in the earlier thread quoted above the bolts nearest the pivot would carry most of the load where in actual fact under these conditions the bolts furthest away from the pivot should see the most load. This is also true of happypaps situation (as he states in his post)as the load on the lid moves closer to some of the fixings they see a greater load than those further away that is because the pivot point for the lid is at the farthest point directly opposite the offset load.

regards desertfox

 

[Lcubed]

Hi desertfox,
We appear to use the same method where a cantilever beam is concerned. Moment times distance from the pivot to the first row of fasteners divided by the sum of the squares of the distances from the pivot to the rows of fasteners, plus moment times distance from the pivot to the second row of fasteners divided by the sum of the squares of the distances from the pivot to the rows of fasteners, etc., all adjusted for the number of fasteners in each row.

But I understood that, in the case at hand, the load was not overhung. Looking at happypap's last entry above, I get the impression that the load is inside a pattern of fasteners and somewhat removed from the centroid of the fastener pattern. The method I posted would work for this problem, and the fasteners nearest the load would pick up more load than those farther away, but I haven't tried to relate this method to the overhung load type of problem. Please let me know if I misunderstood the problem.

happypap, your input??.

Thank you, desertfox, for the solution in your referenced thread, and thanks to Aviat for showing me how to access it.
Regards,
Lcubed

 

[desertfox]

Hi Lcubed

I found an error in the solution to my thread reference which I have now corrected ie I didn't square the distances
between pivot and bolt lines.

Still struggling to get the right results from your method
into my example though, with your method how do you check
equilibruim of the moments as you state to square the distances from the load to each bolt line, if the load on the circular lid is offset then there must be a pivot point
or am I missing something?

regards desertfox

 

[Lcubed]

Hi desertfox,
Hmmmm!?!? I'll need to run some numbers and refresh my memory about this. There is force balance; I have that in my spreadsheet as a check, but I'll need to check on the moment balance. I have not used this equation for a few years, so I don't have a snappy answer. I'll get back to you.

I am not sure my solution will work at all for a cantilever beam; it seems that it should, but I don't recall that I ever tried it on that kind of problem. As I stated earlier, I understood the load in this case to be within a fastener pattern. I know that there is moment, of course, but I've not dealt with it using this method.
I'll work up an answer and respond soon.
Regards,
Lcubed

 

[desertfox]

Hi Lcubed

Thanks for your response.

The way I see it is if your dealing with the distance between external force and bolt line(s) then it should not
matter what shape the plate is or whether the load is within
the fastener pattern or outside it.
I decided to use both methods on a rectangular plate 6" x 4"
with fixing centres of 0.5" , 3.0" , 5.5" from one end of the long side of the plate with 2 holes on each row, the holes on each row being 3" apart equally distributed on the 4" wide dimension.
I now place a 10lb load acting vertically upwards at a distance 4.25" from one end of the 6" dimension and centrally within the 4" dimension ie (2").

using my method :-

10*4.25= 2*u*(la^2+lb^2+lc^2)

42.5= 2*u*(0.5^2+3^2+5.5^2)

finding u = .5379746855

this gives individual bolt loads in each row of

la= .269lb

lb= 1.614lb

lc= 2.959lb

multiply each force by 2 and sum comes to 9.684 lb force
and it should be 10lb but the error is due to rounding.

Using your method I get the following:-

la = 3.75"

lb = 1.25"

lc = 1.25"

K= 17.1875 and M= 32.083333

using your formula I get the bolt load per row=

la = 1.428lb

lb = 4.285lb

lc = 4.285lb

if you sum these the force balance is 9.998lb very close to 10lb again error due to rounding.

However if you divide these figures by 2 to get individual bolt loads then a bolt in row :-

A=0.714lb

B=2.1425lb

C=2.1425lb

So you can see there are differences in the bolt loads which
to me should not be present and both methods should give the same results. With the first method there is a force and moment balance, with the latter there is only a force balance which doesn't seem quite right to me.

Best Regards for now

Desertfox

 

[arto]

The above methods are in Grampa's "Machine Design" by Maleev from the 1940's - assume a rigid base, with bolt force proportional to distance from the pivot point.

 

[desertfox]

Hi arto

Yes we have assumed a rigid base and bolt loads proportional
to the distance from the pivot point.
What I am trying to establish with Lcubed's post in which he takes distances from the external loads to the bolt fixing centres is how he establishes the sum of the moments to be zero. Look at the example of the 2 methods above in the latter method I can sum forces to zero but not the moments, in the earlier method we can establish both moment and force equilibrium.

regards desertfox

 

[Lcubed]

Hi desert fox,
Sorry to be out of touch for so long. I've been traveling, and now I'm trying to catch up with a lot of things. I quickly sketched up a square 4 x 4 plate with fasteners at the four corners and a load P at the center, then moved the load down 1.0 inch. I got loads on the top two fasteners of .3377P and on the bottom two fasteners of .1623P. Taking moments about the horizontal axis through the center, where the load was originally, I got moment balance; i.e., the moment about the axis due to the offset load was 1 P, and 2 x .3377P + 2 x .1623P adds up to exactly that. I haven't taken time to do a more complex problem, nor have I tried to relate this approach to the method which we both seem to use for a row of fasteners extending away from a pivot point, but I think this approach will always be correct. Let me know if you still have problems with it.
Cheers,
Lcubed

 

[desertfox]

Hi Lcubed

Two things here the exanple you just posted you take moments about the centreline of the plate however if you move the load off centre it will pivot not about the plate centreline but the edge furthest away from the load, secondly your original method that I questioned suggests that you use the distance between the load and each row of fasteners. LoL and thirdly if you look at my example in my last post both methods if correct should give the same answer.

 

[desertfox]

Hi Lcubed

I think I have found the answer, your original method will work ie taking the distance between load and fixing centreline but only if there's only two rows of fixings, take a look at this thread:-Thread404-53778.
This is why my last example shows differences because I used more than two rows of fixings.

regards

desertfox