Deleted gearcase thread issue 8

[preload]

On Mike's suggestion, I have asked admin of the forum to delete the thread. But we can discuss that here

Desert forx post
"What grade of bolt is it ie whats its proof load? img511 and 514 don't work.
I assume the bolts are in tapped holes?
Could be a number of things like you have already mentioned
vibration;cyclic loading.
If the bolt pre-load is not high enough the bolts can loosen
off under vibration and subsquently fail in fatigue.
Have you any of the failed screws? if so post a pic.
In addition if your truely getting 80% - 90% of proof load on all bolts it doesn't give you much room for increasing
bolt tension unless you use a higher strength material.
Finally the bolts will not see equal tension , shear etc
for a given engine position: but depending on engine position and external force position at a particular point in time the bolts will share the load unequally.
"

Yes the bolts are in tapped holes and the holes are blind. Bolts dont break, they just come loose or missing.

6 Bolts,joint and proof load info

4 perimeter bolts (Bolt 1-4) – 3/8-16 (1.75 in) unc-2A stainless steel plated(tq spec – 27 ft-lb target)-min proof load - 8370 lbs
1 bolt in center- 3/8-16 (3.5 in) unc-2A stainless steel(tq spec – 27 ft-lb target)-min proof load -8716
1 bolt in center– 7/16-14 (3.5 in) unc-2A stainless steel(tq spec – 47 ft-lb target)-min proof load - 9567


Desertfox you saw one of the pic which is been deleted now right.Did that give u an idea about the joint?

 

[desertfox]

Hi dimjim

Thanks I see my error now I got J =1.656 but I then multiplied the actual engagement length instead of the engagement length obtained from the formula with equal strength materials.
Looks like the engagement lengths aren't to bad then except for the 3/8" centre bolt, I'll check the 7/16" bolt later.

regards

desertfox

 

[preload]

Dijim,

Your assumption of 2A and 2B fit for external and internal is right.

 

[desertfox]

Hi dimjim

There seems to be an error in the maths in machinery's,
we both agree that J is around 1.6xxx depending on whether
you use 100000 for the bolt or 102000.
Now the machinery's goes onto say that if the materials are different strength's then to calculate J as we have done and if this exceeds 1 then the engagement required is J multiplied by the value of Le obtained by formula (1) which is for thread materials of equal strength.
I have checked this several times now and if the threads were of equal strength then for the 3/8" bolt I need only 0.269" of engagement. Now multiply this figure by J and you get a figure for engaged length of 0.454" using 1.689 and using 1.656 we need 0.445", but we know that the engagement
of the 3/8" bolt is 0.596" so it doesn't add up.
I notice that you assume its 1.689*D which assumes that you need at least 1D of engagement to start with.

http://www.boltplanet.com/un/errata.htm

This site talks about some of the errors with the bolt strength calcs, scroll down the site till you see the title
"more errors in machinery's handbook"

interesting is it not?

regards

desertfox

 

[dimjim]

The apparent error in Machinery's Handbook may be considered immaterial in the following sense. Machinery's Handbook (page 1416, 25th edition) provides a formula for calculating a ratio J to be used to determine whether the nut thread will strip before the bolt thread in cases where the nut material is not as strong as the bolt material. This formula involves the nut and bolt thread shear stress areas and so involves a length of engagement. But the areas are used in ratio in this formula so that the length of engagement cancels itself. The formula provides the same result whatever length of engagement is used. In fact, we could use just the nut and bolt thread shear stress areas in a single turn of thread and get the same result. Simpler yet, we could leave the length of engagement out of the calculations altogether which would amount to calculating shear stress area per inch of length of engagement as in Blake's table.
----------------------------------------
That is interesting. You are correct in that I assumed
Le for steel on steel would be 1.00D. When I use the calculation for Le provided earlier in the Handbook I get
a value about .744D or .72D depending on which At calculation you use. If I multiply 1.689 times these
values I get 1.2566D and 1.216D. That is interesting
as I have seen values in bearing catalogue for low carbon
steel to require values of 1.25D to 1.5D for length of
engagement depending on the bolt grade. It is also interesting that the normal range of heli-coils are
either 1.00D or 1.50D for lengths and even provide
2.00D in special cases. Thanks for pointing out my
incorrect assumption. Something that looks suspect
is that in the Le calculation .50 plus .57735n() is used
whereas in the As calculation .50/n and .57735() is used.
Has this been changed in the 25th edition? I only have the
17th edition of Machinery's Handbook. Seems like an error.
I also like to use the equations for the At values as
follows:
At = .7854 (D - .9743/n)^2 and
At = .7854 (Esmin - .32476/n)^2 or see these as

At = Pi (D/2 - .9743/2n)^2
At = Pi (Esmin/2 - .16238/n)^2
just for visualizing or comparing these two formulas.
I guess since this is an Area and we tend to think of a
circular area as Pi r^2 or Pi (D^2)/4 either makes sense.

 

[desertfox]

Hi dimjim

Thanks for the response, I wasn't questioning your assumption which is I feel a valid, one for materials of the same strength.
What concerns me more is that we work out J and it leads us to believe there is not enough thread engagement however when you follow it through and multiply the thread engagement obtained from formula (1) (equal strength materials) you obtain in this case a shorter thread engagement required than what we have actually got ie 0.445"
as compared with 0.596" in practice.

regards

desertfox

 

[desertfox]

Hi dimjim again

I meant to add that the same formula are shown in 17th and 25th editions no modifications at all.


regards

desertfox

 

[desertfox]

Hi dimjim

Have a look at this site I found some more formula for thread engagement:-

http://www.utm.edu/departments/engin/lemaster/Machine Design/Lecture 28.pdf

Its late here and I may of made mistake however I used the
formula and got a thread engagement of 1.974" and its seems
quite a lot.

regards

desertfox

 

[dimjim]

That looks similar and I will run thru it.
Thanks. Will print it out and go thru the
calcs.

 

[preload]

Desertfox & Dimjim

Ultimate Tensile strength of the bolts is 120-160 Ksi.

 

[desertfox]

Hi preload

Thanks for the info

regards

desertfox

 

[preload]

Desertfox & Dimjim

The picture is here.

http://sites.mercurymarine.com/pls/portal/docs/102/4141.JPG

the middle picture is the trimmed out position (our bolts failed in this position). The middle pic shows only trimed out position of gear case, but in or application we even have lift of the gear case till just one prop blade is in water.

Regarding embedment issue, yes the dept agreed that we are embedding the joint . We will be doing some lab test to see how much embedding is caused.


PS

The plan is to strain gauge some fasteners and do field study to see whats the dynamic loads the joint is seeing. If I strain guage the bolts, the dynamic loads I get after experiment is combination of shear+tension right? But I want to know the tensile loads and shear laods separately. How do I need to set the test to get these two loads seperately.

 

[preload]

http://img.products.howstuffworks.com/cctool/PrdImg/images/pr/177X150/00/01/d1/91/c6/30511558.JPG

this is another picture which shows the engine in normal condition and the engine when it is trimmed at an angle.

 

[desertfox]

Hi preload

Thanks I will see the pics on friday.

In the meantime I am still looking at the joint.

regards

desertfox

 

[CoryPad]

If you want to know the shear and tension forces applied to the screw, then you should apply strain gages to the joint members, not the screw. Actually, applying a strain gage to the screw in addition to the parts is the best thing to do - you can see preload response to applied forces (checking for things like embedment, loosening, etc.).

Regards,

Cory

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.

 

[dimjim]

Desert fox,
I think you meant J factor equal to 1.97
rather than inches. I do feel that the value
from the Le equation is the one to use in
the other calculations as entering an actual
value of length in the Q equation would be
wrong. I think it should read Le required
Q = Le(required) times J.
I got J = 1.987 which is a ratio.
I would like to see others respond to these
equations in Machinery's Handbook under the
Bolt and Nuts section. Remember too that they
are using 2 as a safety factor against stripping
vs min cross section of the screw or bolt area.
I think the really critical issue is embedment
of the washers into the mounting surfaces as well
as the painted surfaces.

 

[preload]

Corypad,

Thanks a lot for the response. That was really helpful.
You said "you can see preload response to applied forces (checking for things like embedment, loosening, etc.)" Could you please explain me more how determining applied forces can help me understand embedment?

Dimjim,Desertfox,

Every one agreed theoretically that embedment might be one of the main causes, but how can I prove that? What kind of lab test proves embedment on my joint?
For example" measure the bearing joint hole depth and install screws at specified torque and back off the screws and re measure the hole depth precisely, and if there is change in length then embedment occurred?????" or is there any other sophisticated lab way??

 

[CoryPad]

You can look for peaks in the applied force vs. time data, then look to see if there is a corresponding fastener preload reduction after the applied force peak. This likely means that the joint has yielded/embedded.

Regards,

Cory

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.

 

[desertfox]

Hi dimjim,preload

Yes dimjim I agree the 1.987 should be the factor to multiply the diameter with.
So if we work on a factor of 2 then for all the 3/8" bolts
they should have a thread engagement of 0.75" which gives a small margin of safety.
Haven't checked the 7/16" screw yet will do that later assuming its the same material spec as the other fixings.

Preload it looks like your thread engagement could be marginally short for all the 3/8" bolts this should also be considered in the joint failure.

Dimjim I am ignoring the formula in the machinery's now and using them I found in that link.
I agree also that the paint and embedding issue is probably paramount.

regards

desertfox

 

[preload]

Desertfox,

for the perimeter 3/8 bolts we have thread engagement length of 0.707 in
for center 3/8 bolt we have 0.79 in

if 0.75 inch is the required LE then I am good except for the perimeter 3/8 right?

 

[dimjim]

9300lb/0.0775 = 120000 psi the 0.0775 square inches is the stress area for your bolt
For steels up to 100000 psi .0775 square inches is correct. As = pi/4 (D - .9743/n)^2
For steels over 100000 psi .0747 square inches is correct. As = pi (Dpe/2 - .162/n)^2
Dpe = Min. Pitch Dia. n = No. of threads per inch.

Yields 8976lb/0.0747 = 120000 psi max clamping force.


In order to help lessen the embedment, you might
want to consider using larger diameter washers to
reduce the bearing stresses below the washers.
You also might want to incorporate helical serrated
washers. This may have been mentioned before.

9300/(.7854*(0.72^2 - 0.441^2))= 36555.7 psi
9300/(.7854*(0.72^2 - 0.406^2))= 33490.7 psi

I also wonder if after 24 hours, if you retighten
the bolts, if that would help. I remember seeing
in the literature where someone was recommending
going to 3 tightening cycles. I should keep better
documentation when I am reading thru the literature.