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'[OT]: Mechanical Strength of a Bolt'
2006\01\30@063306 by

I need to know the strength of a bolt.

It is rated 10.9, so i think it's strength is 1000N/mm2 when pulling it.

If i use this bolt to join 2 elements, which try to shift against each
other, how do i calculate the maximum force it can stand?

^
|  F
|
111
111
111222
111222
111222
BBBBBBBBBBBBBBB
BBBBBBBBBBBBBBB
111222
111222
111222
222
222
222
|
|  F
V

B = bolt
1 = element 1 pulling up
2 = element 2 pulling down
F = maximum shearing foce it can stand

thanx!
tino

{Quote hidden}

A quick google reveals that a rule of thumb is that shear strength = 0.6 * tensile strength.  Note that in most assemblies, the plates will be clamped together and (at least some) shear stress will be taken by the friction between the plates.

Regards

Mike

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0.6 seems to be good for practical use.
unfortunately, i need to get the exact value for an M20 shaft screw,
with a shaft diameter of 21mm.
googling around, i wasn't able to get any info on something like that!
tino

************************************************************************
******************************

>{Original Message removed}
That is not very easy to answer.
The maximum torque you can use depends on the size, the surface of the
material you are bolting together, thread type etc.
The easiset way to find it is to look in the manufactores catalogue, here is
an example, in Swedish though, but it should be understandable
http://www.bulten-stainless.se/teknisk_3.htm

If you look at the top "Åtdragningsmoment Mv I Nm" recommended torque for
well oiled straight  surfaces.
Bumax109 is 10.9

"Brottkraft" is the force to break the bolt in kN
"Sträckkraft" is the force to stretch the bolt (making it longer) before it
breaks.

The sheer force depends on the bolt itself and the material you are bolting
together, although the bolt might take the sheer, it is not necessary that
the material will do.

You could probably use the formula for a rivet Ts=F/(n x (pi x d*d)/4)

Where Ts (tau-s) is the force in N/mm2, n number of surfaces, d is hole
diameter.

One more thing that complicates is the friction between the various
surfaces.

With best regards

Sweden

Verus Amicus Est Tamquam Alter Idem

> {Original Message removed}
Though I have numerous engineering books in the office here
I couldn't find the answer. The only answer on the web I could
come up with is 30% of the ultimate tensile stress.
http://www.bigcee.com/reference.html He/she claims
to have data from the AIAA Design Handbook for Aerospace Engineers
of which I don't have a copy

YMMV
Peter van Hoof

----- Original Message ----
From: "Buehler, Martin" Martin.Buehlerkeymile.com

0.6 seems to be good for practical use.
unfortunately, i need to get the exact value for an M20 shaft screw,
with a shaft diameter of 21mm.
googling around, i wasn't able to get any info on something like that!
tino
Mike,

On Mon, 30 Jan 2006 11:55:27 -0000, Michael Rigby-Jones wrote:
>...
>  Note that in most assemblies, the plates will be clamped together and (at least some) shear stress will be
taken by the friction between the plates.

Indeed, up to and including all of it!  Steel-framed buildings are generally designed such that the bolts hold
the plates together, creating friction which takes the shearing loads.  Otherwise the bolts at the bottom of
(say) the Empire State Building would be taking the weight of the whole building, which is quite a tough job
for a bolt!  :-)

Cheers,

Howard Winter
St.Albans, England

> 0.6 seems to be good for practical use.
> unfortunately, i need to get the exact value for an M20 shaft screw,
> with a shaft diameter of 21mm.
> googling around, i wasn't able to get any info on something like
> that!

There *IS NO* exact value.
Reality varies.
What safety factor are you using?
Why?

RM

Russell McMahon wrote:

>> 0.6 seems to be good for practical use. unfortunately, i need to get the
>> exact value for an M20 shaft screw, with a shaft diameter of 21mm.
>> googling around, i wasn't able to get any info on something like that!
>
> There *IS NO* exact value. Reality varies. What safety factor are you
> using? Why?

Intuitively, I'd also say that it depends on the thickness of the pieces 1
and 2 (even though that's then not exactly shear stress, but they are shown
thin enough to possibly make this a factor).

BTW, where do you measure a diameter of 21 mm for an M20 thread? (It's
called M20 for a reason... :)  Check out thread geometry here for example
http://euler9.tripod.com/bolt-database/23.html

Gerhard

Sorry if I already posted this...

from the aerospace structural side of it....

1.  Get the maximum single shear strength of the fastener from the  vendor if possible.  If this data is not available, find the  material the fastener is made of, and obtain its Fsu (utimate shear  stress allowable, b-basis).  Multiply this by the fasteners cross  sectional area: P = Fsu *  pi * r^2 .  This will give you the  ULTIMATE single shear allowable of the fastener.

2.  Find the Fbru (ultimate bearing allowable) for the  sheet/plates material. Assuming the center of the fastener/hole   is at least 2 X D (diameter of the fastener) away from the edge,  multiply Fbru times the sheet/plate thickness times the fastener  diameter:  P = Fbru * t * D.  This will give the maximum  ULTIMATE bearing load for the sheet/plate.  In the case of the  plates are of different matierals or thickness, calculate the lesser of  the two allowables.
If the edge distance < 2D then things get a little more complicated  in that you must know (obtain/find) Fbru for another value of edge  distance to hole diameter ratio, then linearly interpolate between the  two know values.

The ultimate strength of the joint is the lesser of the two.

A few qualifiers:

If the E/D ratio gets much less than 1.5D then the mode of failure in  the sheet/plate is no longer bearing.  The mode ends up being a  combination of tear-out and shear. The calc is somewhat complicated  (though not overly so), but does require that you know a bit more about  both the geometry and properties of the materials.

Things get a bit more complicated also if the plates get too  thick.  Because the loading axes of the plates are not colinear  and the plates have finite thickness, the fastener will necessarily be  in bending.  The allowable amount of bending moment of a  fastener/joint is dependent upon a lot of different variables and is  usual obtained through test of the given material  system/geometry.   Having said that, if the part thicknesses  are less that the diameter (+/- a bit) of the fastener then pin bending  will probably not be an issue (THIS IS NOT A HARD AND FAST RULE!!!).

Also, at ulimate load, because of the degree of relative straining and  moduli/comliance (stiffness) of the associated parts in the joint ,  friction is generally not an issue or considered.

Once again, this is from the viewpoint of an aerospace structural engineer... ymmv

mark rewis wrote:

>   from the aerospace structural side of it....

Should be good enough for most anybody... :)  Thanks!

> Also, at ulimate load, because of the degree of relative straining and
> moduli/comliance (stiffness) of the associated parts in the joint ,
> friction is generally not an issue or considered.

Can you please explain this a bit? Does this mean that when we operate in a
range where shearing is an issue, the shear forces are so strong that the
material friction is not relevant? Is this dependent on the materials in
use?

Thanks,
Gerhard

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