On 1 Sep 2006 09:10:02 -0700, "Bob" wrote:

Hi Everyone:

How much pull can a 316 SS PH Machine Screw ¼ x 20 handle before it
fails?

Basically I want to know how many fasters I need to use to secure some
deck hardware.

///
Bob

Annealed 316 wire yields at a pressure of 34,800 lb.
[see Matweb]

The cross section area of a 1/4 in X 20 tpi screw is about
pi X r X r square inches where r is the effective radius.
Using r = 0.06 inch gives a = 0.011 sq in
so the force at yield is about 0.011 X 34800 lb
That's about 390 lb.
The bolt will break before the nut.
You could give a X5 safety factor and take the maximum
unit load as 75 lb each.

You're welcome

Brian Whatcott Altus OK

Brian Whatcott wrote:

Annealed 316 wire yields at a pressure of 34,800 lb.
[see Matweb]

The cross section area of a 1/4 in X 20 tpi screw is about
pi X r X r square inches where r is the effective radius.
Using r = 0.06 inch gives a = 0.011 sq in
so the force at yield is about 0.011 X 34800 lb
That's about 390 lb.
The bolt will break before the nut.
You could give a X5 safety factor and take the maximum
unit load as 75 lb each.

You're welcome

Brian Whatcott Altus OK


Brian,

Thanks for walking me through that. Seems straight forward with your
explination.
Now time for me to get back to the boat.

Bob

Bob wrote:
Brian Whatcott wrote:


Annealed 316 wire yields at a pressure of 34,800 lb.
[see Matweb]


The cross section area of a 1/4 in X 20 tpi screw is about
pi X r X r square inches where r is the effective radius.
Using r = 0.06 inch gives a = 0.011 sq in
so the force at yield is about 0.011 X 34800 lb
That's about 390 lb.
The bolt will break before the nut.
You could give a X5 safety factor and take the maximum
unit load as 75 lb each.

You're welcome

Brian Whatcott Altus OK



Brian,

Thanks for walking me through that. Seems straight forward with your
explination.
Now time for me to get back to the boat.

Bob


I think this whole analysis, while very good, leaves out an important
part: the threads. When screwed into fiberglass or wood (or perhaps
even aluminum, depending on the length of thread engagement), the
failure mode will be to rip the threads out of the hole, rather than to
break the screw body in tension.

Also, computing the max load on the assumption that it is indeed the
body which fails (like for example, if the screw were to be installed
into a stainless plate), fails to take into account the stress riser
effect of the threads. Because of this it is likely that the screw will
fail well before the calculated value of 390 lb. Thus the 5x safety
factor based on the theoretical which Brian recommended...

bob

The root diameter of a 1/4-20 screw is .201 inches and the min. yield
strength is (18-8 SS screw) 40,000 psi which, when combined with a safety
factor of 5 gives 254 lb safe load (depending on what is screwed into).

On Tue, 5 Sep 2006 19:15:51 -0400, "Dave W"
wrote:

The root diameter of a 1/4-20 screw is .201 inches ...


Oh really? :-)

Brian Whatcott Altus OK

Sure is.... according to my Starrett tap and drill guide. I was a little
surprised so I measured some and found it to be correct.
Dave
"Brian Whatcott" wrote in message
...
On Tue, 5 Sep 2006 19:15:51 -0400, "Dave W"
wrote:

The root diameter of a 1/4-20 screw is .201 inches ...


Oh really? :-)

Brian Whatcott Altus OK

On Wed, 6 Sep 2006 09:01:28 -0400, "Dave W"
wrote:

Sure is.... according to my Starrett tap and drill guide. I was a little
surprised so I measured some and found it to be correct.
Dave
"Brian Whatcott" wrote in message
.. .
On Tue, 5 Sep 2006 19:15:51 -0400, "Dave W"
wrote:

The root diameter of a 1/4-20 screw is .201 inches ...


Oh really? :-)

Brian Whatcott Altus OK


Hmmm...I'm getting an uneasy feeling that I am being pulled
in further than I want to go. Ah well: though machine screw series
are designed to break at the bolt, the failure mechanism isn't quite
via yield at some effective diameter as I suggested.

A slightly better model is to visualize shear failure at the thread
root, so this calls for computing an effective area held by the nut
times the material shear ultimate - but this ignores (at least) two
other factors:
1) you may have seen a model acrylic bolt section stressed at the
threads, and viewed through polarizers - the shading highlights the X2
to X3 stress multiplication caused by the thread root form. This is
where the cracks start.
2) fatigue life is dependent on a smooth transistion between threads
and the shank of the fixing - this too can have a /2 factor on
fatigue life.

So, the carry home message is, nobody ever got jailed for misconduct
for being either too conservative, or for knowing exactly what they
are doing if they design right to the yield limit (as aero engineers
often need to do.) But people do die if the engineering is skimped.
The choice is yours, as always.
I am of course joking about engineers being imprisoned for screwing
up - it just doesn't happen - not in the US at least, as far as know.

I think it would probably be better to pursue this kind of topic on a
group like sci.engr.mech or sci.engr.civil if you want more
Or, you can find perfectly respectable allowable or ultimate
force tables for screws of many sizes and types in web-accessible
form. Probably more than a sailor wants or needs?

Respectfully

Brian Whatcott Altus OK