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