Help on shear rating calc

I need to calculate the shear rating for a bolt being pulled across a glassed core. Not out, but across.

the bolt is 3/8-16" and would have a dimension in the core of pi•0.375"/2, or 0.589"

the core is corecell M200 with a shear alone of 428psi; thickness is 12mm or 0.472" no skin

the skins are triaxial, about 850-900gsm, sorry about the units mixing, vacuum bagged and epoxy and post cured

I can do the calc, but have no way to know how the skin contributes to the rating. I am sure it is pretty significant.

Again, this is for a bolt being pulled through the core n skin and not being pulled out of or into..this the estimate of half the circumference of the bolt.

I am running this calc to establish bridle eye fixture bolting..

Thanks. Dan

 

Hi Dan,

Im a bit confused. If the bolt is being pulled through....what do you mean by not being pulled out?
As this is the same axis as the applied load?

 

The load will be from the anchor bridle and pulled mostly forward. I am going to build a fixture like Barry advised, but need to know that I won't tear my core sideways or in the same plane as the core.

 

Pink - hull
Green - bolt(s)
Yellow - fixture
Blue - force, mostly forward anchoe bridle

 

If I 45/90 degree the bridle, the force direction changes, but I have to spec the bridle yet and planned to go much further out..

 

You appear to be over complicating matters.

It is a simple pull through the material, like so:



The shear area, is the foam depth x length to end of the foam.

So, in this case it is 12mm x "h", where "h" is this:



You then need to decide if the load path is single shear or double shear.
If it is a tight fit, it'll be double shear.

That's it.

 

I want to know the amount of force required to pull this fixture through the laminate..



The intent is to add a bracket to the bottom of the fixture for a bridle rope attachment. I realize the trampoline and beam provide opposing force, but I don't care. The plate supporting the netting beam is only held in place by four 3/8-16 bolts.



Here is a very crude sketch. Credits to @Barry for the concept, not my bad art. The 90 is hard to draw as is the gusset. If I attach an anchor line to this I want to know how much force it'll tolerate before I damage the boat. Again, I realize the opposing forces. The trampoline is imperfect because it is not absolutely tight. The anchor would pull on the bracket.



Here is Barry's sketch. No scaling. My drawing is scaled a bit better.

 

I did calculate the shear rating of the hardware at 2373# each. Since the bridle shares the load; the bolts are enough. But I believe the bolts are stronger than the laminate.

right now, I only have a backing plate of 6mm okume inside, and probably need to make something better than that as well

the load would be applying compression to the laminate; so it is possible I have to use compression data; the compressive strength of the foam is 638psi. I'm sure you love the units...but the glass and epoxy also must add

 

Dan

What you show is a quantum leap from your opening question.
The fact that you have a member attached to the ends of the hull of a multihull, it will be subjected to many differing load scenarios.

If the attachment is as you show... not much!...since the danger is that as soon as you try to stiffen up that joint by making it "fixed" you start attracting all the global loads.

 

Here is another catamaran doing same. Their plate is quite a bit beefier than mine. And so are their four fastenings. But I never added the anchor to the thing until now. And the boat might be 2x as big..

It is a new prototype and probably ought to have asked Richard sooner. But noob here. Originally figured the anchor was off the bow even.

 

Dan,

The question is back to front.
It is not a case of how much for or what is the shear strength. The question is, what forces will that joint be subjected to....then and only then can you design the joint to be fit for purpose.
Since you calculate the joint will fail at 1 tonne or 10 tonne..it makes little difference if the load applied is greater than either.
Thus, you need to establish what the force it will be subjected to, before doing anything else.

 

It is an auxiliary beam and the purpose is to support the forward trampoline. It is not doing anything for the boat et al. The beam is not even bolted to the fixture and is allowed to float. Sorry, just saw the note.

 

If you think there is no load... what is the worry about then?

 

oh, I meant no load related to cabins n such; the beam is there for the netting which could have a people load

I asked Richard if the netting beam was keeping the hulls from separating because the beam is NOT attached to the hulls/fixture mechanically other than the clamps. He said it was only supporting the netting. I was nervous about this and intended to run a bolt through..

If I come off as chippy; not so. I am genuinely worried about damaging the boat with the anchor someday if I hook it to the fixture.

 

Then make it a weak joint, so when/if it fails, it does no damage to the hull.