Limber hole size rules and frame strength?

@Steelboat,

I updated the spreadsheet for the updated parameters.

I cannot make the framing work.
The stringer combined with the hull plating is only good for one frame spacing.

Amidships, are the frames resized, doubled up or is the frame spacing reduced?
I am still hoping for a significant chine stringer.

AlanX

 

Hole/notch, et al... make no difference...it is an opening in a web.

Don't follow... care to explain?

As long as you don't exceed R25mm, as previously noted, you'll be fine.
Just ensure good return on the welds.

 

After looking a bit I see frame to frame spacing varies from 23-28". All the frames look mostly the same, with the exception of the forward and aft engine room bulkheads. Those are steel and quite heavy.

At the intersection of each chine, there is a 3/4" solid bar. The hull has three chines. Is that a chine stringer?

 

Hi Ad Hoc,

To estimating the loss of section modulus of a fixed ends rectangular member with a notch located at the maximum bending moment (at one of the ends):

• Lets say we have a clean 70 mm high (h) and 9.5 mm wide (w) member and we reduce this to 45 mm height (h') by notching 25mm.
• This is a case where we have placed the notch at the "extreme fiber", where it has the greatest impact on the section modulus.
• Lets ignore the effect of stress concentrations here.
• Lets use SM=h^3*w/6.

The reduction of section modulus (strength) and be estimated as ~(h'/h)^2 or ~41%.

If the hull plating is included then the section is a T with a hole in the web adjacent to the flange (or the hull plating).
The width of the flange of the T would be limited to the effective width of the flange (depending on the thickness of the flange and the material etc).
The hole is near the centroid of the T so it has a much reduced impact on the section modulus of the T.
I am not going to work-out the formula for this here.

Unless I miss-understood, are you saying that I should include the hull plating in the section modulus calculation even though there is a significant gap between the hull plating and the frame?

Regards AlanX

 

As I suspected, we are talking at cross purposes and different definitions.

My question is relating to what you are referring to as...."extreme fibre"....

 

Ad Hoc,

Haha, "Extreme Fiber" is just a text-book term for identifying the location on a section that hits the maximum stress first.

I checked google and it is still used by Section modulus - Wikipedia https://en.wikipedia.org/wiki/Section_modulus .
Regards AlanX

 

AlanX

I suspected that was the case. Not seen it used this way since Uni days and very old text books from way way back.
We tend not to use the term "extreme fires" in Naval Arch structural design. Ostensibly because 'fibres' refer to composite structures (like carbon fibre etc), and not isotropic materials which do not have "fibres" per se.

It is generally denoted as the distance form its neutral axis, and if as you cite at the extreme fibre, we just note at max distance from the NA.

However, that, back to my post, this modulus aspect has little relevance in this context. Since the problem is all about the shear, not bending stress. That's what Class rules do not allow greater than 50% depth for cut outs etc....simple bending theory points to why too.

 

@Steelboat

I have to conclude that the internal wooden bulkheads and possibly other "furniture fixtures" are structural.
I am not going to calculate the maximum bulk-head spacing as the section modulus calculation for the Chine Stringers is a bit tricky.

I have revised the draft canoe body to 0.5 m based on your drawing which reduces the design pressure to 17.5 kPa.

I have attached my workings for your reference.

I am quite confident you can enlarge the limber holes to 25 mm radius.

Regards AlanX

 

Ad Hoc,

I agree "extreme fires" (haha) or "extreme fiber/fibre" may be less than an ideal, that was the term that was taught in the Statics Unit back then (40 years ago).
---
Shear vs Bending Moment in Steel Fixed Beams with Uniform Distributed Loading

Never had a problem with shear in a beam design. But okay lets have a look:

• For a both fixed ends and a uniformly load beam (what we usually examine here) the equations for the maximum shear and bending moment at the ends are:

Lets look at the beam equations:

• Vm = p*Lv/2 for Shear
• Mm = p*Lm*Lm/12 for Bending (fixed ends)
• p is the load per m (N/m)

The strength equations for steel (the safety factors) came from AS1250:

• Vm = 0.45*Fy*(W*H) for Shear
• Mm = 0.66*Fy*(W*H*H/6) for bending
• Fy is yield stress (Pa), H is beam height (m) and W is beam width (m)

Rearrange both equation:

• Lv = 0.45*2*Fy*W*H/p
• Lm = sqrt(0.66*2*Fy*W*H*H/p)

Extract a non-dimensional number:

• Lv/Lm = 0.9/sqrt(1.32)*sqrt(Fy*W/p)
• = 0.7833*sqrt(Fy*W/p)

With the following parameters:

• Fy = 250000000 N/m^2
• W = 0.0095 m
• p = 17500 Pa/m^2 * 0.66 m (panel spacing)
• = 11550 N/m

We get:

• Lv/Lm = 11.23

Which means in this beam Lv would need to be 11.23 times longer for shear to be critical. Or Lm is critical long before Lv.

Notes:

• Only section width (W) and load (p) can materially change the Lv/Lm ratio for steel.
• For shear to be important the load (p) has to be very high (i.e. for very short beams) or the width is very small (not very likely as the beam would warp/buckle first).
• We can workout how short the beam has to be for the shear and moment to be both critical at the same time. The load per metre needs to be 1.457 MN/m and the beam length 0.103 m.
• Now I am not saying you should discard your design rules or guidelines.

Regards AlanX

 

Thanks for this, which also explains it.
Interesting code, but i'm sure you realise boats are built according to Class rules, not rules for bridges buildings etc., and have different caveats and fudge factors.

Class (for marine vessels) uses different means at arriving at a minimum scantling for structures.
It takes into account, fixity, material and filler wire. One can of course use first principles, but that requires more caveats to justify to the regulatory body that is approving the structure.

But, the original question from the Op... the %'age cutout - of a web - using R25 is suffice.

 

Hi Ad Hoc,

Okay, understood.

AlanX