Calculating soundboard rigidity and brace stress

[GregHolmberg]

I'm new to this whole engineering thing, so I'm not sure what I'm doing when calculating soundboard rigidity and brace stress, as in Design section 4.4.

Unfortunately, calculating the net second moment of area for the soundboard, Is, and flexural rigidity, EIs, was left as an exercise for the reader.

I tried to calculate for Fig. 4.4-19, and I got a neutral axis x = 3.20 mm, Is = 1.072E-08 m^4, and EIs = 128.7 Nm^2 for Sitka Spruce (E=12.00 GPa).

If I understand Table 4.4-2, the correct answers are x = 2.84 mm, Is = 1.06E-08 m^4, and EIs = 127.2 Nm^2.

Has anyone else tried this? Can we compare answers? Here's my spreadsheet.

I'd like to make sure I'm doing it right before I try to apply it to a new design.

Or try to find a replacement for King Billy Pine in the US. Alaskan Yellow Cedar? Northern White Cedar?

By the way, I maintain a table of wood properties on Wikipedia, which you may find useful.

Thanks,

Greg

[seeaxe]

Here you go.

I cannot see your spreadsheet but I amended my own to match the figure and got the same results as in Tab 4.4-2. Perhaps check your understanding of the parallel axis theorem?

If you want me to send me your email in a pm Ill happily send this to you, with the usual caution about using other people's spreadsheets!!

Having done this its interesting to see the importance of those triangular bits on the top of the main braces. They contribute 58 percent of the I value, because they are the farthest away from the neutral axis.

its also interesting to note that we ignore the curvature on the top in this calculation. I haven't ever checked the effect of that but I will one day. I suspect it wouldn't take much curvature before that became significant. For another day (night)

Cheers
Richard

[GregHolmberg]

Thanks, Richard!

I fixed the permissions on the spreadsheet, you should be able to see it now.

Look at the "Braces" tab.

I need to figure out what the curvature adds to the rigidity, as I plan to make an archtop with a top cylinder radius of 2 m perpendicular to the grain. I think the formula in Section 1.6.5 can help, along with ω = SQRT(K/m) and K = F/y.

Greg

[GregHolmberg]

OK, thanks to Richard, I found my errors. I'd failed to multiply by the quantity of braces (2) in some places. I now get the same numbers as in the book, Table 4.4-2.

I'm looking for alternative species to use for braces. So I entered the 46 lightest species from my list on wikipedia (taken from the wood database), from Balsa (150 kg/m^3) to Australian Blackwood (640), and calculated the peak stress for each. As in the book, Sitka Spruce is the standard (E=12), and the others have their base (width) dimension altered to be equivalent.

You can see all the species in the spreadsheet, sorted by density.

King Billy Pine is indeed the lightest species that doesn't exceed the allowable stress.

Next are Western Red Cedar and Obeche. Although at about 80%, and with the variability in samples, these might best be avoided. Gore used different values for WRC and found it to exceed the limit.

Next are Engelmann Spruce, Sugar Pine, Eastern White Pine, Norway (European) Spruce, Basswood, Redwood, and so on.

In fact, above a density of 350 kg/m^3, the only species that exceeds its stress limit is Mediterranean Cypress.

I suppose that the next thing to figure out is which species are bendable. I'm not sure how to figure that out, so if any of you out there have experience bending any of these species, let me know.


Greg

[seeaxe]

Watch out for shear stresses as well as bending stress. I recall something in the books about balsa failing in shear well before the flexural limit was reached.

[GregHolmberg]

OK, now that I've calculated the rigidity (EI) of the soundboard, it seems like it should be possible the calculate the deflection (y) as a function of the force (F) applied, and therefore the spring constant, K = F/y.

In Equ. 1.3-1 Gore shows the formula for deflection of a beam:

Code: Select all

y = FL^3 / 48EI

Of course, a soundboard is not a beam, and using this formula K = F/y = 48EI/L^3, for a Sitka panel with King Billy Pine braces (EI = 70.1 and L = 0.280), I get K = 153,378 N/m. Which is a lot more than the 42,700 used for the classical in Table 2.4-1, or the 48,000 I estimate for a medium steel-string.

On the other hand, if I increase the width of the panel from 0.280 to 0.390 (as in a medium-sized guitar), I get K = 56,760. So maybe the formula for a beam can work? Not sure.

Any engineers out there? For me, this is uncharted waters.

Greg

[GregHolmberg]

Watch out for shear stresses as well as bending stress. I recall something in the books about balsa failing in shear well before the flexural limit was reached.

Yes, this is discussed in section 4.4.6 (commonly referred to G), but it wasn't shown how to calculate or measure it.

There was a formula for G in a panel on page 4-59, but I don't think that would apply to a beam (i.e. a brace).

At the top of page 4-48, he says that due to the increased torque of the steel strings, the shear strength of balsa in a balsa/CF composite would be exceeded by 2X, and you should use KWP+CF or WRC+CF. But again he doesn't show how to calculate this.

Greg