I'm planning to attach a 6mm countersink bolt to Finnish softwood plywood. I'm worried that the wood could gradually change due to for example moisture changes over time. Thus, I would like to add some mechanism for ensuring the bolt preload stays at some reasonable level.
Finnish softwood is quite soft. I found that according to code, the perpendicular-to-grain compressive strength should be treated as 6.5 MPa (http://rmseura.tkk.fi/rmlehti/1992/nro1/RakMek_25_1_1992_3.pdf) -- and when working with plywood, perpendicular to plywood surface is perpendicular to grain.
I purchased some M6 spring washers. Their measurements are: 11 mm outside diameter, 6.2 mm inside diameter, 3.4 mm spring length, 1.4 mm spring plate thickness (so the spring washer compresses by 2 mm when fully compressed).
The contact area with wood is pi*((11/2)^2-(6.2/2)^2) = 64.842 mm2 so I can never hope to have more than 64.842*6.5 = 421.5 N of preload in the bolt.
I would like to estimate how much preload the spring washer maintains. It's a spring and the only source for spring force I found is from Wikipedia, https://en.wikipedia.org/wiki/Spring_(device) where I see the following formula:
Fmax = (E*d^4*(L-n*d)) / (16*(1+v)*(D-d)^3*n)
...where E is Young's modulus, d is the diameter of round spring wire, L-n*d is the measure of how much the spring compresses (L is uncompressed length, n is loop count and d is spring wire diameter), v is Poisson's ratio, D-d is average of inner and outer diameters of spring (D is outside diameter of spring), and n is loop count.
Unfortunately, that doesn't help me much, because spring washers aren't made of round wire.
So I devised a method to replace d^4 in the formula by something that should describe in a similar manner the rectangular spring washer wire.
The second moment of area (https://en.wikipedia.org/wiki/List_of_second_moments_of_area) of round wire is:
I = pi/4 * (d/2)^4
The second moment of area of rectangular wire in the spring washer is:
I = b*h^3/12 = (11-6.2)/2*1.4^3/12 = 0.5488 (in units of mm^4)
So I'm calculating then which diameter of spring wire has the same second moment of area:
I = pi/4 * (d/2)^4 = 0.5488
d/2 = (4*0.5488/pi)^(1/4)
d = 2*(4*0.5488/pi)^(1/4)
d = 1.8286 mm
So now I'm using the spring formula but replace d^4 by 1.8286e-3^4, and use v=0.3 for steel and E=200e9 Pa for steel.
Fmax = 200e9 * 1.8286e-3^4 * 2e-3 / (16 * (1+0.3) * (11e-3-(11e-3-6.2e-3)/2)^3 * 1)
Fmax = 338.05 N
My question is: how accurate is my method of estimating spring washer preload? Am I replacing d^4 in the formula correctly by using d of a round wire that has the same second moment of area that the real cross section has?
I understand that spring washers aren't usually regarded as good. The reason is that M6 8.8 bolt is recommended to have a preload of about 7200 N, and 338 N of the spring washer is only less than 5% of the recommended preload.
However, I'm not working here with metal but rather with wood. Wood isn't as strong, and wood lives with moisture changes, so therefore, maybe the spring washer could actually be useful here?
