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Old 12th August 2004, 12:04 PM   #1
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Post Panel thickness, flexing and resonance.

Something I just found out last night - might be old hat to the rest of you but it's new to me anyway...

Seems that if you double the thickness of a wooden panel it's stiffness increases by a factor of four (goes up with the square of the thickness) but the increased mass works against an increase in resonant frequency. It does increase though by a factor of the square root of the thickness increase.

E.g. if we have a panel 10mm thick that has a resonance of 100 Hz and needs 10kg to bend it 1mm, then if we make it 20mm thick then it will have a resonance of 141.4 Hz and need 40kg to bend it 1mm. If we go to 30mm thick then we get a resonance of 173.2 Hz and 90kg to bend it 1mm.

This only applies to a single thickness of wood or several thinner sections glued together. If the thinner sections are simply stacked together without glue then the stiffness is the sum of the individual sheets and the resonance will be unchanged.
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Old 12th August 2004, 01:43 PM   #2
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Stiffness increases with the cube of thickness, in accordance to the equation for the area moment of inertia of a rectangular section given by I = (1/12)b*h^3.

What source stated that it increases with the square of thickness? Perhaps that comes from the stiffness of a membrane, not a beam... which is probably appropriate. Have to go grab Roarke's.
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Old 12th August 2004, 01:49 PM   #3
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http://www.mindspring.com/~thayer5/f...ler/euler.html

Notice that thickness is cubed, so double the thickness and stiffness increases 8x.

GM
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Old 13th August 2004, 12:57 PM   #4
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Default The Mechanical Properties of Wood, by Samuel J. Record. 1914.

"From this formulę it is evident that for rectangular prismatic beams of the same material, mode of support, and loading, the load which a given beam can support varies as follows:

(1) It is directly proportional to the breadth for beams of the same length and depth, as is the case with stiffness.

*********
(2) [The load-bearing capacity] is directly proportional to the square of the height for beams of the same length and breadth, instead of as the =cube of this dimension as in stiffness=.
*********

(3) It is inversely proportional to the span for beams of the same breadth and depth and not to the cube of this dimension as in stiffness.

The fact that the strength varies as the square of the height and the stiffness as the cube explains the relationship of bending to thickness. Were the law the same for strength and stiffness a thin piece of material such as a sheet of paper could not be bent any further without breaking than a thick piece, say an inch board."

Source -> http://www.gutenberg.net/dirs/1/2/2/...-h/12299-h.htm

Well well well. You learn something new every day.
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Old 13th August 2004, 01:08 PM   #5
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So then, if my unbraced sub boxes made of 18m MDF and somewhat flexy are covered with an outer layer of 32mm MDF glued on, then the stiffness will be (50/18)^3 = 21.43 times less flexy!
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Old 13th August 2004, 01:10 PM   #6
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Quote:
(1) It is directly proportional to the breadth for beams of the same length and depth, as is the case with stiffness.
Keeping these facts in mind consider the effect of either adding an additional 3/4" of material by doubling up on the panel thickness or adding another 12" or so of material in the flexing plane via a shelf style panel to panel brace and it should be obvious that pound for pound panel to panel bracing is at least four times as effective as the adding of pure mass in vibration control.
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Old 13th August 2004, 04:24 PM   #7
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Default Re: The Mechanical Properties of Wood, by Samuel J. Record. 1914.

Quote:
Originally posted by Circlotron
(2) [The load-bearing capacity] is directly proportional to the square of the height for beams of the same length and breadth, instead of as the =cube of this dimension as in stiffness=.
In case you're interested...

As given previously, the equation for the area moment of inertia (which is directly proportional to the bending stiffness) of a rectangular cross section beam is I = (1/12)b*h^3 , where b = the breadth (width) and h = the height (thickness).

The equation for bending stress is S = M*C/I , where M = the moment (load) applied, C = the distance from the neutral axis to the point at which you wish to calculate stresses (the outer surface in most cases, as this is where bending stresses are maximum), and I = the area moment of inertia given by the equation above.

Combining the two, you see that S = (12*M*C)/(b*h^3). Since, in most cases, load bearing capacity is determined by the maximum bending stress, C = half the thickness of the beam. So S = (6*M*h)/(b*h^3), which simplifies to (6*M)/(b*h^2). So you can see how the stiffness and stress interact, and where the proportionality of stiffness to h^3 and the proportionality of strength to h^2 comes from.

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(3) It is inversely proportional to the span for beams of the same breadth and depth and not to the cube of this dimension as in stiffness.
In the equation above for bending stress, the moment is, in most cases we are concerned about, proportional to the length of the beam. i.e., M = F*L for a simplified case, where F is some applied load and L is the beam (or panel) length. Thus, doubling the beam length also doubles the moment (for a given load) applied, which doubles the bending stress generated and halves the load bearing capacity, in accordance to statement (3) from your posted source. What is important for our purposes though is not necessarily the load bearing capacity, but the deflection under that load (as hinted at by the paper and 1" board example given).

I stated earilier that bending stiffness is directly proportional to the area moment of inertia, thus to the cube of thickness... and it is. The full equation for bending stiffness of a beam loaded by a uniform pressure (as is the case in a loudspeaker enclosure) is given by K = (C*E*I)/L^4 , where E is the elastic modulus, I is the area moment of inertia (above), L is the length, and C is a constant that describes the level of fixity of the beam ends. Thus while the load carrying capacity is inversely proportional to the length, the deflection under that load is proportional to the ^4th of length.

Quote:
Originally posted by BillFitzmaurice
Keeping these facts in mind consider the effect of either adding an additional 3/4" of material by doubling up on the panel thickness or adding another 12" or so of material in the flexing plane via a shelf style panel to panel brace and it should be obvious that pound for pound panel to panel bracing is at least four times as effective as the adding of pure mass in vibration control.
I would tend to agree, but a closer look should always be taken from an energy perspective. While two solutions might yield the same panel stiffness, and thus the same deflection under a given static load, pressure loading of an enclosure panel is by nature a transient event. And the energy required to stimulate a panel to vibrate or resonate at a given amplitude is also dependent on mass, not just stiffness. So while the more massive solution to a given stiffness might be the least "efficient" in terms of material required, cost, shipping weight, outer enclosure dimensions, etc., it would be the most effective in terms of resisting excitation.

However, I suspect that you are absolutely correct in that the differential between added stiffness through panel subdivision and through panel thickening would also yield the most effective from an energy viewpoint at a given weight target.
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Old 14th August 2004, 12:17 PM   #8
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I liked the idea of adding thickness rather than bracing because my boxes are glued up and have a layer of carpet nailed onto the inside surfaces. I imagine it would damage things somewhat to open them up sufficiently to put bracing inside. I have been thinking about it for nearly a year so I didn't exactly rush into this decision... Besides, it's way easier to just bung something on the outside. It also covers over a few port and driver experimentation holes.
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