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I'm an amateur at engineering, working more on musical instruments, and I'm often in confusion regarding what exactly constitutes stiffness of a beam for a load applied on it so it could resist deformation, for example on googling for a rectangular beam I get the basic definition that it is proportional to breadth × square of height. But the moment of inertia is proportional to breadth × cube of height/12.

And for a triangular beam there's even less information (think the general shape of a guitar brace with the base of the triangle taking the load and the point hanging freely downwards).

Please give me some simple formulas to calculate both if possible, the load will not be uniformly distributed on the beam but for the present consideration, the load may be applied either downward at the middle of the beam (like a violin) or pulling the beam upwards (like an acoustic guitar). If possible please provide both scenarios.

I'd like to add pictures of the forces acting on the bracing (or beam if you will) so you could get a better picture of what stiffness I'm seeking. The red colour defines the beam and the blue is the location where the load will be applied on the beam, the arrow pointing towards or away from the blue marker shows the direction of the force being applied to the beam

Guitar construction simplified

Violin construction simplified

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    $\begingroup$ Either the web site you found is wrong, or you are confusing the maximum stress in the beam (which has a $h^2$ factor) with the stiffness (which has a $h^3$ factor.) For a triangular beam (or any other shape), just calculate the moment of inertia of the cross section. $\endgroup$ – alephzero Feb 9 at 16:43
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    $\begingroup$ what type of load are you interested in? Do you want to constrain deformation? $\endgroup$ – NMech Feb 9 at 16:55
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    $\begingroup$ Trying to answer this with "simple formulas" probably won't help, because you need to understand which formula applies to different situations. For example the neck of a guitar is (ignoring the truss rod!) a simple cantilever beam, but a rib glued inside the back of the guitar is different, because the back affects how the beam bends (i.e. the edge of the beam fixed to the back can't "stretch" because of the back) IMO, you really need to study a textbook on strength of materials to make sense of what you want to calculate. $\endgroup$ – alephzero Feb 9 at 17:03
  • $\begingroup$ I've just realised that you might be talking about a variable cross-section beam. Like in this picture? Because the phrasing for engineers meant something else and not everybody (including me) is either a native english speaker and a musical instruments creator, could you clarify if that is the case., $\endgroup$ – NMech Feb 14 at 3:48
  • $\begingroup$ I was thinking of variable cross section but that would be a rather difficult question, for now I'm simply wanting to find how stiff a simple rectangular beam can be (in a measurable number with only height and breadth being the variable) when acted on by the load of the strings and compare with how much a triangular beam is stiff in the same situation. $\endgroup$ – Aecarvalho Mar 1 at 5:29
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For members of same material and same length, the one with larger moment of inertia is stiffer. If material is same, but their lengths differ, compare (I1/L1) and (I2/L2), the larger one is more stiff. Otherwise, the larger one is stiffer by comparing (E1I1) and (E2I2), if L is constant; or (E1I1/L1) and (E2I2/L2), if lengths differ.

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deflection of a beam

Depending if the beam is cantilever, simply supporter or else you have different formulas.

For the simple case of a cantilever beam, with a concentrated load at the end the maximum deflection at the end is :

$$\delta_max = \frac{P \cdot l^3}{3EI}$$

for other cases you can find under deflection of beams tables with equations like the following: enter image description here

The tricky part is (almost every time) to find the second moment of area.

Calculation of second moment of area for triangle

For a general triangle, like the following

enter image description here

the moment of inertia about the centroid is:

  • $I_x= \frac{b h^3}{36}$

  • $I_y= \frac{b h}{36}(b^2 - b\cdot a + a^2)$

if you want to calculate at a different angle you can use the following transformation equations

$$I_u =\frac{I_x + I_y}{2}+\frac{I_x - I_y}{2}\cos(2\phi)- I_{xy} \sin(2\phi)$$ $$I_v =\frac{I_x + I_y}{2}-\frac{I_x - I_y}{2}\cos(2\phi)+ I_{xy} \sin(2\phi)$$ $$I_{uv} =\frac{I_x - I_y}{2}\sin(2\phi)+ I_{xy} \cos(2\phi)$$

To complete the above transformation you'd need the product moment of inertia (which is a bit more complicated). Using the notation in the image below

enter image description here

You could use the following formula, to calculate the product of inertia with respect to the O point (2 right angles) in

$$I_{x0,y0} = \frac{h^2}{24}(-b1^2+b_2^2)$$

To calculate about the product about the centroid you could use the following formula

$$I_{x,y} = I_{x0,y0} - A \cdot dx\cdot dy = I_{x0,y0} - A \left(-\frac{2}{3} (\frac{b}{2}-b_1) \right)\frac{-h}{3} $$

$$I_{xy} = \frac{h^2}{24}(-b1^2+b_2^2)- \frac{1}{2} b h \left(-\frac{2}{3} (\frac{b}{2}-b_1) \right)\frac{-h}{3} $$

Finally, (if my simplifications arecorrect ):

$$I_{xy} = \frac{h^2}{24}(-b1^2+b_2^2)- \frac{1}{9} b h^2 (\frac{b}{2}-b_1) $$

Finally for product of inertia $I_{xy}$

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There is an easy way you can calculate a god approximation to the bending strength of the guitar handle.

You can assume you guitar handle cross-section as a trapezoid and measure the moment of inertia, I of that as per this formula. $I_{Cx} \ $ is the moment of inertia about the trapezoid's y centroid parallel to the x axis, where you are interested in. All you need is the alloable stress of the material.

$$\bar{y}= \frac{H}{3}* \frac{2a+b}{a+b}$$

$$ I_{Cx} = \frac{H^3}{36} * \frac{a^2 +4ab +b^2}{a+b}$$

If you guitar handle section is like a triangle:

$$\bar{y} =\frac{H}{3}$$

$$I_{Cx}= \frac{BH^3}{36}$$

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Isosceles terapezod

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