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I am building a firewood storage cabinet which should support around 100-150 kg of firewood.

The cabinet is made out of wooden boards (20 mm thick for all sides). The length is 550 mm and depth is 400 mm. The wooden logs are 300 mm in length.

150 kg for a wooded shelf of 20 mm in thickness is too much (I want to use 20 mm wood for all sides) and I want to make sure it will not bent. My idea is to reinforce the shelf with hollow steel rods under it. Round rods, because those can be easily drilled into the walls of the cabinet. I know square rods are stronger, but square holes are tricky to make.

The problem is that if the rod is thicker than D=10 mm and the distance from the front is is more than x=50 mm, the rod will be visible under the shelf from a distance (the shelf start at 340 mm from the ground). So I would like to keep the first rod around x=100 mm from the front and use rods of D=10 mm for the aesthetics of it.

I am not sure though if a rod of D=10mm / d=2mm is strong enough for this. Or how many rods will be enough (n).

My question is, what size steel round ronds (D / d) can I use and how many of them (n) will be enough? Another option would be to only use the thinner rod front rod and thicker rods after. But I am also very curious how this can be calculated.

Below is the schematics with sizes and some variables which I used:

       rod length = 550 mm
total wood weight = 150 kg
                D = round rod diameter
                d = rod wall thickness
                n = number of rods needed to support the weight

Schematics of the cabinet

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  • $\begingroup$ is there a reason you want to add steel rods? I suggest that standard wooden joists would work better by having better connection to both the floors and sides of the structure. 20mm x 75 or 100 would seem to me to be the right way to go. You will also need diagonal bracing. $\endgroup$ – Tiger Guy Mar 29 at 22:06
  • $\begingroup$ I thought about that, but I don't want anything to be visible underneath the shelf on which the wood is stacked. And I don't want the shelf to bend after couple of years. That's why I want something which will prevent that + being as thin and possible. $\endgroup$ – ddofborg Mar 30 at 7:21
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If you choose your rods appropriately, a bigger concern is going to be those holes you drill for the rods. For now, I'll leave you with a beam bending approximation approach:

$$Stress = \frac{LeverMoment * CentroidDistance}{MomentOfInertia}$$

With your tube definitions, you get that the

$$CentroidDistance = D/2$$ $$MomentOfInertia = \frac{\pi}{64} * ( D^4 - (D-d)^4)$$

Since the "beam" is supported on both ends, you'd use half the weight you expect on the rod and half the rod length for the LeverMoment. Best case your load is distributed across all rods. (I'd just absorb the factor of uneven distribution into the factor of safety).

$$LeverMoment = Force * Length$$ $$LeverMoment = \frac{75kg}{n}* 9.81 \frac{m}{s^2} * 225mm $$

You'll want to keep the $Stress*FactorOfSafety$ below the Tensile Strength of your material. Substitute it all and you get

$$\frac{TensileStrength}{FactorOfSafety} > Stress = \frac{\frac{150kg}{2n}* 9.81\frac{m}{s^2} * \frac{550mm}{2} * \frac{D}{2}}{\frac{\pi}{64} * ( D^4 - (D-d)^4)}$$

Clean up some numerator-denominator business:

$$\frac{TensileStrength}{FactorOfSafety} > Stress = \frac{8 * 150kg* 9.81\frac{m}{s^2} * 550mm* D}{n * \pi * ( D^4 - (D-d)^4)}$$

I'll leave minding the unit conversions and finding the strength of your rod material and deciding the factor of safety to you. Just make sure the factor of safety is higher than usual to cover uneven weight distribution on top of everything else it needs to cover (like weight of your building materials themselves)

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Given: Total weight = 150 kg, uniformly placed over beam with 0.55m span length.

M = wL^2/8 = (150/0.55)(0.55^2)/8 = 10.3 N-m = 0.0103 kN-m

For A302, A304, stainless steel tubing, Fy = 207, Fa = 0.6 Fy = 124.2 MPa = 124200 kN/m^2

For D = 10 mm, y = D/2 = 5 mm, I_req = My/Fa = [0.01030.005/124200]*1000^4 = 414.6 mm^4

For round pipe with D = 10 mm, d = 1 mm, I = PiD^3d/8 = 3.1416*!10^3)*(1)/8 = 392.7 mm^4

By inspection, you will need 2 tubing under the log, I_2 = 2*392.7 = 785.4 mm^4

Check permissible deflection =< L/360 = 550/360 = 1.53 mm

Actual deflection = 5wL^4/384EI_2, E = 193 GPa

Deflection = 5*(150/550)(550^4)/(38419310^9785.4) = 210^-6 mm, negligible

Check my calculation and number before make your call.

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Generally the maximum bending stress $\sigma$ on a beam under bending is given by:

$$\sigma_{max} = \frac{M}{I}\cdot y_{max}$$

where:

  • I: the second moment of area of the rod (2d, because the internal diameter is by removing two thickness from the extenral ) $$I =\frac{\pi}{64} \cdot ( D^4 - (D-2 d)^4)$$
  • $y_{max} = \frac{D}{2}$ in this case
  • M: is the bending moment on the beam.

The bending moment will be affected from the weight on the rod. Assuming that you have two rods and that the weight is equally distributed (not a very safe assumption), then the force on each rod will be equal to: $$F_{rod} = \frac{F_{total}}{n}$$ where:

  • $F_{total}$: total weight (for 100kg~ 1000N)
  • $n$ : number of rods

Since the weight is distributed along the length of the rod, then the "weight per length is ":

$$ w_{rod} = \frac{F_{total}}{n\cdot l_{rod}} = \frac{F_{total}}{n\cdot 550}[\frac{N}{mm}]$$.

Now, depending on whether the ends can be thought of as fixed or simply supported the Moment is different:

Fixed Simply supported
Loading enter image description here enter image description here
$M_{max}$ $\frac{w l^2}{24}$ $\frac{w l^2}{8}$
x where $M_{max}$ $ x=0 ,x=l$ $x=\frac{l_{rod}}{2}$

It is best to assume simply supported because it is safer. In that case the maximum stress is:

$$\sigma_{max} = \frac{\frac{w l^2}{8}}{\frac{\pi}{64} \cdot ( D^4 - (D-2 d)^4)}\cdot \frac{D}{2}$$

$$\sigma_{max} = \frac{\frac{w l_{rod}^2}{8}}{\frac{\pi}{64} \cdot ( D^4 - (D-2 d)^4)}\cdot \frac{D}{2}$$ $$\sigma_{max} = \frac{\frac{F_{total}}{n\cdot l_{rod}}\frac{ l_{rod}^2}{8}}{\frac{\pi}{64} \cdot ( D^4 - (D-2 d)^4)}\cdot \frac{D}{2}$$

$$\sigma_{max} = \frac{F_{total}l_{rod}}{n}\frac{4D}{\pi \cdot ( D^4 - (D-2 d)^4)}$$

If you use n=2, and D=10, d=2 the rest values then

n $\sigma_{max} [MPa]$
2 402
4 201
6 134
8 100

I would prefer a safety factor of 2 and assuming a yield stress of 240[MPa], I would go for at least 6 rods (for the bending stress).

Other considerations

Normally at the interface of steel rod and wood, you will need to consider

  • Bearing strength
  • shear failure.
  • tear out (this is not an issue due to the long length of the wood plank)

Bearing strength.

The bearing strength will be calculated in this case by:

$$\sigma_{bearing} = \frac{F_{total}}{2n (D\cdot t_{wood})}$$

where:

  • $\frac{F_{total}}{2n}$ is the force at each end of the rod (there is support on both ends)
  • $D$: diameter of hole
  • $t_{wood}$: thickness of wood

for n=6 then $\sigma_{bearing}\approx 0.4[MPa] $, therefore reasonably safe.

Shear strength of the rod.

Shear strength of the rod can be calculated in this case by:

$$\sigma_{s} = \frac{F_{total}}{2n}\frac{1}{\frac{\pi}{4}(D^2-(D-2d)^2)}$$

where:

  • $\frac{F_{total}}{2n}$ is the force at each end of the rod (there is support on both ends)
  • $\frac{\pi}{4}(D^2-(D-2d)^2)$: the rod cross-section

for n=6 then $\sigma_{shear}\approx 1.6[MPa] $, therefore safe.

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  • $\begingroup$ I see a big difference in n from you and @r13. 2 vs. 6. @r13 made a small error using d=1 (it should be 2) and even that gave him n=2. $\endgroup$ – ddofborg Mar 30 at 7:25
  • $\begingroup$ One many occasions we have had disagreements with r13 see (here and here). In the end, I think I came to the conclusion that agree to disagree. So I don't want to comment on his results. In any case, you need to decide for yourself what is right and what is wrong, because you are the one building it. $\endgroup$ – NMech Mar 30 at 7:43

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