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I'm trying to resolve a theoretical problem. If I have 2 steel boxes made from the same gauge steel (3mm stainless steel) & one is 7 meters by 2 meters by 2 meters & the other is say 30% smaller in length / width / height, which would be the more ridgid? I would expect the smaller box but just want to confirm with a professional. :) Thanks for helping!

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  • $\begingroup$ Make a prototype out of any old household stuff to get an intuitive feel for how stiffness changes with size! $\endgroup$ Apr 30 '20 at 19:26
  • $\begingroup$ kamran and I give opposite answers, and I've realized it's because there's a doubt about how you intend to load this. Is it meant to work as, for example, a water tank, having to handle lateral forces (i.e. water pressure)? Or is meant to work like a beam, supported on two ends and handling vertical loads? If the former (water tank), then kamran's answer is correct. If the latter (beam), mine is. $\endgroup$
    – Wasabi
    Apr 30 '20 at 21:54
  • $\begingroup$ Wow! interesting outcomes to what I thought was a simple question, so Wasabi answering your query about function, in theory, I would be putting four wheels on it & a solid mass inside as a load & it would be moved over uneven terrain. $\endgroup$
    – box guy
    May 1 '20 at 6:57
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yes, the smaller one is stronger and stiffer.

Generally, a beam stiffness is $\frac{48EI}{L^3}$ for a simply supported beam with a unit load at the center of the span. most other supports and loadings share the same denominator with EI multiplied by different constants on top.

We can find similar equations in a plate stiffness.

the key concept is the denominator has L to the cube.

That implies by increasing any of the dimensions of the plate we reduce its stiffness by cube power.

Edit

After @Wasibi's misunderstanding of the question and mistaking the fact that OP wanted to use the same thickness. 3mm, stainless steel in both cases I repeat my answer is the correct answer, and @Wasabi, unfortunately, has given the wrong answer.

Just to clarify it to OP.

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  • $\begingroup$ Hey, thankyou for such a quick answer! :) $\endgroup$
    – box guy
    Apr 30 '20 at 18:25
  • $\begingroup$ You forgot that $I$ has dimension $L^4$. Changing the cross-section dimensions will therefore usually have a larger impact than the reduction in span. $\endgroup$
    – Wasabi
    Apr 30 '20 at 21:07
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It depends on the specific dimensions of the different parts.

@kamran's answer has the correct stiffness equation of $\dfrac{k\cdot EI}{\ell^3}$ (where $k$ depends on the loading and boundary conditions). We must note that $\ell$ is raised to the third power and is therefore very influential.

But we must not forget about the moment of inertia $I$, which has dimension $L^4$. For example, a rectangular beam has moment of inertia $I = \dfrac{bh^3}{12}$. So if you had a rectangular beam, it's stiffness would be equal to

$$K_{orig} = \dfrac{k\cdot Ebh^3}{\ell^3}$$

If you reduced everything by 70%, you'd get

$$K_{shrink} = \dfrac{k\cdot E\cdot 0.7b\cdot (0.7h)^3}{(0.7\ell)^3} = 0.7K_{orig}$$

However, you aren't dealing with a rectangular beam, but a box section. In this case, the moment of inertia is equal to

$$I = \dfrac{bh^3-(b-2t)(h-2t)^3}{12}$$

If you were shrinking all the dimensions equally by 70%, then the end result would be the same as for rectangular sections: a 70% reduction in stiffness. However, you've said that the thickness of the boxes will be the same in both cases.

In this case, whether the bigger or smaller beam will be stiffer depends on the proportion of $t$ to $b$ and $h$. I used Wolfram Alpha to get the value for $t$ at which the deflections will be equal, but it's so ugly I can't be bothered to copy it over here. Suffice to say, you can't assume that the normal or reduced box-beam will be necessarily stiffer, you need to do the math to check for your specific dimensions.

For the record, using the dimensions given, we get the following moments of inertia:

$$\begin{align} I_{2000x2000x3} &= 1.5928e+10\text{ mm}^4 \\ I_{1400x1400x3} &= 5.4528e+09\text{ mm}^4 \approx 0.342 I_{2000x2000x3} \end{align}$$

Looking at the stiffnesses and hiding $E$ within the $k$ constant (since it doesn't matter for this analysis):

$$\begin{align} K_{2000x2000x3} &= \dfrac{k\cdot I}{\ell^3} \\ &= \dfrac{k \cdot I_{2000x2000x3}}{\ell^3} \\ K_{1400x1400x3} &= \dfrac{k\cdot I}{(0.7\ell)^3} \\ &= \dfrac{k \cdot I_{1400x1400x3}}{(0.7\ell)^3} \\ &= \dfrac{k \cdot 0.342I_{2000x2000x3}}{(0.7\ell)^3} \\ &= \dfrac{0.342}{0.7^3}\dfrac{k \cdot I_{2000x2000x3}}{\ell^3} \\ &= 0.998\dfrac{k \cdot I_{2000x2000x3}}{\ell^3} \\ &= 0.998K_{2000x2000x3} \\ \end{align}$$

So both beams are almost exactly the same stiffness, but the bigger beam would be a tiny bit stiffer with the dimensions given.

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  • $\begingroup$ I did not forget. it seems you did not read OP.s question carefully. He says "using the same gauge# 3mm stainless steel. He meant changing the length and with of the plat not its thickness. $\endgroup$
    – kamran
    Apr 30 '20 at 21:16
  • $\begingroup$ @kamran the OP says "30% smaller in length / width / height". I admit I'm understanding that as "length and width and height". And he seems to be talking about a box-section (of 2m height and width and 3mm thickness), not a flat plate. $\endgroup$
    – Wasabi
    Apr 30 '20 at 21:22
  • $\begingroup$ he means the length, width, height of the cube, not the plate. Read the question one more time carefully pleas. $\endgroup$
    – kamran
    Apr 30 '20 at 21:26
  • $\begingroup$ @kamran I'm having trouble what you mean by "plate" in this case. As I understand it, the OP is talking about two boxes, both of which have 3mm-thick walls. One box is 2x2x7m in size and the other is 1.4x1.4x4.9m. The smaller box therefore has a smaller span in any direction, but also has a smaller cross-section. What are you understanding? $\endgroup$
    – Wasabi
    Apr 30 '20 at 21:33
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    $\begingroup$ Oh, ok. But as you yourself have said many times at the end of the day the difference between an engineer and a computer is judgment. We cant have paper-thin beams crumbling on us by crushing or wrinkling into explosive collapse. $\endgroup$
    – kamran
    Apr 30 '20 at 21:54

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