A box is being lifted by a crane. Sometimes the box will be completely full, and sometimes it will only be partially full (indicated by the gray shaded regions).

enter image description here

The question is, where is the axis of bending in both cases? Does it follow the center of mass? Or does it always follow the geometric center?

(Of course I hope the axis of bending will always be in the middle so it could take full advantage of the roof and floor to resist bending.)

Also, if it matters, the crane will actually lift the box at 4 points, not 2. And I moved them inwards from the edges just because I thought that would resist bending more, due to a shorter beam.

Once I know where the axis is, I can calculate the moment of inertia and use the deflection formula

$$\delta = \dfrac{5qL^4}{384 E I}$$


As advised, I did this for a U-shape, but also for the square, for comparison. I had to choose concrete numbers so I picked a 3x3x12 box with wall thickness of 1 cm.

enter image description here

For the square, I found the formula here.

$ I = a^4 - (a - 2t)^4 $

$ I = 3^4 - (3 - 0.02)^4 = 2.13849584$

For the U-shape, I found the much more complex formula here and took it about the y-axis since that diagram was vertical.

$ I = \frac{2ta^3 + (a - 2t)t^3}{3} - (2ta + (a - 2t)t)\frac{2ta^2 + (a - 2t)t^2}{2a^2 - 2(a - 2t)(a - t)}$

$ I = \frac{2ta^3 + at^3 - 2t^4}{3} - \frac{(2ta + ta - 2t^2)(2ta^2 + at^2 - 2t^3)}{2a^3 - 2(a^2 - 3ta + 2t^2)} $

$ I = \frac{0.54000298}{3} - \frac{(0.0898)(0.180298)}{54 - 2(8.9102)}$

$ I = apprx. 0.179553$

So now I can finally use the deflection formula with both I's. I'm using steel so E = 200 GPa. The distributed load, q, is for a box 1/3rd full of crushed iron ore which has a density of 2,500 kg/m$^3$. q = 72,523 N/m.

$\delta = \dfrac{5qL^4}{384 E I}$

$\delta = \dfrac{5 * 72523 * 12^4}{384 * 200,000,000,000 * I}$

$\delta = \dfrac{7,519,184,640}{76,800,000,000,000 * I}$

$\delta = 0.00004578$ m for the square.

$\delta = 0.0005452766$ m for the U-shape.

...I have to conclude I did something wrong. It is hard to imagine a box of that size, with just 1-cm thick walls, would deflect by less than a millimeter when holding up iron ore, even at 1/3rd full.


2 Answers 2


If by axis of bending you mean box functioning as a beam and also the top and walls and bottom of the box are integrated and work like a hollow rectangular beam, then the neutral axis is at the mid height of the box.

If the content of the box is liquid, it is evenly distributed along the length of the box and should be considered distributed load. The level of liquid doesn't affect section properties of the box, such as its moment of inertia or location of its neutral axis.

  • 1
    $\begingroup$ I'd probably recommend calculating the box as a U-shape, since the lid is unlikely to be sufficiently connected to the rest of the box to work effectively as part of the structural element. $\endgroup$
    – Wasabi
    Jul 19, 2019 at 2:16
  • $\begingroup$ @Wasabi Good point. Sadly this will most likely cut the strength in half since there's no roof to go into compression. $\endgroup$
    – DrZ214
    Jul 19, 2019 at 3:22
  • $\begingroup$ Not half, because if no top the neutral axis moves down. I'd estimate the I then would be 65%. But then the walls could warp or buckle laterally.. $\endgroup$
    – kamran
    Jul 19, 2019 at 3:30
  • $\begingroup$ @kamran Turns out it's more like an order of magnitude less. The neutral axis moving towards the floor should in theory make it less than 50%, but I did not expect an order of magnitute. I may have calculated wrong since the U-shape formula for I was very complicated. The OP has an update. $\endgroup$
    – DrZ214
    Jul 21, 2019 at 4:07

As implicitly mentioned in @kamran's answer, the neutral axis (which is what I understand by 'axis of bending') is constant regardless of the applied load.

That's because the neutral axis is a property of the structure resisting the load, and in this case the load does not help to "carry itself".

So the neutral axis will merely be equal to the position of the box's centroid. However, it is important to note the box should almost certainly be considered a U-shape. That's because the box's lid is unlikely to be sufficiently "bound" to the rest of the box to truly act monolithically with the rest. So you'd want to consider a lid-less box as your cross-section (though having to carry the weight of the lid, of course).

Unfortunately, U-shapes are very tricky to design under bending, due to the possibility of localized buckling of the "wings". This is especially problematic Given that your U-shape needs to be "belly-down" (that is, looking like a U, not an $\Pi$), which means that the top of the beam (in this case, the wings) will be under compression. And given how low the center of gravity would be in this orientation, it'll be very high compression in the members the least equipped to withstand it.

I'd honestly recommend talking to an engineer to calculate this for you (especially given that there are myriad other aspects we haven't taken into consideration, such as wind and dynamic forces), or perhaps renting boxes made for this situation instead of trying to make your own. Another "solution" would be to use more than two support points on each side of the box. Using many supports would drastically reduce the applied load.

  • $\begingroup$ If I was ever going to build my own box for iron ore, I would definitely consult an engineer. I did the math on the u-shape and square for comparison. However, I got teensy-weensy values for deflection, less than a millimeter in both cases, so I am very untrusting of my math. The I-formula for the U-shape was very complicated. Made update to the OP. $\endgroup$
    – DrZ214
    Jul 21, 2019 at 4:12

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