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I have designed a circular, 800 mm (31.5") diameter tripod table. The top is made of stone and weighs 40 kg (88 lbs). The height of the table is 750 mm (29.5"). There is a central column with tripod feet lower down which touch the floor 108 mm (4.25") in from the outer edge of the table top if looking in plan. The fulcrum created by two of the feet is a line 240 mm (9.5") to the table edge at its widest point.

Question: How much weight can be put on the edge of the table top before the table tips over, and how much can this weight increase if the feet touchdown 90 mm (3.5") from the outer edge? (In this case the fulcrum line would now be approx 230 mm (9"))

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I think (with some simplification) it should be 23 kg.

And this is why:

Assumptions:

  • legs are weightless
  • table's mass is a pointmass in the center of the table
  • table will topple over the connecting line between two legs

First we do a balance of momentum:

$$ \begin{align} \Sigma M &= 0 \\ \Sigma M &= F \cdot (400 mm -d) - 40 kg \cdot d \\ \rightarrow F &= \frac{40kg\cdot d}{400mm-d} \end{align} $$

Now what is $d$? It is the (at the moment) unknown distance to the fulcrum line (considering, the table will not topple over one, but over two legs). Let's calculate that $d$!

We know the table's legs have an angle of 120° between each other. We want the height of the triangle composed of two legs and the connection line between them. Our rectangular triangle is built by the height of the former triangle, half of the connection line and a leg's distance to the center (which is 400mm - 108mm = 292mm and now the hypotenuse). The angle is split in half, so its 60°. We can now calculate the height (now the adjacent leg to the angle) by $$ d = cos (60°) \cdot 292mm =146mm $$ We can now solve the momentum balance:

$$ F = \frac{40kg\cdot 146mm}{400mm - 146mm}=22.992kg $$

Correct me if I'm wrong. (some pictures would help...)

For the new configuration it could be increased by 2.314kg (you can calculate that yourself by now... :P )

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  • $\begingroup$ Some of the mass of the table will be on the other side of the fulcrum and so acting in the other direction. Which I don't think you account for. $\endgroup$
    – nivag
    May 6, 2015 at 13:26
  • $\begingroup$ Nope, it's kept very simple. The simplification from my assumptions don't account for distributed mass... $\endgroup$
    – Knigge46
    May 6, 2015 at 13:31
  • $\begingroup$ Also the OP gives the fulcrum as 160mm from the centre although your calculation here does seem correct. $\endgroup$
    – nivag
    May 6, 2015 at 13:33
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    $\begingroup$ I think that is a significant oversimplfication in this case. Unless I've done my answer wrong I get almost twice the load. $\endgroup$
    – nivag
    May 6, 2015 at 13:37
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    $\begingroup$ @AndyT I think this answer is the correct one. Unless you start talking about moment of inertia (e.g. if we wanted to find out dynamic stability) there is no reason not to lump the table mass into the center of the tabletop. $\endgroup$
    – regdoug
    May 7, 2015 at 14:03
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The question is not quite internally consistent.

The table is 400mm radius, and the feet are apparently on a 292mm radius pitch circle (since they are 108mm in from the edge). By simple trig, the fulcrum line is therefore 292cos(60) from the circle centre, which is 146 from the centre, which is 254mm from the edge (the question states this is 240mm).

However:

If the legs and top are substantially rigid (ie, small deflections - they aren't bending so much that their shape in plan changes) the mass of tabletop acts at the centre, so the moment of mass of the top is 40kg x 146mm = 5840 kgmm about the fulcrum line.

So that much moment will balance the table (put it on the point of toppling over the fulcrum). So the mass we can put there is 5840/254 = 22.99kg.

If 90mm from the edge, feet pitch circle is now 310mm radius, teh fulcrum is 155mm from the centre and 245mm from the edge, so we can carry 40x155/245 = 25.31kg, ie 2.32kg more.

That was the easy way, and is already advanced in one answer (Knigge46's), but you can do the stabilising moment calc the hard way if you like:

Consider an arbitrary strip of tabletop, parallel to the fulcrum line, width delta-x. By pythagoras the length of this strip is 2(r^2-x^2)^.5 where r is tabletop radius and x is perpendicular distance from the strip to the centre.

Let f be the offset between table centre and fulcrum line. Let d be the areal density of the top. Thus the moment of this strip about the fulcrum line is d . 2(r^2-x^2)^.5 . (x-f) delta-x, which you can summate across the full table top, or change from a summation to an integration simply by changing the delta-x to a dx.

If you do that integration for x from -r to r, you get -5.840kgm, or 5840 kgmm, the same answer as treating the table top as a point mass at the centre

If you want to do the stabilising and destabilising contributions from the top you can do that by integrating over different ranges:

stabilising, (-r <= x < f) gives -6.986kgm

destabilising (beyond the fulcrum) (f < x <= r) gives 1.146kgm

So net effect of the top is stabilising by (-6.986 + 1.146 ) = -5.840kgm, again.

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I'll assume you apply the load at a single point at the worst position (on the edge equidistant between two legs). The real loading may be higher than this depending on distribution and position.

In this case you have a simple lever problem, with the load and part of the table balanced by the rest of the table (assuming the legs are relatively light).

The fraction of the table on the loaded side of the table will be a circular segment. Therefore, the fraction is

$$f=\frac{\theta-\sin{\theta}}{2\pi} $$

where $\theta$ is in radians and is given by $$ \theta =2\arccos{\frac{d}{R}} $$ $d$ is the distance from the centre to the fulcrum and $R$ is the radius.

Assuming uniform mass density of the table each segment will act through its center of mass. Therefore there are three components to balance. The table will tip if.

$$(R-d)L + \frac{R-d}{2}fM > \frac{d+R}{2}(1-f)M$$

rearranging

$$L>(\frac{R+d}{R-d}\frac{1-f}{2}-\frac{f}{2})M $$

for your table $R=400$ mm, $d=160$ mm, $M=40$ kg.

In this case,

$$theta =1.59 rads$$ $$f=0.094$$ $$L_{max}=40.4kg$$

(I've probably made some calculation or algebra error here so I would check it).

For this model the height of the table is irrelevant. I reality lateral loads on the table are non-zero, e.g. someone pushing sideways. I which case height will have a significant effect.

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    $\begingroup$ I spotted my error. Mass density (in a 1D sense) is not uniform for a circular table. I will adjust my answer when I have time. $\endgroup$
    – nivag
    May 7, 2015 at 8:18
  • $\begingroup$ Another error is to use d=160mm from the OP, whereas @Knigge46 has correctly recalculated it to be 146mm. $\endgroup$
    – AndyT
    May 7, 2015 at 12:22
  • $\begingroup$ You don't need any of that maths - with the implicit assumption that the legs and top are effectively rigid the weight of the top effectively all acts at the centre $\endgroup$
    – achrn
    May 7, 2015 at 12:37
  • $\begingroup$ The error is the assumption that the centre of mass of the circular segment is at (R-d)/2 and that the centre of mass of the rest of the table top is at (d+R)/2. Neither of these is true. This might be what you mean by the mass density not being uniform, but actually all the answers assume the areal mass density of the top is uniform. $\endgroup$
    – achrn
    May 7, 2015 at 20:43
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As there are 3 feet below a radial surface we need to calculate the maximum moment of area between the feet induced by table weight. This requires calculation of the midpoint of triangular line between two feet which describe an arc of 292mm with an angle between them of 120°. The radius calculates to $292 / 2=146mm$ Using cosine rule we can determine the length of the chord midpoint to be $$L=\frac{\sqrt{ 2(146 ^2)-2(146^2) \cos 120}}{2}$$ $$L=126.44$$ Now we calculate the inscribed radial distance as the perpendicular distance to L calculates by Pythagoras $$ \sqrt{146^2-126.44^2}=r_2$$ $$r_2=73mm$$ Comparing the table to radius to the effective minimum support cantilever radius between the feet to be $\frac {800}{2}-73=327mm$ from the table edge making that centroid quite high up.

In my opinion, this table is not going to be terribly safe to use considering that the support radius is just 73mm against a top heavy design of 400mm $\frac{3}{4}$m above the ground giving an axial load of 13.3kg which is 130.8N self weight.

To calculate the tipping point, we really only need to treat this as a simple beam equation.

The result is that 8.32kg at the most precarious edge would be the limit of tipping. For safety I would factor 25% minimum which would reduce this to 6.24kg.

Considering the weight of the table falling on a small child would likely cause serious injury and probably break a person's foot if tipped over accidentally, I'd add lateral stabilizers to the table feet parallel to the table radius in the shape of triangles pointing outwards. The base of them being 140mm long as isosceles triangles with 2x80mm sides

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