# Calculating tensile strength for aluminium

I'm making a Z (coffee table) linkage with two interlocking C channels and a couple of steel linkage bars. While moving, they aren't load bearing but when extended, they will take a weight of around 300lbs. I need to figure out whether 3mm T66 aluminium can take that weight without tearing.

The entire weight will be spread across two bolts (10mm with bushing) and the top half of that so that's Pi x 10mm diameter / 2 because the weight is all on the top half x 2 bolts - around 31.4mm x 3mm thick = 94.2mm squared. I've found tables of properties for T66 which give figures in MPa (megapascals?) but that's where my half-remembered high school physics gives out.

How do I calculate the maximum load across a given area from these properties?

Edit: To include sketch. The block to the left is the (fixed) seat base, along the top are Unwin rails along which the seat can slide. When not needed, the linkage around the 4 bolts can rotate it away into the floor. When they are raised (as shown), the seat+occupant, for which I am allowing 300lbs) can sit directly over the hinged section. Ignoring the rail overlap, that means that the full weight is supported by the top of the two round pivots (actually, since it's made from C channel and there are one of these on each side, that's 2x2x2x10mm bushings in 3mm aluminium).

Hope this makes the question clearer.

• a sketch would be nice. Commented Apr 3, 2020 at 20:50
• I'll do my best (just trying to figure out CAD) so bear with me and promise not to laugh!
– Mike
Commented Apr 3, 2020 at 21:09
• Ok, as the saying goes, the devil is in the details. Commented Apr 3, 2020 at 21:11
• even a hand sketch will do. Commented Apr 3, 2020 at 21:26
• In my opinion, when designing/verifying these components, the major concerns include tearout/failure of the bolts and the strength of the individual components. In terms of analysing the frame, it should be relatively straightforward by analysing each member individually and then combining the results.
– Amit
Commented Apr 4, 2020 at 3:21

I confirm that in your case the yield stress is an adapted limit value. However, there is a unit error in your computation. You forgot to convert the area in m2 which means that the results is not 362 Pa but 362x106 Pa = 362 MPa. Assuming that your reasoning is correct to compute the effort, your are above the limit of 170 MPa.

• Okay, my maths is a little rusty too but I thought that was what I was doing when I divided by 0,376m instead of 376mm?
– Mike
Commented Apr 4, 2020 at 13:33
• My bad. However your conversion is not correct: 376mm2 = 376.10^-6 m2 (and not 0.376, since it is an area) So the correct result is 0.362 MPa Commented Apr 4, 2020 at 14:43
• Ah, yes so that's dividing by 0.000376 which gives the same result but in KiloPascals. Thanks. Still, it looks like I have an order of magnitude to play with so I'll give it a shot. I might even run the figures for 1mm and 2mm channels.
– Mike
Commented Apr 4, 2020 at 15:37

Okay, trying to refresh my knowledge with a trawl through Wikipedia and online physics text books.

The yield strength is the point where the material can return to its normal shape after deforming under stress so I guess that's what I'm looking for. For this material, it's 170 MPa. While we're at it, let's go metric on the mass to avoid mixing units so 300lb = 136Kg.

As stated earlier, all of the weight is applied to the top half of 2x10mm x 2 sides per C channel x 2 units per seat. Circumference = pi x d Half the circumference = pi x d / 2 = pi x 10mm / 2 x 3mm thick = 47mm2 per hole (approx). 2 holes per side per mount = 8 bolts = 376mm2 total.

So the question now is "Which is greater, 170MPa or 136Kg spread across 376 square millimeters?"

According to Wikipedia, 1 Pascal = 1Kg/m2, so the load is 136 / 0.376 = 362Pa.

So by my reckoning when the load is measured in units and the capacity is measured in hundreds of mega units, I should be fine.

Right? I'm open to corrections on any of this. Google-reckoning can only get you so far.