New answers tagged

0

Buildings are typically designed with a 50-year lifespan. So a 2% yearly chance of passing the limit makes intuitive sense. But you're right: that'd basically mean we expect the structure to collapse sometime during those 50 years. So that load with a 2% yearly chance of being surpassed makes sense... but only as a starting point. The thing to remember is ...


1

If you take a read more carefully the Eurocode, it says "2% annually". That should be translated that in a full year, there is only a 2% chance to get to that wind level. So one simple way to think about this is that there is one chance in 50 years for that failure to occur (which is considered a reasonable scenario. The more correct mathematical ...


0

The equations found do match the equations above, they are just written differently but will yield the same outcome.


0

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 ...


0

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 ...


0

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 - ...


Top 50 recent answers are included