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You can easily measure this by dangling a weight from a string and winding that around a pulley of known diameter. This arrangement will produce a fixed torque on the pulley, which you can adjust by changing the weight. Now all you have to do is attach a knob to the the pulley and you can feel what a given torque is like. Adjust the weight until the torque ...


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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 ...


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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 ...


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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 - ...


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