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Iron bar bending force 
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#1
Jun1308, 08:05 PM

P: 1

1. The problem statement, all variables and given/known data
Given an iron bar, of round crosssection, fixed by its extremities (bar measures 152 mm in length and 4.8 mm in diameter), a weight of 88.5 kgs is hung from its middle, bending it at least 90 degrees. How much weight/force would be needed to bend another bar of the same material and size, but of hexagonal or square section? 2. Relevant equations Torque = Force * R Sin(90°) * R 3. The attempt at a solution Haven't got any idea. There should be something lacking in the text or more probably some calculus skills given as knows and hence omitted. Hope you could clear up this question, thank you in advance 


#2
Jun3008, 01:27 PM

P: 208

I'm not sure, but I think that you need a second moment of inertia over here, which is defined as
[tex]I = \int r^2\, dA[/tex] where the are is perpendicular to the axis of bending. The bigger this moment of inertia is, the harder it is to bend or deform the beam, so you just use proportion to find what you need. 


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