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Calculating the force to bend steel 
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#1
Apr312, 04:07 PM

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Picture a game of hangman. You have a vertical pole, a horizontal bar extending out from the pole and then the poor guy being slowing hanged hanging at the end of the horizontal bar. I'm trying to calculate much weight the pole can support before it bends.
Well, actually I'm trying to hang a TV on the horizontal bar, and I need to know how to size the a piece of steel for the "vertical pole" to support the TV so it won't bend. It should be a fairly simple statics problem. But it's been 20 years since my statics class. TV is 22 lbs and the horzontal bar is 2 ft long. The vertical pole is 18" high. I beleive that gives me a force of 66lbs that is being extered on the "vertical pole" were it is ancored at the bottom. Is that correct? How do I think figure out the size steel that will resist bending given this force? Any help is appreciated. 


#2
Apr412, 01:10 PM

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There will always be some bending (deflection) of the vertical and horizontal members, which will be quite small if the members are sized properly to take allowable bending stresses and slight deflections. For the vertical member, it is subject to an axial compressive load of 22 pounds, and a bending moment at its top of (22)(24) = 528 inlb.
Neglecting the axial load and using 36 ksi steel, max stress is Mc/I and max deflection is ML^2/2EI. I'm not sure what shape member you have in mind, but if it is say a 1'' diameter steel rod, c= 1/2", I = .05 in^4, E =30x10^6 psi, and M = 528inlb, solving, max stress = 5000 psi which gives a safety factor of about 7, no problem, and the deflection at the top is about 1/16''. Assuming your base is fully fixed without any rotation. You can try different sizes and shapes...if you tried a 1/2" rod, then the max stress is way too high. 


#3
Apr412, 02:03 PM

P: 2,048

Triangulation.



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