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Q: bending strength of 1" x 2" rectangular steel tubing ( Answered ,   0 Comments )
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 Subject: bending strength of 1" x 2" rectangular steel tubing Category: Science > Physics Asked by: km2000-ga List Price: \$5.00 Posted: 15 Nov 2005 17:29 PST Expires: 15 Dec 2005 17:29 PST Question ID: 593504
 ```Visualize with me if you will... I am making a nice long workbench for a video editing suite. The bench is to be 9 feet long x three feet wide, with a shelf above it 9 feet long x 2 feet wide. Both must be able to support about 300 pounds at the center of the span. I would prefer to do this in a single (9 foot) span per shelf. I think using two lengths of 11 or 14 guage 1" x 2" rectangular steel tubing across each span would be more than enough. I have a conference table with similar reinforcement. I don't know how to calculate the bending strength of my materials. Would 14 guage (.083") 1' x 2" be strong enough? Will 11 guage (.120") be stong enough? I would prefer the "center sag" to be less than 1/4".```
 ```Hello km2000, what you have is a Simple Beam with a concentrated load at the center. We will assume that the weight of your shelf material is negligible. The formula for deflection is: D = P x l^3 / 48EI Where P is the load in pounds. In your case since you have 2 pieces of tubing they share the load equally. So, P = 300/2 or 150# l = 9 feet or 108 inches E is a constant for steel = 30,000,000 psi I is the moment of inertia of the rectangular tubing. From this table: http://www.alliedtube.com/pdf/section.pdf I = .2379 inches^4 for 1 x 2 x 14 Ga. tube Plugging in the numbers we get: D = 150 x (108)^3 / 48 (30,000,000)(.2379) = 0.55 inches Since this over twice the deflection you want we have to find a tube with I = .55/.25 x .2379 = .523 Going to .120 wall will not get enough increase in I to achieve the 1/4 inch deflection. I see that there is not a 1 x 2 x .120 wall tube listed in the chart, but it would be approximately equal to (.120/.083).2379 or .34 . Looking in the chart 1 1/2 x 2 1/2 x 14 Ga. has an I of 0.5397 which would give you what you are wanting. D = 150 x (108)^3 / 48 (30,000,000)(.5397) = 0.24 inches If you want a little safety factor, you could go slightly heavier wall thickness. If there is anything about this that you don't understand, please ask for a clarification. Good luck, Redhoss```
 km2000-ga rated this answer: ```Well redhoss, you have provoded me with EXACTLY what I need to know, in EXACTLY the terms I need to know it in. Thank you.```