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Q: bending strength of 1" x 2" rectangular steel tubing ( Answered 5 out of 5 stars,   0 Comments )
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".
Subject: Re: bending strength of 1" x 2" rectangular steel tubing
Answered By: redhoss-ga on 15 Nov 2005 18:58 PST
Rated:5 out of 5 stars
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:

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:5 out of 5 stars
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.

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