deflection of steel rectangular tubing
Category: Science > Physics
Asked by: mikeinmesquite-ga
List Price: $10.00
21 Dec 2005 16:04 PST
Expires: 20 Jan 2006 16:04 PST
Question ID: 608669
I am building an open staircase with a single stringer using 6in. x 4 in. rectangular steel tubing, with the long dimension vertical. The overall tube length will be approx. 166 inches. It will be supported at the floor and welded to a steel I beam at about 96in. off the floor, so that the angle off horiz. will be about 35 degrees. The dead load will be about 900 pounds (evenly supported) and the live load will be 300 pounds center-loaded. Question: what tube wall thickness is required to keep the beam (tubing) deflection safe and in an acceptable range so the stair does not feel ?bouncy?(my guess is less than ¼ in.). I had planned to use 6x4x¼ inch wall tubing with a moment of inertia of 11.1 (in)4. Is this acceptable? What is the calculated deflection? Doesn?t it help that the ?beam? is angled, not horizontal?
Re: deflection of steel rectangular tubing
Answered By: redhoss-ga on 22 Dec 2005 09:19 PST
Hello mike, it is a crappy day here and because I have nothing better to do we will work on your project. First we need to get the component of the loads in the direction perpendicular to the neutral axis of your rectangular tubing. As you guessed it "helps" that the beam in on an angle. Both loads need to be multiplied by the cosine of 35 degrees or 0.82: Point load = 300 x 0.82 = 246 lb. Distributed load = 900 x 0.82 = 738 lb. Each side of the staircase supports 1/2 of the point load and the distributed load. Even though the distributed load is applied evenly along the length of the beam there is very little difference between that case and assuming that the entire load is applied at the center of the span. Also, the formula for a point load in combination with an evenly distributed load gets much more complicated. For our simple (worst case) solution we will assume: P(1) = 246 x 1/2 = 123 lb. P(2) = 738 x 1/2 = 369 lb. P(total) = 482 lb. Deflection = (P x L^3) / (192 x E x I) Where: L = 166 in. and L^3 = 4,574,298 in.^3 E = 30,000,000 psi (a constant for any steel) I = 21.6 in.^4 (your value of 11.1 is for the wrong axis) D = (482 x 4,574,298) / (192 x 30,000,000 x 21.6) = 0.01 inch Your design looks very good to me and you should feel no bounce. Please ask for a clarification if you have any questions about what is shown here. Good luck with your project, Redhoss
rated this answer:
excellent answerer! very fast response as well
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