Calculating the ability of a Steel Tube to support a Load
Asked by: brymbo-ga
List Price: $40.00
13 Oct 2006 12:15 PDT
Expires: 12 Nov 2006 11:15 PST
Question ID: 773254
I have 3 load-bearing lintels coming together, and resting on, a brick pillar. This brick pillar is bulky and I wish to replace it with a 4" dia vertical steel pipe with a triangular plate welded on the top for the lintel ends to rest on. (This pipe will be 7ft. long) Such a pipe would, I think, have a moment of inertia of 6 ins^4 and section modulus, A, of 3 ins^3. In calculating the loads imposed by each lintel I have used the following values. Wt. of roof = 15 lbs/squ ft, Wt. of bathroom foor + people + furniture = 33 lbs/squ ft. Density of concrete slab roof = 4042 lbs/cu yard. Wt. of brick = 10 lbs. I would like confirmation that these values are reasonable. Using the above values I have calculated the position and value of the centre of effort of these 3 loads. The value is 8561 lbs. If the steel tube pillar is not positioned directly below the centre of effort as calculated but (in error) displaced laterally by a distance r, this will create a bending moment on the pillar. What is the max. permissible value of r? i.e. how accurate do I need to be in my positioning of the pillar? What thickness should the triangular top plate be. The "worst" lintel load on it is 3333lbs at a distance of 13 ins. from the correct positioning of the pillar (I assume I'll need to use triangular gussets welded to plate and pipe)
Re: Calculating the ability of a Steel Tube to support a Load
Answered By: redhoss-ga on 17 Oct 2006 13:01 PDT
Okay brymbo, I have scratched my head long enough and have a good answer for you. Your question caused me to go way back and look at some things I haven't looked at since engineering school. When I mentioned Euler's equation I somewhat mislead you. The version of Euler's that you solved gives an answer that is theoretical and should not be used in your case. If you are interested, read this: http://www.efunda.com/formulae/solid_mechanics/columns/inelastic.cfm What we are looking for is a more conservative solution that allows some factor of safety. In my AISC (American Institute of Steel Construction) manual there are tables for column selection. There are also formulas from which these tables are derived, but they are very boring and not necessary when you can use the applicable table. For your particular column (3 1/2 extra strong steel pipe) and length (7 ft) the value for the allowable concentric load is 63,000 lbs. There is also a "bending factor", B= 1.17. What we need is P' which is defined as the equivalent axial load which would be required to produce a maximum bending moment (M) in the column. I hope this hasn't confused you too much. Here is what we need to calculate: P' = B x M and P + P' = 63,000 lbs We know that in your case P = 8,561 lbs Therefore: P' = 63,000 - 8,561 = 54,440 lbs M = P'/B = 54,440 / 1.17 = 46,500 in-lbs So, for your 8,561 lb load to produce this moment the maximum offset is: 46,500 in-lbs / 8,561 lbs = 5.43 inches This means that you could miss the center of the column by 5.43 inches to either side. Now for the top plate. A good rule-of-thumb is to not use a plate thickness greater than the wall thickness of the column. This is true for two reasons. It is never good to weld a thick section to a lighter material. You will not get good penetration into the thicker material. Also, you will fail the lighter material anyway. Your gussets and top plate should be cut from 3/8 inch plate. I believe that this is what you are looking for, but please ask for a clarification if I have missed something or you don't understand. Good luck with your project, Redhoss
rated this answer:
Very pleased with the answer. An added dimension that made the process all the more agreeable was the pleasant way Redhoss asked for various clarifications and gave his answer.
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