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Q: Calculating the ability of a Steel Tube to support a Load ( Answered 5 out of 5 stars,   0 Comments )
Subject: Calculating the ability of a Steel Tube to support a Load
Category: Science
Asked by: brymbo-ga
List Price: $40.00
Posted: 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)

Request for Question Clarification by redhoss-ga on 13 Oct 2006 16:32 PDT
I understand your question completely. I can't dispute any of your
numbers, but without knowing the exact layout of your supported load I
have no way of confirming if the 8561 number is correct. You seem to
know what you are talking about and I trust that you are correct in
everything you say. The remaining question is the application of
Euler's equation for column buckling using your numbers. The values
you use for the pipe columns are consistant with 3 1/2 inch extra
strong pipe which has a wall thickness of .318 inches. If this is
true, I can recommend a top plate thickness and gussets. If this
sounds like what you are looking for, I will be glad to help you.

Clarification of Question by brymbo-ga on 14 Oct 2006 01:28 PDT
It is difficult to convey details of the layout to you without a plan
but if the values I have used for floor/roof loading etc are correct
then I am reasonably happy with the value and positioning of the 8561
resultant.(I have placed proportionate loads on a scale model and it
balances about the centre of effort)
Do please go ahead with your answer. If you can show me the workings
it would be useful in case I want to use alternative sizes of
pipe/plate. It would also be useful to me if you would confirm that
Euler's equ does apply to this problem and would give a loading
capacity of 250,000 lbs!

Request for Question Clarification by redhoss-ga on 14 Oct 2006 06:40 PDT
I don't understand where you get the "loading capacity of 250,000
lbs!". What does this have to do with your problem.

Clarification of Question by brymbo-ga on 14 Oct 2006 08:17 PDT
When you mentioned Euler I did the equation F = E*I*Pi^2/L^2
i.e. F = 30,000,000*6*3.14^2/84^2 lbs = 251,520 lbs. This is extra to
the original question but I'd still like to know whether  it is
correct. Sorry for the confusion.

Request for Question Clarification by redhoss-ga on 14 Oct 2006 14:57 PDT
It seems to me that you are well capable of answering your own
question. I am very impressed with your knowledge of engineering. I
can help if you want, but you really don't need any help at all.

Clarification of Question by brymbo-ga on 14 Oct 2006 16:13 PDT
I would still like you to answer if you would. I need to know from
someone I trust  1. Will such a pipe support 8561 lbs? 2. How much
tolerance is there in the lateral positioning of the pillar? 3.What do
I use for a top plate? I have only done these calcs. by looking at
previous answers of yours. I have no experience and I may be missing
something important so its well worth the fee to me to have someone
hold my hand. Thanks in anticipation.
Subject: Re: Calculating the ability of a Steel Tube to support a Load
Answered By: redhoss-ga on 17 Oct 2006 13:01 PDT
Rated:5 out of 5 stars
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:

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

Request for Answer Clarification by brymbo-ga on 18 Oct 2006 09:35 PDT
That is a first class answer Redhoss. Thank you v. much. As you've
never contradicted the values I used for roof/floor loading etc. I
assume they are reasonably O.K. I'll feel a lot easier going ahead
with the job now and the advice about the top plate is also v.
welcome. Thanks again

Clarification of Answer by redhoss-ga on 18 Oct 2006 10:13 PDT
Thank you for the kind words. Yes, you did a good job in calculating
the loads and I see nothing that looks the least bit out of line.
brymbo-ga rated this answer:5 out of 5 stars
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.

There are no comments at this time.

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