The orientation of steel tubes significantly affects their load-bearing capacity, with vertical tubes generally able to support more weight than horizontal tubes due to their resistance to buckling. Horizontal tubes are more prone to crushing and bending, especially if not adequately supported. Experiments with simple materials like drinking straws illustrate these principles, showing that vertical configurations typically hold more weight. Calculating the maximum load involves determining the cross-sectional area and yield strength of the material, while also considering potential side loads and structural integrity. Proper assessment of both vertical and horizontal members is crucial for ensuring the overall stability of the structure.
#31
blake92
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SteamKing said:
The W6"x20# beam is A36 steel, so the allowable stress will not be 30.36 ksi. At best, it will be 0.66*36 = 23.76 ksi in bending, according to AISC.
Thats right!
okay i redid the calculations and instead of406.8kip*in i now get 321.6 kip*in
and it still ends up working in the end. I just wasnt sure if it was okay for me to drop the negative symbol from the Mw above?
am i using the correct equations to determine the uniform load on the I-beam?
You have also assumed in your calculations that the ends of the I-beam are simply supported, which IMO is not entirely accurate, due to the construction layout of the rack. The ends of the beam are not free to rotate, so a bending moment develops there.
I've taken a quick look at the die rack and made some hand calculations of the maximum loading which it can support. These calculations are attached below:
If the two HSS tubing cross-members in the middle are loaded to 250 lbs./in, and there is no distributed load applied directly on the I-beam, the bending stresses at the ends of this beam reach the allowable bending stress for A36 steel. (the self-weights of all members have been neglected)
Of course, these calculations are extremely rough. If you are concerned that the rack may be overloaded, I would recommend that a structural model of the entire rack be analyzed by an experienced structural engineer.
As I have mentioned in a previous post, mixing HSS and mild steel members in the same frame means that you will load up the mild steel member to its stress limit before the HSS members.
Of course, none of this analysis has considered the welding used to fabricate this rack, which is something one can do only be examining the actual rack.
#35
blake92
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SteamKing said:
blake92:
I've taken a quick look at the die rack and made some hand calculations of the maximum loading which it can support. These calculations are attached below:
If the two HSS tubing cross-members in the middle are loaded to 250 lbs./in, and there is no distributed load applied directly on the I-beam, the bending stresses at the ends of this beam reach the allowable bending stress for A36 steel. (the self-weights of all members have been neglected)
Of course, these calculations are extremely rough. If you are concerned that the rack may be overloaded, I would recommend that a structural model of the entire rack be analyzed by an experienced structural engineer.
As I have mentioned in a previous post, mixing HSS and mild steel members in the same frame means that you will load up the mild steel member to its stress limit before the HSS members.
Of course, none of this analysis has considered the welding used to fabricate this rack, which is something one can do only be examining the actual rack.
ill look over your calculations.
im not concered that its overloaded, i just needed to know at what point will it be overloaded or roughly what point.
i also tought about the welds but 1) we don't know who welded it or how well they welded but assuming they did a satisfactory job i do have calculations to determine the weld strength.
#36
blake92
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can you also explain what you did here,
Beam Forces and Moments:
P = 12,000 lbs. from loading on cross-member
a = 35 in.
b = 109 in.
R1 = 10,218 lbs.
R2 = 1,782 lbs.
mainly what is the 109in and where is it measured from?
and the significant difference in your reactions at 1 and 2
If the beam is 144 in long, and one cross member is located app. 35 in from one end, then 144-35 = 109 in.
See p.6 of the calculations, Load Case 17 for a layout of the beam and the formulas used to calculate the bending moments. There are two cross-members tying into the mid-span of the I-beam, so I used Load Case 17 to calculate the moments due to the reaction coming from one cross member, and then flipped them w.r.t. the center of the I-beam and combined the moments to get the total load due to the two center cross-members. An example of superposition, if you will, to calculate a complex beam loading from a simpler one found in a table.
On that same page, Load Case 15 was used to calculate the moments and reactions in the
cross members under the distributed loading.
#38
blake92
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SteamKing said:
If the beam is 144 in long, and one cross member is located app. 35 in from one end, then 144-35 = 109 in.
See p.6 of the calculations, Load Case 17 for a layout of the beam and the formulas used to calculate the bending moments. There are two cross-members tying into the mid-span of the I-beam, so I used Load Case 17 to calculate the moments due to the reaction coming from one cross member, and then flipped them w.r.t. the center of the I-beam and combined the moments to get the total load due to the two center cross-members. An example of superposition, if you will, to calculate a complex beam loading from a simpler one found in a table.
On that same page, Load Case 15 was used to calculate the moments and reactions in the
cross members under the distributed loading.
i wish i could show you the book I am using, It has a page very similar to page 6 of your document except it has a case where there is a simple beam with two equal concentrated loads symmetrically placed.
So it shows an I-beam with two "P"s coming down at equal distanced from each other then i would combine this with the case for a simple beam with uniform loads
thus my V=P and V=wL/2 so for the combined situation id have P+(wL/2)
#39
blake92
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there is even an example in the book that gives literally the exact same situation that i have except it uses metric and they are obviously different numbers.
so i just followed the steps it shows but with my numbers and i never get a reasonable answer. almost triple the allowable stress.
#40
blake92
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Here i was able to attach a picture of the problem i was using as an example.
The moment calculation formulas are taken from the AISC Manual for Steel Construction (I forget the exact edition, but probably the Eighth edition or so).
I would also like to point out that the properties of the HSS 6"x6"x1/4" tube, which I took from the AISC database, are slightly different than your values, particularly the cross-sectional area. This is why I included a sheet of section properties for the tube and the I-beam in my calculations.
Where I disagree with the example problem you posted, the beam is simply supported in the example, which means that the ends are free to rotate and thus cannot accept a bending moment. I have treated the ends of the I-beam rather conservatively by considering them to be fixed, that is, the ends are not free to rotate, thus a rather sizable moment develops there.
The truth lies somewhere in between our two approaches, which is why I suggested that a structural engineer develop and analyze a model which includes the effect of the stiffness of the cross-members on the response of the I-beam to various loads.
Beam Forces and Moments:
P = 12,000 lbs. from loading on cross-member
a = 35 in.
b = 109 in.
R1 = 10,218 lbs.
R2 = 1,782 lbs.
mainly what is the 109in and where is it measured from?
and the significant difference in your reactions at 1 and 2
The 12000 lb. load was the maximum load which could be applied to the I-beam by the middle cross members, each located 35 in. from the end of the I-beam. R1 and R2 are the reaction at the ends of the I-beam due to the application of a single 12000-lb. load as described. I did not use these reactions further in the beam analysis because the fixed-end moments produced much greater bending stress in the I-beam.
Obviously, the stresses in the beam ends will be greater due to these reactions, but I neglected them to estimate the max. loading which could be permitted on the cross-members without over stressing the I-beam.
#43
blake92
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SteamKing said:
The 12000 lb. load was the maximum load which could be applied to the I-beam by the middle cross members, each located 35 in. from the end of the I-beam. R1 and R2 are the reaction at the ends of the I-beam due to the application of a single 12000-lb. load as described. I did not use these reactions further in the beam analysis because the fixed-end moments produced much greater bending stress in the I-beam.
Obviously, the stresses in the beam ends will be greater due to these reactions, but I neglected them to estimate the max. loading which could be permitted on the cross-members without over stressing the I-beam.
okay i see what you're saying. Let's say i had this all figured out, and say we determined the uniform load of the I-beam to be 10kips/ft. where would we go from there? like what would the next step be in determining the max load of the rack?
we know how much the legs can support vertically and we know the uniform loads for all the horizontal beams. would the max load essentially be the max load of the legs?
#44
blake92
50
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would i be able to say,
the horizontal beams together can hold,
24kips (purple)
24kips (purple)
24kips (purple)
24kips (purple)
16kips (blue)
10kips (i-beam)
which totals 122kips?
of course it wouldn't be exact but as a rough idea?
The loads in the legs are not the limiting factor in the capacity of this rack.
The limiting factor is the 144"-span I-beam made out of mild steel (A36). The HSS tubing has a much higher yield stress that the I-beam. If you load the cross members to 24 kips, that's the limit of what the I-beam can support, without any additional load distributed along the length of this beam. (in other words, no 10 kips or 10 kips/ft., or whatever).
IDK what your desired loading is for this rack, but the W6"x20# mild steel beam is not cutting it, IMO. If you want to add additional vertical supports in way of the cross members' intersection with this beam, as is shown on your revised sketch under the longitudinal tube beam at the back of the rack, that would reduce the bending stress in the I-beam. If you wanted to replace the mild steel I-beam with a HSS tube, that would probably increase the allowable load of the rack.
A beam with an unsupported span of 144" (12 feet) can only support so much load.
#46
blake92
50
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SteamKing said:
The loads in the legs are not the limiting factor in the capacity of this rack.
The limiting factor is the 144"-span I-beam made out of mild steel (A36). The HSS tubing has a much higher yield stress that the I-beam. If you load the cross members to 24 kips, that's the limit of what the I-beam can support, without any additional load distributed along the length of this beam. (in other words, no 10 kips or 10 kips/ft., or whatever).
IDK what your desired loading is for this rack, but the W6"x20# mild steel beam is not cutting it, IMO. If you want to add additional vertical supports in way of the cross members' intersection with this beam, as is shown on your revised sketch under the longitudinal tube beam at the back of the rack, that would reduce the bending stress in the I-beam. If you wanted to replace the mild steel I-beam with a HSS tube, that would probably increase the allowable load of the rack.
A beam with an unsupported span of 144" (12 feet) can only support so much load.
this set-up is supposed to be able to hold a total of 20,000lbs evenly distributed across the entire top surface.
but of course that's the bare minimum that we want it to hold. we really want it to hold 1.5x to 2x that much weight
blake92: I currently would assume the beams are simply-supported (for checking the midspans). If we blindly ignore dynamic loading, uneven loading, and weldment stresses, then I currently would say the horizontal beams together could hold,
Code:
26 900 N (purple)
89 120 N (purple)
89 120 N (purple)
26 900 N (purple)
0 N (blue)
0 N (I-beam)
which totals Ptotal = 232 040 N = 232.04 kN.
blake92 said:
okay, I redid the calculations and ... I now get [Mallow =] 321.6 kip*in.
Agreed (almost). I currently got Mallow = 36 220 N*m. Next, from your other thread ...
blake92 said:
... for a 6 ft section of the tubing, max load is [Pallow =] 131 kips. Does that sound correct to you?
Agreed (close enough). I used your given length L = 1613 mm, and currently got Pallow = 600 950 N, using your assumed boundary conditions. However, I would rather use ke = 2.10 for the leg, which instead gives Pallow = 530 230 N.
Therefore, we see that Pallow does not govern, because Pallow (530 230 N) is much greater than Ptotal/(4 legs) = 58 010 N. Therefore, the governing stress is currently bending stress at the midspan of the I-beam, where Mallow = 36 220 N*m. Therefore, Ptotal = 232 040 N.
However, perhaps they set Ptotal = 88 960 N due to dynamic loading, uneven loading, and/or weldment stresses. Unless you analyze everything carefully, including weldment stresses, etc., then it might be advisable to not exceed the previous load, Ptotal = 88 960 N.
Last edited:
#48
blake92
50
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nvn said:
blake92: I currently would assume the beams are simply-supported (for checking the midspans). If we blindly ignore dynamic loading, uneven loading, and weldment stresses, then I currently would say the horizontal beams together could hold,
Code:
26 900 N (purple)
89 120 N (purple)
89 120 N (purple)
26 900 N (purple)
0 N (blue)
0 N (I-beam)
which totals Ptotal = 232 040 N = 232.04 kN.
Agreed (almost). I currently got Mallow = 36 220 N*m. Next, from your other thread ...
Agreed (close enough). I used your given length L = 1613 mm, and currently got Pallow = 600 950 N, using your assumed boundary conditions. However, I would rather use ke = 2.10 for the leg, which instead gives Pallow = 530 230 N.
Therefore, we see that Pallow does not govern, because Pallow (530 230 N) is much greater than Ptotal/(4 legs) = 58 010 N. Therefore, the governing stress is currently bending stress at the midspan of the I-beam, where Mallow = 36 220 N*m. Therefore, Ptotal = 232 040 N.
However, perhaps they set Ptotal = 88 960 N due to dynamic loading, uneven loading, and/or weldment stresses. Unless you analyze everything carefully, including weldment stresses, etc., then it might be advisable to not exceed the previous load, Ptotal = 88 960 N.
no, the 88,960 N (20,000lbs) is just the weight of the heaviest piece of equiptment they usually put on the rack. So i know it can atleast hold that much.
