Horizontal Vs Vertical Steel tube

[blake92]

How does the orientation affect how much a steel tube can hold? or does it not really have an effect?

 

[Simon Bridge]

It's to do with the loading.
The horizontal tube will crush more easily, and can bend in the middle (depending on how it is supported).
The upright tube can still buckle in the middle - but only if it is not perfectly upright or already has a defect along it;s length.

Experiment with drinking straws, or cardboard tubes - which way up do they hold the most weight?
(You'll need more than one straw.)

 

[blake92]

do the vertical tubes have a completely different weight capacity than the horizontal? or just different ways it can fail?

 

[Simon Bridge]

They have different load-bearing capacities - just like any change in geometry does.
Try the experiment suggested - get some cardboard tubes and lie them horizontally and see how much weight they support ... try again vertically.

 

[blake92]

do you know how to determine the capacity of the vertical tube?

 

[Simon Bridge]

You pile stuff on it until it breaks.
Usually someone has already done it - you can look it up in tables supplied by the manufacturer.
If you are buying it, then give your specs to the sales rep and they will select a range for you to choose from.

You may want to look up "tensile strength" and "compressive strength".

Note: a steel tube, end on, with a LOT of weight on it, will usually be driven into the ground.
Consider: a nail is basically a steel cylinder. It will take a lot of pounding without bending provided it is hit end on. But hit it skew and it bends easily.

 

[eistein1994]

How do you determine the Wall thickness for square steel tubing with a length of 5300mm that must be able to bare a work load of 10-12ton?

 

[blake92]

How do you determine the Wall thickness for square steel tubing with a length of 5300mm that must be able to bare a work load of 10-12ton?

i don't know a whole lot but i do know youre going to need more information than that. If you don't have to have certain width and height dimensons you could have litterally hundreds of possiblities. Is there certain width/height dimensions that the tube needs to be? or maybe a certain weight?

I attached a document that i found very helpful and as you can see there is countless possibilites.

 

[Baluncore]

Firstly ,as height gets greater you will reach a point where column buckling occurs spontaneously.
http://en.wikipedia.org/wiki/Buckling#Columns

Secondly, what side loads such as wind are expected?
Thirdly, how will you erect it without failure of the structure?

 

[blake92]

Firstly ,as height gets greater you will reach a point where column buckling occurs spontaneously.
http://en.wikipedia.org/wiki/Buckling#Columns

Secondly, what side loads such as wind are expected?
Thirdly, how will you erect it without failure of the structure?

We will be assuming no other outside forces like wind are acting on it.

Here is a rough idea of what I am talking about. I just wanted to find how much the vertical supports could roughly hold.

 

[Baluncore]

Find the total cross sectional area of material in the leg wall.
Find the minimum yield strength of the material and so compute the maximum safe compressive force.
Divide by g to convert that force to a weight.
Don't forget to subtract the weight of the leg and the frame above from that maximum.

 

[blake92]

Find the total cross sectional area of material in the leg wall.
Find the minimum yield strength of the material and so compute the maximum safe compressive force.
Divide by g to convert that force to a weight.
Don't forget to subtract the weight of the leg and the frame above from that maximum.

i did most of that,

A=5.75in^2
σ=30.36ksi
I=31.75in^4
L=5ft
radius of gyration= 2.349in
K=.5 (i just assmued fixed ends since its secured at the top and bolted to the floor)

i then used this equation- Pcritical = (∏^2)(E)(I)/(KL)^2

but my answers never made any sense so i wasnt sure what i was doing wrong.

 

[Baluncore]

but my answers never made any sense so i wasnt sure what i was doing wrong.

I am not at all surprised that you are not sure.
If you insist on using archaic measurement systems or don't show your working, then I really can't help you.
Use SI.

 

[SteamKing]

i did most of that,

A=5.75in^2
σ=30.36ksi
I=31.75in^4
L=5ft
radius of gyration= 2.349in
K=.5 (i just assmued fixed ends since its secured at the top and bolted to the floor)

i then used this equation- Pcritical = (∏^2)(E)(I)/(KL)^2

but my answers never made any sense so i wasnt sure what i was doing wrong.

The critical buckling load of the legs of your structure may not be the limiting feature of your frame.

I refer you to this thread, where the buckling of a column similar to your frame was discussed:

https://www.physicsforums.com/showthread.php?t=763056

In addition to the strength of the legs, you also have to check the upper part of your frame if loads are applied on the horizontal members. In general, you want to examine the longest horizontal members in bending and shear, then check the shorter members, until everything is checked. Once you have identified the horizontal member or members whose loading is most critical, then you can check the loading in the legs to see if any problems arise there.

Because it appears you are using open structural tubing in some of the horizontal members of your frame, a more detailed investigation of the tubing structure may be required, depending on the max. loading of the frame.

Once you have the loading of the structure determined, then the weld design can commence.

Open structures generally do not attract much in the way of wind loads, unless you are designing the frame to withstand a tornado or some other type of cyclonic wind.

If you expect some significant side loading, say from seismic activity, or if the frame is to be mounted on a moving platform, then additional investigation will have to be carried out, to check the strength of the frame under combined vertical and lateral loading.

Don't worry about the units for your calculation so much, as long as you use consistent ones. Converting units back and forth just provides another source for errors to creep into a calculation.

 

[blake92]

The critical buckling load of the legs of your structure may not be the limiting feature of your frame.

I refer you to this thread, where the buckling of a column similar to your frame was discussed:

https://www.physicsforums.com/showthread.php?t=763056

In addition to the strength of the legs, you also have to check the upper part of your frame if loads are applied on the horizontal members. In general, you want to examine the longest horizontal members in bending and shear, then check the shorter members, until everything is checked. Once you have identified the horizontal member or members whose loading is most critical, then you can check the loading in the legs to see if any problems arise there.

Because it appears you are using open structural tubing in some of the horizontal members of your frame, a more detailed investigation of the tubing structure may be required, depending on the max. loading of the frame.

Once you have the loading of the structure determined, then the weld design can commence.

Open structures generally do not attract much in the way of wind loads, unless you are designing the frame to withstand a tornado or some other type of cyclonic wind.

If you expect some significant side loading, say from seismic activity, or if the frame is to be mounted on a moving platform, then additional investigation will have to be carried out, to check the strength of the frame under combined vertical and lateral loading.

Don't worry about the units for your calculation so much, as long as you use consistent ones. Converting units back and forth just provides another source for errors to creep into a calculation.

I determined the allowable load for each of the beams so i know how much they can hold individually assuming the load was evenly distributed on their top face. I also determined that each of the four vertical beams can hold 131,000lbs before failure( assuming the load is applied directly straight down and no other forces are acting on it) i think that is correct but not 100% positive. I feel it makes sense but I am not sure.

I would they like to know even if the 131,000 is wrong how would i go about determining the max load for the entire structure? would i just say its roughly 131,000*4 (for each leg) or is there more to it than that?

 

[SteamKing]

I determined the allowable load for each of the beams so i know how much they can hold individually assuming the load was evenly distributed on their top face.

Great! What are the allowable loads you determined for these members? From your sketch, it doesn't look like the horizontal members are all the same cross section. Can you provide details?

I also determined that each of the four vertical beams can hold 131,000lbs before failure( assuming the load is applied directly straight down and no other forces are acting on it) i think that is correct but not 100% positive. I feel it makes sense but I am not sure.

The max. load in the vertical supports may not be the limiting feature of this frame.

I would they like to know even if the 131,000 is wrong how would i go about determining the max load for the entire structure? would i just say its roughly 131,000*4 (for each leg) or is there more to it than that?

There's probably more to it than just the max. allowable load in each leg, but I can't say because I don't have all the details of your structure and your calculations.

 

[blake92]

Great! What are the allowable loads you determined for these members? From your sketch, it doesn't look like the horizontal members are all the same cross section. Can you provide details?
The max. load in the vertical supports may not be the limiting feature of this frame.
There's probably more to it than just the max. allowable load in each leg, but I can't say because I don't have all the details of your structure and your calculations.

The following are the max allowable UNIFORM loads on the horizontal beams:
*assuming a weight is applied evenly across the top surface to each beam*

HSS 6x6x1/4in Square Steel Tubing

for the four equal length 8ft beams (purple if the color shows in the picture):
i got an answer of 24kips or 24,000lbs for each.

for the baby blue beam 12ft that is perpendicular to the first four:
i got an answer of 16Kips or 16,000lbs
i felt this made sense bc as a beam gets longer it would be able to hold slighlty less.

Lastly the front beam (dark green). This beam is actually an 12ft I-beam so i had a harder time figuring out the uniform load but i was able to determine a max point load at the mid point of 8.6kips so i know it can at the very least hold that much.

i even went on to determine the weight of each beam which is about 19.02lb/ft for the square tubingAlso it might be hard to tell in the picture but the four purple beams are notched and fit inside the I-beam where they were then welded.

 

[SteamKing]

The following are the max allowable UNIFORM loads on the horizontal beams:
*assuming a weight is applied evenly across the top surface to each beam*

HSS 6x6x1/4in Square Steel Tubing

for the four equal length 8ft beams (purple if the color shows in the picture):
i got an answer of 24kips or 24,000lbs for each.

for the baby blue beam 12ft that is perpendicular to the first four:
i got an answer of 16Kips or 16,000lbs
i felt this made sense bc as a beam gets longer it would be able to hold slighlty less.

Lastly the front beam (dark green). This beam is actually an 12ft I-beam so i had a harder time figuring out the uniform load but i was able to determine a max point load at the mid point of 8.6kips so i know it can at the very least hold that much.

i even went on to determine the weight of each beam which is about 19.02lb/ft for the square tubingAlso it might be hard to tell in the picture but the four purple beams are notched and fit inside the I-beam where they were then welded.

Because of the layout of the horizontal members, you can't assume that all of the loads are evenly distributed. For example, the two purple beams in the middle of the frame, which connect the two 12-ft beams, impose point loadings on the longer beams at the points of connection. Even though the loads on the purple beams may be evenly distributed, the same cannot be said for the longer beams. You have to analyze the longer beams with a mix of point loadings and distributed loads.

In other words, the longer beams must support the loads directly imposed on them as well as the load imposed on the shorter beams.

BTW, which section did you select for the I-beam?

 

[blake92]

Because of the layout of the horizontal members, you can't assume that all of the loads are evenly distributed. For example, the two purple beams in the middle of the frame, which connect the two 12-ft beams, impose point loadings on the longer beams at the points of connection. Even though the loads on the purple beams may be evenly distributed, the same cannot be said for the longer beams. You have to analyze the longer beams with a mix of point loadings and distributed loads.

BTW, which section did you select for the I-beam?

i realize the point load part, and in my hand drawn sketchs of this set-up i made the two middle purple beam have a point load on both end beams. The one thing i was not sure about is how large that point load is? let's say i assume the two middle beams can hold a max of 24kips (uniform load) then does that mean i can split the difference between the front and back beam? for example each middle purple beam would apply a point load of 12 kips on the front and back horizontal beam.

also i used a W 6x20 A36 I-beam

 

[SteamKing]

i realize the point load part, and in my hand drawn sketchs of this set-up i made the two middle purple beam have a point load on both end beams. The one thing i was not sure about is how large that point load is? let's say i assume the two middle beams can hold a max of 24kips (uniform load) then does that mean i can split the difference between the front and back beam? for example each middle purple beam would apply a point load of 12 kips on the front and back horizontal beam.

also i used a W 6x20 A36 I-beam

Yes, if you create a free body diagram of one of the middle beams with a distributed load, you'll see that the end reactions of this beam are each half of the total distributed load. These reactions must then be applied to the long beams so that they can be analyzed in turn.

 

[blake92]

Yes, if you create a free body diagram of one of the middle beams with a distributed load, you'll see that the end reactions of this beam are each half of the total distributed load. These reactions must then be applied to the long beams so that they can be analyzed in turn.

now since two of the four purple beams rest over top the vertical beams do i only really need to consider the middle two purple beams?

Also this is what i have so far,

A= 5.75in^2
S= 10.58in^3
wt=19.02lb/ft
σy= 46ksi = σallow=30.36ksi

Mallow= Sσallow
M = 10.58*30.36ksi = 321.2kip*ft

Mp=Pa
Mp = 12kips*35in= 420kip*ft (35in is the distance from the edge of the back beam to the middle of one of the middle purple beams)

Mw=Mallow-Mp
Mw= 321.2-420= -98.8kip*ft (which can't be right because its negative but that's what i keep getting)

When iv done similair problems that answer is alwasy positive. i don't know if maybe I am using the wrong equations or what.

the ones I am using say they are for "Determining the allowable uniform load ω that can be applied to the beam that is also subjected to a pair of loads."

 

[blake92]

Yes, if you create a free body diagram of one of the middle beams with a distributed load, you'll see that the end reactions of this beam are each half of the total distributed load. These reactions must then be applied to the long beams so that they can be analyzed in turn.

i tried one other way,

A=5.75in^2
S=10.58in^3
wt= 19.02lb/ft
L=12ft
a=35in=2.916ft

Mmax= Pa+wL^2/8
(12,000lb)(2.916ft)+(19.02lb/ft)(12ft)^2/8
= 35334.36lb*ft

σmax= Mmax/S
(35334.36lb*ft)/(10.58in^3)
= 40,076.77psi

Flexural stress for HSS 6X6X1/4 =30,360psi
40,076psi > 30,360psi thus the bending strength of the beam is not satisfactory!

the only reason why i doubt this is because this current racks are in use with sometimes up to 20,000lbs on top of them and not one has ever failed in the decades they have been in used.

 

[SteamKing]

I'll take an in-depth look at your calculations and get back to you.

 

[blake92]

I'll take an in-depth look at your calculations and get back to you.

Okay, thank you!

 

[blake92]

I'll take an in-depth look at your calculations and get back to you.

I apparently made a mistake when drawing up the design and there are these extra supports in the back as displayed. These new vertical beams are directly beneath the two middle purple beams.

because of this, could i assume that where those vertical beams come up is the end of the beam, and solve for a beam with two equal loads on above where the reactions are at the end?

 

[SteamKing]

I apparently made a mistake when drawing up the design and there are these extra supports in the back as displayed. These new vertical beams are directly beneath the two middle purple beams.

because of this, could i assume that where those vertical beams come up is the end of the beam, and solve for a beam with two equal loads on above where the reactions are at the end?

You still have to analyze the longitudinal I-beam for max. load, if for no other reason because it is supported only at the ends.

The longitudinal HSS tube in the back has supports under all points where the cross beams are welded to it, so it is not as critical as the I-beam. However, the vertical supports for the middle cross beams appear to be quite a bit smaller than the end supports, so they should be checked to determine max. loading as columns. Because the two cross members in the middle of the frame are supported only a the ends and have the largest span, they still need to be checked for max. distributed loading.

 

[blake92]

You still have to analyze the longitudinal I-beam for max. load, if for no other reason because it is supported only at the ends.

The longitudinal HSS tube in the back has supports under all points where the cross beams are welded to it, so it is not as critical as the I-beam. However, the vertical supports for the middle cross beams appear to be quite a bit smaller than the end supports, so they should be checked to determine max. loading as columns. Because the two cross members in the middle of the frame are supported only a the ends and have the largest span, they still need to be checked for max. distributed loading.

I figured you would say that so i already figured out that the shorter orange beams(2ft) can hold 89kips and the longer red beams(3ft) can hold about 85kips before failure (once again assuming its standing perfectly stright with no other forces acting on it) they are also both HSS 4x4x1/4. I am currently working on a point load for the I-beam.

 

[blake92]

For the point load on the I-beam

W 6x20
A=5.9in^2
S=13.4in^3
depth(d)=6.20in
thickness(tw)=.260in
wt=20lb/ft=.02kips/ft

Mallow= Sσ
M= (13.4in^3)(30.36kip/in^2) = 406.8kip*in

Mp= Pa
Mp= (12kips)(35in) = 420 kip*in

Mw=406.8-420 =13.2kip*in

w=8M/L^2
w= (8)(13.2kip*in)(12in)/(12ft)= 1.1kips/ft

Vmax= P+wL/2
Vmax= 12kips+(1.1)(12)/2 = 18.6 kips

Average Shear Stress = Vmax/dtw
shear stress = 18.6/(6.20)(.260)= 11.5ksi

11.5 is within the allowable stress of 14.5ksi for A36 steel



i think i did this all correctly, the only part was when i solved "Mw" i didnt make my answer negative because since its a moment i thought sign only meant direction and i didnt think that mattered and i only needed the value.

 

[SteamKing]

For the point load on the I-beam

W 6x20
A=5.9in^2
S=13.4in^3
depth(d)=6.20in
thickness(tw)=.260in
wt=20lb/ft=.02kips/ft

Mallow= Sσ
M= (13.4in^3)(30.36kip/in^2) = 406.8kip*in

Mp= Pa
Mp= (12kips)(35in) = 420 kip*in

Mw=406.8-420 =13.2kip*in

w=8M/L^2
w= (8)(13.2kip*in)(12in)/(12ft)= 1.1kips/ft

Vmax= P+wL/2
Vmax= 12kips+(1.1)(12)/2 = 18.6 kips

Average Shear Stress = Vmax/dtw
shear stress = 18.6/(6.20)(.260)= 11.5ksi

11.5 is within the allowable stress of 14.5ksi for A36 steel



i think i did this all correctly, the only part was when i solved "Mw" i didnt make my answer negative because since its a moment i thought sign only meant direction and i didnt think that mattered and i only needed the value.

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.

 

[SteamKing]

This rack is limited in its capacity by the W6"x20# longitudinal beam. Out of all the other members, this beam is the only one made out of mild steel, it has the longest unsupported span, and the W-section, being open, is not as strong as the HSS tubing. That's why you cannot just load up the cross members arbitrarily without checking what happens to the W6".

 

[blake92]

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?

 

[blake92]

i take that back. it doesn't work.

am i using the correct equations to determine the uniform load on the I-beam?

 

[SteamKing]

i take that back. it doesn't work.

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.

 

[SteamKing]

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:

View attachment Rack-Calc.pdf

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.

 

[blake92]

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:

View attachment 71733

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.