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Stress for tubing 3
- Thread starter DON54660
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Don54660...it would seem that your question would be a simple one; however, it is not!
For most static structural applications such as buildings and components, the allowable stresses for materials have been well established for many years. In most instances, the allowable stresses for steel will be in the range of 60 to 66 percent of the yield strength for tension and bending induced tensile stress, and about 40 percent of the yield strength for shear stresses.
These values accommodate the variabilities of materials, minor load variability, load sharing/distribution not considered in typical analytical techniques, etc.
For applications other than buildings, the values can vary a great deal, depending on application, margin of safety, load reversals, fatigue, etc.
For most static structural applications such as buildings and components, the allowable stresses for materials have been well established for many years. In most instances, the allowable stresses for steel will be in the range of 60 to 66 percent of the yield strength for tension and bending induced tensile stress, and about 40 percent of the yield strength for shear stresses.
These values accommodate the variabilities of materials, minor load variability, load sharing/distribution not considered in typical analytical techniques, etc.
For applications other than buildings, the values can vary a great deal, depending on application, margin of safety, load reversals, fatigue, etc.
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- #3
Thank you for responding, I am aware of a safety factor. My concern is in one reference book it list 22,000 for tube (Mild steel) and another for tensile and compression 65,000 psi for mild steel??? or am I missing something. This is for general construction and they do list a safety factor for certain conditions. My question is why a spec for tube and another for steel beams and sheets.
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- #5
Hello Lcubed, Actually two seperate references (Books)
The first is:
1.Machinery's Handbook H.Rourge Chapter title section is 'Strength of Materials'.
It was given in a chart called Stresses at yield point of different materials. examples
A. Machine steel...tensile 47,000 shearing 30,000 lb per sq in
B.High carbon steel... tensile 60,000 shearing 27,000
C.Steel tubing...22,000 tensile and shearing 13.000.
The caption above this table says this,"The accompanying table gives values for the unit tensile and unit shearing stresses at the yield point of a number of different matertials. By using a suitable factor of safety , the permissable working stresses ,may be obtained from this table", meaning the information given above and also meaning add in a safety factor for the figures shown.
The next was given in a book on building 'space frames' for sports car chassis (A British book). it listed the following for tubes.
A.20 ton tensile for welded seam mild steel tubing BS 980 ERW1.(Electrical seam welded tubing)
B.25 ton tensile for welded seam mild steel tubing BS 980 ERW2
C.Cold drawn BS 980 CDS1 20 Ton tensile
D.Cold drawn BS CDS2 28 Ton tensile
C.Other Cold Drawn.
TI 35 ton
T45 45 Ton
T50 50 Ton.
This last book is talking about square and round tubing only, that is, no shapes or sheet material,however the specs are close to the one given in the Machinery's Handbook for tube .
Hope this help clarify what I am looking for, which is, what stress should I use for tubing? I can figure in a safety factor for different condition, those are given in the Machinery's Handbook. Don
The first is:
1.Machinery's Handbook H.Rourge Chapter title section is 'Strength of Materials'.
It was given in a chart called Stresses at yield point of different materials. examples
A. Machine steel...tensile 47,000 shearing 30,000 lb per sq in
B.High carbon steel... tensile 60,000 shearing 27,000
C.Steel tubing...22,000 tensile and shearing 13.000.
The caption above this table says this,"The accompanying table gives values for the unit tensile and unit shearing stresses at the yield point of a number of different matertials. By using a suitable factor of safety , the permissable working stresses ,may be obtained from this table", meaning the information given above and also meaning add in a safety factor for the figures shown.
The next was given in a book on building 'space frames' for sports car chassis (A British book). it listed the following for tubes.
A.20 ton tensile for welded seam mild steel tubing BS 980 ERW1.(Electrical seam welded tubing)
B.25 ton tensile for welded seam mild steel tubing BS 980 ERW2
C.Cold drawn BS 980 CDS1 20 Ton tensile
D.Cold drawn BS CDS2 28 Ton tensile
C.Other Cold Drawn.
TI 35 ton
T45 45 Ton
T50 50 Ton.
This last book is talking about square and round tubing only, that is, no shapes or sheet material,however the specs are close to the one given in the Machinery's Handbook for tube .
Hope this help clarify what I am looking for, which is, what stress should I use for tubing? I can figure in a safety factor for different condition, those are given in the Machinery's Handbook. Don
I couldn't find the table in my 24th Edition of MH, but tubing made of a material should exhibit the same mechanical properties as any other form of the same material. If the tubing is welded, there may be a reduction in strength due to the welding, and that might account for the low published strength. Other than that, the same steel should have the same strength.
Since buckling is a common failure mode for tubing, my first thought in my earlier post was that buckling might be taken into account in your references, but I don't think that is the case.
Given only the information cited, I would lean toward the conservative approach, and use the low value. Sorry I can't be of more help.
Since buckling is a common failure mode for tubing, my first thought in my earlier post was that buckling might be taken into account in your references, but I don't think that is the case.
Given only the information cited, I would lean toward the conservative approach, and use the low value. Sorry I can't be of more help.
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- #7
Thanks all, I am not making anything, I am trying to understand some things about tube stress and design. Here is a sketch I made of a space frame that uses the tube I have questioned.
Later I shall post a calculation I made for bending to see if I have the concept right. Of course I am aware to design such a space frame is beyond the average person but I would like to look at something and understand (Roughly) what is going on. Don
Later I shall post a calculation I made for bending to see if I have the concept right. Of course I am aware to design such a space frame is beyond the average person but I would like to look at something and understand (Roughly) what is going on. Don
Don54660...
Your reference showing the steel tube at 22,000 psi for tensile and 13,000 psi for shear is typical of a mild steel, similar to A36 or other mild steels (SAE 1040 etc.). Most steel tubes are going to be in the mild steel range, whether round, rectangular, or square, though there are some high strength tubes available.
For bending, the 22,000 psi allowable tensile (extreme fiber stress) is probably close.
Your reference showing the steel tube at 22,000 psi for tensile and 13,000 psi for shear is typical of a mild steel, similar to A36 or other mild steels (SAE 1040 etc.). Most steel tubes are going to be in the mild steel range, whether round, rectangular, or square, though there are some high strength tubes available.
For bending, the 22,000 psi allowable tensile (extreme fiber stress) is probably close.
- Thread starter
- #9
Thank you,
Is the following correct?
A chinning bar (tube) is supported at each end at a distance of 72 inches (resting on its edges) and an 85 pounds load is placed exactly at the center of this bar. The chinning bar is actually a tube supporting this load and is 1x1x.062wall, a hollow square tube(Properties 22,000).
I want the stress and deflection for the tube. Also I want to know if this load is acceptable for the 22000 (Is this yield strength) spec from earlier posts.
So...
85x72/4=1530
1530x.5/moment of Inertia for the 1 inch tube (.03448)
1530x.5/.03448=22186psi
Question:is that too high for the tube for a static load.
The deflection for the tube would be .63 with the conditions indicated, that is, 72 inches and an 85 pound load.
Therefore.
85x72Cubed/48 (30,000)(.03448)=.63 deflection
Is this right? Is it yield or ultimate stress?
Is the following correct?
A chinning bar (tube) is supported at each end at a distance of 72 inches (resting on its edges) and an 85 pounds load is placed exactly at the center of this bar. The chinning bar is actually a tube supporting this load and is 1x1x.062wall, a hollow square tube(Properties 22,000).
I want the stress and deflection for the tube. Also I want to know if this load is acceptable for the 22000 (Is this yield strength) spec from earlier posts.
So...
85x72/4=1530
1530x.5/moment of Inertia for the 1 inch tube (.03448)
1530x.5/.03448=22186psi
Question:is that too high for the tube for a static load.
The deflection for the tube would be .63 with the conditions indicated, that is, 72 inches and an 85 pound load.
Therefore.
85x72Cubed/48 (30,000)(.03448)=.63 deflection
Is this right? Is it yield or ultimate stress?
Don,
E is 30,000,000, but that appears to be just a typo. Deflection and stress calculations look correct. Stress is simply stress, not yield or ultimate. Having calculated the stress, you need to compare the stress to the limit which you do not wish to exceed. In this case, you presumably do not want any permanent deformation after the application and removal of the load, so yield strength of 22 ksi is the upper limit of allowable stress, and it is being slightly exceeded.
Also, the load on a chinning bar is not static. In theory, a "suddenly applied" load causes twice the deflection of a gradually applied static load. The load on a chinning bar is probably subject to a much larger multiplier. I don't know what standards are used by the sporting goods industry, but I think you should apply a factor of at least 5.0 to the load to get the equivalent static load of 425 pounds.
Hope this helps.
E is 30,000,000, but that appears to be just a typo. Deflection and stress calculations look correct. Stress is simply stress, not yield or ultimate. Having calculated the stress, you need to compare the stress to the limit which you do not wish to exceed. In this case, you presumably do not want any permanent deformation after the application and removal of the load, so yield strength of 22 ksi is the upper limit of allowable stress, and it is being slightly exceeded.
Also, the load on a chinning bar is not static. In theory, a "suddenly applied" load causes twice the deflection of a gradually applied static load. The load on a chinning bar is probably subject to a much larger multiplier. I don't know what standards are used by the sporting goods industry, but I think you should apply a factor of at least 5.0 to the load to get the equivalent static load of 425 pounds.
Hope this helps.
Don,
I ran across your thread on this subject in another forum, and by the time I had composed a response to a comment by ADK, I had lost track of the location of that thread. Here is what I wanted to say, but it was prompted by ADK in the other forum, so it doesn't connect with anything in this thread.
Moment over section modulus = M/(I/c) = Mc/I, so ADK is just restating the equation, which can be useful sometimes, particularly in selecting structural sections. Your stress calculations are correct.
As long as instability doesn't enter the picture, allowable bending stress can safely exceed pure tensile strength, because the maximum stress is not imposed on the entire cross section. For your purposes, however, just take the yield strength as the maximum allowable tensile stress.
Plastic buckling (crippling) due to excessive compression on the upper part of the tube is the most likely failure mode if you were to overload this tubular beam to failure. It is not likely to occur under the conditions you have described, but be sure not to exceed the COMPRESSIVE yield strength of your material, as it is just as important as the TENSILE strength.
Hope this helps.
I ran across your thread on this subject in another forum, and by the time I had composed a response to a comment by ADK, I had lost track of the location of that thread. Here is what I wanted to say, but it was prompted by ADK in the other forum, so it doesn't connect with anything in this thread.
Moment over section modulus = M/(I/c) = Mc/I, so ADK is just restating the equation, which can be useful sometimes, particularly in selecting structural sections. Your stress calculations are correct.
As long as instability doesn't enter the picture, allowable bending stress can safely exceed pure tensile strength, because the maximum stress is not imposed on the entire cross section. For your purposes, however, just take the yield strength as the maximum allowable tensile stress.
Plastic buckling (crippling) due to excessive compression on the upper part of the tube is the most likely failure mode if you were to overload this tubular beam to failure. It is not likely to occur under the conditions you have described, but be sure not to exceed the COMPRESSIVE yield strength of your material, as it is just as important as the TENSILE strength.
Hope this helps.
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- #12
Thanks that is reassuring as ADK post said I incorrectly used Moment of Inertia instead of the Section Modulus. I wrote this reply to his post but as you have noticed it is gone.
For the records here was the post I prepared to ADK post in reply. LCubed, please correct me if my thinking is wrong. Thanks. Don
Here is what I posted that disappeared.
Hello ADK,thanks for your response. I would like to respond to your reply.
You say I have "Incorrectly used moment of inertia rather than section modulus". I went back and worked the bending formulas again and what I have found is I get the exact same answer if I use moment of inertia or section modulus, Because...
Fb=My/I=M/sx, meaning both give the very same answer.
and Sx=I/y or that is my understanding.
If Fb=My/I=M/sx is true I couldn't be wrong,because it is just a different way to arrive at the same answer.
To explain the above.
Y =distance from axis to extreme fiber,in my case, that is, .5 or 1/2 on an inch.
I=moment of intertia.
S=stress in tension and compression in psi. Bending in my case
therefore,the following examples.
Fb=M/sx = 1530/.06896= 22,186, using your suggestion(Section Modulus)
Fb=My/I = 1530x.5/.03448= 22,186, my calculation.(Inertia)
So how am I incorrect, these are both the same answers.
If Sx=M/Fballow than 1530/22.186=.06896 or Sx=.06896
is the same as Sx=I/y or .03448/.5=.06896 one is the Section Modulus and the other is Moment of Inertia.
So please help me and explain how I went wrong as I can't find it.
I have said I am only interested in learning how things work and do not plan to use this information to design anything, just to make sure I understand things correctly.
It would help if you could please explain also what it means to "review variable,and define them", that is, in the case given. Thanks again, Don, P.S. I do have three books and other reference material, I am merely checking to see if I understand the formulas given in those books.
Thanks Don
For the records here was the post I prepared to ADK post in reply. LCubed, please correct me if my thinking is wrong. Thanks. Don
Here is what I posted that disappeared.
Hello ADK,thanks for your response. I would like to respond to your reply.
You say I have "Incorrectly used moment of inertia rather than section modulus". I went back and worked the bending formulas again and what I have found is I get the exact same answer if I use moment of inertia or section modulus, Because...
Fb=My/I=M/sx, meaning both give the very same answer.
and Sx=I/y or that is my understanding.
If Fb=My/I=M/sx is true I couldn't be wrong,because it is just a different way to arrive at the same answer.
To explain the above.
Y =distance from axis to extreme fiber,in my case, that is, .5 or 1/2 on an inch.
I=moment of intertia.
S=stress in tension and compression in psi. Bending in my case
therefore,the following examples.
Fb=M/sx = 1530/.06896= 22,186, using your suggestion(Section Modulus)
Fb=My/I = 1530x.5/.03448= 22,186, my calculation.(Inertia)
So how am I incorrect, these are both the same answers.
If Sx=M/Fballow than 1530/22.186=.06896 or Sx=.06896
is the same as Sx=I/y or .03448/.5=.06896 one is the Section Modulus and the other is Moment of Inertia.
So please help me and explain how I went wrong as I can't find it.
I have said I am only interested in learning how things work and do not plan to use this information to design anything, just to make sure I understand things correctly.
It would help if you could please explain also what it means to "review variable,and define them", that is, in the case given. Thanks again, Don, P.S. I do have three books and other reference material, I am merely checking to see if I understand the formulas given in those books.
Thanks Don
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- #15
To LCubed and Bradh,
as that was my first posting on Eng-tips I didn't know where to put my question, so I posted in groups that seemed to relate to the question.
As far as what happen with the thread. I contacted admin. and they said they have no idea what happened and said they don't delete any posts.
So my quess is the processor (Computer) saw it as a duplicate (Titled the same)as this happened during a period when the site was down and then reopened, (No one told me that you shouldn't cross post, so I didn't see a problem doing this, anyway I will only post one at a time in the future.
Lcube, I do have another question and I am in the process of preparing a sketch for it, it involves shear loads in thin panels. I hope to have it ready tomorrow. Thanks to all for generous advice and help, as it really did. Don
as that was my first posting on Eng-tips I didn't know where to put my question, so I posted in groups that seemed to relate to the question.
As far as what happen with the thread. I contacted admin. and they said they have no idea what happened and said they don't delete any posts.
So my quess is the processor (Computer) saw it as a duplicate (Titled the same)as this happened during a period when the site was down and then reopened, (No one told me that you shouldn't cross post, so I didn't see a problem doing this, anyway I will only post one at a time in the future.
Lcube, I do have another question and I am in the process of preparing a sketch for it, it involves shear loads in thin panels. I hope to have it ready tomorrow. Thanks to all for generous advice and help, as it really did. Don
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