Crush strength calc for short steel member

[AMMoyer]

Our customer has asked for an axial crush/column strength calc for a short steel tube we are developing for them. The tubing is 1026 DOM with an OD of 0.945" and ID of 0.708" and is 2.028" long. They want to know the load at which 3% yield occurs. Here's what I've used:

total elong. = 3% x 2.028" = 0.0608"
F = unknown
l = original length = 2.028"
E = 30,000,000 psi
A = 0.308sq in

total elong. = Fl/AE

Solve for F I get 277,000 lbs. That seems high to me, but I checked that the length is short enough to use a purely compression calculation. I just need a simple calc model for this. Have I totally missed something or am I on the right track?

 

[trainguy]

AMMoyer,

You question is difficult to understand. I believe you are not on the right track.

1) FL/AE will calculate an elastic deflection, assuming the material remains elastic. This is unrelated to yielding.

2) 3% yield is a term I do not understand. Does your client require the load at which the total strain (elastic + plastic) is 3%? Or the load at which the plastic strain alone is 3%?

3) Usually, a crush strength involves the member shortening, so I'd stay away from terms like "elongation". I know it sounds picky, but we engineers are a picky bunch...

tg

 

[TheTick]

Wouldn't this be a buckling problem?

 

[CoryPad]

As trainguy stated, your initial equation is for 3 % elastic deflection/strain, which is not valid for steel. The 3 % likely is an offset deflection/strain. A simple model would assume perfectly plastic behavior or power law hardening. Physical testing usually is where this behavior is determined.

 

[djm883]

If you are putting the short tube between two surfaces, be careful of barreling (unless you are want it). If you want the ends to be able expand as the length compresses you will want to lubricate the cylinder ends.

 

[AMMoyer]

I have requested a clarification on whether the 3% "yield" is total strain or only plastic strain. My apologies for the elongation reference when its supposed to be compression.

CoryPad, why is this not valid for steel in the elastic range?

I don't believe this is a buckling problem since the slenderness ratio is so small.

If they want to know the load at which the plastic strain is 3%. How do I calculate that?

 

[CoryPad]

The elastic strain for steels is small, usually less than 0.005 or 0.5 %.

To determine plasticity, you need a method to measure strain independent of the load train. This is done with strain gages, extensometers, or video methods, among others.

 

[chicopee]

AMMoyer, your equation is not valid in the plastic region because the stress-strain relationship is no longer assumed linear in the plastic region. Your equation is a linear relationship only assuming the cross sectional area to be constanst. Also, in the plastic region the cross section area will increase during compression to the point where the piping will bulge.
Being a short member, forget bending.

 

[desertfox]

Hi AMMoyer

Why doe the customer require this calculation would it not be easier to supply a material certificate with guarenteed minimum tensile strength an yield on it.

desertfox

 

[Ron]

Someone check my math, but from a quick buckling check I get a buckling load of 2855 lbs.

This is a buckling issue...

 

[Zapster]

Ron I believe your math is wrong.

 

[Ron]

Zapster...you're right...not even close!

kl/r is less than 15 so buckling not an issue.

AMMOYER...I approached from different direction and got your same answer...277k. Does seem extraordinarily high. I can't see how you could mobilize that kind of load on such a small tube....except in a testing machine.

I think the magnitude is so far off because we are assuming the stress-strain relationship is the same for tension and compression in the steel...Not so. I'm not sure what the compression modulus is for your material, but it might behave more like a secant modulus rather than a tangent modulus. That could change the answer by a factor of maybe 10?

You might pursue the variation in tension-compression moduli to solve this.

 

[Bribyk]

Local buckling of the tube wall, like crushing a pop can, may be your failure mode.

The numbers sound high but, assuming a 50ksi yield strength, your tensile yield load would be about 162 kip and, if it doesn't buckle locally, it should hold more in compression.

 

[kslee1000]

They want to know the load at which "3% yield" occurs.

This request is vague.
1. Load at 3% of yield strength of the metal?
2. Load at 3% of yield strain of the metal?
3. Load at the metal when it reaches 3% strain?

(1) does not make sense (a straight cal).
(2) what type/grade is the metal?
(3) agree with others, the strain is beyond elastic range. Unless the grade (fy) is greater than 870 ksi (29000*3%).

 

[AMMoyer]

They are specifically asking for the load in compression when 3% strain is reached. For this material the elastic limit is reached at 0.24% strain so we are well into the plastic region. The load at the yield point is then 22.2ksi

Is there a simple way to estimate the load at the 3% strain in compression? I understand that steel behaves roughly the same in tension and compression below the yield point, but what about above there? The actual failure mode or failure point is not critical unless it will help me determine the load at 3%.

I appreciate all the input.

 

[kslee1000]

The approx. load at 3% yield is the same as the load at yield (assume a straight line interpretation on stress-strain curve). The true load can be higher, but nobody can judge it without the corresponding stress-strain curve for that material.

 

[desertfox]

hi AMMoyer

Why does the customer want to know a load beyond the elastic limit?

desertfox

 

[kslee1000]

Watch out, someone may have knowldge on the very soft metal. Per info provided, the young modulus E = fy/Sy = 22.2/0.24% = 9,250 kai << E = 29,000 ksi, common for metals used for construction.

 

[cloa]

Its strange for a 1026 DOM, its yield stress would something like 70ksi e.g.

http://www.aedmotorsport.com/MildSteel/1020SteelDomTube.htm

 

[TVP]

cloa,

AMMoyer made an error with the units in his last post: the load (force) is ~ 22,000 pounds for compressing this tube/sleeve to a plastic strain of 3% using 1026 steel. This is very close to the value obtained using the approach described by kslee of just calculating the load based on a stress at yield (perfectly plastic behavior): 18,500 pounds for a yield strength of 60ksi; 21,500 for 70ksi yield; etc.

AMMoyer,

Yes, there are methods for estimating the performance in compression. I will have to look through my files to find some. Usually this type of behavior is confirmed through part testing, so even if you have to start by machining samples, you could relatively quickly get some data.