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How to calculate elongation of member at yield 1

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BAGW

Structural
Jul 15, 2015
392
Hi,

I wanted to know how to calculate the elongation of the member at yield and fracture? Elongation at yield can be calculated using the stress-strain equation E = (P/A)/(delta l/L), where P/A = Fy.

For plates, how do I calculate the maximum elongation of plate before it ruptures. How much can the plate elongate before it ruptures?

Thanks
 
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What's the purpose of the exercise? If you need actual information for a specific piece of steel, consult the mill supplying the material - they'll have test data showing elongation at failure. If it's more of a theoretical exercise, pull the ASTM standard for the material and review the minimum elongation at failure as well the other minimum mechanical values. Then, it's pretty much the same as what you said above, though it will depend on the boundary conditions of the plate.
 
I was looking at ASTM standrads from some steel Fab

It says min 8" elongation 20%. what does this mean? Is it 20% of 8", 1.6"? So plate can elongate upto 1.6" before it fractures?

I am looking at A36 steel
 
Your question is highly theoretical, I think it involves both testing and energy method. Note the Δ = PL/EA is only valid up to the initial yield, after that, stress-stain relationship is non-linear.

You haven't answer the question,

phamENG said:
What's the purpose of the exercise?
 
If you are looking at a beam, one usually wants to know the deflection perpendicular to the beam, not a change in length. Mighty small.
 
as asked previously … what's the point of the exercise ?

elongation strictly (IMO) only applies to axial elements (like tension test coupons).

beam deflect out of plane (principally).

plates can deflect in-plane or out-of-plane, depending on loading.

possibly the spec means the minimum of either 8" or 20% (so cross-over when L = 40").

another day in paradise, or is paradise one day closer ?
 
The question is not specific enough and the proper solution will vary tremendously based on initial assumptions. That being said, you can investigate this yourself by learning finite element methods. Topics like crack initiation/propagation, element failure, and complex plastic analysis have made leaps and bounds in recent history. Use the right tool for the right job.
 
The 20% in 8" refers to the stretch in the test coupon - typically a "dog bone" with an 8" narrowed center. That test piece has to elongate a minimum of 20% before the piece fractures. This value is not necessarily linear (A992 minimums are something like 23% in 2 inches but 18% in 8 inches), but it will give you a good idea of the material's behavior.
 
That being said, you can investigate this yourself by learning finite element methods.

Beware of reverse learning process, especially the beginners.
 
I wanted to know how to calculate the elongation of the member at yield and fracture? Elongation at yield can be calculated using the stress-strain equation E = (P/A)/(delta l/L), where P/A = Fy.

If we are talking a ductile material.....you'd figure it based on the stress-strain diagram. Up to the yield point you'd get it via the modulus of elasticity.....after that you would figure it based on reading the diagram and sum the elongation(s). For ductile steel, you'd have to break it into about 4 numbers. (Elongation at the elastic limit, at the end of the yielding region, at the peak of the ultimate stress, and at the fracture point.)

Say (for example) you had something that went into the plastic region.....the strain (ε) at the elastic limit would be yield stress/modulus of elasticity. The strain into the plastic region would have to be read off of a accepted stress-strain diagram.

From all that you would figure the (total) elongation at the desired point(s) of stress would =Σ(εL)

Keep in mind, if it is unloaded prior to fracture.....and has been strained into a plastic region.....there will be some permanent deformation but also some elastic strain recovery. (But that is another topic.)

 
Retired. I'm the first one to rail against improper application of FEA, but a hypothetical question like this would be a great time to learn or apply it. Diving into continuum mechanics and it's implications in elastic and plastic material behavior is very enlightening, but painfully dull without accompanying case studies and visualization.
 
DrZoidberWoop,

Don't misunderstood my comment. Under proper guidance from a knowledgeable person, computer is an excellent tool for learning. I do utilize the programs to verify the correctness of my calculations from time to time. Yes, visualization made judgement calls much easier. But I do have quite experience from which I can most likely to identify the mistakes in the inputs/modeling, and call the output junks. The OP obviously not.
 
I should have put this first, apologies.

Trying to understand/prove to myself double angle type 2 connection for trusses. Type 2 connection always has been a myth to me. I have always been told, the angle yields under the heavy gravity loads and behaves like pinned for the gravity loads. It starts behaving like fixed under smaller wind lateral loads. Obviously this concept works only in low seismic areas where the connection is not stressed for loading and unloading.

The above being said, this can only be achieved with small angle connection by allowing it to yield under gravity loads. Where is the line between yielding and fracture? How do I know how much can a connection yield before it fractures? Trying to get my head around this. I could not find any good research which explains this.

 
BAGW said:
It says min 8" elongation 20%. what does this mean? Is it 20% of 8", 1.6"? So plate can elongate up to 1.6" before it fractures?

I am looking at A36 steel

A36 steel is produced per A6 which gets its testing methods from A370, and clause 13.4.1 says "Fit the ends of the fractured specimen together carefully and measure the distance between the gage marks ...", so you are correct. The gage marks that were initially 8" apart must be at least 9.6" apart after fracture (20% is the minimum elongation).

However some of the elongation occurs after the maximum tensile strength so you could never get this much elongation without also having failure.
Kannan-Material-1-min-486x420_chywta.jpg
 
BAGW - I believe the "type 2 connection" was replaced with the "Flexible Moment Connection" in more recent editions of the Steel Construction Manual. I'm still in the 14th edition - see page 11-2, Load Determination, 2nd paragraph. In that case, the determination of elongation is not required (pin for gravity, fully fixed for lateral). It sounds like you're trying to design a true partially restrained moment connection, which is a bit more involved. If that's what you're doing, I think it would be wise to keep it in the elastic range with more predictable behavior.
 
Yes something similar to 11-2. But not much guidelines provided in the manual for such connections
 
As I understand the FMC, you size the beam as if it were simply supported so it has sufficient stiffness to support the loads within the bounds of acceptable serviceability limits regardless of end restraint (because the long term ability of the moment connection to resist medium to long term static loading is questionable), and then size the moment connection for wind load only.

Check out this article - it should point you in a better direction.
 
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