Ultimate Tensile Strength doesn't equal F/A? 2

[tydguy]

Hi, I'm stumped by a recent tensile test and I need someone to explain this to me. I feel like I'm missing something super basic.

We have a steel component that is designed to fail at a specific tensile load, 40,000 lbs. The design is simply a threaded joint with a necked down radius that is loaded only in tension (See attachment). The neck diameter is .546" (Area = .234in^2). UTS of the 4140 steel is 140 KSI. Force = UTS*Area = 140000 * .234 = 32800 lbs. When pulled to failure, we get 40,100 lbs. MTR showed UTS was correct. What am I missing? Is there some effect because of the profile, or the some other factor at play here? Thanks for any help or resources you can provide.

 

[MintJulep]

The 140 KSI ultimate tensile strength is a MINIMUM required value.

It is common for the actual strength of any given batch to be higher than the minimum.

 

[tydguy]

The reported value on the MTR was just over 140KSI, maybe 141KSI (can't remember exact value now). It doesn't account for the large difference, since UTS would have to be 170KSI to pull at 40000 with F/A. Any other suggestions?

 

[Compositepro]

Do you understand the difference between engineering stress and true stress? Engineering stress is calculated by dividing the measured stress by the original measured cross-section. True stress is calculated dividing the measured force by the actual cross-sectional area at the time the force was measured. Your test specimen necks down in diameter as you stretch it so the true cross-sectional area goes down. True stress will be higher than engineering stress.

 

[EdStainless]

The notch geometry matters, a lot.
If you change radius the values will change.
The radius will restrict the amount of necking before fracture (tensile bars are smooth).
Look at the difference in reduction of area between the tensile test and your breaks.
When you correct to true stress using the RA values you will get matching numbers.

= = = = = = = = = = = = = = = = = = = =
P.E. Metallurgy, Plymouth Tube

 

[tydguy]

Hmmm...I don't remember much about engineering stress vs True Stress. How do I apply that to this situation? I measured the diameter of the neck post fracture. It's .434" (Area = .148in2). Fractured at 41,100 lbs. Now F/A stress at fracture = 277700psi? How does that relate to UTS?

 

[EdStainless]

That is the true UTS.
Now if you have an MTR that has the RA listed on it I'll wager that if you adjust the reported engineering UTS by the RA you will get something close to your 277ksi.

= = = = = = = = = = = = = = = = = = = =
P.E. Metallurgy, Plymouth Tube

 

[Compositepro]

https://en.wikipedia.org/wiki/Stress–strain_curve

 

[tydguy]

@EdStainless: RA from the MTR is 56%. So are you saying that True UTS = Engineering UTS/RA%? In this case, that would give 252000psi. Definitely closer, but not quite where I feel I could calculate the fracture point from the MTR. We need to make a bunch of these at various pull values, so my goal is to calculate what the neck diameter needs to be for each batch based on the MTR and the required pull load. Is there going to be some other empirical factor based on the shape of the notch?

@CompositePro: Thanks! I'll read through the wikipedia page.

 

[metengr]

This is really a problem related to notch geometry to result in tensile failure at a designated load. No different than rupture discs that are pre-notched for failure to ocurr at a designated pressure.

Your essential variable will be the notch geometry and duplicating this geometry to achieve tensile failure at a designated force given a UTS. The local stress concentration can be determined using a graph showing stress concentration factor as a function of the notch radius.

 

[tydguy]

@metengr: Do you have any good resources regarding the effect of notch geometry on ultimate tensile strength? (I thought stress concentrations were more for yield than full fracture.)

I still don't feel I can get from the inputs to a calculated failure that matches my pull value without some fudge factors. Is there no other way than just testing the specific geometry each time?

 

[EdStainless]

No there isn't, unless you make a whole range of parts with different properties and different depth and radius of notches and come up with correction factors that apply to your application.

= = = = = = = = = = = = = = = = = = = =
P.E. Metallurgy, Plymouth Tube

 

[tydguy]

Thanks for the help everyone. More reading required on my end. At least it doesn't appear that I'm missing something stupid/easy.

 

[MagBen]

You made a mistake when calculating the true UTS. It need to be divided by (1-RA), not RA, so true UTS=141/(1-56%)= 320ksi!

When force = 40,100 lbs, your necking area = .148 inch^2, the true UTS for your part = 40,100/.148 = 271 ksi. Notch factor = 271/320 = .85. In other words, your geometry makes material "weak", this is because your part necked less than the consistent/smooth tensile specimen, RA decreases from 56% to 36.7% [=(.234-.148)/.234]

 

[EdStainless]

Thanks Ben, I knew that but overlooked it.
Yes, the notch makes the parts relatively weaker.
The RA is lower in the notched parts because the notch restricts the part from necking, it forces a concentration of the deformation.

= = = = = = = = = = = = = = = = = = = =
P.E. Metallurgy, Plymouth Tube

 

[realtime2]

Your part has a severe radius, much like a diametral test specimen.
I.e.: it is already "necked" from first load application.
One needs to apply Bridgeman's correction for necking.
See figure 1.10 in

http://libres.uncg.edu/ir/uncc/f/RomeroFonseca_uncc_0694D_10584.pdf

or search web for Bridgeman correction.
It might be a good idea to put a diametral extensometer on the test part
to measure the shrinkage during the tensile test.