crack growth and cyclic fatigue failure example problem
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- เผยแพร่เมื่อ 25 ส.ค. 2024
- Using crack growth equations to solve for cyclic fatigue when stress is alternated in two different ways: first a series of low stress cycles followed by a series of high stress cycles.
Very much helpful, thank you very much man.
Help me a lot to understand this topic. TQ very much
thank you for my final exam paper :)
This was very helpful. Thanks for sharing!
Glad to help. Let me know if there's something else you'd like to see
is this for an internal or surface crack, as for internal dont we need to use half the crack length, as length = 2a not a
where did you get the value of da?
Can you help me how to do fatigue analysis of a offshore jacket structure using fracture mechanics
Why isn't the change in stress multiplied by 10^6 (as in 'M'Pa)? Thanks
Good question. In this question it states what the crack growth constants, n & A, are and it says for these units. If they had given the full units of the crack growth constants we would see that it was given in megapascals, not pascals. Therefore we don't need to multiply by 10^6 in order to convert it to pascals.
Ahh makes sense thanks a lot :)
You are the Ron Burgundy of fracture mechanics
Pat Ginnetti hahaha. Absolutely made my day
I appreciate your help. The formula I have in my textbook (Hertzberg) shows that Nf = 2/[(m-2)*A*Y^m*sigma^m)] * the integral of ao to af. I get a different result. I noticed you have Pie in your formula wheras I don't. Any idea why?
Pat Ginnetti so your textbook is giving you a generic solution where they did the integral for you. Whether or not you use pie or not is something I don't fully understand. Some textbooks included whereas others do not. I think that it should be included., but if yours does not then you should probably leave it out since your professor is using that textbook
UR A LEGEND
abdulla alsuwaidi thanks! Subscribe to help me grow the channel and see new problems that I post.
where did the -1.8182 come from if u can clarify please :)
Sure, the integral originally had 1/a^(3.1/2) which is just 1/a^(1.55) which is just a^(-1.55). When you take an integral you know you need to add 1 to the exponent and then multiply your term by 1 divided by the new exponent. In our case, when we add 1 to (-1.55) it becomes (-0.55) and then we remember to multiply our term by 1/(-0.55) which is equal to -1.8182. Does this clarify?
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i make this out to be... -n x 1/-n+1 so -1.55 x 1/-0.55 gets me 2.8181, i must be getting close. what am i doing wrong. great video btw.@@TaylorSparks
if there is a residual stress presence where crack is and your trying to calculate the critical crack length do you add the residual to the yield stress of material or do you subtract. ( general question)
abdulla alsuwaidi again, this residual stress probably means the fracture is multimodal and more complex than our simple scenario here. However, whether you add or subtract this contribution depends on the nature (sign) of the residual stress.
what if your calculated critical crack length is bigger than the physical thickness of vessel, dose the failure mode become plastic collapse ?
abdulla alsuwaidi good question. Remember, the fracture I am calculation g here relies on a number of assumptions. Most importantly, mode I fracture. As flaw size gets larger and approaches thickness we can adjust geometric parameter but it has no solution for when the flaw is larger than the vessel thickness.
how do u get (-1.8182) from the integration?
I explain that in the comment to Abdullah below :)
Your tea is boiling
Lol. Got a new mic for future vids since I made this one :)