I'm a structural engineer, and this is incorrect. For structural steel (e.g., A992, 350W, etc.), the tensile strength is the same as the compressive strength. That's why we only need one value for both.
And yes, when we design beams, we check the stresses in both tension and compression. For a symmetrical shape like an I-shaped beam, both verification will be the exact same formula so we don't need to calculate both explicitly.
For columns, we do check the compressive stress, and it is the controlling failure mode for short columns. Long, slender columns will buckle before the compressive stress exceeds the allowable limit.
It depends on the geometric stability of the column shape, which is why I said what I said. The reality is that compressive stress limits are much lower in steel than tensile, as you’re noting.
They are only the same in an abstract theoretical sense, not an actual one.
And bucking within the column (and/or the resistance too it) is primarily resisted by tensile strength.
>It depends on the geometric stability of the column shape, which is why I said what I said.
What you said was like 99% completely false.
>that compressive stress limits are much lower in steel than tensile
No they are not.
> as you’re noting
I never acknowledge that. Your lack of understanding in what column buckling is doesn't change anything. Unironically, you should ask chatgpt, it can probably give you a high level explanation that is more compatible with your understanding of physics than what I can give you.
Buckling is a stability problem, not a resistance problem. The compressive stress resistance is literally not part of the formulas that we use to verify it. The column could have 100 MPa, 350 MPA, or 359918 MPa compressive resistance and it would change nothing when it comes to buckling. Only the elastic modulus and the physical shape of the column (length, inertia, etc.) is relevant for buckling verification.
>And bucking within the column (and/or the resistance too it) is primarily resisted by tensile strength.
Than please explain to me why we don't need the tensile (or compressive) resistance of the material to know the buckling resistance of the column. I really want to hear that one.
Ah engineers who do not understand the fundamentals of the values they are using. I expect this will piss you off, but I expect it is important you know.
The easiest way to visualize Young’s modulus is typically using a stress/strain curve from a tensile force test, but can also be visualized by a compressive force test.
Young’s modulus is the elastic region that is ‘under’ both curves, since it’s a value for how much something can safely bend* and return to it’s original shape without permanent damage, and bending causes both compressive and tensile forces in a material.
The lowest value of the two upper limits - of course - will set the upper elastic limit!
Specifically, to a very simple approximation, the ‘inside’ of a bent beam will be limited by compressive strength, the ‘outside’ by tensile. The overall beams strength will be whatever the lowest of the two values is, as that is when the beam will fail somewhere, hence, that what you want to use in civil engineering, eh?
*at a first approximation. It is of course much more complex than that.
For a column, the limit is the lowest of the ability of the column to prevent deflection (tensile) and bear the load without ‘pancaking’ (compressive strength). Pancaking is resisted by the full cross sectional area of the column, while deflection is generally only resisted by (to a rough approximation - greatly depending on the actual geometry, as I noted in my prior comment!) half the thickness of the column, so tensile strength of the material is usually the limiting factor for most simple columns of non-trivial length, where unsupported column deflection is the dominant failure mode.
Notably, this is why concrete often uses rebar in civil engineering, as concrete has trash tensile strength on it’s own and requires massive volumes to have sufficient tensile strength to form acceptable beams or unsupported columns, or use only very limiting forms to ensure only compressive forces actually occur. Unreinforced Concrete’s Young’s modulus is trash for this reason.
Which for civil engineering is the best bet!
Also, add a good safety factor on top due to all the other materials fundamentals that they apparently don’t teach in civil engineering school?
Which hey, I get, because they’d be too overwhelming eh? And they’re generally drowned in the noise at that scale anyway.
Depending on the sub discipline, mech-e, aero-e, etc. will of course have to know these things at a finer level of detail.
Someone computing fuselage thickness, designing an engine connecting rod or turbine blade, etc. needs to know what is going on at a much finer level eh?
Those folks will also have to quantify the type of failure modes they expect (fatigue limit, tensile failure, compressive failure, wear, etc.) and design around it.
Young’s modulus isn’t fully useless in those usages, but more specific values tend to be far more useful, as Young’s modulus is fundamentally the lowest value of a mix of material properties.
>it would apply even if every data center built an entirely renewable dedicated solar farm to power it. After all, energy is fungible: the newly built solar farm could be going to help consumers transition to renewable.
No, because that solar farm would have never been built in the first place without the datacenter. You either get both or you get none of them. It's completely different than using an existing solar farm that was already there before anyone started planning for that datacenter
>Bitcoin attempted to replace cash, but failed because the transaction costs are orders of magnitude too high.
The current fees are less than 0.40$. It may be too high for a starbuck coffee, but that's way lower than the fees charged by a credit card provider if you are purchasing something over 50$. On a 2%+0.10$ structure, you only need to transfer around 15$ before your credit card fee is higher than the current average BTC tx fee.
honestly, I think it makes no sense to spend more than 30$ on a calculator if it can't do symbolic math.
The way you input things like division, integrals, matrix, etc. on newer calculators like the nspire is far superior than the older calculators (eg. ti-84, ti-89, etc.). They look like how you write them on a blackboard instead of relying on purely parentheses or "," and ";" to separate parameters. It's like going from Excel to Mathcad
The US has roughly 10x more population than Canada. The solution is really simple, just hire 10x more humans to manage the vote counting.
Paper voting worked for thousands of years and was at the core of the foundation of this country.
There is no need to compromise the results of the election just to scale in a slightly more efficient way. If you need 10x more people because the volume is 10x higher, just hire 10x more people.
>I don't understand why voting machines can't just print your vote on a piece of paper behind a plastic window for you to see while also recoding the vote in a database
If it's counted electronically from the database, the piece of paper is completely worthless. Unless you can get the entire voting population to give you their paper and then count them, you will never know if the count is right. If a hacker switched 15% of the vote from one party to another, how could you tell from a piece of paper that tells you who you voted for?
you can count the paper votes only in your voting point/building. If there are abnormalities you can alert other people to trigger the global recounting
Yes, it's not foolproof, attacker can just modify the electronic voting data in places where he knows people don't usually do recounting. But it makes his job harder
We had front-page news about how the election was "hacked by Russia" and trump cheated for over a year after his first win in 2016 (let's not pretend that keyword was chosen accidentally); They tried to put him in jail for it. In 2020, trump did the exact same thing and went even farther with it. And in 2024, the DNC tried again to claim cheating happened.
How many cycle of this BS do we need to go through before we accept that elections need to be done properly and safely?
The entire point of a democracy is that elected leaders get their legitimacy and their acting power from the certainty that it was voted by the population. Not everyone will agree with their ideas, but the majority do and we all agree to follow their lead because that's what the population want. If the vote is compromised, everything falls apart.
If the "will of the people" turn into the "will of an intern at Dominion who fucked with the code and rigged the election" or "the will of Pakistani hacker", it breaks the entire system.
I have to seriously disagree on the particulars, here.
The Russia allegations ranged from "Russia hacked DNC servers/accounts to interfer in favor of Donald Trump" (demonstrably true in several instances) to "Russia hacked voting machines" (very probably false). And then in 2024 the DNC quickly accepted election results.
By comparison, Donald Trump still claims that he legitimately won the 2020 elections, the majority of his base still believes it, Fox News spent years spreading that message even though their own journalists thought it was bullshit, etc.
I maintain that this is a systemic problem and a better system would not have given Trump the leeway to do this, but let's not pretend it's a bipartisan issue.
Just to be clear, the 2024 election was indeed compromised. Salt Typhoon (China) hacked the communications of both campaigns due to massive cybersecurity failures in law enforcement portals of all major US telecommunications companies.
And yes, when we design beams, we check the stresses in both tension and compression. For a symmetrical shape like an I-shaped beam, both verification will be the exact same formula so we don't need to calculate both explicitly.
For columns, we do check the compressive stress, and it is the controlling failure mode for short columns. Long, slender columns will buckle before the compressive stress exceeds the allowable limit.