14

In short the heat treatments in steel change the phase of iron between the following phases: Austenite Cementite Martensite Bainite Ferrite Perlite. (Actually quenching does not allow low temperature phase changes to occur, so effectively the phases are sort of "frozen" in their high temperature equivalents). Figure 1 : example of continuoous ...


12

For the explanation of the loss of stress from E to F, I'll refer you to my answer to another question. Basically, the drop in stress seen in the diagram is a consequence of the fact that we use engineering stress (not taking the loss of cross-sectional area into consideration) instead of true stress. In a true stress diagram, there is no such drop. See this ...


11

Flexural capacity is based on the stress at the extreme fiber (the point farthest vertically from the neutral axis). $$\sigma = \frac{My}{I}$$ Or, rearranging $$M = \sigma \frac{I}{y} = \sigma S$$ M = moment capacity $\sigma$ = allowable stress S = I/y = section modulus So the relative capacity of the tubes will be a function of their section moduli. ...


9

To add to the other answers: interstitial solutes effectively "lock" dislocations in BCC metals (such as with carbon and nitrogen dissolved in iron), by forming what are called Cottrell atmospheres (Wikipedia). The lowest free energy is achieved when the interstitial atoms are in the dislocation, because the dislocation has a larger interatomic spacing due ...


9

The effects of heating-quenching a metal is explained below Transformation hardening is the heat-quench-tempering heat treatment cycle addressed earlier in this article. It's used to adjust strength and ductility to meet specific application requirements. There are three steps to transformation hardening: Cause the steel to become completely austenitic by ...


8

There are two methods to test flexural strength, three point and four point method. Three point method: and the corresponding moment line: four point load: and its moment line: The maximum flexural stress happens to concentrate only in a point in three point test, if you are testing a non-homogeouse material as it's always so in reality, it's not ...


8

As your construction is 3D printed, increasing the strength at the point of current breaking will more likely transfer the damage to a new location. The obvious and possibly impractical solution is to solve the jamming problem, rather than to try to power through it. If it's not practical to prevent the jamming, consider to convert the jamming related ...


6

This a table of a search done on research that was conducted between June 1959 and October 1960 by the Controls Section, Engineering Psychology Branch, Behavioral Sciences Laboratory, 6570th Aerospace Medical Research Laboratories Engineering Branch, Electronics Technology Laboratory. Units are ounces-inches. They have a variety of knob sizes and textures. . ...


6

You are mixing apples and oranges. Many steels harden by rapid cooling, but very few other metals do that; specifically, only aluminum bronze and certain titanium alloys. Many metals will strengthen by age hardening; Rapid cool softens, and then time at a lower temperature strengthens them. There are a myriad of combinations, like HSS (high-speed steels). ...


5

Start out by doing some basic calculations. Air is mostly N2, with a atomic mass of 28, and O2, with a atomic mass of 32. There is more nitrogen than oxygen, so let's say air has a average atomic mass of 29. At 0 °C and 1 atm, one mole of ideal gas occupies 22.4 liters. Let's say 20 °C is more realistic, so we'll scale that by the absolute ...


5

When the load is removed, the elastic strain is recovered completely (by definition); the plastic strain is not recovered at all (by definition): If the load is reapplied, the curve picks up where it left off (i.e., the system moves back up the unloading line and resumes plastic deformation). This point is explained here.


5

The incorrect posts are weaker as it is the dimension perpendicular to the fence which is critical. It sounds like you have a breach of contract here so you should ask them to replace with the originally specified posts without further charges to you.


5

Material Properties For the linear-elastic analysis you describe, the key pieces of material information are the elastic modulus (E) to calculate deflection and the yield stress (Fy) to check if the material remains elastic under the given loading. For steel, E is commonly around 200 GPa. The Fy can vary significantly depending on the grade of steel, so you'...


5

Shear buckling of beam web happens when the shear at a section of the beam under consideration surpasses the controlling combination of factored shear. The following three equations are the LFRD method for: no web instability Inelastic web buckling Elastic web buckling These depend on the ratio $h/t$ of the web. If this ratio is bigger than 260 web ...


5

Build the auger around a stronger shaft which has a better connection to the motor shaft. Perhaps a carbon fibre or even a steel centre. But then consider what will be the next breaking point. Or make a joint between the auger and the motor that fails if the mechanism gets jammed - just make it easy to replace.


5

You might consider dealing with it in software. Sense the stepper motor current, and when it exceeds a certain threshold that indicates jamming, perform a high speed reverse rotation for one or more revolutions. Then proceed in the forward direction.


5

I just found an answear, the keyword is "stress concentration". Sometimes, the material is removed to make a stress relief in an element. Some examples: The origin source I found it: https://www.youtube.com/watch?v=QtSki5nfO2g


4

Strength can mean different things in different contexts and technical definitions can vary significantly from what is commonly understood by the word 'strength'. For structures where actual forces are more relevant you might talk about rated loads, safe working loads or design loads in conjunction with factors of safety but this usually needs to be ...


4

This has been asked and answered numerous times before: https://www.quora.com/Why-is-strain-energy-equal-to-1-2*force*displacement-What-about-the-remaining-half basically, it's because you're calculating the area under a triangle, because the strain energy increases linearly as the displacement increases. The Work term here refers to the final position, but ...


4

For me a good (compromise of a) solution would be to redesign the shaft with an rectangular slot (although I am still unclear as to the fracture type, so I suspect is torsional). Like so Then you can use a rectangularly shaped metallic shaft along the length, which will be able to transmit more gradually the torque. *Figure 2: rectangular shaft image ...


4

The other answers describe the "materials science" mechanisms of iron vs. temperature. I'm going to add this: Matter "tries" to reach a minimum energy state whenever possible. In general, then, if you cool something as slowly as possible, you'll come closest to a solid which is a perfect crystalline structure. See "annealing."...


3

Disclaimer - No contract exists between us. This post is provided as information and opinion only and does not form "advice" in any way. Although I am a Chartered Civil Engineer I have zero professional experience with masonry. My views here should be treated as those of a keen DIYer (which I also am). I suspect it'll probably be ok without any lateral ...


3

There is another safety issue - the way the bowl is attached to the structure of the building. In the UK, the only structural attachment is by two (large) brass screws passing through cast holes in the bowl and screwed into the wooden floor. Since the top of the bowl usually overhangs the base, applying a large off-centered load while climbing on and off ...


3

With the exception of anything extremely unusual: the concrete will continue to gain strength until 28 days. In fact, it will continue to gain strength for something like 50 years, but the main strength gain happens early on. The motivation for waiting for the 28-day test is therefore nothing to do with strength. Far more likely: the later you are paid, the ...


3

The simplest way to tell if a metal is ductile or brittle is to look at a fracture surface (e.g. from a tensile or impact test). To have either a ductile or brittle metal defines the failure mechanism which you can tell by having either a ductile fracture or a brittle fracture. Elongation and reduction of area as well as notch toughness (impact test) and ...


3

A shell structure under external pressure is likely to fail in buckling before it reaches the compressive strength of the material. Buckling of curved shells is a complex problem in general, but for a complete sphere a simple formula (source here) is $$P = \frac{2E}{\sqrt{3(1-\nu^2)}}\frac{t^2}{r^2}$$ Where $E$ and $\nu$ are Young's modulus and Poisson's ...


3

Shot peening improves the fatigue life , mostly used for steel springs. It does not change the bulk strength much . Shot peening is also used for aluminum. A very smooth ,polished surface can also improve "strength" in VERY highly stressed high strength steels ( this is likely of little use in ordinary applications ; I measured the effect in hydrogen stress ...


3

What you want to look for is Shear buckling or Web buckling. It is quite an advanced subject though. In order to determine mode of failure, critical loads etc, you need to specify explicity the boundary conditions, and then solve the problem for the dimensions. In case that you want to determine the permissible load of the member outside the context of a ...


3

Yes, you are right, we ignore the axial deformations because in typical frames like multi-story structures they are of second-order magnitude compared to deformations due to bending moment and shear. Usually, the strains due to moment are in the order of 100 times more than axial strains, especially complying with the code's recommended stronger columns ...


3

In machine tools where an auger is used to remove scrap metal chips, the current in the auger motor is sensed. If it exceeds a certain level due to jamming, the electronics temporarily reverses the auger direction which can clear the jam. Lubrication of the material-auger interface is very important with augers, which have a huge swept surface area.


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