11

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. ...


8

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 ...


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 ...


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. . ...


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'...


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 ...


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

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 ...


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 ...


2

Further adding to Wasabi answer, The yield point phenomenon occurs due to the segregation of impurity solute atoms(C and/ or N in Fe) around dislocations so as to reduce the strain energy associated with the distorted atomic arrangement (maximum in bcc metals, less in hcp metals and least in fcc metals). This additional stress required to free the ...


2

In my opinion the comment of Ethan48 should give you enough information to answer your question. Hereby a bit more elaboration. With a column under axial compression the stress is uniformly distributed*. Thus the stress is indeed given by $\sigma = F/S $ When designing a column under axial compression, you do not only check whether the stress is not too ...


2

This is from Wikipedia, https://en.wikipedia.org/wiki/Shear_force : "Shearing forces are unaligned forces pushing one part of a body in one direction, and another part of the body in the opposite direction. When the forces are aligned into each other, they are called compression forces." In engineering terms, it is the force that forces the body to be ...


2

Toilet bowls aren't all the same so it impossible to say. Some may take it some may not. There are several factor at play. When climbing on the bowl you are putting load onto one leg and have some impulse going on for lifting the other leg. Depending on where you step this increased load can be stressful for the bowl. While the bowl may generally be able to ...


2

There's a rather exhaustive dissertation regarding this particular type of crane, Non Slewing Articulated Mobile, based on a study of apparent out-of-balance situations in Australia. Formulae are provided as well as diagrams of center of gravity calculations on flat and sloped surfaces. The above link also includes a download of the entire document in PDF. ...


2

There are correction factors in academic table books. A very detailed explanation is in the "FKM Richtlinie" chapter 1.2 material characteristic. But you mentioned that you have the book Roloff / Matek: Look at formula 3.7: $R_m=K_t*R_{mN}$ and $R_p=K_t*R_{pN}$ $K_t$ is the technological size influence factor – not sure if this is the right translation ...


2

The tensile strength is diameter dependent. How greater the diameter, how longer it takes to discharge the heat, so, the time it takes to cool down the material (here is steel) has an impact on the microstructure and consequently the mechanical characteristic of the material. The tensile strength would be lower than what is already mentioned in the table, ...


2

A good place to start is Ogden's book Nonlinear elastic deformations, section 5.2.3: "Extension and inflation of a circular cylindrical tube". For a material with a strain energy function: $$ W = F(I_1) - A I_2 + B I_3 $$ where $I_1$, $I_2$, $I_3$ are the three principal invariants of the stress tensor, Ogden shows that for the situation where a stretch $\...


2

Material strength is generally expressed in the form of stress but for a component using a load is perfectly logical. Equally while using Newtons would be more precise quoting an ultimate load as a mass is often more practically useful. What is less clear is whether this is a 'safe working load' (ie you can hang a mass of 18kg off it and it won't break) or ...


2

The bullets won't actually 'shatter' as they are soft lead (in a copper jacket). Bullets normally deform on impact with a hard surface (such as armour) and will spread out (like plasticine). Some of the energy normally goes into deforming the armour and denting it, but if the armour was hard enough and stiff enough, it could resisit the bullet enough making (...


2

The second moment area of a hollow square section, I, can be calculated by subtracting the I of hole from the I of square. We call the outer side ,a and inner side b. $$ I = \frac {(a^4-b^4)}{12} $$ And $$ S = \frac {(a^4-b^4)}{6a} $$ So roughly, you need three of the smaller size beams for each 40x40 mm. But since the deflection of the smaller beam ...


Only top voted, non community-wiki answers of a minimum length are eligible