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20

Because bridges and other structures are not static objects. They must be allowed to flex under varying loads and also accommodate changes in length from thermal expansion. The hinge pin allows changes in angle. and the sliding joint between the upper hinge plate and the flat plate on the bottom of the beam allows changes in length. If the connections were ...


20

I believe it has less to do with strength and more to do with stiffness. A rod of aluminum of the same length and weight as a a steel rod will be just as strong (force required to break) but have three times the cross-sectional area which hugely increases the second moment of inertia by nine times (it scales to the fourth power of distance/length which is ...


19

It seems realistic to me. This is an undersea pipeline at depths of over 2 km. The pressure would be considerable at those depths (on the order of 20 MPa or 200 atmospheres). The pipe would need to be thick enough to withstand these (very high) pressures. The Nord Stream pipeline under the Baltic sea is at depths of up to 210m and uses 1220 mm diameter ...


19

While not an ideal situation, it is common enough that this type of cut/reduction of the beam as it comes to its support actually has a name. This is more often referred to as a coped or dapped steel I beam. There are various ways to transition from the full depth of the beam to the depth you may require at your support. Some examples are: Sometimes the ...


18

Steel is defined as an alloy of iron and carbon; there is no such thing as a non-ferrous steel. If you alloy some other metal with carbon, it becomes something other than steel. Looking for a steel without iron in it would be like looking for brass or bronze without copper. You can alloy things other than copper with zinc, tin, or aluminum, but those would ...


17

There are a few reasons. I'm firstly going to assume you're talking about replacing a bunch of small rebars by a single reasonably-sized one: i.e. instead of $15\phi8$ (7.54 cm2), using $1\phi32$ (8.04 cm2). One reason is to improve ease of constructibility. Reinforced concrete beams also have transversal reinforcement, and it's very common to place rebars ...


16

This is a textbook example of what not to do. We don't get into stress concentration at the cut off of the corner of the beam, or the fact that the two very different stiffnesses of the beams are a constant cause of differential deflection and vibration. The thin edge of the web sitting on the column cap is an unstable mechanism waiting to either kick the CC ...


15

The stiffness of a rectangular cross section, be it steel, concrete, wood, or any other material, is related almost entirely to it's modulus of elasticity, $E$, and it's moment of inertia about the axis of bending, $I$. Since you already have your material set, steel, you cannot change $ E $. What you can change is your $ I $. The moment of inertia for a ...


15

Keeping things simple, steel is an alloy of iron and carbon, whereas stainless steel is essentially an alloy of iron, carbon and chromium or iron, carbon chromium and nickel. All forms of steel, whether they be ordinary iron and carbon alloy or stainless steel are made from a melt in furnaces. Because of this stainless steel cannot be plated to ordinary ...


14

This will at least depend on the: Rate of Cooling Magnetic field strength Exact composition The magnetic field will alter the microstructure as you can read in, for example, Yudong Zhang, Nathalie Gey, Changshu He, Xiang Zhao, Liang Zuo, Claude Esling, High temperature tempering behaviors in a structural steel under high magnetic field, Acta Materialia, ...


13

In general, you want to stay below the recrystallization temperature. Steel is composed of grains, and different types of steel have different grain sizes. The size of these grains affects the steels behavior once it gets past the yield point. At the recrystallization temperature, new grains will nucleate and grow, which undoes any sort of hardening that the ...


13

Is such wall thickness realistic for an industrial gas pipeline? Looking through ASME B36.10M-2004, Welded and Seamless Wrought Pipe, there are plenty of pipes that have thicknesses in that range. For example, 16" diameter, Schedule 160 pipe is 1.594" thick: This translates to about 40.5mm thickness. As another extreme example, 24" diameter, Schedule 160 ...


13

Yes, steel (even some stainless) may be cheaper than aluminum, but the material cost of an item is seldom the majority of the total cost, especially a small item such as a potato peeler. Making aluminum parts with complex, curved shapes can be fairly easily done by casting. Aluminum pours at around 1500 °F, which can be achieved with low-cost furnaces and ...


12

Based on it being an EN steel grade: The first number is 100x the carbon content percentage (so 0.11%), the letters are added elements (sulphur and maganese), and the last number is the sulfur content (0.30%). You can see the full details here. The full format seems to be: [X][% carbon][added elements][% of added elements, hyphenated] Note that the X is ...


11

That is correct, there are a number of unwanted, or tramp, metals (Cu, Sn, Sb, As) that enter the recycling stream from, for example, car bodies that are ground into scrap without removing all the copper wiring, or tin-coated steel cans. Antimony and arsenic tend to creep in from low-quality and low-cost primary iron sources. The answer to the question is no....


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. ...


10

BS5950-1:2000 Clause 1.3.23 defines an H-section as having "an overall depth not greater than 1.2 times its overall width", and Clause 1.3.25 defines an I section as having "an overall depth greater than 1.2 times its overall width". Note that at exactly a ratio of 1.2, it would be an H section not an I section.


10

The main purpose of rebar is to improve the tensile strength of concrete and in practice most of these loads come from bending rather than pure tension. When a beam in subject to bending forces the greatest stress are at the edges and faces of the beam so just having one big bar running down the centre wouldn't do very much as this part of the structure ...


10

The reason is pretty simple. Steel is significantly stronger than concrete. Nowadays we have high-performance concretes with $f_c > 100~\text{MPa}$ (and ultra-high-performance, which is substantially higher), but most ordinary structures don't use such high strength concrete. This bridge seems relatively weathered, so the concrete is probably at most $...


10

One point of note is that the yield region is not as cleanly defined as BCD is in the image (although most books have it that way). In reality the yield region looks like The following image is one from many actual measurement that I took years ago (its Force Displacement). You can see the jagged yield region after the elastic region. The point is that the ...


9

Here's the formula(s) we use Beam Bending (available on Wikipedia) $$EI\frac{d^4\,\delta y}{d\,x^4}=q(x)$$ $$I=\int (y-\bar y)^2 dA$$ $$\bar y= \frac1{A}\int y \;dA$$ $$\sigma_{max} = y_{max}E\frac{d^2\,\delta y}{d\,x^2}_{max}$$ Where $A$ is the cross-sectional area of the beam, $y$ is the position along the axis in the direction of the beam loading, $\...


9

There are three (or four) possible terms for what you are asking: resilience, ductility or toughness (a steel that is resilient, ductile and/or tough) A material's resilience describes its ability to absorb energy by deforming elastically. Its ductility (or malleability, there is a slight difference between the terms, but its not relevant here) represents ...


9

Stating categorically that bamboo is stronger than steel is a bit like stating that automobiles are faster than planes. On its face, it is a bit shocking, and seems wrong. But a rocket-powered automobile is certainly going to be faster than a one-seat propeller plane over a short distance on a controlled track. Then again, the same rocket car is going to ...


9

TL;DR: Materials perform differently under different loading conditions. Some applications are more suited for steel others for aluminium I will try to give another more general perspective/approach, using the concept of material indices (mainly other users gave perfectly adequate explanations of the differences between bending and tension and stiffness vs ...


8

For structural applications (in the US), the most common bolt for weathering steel is ASTM A 325 Type 3. Type 1 is a plain steel bolt that can be galvanized, but in this situation the zinc in the galvanizing will quickly be used trying to protect the rest of the structure. Update for British bolts Interestingly, the only option for UK seems to be to get ...


8

Meet Vladimir Shukhov, a Russian architect who first developed hyperboloid structures. He was born in 1853, died in 1939, and created over 200 hyperboloid structures in the intervening years. He was the reason hyperboloids gained the popularity that they did. His first design, the first hyperbolic structure ever, was the Shukhov Tower in Polibino, pictured ...


8

First, I want to say that I'm suspect of that material actually being ASTM A311 (though I don't have a copy of that standard available to verify.) Looking on Matweb, this material is the closest I found in the 1050 steels. This search turns up quite a few results, but from what I can tell of A311, it's for cold-drawn and stress relieved bars, but doesn't ...


8

The simple answer is: "yes, you can use the given equations with Pr=0." There are three reasons for my answer: The code section (AISC 360, Chapter H) doesn't specifically state that Pr cannot be equal to zero. The equations in code section H1 are referenced by other sections that apply to both compression and tension (e.g. H2), so there must be some ...


8

If your beam design is governed by yielding in bending (not lateral-torsional buckling/plate buckling, etc) then you need to increase the second moment of area (I) to increase the bending capacity. Usually this is done by fastening additional plates to the beam, typically onto the flanges. Here are some examples of such a strengthening strategy: Bolting: ...


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