33

I will expand on DKNguyen answer, because to my knowledge also the two reasons are: reduce contact/bearing stresses (having a significant effect on thin finishes live galvanisation) change the joint tightening characteristics (see joint diagram). reduce contact stresses on surfaces. The basic idea is that since contact stress is defined as: $$\sigma = \...


27

It is for spreading out the stress. But it is also for giving the bolt a bearing surface to turn on. The washer always goes on the side (nut or bolt) that is being turned. It prevents it from marring up the work surface and also changes the tightening characteristics. I don't know the specifics of that though but that's what I was told by a toolmaker. Always ...


16

Except for special applications, most washers are made of dead soft steel, which deforms under the compressive load imposed by a tightened bolt head. As the washer smooshes, it minimizes stress concentrations caused by bumps under the bolt head and surface flaws in the part the bolt is running through.


15

To visualize part of Nmech's answer: in the image, the washer actually greatly increases the contact area of the bolt head. The bolt head looks pretty big: But most of that is the shaft, which obviously does not spread out load on the material. So the actual contact area looks like this: Comparatively, the bolt head on the washer looks like this: That's a ...


14

While some of these answers are close, they are (at the time this answer is written) all incorrect to some degree. Pressure and stress are very closely related -- in fact, one could argue that pressure is, in a sense, a subset of stress. To be specific, the pressure in a material is the isotropic part of the total stress in a material. Pressure is a ...


14

It is done carefully. Lots of other rotating machinery has the same problem, and whole systems exist just to deal with it. For example, jet engines are usually smaller, but also usually spin much faster. Balancing a jet engine is something that gets lots of attention at manufacturing, and again any time when it is put back together after having been ...


13

We use engineering strain even though it is not the "correct" value because in most cases, specifically in the elastic regime, engineering strain differs negligibly from true strain. For linear elastic, Hookean materials, it is generally the case strain at the elastic limit is very small. Even the strongest steels, for example, have an upper limit when cold ...


13

As grfrazee said, you won't know for sure until you do a finite element analysis. I was intrigued by this question as a colleague and I got into a discussion about this. While we both agreed the diagonal bracing would be better at resisting deflection, we wondered by what factor it would be better. We were really curious so we settled the debate and did a ...


12

Assuming the joints are welded, for the top gate to deform as you draw it the vertical bars will have to bend into an "S" shape. The flexibility in bending will be proportional to the cube of the length, if everything else is the same. The stiffness of the three sections of the top gate will be proportional to $1/1^3 = 1$, $1/0.6^3 = 4.6$, and $1/0.4^3 = ...


12

Sorry for not taking your "Thin Hollow Brick-Shaped Object" as an example - I felt a standard notched test specimen would illustrate the principle more clearly: It can be seen that adding external radii to the test sample has no effect on the stress. In this example, cutting even more material away from the corners would make no difference to the failure of ...


11

How To Read the Graphs The graphs are plots of stress versus position. Stress is the force per unit area exerted on a material. Positive values are tension and negative values are compression. The first (leftmost) plot shows the normal stress vs. position that would be expected from bending if there was no pre-stress. The middle plot shows the pre-...


11

That is due to necking. When a bar is under tension, two opposing mechanisms take place: The bar loses cross-sectional area since it attempts to retain its volume while being stretched (see Poisson's ratio). Once the bar begins to suffer plastic deformations, it undergoes stress hardening and it's elastic modulus rises. So long as the increase in elastic ...


10

The torsion constant $J_T$ relates the angle of twist to applied torque via the equation: $$ \phi = \frac{TL}{J_T G} $$ where $T$ is the applied torque, $L$ is the length of the member, $G$ is modulus of elasticity in shear, and $J_T$ is the torsional constant. The polar moment of inertia on the other hand, is a measure of the resistance of a cross ...


9

Adding to @starrise's answer: With regards to your dismissal of reasons 1 and 2, you are forgetting to consider the cost-benefit analysis regarding them. As @starrise showed in their answer, the difference is usually not substantial (though other materials will usually have larger differences). On the other hand, the materials always show intrinsic ...


9

Things scale, but not always in a nice simple way. For example consider the units of stiffness of part of the structure are $E$ (Young's Modulus) times some length, but the units of weight are $\rho$ (density) times some length cubed. So a scaled-down model structure will deform much less under its own weight than the original. (This explains why elephants ...


8

Assumptions The angle between the wall and the strut is $\theta$ $a$ is the depth of the table top $P$ is the weight on the table top, applied at the edge furthest from the wall The strut will fail when it buckles, which implies $F_{\text{max}}=\frac{\pi^2EI}{L^2}$ where $L$, $E$ and $I$ are the length, the elastic modulus, and the moment of area, ...


7

While you've described your problem pretty well, I don't think you're going to find a satisfactory answer without having to run a fairly complex finite element analysis on both structures. The first gate structure will behave similarly to a Vierendeel truss since you have all of the pieces essentially moment-connected. The second gate structure will likely ...


6

Splitting a uniform load into separate pieces that are still continuous will have no effect. This is frequently done. As far as your question about bending in HE and EB, there shouldn't be any bending because all of the forces are balanced. A sum of the moments at H or E will show that the moments from the beams on either side are opposite and equal. That ...


6

This is something that I've looked at with tunnels, rather than pipes, and arguably with smaller deflections. Hopefully it'll be some help, however. If you can satisfy yourself that the rate of curvature is relatively small- then you can approximate the stress generated by the lowering using simple beam theory: $$\sigma=\frac{Ey}{R}$$ Where: E is the ...


6

You're looking for $\tau_{xz}$, which pertains to horizontal shear flow along the top flange. (Rather than the vertical shear flow considered in $\tau_{xy}$.) As Mark noted, the shear flow starts from the center of the flange and flows outward/down. The formulation for the shear stress calculation remains essentially the same. Only the area changes -- we're ...


6

Firstly, the shear center is the point at which an applied load produces no torsion on the section. For a singly or doubly symmetric section, the shear center will lie on the axis of symmetry. For the unequal leg angle shown here, we can say by inspection that the shear center must be at the intersection of the two legs because the shear flow will produce no ...


6

All these terms refer to the effect of loading on the deformation of materials. Let us assume that we start with zero load and zero deformation. Elastic deformation If you increase the load you get an increase in deformation. During the process of elastic deformation, if you decrease the load to zero you will not have any residual non-zero deformation. ...


6

Using a tie-and-strut model makes it easier to explain the difference here. A tie-and-strut model is a simplification of the structure which describes where the structure is under tension, and where it is under compression. The model gets effectively simplified into a truss. Below we have two examples of the axial load on members of two different trusses ...


6

The main reason from a designer's point of view is the consequence of a failure. Almost always, something like battery failure or a puncture doesn't cause any collateral damage - you just replace the failed part. (Even a tyre blowout when driving at high speed rarely causes much secondary damage - and that is a rare type of "tyre failure" compared with the ...


6

For impact force calculations you need some more information such as the shear modules and the material properties like toughness and tensile yield and shear yield. Also you need to define a physical object that will impact the sheet. If the object is much more rigid than the sheet and pointed it will go through like a bullet. Imagine an impact force has ...


6

I think you might have a misconception regarding to how far the pressure from the fasteners extends. One subject you might want to have a look into is "bolt joint stiffness". The most popular is the "Rotscher’s pressure-cone method". Essentially there is a pressure cone which radiates outwards with an pressure cone angle a. According to ...


6

The Load-Displacement (or Load Extension) and stress strain diagrams are two diagrams identical in shape. See below. The main visible difference is the values on axis (which are at first glance neglected). So, it is natural when you first encounter them to question why do you want to learn about a stress and strain diagram which has obscure quantities (as ...


6

Another important part of the answer is the symmetry of the stress pattern. The stress caused by a bolt head varies greatly between the points of the bolt head and the straight sides. As a result local stresses, which are what you really care about because those are what the materials have to withstand, can be much higher than the average stress. A washer's ...


5

You are asking for an unrealistic material. If you peruse the engineering toolbox you'll find that the Young's modulus of typical materials is measured in 109 Pa so the value you are asking for is 7 orders of magnitude smaller than typical materials. Even the Young's modulus of the rubber in rubber bands is ~107 Pa. Another simple thought experiment will ...


5

It's fairly complicated. I found an article in The Bone Journal Comparison of three-point bending test and peripheral quantitative computed tomography analysis in the evaluation of the strength of mouse femur and tibia where they used tomogrphy to calculate inertia. However if you don't have CT you can try the hard way. You'll have to calculate bending ...


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