9

You are worrying about the numbers and forgetting what the numbers mean. This is for the PE exam, so this is a very important topic to make clear in your mind. You solved the problem. You came up with an answer for what is required. You now have two options: Choose a small footing that your own solution just proved is too small for the requirements. Choose ...


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 ...


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

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

Modern building failure due to tipping is rare but did occur, usually under strong earthquakes and loss of foundation (soil and structure), as the tilt will shift the gravity center of the building that induce a tremendous amount of stress on the foundation piling and the soil mass surrounding the piling. So, how much is too much for a building to handle (...


3

An extended discussion on internal vs. external forces We usually like to describe hinges as "places where the moment is always zero." But, wait a minute, the moment is always zero anywhere in a stable structure. Don't believe me? Let's take a look at the most trivial structure ever, a simply-supported beam with a uniform load: So, it has a span ...


3

Tl;DR: yes it is correct (disregarding rounding errors) assuming points at A: ( $x=0$) B: ( $x=\frac{1}{3}L$) C: ( $x=\frac{1}{2}L$) D: ($x=\frac{2}{3}L$) E: ( $x=L$) It is obvious that there symmetry in the problem. Therefor point C (in the middle will not move at all). For the half problem (ABC) you can write the equilibrium of forces. $$F_A + 14.4 +F_C=...


3

First thing first, you need to get hand on the ASME Boiler and Pressure Vessel Code (BPVC Code). Starting division 8, section 1, for safety and quality of your pressure vessel. After checking on all relevant provisions/standards, you can use the equations to derive the required wall thickness for your application. Notation σ_H = hoop stress, psi or MPa D = ...


3

On top of flexural deflection, the Timoshenko Beam Theory includes a term called "Shear Deformation" that is to account for the additional deflection and shear stress due to shear effect, which is quite samall compared to the effect from the flexural, thus it is usually ignored except higher accuracy is desired. (See figure below) The slope of the ...


3

Here be dragons. Or, to paraphrase a hadith of the Prophet Mohammed, "There are seventy ways to compute element stresses, and all of them are wrong." The naive approach is to differentiate the element shape functions to find the strain at a point, and then use the stress-strain relationship for the material. The problem is that since the shape ...


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.


3

TL;DR: Yes, any structure deforms if you put a load on it. Even adding an ant on top of a granite mountain will change (lower) the height of the mountain - imperceptibly so but it will still change it. The problem is that its not possible to measure it. That is the whole idea behind Young's Modulus (modulus of elasticity). Essentially, all materials behave ...


3

None of the three would occur if the material of the element holding the rod is "rigid", or is much stiffer than the rod. For such cases, my assessment is as shown below. For the case that the rod is much stiffer than the element holding it, the stress distribution is based on the flexibility of the rod. You can verify the result using a model ...


3

A good reading about that is the: History of strength of materials by Timoshenko. It describes the history behind Galileo Galilei and Hooke until the era of Cauchy, and the intermediate steps Galileo (sometime between 1592 and 1610) figured out that for tension, when you double the cross-section the load to break also doubles. But run into problem with ...


3

The stress is not uniform near the grip because there is a biaxial state of stress. There is the axial force the compression force from the grips. That is the reason that if the Cross-section was uniform the failure would start from the grip. the mental image is the following: If you have a ballon that you apply force from one end, and at the same time ...


3

This might change depending of what discipline you are investigating but: homogeneous: refers to a material that you cannot distinguish different phases in it. Usually it refers to the density properties, and it's an indicator that the density is uniform when you look at it at various scales (from m to sub mm). (some etymology: the word comes from the ...


3

Bearing stress is essentially the contact pressure between two surfaces (the bolt and the joint in this example). I find illuminating the following picture that shows the force flow. figure (source chegg) how it is transferred though one part of the joint to the bolt and then to the other part of the joint. The bearing stress in this bolt correspond to ...


3

not so much a formal answer (more like an answer for the intuition), is that during torsion on non circular shafts, the stresses tend to deform the material in such a way that the cross-section resembles more that of a circular. In order for that to happen , material needs to be deformed/pulled from adjacent section. The end result is that it contracts.


3

A non-circular section under the torsion deforms in both St Venant shear and also by warping. For example, an I beam deforms both by warping and shear deformation. Warping creates tension and compression in flanges. so the plane of the flange does not remain flat anymore. source .


3

The hinged region at B is not a support, actually. Thats just a connection (or to be more precise, a pinned connection). A pinned support is something different; a pinned support is something what you see on the right end C. The support itself is connected to the ground (for example), and cannot translate in any of the directions (which means that the right ...


2

I would have thought the nut and screw screw thread was more likely to be the failure point, but you are looking at the part and I'm only looking at a sketch of it. The yield strength of grade 5 steel is around 600 MPa. The area of the pin is about 47 mm^2. So to permanently bend the pin by shearing it would take about 28000 N or about 6300 lbf. (Or arguably ...


2

Like alephzero already mentioned the pull up part is usually the less stresses part. The problem is with the deceleration forces when you are at the top of the pull up bar and suddenly you let your self down. Although that seldom happen (i.e. there is a constant force for deceleration), it is best to assume the worst case scenario. I'll make the following ...


2

Yes, That is a good estimate. Another reasonable way is to measure how fast you can pull yourself up. Say if you can time your lift acceleration and it is $12m.s^2$ you are applying, $mg+1.2mg=2.2mg \ , $ 2.2 times your weight. Of course, If you have access to one of those benches with lift up weights, the ones that have rubber bars locking you in, you can ...


2

Using conditions given in the sketch above, let's do a parameter study. M = AsFy(d - a/2) a = Asfy/0.85fc'b, note, since As, b, fc' and fy are constant throughout this exercise, "a" is constant as well. Let's assume a = 50mm. d = 65, M1 = AsFy(65-50/2) = 40Asfy d = 115, M2 = AsFy(115-50/2) = 90Asfy d = 165, M3 = AsFy(165-50/2) = 140Asfy M2/M1 = ...


2

It is important to understand what is an internal hinge, its behavior, and assumptions around it. An internal hinge represents a discontinuity in the beam, it carries a pair of equal but opposite force couple, one to each side of the hinge, carried over from the beam segment it supports/connect to, so sum Fy is zero. Another important assumption is the hinge ...


2

I would say that you can divide them into two groups: At nodes with two member and no external force Node C: CD,CB, Node G: FG, GH Node K: KJ, KL If you write down the equilibrium at node C for x and y you would get: $$x: \quad \sum F_x = 0 \rightarrow CD = 0 $$ $$y: \quad \sum F_y = 0 \rightarrow CB = 0 $$ Same at the other nodes. Note that nodes A, and ...


2

Usually $\kappa = \dfrac{\text{average shear strain on section}}{\text{shear strain at centroid}}$ but the problem with this definition is that it depends on external loading so you can't actually make a table for $\kappa$ values. However, we could make an assumption and define $\kappa$ purely in terms of the shape of the section: $\kappa_i = \frac{1}{\...


2

You formulate the matrices in a local coordinate system where local $x$ axis is along the length of the beam, and the local $y$ and $z$ axes are the principal axes of the beam section. In that coordinate system, $I_{yz} = 0$ by definition. ($I_{xy}$ and $I_{xz}$ are also zero). Finally, you transform the equations in the the global coordinate system using ...


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