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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 ...


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 ...


14

The strength of a horizontal beam is defined based on a property called the section modulus. Shapes that have more material distributed near the top and the bottom have a higher section modulus. This is why I-shaped beams or tall rectangles are common choices. Round shapes, on the other hand, have most of their material concentrated around the center of the ...


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

A cylindrical beam is probably not the best beam for given situation. This doesn't mean that there are no specific situations where a cylindrical beam would be useful. Pros Can resist the same load in any direction or combination of directions. It is as strong in the horizontal direction as the vertical direction. No need to worry about or check the ...


12

Whenever one speaks about removing a structural element (column or beam), the initial hypothesis must always be: NO! GOD NO! WHAT ARE YOU DOING! The second hypothesis must always be: No. No. No. No. That being said, it can be done. But it is not trivial and requires a serious investigation by a structural engineer. This is a "last resort" issue. ...


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.


9

The way to derive the deflection is via the relation $w = EI\dfrac{d^4\delta}{d^4x}$, which means that the deflection is the fourth integral of the loading (and rotation can be taken as the derivative of the deflection and therefore equal to the third integral of the loading). So, let's begin with the loading: $$w = w_{max}\dfrac{x}{L}$$ The first integral ...


9

For a simple visual demonstration, take one of the spans in your example. If it is fully hinged, then each span can be represented as a simply-supported beam. If the beam is continuous, then each span can be represented as fixed-and-pinned (or at least pinned and partially fixed). The only difference between these two models is the moment reaction at the ...


8

The dimensions of the beam and magnitude of the deflection are important here. In most structural applications, it's reasonable to assume the length of a beam is unchanged by a small deformation. One of the basic assumptions of beam theory is that there is some internal surface of the beam called the neutral axis that holds no tension or compression stress, ...


8

If you are looking at the column as being supported at the ends, you are correct that the n=1 mode gives the lowest buckling load. The other modes (n=2,3,...) aren't useless though. Long columns are often braced at regular intervals to reduce the unbraced length of the column. For a given length of column, these braces force the column to buckle under a ...


8

This rule is typically applied when studying statics. Static means that your structure or object does not move. If the moments didn't all add up to zero, that would mean there was a net force action on the object, which would cause it to accelerate and move. Since it is static and not moving, we know that whatever forces are acting on the object, they must ...


8

I'm going to run with this assuming the arm looks like the following diagram (I'm ignoring the 11.4 pounds of the bar for now to make the concepts easier to explain - that can be added later by assuming all 11.4 pounds run through the center of the bar.): This is a pretty simple setup. To figure out the loading on the shaft, we have to move the load from ...


7

Moment of inertia is not a material property, it is a geometric property. Regardless of the material you are dealing with, a member that is 2" wide by 10" tall in cross section has the same moment of inertia. Therefore, you should use the standard equations for calculating your beam section's moment of inertia. To address the portion of your question ...


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 ...


7

Whether or not buckling modes with $n>1$ exists depends on how you look at the structure. As @hazzey notes in his answer, columns with bracing may display buckling modes with $n>1$. These buckling modes, however, are simply equivalent to the $n=1$ modes of the individual segments that compose the column. To be clear, this doesn't mean that the ...


7

You're hanging it from a rod supported by both ends - and need to use the bending equations. For this case (case 7 in the link), the max weight is: $$W_{max} = \frac{\pi d^3 \sigma}{8L}$$ $\sigma$ is the tensile stress, $L$ is the rod length, $d$ is the rod diameter. The sag is: $$\delta = \frac{4W_{max}L^3}{3E \pi d^4}$$ where $W_{max}$ is the weight, ...


7

in a cantilever beam the deflection is $$\delta_{max} = \frac {PL^3}{3EI} $$ In this case assuming free sliding between the planks the load P is going to be supported equally between the 3 planks. So the deflection will be $$ \delta_{max} = \frac {(P/3)L^3}{3EI_{\text{single board}}} $$ Because $I$ is proportional to the cube of the board's height (in ...


7

Stiffness is a murky term frequently used ambiguously in engineering. However, the most common definition of stiffness is the product of a beam's Young's Modulus $E$ (which is a function of its material) and its moment of inertia $I$ (which is a function of its cross-section). So $\text{Stiffness} = EI$. Loading has nothing to do with stiffness according ...


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

In case Eurocodes do not provide enough information, some sources exist. In the case of elastic critical moment for lateral-torsional buckling, an NCCI (Non-contradictory, complementary information) document exists. The document code is SN 003, and one version (maybe not the latest) can be accessed here. Hopefully, this will cover your current needs. In ...


6

Deflection A deflection is a movement in a direction. It is up to you to determine your frame of reference. It sounds like the calculations that you have done yourself (all positive) are shown for a force also in the positive direction. This makes sense logically. The deflections shown in the graph that you attached (all negative) may be in a different ...


6

Adding to @Air's answer, there's also the issue of boundary conditions. A simple span where neither support allows for axial displacements will have a slight gain in length, including along the "neutral axis". This is because, in this case, the "neutral axis" will hold a tensile stress. Since the beam deforms, the neutral axis changes from a horizontal ...


6

To put it simply your loads are causing the beam/bar to bend right? So at the cut we need to consider the internal bending moment ($M$). Like you said: the professor has cut the bar to calculate the INTERNAL forces experienced by the member. In any 2D problem (say in the x-y plane) the first thing that you learn is that whenever you take a cut there's three ...


6

SHEAR CENTER Why do we care? Because we need to know if the beam is subjected to torsion in addition to flexure. What is it? The point in the cross section (or outside the cross section) where we can apply load and produce beam bending without twisting. Loads applied anywhere other than the shear center will produce both bending (moment) and twisting (...


6

This means that the member is considered to be fully loaded for a failure mode which is being analyzed by that stress calculation. Some things to (probably) keep in mind include: Some conservatism likely exists in the analysis/design calculation. A member reported to be at 100% capacity is (probably) not actually at 100% capacity. This conservatism could ...


6

If the white beam on the left is adequate, the one on the right is much bigger than it needs to be, so hacking a piece out of it might not matter. In general this idea is a horrible example of engineering "design". Still, if you follow the sticker on the wall and "maintain social distancing," that will reduce the number of deaths when it ...


5

The answer to the headline question is yes, it is equal. It physically has to be equal, since you can't have two different values of a particular directional stress at one point. The quoted equation is for shear stress, where V = total shear force at the location in question; Q = first moment of area 'beyond' the point considered; t = thickness in the ...


5

Beams Only First off, Lateral Torsional Buckling (LTB) is only applicable to beams. Beams are members that are loaded so that they have moment. This is what the first image shows. The members in the second image are not beams. They are struts that are loaded primarily axially. They are holding the walls of the excavation apart. This means that LTB is not ...


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