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In Euler buckling theory, buckling occurs if a small (infinitesimal) lateral deflection causes a reduction in the internal strain energy of the structure. That calculation is problematic when $E$ is not constant. In your example, if $\sigma > 240$ MPa, the strain energy in the unbuckled state depends on both $E_1$ and $E_2$, and that is easy to calculate ...


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Regarding the first part of your question " The buckling now happens when the material reaches the lower limit, which is also when its modulus changes" If I understand correctly your problem is what happens is when the buckling stresses are not high enough. Obviously there is a chance that there might not be buckling. However, keep in mind that ...


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IMHO, any lateral force would make this collapse. Nodes 1234 indeed create a statically determinate body. However, the rod 3-5 and 4-6 are parallel and of equal length. Therefore body 1234 can freely rotate. Regarding how you show that mathematically, (without knowing the context of your assignment): 1234 is a rigid body you can substitute 1234 with a ...


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There's a reason why they're called "factors of safety" The answer to your question is in its title. When designing a structure, we need to make sure of two things: that the structure will remain a structure and not end up as a pile of rubble (i.e. that the structure is safe) that the structure can actually be used as intended. The first is easy ...


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Because the service load system, which is the old system being replaced by the LFRD method, limits the strength of the structural members like steel or concrete to a factor of usually around 60%. Roughly a safety factor of 150%. This system does not differentiate between live loads or dead loads or transient loads such as wind loads or earthquake loads. The ...


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For starters, what makes you think the concrete doesn't crack at the top? Let's assume for a second that it does crack. Well, if we were dealing with a pure concrete beam, that'd be a big problem since the crack would propagate down the height of the beam until it falls apart. Thankfully, prestressed concrete beams aren't just concrete and a prestressing ...


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your question has many variables. what are the spans, equal or one is longer? just as a first step if they are same length and carry the same load it is trivial and you just need to consider the nominal connection forces. If lengths are different or one is carrying more load, then we have to calculate individual deflections and assum the larger deflection ...


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Note that the pretension $P$, besides causing a bending moment $(M=P\cdot e$, where $e$ is the eccentricity) to counteract the expected loads, also inflicts a normal (compressive) force ($N=-P$). Usually, a stress analysis needs to be made for the different loading setups to confirm that tensile stresses either do not occur or exceed a certain value. This ...


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Because there are rebars designed for this exact task on the top layer of the beam rebars. These are in addition to other rebars required for other loads, such as negative moment rebars at the supports. the codes allow certain bars to share loads.


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ASCE 7-16 Section 11.4.5 provides the equations for the 5 percent damped spectral response acceleration, Sa, relative to period, T, in the following range. 0 ≤ T < T0, T0 ≤ T ≤ TS, TS < T ≤ TL, and TL < T here is a link to a detailed solved example. look at Example 2, how to create design spectra with 5% damping. Except that the link, they have for ...


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The compressive stress is, of course, a factor, but if we consider a simple brick wall, then this H calculation would not suffice since there is lateral stability to consider as well surely, any misalignment or lateral force e.g. crosswind would compromise height a lot I think.


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We calculate the deflection $\delta_a$ and $\theta_a \ \delta_a=\frac{Fa^3}{3EI} \\ \theta_a=\frac{Fa^2}{2EI}$ . For any point x, a<x<L we add the slope times*(x -a): $\ \theta*(x-a)$ , while noting the slope doesn't chang past point a, along the beam,to the deflection and we get: $$\delta= \frac{Fa^2}{6EI}(3x-a)$$


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I would first calculate the deflection $\delta$ and rotation$\theta$ for The point at the end of the double-layer assuming the rest of the cantilever is rigid, using the equivalent area method, like in concrete beam design, Wasabi explained it. Then calculate the deflection and rotation for the rest of the beam and add it to the first part. This method of ...


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Since Wasabi already provided the analytical way(+1), I will only provide a shortcut, because the analytical can be very painful. Since you already have the geometric model, what you can do is you can perform a FE simulation with a known Force at the location you are interested. Then just measure the displacement that FE predicts (lets assume its $\Delta s$),...


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This is a case of a variable-cross-section beam. What you need to do in these cases is separate the cantilever into two parts: near the support it has one cross-section, near the free end, another. The analysis can then be done by hand, but it's a pain and I honestly wouldn't recommend it. Just use some analytical software to get the solution. You also ...


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"split lock" spring washers are nearly worthless for serious torque or vibration. Belleville type lock washers are better, especially if serrated. The best by far, amazing actually, at least for resisting torque (but not length changes) are the nord-lok. They are expensive. If the screw is very long and there is thermal expansion or you are ...


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The loctite is only good for one kind of rootcause. Please try the loctite + spring washer combo. If it is still loose, use 2 pcs spring washer. These should do the job. Or worst case use mechanical fixation washer with a lip what is possible to bend on the nut side. I hope this help. Br.: Peter


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Hooks are used in many situations in RC. They are used for example in column beam intersection to give ductility and confinement of the joint in case of an earthquake. They are used in beams and columns for transition or when there is not enough bonding strength length available, by developing tension in confined areas, or they are used at the end of the ...


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Not only there may be wight of the concrete above the bar, but there may also be other loads as well like floor machinery. (we always mind the punching shear though). but all those loads will translate into tension by the bending of the beam under the moment. Except at supports where the beam can not bend, hence the sheer tie bars. In columns on the other ...


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As in @alephzero's comment, the formula you refer to is wrong. The analysis of a column of nonhomogenous, composite material can be a bit complicated. Empirically the ancient masons had some secret codes and formula's as to how to treat tall masonry or arches and big domes. In Iran, they have even built ancient masonry towers designed to shake per demand. My ...


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It can't be done Indeed, this problem is unsolvable using only equilibrium equations. Broadly speaking, structures can be split into three categories: hypostatic: when there are too few supports to keep the structure stable. These are commonly called "mechanisms" isostatic or statically determinate: when there are exactly as many supports as ...


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Consider this simplified analogue to a suspension bridge: we model the roadway as a single point mass $m$, which is suspended from a rope attached to the cables. Now think about what's the best angle for the cable to make at the attachment point: completely straight, or hanging through a bit? The answer is, it must hang down at least a bit. Recall that ...


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A suspension bridge is an efficient use of material, because all the forces in the structure are carried by tension, and not by the parts of the structure resisting bending. Beams which resist bending are often useful components of structures, but they are an inefficient way to use material because the stress in the beam is not constant through its depth, ...


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It's not possible for the cable to be perfectly horizontal. Cable can only pull; that is, the force they apply at a connection point is along the direction of the cable at that point. So a horizontal cable would be able to exert only a horizontal force. But the cable needs to transmit the weight of the bridge, and to do that it must apply a force with a ...


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A suspension bridge cable is the most effective use of material. Because it uses high tensile cables in a position where they are the most effective material: tension. The parabolic shape is the closest natural geometry a cable assumes under uniformly distributed horizontal load ( not the self-weight, that is a catenary curve). It has reduced the structure ...


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The next time you see some kids playing jump rope, go out and ask them to hold the rope perfectly straight. If two kids hold the two ends of the rope with it dangling until it is almost on the ground, they have to hold the ends of the rope up but they don't have to pull much, end to end. The straighter the rope is, the more they will have to pull. If you ...


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I'd recommend first making a hole to fill the channels in the brick with mortar/concrete locally. This way you will effectively have a solid brick which allows for a much stronger fastening and a much larger range of applicable anchors. The anchors you show above are all mechanical anchors, which is not the optimal solution when a waterproof result is ...


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If you are a beginner and you already use solidworks I would suggest sticking with it. It has a decent FEA/CFD addon that you can use in a single piece of software. That alone will save you loads of time. The main drawback is that you might not have the material models and customizability of other pure bred FEA Tools. Ansys, (and a lot others) is/are more ...


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I don't think of a screen protector as a mono layer, but a lamination. Deformable (softer) materials (possibly above and also) below the rigid layer (this layer also generally provides adhesion). That way point loads become distributed (potentially to the protector and also) to the device. Example: a thin clear sticky low molecular weight polymer (1), an ...


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I would guess that your build material has absorbed moisture from the environment and is boiling as you heat the build material. Recommend you place the build material on the floor in a car, leave the windows cracked, and place the car in the sun. The warm air will draw out the moisture and the moist air will be lost through the cracked windows. Think of the ...


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If the ring were free-standing it would support the pressure P by tangential tensile stress. $\sigma_{tensile}=P*r/t*\Delta r$ We need the elastic properties of the outer layer. Just for sake of argument let's say the $E_{big \ ring}=1.5*E_{small\ ring}$ And say the thickness of the bushing is the same as the interior ring and the middle ring could be ...


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In this particular scenario, you can have two types of fixtures with two different a) fixture to the right (See red) In that case you only need to test for the shear failure of the ring. The shear stress will be: $$\sigma_s = \frac{F}{A} $$ where: $F = \pi (r_o^2 - r^2) P_0$ : the force on the shear surface $A = 2\pi \cdot r \cdot t$: the shear surface $$...


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Fitted 3 screen protectors to my phone - replaced 2 and no damage to screen, therefore two things are evident: dropped phone and damaged two protector screens no guarantee that dropping phone in those situations would have broken the original screen My conclusion is that fitting the screens may reduce the risk of breaking the original screen. However, ...


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