8

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 ...


4

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 ...


3

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 ...


2

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


2

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 ...


2

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 ...


2

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 ...


1

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: $$\delta= \frac{Fa^2}{6EI}(3x-a)$$


1

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 ...


1

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 $\Delta s$),...


1

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 ...


1

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 ...


1

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, ...


1

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 ...


1

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 $$...


1

When the beam bends, the tension and compression at the outer (top and bottom) surface of the flanges is higher than at the inner surfaces. That follows from the basic equation giving the axial stress in the beam = $My/I$. If Poisson's ratio is not zero, this non-uniform stress will bend the flanges from the web of the beam to the extremity of the flange. ...


1

$\lambda_1$ is the minimum slenderness ratio at which buckling will be the dominant condition for a purely axially loaded element. Interestingly, it is an intrinsic property of the material, not the geometry. It's derivation is simple: $$\begin{align} P_{crit} &= \dfrac{\pi^2 EI}{L^2} \\ \dfrac{P_{crit}}{A} &= \pi^2 E \cdot \dfrac{I}{L^2A} \\ \sigma_{...


1

I'm not an engineer by any stretch of anyone's imagination, but I have 30 years in aluminum extrusion. In the conditions you have, I would agree with going to steel. You probably have a 6063 T6 tube. With the amount of force generated by wind, you'll have enough cold-work in the aluminum to make it brittle, then it'll simply snap. If you're dead-set on ...


1

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.


1

"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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