New answers tagged

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It may not be exactly what you were looking for but the Idea could be useful. Almost all cordless window shades could be modified to do what you need. They have weighted (not spring-loaded) pawls that due to centripetal force release the clutch at a certain slow speed allowing the shade to roll but locks if you stop.


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In the flutter, actually, we have a new system with different dynamic behavior, namely the frequency, damping, and also the mode shapes are changed. In the flutter, the source of vibration is a disturbance and the system fulls on a self-excitation. For example in an aeroelastic flutter, the frequency and damping of the wing change as the airplanes speed ...


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American Wood Council has standards including design examples for all sorts of wood screws, bolts, lag-screws, etc. It addresses many factors such as species of the wood or woods if different species are to be connected, their density, moisture, and others. source Here is an example of designing the allowable withdrawal of a lag screw. .


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To optimise efficiency with multiple locomotives, none of the driving wheels must slip. This means that coordination is necessary between the locomotives. Passenger trains have been worked as multiple units for more than a century with driving motors (electric or diesel with mechanical transmission) distributed along the length of the trains, as you suggest. ...


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I think you meant to say the spring is massless, correct me if I'm wrong. Force on the mass by string if we assume down direction as positive ( to follow your sign -kx pulling up) is $kx, \ $ it is pushing the mass down. At the same time, the force exerted by the mass on the spring is $-kx$. The mg component is permanently pulling the mass down and as far ...


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Can you even find a gear big enough with the appropriate flat center surface? Most gears have center holes in them after all unless you plan to EDM one yourself out of a plate. With your new method you only need one driven gear if you wish. No need to drive all three gears, gear tooth strength permitting. If you choose to drive multiple planets with multiple ...


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The concept behind the friction force should always be greater than the tangential force is that the maximum friction force should be greater to the force that causes the rotation of the wheel. If you try to apply more torque, then there will be slipping. This is very similar to the following concept: If a small force $F_A$ is applied, then the friction ...


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Assuming that: x=0 is the point where the spring is neither compressed nor extended x is positive upwards gravity plays a role then the motion equation is: $$m\cdot \ddot{x} + k\cdot x = -m\cdot g$$ By redistributing the terms: $$m\cdot \ddot{x} = -m\cdot g - k\cdot x $$ Because the first term ($m\cdot \ddot{x}$) is usually considered inertial force then ...


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Let's sum forces about point "G": $\sum F_x = 460cos15 - 120cos70 = 444.3 - 41 = 403.3$, pointing lt to rt. $\sum F_y = 460sin15 + 120sin70 - 100 = 119.1 + 112.8 - 100 = 131.9$, pointing down. Let clockwise rotation be positive: $\sum M_G = 444.3(0.47) - 119.1(0.05) - 41(0.47) + 112.8(0.19) - 100(0.59) = 146$ - rotate clockwise. Because there is a ...


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You are right. What you are referring to maybe the property called Thermal diffusivity, $\alpha$ $$\alpha =\frac{k}{\rho C_p}$$ Cp = specific heat k = thermal conductivity $\rho $= density So Thermal diffusivity is inversely related to specific heat. However in your example as soon as slab A reaches the T2, a bit after slab B, on the right side they both ...


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In Siberian mountains, it is very usual to have full-time "double pull" due the steep inclines, and "extra push" loco attached at some complex segments of a route to make sure the train to stay on its schedule. Like 2 x 2TE10 (4 sections in sum) as pullers and trailing CHME3 as a pusher.


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Your structure would be a truss and you could apply any conventional truss method to solve it if the support at E would be pin support. You would then get the BD as a zero force member. As it is both AC and CE are acting as a beam with both moment and axial load. It is not a truss and as you noticed you will get wrong answers if you try to solve it like a ...


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You can use a mechanism like a bicycle pedal. Connect your bushing where normally the pedal is mounted. Either have it turned by the chain as in a bike or a servo. .


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Not really a true answer to the question as posed (and others have answered quite sufficiently that the premise is false). But, for additional background information, there are several other reasons beyond what's already been mentioned that have lead to distributed power (at least in the USA) being much more common now than in the past. Longer trains allow ...


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The method of joints is usually be applied to trusses. The structure you have here is a frame and its members behave differently to trusses rods. More specifically a rod (in a truss) develops/transfers only axial loads. This is done because forces are only applied at the end of the rod ( only two forces are applied and that requires that they are equal and ...


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I question why you do not want the bushing to rotate relative to the thing passing through it because if there's no relative motion then there's no point for the bushing. You're going to need more detail. Because based on what you little wrote inserting the bushing into a off-center circular slot on a rotating disc would be acceptable. Sure the thing passing ...


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It hasn't asked you for a section view. It clearly has asked you for a left side view which means the view from outside the object. There is no requirement in the question to show hidden lines but it would be normal to include them for clarity to show the depth of each hole - through holes in this case.


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The front view differs from a section view. As the section view will see solid lines representing the edges of the holes at the cut; the lines are dashed in the front view to indicate the depth of the bores.


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There is no unique definition for characteristic length. The characteristic number is a number which is related to both the problem and the geometry. So the characteristic length can be defined as any number that returns length units e.g. the square root of the surface area, as the third root of the volume, as the volume-to-surface ratio, etc. At the end of ...


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I'm going to ignore molecular collisions, etc. and stick to the empirical world, which is where we engineers excel. Energy content of flour is 353 Kcal/100g. This equals 1.477 MJ/100g Energy content of gasoline is 46.5 MJ/kg, or 4.65 MJ/100g So, what this tells us is that gasoline has a much higher energy density than flour. In practical terms, this means ...


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Other answers have very valid points about speed of connection, cost effectiveness of using helpers only where necessary, communications, etc, but Mazura hit on a VERY KEY point in his comment that's been missed in all the answers: You cannot push a long string around a curve without it falling off the rails. As consists have grown larger (mile plus long ...


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I'll be combining the two other existing answers (so far). If you are interested more in ease of implementation and low cost then the way to go is connect each rod to a generator produce electricity and then use a motor to get the power at the rpm you want (Eric S solution). If you are interested in a solely mechanical system or if (at a lesser extent) ...


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For long freight trains and those that will be climbing to stations at higher altitudes, an extra or two locomotives are attached to the front. I've always wondered why. As usual there are multiple issues. Most important is landscape. If more power than a single locomotive can provide is not needed for the whole trip, but only a short stretch, like climbing ...


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My intuitive attempt of reasoning, to be taken with a grain of salt since it's basically just an educated guess: Couplings naturally have some amount of play. As long as all couplings are under constant pull force (all locomotives at the front), this play doesn't come into effect and the whole train moves continuously as a single unit. But if we also put ...


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Use adjustable gearing combined with a pawl and spring coupling similar to a bicycle casette. When the target output spins faster than the input, the ratcheting mechanism lets the output freely (a bit of energy lost from the ratchet) turn. You'll have to figure out the adjustable gearing bit and how to increase/decrease the ratios based on what is going on. ...


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You could have each rod drive a generator and then combine the electrical output of the two generators.


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Very large trains are not only powered from the front. The front is preferred, though, since it is the easiest to couple and uncouple extra locomotives and a wired connection is more reliable than radio. The crew travels in the cab, so there is no walking or backing the consist up to pick up the person who made the coupling. Many territories will have all ...


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Each must be applying the exact same pull else if one is pulling harder than it is effectively taking on all the work with the other 2 idling. Plus, the load on the first coupling is 300 tons, on the second 290, and so on. The locomotives are controlled by power or throttle setting, not by speed setting. If power is set at 65% and there is a few percent ...


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Tentatively, this looks like it's analysing a sound wave, $\mathsf{d}x/\mathsf{d}t$ is the phase velocity of the wave in the lab frame, $\pm c$ is the phase velocity of the wave in the rest frame of the fluid, and $u$ sometimes represents the local, instantaneous flow velocity, and sometimes represents the average flow velocity over the fluctuations in the ...


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I think U is the flow velocity field. it can be pointing in any direction, but $\frac{dx}{dt}$ is only the x component. Like in vortex U is turning but $\frac{dx}{dt}$ is not.


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For engineering stress-strain curve, the only region where you would see basically two discrete strain values for one single value of stress is the necking, i.e. the region beyond Ultimate point on the curve. What happens there? The structure looses its stiffness and load bearing capacity, and just keeps on shrinking (in the necking area) without taking a ...


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The stress-strain curve isn't created using a incremental load. Instead what most standards for tensile testing nowadays require a constant displacement rate. I.e. you have a setup similar to the following (it may vary of course) Figure: Tensile test setup (source engineeringarchives) The crosshead (horizontal bar that the load cell is attacheds) moves at a ...


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An internal hinge is a special device used to link two beam segments. Similar to the typical pin support, it can generate reactions in the direction opposite to the load (applied force), with the condition that structural equilibrium must be maintained within the support, $\sum F_x = 0$ and $\sum F_y = 0$. Both the systems above are considered as "...


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One way to imagine the force on the beam AB is to substitute the BC with just its tributary load P/2, (the support it gets from the hinge). Now we have only a cantilever beam AB with a concentrated load of $\ \frac{P}{2} $ applied at the left hand. So the reaction moment is as the book says: $\ M=\frac{P}{2}*L=\frac{PL}{2}$


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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 ...


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Like you said at point B there will be a reaction force from the left. However, if you look that system in isolation, that reaction needs to be supported at the end. And that force at B creates a reaction moment on B. Since the reaction at B will be equal to P/2 (just take the FBD of the right part and calculate the reactions), then the bending moment at ...


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"I want to be able to hold the arm in place against the force of gravity... Both arms can rotate in any direction around their respective attachment points." Your system is unstable, which is caused by the fact that there is no fixity against rotation. Note that the "probable force" in the lower sketch can be induced by the deflection ...


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If it really is the case that $F_{\textrm{d}} = -c_1V$, with constant $c_1$, then there's an acceleration $\mathsf{d}V/\mathsf{d}t =-\left(c_1/m\right)V$. That differential equation is solved by an exponential decay with decay constant $c_1/m$. So it looks like, by the definition of $c_2$ you provide, $c_1 = c_2/2$.


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there is a big difference if you are considering the static or the dynamic case. static case In the static case the simplest way is to add the torque from all the forces. (Although this might sound weird) because you are working in a 3D problem the simplest way would be to use the cross product form of the moment (for all load in the system including self ...


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You are correct; "why not". Oil country pipe uses more than a hundred threads; most are proprietary. Each manufacturer will tell you why his thread is best. For highest strength in the axial direction, load flanks will be 90 degrees, or even negative. If interested in strange threads, look up Hydril- Tenaris Wedge threads


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Finer threads are stronger. When you strip a thread you shear the threads themselves off the fastener. Say you need a thread that will ball through a 10mm hole with 2mm deep threads you have a 6mm diameter core they are attached to. 1mm threads leave 8mm. That's almost a 3rd more material. Added to that, for a given load it will stretch less so the load will ...


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I believe you had a minor error on you left simplified result. you 2*w1 should be reduced by w1 resulting in just w1. I came up with the following diagram. The key to the right varying load, is that the peak values above and below line up and are equal but opposite so they will sum to 0. Since the bottom force becomes 0 at the midway point, you just need ...


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Because (to my understanding your problem is more in the interpretation of derivatives rather structural mechanics) I will try to explain it in parallel example. Hopefully it will make sense. Imagine a car accelerating with $a [m/s^2]$, and at some point in time t suddenly decelerating at $- a [m/s^2]$. The acceleration in time t will abruplty change from ...


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The top bucket will fill up first because the water needs less time to fall from the reservoir to the bucket.


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A similar question is asked about conservation of mass in water from a faucet here and is partially answered in the comments. The buckets should fill up at the same time, since the mass flowrate at any point in the falling stream is equal. As velocity increases further down the stream, the cross-sectional area of the stream will decrease so that $\rho ua$ ...


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This might explain. Turbulent boundary layer consists of three main layers formed in the direction normal to the wall: Viscous Sub-layer, Buffer Layer, Turbulent Region. Friction velocity is calculated using the wall shear stress and fluid density U* = friction velocity = sqrt (wall shear stress/density) , m/s Non-dimensional distance and velocity are ...


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The device will not work at all under the stated conditions. The system will result in unstable conditions under the given design situation. To ensure proper rotation, the guiding rollers need to have a double sided wheel flange (here A) and a rail guide to keep the whole device on track:


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The weight of the beam is frequently ignored in homework and at the lecture or the class, for simplicity. Unless it is given in the question. And for the second part of your question, I would say first do the test without self-weight and then with that, with a comment explaining the difference, this way it will be more clear. As you already know the weight ...


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Macaulay and UDL If by Macaulay's method you mean the use of singularity function (the use of Macaulay's bracket <>), then the method can account for distributed load. For example the moment for a Uniform distributed load (UDL) starting at point x=a and carrying until the end of the beam is : $$M(x) = \frac{w}{2}\langle x-a\rangle^2$$ Discrepancy in ...


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Constraint Method for Pulley Problems Assumptions: The string is taut and inextensible at each and every point of time. The string is massless and hence the tension is uniform throughout. Pulley is massless. Freebody Diagram - Equations: $4t_1 = m_1a_1$, $a_1 = \dfrac{4t_1}{m_1}$ -----(1) $2t_2 = m_2a_2$, $a_2 = \dfrac{2t_2}{m_2} = g$ -----(2) ...


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