Datasets:

bshada

/

engineering.stackexchange.com

like 1

2024-01-02T14:15:41.287

<p>In Autodesk Inventor Professional 2024, I am failing to add a flange to a sheet metal cone (300° for now, eventually the tabs will be used to mount and close the funnel), that is created in a sheet metal part file by</p> <ol> <li>Sketching a line and an axis for a Contour Roll (Sheet Metal tab)</li> <li>Rotating a rectangle (3D Model tab)</li> </ol> <p><a href="https://i.stack.imgur.com/T0avr.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/T0avr.png" alt="open sheet metal funnel modeling attempt" /></a></p> <p>For either there is no preview in the 3D view.<br /> For 1. the flange dialog never offers OK or Apply, regardless of the settings.<br /> For 2. if I adjust the settings provided by the Bend edit &quot;glyph&quot;, the Dialog can be accepted but results in an error that has no further explanation.</p> <p>It works straightaway when adding a flange to a plain Face.</p> <p>What do I have to do to add flanges to this part?</p> <hr> Edit 1: <p>How can flanges be created so the cone is closed? The features (Thicken, Flange) intersect.<br /> Rolling only 180 deg. and mirroring works in principle but also causes an intersection. Thickening inwards does not work.</p> <p><a href="https://i.stack.imgur.com/h2kMn.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/h2kMn.png" alt="enter image description here" /></a></p> <hr> Edit 2: Can't get Bend or Lofted Flange to work either. <p><a href="https://i.stack.imgur.com/tOLUJ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/tOLUJ.png" alt="create bend between contour roll and face" /></a></p>

|cad|autodesk-inventor|

<p>The reason for this seems to be the behavior of the top and bottom faces of the cone, they would make the ends of the flanges weird so that Inventor does not know what to do there. Creating the cone as a 3D model also seems to cause issues with the flange tool. You can solve it by forming the cone via sheet metal tools and adding a little straight extension, just long enough to accept the bending radius.</p> <ol> <li>Create the cone from a single tilted line with the proper measurements and by using the sheet tool &quot;Contour roll&quot;. This lets you create the cone as sheet metal and enables the next steps. <a href="https://i.stack.imgur.com/n5OVkm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/n5OVkm.png" alt="Setup for contour roll" /></a></li> <li>Create a work plane on the edge that you want to bend the flange around, perpendicular to the future flange.</li> <li>Add a little straight extension that is just wide enough to accept the desired bending radius with the standard sheet tool, this allows Inventor to create a well-defined flange and will vanish later in the bend.</li> <li>Add the flange on the end of the extension.</li> <li>Repeat on other side</li> </ol> <p>Final result including the working planes:</p> <p><a href="https://i.stack.imgur.com/fUogym.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/fUogym.png" alt="Flange_Result" /></a></p> <p>Concept of the flange extension:</p> <p><a href="https://i.stack.imgur.com/fwLW7m.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/fwLW7m.png" alt="Flange_Base" /></a></p> <p>I hope the result turns out the way you imagined, let me know if some steps are unclear, then I add more screenshots.</p>

58521

How to add a flange to sheet metal cone (funnel) in Autodesk Inventor?

2024-01-09T02:13:56.100

<p>I just can't seem to get it through my head and I don't understand the difference. As far as I can understand, strength is a materials resistance to permanent fracture while stiffness would be resistance to temporary fracture. If possible could I also get a brief explanation of ductility and resistance to fracture, and the differences between all of these. I'm familiar with chemistry but new to engineering.</p>

|structural-engineering|materials|material-science|

<ul> <li>Stiff and strong: you put a load of one pound on it and it moves 0.000001&quot;, you have to put a load of 1000 pounds on it to break it.</li> <li>Springy* and strong: you put a load of one pound on it and it moves 1&quot;, you have to put a load of 1000 pounds on it** to break it.</li> <li>Springy and weak: you put a load of one pound on it and it moves 1&quot;, you put a load of two pounds on it and it breaks.</li> <li>Stiff and weak: you put a load of one pound on it and it moves 0.000001&quot;, you put a load of two pounds on it and it breaks.</li> </ul> <hr /> <p>* i.e., the opposite of stiff</p> <p>** presumably in tension, in a very long room.</p>

58567

Can someone help explain the difference between strength and stiffness?

2024-01-09T17:14:58.373

<p>As far as I understand they both relate to the amount of deformation based on a given amount of stress.</p>

|mechanical-engineering|materials|material-science|

<p>Stiff and flexible are opposing qualitative metrics in the elastic regime. Under an applied stress, a stiff material will deform less than a flexible material. When the stress is released, both will return to their origin point (no deformation).</p> <p>Ductile and brittle are opposing behaviors in the failure regime <em>after the elastic behavior has occurred</em>. Under an applied stress, a ductile material will deform permanently while a brittle material will break. Both modes are failure. The transition for ductile materials is the yield stress. The transition for perfectly brittle materials is the failure stress. A second mode of breaking called fracture is not identical to brittle failure. Fracture occurs due to macroscopic flaws in the material.</p> <p>Here are general examples for the qualifications of elastic behavior (flexible to stiff) versus plastic (failure) behavior (ductile to brittle) across the various classes of materials by chemistry.</p> <ul> <li>Metals - flexible to stiff, ductile to brittle</li> <li>Ceramics - stiff, brittle</li> <li>Semiconductors - stiff, brittle</li> <li>Thermosets - stiff, brittle</li> <li>Thermoplastics - flexible to stiff, highly ductile to brittle</li> <li>Elastomers - highly flexible, brittle</li> </ul> <p>As you see, we cannot simply put materials in one basket or another using only the two terms stiff and ductile. The two terms are not opposites in the same way, they are qualitative metrics for behavior in two different regimes (elastic or reversible versus plastic or irreversible).</p> <p>Finally, elasticity (flexible or stiff) is quantified by the Young's modulus <span class="math-container">$E$</span>, measured as the stress to cause a specific strain (relative change in length), while ductility is quantified by the relative deformation in the material at breaking <em>after removing the elastic deformation length</em>. For change in length <span class="math-container">$L$</span>, with final and initial at transition, we write <span class="math-container">$(L_f - L_o)/L_o$</span>. A material will higher ductility will have a greater relative change in length at the point that it breaks. A material with zero ductility is perfectly brittle but may still have deformed elastically (been flexible). A rubber band is the best example of this case.</p>

58575

What's the difference between ductility and stiffness?

2024-01-10T00:04:07.717

<p>Strength is just the amount of force that an object can take before breaking, so what is fracture toughness?</p>

|mechanical-engineering|materials|material-science|

<p>Fracture toughness is a measure of the amount of applied work required to initiate a crack in a test sample and then &quot;walk&quot; it all the way across the sample cross-section, so as to tear the sample in two.</p> <p>In this type of test the load is applied suddenly by smacking the test sample with a heavy hammer rather than by gradually pulling the sample apart until it breaks, as in a tensile strength test.</p> <p>The standard impact toughness test is called the <em>Charpy impact test</em>.</p>

58580

What is fracture toughness and how is it different from the strength of a material?

2024-01-12T05:10:48.693

<p>I live in a 14 story apartment complex built in the 70s that has three underground parking levels.</p> <p>Today it looks like there's water leaking through cracks in the concrete on the second and third levels of the parking garage in multiple places throughout (not just one concrete block affected).</p> <p>Is this a I-should-have-left-as-soon-as-it-happened-why-are-you-posting-about-it-online-and-not-runnning situation? Or is this a problem 2-3 years down the line? I told security about it, and I think management knows as they shut the water off for about an hour today. How bad is this if it goes on for more than a day?</p> <p>I'm hoping it's fine if they fix it shortly, but if I shouldn't worry now, at what point should I worry? If this isn't fixed in a week? Or a month? Or if it does get fixed, should I worry if it happens again in another month?</p> <p>Thanks so much for your help!</p> <p><a href="https://i.stack.imgur.com/YIz3b.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/YIz3b.png" alt="Leak 1 center right" /></a></p> <p><a href="https://i.stack.imgur.com/5Sips.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/5Sips.png" alt="Leak 2 constant leaking at concrete edge" /></a></p> <p><a href="https://i.stack.imgur.com/oGjbZ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/oGjbZ.png" alt="Leak 3 over cars" /></a></p> <p><a href="https://i.stack.imgur.com/NC6Zd.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/NC6Zd.jpg" alt="Leak 4, notice the large puddling occurring" /></a></p>

|civil-engineering|concrete|waterproofing|

<p>A pipe or connection to an appliance has cracked and needs to be repaired. Even a defective toilet can leak a lot of water.</p> <p>Concrete structures can likely take the leak without too much damage. But the source of the leak must be located urgently and fixed. After that, the structure needs to be inspected for any damage, mold, or fissures!</p> <p>Building departments and insurance companies accept certificates from companies that are licensed and specialize in leak detection and water damage repair!</p>

58608

How bad is it if water is leaking through concrete in the lower parking levels of my apartment building?

2024-01-15T12:47:28.813

<p>The maximum reachable efficiency for a heat pump can be calculated using Lorenz/Carnot cycle:</p> <p><span class="math-container">$COP = \frac{T_H}{T_H - T_C}$</span></p> <p>Where as the <span class="math-container">$T_H$</span> is the target temperature and <span class="math-container">$T_C$</span> is the reservoir temperature (I might be wrong about the exact terminology).</p> <p>In case of heating houses with brine-water the temperature of brine can exceed 50°C where as the target temperature is max 25°C.</p> <ol> <li>How does the formula translate for this scenario, is it even considered heating?</li> <li>If its not considered heating, would it be logical to step up the brine temperature and store the energy in containers for house use?</li> <li>What is the typical method of incorporating heat pumps with high temperature brine-water sources?</li> </ol>

|heating-systems|heatsink|

<p>Heat pump efficiency is the heat output over the power input. For heat pumps and refrigeration systems, this efficiency is a positive number because you can move more heat than the energy you put in. So COP is used because calling a system 300% efficient just raises more questions.</p> <p>Since heat pumps are made to move heat from cold to hot (from outside at 5C to inside at 22C), the equations fall down if you are trying to use one to move heat from hot to cold (outside at 50 to inside at 22). People don't build complicated systems to move heat from hot to cold because heat flows this way naturally. All it takes is a heat exchanger. In this case, you would use your 50 degree ground temp water and blow air over a coil with that 50 degree water in it. My guess is that you would be less efficient than this by doing it with a system that evaporates and condenses fluids. You would only need to pay to pump the water and blow the air.</p>

58642

Carnot COP of Brine-water Heatpump

2024-01-16T19:51:39.663

<p>If we have a mixture of ideal gases can we consider the mixture also an ideal gas?</p> <p>This holds true in air for example where we consider it an ideal gas, when oxygen and nitrogen are considered ideal.</p> <p>But does this always hold true?</p>

|mechanical-engineering|thermodynamics|chemical-engineering|gas|

<h2>A mixture of gases acts like a mix of each individual gas, not like a new entity.</h2> <p>So yes, it will still act like an ideal gas since it's just the total of all the ideal gases in the mixture. For example, the total pressure of a mixture of gases is equal to the sum of each gas' partial pressure.</p> <p>Note that some liquids act differently than this when mixed and form an azeotrope, which has different characteristics than its components, which is why water and alcohol will not stratify once mixed.</p>

58670

Behaviour of a mixture of ideal gases

2024-01-20T15:12:33.133

<p>What size of drilling screw diameter do I need for a worn bore thread? (In my case, it is related to eyeglasses.)</p> <p>The original screw thread diameter is 1.4mm, but since the bore thread is worn, I need to use a drilling screw that, I assume, should be larger (but not equal) than the original one. If it should be larger indeed, how much larger should it be? +0.1 mm or less?</p>

|threads|

<p>I am fairly certain that neither self-tapping or self-drilling (not the same thing) screws exist for diameters that small. Furthermore, self-tapping threads that are that fine which are required for lengths that short do not seem workable.</p> <p>You are also probably not equipped to drill it and tap it even if you had the drill and tap. It's just so small and short your that the alignment is going to be off or something is going to cross thread or break.</p> <p>If you did drill out and re-tap the hole, you would use a 1.25mm drill (which is the tap drill size for a M1.6 screw) and then follow that with a M1.6 tap. And you would not use a power drill, because if you had an appropriate drill and setup you wouldn't have asked your post. You would use a pin vise and do it manually by hand but that makes alignment much more difficult.</p> <p>Might I suggest pinning it? You would not necessarily need to drill out the hole which would giving you more attempts at it if you mess up. If you can't get a press fit to work, then you cal also flare the pin on both ends of the pin with a punch to get it to stay in.</p> <p>Also, how did you get that 1.4mm measurement? For the precision you require and the diameters you are working with, caliper measurement should not be trusted.You would need a micrometer for that.</p> <p><strong>UPDATE:</strong></p> <p>I did look at on McMaster-Carr and there are M1.4 &quot;self-forming screws for soft metal&quot;. Forming a thread does seem like it would work better than cutting a thread for something this small.</p> <p>So if you trust that M1.4 measurement, then get M1.4, M1.6 and M2.0. I would try the M1.4 first to see if you can't just reform the thread. But it will probably not be removable safely after that. If that fails, then eyeball the M1.6 and M2.0 for which one you think will work. If that fails then you can go with the pinning approach previously mentioned. If you go to M1.6 or M2.0, you would not be drilling it out because you're probably going to mess that up, even if you use a pin vise and to it manually. You're going to be trying to force it in so it will take more effort than normal, but it might also not work and break or strip so you need to judge when it is not working and back off to go with a different method.</p>

58698

What is the correct drilling screw diameter for worn bore thread?

2024-01-26T22:46:19.473

<p>Suppose I have a room, 10x10x10m with insulated walls (R-value: 4) and a single double-pane window 1x1m (R-value: 0.35).</p> <p>Now the total thermal power required to keep the room at a desired temperature is given by:</p> <p><span class="math-container">$$ P = \left( \frac{6\cdot 10\cdot 10 -1\cdot 1 }{4} + \frac{1\cdot 1}{0.35} \right) \Delta T $$</span></p> <p>I am wondering what would happen if the window is left open? In that case, wouldn't the R value be zero and power go to infinity? Clearly this can't be right... I should be able to assume a finite R value for this case. But which and why?</p> <p>EDIT: I am somehow suspicious in this case the hole is completely ignored in this formula but taken into account as convection term?</p>

|thermodynamics|thermal-conduction|thermal-insulation|

<p>If this hole is the only opening in the room, there will be a buoyancy-driven exchange flow through it (if the interior is hotter than the exterior, in through the bottom half of the hole and out through the top half of the hole, and vice versa). This exchange flow <a href="http://maeresearch.ucsd.edu/linden/pdf_files/75l99.pdf" rel="nofollow noreferrer">is known as &quot;mixing mode ventilation&quot;</a>. Since the outgoing air is at a different temperature from the incoming air, this means a net energy flow. I'd expect the pressure differences driving the flow to be proportional to <span class="math-container">$\mathrm{\Delta}T$</span>, and therefore the flow rate (through the usual formula for an orifice flow) to be proportional to <span class="math-container">$\sqrt{\mathrm{\Delta}}T$</span>. The rate of energy transfer <span class="math-container">$P$</span> will be proportional to the product of flow rate and <span class="math-container">$\mathrm{\Delta}T$</span>, i.e. <span class="math-container">$P$</span> will be proportional to <span class="math-container">$\mathrm{\Delta}T^{3/2}$</span> and your <span class="math-container">$R$</span> value proportional to <span class="math-container">$1/\sqrt{\mathrm{\Delta}}T$</span>, with a constant of proportionality that depends on the top-to-bottom height of the opening. I'd guess a good place to look for a detailed formula for that constant of proportionality would be volume A of the CIBSE guide (or ASHRAE's equivalent if you're of the American persuasion), but I don't have access to a copy right now to check.</p> <p>If there's another opening somewhere at a different height, there's more likely to be a one-way flow through each hole <a href="http://maeresearch.ucsd.edu/linden/pdf_files/75l99.pdf" rel="nofollow noreferrer">(&quot;displacement mode ventilation&quot;)</a>, and the overall exchange flow (and therefore the <span class="math-container">$R$</span>-value) will be a combined property of the pair of openings, so there's not a well-defined <span class="math-container">$R$</span> value for one of the openings in isolation. Nevertheless, I'd still expect the proportionality of the combined <span class="math-container">$R$</span> value to <span class="math-container">$1/\sqrt{\mathrm{\Delta}}T$</span> to hold up, just with a different constant of proportionality that depends on the height difference between the two openings and the ratio between their areas. Again, volume A of the CIBSE guide is probably the place to look for a detailed formula.</p>

58756

R value of a hole?

2024-01-27T06:53:18.613

<p>I am working with FreeCAD 0.20 and Sheet Metal work bench 0.2.63, but universal CAD tips are also welcomed. As shown in the image, I would like to bend sheet metal with smoothly chancing width along the bend. So, I cannot use for example negative extend values causing stair-like features, unless the sharp transitions could be smoothed somehow afterwards. Any ideas?</p> <p>Edit: I also tried the Sketch On Sheet metal tool, but it removes material appropriately only if the sketched cuts are fully within the sheet. If the cuts go over the edges the cuts become erratic.<a href="https://i.stack.imgur.com/ji7MP.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ji7MP.png" alt="Visualization about intended sheet-metal bend." /></a></p>

|design|cad|bending|metal-folding|metalwork|

<p>A tip I got elsewhere is to first make the flat sheet with appropriate width transitions, then bend it afterwards. I'll give this approach a try. Other tips are still welcomed, just in case.</p>

58758

How to CAD: smooth width transition in sheet-metal bends

2024-01-31T09:53:06.993

<p>I am trying to create a device that can measure the rotation of a ring on a finger. Basically there's going to be two rings on the finger, one that rotates and one that is stationary. The idea is that the stationary (or rotating ring, if that's an option) will be able to measure how much it's rotated relative to the stationary ring.</p> <p>I obviously can't have anything in the center of rotation so there has to be some other option. I had the idea of making my own potentiometer which might be possible but I was wondering if there are any other options.</p> <p>My idea for the project is to rotate the ring with a string attached to another finger, that way getting an idea for how far away the finger is from the ring.</p>

|mechanical-engineering|electrical-engineering|distance-measurement|

<p>Have 3 magnets on one, irregularly spaced which will show rotation.</p> <p>Then Hall effect sensors on the other, as the signals are recorded the relative spacing shows rotation and even direction.</p>

58796

How to measure the rotation direction and speed of a ring on a finger?

2024-02-01T13:41:58.913

<p>I am trying to figure out how to compute torques in a network of rotating objects. I assume that each object has a given moment of inertia, and some objects are accelerated by external means. The objects are connected mechanically by zero-mass rigid connectors. Due to the rigid connections, all objects are forced to rotate at the same speed at any time.</p> <p>For example, I could have three objects: A, B and C. All have a moment of inertia of 1kgm^2. A is connected to B, B is connected to C, and C is connected to A.</p> <p>If we now accelerate A with 15W of power, then due to the equal moments of inertia, each object will gain kinetic energy at a rate of 5W. This will result in torques experienced by the connectors between the objects. I assume that I could compute the torques based on the power transmitted through a connection. In this simple example, due to the symmetry, there should be 5W transmitted from A to B, and 5W from A to C, while B to C transmits no power.</p> <p>While in this simple example this is easy to figure out, I am interested in how to compute these transmitted powers in the general case, with many more objects.</p> <p>If there was no cycle, then it would be easy. Any object that is connected to only one other object would receive the power P_1 it accelerates at. The next object would receive the power P_2 itself accelerates at, plus P_1, so P_1 + P_2. And so on. But with a cycle, there is no starting point from which I can compute everything else like this.</p> <p>How can this problem be solved, given that there are cycles in the network?</p>

|mechanical-engineering|statics|kinematics|

<p>This isn't really an answer, sorry in advance.</p> <p>Not sure the exact language to use, but it seems to me that a cyclic network of rigid connections would be is overconstrained in some sense, and so I don't think we could know how much energy travels in each possible path.</p> <p>What would be necessary to bring the network of flow in to a balance?</p> <ul> <li>Would it be sufficient if we allow some elasticity (&amp;damping?) ?</li> <li>Or would it be necessary to require that the links be equally stiff?</li> </ul> <p>If we can establish a balance principle, then we can use &quot;0 flow across the balance point&quot; to resolve the cyclic network.</p> <p>Example below: 25W going into 5 objects, each of which absorbs 5W into its inertia. By symmetry, there would be no flow of energy at the opposite side of the circle in the diagram.</p> <p><a href="https://i.stack.imgur.com/7RCcu.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/7RCcu.jpg" alt="a" /></a></p>

58813

How to compute the torques of a network of rotating objects with cycles?

2024-02-07T12:08:01.513

<p>Most optical tables have holes. Are there also optical tables without holes? Or do these things have a different name?</p>

|mechanical-engineering|materials|control-engineering|optics|holes|

<p>This is called a <a href="https://en.m.wikipedia.org/wiki/Surface_plate" rel="nofollow noreferrer">surface plate</a>.</p>

58847

Optical table without holes

2024-02-10T00:28:59.433

<p>I am trying to build a tool that could help someone/something weld.</p> <ul> <li>Being very close to a MIG or TIG welding gun may be dangerous/undesirable.</li> <li>Welding needs to be done in a straight line (which can be difficult for some people/robots).</li> <li>Welding needs to be done pretty slowly (which can be difficult for some people/robots).</li> </ul> <p>As a Bear of Very Little Brain I thought to myself:</p> <ul> <li><p>With <a href="https://www.youtube.com/watch?v=zH3wK5gEWPw" rel="nofollow noreferrer">Scott Russell's linkage</a> I can create some distance between the user and the object being welded and this will also ensure a perfect straight line.</p> </li> <li><p>With <a href="https://www.youtube.com/shorts/w6y4I4OHiok" rel="nofollow noreferrer">a pantograph</a> I can turn a large movement into a much smaller movement.</p> </li> </ul> <p>So I basically have these 2 mechanisms:</p> <p><img src="https://i.imgur.com/qJBSW3o.png" alt="Text" /></p> <ul> <li><p>I think the user/robot should use (C).</p> </li> <li><p>(J) can hold the welding gun</p> </li> <li><p>(B) can be connected to (H).</p> </li> </ul> <p>I think that would achieve these goals, but it would be better if I could combine them without sticking one on top of the other and I'm sure this design can be improved. Do you have any suggestions how to simplify and/or improve this design?</p>

|mechanical-engineering|

<p>Section ABD is same as section GHI.</p> <p>Extend the BD link to match HIJ.</p> <p>Constrain C, instead of B.</p>

58859

Welding bot arm

2024-02-11T09:54:38.437

<p>My Inventor model somehow ended up with its coordinate system very misaligned with its faces so that when I try to create a drawing, the orthographic projections I get aren't the best (the top right one is the 'isometric' view): <a href="https://i.stack.imgur.com/WWLYD.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/WWLYD.png" alt="enter image description here" /></a></p> <p>I created a new custom UCS in the model so that the front, top, and side views I get from it look good. But how do I place the model view in the drawing file using the UCS as the default coordinate system?</p>

|cad|technical-drawing|autodesk-inventor|

<p>I solved my problem : I reset the view cube's &quot;top&quot; and &quot;front&quot; orientations by rotating it to the correct ones, right clicking the cube, and selecting &quot;Set Current view as&quot;.</p>

58870

How to place object in Inventor drawing using custom UCS

2024-02-12T19:08:48.573

<p>I'm doing a static analysis of a square thread lead screw. I think this site gives a good description of the problem.</p> <p><a href="https://engineeringstatics.org/Chapter_09-screw-friction.html" rel="nofollow noreferrer">Screw threads statics</a></p> <p>To calculate the moment on the thread the mean radius of the thread is used.</p> <p><span class="math-container">$$M = W \space R \space tan(phi_s + alpha)$$</span></p> <p>Where M is the moment required to raise the screw to impending motion, W is the force load on the screw, R is the mean radius of the screw, phi_s is the screw friction angle, and alpha is the screw thread pitch.</p> <p>Further down on the same site a disc friction problem is looked at.</p> <p><a href="https://engineeringstatics.org/Chapter_09-disc-friction.html" rel="nofollow noreferrer">Disc Friction Statics</a></p> <p>Here the moment is calculated by integrating with respect to the radius.</p> <p>Adapting this approach to analyze the screw thread, the moment required to raise the screw would be.</p> <p><span class="math-container">$$M = \frac{2}{3} \frac{(R_o^3 - R_i^3)}{(R_o^2 - R_i^2)} W \space tan(phi_s + alpha)$$</span></p> <p>Where R_o is the outside radius and R_i is the inside radius.</p> <p>These yield similar but different results. My question is, is the first approach using the mean radius simply a simplification/approximation of the disc friction method of integrating over radius and the later will be more accurate? Or is it incorrect to integrate over the radius when looking at a screw thread analysis? If so, can you please elaborate on why.</p>

|mechanical-engineering|statics|machine-design|

<p><em>My question is, is the first approach using the mean radius simply a simplification/approximation of the disc friction method of integrating over radius and the later will be more accurate?</em> Yes</p> <p><em>Or is it incorrect to integrate over the radius when looking at a screw thread analysis?</em> This would be the most accurate</p> <p><em>If so, can you please elaborate on why.</em> The area of friction is a function of the radius squared, so a simple average won't be quite correct.</p> <p>Recognize that neither is likely to product exactly real-world results, other factors come into play, namely what exactly is the coefficient of friction.</p>

58876

Static analysis of a lead screw and disc friction

2024-02-12T20:32:30.990

<p>Do I need to perform all the lower IPx(1-7) rating tests to achieve an IPx8 rating or covers the IPx8 rating test all the other ones.</p>

|mechanical-engineering|waterproofing|

<p>The following comes from the <a href="https://assets.new.siemens.com/siemens/assets/api/uuid:7f2844f1-502a-4c23-b123-117894c39a8c/ip-vs-nema-whitepaper-icp-us.pdf" rel="nofollow noreferrer">Siemens summary of the IEC 60529 IP rating</a>:</p> <blockquote> <p>Note on the second characteristic numeral (protection against ingress of water): Up to and including characteristic numeral 6, the requirements of all lower characteristic numerals are also met. The same is not true for characteristic numerals 7, 8 and 9. If an enclosure meets several requirements, this must also be indicated as appropriate.</p> </blockquote> <p>So it looks like the answer is not quite. In particular, IPx6 (strong water jet) and IPx7/8 (water immersion) might be independent of each other.</p>

58878

IPx8 rating test

2024-02-13T09:56:32.100

<p>In a fully connected truss (each node connected to all other nodes), the number of members (bars) is equal to <span class="math-container">$n(n−1)/2$</span> where <span class="math-container">$n$</span> is the number of nodes.</p> <p>I don't understand the derivation of this formula. Can you please explain it to me?</p> <p>reference of the <a href="https://www.sciencedirect.com/science/article/pii/S0965997821000612#bib0001" rel="nofollow noreferrer">formula</a></p>

|structural-engineering|structures|

<p>There are n nodes. Each connects to all the others, so n-1 connections to each node. But if node A is connected to node B then member AB is the same as BA so you've counted each member twice. Hence /2.</p> <p>So, (n nodes) x (n-1 connections to each node)/(2 because you have doubled up)</p>

58890

How is the number of memebers in a truss calculated?

2024-02-14T01:39:07.010

<p>What is acceptable connectivity for an Elevator Phone system. As analog systems are going by the way side, Cellular is one option, however is it acceptable to use point to point WIFI bridging to accomplish this task and where in the code regulations is the information regarding this ?</p>

|wifi|regulations|

<p>The standard seems to specify 2-way communications, with requirements for testing, alternate power, etc., but does not specify how that should work.</p> <p>There are a lot of requirements there, but they seem to leave to you how to make it work. The applicable section is 2.27.1 Car Signaling Devices.</p>

58896

In Elevator Phone Connectivity Reuirements ASME 2019 is WIFI bridging Acceptable?

2024-02-14T05:58:54.570

<p>What do the following annotations mean in this AutoCAD drawing mean?</p> <pre><code>R1.25 TYP (circle/hole Left) 2XØ1.0 (circle/hole Right) Ø1.25 (circle/hole Bottom) </code></pre> <p>Based on those annotations, what is the diameter (or radius) of each of the three circles/holes in this AutoCAD drawing?</p> <p>Also, if I'm reading correctly, the FILLET command can be used for &quot;rounds&quot; and &quot;fillets&quot;, correct? In reference to the dimension annotation R.5 (meaning radius 0.5).</p> <p><a href="https://i.stack.imgur.com/hrQPb.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/hrQPb.jpg" alt="enter image description here" /></a></p>

|mechanical-engineering|autocad|

<p>The outside circle like you said has a radius of 1.25. The two holes at the top have a radius of 0.5 as you said and the bottom hole has a radius of 0.625. What the drawing doesn't say I think is the &quot;outside circle&quot; of the bottom and right hole.</p>

58898

AutoCAD circle dimension annotation meanings

2024-02-15T16:08:45.620

<p>I have an axisymmetrical part that is subject to an internal (non-constant along the inner border, but constant in time) pressure (among other loads, but this is the one I am interested about) and I have to analytically evaluate the stresses across its cross section.</p> <p><a href="https://i.stack.imgur.com/t74Z4.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/t74Z4.png" alt="enter image description here" /></a></p> <p>Due to the geometry of the part and the distribution of the load, I assumed symmetry about the central axis. I also chose to simplify the cross section like this:</p> <p><a href="https://i.stack.imgur.com/gSgr6.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/gSgr6.png" alt="enter image description here" /></a></p> <p>The pressure can be decomposed into radial and axial pressure. I know how to deal with the hoop stresses generated by the radial pressure, I can use the thick wall model and get a good estimate from that. The axial pressure will compress the ring, this too looks like a simple calculation.</p> <p>The problem comes when I want to evaluate the effects of the torque induced by the pressure about the centroid of the section.</p> <p>I did some research and found this case covered in the book &quot;<em>Strength of Materials - Part 2</em>&quot; by Timoshenko, pages <strong>138-143</strong>, but there are still some things I don't understand, because this topic is covered very briefly, and only for rectangle cross sections, so I still have some questions that I can't figure out. I've been stuck on this problem for some days, and I can't seem to find any other source that covers this particular topic in depth.</p> <p>I also checked Roark for tables of various cross sections, but didn't find anything about this particular case.</p> <p>I am also uncertain about how to exactly calculate the resultant forces and torques due to pressure, because the book calls this problem &quot;<em>Twisting of a circular ring by couples uniformly distributed along its center line</em>&quot;, but I have the doubt that since the cross section rotates around the neutral fiber (similarly to how curved beams bend around the neutral axis) shouldn't the forces and moments be calculated about the neutral fiber ,and not about the centroid?</p> <p>I also have no reason to think the centroid and the neutral fiber coincide because I am working in radial coordinates (which could generate radial offset, like in curved beams), and there's no symmetry about the horizontal plane in this case (which could generate vertical offset), like in the following picture, where G is the centroid, N is the neutral fiber:</p> <p><a href="https://i.stack.imgur.com/sx5Z6.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/sx5Z6.png" alt="enter image description here" /></a></p> <p>My questions are:</p> <ul> <li><p><strong>Should I decompose the pressure about the centroid or the neutral fiber?</strong></p> </li> <li><p><strong>Are there manuals/handbooks/tables about this particular load that contain properties about various cross sections?</strong> I am in the process of finding these properties manually for a trapezoidal cross section, but it's quite a slow process and it's easy to miscalculate.</p> </li> </ul> <p>I could as well have misunderstood something, these topics about axial symmetry are something I am trying to study by myself, we didn't cover this at uni for now, but I need it for a student project. I am still trying to wrap my head around this topic and may have missed something obvious.</p> <h2>Additional information:</h2> <p>The ring is bolted to an external thin walled cylinder, which is also subject to internal pressure in the radial direction (this is a combustion chamber):</p> <p><a href="https://i.stack.imgur.com/vA8YG.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vA8YG.png" alt="enter image description here" /></a></p> <p>The cylinder provides the vertical equilibrium, and it is long enough so that what happens on the other side (top) of the cylinder won't affect what happens to the ring, apart for vertical forces that make the vertical equilibrium of the ring possible.</p> <p>I chose to ignore the screw holes (8) at first to simplify the problem, I'll include their effects later after I'll manage to get a hold of the simpler case first. I also want to assume that the screws provide a distributed, rather than concentrated, reaction force. Again, I'll try to see if I can remove this simplification later in the analysis.</p> <p>I am also thinking that since the ring is visibly stiffer (thicker) than the thin cylinder, I can assume that the radial component of the pressure is unloaded on the ring only, at least in this zone, so what I mean to say is that the cylinder doesn't help the ring much in resisting tangential elongation. The same can be said for the rotation that I am investigating I think, I can assume that the only thing resisting the rotation of the ring are only its fibers and not those of the thin cylinder. This also puts me in a safety condition, as the cylinder will surely help a bit somewhere.</p> <h2>EDIT</h2> <p>Alright thanks to everybody who commented, I think I am getting a hang of this now. I had to go back to basics and follow carefully the derivation of the stress formula in the case of bending of a curved beam, to better understand the logic behind choosing the centroid as the point of reference. For this case the logic is similar, I am not forced to choose the centroid, but it's useful to choose it as reference point when deriving formulas and decomposing loads because there will be other loads that will cause other deformations in this part, so it's better to use the centroid as it's a point that is easy to utilize for other loading conditions too. If say I decomposed the loads with respect to the neutral fiber rather than the centroid, I would easily calculate the rotation of this ring, but the choice of the point of reference would make it harder to calculate stresses for the other components of the given load.</p>

|pressure|statics|solid-mechanics|strength|

<h1>Solution using ring</h1> <p>Using figure from the question:</p> <p><a href="https://i.stack.imgur.com/gSgr6.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/gSgr6.png" alt="figure from the question" /></a></p> <h2>Ring rotation stiffness</h2> <p>Section <span class="math-container">$\Gamma$</span> rotates by angle <span class="math-container">$\varphi$</span> around point <span class="math-container">$S$</span> located at height <span class="math-container">$s$</span> from the bottom. Axial coordinate <span class="math-container">$z$</span> starts at the bottom and inner boundary of the section can be expressed as: <span class="math-container">$$r_i(z) = R+(B-b)/h\cdot z_i$$</span> with <span class="math-container">$k=(B-b)/h$</span>: <span class="math-container">$$r_i(z) = R+k\cdot z_i$$</span></p> <h3>Hoop strain and stress</h3> <p>Suppose, that the section will just rotate around a point as reaction to moment. The points above the line will get closer to axis and points below the line will get further from axis: <span class="math-container">$$\epsilon_t = \frac{2\pi \cdot(r - (z-s)\cdot \varphi- 2\pi\cdot r)}{2\pi r} = -(z-s)\cdot\frac{\varphi}{r}$$</span></p> <p>The hoop stress is then: <span class="math-container">$$\sigma_t = E\cdot \epsilon_t = -(z-s)\cdot E\cdot\frac{\varphi}{r}$$</span></p> <h3>Relation between moment and stress</h3> <p>To understand relation between moment and stresses, you need to look at the situation from above. Imagine small angle <span class="math-container">$d\vartheta$</span> part of the ring. The hoop stress has a radial component with respect to central section: <span class="math-container">$$dF_r = -2\sigma_t\cdot dA\cdot \sin(d\vartheta/2)$$</span> which for small angle <span class="math-container">$d\vartheta$</span> is: <span class="math-container">$$dF_r = -\sigma_t\cdot dA\cdot d\vartheta$$</span></p> <h4>Position of the &quot;neutral axis&quot;</h4> <p>The moment should of course not result in net radial force on the section <span class="math-container">$\Gamma$</span>, so it must be true, that: <span class="math-container">$$0 = \int\limits_{\Gamma} dF_r = \int\limits_{0}^h\left(\int\limits_{R+k\cdot z}^{R+B} \left((z-s)\cdot E\cdot\frac{\varphi}{r}\right)dr\right)dz = E\cdot\varphi\cdot\int\limits_{0}^h(z-s)\left(\int\limits_{R+k\cdot z}^{R+B} \frac{1}{r}dr\right)dz$$</span></p> <h5>Mean radius simplification</h5> <p>In case of rectangular section, integral of <span class="math-container">$1/r$</span> is very simple and does not affect the second integral, positioning neutral axis at the center of mass. In your case however, it would be quite complicated. Luckily, the term <span class="math-container">$1/r$</span> can be simplified as a constant using a mean radius value when (radial) dimensions of the ring are much smaller then its distance from the axis. I will use weighted mean radius here: <span class="math-container">$$r_m \cdot A = \int\limits_{0}^h\left(\int\limits_{R+k\cdot z}^{R+B} r dr\right)dz = \frac{1}{2}\int\limits_{0}^h\left((R+B)^2-(R+k\cdot z)^2\right)dz$$</span> <span class="math-container">$$r_m\cdot \frac{B+b}{2}\cdot h = \frac{1}{2}\left(\left(2RB+B^2\right)h-Rkh^2-\frac{k^2 h^3}{3}\right)$$</span> finally: <span class="math-container">$$r_m = \frac{\left(2RB+B^2\right)-Rkh-\frac{k^2 h^2}{3}}{B+b}$$</span></p> <h5>Back to the neutral axis</h5> <p>With the mean radius simplification, condition for neutral axis position has following form: <span class="math-container">$$0\cdot r_m = \int\limits_{0}^h(z-s)\left(\int\limits_{R+k\cdot z}^{R+B} dr\right)dz$$</span> From this (<span class="math-container">$s$</span> is basically at the centroid): <span class="math-container">$$s = \frac{3B-2kh}{6B-3kh}h$$</span></p> <h4>Moment due internal stress from ring rotation</h4> <p>Whole moment acting on the <span class="math-container">$\vartheta$</span> part of the ring is sum of forces <span class="math-container">$dF_r$</span> acting on lever arms <span class="math-container">$z-s$</span>: <span class="math-container">$$dM = \int_\gamma dF_r\cdot (z-s) = \int\limits_{0}^h\left(\int\limits_{R+k\cdot z}^{R+B} \left((z-s)^2\cdot E\cdot\frac{\varphi}{r}\right)dr\right)dz$$</span></p> <p>Using the mean diameter <span class="math-container">$r_m$</span>, this can be simplified to: <span class="math-container">$$dM = \frac{E}{r_m}\cdot \varphi\cdot \int\limits_{0}^h\left((z-s)^2\int\limits_{R+k\cdot z}^{R+B} dr\right)dz$$</span> Finally: <span class="math-container">$$dM = \frac{Eh}{r_m}\cdot \left(Bs^2-\frac{2Bs+ks^2}{2}h+\frac{B+2ks}{3}h^2-\frac{k}{4}h^3\right)\cdot \varphi$$</span></p> <p>Full moment acting on the whole ring is sum of all the moments on <span class="math-container">$d\vartheta$</span> parts: <span class="math-container">$$M = 2\pi\cdot dM$$</span></p> <h2>Moment caused by the external loads</h2> <p>First, all of the loads acting on the ring are required. The pressure indicated in the figure will be in axial equilibrium with a distributed axial force from the connected cylindrical shell: <span class="math-container">$$F_s = p\cdot \pi\cdot \left((R+B-b)^2-R^2\right)$$</span></p> <p>For a full moment calculation, focusing just on a small angle <span class="math-container">$d\vartheta$</span> part of the ring might be clearer than using whole ring. Contribution from the force of attached shell (with thickness <span class="math-container">$t$</span>) is straightforward: <span class="math-container">$$dM_F = F_s\cdot \frac{d\vartheta}{2\pi} \cdot (R+B+t/2-r_m)$$</span></p> <p>Contribution of the pressure is trickier, because it acts on a trapezoidal section and local lever arm should be expressed with respect to center of rotation <span class="math-container">$[r_m, s]$</span>.</p> <h1>Solution using equivalent shell</h1> <p>From the first figure you have shown, it seems, you might be able to use a cylindrical shell with equivalent thickness (somewhere between <span class="math-container">$B$</span> and <span class="math-container">$b$</span>) or even multiple such shells to capture bevelled geometry more accurately. This I think is at the edge of feasibility of available analytical solutions.</p>

58913

Twisting of thick ring due to internal pressure

2024-02-17T11:16:57.790

<p>I am currently designing an axial air compressor for a project, and I require the stator blades to reduce the velocity from 50m/s to 2.5 m/s in each stage of the stator. This requires a ratio of the areas to be 20. Hence, the circumferential gap between the edges of consecutive blades need to have a ratio of 20 throughout to be able to achieve the same. The blades start at a radius of 30mm and end at 50mm. This makes the ratio of 20 seem difficult to achieve.</p> <p>Hence, I was wondering if it would be possible to divide the ratio of 20 into two stages of stator blades, one of ratio 4, and one of 5, hence achieving a more practical design. However, I have not seen such a thing done anywhere.</p> <p>I believe it would make sense because the stators essentially convert kinetic energy into pressure energy, and hence the different stages shouldn't matter much.</p> <p>However, I am struggling to understand how the air would behave at the interface of the two stators. There would be a sudden change in area at that point, which is something I don't understand how the behavior would be in.</p>

|mechanical-engineering|fluid-mechanics|thermodynamics|compressors|

<p>You are completely missing the point of how an axial compressor operates. Let's say you design it so the axial velocity everywhere in the machine is a constant 2.5 m/s. The area of the flow will reduce in proportion to the density increase. All of the velocity change will occur in the transverse plane as changes to swirl velocity. So the rotor adds about 50 m/s of swirl to the flow and the stator converts about 50 m/s of swirl kinetic energy into PV work. The cross-section changes are whatever you want them to be for cost and convenience.</p>

58929

Is it possible to use two sets of stator blades after each set of rotor blades in an axial compressor?

2024-02-21T16:57:28.000

<p>This is a question that has been haunting me my entire career. The pressure loss caused by friction is not the same as the pressure &quot;loss&quot; caused by vacuum, right?</p> <p>I mean, if I have a gas flowing though a pipe, the more restricted the way is, the bigger is the pressure drop caused by the pipe and therefore it should be more difficult for me to push the gas through said pipe.</p> <p>Maybe the best way to put it is: what is the difference between a 2000 Pa pressure drop caused by friction compared to using a pump to pull a gas with a suction pressure of -2000 Pa? Why does one of them block my gas flow and the other helps it flowing?</p> <p>Thanks in advance, this question drives me crazy</p>

|fluid-mechanics|

<p>Consider the fluid as a control volume -- a defined size of the fluid. Pipe friction works at the surface of the control volume to prevent the entire fluid as a unit from moving through the pipe in its desired direction. Pumping (suction as you phrase it) works at the surface of the control volume to move the fluid in a specific direction.</p> <p>Alternatively, recall from physics that a friction vector goes in the opposing direction to the motion of the object. You push (or pull) a block along a surface, and friction pushes <em>in the opposing direction</em>. Both forces do work on the object. But one does work in the direction of the force (pumps), while the other does work in the opposing direction (friction).</p>

58963

Difference between pressure drop and vacuum

2024-02-21T18:55:44.203

<p>So I understand how H infinity control can be used to synthesize a controller that is in the standard error-feedback framework shown here:</p> <p><a href="https://www.researchgate.net/figure/1-DOF-feedback-control-system_fig1_313128407" rel="nofollow noreferrer">https://www.researchgate.net/figure/1-DOF-feedback-control-system_fig1_313128407</a></p> <p>but I was wondering can it be used for different architectures like a &quot;2 dof&quot; controller where the reference and the output are passed in separately to the controller:</p> <p><a href="https://www.researchgate.net/figure/General-2DOF-PID-controller-structure_fig3_327690406" rel="nofollow noreferrer">https://www.researchgate.net/figure/General-2DOF-PID-controller-structure_fig3_327690406</a></p> <p>or can it be used to design a feedforward controller like shown here:</p> <p><a href="https://apmonitor.com/pdc/index.php/Main/FeedforwardControl" rel="nofollow noreferrer">https://apmonitor.com/pdc/index.php/Main/FeedforwardControl</a></p> <p>As I understand it, the only requirement for the synthesis of an H infinity controller is that the weighted system must be representable by a linear fractional transform with the controller being connected to a generalized plant like so:</p> <p><a href="https://www.researchgate.net/figure/Closed-loop-system-TzwK-LF-T-P-K-formed-via-lower-linear-fractional-transformation_fig1_251939874" rel="nofollow noreferrer">https://www.researchgate.net/figure/Closed-loop-system-TzwK-LF-T-P-K-formed-via-lower-linear-fractional-transformation_fig1_251939874</a></p> <p>It seems like for all of the other controller frameworks we can specify the system as this linear fractional transformation, so is it safe to assume that we can apply standard H infinity control synthesis to these other frameworks (like 2 dof or feedforward)? I thought it would be but I cannot find any examples doing this and I was wondering if there were any aspects to these systems that made the synthesis infeasible. Sorry for the links, I don't know how to directly link picture diagrams, and thanks!</p>

|control-engineering|control-theory|

<p>You can use H-infinity design to optimize any control system that can take the form</p> <pre><code> ┌─────────┐ z◄───┤ │◄────w │ P │ y┌───┤ │◄───┐u │ └─────────┘ │ │ │ │ ┌───┐ │ │ │ │ │ └─────►│ K ├───────┘ │ │ └───┘ </code></pre> <p>If you include your references in the signal <code>w</code>, and the error between the reference and the output in <code>z</code>, you are designing a tracking controller.</p> <p>All the H-infinity optimization does is to minimize the H-inf norm of the transfer function between <code>w</code> and <code>z</code> with respect to <code>K</code>. The generalized plant <code>P</code> can also be designed such that the reference is fed straight into <code>K</code> through <code>y</code>, thus allowing <code>K</code> to contain a reference filter. If you have other external signals that are known, you can do the same for those in order to have <code>K</code> include feedforward also from those.</p>

58966

Does H infinity control have to be limited to a simple 1Dof feedback architecture?

2024-02-22T09:18:50.123

<p>I'm looking for a type of screw like this:</p> <p><a href="https://i.stack.imgur.com/Z1FWc.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Z1FWc.png" alt="enter image description here" /></a></p> <p>but with even smaller diameter (pin size) smooth end. Preferably M6 thread and less than 2mm diameter end, preferably smooth. If specific part recommendations are not allowed on this forum, just a general clue of where to find something like this would be appreciated.</p>

|fasteners|

<p>It's called a dog point grub screw and you may be able to find your desired object at <a href="https://www.mcmaster.com/products/grub-screws/" rel="nofollow noreferrer">McMaster-Carr</a>. The site has a selector panel and one of the entries in the panel includes the ability to specify the tip diameter. Notice also that tip length is in the selector panel. Image from linked site:</p> <p><a href="https://i.stack.imgur.com/ZBgy9.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ZBgy9.png" alt="McMaster-Carr selector panel" /></a></p>

58970

Looking for a type of the screw with very narrow end

2024-02-22T20:27:35.950

<p>I need to design an aluminum enclosure for some electronics, but I have no idea how to screw the different pieces together. I have settled for a folded box design with a top, but I can't figure out how the lid is supposed to be screwed with the rest.</p> <p>By checking on metal enclosures I have access to, it seems that it is frequent to have a hole, kind of &quot;punched&quot; towards the inside, that is made to be exactly the right diameter to be tapped, allowing a screw to be screwed inside. However, I can't find any info about how that is actually done, and in details : which tools are required, what sizes the holes need to be for the formed cylinder to be the right diameter for tapping...</p> <p>I have been checking on self-tapping metal screws too, but could not find very good references specifying the required diameter of the hole for the screw to tap correctly, and the correct sheet metal thickness required to ensure the screw won't snap out of the sheet due to having not enough fillets fitted inside the hole. Or are rivet nuts something that could be considered using for this application ?</p> <p>So which technologies are available for this purpose, available without expensive tooling (hydraulic presses...), especially for 2mm aluminum sheets ?</p> <p>P.S. : I have never worked with sheet metal and could not really find much info on the internet nor have any people around me with some experience about sheet metal enclosures.</p>

|metal-folding|

<p>If the enclosure is going to be opened and closed periodically and you wish to have longer life integrity for this action, consider an alternative method to sheet metal screws.</p> <p>Self-tapping screws can be found in self-drilling form but that aspect makes the longevity of the fastener even more reduced if repeatedly cycled.</p> <p>Your consideration of using rivnuts is a far better choice. The threaded portion remains within the flange at the top opening (or front, as the case may be) while the cover need only to have matching holes of the correct diameter for the machine screws or bolts to pass.</p> <p>Rivnut tools are far less expensive than when I first learned of them, as are the individual rivnuts. Repeated open and close cycles will not be a problem. Ensure that the rivnut is well seated and do not over-torque the fastener (causing the rivnut to spin) and you should find it a suitable solution.</p> <p>One determines the fastener size desired then uses that information to find the appropriately threaded rivnut, which will be followed by a reference to the hole size to drill to accept the rivnut.</p> <p>One last caution: determine the appropriate pulling force/limit, as it is far too easy to apply sufficient force to pull the threads from the rivnut, creating a failure point. My first inexpensive tool required &quot;that's about right&quot; for determining the force to be applied, while the replacement tool has a travel stop preventing the tool from closing too far.</p> <p>Amazon has a number of tool sets which can also include the rivnuts. One can also search for a <a href="https://www.youtube.com/watch?v=SzWj5y00sVg" rel="nofollow noreferrer">DIY rivunt setting tool</a>, which is effectively a compatibly sized bolt with a pressure sleeve and a couple of nuts. If you have only a few dozen to embed, it may be more cost effective to replace dollars/euros/what-have-you with elbow grease. Oh, yeah, apply lube to moving parts of any tool you create and make your life that much easier.</p>

58975

How are sheet metal enclosures closed with screws?

2024-02-23T00:12:47.270

<p>Consider an almost-rigid part on a vacuum table. There is a flat gasket underneath the part that exposes some portion of the bottom side of the part to vacuum. Let's say 50% of the bottom surface is exposed to the vacuum. Assume the body is flexible enough to seal against the &quot;islands&quot; in the gasket, preventing the vacuum from reaching those areas. To a first approximation, how much downward force does atmosphere apply to the part? Is it different than if the entire bottom surface was exposed to the vacuum?</p> <p><a href="https://i.stack.imgur.com/ly9Y7.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ly9Y7.jpg" alt="vacuum gasket for holding part" /></a></p> <p>The picture shows a custom gasket. I'm curious about this issue not only for my custom gaskets, but there is also a more extreme case.</p> <p>A vendor sells what they call &quot;tile gaskets&quot;, which are rubberized gaskets with 1/4&quot; holes on a 1&quot; triangular grid (they say roughly 95% of the board is covered with gasket). Their claim is that if we cover the entire porous spoilboard with these tiles, it limits the loss of vacuum on the non-covered parts of the table and improves hold-down. The loss of vacuum thing is true, but my question is, &quot;How much downward force on the part is there when only 5% of the bottom surface is exposed to vacuum?&quot;</p> <p>I've used the tile gaskets and, in many situations, they seem to do a good job when compared to a bare MDF spoilboard. Part of that, of course, is due to the sponginess and grip of the gasket material.</p> <p>I've been searching for months for insight to this question. Anyone want to give it a shot?</p> <p>UPDATE - Thanks for the answer.</p> <p>The clamping requirements for CNC routers depend on the depth and width of cut, the material, the bit, etc. Although I'm not a machinist, I believe the requirements are significantly lower than for steel machining. Soft metals like Al and brass are easily machined on a router using vacuum only.</p> <p>The biggest issue is the size of the part - 3&quot;x3&quot; parts are kind of hard to hold with no gasketing, especially on an MDF spoilboard. MDF is used for spoilboards due to its porosity - the vacuum essentially flows right through it, but leakage becomes an issue as soon as the entire spoilboard is not covered. Thus the tile gaskets I mentioned above.</p> <p>A lot of the parts we might cut on the router are large, though. An 8&quot;x14&quot; door panel is easy to hold with vacuum. A decent vac pump can easily get 10psi on the part, resulting in over 1000 lbs of downforce.</p>

|cnc|vacuum|

<p>It should always be atmospheric pressure x area exposed to the free space of the vacuum.</p> <p>If MDF is as porous as you say then the MDF itself is acting as a breather material. Incidentally, open-celled foams allow gas to pass through them so if these rubber tiles are made of open-celled foam, these would behave similarly. The area exposed to the free space of the vacuum would be reduced compared to if the materials were not there, but not as much as it might appear.</p> <p>I think the closed cell foam relies on the longer distance in the lateral direction through the material to reduce leaks, well as compression under vacuum to reduce leaks around the footprint of the material, but otherwise allows the vacuum to pass through to the workpiece.</p> <p>However, I have had cases where I was too lazy to cut a proper ring gasket for vacuum pot lids and instead just placed a sheet of This was a closed cell foam (that should not have allowed gas to pass through) across the entire opening and placed the hard lid on top. I wasn't sure if this would work at first but it did, though I can't confirm if the seal was any weaker.</p>

58977

Workholding vacuum only on a portion of the workpiece

2024-02-23T05:18:20.060

<p>I have a compressed gas duster with difluoroethane as the gas. When I shake the bottle, I feel sloshing, so I assume that it is a vapor liquid mixture inside.</p> <p>When I expel the gas, the can gets cold to the touch.</p> <p>I thought that liquid vapor mixtures have constant temperature during phase transition. Why does the bottle get cold?</p>

|thermodynamics|multiphase-flow|

<p>I will expand on <a href="https://engineering.stackexchange.com/a/58981/24351">Solar Mike's answer</a>.</p> <p>The physics/chemistry here is based on the <a href="https://en.wikipedia.org/wiki/Enthalpy_of_vaporization" rel="nofollow noreferrer">Latent heat of Evaporation</a></p> <p>As you let you air out of the can, the pressure drops, which causes <a href="https://en.wikipedia.org/wiki/Boiling_point" rel="nofollow noreferrer">Boiling Point</a> to decrease. This causes the liquid at the bottom of the can to boil into gas. Boiling consumes heat, which cools the can.</p>

58979

Why does a compressed gas duster get cold?

2024-02-23T10:48:56.440

<p><strong>Background</strong></p> <p>I randomly had this idea and i was wondoring if could it be a better alternate technology than what we use in cars now, but as i have little knowledge in the enginnering field(i'm a physics post graduate student) i had to ask this question here. I had this idea when i was learning transistors in which for a particular fixed <span class="math-container">$V_{be}$</span> voltage (between base and emitter in common emitter configuration) <span class="math-container">$I_c$</span> increases with increasing <span class="math-container">$V_{ce}$</span>,but it has a maximum value limit, to increase the limit <span class="math-container">$V_{be}$</span> is increased thus the analogy being <span class="math-container">$V_{be}$</span> acts like a gear and <span class="math-container">$V_{ce}$</span> is the accelerator while <span class="math-container">$I_c$</span> is the speed.</p> <p><strong>The Idea</strong></p> <p>We have descrete gear system in cars now which changes like 1&gt;2&gt;3&gt;4&gt;5. But Like i mentioned in transistor analogy <span class="math-container">$V_{be}$</span> is continuos and can be changed infinitesimally, cant we have the same system in gears used in cars, i.e a <em>Continuous Gear</em></p> <p><strong>Advantage-</strong></p> <p>Well i can't think of any significant advantage in the ease of driving but it will be pretty fun.</p> <p><strong>Question</strong></p> <p>Is this idea even possible? Or we already have something like this in automatic cars?If this idea is somehow possible will it help in ease of driving a car so that it becomes a piece of cake?</p>

|gears|car|

<p><a href="https://en.wikipedia.org/wiki/Continuously_variable_transmission" rel="nofollow noreferrer">Continuously Variable Transmissions</a> are a thing, and often found in things like snow mobiles and other light vehicles.</p> <p>They are not as common in heavy vehicles because they don't handle high torque very well, because they tend to rely on friction of smooth surfaces to transfer power rather than interlocking gears.</p> <p>For maintenance you need to account for the uneven wear induced in in the friction surfaces which need to be rather large and fairly precise.</p>

58984

Idea of a continuous Gear system in Cars

2024-02-24T23:13:44.307

<p>Is there a way to identify the source of low frequency (166Hz) noise coming from the street (I suspect from the opposite building in the closed yard in the city)?</p> <p>The noise comes hearable in the yard as well and comes, maybe from the root or vent system, I can’t be quite sure since I don’t have access there and housekeepers say that nothing there could produce this noise.</p> <p>Can I without a special expensive equipment somehow localize the direction to the source?</p> <p>I have good microphone, PC and some other household appliances.</p> <p>Without getting straight to the roof, just from my window, can I somehow get the exact direction to the source?</p> <p>I am thinking about:</p> <ol> <li>Two moving microphones, parallax, interference and computing direction (and even, possibly the distance) to the source.</li> <li>Cone tube with microphone to make in somewhat “directional microphone”.</li> </ol> <p>Are there any better ways to do so?</p> <p>I live in the city, so the environment is noisy, but in the night, I can choose some quiet time for measurement.</p> <p>The problem I am trying to solve is the 166Hz noise that about 25 Db louder than other sounds which passes through the windows (3 glasses of 4 mm) like through a filter that leaves only this noise while cancelling other sounds and working around the clock this drives me crazy.</p> <p><strong>Update</strong></p> <p>Based on comment below here is the spectrogram of the range 0-200Hz:</p> <p><a href="https://i.stack.imgur.com/TmwV4.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/TmwV4.jpg" alt="Spectrum with closed window 0-200Hz" /></a></p> <p>(Made by Android <a href="https://play.google.com/store/apps/details?id=org.intoorbit.spectrum&amp;hl=en_US&amp;gl=US" rel="nofollow noreferrer">Spectroid</a> application).</p> <p><em>(It shows 165Hz, but usually it is 166Hz, I don't think this difference is important).</em></p> <p>On the 50Hz peak, its loudness varies up to 30Db (can be less or more than on the picture) while loudness on 166Hz usually vary not more than 5Db until there is a strong wind or loud noise (cars, planes, street cleaning, etc.) and I created both frequencies on my PC and I don't bother with 50Hz sound (not so pressing) while I immediately recognized the 166Hz as &quot;the same one annoying sound&quot;.</p> <p>Here is the range 0-1000Hz based on the questions in comments.</p> <p>With open window:</p> <p><a href="https://i.stack.imgur.com/bAnle.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/bAnle.jpg" alt="With open window" /></a></p> <p>and with closed window:</p> <p><a href="https://i.stack.imgur.com/4X3ta.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/4X3ta.jpg" alt="With closed window" /></a></p> <p>In dynamics all the &quot;possible spikes&quot; after 200Hz are moving up and down and actually there is no stable spike like this at 166Hz. This could be seen from the bottom part of each chart, not sure what to call it, let me name it a &quot;legend&quot;.</p> <p>I have recording for the sounds, I will share if it would help.</p>

|sound|

<p>Several important clues here, as follows.</p> <p>First we note the very strong 50Hz spike. I will assume from the spelling of your name that you live in a place with 50Hz AC mains. Many different types of electrical machinery running on 50Hz will produce 50HZ noise- in particular, power transformers in which the core laminations are loose will &quot;hum&quot; audibly at the line frequency, and synchronous AC motors will hum at line frequency too as well as at their rotating speed.</p> <p>Next we note that the amplitude of the 166Hz signal is almost the same as that of the 50Hz spike, and just as sharp. I don't think this is a coincidence, which means the source of the 166Hz signal is coincident with the source of the 50Hz noise. This implies a large electric motor (perhaps ~several kilowatts).</p> <p>To radiate that noise across a street and into your yard means the source is most likely an air-circulating fan of some sort which is using its intake or discharge duct as a <em>loudspeaker</em>, since for example a water pump in a building would be much harder to hear since it isn't well-coupled to the air outside of the building.</p> <p>166Hz translates into a spin speed of 8300RPM for the fan- if it had only one blade, which fans never do. So if we postulate that the fan has six blades, for example, we get a fan speed of 1400RPM which would be in the typical range for a single-phase, one horsepower electric motor driving an HVAC fan. Note also that it is common to drive a big fan with a V-belt and pulleys; the fan turns more slowly than the motor shaft in this case which alters the relationship between motor speed, fan speed, and blade number.</p> <p>Now, how to trace the sound back to its source? This is hard for low-frequency sounds because they radiate in all directions and bounce off of large objects like apartment walls. This creates something called <em>multipath</em> in which a reflection can be mistaken for the source, and <em>phase cancellation</em> in which a reflection will destructively interfere with another reflection reaching your microphone from a slightly different direction.</p> <p>You can improve your chances of getting useful information from a microphone analysis by using a very long, skinny <em>shotgun microphone</em> which is highly directional, but the problem with a shotgun mic is that they are most effective at higher frequencies and less so at lower frequencies. To make a shotgun mic work well at low frequencies requires a mic extension assembly that is several meters long.</p>

59999

How to identify the source of low frequency noise from the street?

2024-02-27T07:05:58.233

<p>I have seen the technology for liquid detergent Auto - Dispensing in washing machines where we load the detergent at one go only and the machine automatically dispenses it according to its need for different load cycles. Now I want a concept and mechanism for the same thing for powdered detergents wherein we load like 1 kg of detergent in the machine and it automatically takes it and uses it acccording to the load. I need help for this concept can someone advice me with how can I do it</p> <p>Guys please provide any/smallest of ideas you have as it is only in the concept building phase anything can be of help for me</p>

|mechanical-engineering|mechanisms|

<p>Industrial applications use augers to feed powder.</p> <p><a href="https://i.stack.imgur.com/2veNs.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/2veNs.png" alt="enter image description here" /></a></p> <p><em>Figure 1. An auger system. Image source: <a href="https://www.parkerionics.com/powder-coating-equipment/powder-handling-equipment/auger-feed--ace-feed--system_pt30.html" rel="nofollow noreferrer">Parker Ionics</a>.</em></p> <p>In your case you just need the powder hopper and screw.</p> <p>You could experiment using a horizontally oriented drill bit (say 6 - 8 mm?) in a snug-fitting tube or block. Powder could be feed through a hole in the top side of the tube / block at the end of the drill helix and outlet would be near the tip. Run the drill anti-clockwise for powder feed.</p> <p>Considerations:</p> <ul> <li>This arrangement could be driven by a stepper motor to ensure repeatable dosing.</li> <li>You may need to consider mounting both ends of the drill bit in bearings to ensure that the drill doesn't gouge out material from the tube / block and add an unexpected ingredient to the washing machine.</li> <li>You need to prevent water vapour getting into the feed screw when not dispensing. This is another design challenge as it needs to have free fall-through for dispensing and by quite air-tight otherwise. A butterfly valve might be best. (On the other hand, you might find that dispensing into the machine's powder dispenser is far enough away from the washing drum that humidity isn't a problem in there.)</li> <li>You could 3D print an auger.</li> </ul> <p>Other material:</p> <ul> <li>[Screw Conveyor vs. Screw Feeder](micro auger feed).</li> <li><a href="https://www.youtube.com/watch?v=pJi4NlWWm48" rel="nofollow noreferrer">Fish tank feeder</a>.</li> </ul> <p><a href="https://i.stack.imgur.com/5Zr0w.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/5Zr0w.jpg" alt="enter image description here" /></a></p> <p><em>Figure 2. An auger made from washers.</em></p> <ul> <li><a href="https://www.youtube.com/watch?v=_JkVqEtmaOg" rel="nofollow noreferrer">Making an auger from washers</a>.</li> <li><a href="https://www.youtube.com/watch?v=TvXFsVSCw5Q" rel="nofollow noreferrer">Screw feed animation</a>.</li> </ul>

60018

Automatic Powder Detergent dispensing in washing machines - Concept Building

2024-02-27T10:12:43.433

<p>I came across some of these parts and have no idea what they are. It could be related to bearings, but most likely not. I also have other, similar looking parts but of all different dimensions (with the same shape). Material feels quite hard, almost like ceramic, and is cold to the touch. I am based in the United Kingdom (if that helps). Seems to me custom made to required size (as there is a customer identification number). I have also included a photo of how they are packed and the label on the box</p> <p>Text on the part itself and the label on the box:</p> <ul> <li>GC800-70GV2S</li> <li>1909177</li> <li>50/30 x 45 x 16mm</li> </ul> <p>I am looking to find out what these parts are and their application.</p> <p><a href="https://i.stack.imgur.com/pDysq.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/pDysq.jpg" alt="enter image description here" /></a> <a href="https://i.stack.imgur.com/hGHEr.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/hGHEr.jpg" alt="enter image description here" /></a> <a href="https://i.stack.imgur.com/UwX38.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/UwX38.jpg" alt="enter image description here" /></a> <a href="https://i.stack.imgur.com/D0ugd.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/D0ugd.jpg" alt="enter image description here" /></a></p>

|ceramics|part-identification|

<p>The p/n GC800-70GV2S is listed here:</p> <p><a href="https://darmann.com/specifications/product-selection" rel="nofollow noreferrer">darmann.com/specifications/product-selection</a></p> <p>Looks like some kind of abrasive shaping or surface finishing tool for production of hip implants</p>

60022

Identification of cylindrical part with unknown purpose

2024-02-27T13:49:26.950

<p>I have a few materials that I would like to get cut (in precise shapes that I cant do on a saw), these are Quartz glass (1mm thick) alumina wafer (1mm thick) (AL2O3) copper coated alumina (2mm thick)</p> <p>I was wondering if it was safe to attempt to cut these in a 150W CO2 laser, or if I am risking damaging myself or the laser.</p> <p>edit: I should have specified, this is a Universal PLS6.150D 150W Laser Engraver</p>

|cnc|cutting|glass|manufacturing|

<p>If you have a safe enclosure as described by Pete W in a comment, then you reduce the risk to damage yourself or others. Risking the laser is usually avoided by tilting the processing head such that reflected radiation does not return straight into the resonator.</p> <p>If those worries are taken care of, you can <em>attempt</em> to cut these materials, but there are a few points why it might get difficult with your setup:</p> <ul> <li>Aluminium and copper are <strong>highly reflective</strong> for the wavelength of a CO2 laser, somewhere around 1-2% absorption, which makes protection for you and your environment even more important, but also minimizes the amount of energy that actually melts the material</li> <li>Cutting aluminium with lasers requires process gas like nitrogen because molten aluminium and oxygen create a ceramic which tends to clog the cut</li> <li>Cutting aluminium with lasers requires higher power than steel because you can't use oxygen to help melt the material (see reason above) and the power losses in the material are higher</li> </ul> <p>That being said, you can surely <em>try</em> to cut these materials, I am just from a rough estimation not sure if the laser will get through.</p>

60024

150W laser cutting various materials

2024-03-01T10:26:52.677

<p>In structural optimization, we try to design our structure in a way to sustain loads in the best way possible. An example could be thinking about transmitting loads to fixed supports in an optimal way.</p> <p>I am thinking of the instant when a load reaches a support. First, before the load reaches the support, it has already traveled inside the structure, so I don't know why we care to transmit it to the support. Second, I suppose that the support for the load is considered as its &quot;end life&quot;. So, why do we want the load to vanish at the support and not within the structure itself?</p>

|structures|

<p>Because the force of the load needs to be counteracted (remember Newton's 3rd law?).</p> <p>A force doesn't just disappear, it needs to be counteracted or things will start moving (remember Netwon's 2nd law?).</p> <p>A support is a great way to create a force in the opposite direction that will prevent the force from acting on elements that cannot handle that force.</p>

60057

Transmitting load to a fixed support of a structure

2024-03-02T18:52:32.803

<p>I want to cut some 6&quot; diameter holes in 1/8&quot; thick lead sheet that is 12&quot;x12&quot; up to 12&quot; x 24&quot; but would like to minimize dust and debris.</p> <p>I was thinking that the best approach to minimizing dust and debris would be to use aviation shears to cut out the hole. However, the shears would need a starter hole to be made and this is mainly the hole I am asking about. Power drilling seems to not be an option.</p> <p>Is 1/8&quot; lead sheet soft enough that I could use a 1/2&quot; drill in a pin vise to manually work it through? That would create debris but at least it would not get kicked up.</p> <p>Alternatively I could use a 1/2&quot; or 1&quot; diameter hole punch on a 5-ton arbor press. That would probably produce even less dust but, Again, I am not sure if that could make it through a 1/8&quot; sheet. I've been able to punch 2&quot; diameter holes in 1/4&quot; thick teflon, but don't have a good reference between lead and teflon.</p>

|metalwork|

<p>Use a chisel and hammer. A narrow chisel will allow you to make a hole within the tolerance you need.</p>

60068

Making hole in 1/8" Lead sheet with minimal dust/debris

2024-03-09T19:54:52.157

<p>A general block diagram of a feedback controlled process with a disturbance in the output is:</p> <p><a href="https://i.stack.imgur.com/nC1Qk.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/nC1Qk.png" alt="enter image description here" /></a></p> <p>If we assume that the following blocks are equal to 1, and therefore we remove them:</p> <p><span class="math-container">$$ K_m = G_v = G_m = 1$$</span></p> <p>We have the following expression in frequency domain:</p> <p><span class="math-container">$$Y = G_dD + G_pG_c(Y_{sp} - Y)$$</span></p> <p>If we rearrange this we reach:</p> <p><span class="math-container">$$ Y = \frac{G_pG_c}{1+G_pG_c}Y_{sp} + \frac{G_d}{1+G_pG_c} D$$</span></p> <p>All books insist that to check stability, we have to check the roots of the characteristic equations alone, i.e., the roots of <span class="math-container">$1 + G_pG_c$</span>, since this appears in both denominators. However, shouldn't we also check the actual numerators? For example, if the disturbance is unstable, and:</p> <p><span class="math-container">$$G_d = \frac{1}{s-1}$$</span></p> <p>Then the system is not stable at all. It does not matter what the roots of <span class="math-container">$1 + G_pG_c$</span> are, since we are adding a positive root to the denominator. It would be: <span class="math-container">$$Y = \frac{G_pG_c}{1+G_pG_c}Y_{sp} + \frac{1}{(s-1)(1+G_pG_c)} D$$</span> The literature i've seen always refers only to the characteristic equation and not the numerator... is there a reason for this? Shouldn't we always substitute the transfer functions first?</p>

|control-engineering|stability|feedback-loop|

<blockquote> <p>For example, if the disturbance is unstable... Gd = 1/(s-1)</p> </blockquote> <p>In that case, the (s-1)^(-1) would be <strong>part of the denominator</strong> after simplification; and hence part of the characteristic equation.</p> <blockquote> <p>The literature i've seen always refers only to the characteristic equation and not the numerator.</p> </blockquote> <p>No, <span class="math-container">$G_d$</span> is a ratio of polynomials and itself contains a numerator and a denominator. The denominator needs to be shifted to the <em>main</em> denominator.</p> <p>e.g.</p> <p><span class="math-container">$$ \frac{(s+1)/\color{red}{(s+4)}}{(s+2)/(s+7)} = \frac{(s+1)(s+7)}{\color{red}{(s+4)}(s+2)} $$</span></p> <blockquote> <p>All books insist that to check stability, ...</p> </blockquote> <p>If they are well written books, they will most likely contain some fine print at the beginning of the chapter or some kind of foot note that says something like: &quot;<em>The individual blocks in the block diagram are assumed to be stable systems</em>&quot;. Since you have not given a specific (reputable) reference text book, it not possible to know what the author's stated assumptions were.</p>

60131

Stability of a closed-loop with disturbance in output. Characteristic equation alone or substitution

2024-03-11T10:42:17.183

<p>I have seen this lead/ball screw replacement device in a DIY video, possibly related to 3D printing, but lost the link and could not find how it is properly called.</p> <p>It consists of a smooth rod clamped tightly by three bearings (ball bearings?) arranged around the rod and fixed to the bed such that rotating the rod (possibly via a friction drive) forces it to advance due to the line of contact of the rod with the bearings being inclined. The sketch below shows only one bearing out of three for clarity.</p> <p>Does this setup actually work and what is it called?</p> <p><a href="https://i.stack.imgur.com/ogz23.png" rel="nofollow noreferrer" title="Diagram of smooth rod clamped by inclined bearings"><img src="https://i.stack.imgur.com/ogz23.png" alt="Diagram of smooth rod clamped by inclined bearings" title="Diagram of smooth rod clamped by inclined bearings" /></a></p>

|mechanical-engineering|

<p>You seem to be referring to this variation of <em>ball screw</em> where a smooth rod is used, instead of a lead screw.</p> <p>It is (apparently) called a <em><strong>threadless ballscrew</strong></em>.</p> <p>From <a href="https://hackaday.com/2016/04/28/threadless-ballscrew-for-3d-printer/" rel="nofollow noreferrer">Threadless ballscrew for 3D printer</a>:</p> <p><a href="https://i.stack.imgur.com/PHTxP.jpg" rel="nofollow noreferrer" title="Threadless ballscrew for 3D printer"><img src="https://i.stack.imgur.com/PHTxP.jpg" alt="Threadless ballscrew for 3D printer" title="Threadless ballscrew for 3D printer" /></a></p> <blockquote> <p>A <a href="https://www.thingiverse.com/thing:124706" rel="nofollow noreferrer">threadless ballscrew</a> that turns rotational into linear motion with no backlash. It works by pressing the edge of three bearings fairly hard up against a smooth rod, at an angle. The bearings actually squeeze the rod a little bit, making a temporary indentation in the surface that works just like a screw thread would. As the bearings roll on, the rod bounces back to its original shape. Watch it in action in the video below.</p> <p>The two benefits of these pseudo-threads is that they fit tightly so there’s no backlash, and they give when too much force is applied, rather than jamming. Eliminating backlash is awesome for a 3D printer, but it’s not obvious how a thread that gives under excessive load is a plus, unless you’ve crashed your printhead into the bed of the printer. Generally speaking, 3D printers don’t subject their screw drives to all that much force, making this an interesting option.</p> <p>A <a href="http://www.amacoil.com/rolling-ring-how.html" rel="nofollow noreferrer">professional version of the same mechanical idea</a> uses special bearings with a ridge in the center, and tips them side to side to change the contact angle, and thus speed of travel (per rotation of the shaft). It’s even got a provision for flipping the bearings over, causing the tram to move in the opposite direction. Pretty cool.</p> </blockquote> <ul> <li>YouTube: <a href="https://www.youtube.com/watch?v=bXX6AXUATD0" rel="nofollow noreferrer">Threadless Ballscrew Test</a></li> <li>Thingiverse: <a href="https://www.thingiverse.com/thing:124706" rel="nofollow noreferrer">OpenSCAD Threadless Ballscrew - Highly Customizable</a></li> </ul> <hr /> <p>For completeness...</p> <p><strong>Ball screws</strong></p> <p>These images are taken from the video <a href="https://www.youtube.com/watch?v=WoPPwGxgWEY" rel="nofollow noreferrer">Chasing Micrometres with the best Ball Screws</a></p> <p><a href="https://i.stack.imgur.com/ClLLy.jpg" rel="nofollow noreferrer" title="Cut away diagram"><img src="https://i.stack.imgur.com/ClLLy.jpg" alt="Cut away diagram" title="Cut away diagram" /></a></p> <p><a href="https://i.stack.imgur.com/OpGlg.jpg" rel="nofollow noreferrer" title="Photo of internals of ball screw"><img src="https://i.stack.imgur.com/OpGlg.jpg" alt="Photo of internals of ball screw" title="Photo of internals of ball screw" /></a></p>

60141

Lead screw alternative with smooth rod clamped by inclined bearings?

2024-03-11T13:23:17.447

<p>When we have a T or I section under a shear force <span class="math-container">$V_y$</span> I usually must account for a horizontal shear force across the section's Z axis (<span class="math-container">$_{xz}$</span>). I understand why a transverse shearing force would obviously cause a sliding action downwards. Mathematically it is possible to show that <span class="math-container">$Q$</span> is non zero in the flange for <span class="math-container">$_{xz}$</span> but is there any physical reasoning for why there is <span class="math-container">$_{xz}$</span> present in the flanges of I and T beams?</p>

|structural-engineering|structural-analysis|stresses|solid-mechanics|

<p>If you look at the shear diagram of I section you see shear in the web starts from the top flange at a non-zero value , calaculated by the <span class="math-container">$q=\frac{VQ}{I} \ .$</span> That q on top and bottom are coming from the horizontal tranguluar shear flow in the flanges. The diagram shows how different sections distribute the flow. <a href="https://hthttps://engineerexcel.com/shear-flow/tps://" rel="nofollow noreferrer">source</a></p> <p>.</p> <p><a href="https://i.stack.imgur.com/kmQUW.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/kmQUW.png" alt="shear flow" /></a></p> <h1>Edit</h1> <p>After OP's comment</p> <p>If you can intuitivly see how the shear flow works on a C-channel loaded vertically parallel to its web with P, then an I section could be imagined as two C-channels loaded each with P/2 glued to each other.</p> <p>,</p> <p><a href="https://i.stack.imgur.com/2CuuS.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/2CuuS.png" alt="two c-channels" /></a></p>

60144

$_{xz}$ in the flanges of T and I Sections

2024-03-12T15:33:16.467

<p>I found a 40ft. length of rope for sale at the dollar store. Not having high expectations to start with, I examined the packaging closer out of curiosity. The front of the label states that the rope has 200lbs of tensile strength... Not all that much, but okay, in line with expectations. Turning it over to read the fine print on the back, &quot;misuse... serious injury or death&quot; mhm, alright... &quot;Do not use to...&quot; uh huh, okay, nope, no, nuh uh, wasn't planning to, and wh...? &quot;The tensile strength and minimum breaking strength of this product have NOT BEEN RATED.&quot; (Emphasis original)</p> <p>From a professional engineering standpoint, what must occur for a product such as this to be formally rated? As it stands with the disclaimers, that 200lbs figure on the front sounds like it was pulled out of someone's wazoo.</p>

|standards|mechanical-failure|tension|

<p>&quot;Rated&quot; means it has been proof-tested by an independent lab (like UL or CSA) according to a published standardized test protocol invented by a standards organization (like ASTM) and sufficient trials have been conducted to achieve statistical significance.</p> <p>This costs money, and cheap product manufacturers skip this step whenever possible. Note that with electrical consumer goods in the USA, you <em>cannot</em> skip this step and <em>must</em> apply for, pay for, and obtain UL approval.</p>

60158

What does it mean for an item's tensile strength to be "rated"?

2024-03-12T18:37:55.253

<p>Can someone provide the derivation for the following equation:</p> <p><span class="math-container">$$\int_l Q_Z(x) dx = M_y (x) \tag1$$</span></p> <p>or equivalently :</p> <p><span class="math-container">$$\frac{dM_y}{dx}=Q_Z\tag2.$$</span></p> <p>I am interested in the <span class="math-container">$1D$</span> isotropic material version.</p> <p><a href="https://i.stack.imgur.com/BRXde.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/BRXde.png" alt="enter image description here" /></a></p> <p><em>Appendix</em></p> <p>When drawing the <a href="https://en.wikipedia.org/wiki/Shear_and_moment_diagram" rel="nofollow noreferrer">shear and moment diagram</a> for this case, we would get something like this: <a href="https://i.stack.imgur.com/q1gWV.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/q1gWV.png" alt="enter image description here" /></a></p> <p>Since the <span class="math-container">$Q_z$</span> is positive, the slope of <span class="math-container">$M_y$</span> has to be positive, because of equation <span class="math-container">$(2).$</span></p> <p>But if <span class="math-container">$M_y$</span> has to have a positive slope and has to be zero at the free end, than it has to look like in the picture, i.e. it has to have a negative value.</p> <p>Can that be a priori recognized from the equation <span class="math-container">$(1)$</span> or is it just a convention that when a beam curves downwards that it will be negative, and when it curves upwards, it will be positive, like explained <a href="https://en.wikipedia.org/wiki/Shear_and_moment_diagram" rel="nofollow noreferrer">here</a>?</p>

|mechanical-engineering|structural-analysis|applied-mechanics|beam|

<p>We just integrate and evaluate <span class="math-container">$x$</span> from <span class="math-container">$0$</span> to <span class="math-container">$l$</span>.</p> <p><span class="math-container">$$\int_l Q_Z(x) dx = Q_z(x) \ x\biggr\rvert _o^l = Q_z(x)l= M_y(x)$$</span></p>

60160

Derivation of the fundamental principle in the mechanics of beams

2024-03-13T16:56:35.900

<p>I am working on a school project where I have to choose a mechanical motion device and create a 3D model of it.</p> <p>I have chosen <a href="https://507movements.com/mm_090.html" rel="nofollow noreferrer">https://507movements.com/mm_090.html</a> because it seems adequate for my modeling skills. However, I am unable to find any approach for designing such type of device.</p> <p>I am an electrical engineer and this is completely out of my branch. Are the any design equations that would provide me basic set of dimensions? I would be also glad for any book recommendation on this topic.</p>

|mechanical-engineering|modeling|

<p><a href="https://i.stack.imgur.com/wbGcOm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/wbGcOm.png" alt="enter image description here" /></a></p> <p><em>Figure 1. A Scotch yoke mechanism. The blue part rotates around the small circle. The red part oscillates left and right with simple harmonic-motion.</em></p> <p>The side to side oscillation of the red part will be <em>2 × e</em>. (<em>e</em> is the eccentricity of the cam relative to the shaft.)</p> <p>The slot in the yoke needs to have a linear dimension between the top and bottom arc centres = <em>2 × e</em>.</p>

60171

Designing Eccentric Motion

2024-03-15T10:06:28.903

<p>I am interested in creating a water dispenser myself for cooling water, and am looking for a cheap and not too engineering complex concept. I found a water dispenser on Aliexpress, for a very cheap price, and am wondering whether you know how the cooling system works? Link to the water dispenser: <a href="https://de.aliexpress.com/item/1005002611932877.html" rel="nofollow noreferrer">https://de.aliexpress.com/item/1005002611932877.html</a> The water dispenser looks like this:</p> <p><a href="https://i.stack.imgur.com/nngMv.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/nngMv.png" alt="Water Dispenser at 550 W rated power" /></a></p> <p>Is it a small refrigeration cycle, or does it contain Peltier elements? If someone would know that would help me out greatly. How is it possible to actually make a working cold water dispenser as cheap as this? Of course I don't know how well it cools the water and to what temperature, but still.</p> <p>Thank you very much for your answers!</p> <p>FBMeca</p> <p>N.B. Not entirely sure in which community this post fits best, feel free to move it if it's misplaced here.</p>

|cooling|heat-exchanger|water-resources|product-engineering|

<p>This device only heats. The &quot;cold&quot; water is at room temperature. Check this alternative website:<a href="https://ma2030.com/Home-Appliances/dmwd-550w-household-electric-water-dispenser-desktop-water-heater-mini-water-boiler-drinking-fountain-constant-temperature-95%E2%84%83/" rel="nofollow noreferrer">link</a> It gives this info: &quot;room temperature and heating&quot;</p> <p>You can see there a picture of the &quot;dual temperature control&quot; which is only two thermostats for hot water temp - which is usual. One for control, one for safety.</p> <p>Warm and hot refer to the two water temperatures available -no cold.</p>

60184

Water Dispenser Cooling System

2024-03-23T12:47:17.357

<p>There seems to be a lot of information around calculating the forces exerted on a structure. When joining wood, once the structural analysis is done, <strong>how is the performance limit of a certain type of joint determined?</strong></p> <p>E.g. building a wooden frame in the garden</p> <ol> <li><p>Building a pergola - Relatively weak joins may be used as the loads are negligible compared to strength. This relatively weak joint is fine for supporting the &quot;roof&quot; of the frame <a href="https://i.stack.imgur.com/6zyQJ.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/6zyQJ.jpg" alt="Image of a pergola with a highlighted joint" /></a></p> </li> <li><p>Putting solar panels on top of the same pergola - Now there is in increased load the frame has to deal with, as well as even greater forces when there are strong winds. Nobody will be recommending the pergola above for that purpose</p> </li> </ol> <p>I.e. How could an engineer determine if a mortice and tenon joint is sufficiently strong for a given joint or if it would be better to use a butt joint with screws?</p>

|structural-engineering|structural-analysis|wood|

<p><strong>By destructive testing of the different kinds of joints under different kinds and magnitudes of loads</strong>.</p> <p>This means that joints of various kinds are constructed, then loaded in a particular fashion until they break, taking measurements of the forces needed to deform the joint and/or separate the parts. Then, some sort of analysis of the results is used to produce structural ratings for the different joints in specified conditions; and if a consensus can be reached for some application, a committee is likely to draft a standard to make it easier to specify requirements for a certain design. An example of such a specification is (in the USA) the <a href="https://awc.org/publications/2018-nds/" rel="nofollow noreferrer"><em>National Design Specification for Wood Construction</em></a>.</p> <p>Here, for example, is an excerpt from a technical handbook referencing empirical methods:</p> <blockquote> <p>Maximum lateral resistance and safe design load values for small-diameter (nails, spikes, and wood screws) and largediameter dowel-type fasteners bolts, lag screws, and drift pins) were based on an empirical method prior to 1991. Research conducted during the 1980s resulted in lateral resistance values that are currently based on a yield model theory. This theoretical method was adapted for the 1991 edition of the National Design Specification for Wood Construction (NDS). Because literature and design procedures exist that are related to both the empirical and theoretical methods, we refer to the empirical method as pre-1991 and the theoretical method as post-1991 throughout this chapter. Withdrawal resistance methods have not changed, so the pre- and post-1991 refer only to lateral resistance. <a href="https://www.fpl.fs.usda.gov/documnts/fplgtr/fplgtr113/ch07.pdf" rel="nofollow noreferrer">Wood Handbook, Forest Products Laboratory. 1999. Ch. 7</a></p> </blockquote> <p>Not just engineers use this method: carpenters often consider themselves, with varying degrees of justification, as engineers of sorts.</p> <blockquote> <p>Most woodworking joints have evolved over thousands of years. Woodworkers learned to make them based on their own experience, and the experience of the people who taught them. To make successful joints you need a little experience, a basic knowledge of the workings of wood and glue, and a bit of common sense. But in the 21st century we want a definitive answer. We want numbers and we want proof. We want something in writing we can point to when we’re not sure of ourselves.</p> <p>Like other magazines, we have tested joints to destruction. <a href="https://www.popularwoodworking.com/tools/testing-wood-joints-to-failure/" rel="nofollow noreferrer">Popular Woodworking</a></p> </blockquote>

60243

Determining the suitability of a timber joint in a given situation

2024-03-23T14:17:45.830

<p>How is the fan efficiency zero at the highest flowrate point (at which there is zero pressure rise)? I know the fan efficiency is given as</p> <p><span class="math-container">$fan\_efficiency = output\_power/input\_power$</span></p> <p>where output_power of the fan is given</p> <p><span class="math-container">$Output\_Power = Pressure\_rise \ \times \ Flow\_rate$</span></p> <p>But even if there is no pressure rise across a fan, the fan does cause the flow to happen by transferring energy (or power) to the fluid. So why is the efficiency zero (and also the <span class="math-container">$fan\_output\_power$</span> zero) at the highest flow rate point?</p> <p><a href="https://i.stack.imgur.com/tQM2c.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/tQM2c.png" alt="Ref: https://www.linquip.com/blog/fan-efficiency/" /></a></p>

|energy-efficiency|fan|

<blockquote> <p>But even if there is no pressure rise across a fan, the fan does cause the flow to happen by transferring energy (or power) to the fluid.</p> </blockquote> <p>Don't assume that the flow in that graph is driven by the fan -- if the fan efficiency is going to zero, that means it is not exerting any net axial force on the airflow. That, in turn, means that the air is being driven by some other mechanism.</p> <p>The fan may be inducing turbulence in the fluid, and that is, indeed, adding energy to the flow.</p> <p>However, in that context, the authors clearly mean that the <strong>useful</strong> action of the fan is to induce flow -- if there's no pressure rise across the fan, then <em>for the purposes of inducing flow</em> it may as well not be there -- so it's consuming a finite amount of power to turn, and generating zero effect. Zero effect for finite effort means zero efficiency.</p>

60244

Why is the fan efficiency zero at high flow rate?

2024-03-25T20:36:25.613

<p>The Schlage FBE365 electronic deadbolt uses a DC motor attached to a helical screw to raise and lower the C-shaped piece shown. The way the rest of the mechanism works is not important for this question - all that matters is that the C-shaped piece needs to go up and then down and it does that by rotating a motor and spring.</p> <p>Additional detail: The C-shaped piece has a tang on its back that inserts between coils. When the rotation happens, the portion of the spring below the tang becomes compressed and the portion of the spring above the tang becomes elongated, which causes the C-shaped piece to be pulled up. Its held there for a couple seconds, then the motor rotates back to its original position, causing the C-shaped piece to be pushed down. Figure 1 shows the C-shaped piece in the down position. Figure 2 shows the C-shaped piece in the up position. Figure 3 shows the C-shaped piece removed - the spring and tang are visible. Figure 4 shows a close up of the tang. See here for a video of it operating: <a href="https://www.youtube.com/watch?v=OqjcM-eb8ZY" rel="nofollow noreferrer">https://www.youtube.com/watch?v=OqjcM-eb8ZY</a></p> <p>Prior to seeing the insides, I would have guessed there is a solenoid moving parts around as is common in many locks. It seems like that could definitely work here - have a solenoid pull up on the C-shaped piece and then have a spring return it to its original position. Any ideas why they chose to use a motor rather than a solenoid? One idea I had is that the solenoid plunger would have to be ferromagnetic, leaving the lock susceptible to someone using an external magnet to lift up the C-shaped piece and open the deadbolt. Any other possible reasons? Has anyone else ever seen a spring used as a screw drive in a design?</p> <p>Figure 1: <a href="https://i.stack.imgur.com/0ze4n.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/0ze4n.jpg" alt="Figure 1" /></a></p> <p>Figure 2: <a href="https://i.stack.imgur.com/2eoO7.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/2eoO7.jpg" alt="Figure 2" /></a></p> <p>Figure 3: <a href="https://i.stack.imgur.com/wOB3O.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/wOB3O.jpg" alt="Figure 3" /></a></p> <p>Figure 4: <a href="https://i.stack.imgur.com/AGmXv.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/AGmXv.jpg" alt="Figure 4" /></a></p>

|mechanical-engineering|mechanisms|electromagnetism|springs|magnets|

<p>The reason drive screws are used is because they give mechanical advantage.</p> <p>Solenoids are linear devices and don't have inherent use leverage for mechanical advantage. A rotary motor with a small lever arm produces more force and somewhat decouples the actuator length from the motor size. A solenoid gets only one stroke to do the same amount of work where the rotary motor gets to spin many many times. That lets you use a stronger spring or a smaller actuator.</p> <p>I've never seen a spring used as a drive screw before though. Not sure if they used that because they wanted the compliancy or because a hollow threaded would require custom manufacture and be very expensive by comparison. It certainly neat how it reduces size since you can shove the motor inside the spring.</p> <p>Seems to me you don't strictly need the compliancy of the spring. To me it seems like the tang just runs inside the spiral space of the spring and acts like a normal lead screw. A hollow, rigid lead screw could also do the job.</p> <p>Where a spring would help though is at the end points. I don't see anything that looks like limit switches to indicate motor endpoints for stopping. The spring's compliance would make it be more forgiving on the motor with regards to slamming into the end points and abruptly, and immediately stalling. With the spring, you could more effectively get away with using a safer, more gradual onset of motor overcurrent to know when to shut the motor off. Also less backlash.</p>

60263

What are the benefits of using a spring as a drive screw in this electronic deadbolt?

2024-03-26T09:18:08.207

<p>If I have a 40A 3 pole Miniature Circuit Breaker (&quot;mcb&quot;), does this mean 40a per pole or 40A altogether? The quality standards made me write some more so I do it.</p>

|electrical-engineering|

<p>40 A per pole.</p> <p>That's the way three-phase loads are specified so that's the way the breakers are specified.</p> <p>It also means that a single phase exceeding the trip current (due, for example, to a partial earth fault on that phase) will trip the whole circuit.</p>

60267

40A mcb. 40A per pole or 40A altogether?

2024-03-30T07:20:34.280

<p>I am still pretty new to control theory, but I have been studying both classical and modern state space/optimal control ideas. I am having trouble understanding what the frequency domain/transfer function view is on some of the challenges of optimal control, for example the inverted pendulum swing-up problem.</p> <p><em>Problem definition</em>:</p> <p>So in optimal control I may have an inverted pendulum system where the pendulum is currently facing down--at a stable fixed point. The objective is to move the pendulum from the stable fixed point to the unstable fixed point where the pendulum is pointing straight up. Now I have some actuation on the pendulum, but there are some torque limits.</p> <p><em>Trajectory Optimization:</em></p> <p>There are a few ways to solve this inverted pendulum problem. One way is &quot;trajectory optimization&quot; which uses convex optimization ideas to find the control policy to drive the pendulum to the top location, meaning the unstable fixed point. I might also be able to use something like fitted value iteration to find a swing-up policy, but generally the fitted value iteration solution is not very robust.</p> <p><em>Question</em>:</p> <p>So I can understand why the &quot;swing-up&quot; problem seems difficult in optimal control, but I don't have a clear sense of how to formulate this same problems as a transfer function or in the frequency domain? I mean I could compute the transfer function for the pendulum equation. However, I am not sure if a PID controller could find a suitable control law to bring the pendulum to the top? Of course, as the pendulum approaches the linear region of the unstable fixed point at the top, I could linearize and use LQR control to stabilize the dynamics. But I am still not clear how this swing-up problem is modeled in the classical control sense.</p> <p>Thanks.</p>

|control-theory|pid-control|optimal-control|nonlinear-control|linear-control|

<blockquote> <p>how to formulate this same problems as a <strong>transfer function</strong> or in the <strong>frequency domain</strong>? I mean I could compute the transfer function for the pendulum equation.</p> </blockquote> <p>The range of motion involved in bringing a pendulum from <em>down</em> position to <em>up</em> position means that the systems needs to be treated as a <em>non-linear</em> system unlike the <em>stabilise-in-the-up-position</em> for angles <em>near</em> to the equilibrium case. So transfer function approaches may not exist as they are intended for <em>linear systems</em>.</p> <blockquote> <p>I am not sure if a PID controller could find a suitable control law to bring the pendulum to the top?</p> </blockquote> <p>IMO, this should be possible by intentionally designing a controller to make the closed system <strong>unstable</strong> for the <em>down</em> position. The oscillations of the the unstable system would grow (if designed properly) <s>until the amplitude reaches 180 deg; i.e. <em>up</em> position</s>^{See comment below.}.</p> <blockquote> <p>But I am still not clear how this swing-up problem is modelled in the classical control sense.</p> </blockquote> <p>This may be because of the range of motion prevents it from being represented as a linear system. I don't know if there exists any literature which tackles swing-up problem using classical methods.</p>

60294

What is the relationship between classical control: transfer functions/frequency domain and the swing-up problem of an inverted pendulum