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2016-06-13T15:07:50.037 | <p>I would like to get my hands on a mechanically-set timer (wind up is fine, so that it's self-powered) which sends a small electrical pulse to a mag-lock once the timer goes off, thus causing the mag-lock to open. Is there anything out there that does this?</p>
<p>If not, what would be the cheapest / easiest way to replicate such function, i.e. some mechanical way to produce a small electrical impulse once a mechanical timer finishes. I don't mind electronic methods, but since this is for heavy outdoor use I'd prefer a purely mechanical setup.</p>
| |electrical-engineering|magnets|timing| | <p>In principal it should not be very difficult to adapt an off the shelf mechanism like a mechanical alarm clock or kitchen timer. They will already have some sort of latching mechanism to activate the bell after the elapsed time so it should be a simple matter to adapt this to close an electrical contact.</p>
<p>Having said that I'm not entirely convinced that a mechanical system is necessarily going to be any more rugged or reliable than an electronic one. As already mentioned in other answers a mag-lock will need a reliable power supply and a timer circuit is simple to achieve with either digital or analogue electronics and can be potted of placed in an IP rated enclosure to make it pretty much indestructible compared to any kind of clockwork system. </p>
| 10288 | Is there a device like this? |
2016-06-13T16:26:19.057 | <p>For reinforcing in concrete, anchorage / lap lengths are determined by the multiplying the diameter of the rebar with a factor. Generally a factor of 40-45 is accepted as the de-facto standard. According to EC2, the factor falls within the range from 34 and 66, depending on concrete quality, etc.</p>
<p>It is my understanding that the factor is in linked to the amount of shear resistance required to act against a force pulling out a rebar at its ultimate tensile stress. The factor is thus the height of the shear cone created. </p>
<p>When lapping a bar of larger diameter with a smaller one, my explanation above would be used to argue the case that the smaller diameter is taken to determine lap length.</p>
<p>I have found various sources where people have said the smaller bar should be used. I can find no code references defining it though. I would assume that one goes for the larger diameter to be safe, but the extra steel obviously increases cost.</p>
<p>Any advice would be greatly appreciated.</p>
| |structural-engineering|reinforced-concrete|reinforcement| | <p>You can use steel reinforcement bar couplers to connect. Do not you think that this will be much easier? After all, you do not need to use welding equipment, only portable hydraulic pressing equipment. Simplicity and high assembly speed are advantages.</p>
| 10289 | What is the splice length for two reinforcement bars in tension of different diameter? |
2016-06-14T12:35:35.440 | <p>I have seen this video pretending to decrease by 5° the temperature inside home using an air-conditioner based on the <a href="https://physics.stackexchange.com/questions/7868/why-does-the-air-we-blow-exhale-out-from-our-mouths-change-from-hot-to-cold-depe">air entrainment</a> principle (see the <a href="https://youtu.be/jPuh8IFbIzQ?t=1m6s" rel="nofollow noreferrer">video here at 1m6s</a>). </p>
<p><strong>EDIT : The video is now private but you can see a picture of the air-conditioner <a href="http://www.iflscience.com/environment/these-diy-ecofriendly-air-conditioners-are-cooling-down-bangladesh/" rel="nofollow noreferrer">here</a></strong></p>
<p>Does such an air-conditioner really work?</p>
| |heat-transfer|airflow|temperature| | <p>This device is not really working using the principles referenced in the other SE article. In this case the air inside the hut is being heated by the sun and this is really about improving ventilation and letting that heated air out.</p>
<p>I think, however, it would perform worse than a fully open window from a heat transfer point of view, however, it would also act as a screen for privacy/exclusion reasons. </p>
<p>I also have doubts it works much better with the bottles than without them (ie. just a series of holes). The article mentions the bottles are used to funnel cold air in but holes would allow that too. Also, the bottles would cause more interference with the drawing out of hot air if the wind is blowing parallel to the wall than simple holes would.</p>
<p>TL:DR It would certainly help cool the air inside, but I think simpler methods would work better.</p>
<p><strong>EDIT: Apparently this question has appeared on multiple SE see: <a href="https://physics.stackexchange.com/questions/261284/how-does-the-eco-cooler-air-conditioner-really-work">Physics.SE</a> and <a href="https://skeptics.stackexchange.com/questions/34302/does-the-eco-cooler-device-cool-the-air-in-a-home-without-electricity">Skeptics.SE</a></strong></p>
| 10299 | Is it possible to use air entrainment for air conditioning |
2016-06-14T14:35:00.417 | <p>I only have basic metal fabrication tools and do not own a mill or a lathe, but would like one. I took apart my cheap drill press to study whether or not I can convert it into a milling machine. The spindle features a 4 tooth 9mm x 9mm tooth male splined section that mates with a female splined bushing press fit into the pulley's bore (see parts numbers 62 and 71 in diagram). The entire spindle fits inside the quill tube (see part 68) creating the quill assembly (see photo below). The spline allows the spindle to extend downwards and upwards in the head casting while also allowing the spindle to rotate with maximum torque.</p>
<p>Currently, I am considering replacing the entire spindle with an 16mm O.D. ER20 tool holder. This will give me a more ridgit hold on the cutting tool as Jacobs Taper cannot handle axial forces required for milling. I am also planning on replacing the existing bearings with either angular contact bearings or tapered bearings. </p>
<p>I would like to maintain the quill's retract and extend mechanism. To do this, I will have to somehow jointhe end of the ER toolholder to a splined shaft that will fit in the female end. Currently, I can't seem to find any off the shelf spline to 16mm coupling. I am not even sure what they are called.</p>
<p>Are there any solutions for coupling or attaching a spline shaft to a plain old shaft? What is it called and do they exist? Finally, I am not completely married to using a spline drive mechanism in order to extend and retract the quill, so I welcome any suitable alternative mechanisms. I saw a ball spline, but wasn't sure if it works the same way as the existing mechanism.</p>
<p>Here's a photo of a similar quill:</p>
<p><a href="https://i.stack.imgur.com/F5fCS.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/F5fCS.jpg" alt="enter image description here"></a></p>
<p><a href="https://i.stack.imgur.com/N5KkC.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/N5KkC.png" alt="enter image description here"></a></p>
<p><a href="https://i.stack.imgur.com/vu1Ti.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vu1Ti.jpg" alt="enter image description here"></a></p>
| |mechanical-engineering|gears| | <p>I would tend to agree with Chris Johns on this one - it's probably not a great use of your time and may not provide satisfactory results - but that's not what you asked... </p>
<p>To directly answer your questions: </p>
<blockquote>
<p>Are there any solutions for coupling or attaching a spline shaft to a plain old shaft? What is it called and do they exist? </p>
</blockquote>
<h2>Yes, there are. A press-fit, or interference fit, may be a suitable means of attaching the two components.</h2>
<p>If you're not familiar with the concept, a press fit (aka <a href="https://en.wikipedia.org/wiki/Interference_fit" rel="nofollow">interference fit</a>) means you press a pin into a hole that is slightly smaller than the pin (creating interference between the components which holds them together). In the case of metal-to-metal were talking a size difference of few tenths' of a thousandth of an inch [.0001] to a few thousandths [.001] of an inch. </p>
<p>If you provide more information regarding the materials of construction (presumably some sort of hardened tool steel for the shaft, not sure about your collar) and actual dimensions (actual meaning you measured the parts with a micrometer, or other suitable measuring device, and not the nominal dimensions provided by the manufacturer or parts description) then you could use an online fit calculator such as <a href="http://www.engineersedge.com/calculators/mechanical-tolerances/force-fit-tolerances.htm" rel="nofollow">this one</a> provided by Engineers Edge to get an idea of the interference needed - maybe look to the FN5 class for example. </p>
<p>I'm suspecting you're going to need a significant interference to withstand the loads expected, so you'll probably need a hydraulic press or a good source of heat, or maybe a combination of the too. </p>
<p>Again, if you know the material of construction, you can use the coefficient of thermal expansion to determine the temperature change necessary to cause the female member to expand large enough to slip the male member in, and then when it cools and contracts it will create the interference necessary to hold the parts in place. </p>
<p>Another option you might consider, is that you can press fit the pieces together (I recommend the shrink fit method above to maintain concentricity) and since this is a splined shaft fitting into a round hole, you may be able to braze the shaft to the coupling (filling in the gaps in the teeth with brazing material) which would be extremely strong. Again without knowing more about the base materials I don't know if this is a suitable suggestion, but it's worth considering. Many cutting tools used for milling operations which have carbide inserts are brazed, and it's proven to be a reliable and strong connection for the job. </p>
<p>Standard disclosure: you should consult an engineer experienced with this type of application (someone specializing in machine design, or machine maintenance might be a good place to start) before taking numbers off some website or the words of some guy who doesn't have all the information about the application. For one - the numbers suggested on the site I linked to are based off an ANSI standard that provides recommendations for common applications of holding a round pin in a round hole - none of which describes your application. And for two - you're playing with some pretty large forces and heavy loads in a milling operation, and a catastrophic failure could send an extremely sharp large chunk of carbide, cobalt, or high-speed-steel straight at your face, eye, throat, or child-rearing organs which would not be very fun at all. </p>
<blockquote>
<p>Finally, I am not completely married to using a spline drive mechanism in order to extend and retract the quill, so I welcome any suitable alternative mechanisms. I saw a ball spline, but wasn't sure if it works the same way as the existing mechanism.</p>
</blockquote>
<h2>Bridgeport mills typically use an R8 quill.</h2>
<p>I don't know if it's feasible, but if you're going down this road you might look at adapting one of these to your drill press. </p>
<p>Lastly - my personal opinion only - is that the amount of time and money you'll spend on this project, which may or may not have a good outcome, is not going to be worth it. Have you considered just purchasing a used bridgeport (or cheaper chinese knock-off) mill? I'm not sure where you're located, but I just did a quick search on craigslist and in my area there are at least 5 ad's for 2500 USD or less, and a couple for 1200 USD or less. I would seriously consider what you want to do with your mill, and decided when it's worth spending the money on a proper machine, and until then, you can always outsource your milling work to a local shop or even an online company like rapid-machining. </p>
<p>Best of luck to you!
-Steven</p>
| 10300 | Are there solutions for coupling a spline shaft to a round diameter shaft |
2016-06-14T23:16:16.193 | <p>I am learning about Variable Reluctance ( VR ) Stepper Motors. As stated in my reference, they "have no detent torque." What does this mean?</p>
| |torque| | <p>Detent torque is the torque while coils are not energised. Permanent magnets and the iron stator hold the motor in a given position without any power needing to be applied, although this force is usually much lower than the full torque - in my experience about 10%.</p>
| 10309 | What is 'No Detent Torque'? |
2016-06-15T14:03:34.193 | <p>How do I recognize a good Hydraulic Servovalve? It is probably the frequency response but what do I have to look for in these diagrams?</p>
<p>Is it correct to assume that the higher the frequency at a given amplitude ratio (i.e. -6dB) the better the dynamic behaviour?</p>
<p>What about the phase shift?</p>
| |hydraulics| | <p>You can consider the frequency response as response of a low pass filter. </p>
<p>For one rough example, </p>
<p>When cut-off freq(freq at -6dB) is 50Hz, If you want to control the valve as 100Hz speed(roughly open and close in 0.01 second), the valve is not suitable for your application, because the valve will filter your input from near 50Hz. </p>
<p>If you want to control the valve on 5Hz(open and close in 0.2 second), the valve will be sufficient, in terms of valve response, because 5Hz is in the bandwidth of the filter.</p>
| 10314 | How do I recognize a good Hydraulic Servovalve? |
2016-06-15T14:47:56.670 | <p>Gypsum is used to prevent immediate stiffen and reduce heat loss of cement. If gypsum has already been added into cement,do we still need to undergo curing process after that?</p>
| |civil-engineering|construction-management| | <p>This a bit late but I will add to the answers already given, with some additional technical and historical perspective.</p>
<p>Portland cement contains four Bogues compounds - C3A, C2S, C3S and C4AF. Of these, one is not hydraulic - C3A. C3A (tricalcium aluminate) was introduced to Portland cement during the 1920's as it was found to reduce the heat required for production, thereby reducing the cost of manufacture. It was <em>not</em> introduced for engineering purposes, as some may suggest. Unfortunately, as it is not hydraulic, C3A will dissolve with time if immersed in water, and it is susceptible to sulphate attack.</p>
<p>During hydration, however, C3A produces about 3 or 4 times more heat than other Bogues compounds - leading potentially to flash setting and cracking problems in the hardened concrete. Accordingly, gypsum, or CaSO4 (calcium sulphate), is inter-ground with the clinker to control this flash setting.</p>
<p>The addition of CaSO4 (particularly the sulphate portion), however, is again detrimental. Accordingly, Portland cement standards limit the maximum SO4 content. Thus we see that the addition of C3A necessitates the addition of gypsum and both are detrimental to long term performance, in my view.</p>
<p>Some will argue that the addition of C3A will have benefit with respect to binding chlorides and resisting steel corrosion. I have not found this to be the case in my projects and research. But this view is subject to some controversy. I prefer to eliminate C3A and gypsum when possible but this will increase the cost of Portland cement.</p>
<p>Now we return to your core question: adding C3A, which will also require the addition of gypsum, will <em>make the concrete more sensitive during construction and will require greater attention to curing, not less</em>.</p>
| 10315 | If gypsum has been added into clinker during the production of cement,is curing process still needed to be carried out? |
2016-06-15T21:52:51.107 | <p>My translation:</p>
<blockquote>
<p>Internal pipe diameter inspection</p>
<p>This kind of inspection is carried out on a by-order basis using a certified gauge (for sizes 40 mm and above) of the go/no go type, with both sides of the pipe subjected to measurement along the length specified in the regulatory documentation. Pipes intended for use in shock-absorbers are inspected using the special instrument called a bore gauge. Instrumental inspection of internal pipe diameter is carried out for <strong>critical pipes</strong> and relies on the use of ultrasonic equipment.</p>
</blockquote>
<p>The original Russian phrase is "трубы ответственного назначения" and could be translated as "pipes used in critically important roles (applications)": when a malfunction may lead to injuries, deaths or major industrial accidents, or to the shutdown of the whole plant, with the resulting financial losses.</p>
<p>Is the phrase "critical pipes" a proper choice, or are there some other phrases to signal the same meaning? Would "critical duty pipes" fit?</p>
<p>The word-for-word of the original is "truby (pipes) otvetstvennogo (of high responsibility) naznacheniya (purpose)".</p>
| |terminology| | <p><a href="http://www.dictionary.com/browse/critical?s=t" rel="nofollow">Critical (meaning 8)</a>: <em>of essential importance</em></p>
<p>"Critical pipes" sounds fine to me, meaning "the ones which are absolutely essential". It doesn't clash with any engineering meaning (which is, I assume, your reason for asking here).</p>
| 10318 | Is "critical pipe" a proper term for a pipe used in a critically important role? |
2016-06-17T09:33:47.453 | <p>In a recent incident in London</p>
<blockquote>
<p>Network Rail said an empty train had travelled past a red signal, which resulted in an automatic derailment. No one was injured. <a href="https://www.theguardian.com/uk-news/2016/jun/16/derailed-train-causes-severe-disruption-at-paddington-station">[link]</a></p>
</blockquote>
<p>The derailment has caused quite a bit of damage, and a lot of travel disruption along this track.</p>
<p>My reading of the National Rail statement is that the derailment was a feature of the system, a response to the signal being passed at danger. While I'm sure it caused less damage than a train collision, it still seems dangerous and expensive. </p>
<p>Things like <a href="https://en.wikipedia.org/wiki/Train_stop">train stops</a>, to trigger the brakes exist, or one could imagine diverting the train into a sand trap. Why weren't options like these used instead of derailing?</p>
| |safety|rail| | <p>The short answer is because it's cheaper and there's less loss of life. </p>
<p>Let's look at a simple example. Now, I only really know about US trains, so forgive me if some of this doesn't translate exactly correct. </p>
<p>First a train can only go where the rails are. It's not an automobile where you can pull off to one side. </p>
<p>Usually there are "side lines" that allow trains to pass, or allow cars to sit without blocking traffic. </p>
<p>Signaling is two fold (mostly). There are the lights next to the track that signal, essentially, that the next segment of track is clear and to proceed. There are "in cab" signals in the engines that signal that the next segment of the path is clear and to proceed. </p>
<p>Different areas have different coverage with signals. Some well traveled lines have the nice in-cab automated signals. Some areas have the lights by the side of track. </p>
<p>Signals are places far enough apart that (with speed limits) a train should be able to stop before sitting the next segment</p>
<pre><code>====|==<T=|=====|=<T==
G R G
</code></pre>
<p>In that figure the R (red) signal should stop a train from entering the section of track that is occupied by a train. The G (green) signals that the next segment is clear. </p>
<pre><code>====|<T==<T|=====|====
G R G
</code></pre>
<p>This figure is an accident waiting to happen. The second train WILL run into the first (as far as the system is concerned).</p>
<p>Next, there are many safety system in place to make sure that the rails stay clear and there are no accidents. </p>
<pre><code> =================================\=/==========================
====|====T>=====|===========|=====X=====|=========|==<T=======
R R R R R
5 4 3 2 1
</code></pre>
<p>Let's say this is a normal track segment and were looking at the signals from the train headed left. </p>
<ol>
<li><p>First, were all ready in an "high alert situation" because (for our example) <code><T</code> should not be on that track. It's true it could divert at the junction but (for this example) that's not normal, the trains are too close and something is already wrong.</p></li>
<li><p>Signals are turned all Red. Meaning STOP NOW. Auto brake systems kick in. </p></li>
</ol>
<p><code>
=================================\=/==========================
====|====T>=====|===========|=====X=====|===<T====|===========
R R R R R
5 4 3 2 1</code></p>
<ol start="3">
<li>On crap, <code>T></code> stopped, but <code><T</code> isn't stopping. </li>
<li>Divert <code><T</code>!!!! (note sometimes the diversion isn't possible)</li>
<li>Evacuate <code>T></code> (note, not possible unless it's a passenger train)</li>
</ol>
<p><code>
=================================\=/==========================
====|====T>=====|===========|===<TX=====|=========|===========
R R R R R
5 4 3 2 1</code></p>
<ol start="6">
<li>OMG, It didn't divert. </li>
<li>Call emergency crews, get them to <code>T></code></li>
<li>A crash WILL happen. Do everything you can to minimize damage.</li>
</ol>
<p><code>
=================================\=/==========================
====|====T>=====|===||==<T==|=====X=====|=========|===========
R R R R R
5 4 3 2 1
</code></p>
<ol start="9">
<li>Auto derailment at ||</li>
<li>No lives lost, Damage in money, crappy press, but no loss of life. Clean up crew is a repair crew not a hearse.</li>
</ol>
<p>Some notes:</p>
<ul>
<li>Usually, an engine under power would not have this problem. They are basically set to stop at any serious failure. </li>
<li>Wagons may have this problem. A train can be some loose wagons that got away from a rail yard. The brakes on wagons are like on semi-trucks it takes air pressure to disengage them, with no pressure they should engage. So if there moving without an engine there are no brakes.</li>
<li>Trains usually take a very long time to get up to speed, and a very long time to stop. Miles, not feet.</li>
<li>When trains are running in the same direction they can run closer together. The trailing train just has to be going slower than or at the same speed as the leading train.</li>
<li>There is always enough distance between trains to stop, but in a crammed line, stopping may take several segments, and can be done at the same time. Train 1 slows and Train 2 slows. This allows more trains per line.</li>
<li>With two trains going towards each other on the same line, you are supposed to keep enough distance between the trains that both can stop if the junction is missed.</li>
</ul>
| 10335 | Why would a train automatically derail if a signal is passed at danger? |
2016-06-17T10:09:59.127 | <p>My translation of <a href="http://www.percey.ru/upload/pntz_tech_catalog.pdf" rel="nofollow noreferrer">a document published by the Pervouralsk New Pipe Plant</a>:</p>
<blockquote>
<p>Inspection of pipe-end root face (“face ring”)</p>
<p>Inspection of this type is carried out as required by the regulatory documentation using a ruler or a gauge.</p>
<p>Bevel angle inspection</p>
<p>Inspection of this type is carried out as required by the regulatory documentation using a [<strong>protractor</strong>? <strong>goniometer</strong>?] or a gauge.</p>
</blockquote>
<p>What tool do we usually use to inspect bevel angles at the end of a pipe?</p>
<p>The Russian word used is "угломер", literally "angle-measurer". According to Multitran, it could be translated in dozens of ways.</p>
<p>(I'm translating the word "truba" as pipe despite some saying that "tube/tubing" is a more proper word: the reason is, the plant uses "pipe" throughout its English-language website)</p>
| |terminology|piping| | <p>That mill mostly makes oil field pipe (OCTG ); at least when I visited years ago. They pretty much follow API standards which are the standards of world oil production. So you want API 5 CT, and 5 L, maybe 5C4. (Name may have changed ,but it will be same technology ). Sorry , but other answers are misleading. It may be helpful to get the GOST ( Russian spec) to compare to API to look for variations.</p>
| 10336 | What tool does one use to inspect pipe bevel angles? |
2016-06-17T18:53:02.733 | <p>Why are nuts 6-sided? Why not 4 or 8?</p>
<p><a href="https://i.stack.imgur.com/YA5ZH.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/YA5ZH.jpg" alt="standard 6-sided nut"></a></p>
| |mechanical-engineering| | <p>There are 3, 4, 8, and 12-sided nuts, and most exotically, 5-sided nuts all in use for specific applications, but the 6-sided hex nut is most popular because it offers a good trade off between a bunch of factors:</p>
<p><strong>Ease of tightening/loosening, especially in tight spaces:</strong>
There are six distinct orientations you can grip a hex nut in, each 60 degrees apart. This means that if you can swing your wrench through a 60 degree arc, you can tighten the nut enough to remove the wrench and place it on the next pair of flats. In fact most open end wrenches have the jaws offset 15 degrees form the handle so that by flipping the wrench you can tighten the nut with as little as 30 degrees of access. (Box wrenches often have 12 points, allowing the same principle without the flipping.) A four sided nut would have a significant drawback that even with an offset wrench head, you would need 45 degrees of access to tighten the nut. Additionally, it's much easier to put an open end wrench on a hex nut as the two adjacent faces on the approach side guide the wrench towards the flats. By contrast, if you approach a square nut with the wrench flats not quite parallel to the nut flats, the wrench can very easily jam. Of course when there is space for access with a ratchet these considerations don't make much difference, but as people try to make machines smaller and lighter, designers often leave only the minimum access required to reach fasteners.</p>
<p><strong>Torque transmitted</strong>
The more points the nut has, the more likely it is that the bearing surface of the point will fail in compression and 'round over.' This problem is exacerbated by the fact that some gap has to be left between the wrench and the nut to allow for manufacturing and alignment tolerances, which significantly decreases the contract area between the wrench and the nut. In practice, it's pretty hard to round over a hex nut unless you are using an adjustable wrench (where slop in the mechanism can lead to a bigger gap between the nut and the wrench and less parallel wrench faces.) A square nut is even harder to round over, and in this aspect is more desirable than a hex nut. 12-point nuts do exist, but if the faces were flat they would round over very easily, so instead they are manufactured with points that increase the size of the torque-transmitting face.</p>
<p><strong>Joint Properties</strong>
Standard square nuts have a larger bearing area than their standard hex counterparts. This sometimes makes them preferable for a connection to a soft substrate like wood as they are less likely to pull through the material. Because square nuts can accommodate a greater gap between the wrench and the nut, they can be used in captive situations like weld-on cage nuts, or self-aligning server rack nuts. One common use for square nuts is on blind, low strength connections to soft wood. The bolt can be tightened from one side and the nut will embed itself in the wood, with the large faces preventing it from cutting a counter-bore into the wood. By contrast a hex nut would typically cut a circle in the wood and spin. Oversized hex nuts, thick washers, and tension control bolts all allow alternate solutions to these problems using hex nuts, but generally at a higher manufacturing cost.</p>
<p><strong>Material Efficiency</strong>
In order to maintain the minimum amount of material between the threads and the edge of the nut, square nuts have a fair amount of 'wasted' volume in the corners that increases the amount of metal per nut for a given screw size. As you increase the number of flats, the shape of the nut becomes closer to a circle and more material efficient. (By the same logic, when laying out a connection hex nuts take up less space whereas square nuts have to be placed further apart.) For a few nuts, this difference is negligible, but for a company that buys or produces nuts by the truckload, it would add up very quickly.</p>
<p><strong>Ubiquity</strong>
Even if these trade-offs changed, the simple fact that most people who work on machinery and structures have tools designed around 6-sided nuts would provide a big dis-incentive to change. Auto mechanics, for example have to buy a whole special set of tools when a car manufacturer decides to use an unusual fastener. In the case of the 5-sided nut, the very reason it is used is to be tamper resistant because very few people own wrenches that fit it, and adjustable tools designed for even-numbered polygons won't work. It is mainly used for fire hydrant fittings and valves.</p>
| 10349 | Why are nuts 6-sided? |
2016-06-18T16:20:02.683 | <p>Let's say that the engine is spinning at 7000 rpm and the transmission is spinning at 2000 rpm.</p>
<p>How do you calculate the rpm at which the two will meet when the clutch pedal is released and how much rpm will be transferred each second?</p>
| |mechanical-engineering|automotive-engineering| | <p>You need a speed-vs-torque curve for the engine, which will generally have a negative slope. Then you need a speed-vs-torque curve for the transmission and whatever load is connected to it, which will generally have a positive slope. Plot both curves on the same graph, and the point where they cross is the RPM you'll have when the clutch is engaged.</p>
<p>If either of the curves is not fixed — e.g., throttle changes on the engine, or tire slippage on the transmission, the problem becomes much more difficult.</p>
<p>In order to determine how the RPMs of the two parts change <em>as the clutch engages</em>, you'll need information about the moments of inertia of the various parts, along with data about how the clutch pressure is changing with time.</p>
| 10363 | Car clutch rpm transfer |
2016-06-18T20:58:48.550 | <p>I'm working on building a 3D Zoetrope like <a href="https://www.youtube.com/watch?v=RjSxrVXsfVM" rel="nofollow">the Disney Zoetrope in this video</a>. The part that has me stumped is the bearing. I bought a lazy susan bearing but it's a bit loud, doesn't have any mount options, and doesn't spin as freely as I'd like. </p>
<p>I also see turntable bearings that seem like essentially the same thing but with better mounting options. My worry with those is their size. I have a 4 foot platter and most bearings are quite small, which is why I went with the lazy susan bearing at 24" diameter. </p>
<p>Thrust bearings have also come up but the ones I've seen don't have any mount options. They seem constructed exactly as I need them with the bearings being sandwiched vertically (top to bottom), instead of horizontally between an outer and inner ring.</p>
<p>I've also been reading that certain ball bearings have a a radial and/or angular thrust capacity. And that this may provide the rotation I need, though again the mounting options seem really limited for a turning table.</p>
<p>I've been googling for days and seem to have a mental block. I don't know how I would stabilize a 4 foot platter if the bearing is small and centered, like the turntable bearings I found. So, what kind of bearing would be best for this? How would I mount it (if its not super obvious)?</p>
<p>I'm thinking of using a pulley to drive the zoetrope, though I'd prefer a direct connection with the motor drive shaft in the middle of the platter.</p>
| |mechanical-engineering|bearings| | <p>I would suggest getting an unbraked hub assembly intended for a lightweight trailer. You can get the stub axle, bearings and hub as a set and it gives you a convenient flange mounting for your main platter. As long as the platter itself is reasonably stiff and well balanced it shouldn't need any additional support. </p>
<p>I recently used a similar arrangement for a rotating plinth for a trio of bronze and glass statues with a total rotating mass of around 150 kg with a diameter of 1.0 metres. <a href="https://i.stack.imgur.com/GQXYX.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/GQXYX.jpg" alt="turntable "></a> In this case I just cut the stub axle to the required length and welded it to a 6mm thick steel plate which was bolted into a corresponding plate in the framework of the base, this allowed it to be squared with shims. Note that the castors visible are for moving the whole thing and not related to the rotating mechanism. I was able to stand on the end of one of the rotating arms without any problem. </p>
<p>For the drive you could have a pneumatic or rubber wheel directly driving the underside of the platter. This also gives you to opportunity to fine tune the gear ratio according to where on the radius you put the drive wheel. This is probably the easiest drive solution to implement. You would probably want the motor and drive wheel on a sprung or counter-weighted pivot a bit like the belt drive on a lathe. Here a motor with an integrated gearbox would be make life easier as you won't need to provide a large reduction ratio. </p>
<p>Alternatively you could fix a pulley to the hub itself (note that the axle is fixed and it is the hub which rotates). </p>
| 10368 | Quiet Turntable Bearing for 3D Zoetrope |
2016-06-18T23:49:12.970 | <p>What are some basic, open source aerodynamics or CFD simulation software that can import Blender or Sketchup models? I am looking for software that is easy-to-use, free, and suitable for educational purposes.</p>
| |fluid-mechanics|aerospace-engineering|simulation|computer-aided-design|aerodynamics| | <p>CFD requires a mesh of the model. To my knowledge, blender and sketchup do not do this. You may need an intermediary mesh generator to realize any cfd simulations, freeware or otherwise. Beware that mesh generators produce a plethora of different file formats which may or may not be readable in your cfd simulator of choice.</p>
<p>If you're looking for a good open source cfd simulator, i'd recommend OpenFOAM. Beware: it is neither simple to understand or use and i don't recommend it unless you really have a good grasp of fluid dynamics snd numerical discretization and solver methods</p>
| 10371 | Simple, open source aerodynamics simulation software that can import Blender or Sketchup models? |
2016-06-19T11:38:42.577 | <p>On the pictures below is a tool called "bridge cam gauge", mentioned <a href="https://engineering.stackexchange.com/a/10352/349">in this answer</a>. I wonder why the words "bridge" and "cam" are used in its name. </p>
<p>I <a href="https://en.wiktionary.org/wiki/cam#Noun" rel="nofollow noreferrer">found a definition</a> of "cam" that might be relevant:</p>
<blockquote>
<p>A curved wedge, movable about an axis, used for forcing or clamping two pieces together.</p>
</blockquote>
<p>As I understand, the 'cam' here is the flat roundish beaky plate that you can rotate about its hinge. </p>
<p>But what about "bridge"? <a href="https://en.wiktionary.org/wiki/bridge#Noun" rel="nofollow noreferrer">Wiktionary offers a variety of senses</a>, which makes it hard for a non-native speaker of English. </p>
<p><a href="https://i.stack.imgur.com/UMgy9.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/UMgy9.jpg" alt="enter image description here"></a></p>
<p><a href="https://i.stack.imgur.com/PjDAG.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/PjDAG.jpg" alt="enter image description here"></a></p>
<p><a href="https://english.stackexchange.com/q/333286/48335">Also asked on ELU SE</a>.</p>
<p>P.S. It is also called "cam type weld gauge"</p>
| |terminology| | <p>See here <a href="http://www.newmantools.com/gauge/wghowto.htm#wg4" rel="nofollow">http://www.newmantools.com/gauge/wghowto.htm#wg4</a> (gauge type WG-4) for how it is used. </p>
<p>It works the same way as a cam. The rotating part (marked "undercut or reinforcement" in your second picture) has a pointed end (marked with the arrow) that "follows" the profile of the parts to be welded like the follower on a conventional cam, as in the graphic in your Wikipedia link.</p>
<p>I think the "bridge" part just means that the gauge has two "feet" that are in contact with the part being measured, and "bridges the gap" between them. See <a href="http://marinenotes.blogspot.co.uk/2012/08/sketch-and-describe-bridge-gauge-how-is.html" rel="nofollow">http://marinenotes.blogspot.co.uk/2012/08/sketch-and-describe-bridge-gauge-how-is.html</a> for a different type of "bridge gauge", which doesn't have a cam.</p>
| 10377 | What is the etymology of "bridge cam gauge"? |
2016-06-19T13:42:21.757 | <p>Why do we design foundations at SLS and not ULS?</p>
<p>If we use ULS wouldn't we be taking a higher factor of safety?</p>
<hr>
<p>SLS = Serviceability limit state. It considers how the structure is actually expected to behave, generally by using partial factors of 1.</p>
<p>ULS = Ultimate limit state. It considers complete structural failure, utilising partial factors to increase loads and decrease strength.</p>
| |civil-engineering|foundations| | <p>No. In fact, SLS frequently leads to a higher factor of safety than ULS.</p>
<p>This is because ULS deals with whether or not the structure will collapse. It does not care whether the structure will stand by generating absurd deformations, massive cracking (in the case of reinforced concrete structures), or if it will vibrate like a guitar string whenever you take a step.</p>
<p>SLS, on the other hand, does care about such things. It makes it so that your structure must not only stand upright, it must also not have excessive deformations, cracks or vibrations. This frequently means that a larger cross-section or more steel reinforcement is necessary to satisfy SLS than would be necessary according to ULS. This therefore leads to a larger safety factor.</p>
<p>Obviously, this isn't always the case. After all, ULS uses a reduced material strength and increased loads, while SLS uses the characteristic values for everything. Though SLS's underlying assumptions are quite conservative, in many cases ULS is still the controlling factor.</p>
<p>Foundations, however, deal with soil. The main concern here tends to be how the soil will behave under load. Once fully loaded, the structure will sink into the soil due to the applied compression. That is an unavoidable fact. The question is whether your structure will sink by one meter or one millimeter. That is the sort of question that is best answered by SLS. After all, so long as the structure is still standing, sinking by one meter is fine according to ULS.</p>
<p>And it's not as if foundations have no factor of safety. In fact, theirs is usually the highest safety factor in the entire structure. One usually defines that the soil's safety factor (ultimate strength divided by applied stress) must be greater than 2, at the very least. Concrete structures (other than their foundations) usually have a total safety factor of around 2, and steel structures of around 1.7.</p>
| 10379 | why are foundations designed at SLS |
2016-06-19T17:28:00.140 | <p>I have a heating system with which I heat up 10 liters of water by 24.2°C in 610 seconds.</p>
<p>My system consumed 0.5 KW of electric energy in this time.</p>
<p>Now I used this formula:
Specific heat of water * volume * temperature difference / time</p>
<p>and I get 1,66 KW (created)</p>
<p>efficiency 332%</p>
<p>Is this correct? Or must I use some other formula to calculate efficiency?</p>
| |heat-transfer|energy|energy-efficiency|heating-systems| | <p>You are confusing units, which is making your question difficult to understand.</p>
<p>Electrical energy is measured in kW-<em>hours</em> (or Joules), not kW. If your heater consumed 500 W of <em>power</em> for 610 s, then you used 305 kJ of <em>energy</em> to heat your water.</p>
<p>Similarly, if 10 l (10 kg) of water was raised by 24.2 °C, then it absorbed</p>
<p>$$4.186 \frac{J}{g ^\circ C} \cdot 10 kg \cdot 24.2 ^\circ C = 1013 kJ$$</p>
<p>of <em>energy</em>. Clearly, something is wrong with your description of what actually happened.</p>
<p>If the heater actually used 0.5 kWh of <em>energy</em> (i.e., 2.95 kW for 610 seconds), this corresponds to 1800 kJ, which would indicate that the process is only about 1013/1800 = 56.3% efficient. You lost quite a lot of the incoming energy in the form of heat that ended up somewhere other than in the water.</p>
| 10383 | Calculate efficiency of water heating |
2016-06-21T14:17:18.867 | <p>I have an idea for a sports equipment innovation "a new product" that a track and field athlete can use to practice and train his sport. The technology involves a laser beam. I'm wondering if my idea is feasible and/or if it has already been done because we think the idea is good. </p>
<p>The idea is that when training pole vault or height jump, the athlete uses a laser beam instead of a bar, and the laser beam measures if the athlete succeeds. Has it already been done or is it not feasible?</p>
| |mechanical-engineering|design|product-engineering| | <p>It's certainly feasible. I don't know if it has been done. I don't think it's a great idea, for the following reasons:</p>
<p>1) What would the athlete be aiming for? It won't be as visible.</p>
<p>2) In the actual sports, you're allowed to touch the bar, as long as it doesn't fall.</p>
<p>3) You may be affected (put-off) by the presence of a physical bar in a competition, when you trained without one.</p>
<p>It would potentially shorten the time between attempts, but it would be less representative of the real thing.</p>
| 10414 | Feasibility of product idea |
2016-06-21T14:37:41.353 | <p>I am working with a piece of equipment that has a vacuum sealed chamber.
The outer wall (ring) of the chamber is shown in the image below.
This ring is bolted to a bottom plate and a top plate is raised up and down using pneumatics.
When the chamber is sealed the differential pressure between the inside and outside is nominally maintained at 10 psid.
The system also has a laminar flow meter and once vacuum is pulled and the chamber is for the most part isolated, we can read about 1.5cc/min flow (leakage) in the chamber.
Maximum flow readings for all other system components other than the chamber seal is below 0.0.03 cc/min, so I am convinced the issue is with the sealing of the chamber.</p>
<p>In the image can be seen a gland for an o-ring. This is an existing design, not of our own, so only minimal alterations to the design may be allowed.
We will be looking at the flatness and finish on the top plate, but there has always been some variability on leakage from cycle to cycle.</p>
<p>So, this is my question...
<strong>What is the likelyhood that adding a second o-ring in series with the first, with a minimal gap between them, may reduce the total leakage rate of the chamber?</strong></p>
<p><a href="https://i.stack.imgur.com/U4Lhu.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/U4Lhu.jpg" alt="enter image description here"></a></p>
| |mechanical-engineering|vacuum|seals| | <p>I'm assuming that the piece with the groove and the mating pieces are metal or a dense molded plastic. If this was a 3d printed piece, I'd be worried about porosity of the parts too. </p>
<p>A second O-ring won't do much for you honestly. If your leak rate is that high, I'd question the design of the O-ring groove and O-ring selection. Does the O-ring sit proud of the groove when installed? If so, by how much? How much stretch is in the O-ring when installed? What's the ratio of the groove cross section to the nominal cross section of the O-ring when installed? </p>
<p>A common mistake is to design such a groove based on the nominal cross section of the O-ring and not accounting for the fact that this shrinks as the O-ring stretches. </p>
<p>The mating surfaces must be relatively flat, in order to control the gap. if there's an excessive gap and the O-ring material is too soft, the pressure can push the O-ring material into the gap and cause leaks. The best resource I've found for designing this type of static seal interface is the <a href="http://www.parker.com/portal/site/PARKER/menuitem.6c94058cc6466c6315731910237ad1ca/?vgnextoid=a141a35c7575e210VgnVCM10000048021dacRCRD&vgnextfmt=default" rel="nofollow">Parker O-Ring Handbook</a> </p>
<p>I'd also like to ask; Why use an O-ring? Why not a flat gasket? Or even a liquid gasket replacement like Loctite 510 or one of the hundreds of other sealants?</p>
| 10415 | Series O-ring in Vacuum Application |
2016-06-21T23:27:49.040 | <p>I'm writing code for a microcontroller that will ramp up the speed of a stepper motor as quickly as possible, for a jig that I'm building, that moves a workpiece from one position to another along a linear rail.</p>
<p>This question is not about modifying the mechanical or electrical system around the stepper motor, but simply the rate at which the stepper motor is stepped.</p>
<p>Given a desired speed, it will ramp up the speed of a stepper motor (and so the speed of the workpiece) from stationary. (The stepper motor will not have any kind of feedback to the controller other than a switch at each end so it can periodically establish its position, and stop if it's lost/gained too many steps in one direction).</p>
<p>So, I'll need to implement/write a function which gives the optimal speed $s$ (revolutions per second) at time $t$ for maximum acceleration. (Where $s(t=0)=0$) which the microcontroller will use to determine how fast to accelerate the workpiece / stepper motor.</p>
<p>The function might look like this: </p>
<p>$$s = kt$$</p>
<p>Where $k$ is an optimized constant, $t$ is in seconds, $s$ is revolutions per second.</p>
<p>The maximum maximum acceleration will be specified by choosing a value of $k$. It will be determined by experiment rather than calculation.</p>
<hr>
<h2>What I want to know is:</h2>
<ul>
<li><strong>One (or more) formulas that I might use. I would then attempt various contants, varying them so I find values that result in good acceleration.</strong></li>
<li><strong>A strategy for determining those constants.</strong></li>
</ul>
<hr>
<p>E.g. using the above formula: </p>
<ol>
<li>start with some value of $k=k_0$</li>
<li>Attempt #1: Run the motor with that value of $k$, starting from $t=0$. Record the speed at which the motor failed to accelerate.</li>
<li>Attempt #2: Double $k_0$, call it $k_1$. Repeat step 2.</li>
<li>Attempt #3: Use half of $k_0$ as $k_{-1}$. Repeat step 2.</li>
<li>If $k_1$ gives the highest speed, double it as $k_2$.</li>
<li>If $k_{-1}$ gives the highest speed, half it as $k_{-2}$.</li>
<li>If $k_0$ gives the highest speed, followed by $k_1$, take the average as $k_{0.5}$</li>
<li>If $k_0$ gives the highest speed, followed by $k_{-1}$, take the average as $k_{-0.5}$</li>
<li>Repeat step 2.</li>
<li>And so on</li>
</ol>
<h2>Further thoughts</h2>
<p>I think the above formula might work well assuming there is the equivalent of a flywheel attached to the motor, and:</p>
<ul>
<li>there is zero friction</li>
<li>the motor exerts constant torque at any speed</li>
</ul>
<p>Is that correct?</p>
<p>Similarly assuming:</p>
<ul>
<li>the motor and system has zero mass / inertia</li>
<li>the motor exerts constant torque at any speed</li>
<li>the only resistance to movement is friction</li>
</ul>
<p>Then the motor can still accelerate to full speed immediately. (So $k$ could be infinite). And if there is inertia, assuming friction forces are constant, (ie independent of speed), assuming I'm correct so far, then the formula is still good; $k$ will just be a lower value.</p>
<p>But that leaves questions:</p>
<ul>
<li>What is a typical relationship between stepper motor speed and torque? Clearly it tails off as speed increases, otherwise there would be no limit to its speed. So it clearly won't be taken into account by the above formula. I am investigating <a href="http://www.orientalmotor.com/technology/articles/article-speed-torque-curves-for-step-motors.html" rel="nofollow">this link</a> which goes into much detail, but perhaps there are factors which are more significant than others, or there are more basic models that give good approximations.</li>
<li>What other factors could be significant that would can not be modeled within the above formula?</li>
</ul>
| |mechanical-engineering|applied-mechanics|stepper-motor|optimal-control|acceleration| | <p>One way might be to try to model and understand the physical problem. You are however clear in your question that you would prefer an empirical approach where constants are tuned to experience data.</p>
<p>I would suggest that the easiest and most general way is to use a piece-wise approximation of the velocity curve. Generically this may be made by using <em>basis</em> functions $b_i(t)$ such that the velocity $s$ at any time is given by</p>
<p>$$ s(t) = \sum_{i=0}^{N}b_i(t)$$</p>
<p>A simple, and in this case likely sufficient, approach may be to use piese-wise linear velocity (corresponding to constant acceleration over a time interval). That is, divide the time into a finite number of time steps $t_0, t_1, ..., t_N$ and at each such time define a <em>parameter</em> $v_i$ which is the velocity that should be used at that time.</p>
<p>Your algorithm for the engine could then, for example, at any given time $t$ do a <em>bisection search</em> in your $t_i$ array and do <em>linear interpolation</em> between $(t_j, v_j)$ and $(t_{j+1},v_{j+i})$ to calculate the output $s(t)$ at any time.</p>
<p>You are likely to find that you do not need to divide the timeline into more than a few (order of 10 steps and thus order of 10 parameters) pieces in order to achieve a solution that is acceptably close to the optimal. You may also experiment with having non-constant time steps such that for example the time step increase when the time increases and there is less change in the acceleration the system can achieve.</p>
<p>You can also see <a href="https://calcopedia.com/s/demo/accelerationofamass" rel="nofollow">this sample example</a> for an illustration of how close to an optimal solution it may be possible to get only with a few (10) time steps.</p>
<p>The additional advantage of such an approach is that you can easily tune your velocity in one part of the process without having to adjust the parameters to maintain the response at other times. This leads us to a suggested approach for tuning the parameters:</p>
<ol>
<li>Use a constant acceleration of the entire range ($v_i$ increase linearly) and tune it such that you can reach close to the terminal velocity (see the sample).</li>
<li>Tune each $v_i$ parameter individually, that is, first adjust $v_1$ such that you <em>just about</em> can sustain that acceleration in the time interval $[t_0, v_1)$.</li>
<li>Next you can do the same process for $v_2$ etc. etc. (and notice that once you have configured $v_i$ you can go on to tune $v_{i+1}$ without ever having to go back and adjust $v_i$</li>
<li>You have now constructed a linear approximation of the optimal velocity.</li>
<li>If you have areas where there are large changes, or the given velocity is ok in the start of a block but fails in the end you can always refine the "mesh size" in that area by introducing one additional $(t,v)$ pair.</li>
</ol>
<p>The tuning of each parameter based on experiment may be based on a procedure as you suggest, i.e. doubling the $v_i$ if you encounter no failure, else bisecting between the highest <em>know ok</em> and the lowest <em>known not ok</em>.</p>
| 10421 | Optimising driving speed of stepper motor for maximum acceleration by trial and error |
2016-06-23T16:20:16.697 | <p>I'm designing an environmentally sealed chamber which will need to have its oxygen level controlled. </p>
<p>The chamber has a volume of roughly 1.5 liters. I don't expect pressure to vary too much ... maybe .8 to 1.2 bar. I'm not sure about flow rates, but we would want fairly fine control over the O2 concentration, so maybe a maximum of around .1 L/s?</p>
<p>To that end, our intent is to have a supply of nitrogen (or other inert gas) and a supply of oxygen. I've never worked with gas canisters of any sort before, but I need to be able to feed controlled amounts of nitrogen and oxygen into the chamber as well as control a relief valve of some sort, all electronically. What sort of valves/controls would I need to allow a controlled amount of gasses in/out of the chamber?</p>
| |design|compressed-gases| | <p>It sounds like you might be looking at a similar setup to purge gas for arc welding. </p>
<p>Here you would typically a <a href="http://www.r-techwelding.co.uk/argon-regulator/" rel="nofollow">regulator</a> which screws into the valve fitting of the gas bottle which provides coarse control for flow-rate and also has a pressure gauge so you can see how much gas is left in the bottle. For fine control applications like TIG welding it is also common to add a <a href="http://www.r-techwelding.co.uk/argon-flow-meter/" rel="nofollow">combined flow meter and valve</a> downstream of the regulator (usually screwed into the regulator outlet) these are typically of a 'peashooter'type in the range of 0-25 litres per minute. In my experience this will allow for straightforward adjustments in the range of +/- 0.5 litre per minute with good consistency. </p>
<p>All of this is manually set but with welding the gas flow is turned on and off by a solenoid valve so you could simply add an off the shelf welder solenoid into the loop and hook it up to your control system.
<em>Note that these solenoid valves can be a bit leaky over time so you may also want a more positive manual isolation valve in the system, although turning off the bottle vale will achieve the same thing</em> . So you preset the flowrate and control the duration of pulses of gas on demand rather than attempt to precisely adjust a constant flow-rate. </p>
<p>This has the advantage that it is all off the shelf and easily available components and the requisite fittings and hoses are readily available through retail outlets. Without knowing your exact requirements it is hard to know if this sort of setup would be ideal but it does have the advantage of being a cheap and convenient way to set up a proof of concept prototype at the least. </p>
<p><strong>One thing to be wary of</strong> is that the reactivity of oxygen usually demands that oxygen specific parts (especially hoses, seals and regulators) are used. But if you are using inert gas to fine tune the mixture this shouldn't be a major problem. Oxygen is routinely used in industry and it's not absurdly dangerous but you should certainly familiarise yourself with the risks and proper precautions associated with oxygen specifically and bottled gas in general as some of the hazards are not intuitively obvious. </p>
<p>This also applies to the chamber itself, any seals, adhesives, hoses etc should be rated for exposure to pure oxygen and you may need to employ a check valve to keep oxygen out of your inert gas lines. </p>
| 10441 | Device (electronic valve?) for controlling O2/Nitrogen levels in sealed chamber |
2016-06-24T04:42:01.133 | <p>In engineering classes we made the assumption that structural supports (fixed, roller, etc.) are rigid. Thus, the displacements of a beam ,for example, at a fixed support are set to zero in the horizontal and vertical directions as boundary conditions. However in real situations shouldn't there be some displacement due to deformation in the support? Wouldn't that affect stresses/strains in both the beam and the support considerably? </p>
| |structural-engineering| | <p>I recently saw a picture on Facebook of a t-shirt quite similar to this one.</p>
<p><a href="https://i.stack.imgur.com/9FjjF.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/9FjjF.png" alt="enter image description here"></a></p>
<p>This is as good a definition of engineering as any. And that's because you are correct.</p>
<p>To say that any support is perfectly rigid is a <strong>lie</strong>, since there will always be some deformation of the support. But, then again, stating that a material is perfectly elasto-plastic (<a href="https://en.wikipedia.org/wiki/Hooke%27s_law" rel="nofollow noreferrer">Hooke's Law</a>) is also a <strong>lie</strong>, since micro-imperfections will always make a member deform in a slightly chaotic (as in, not perfectly predictable) manner. <a href="https://en.wikipedia.org/wiki/Euler%E2%80%93Bernoulli_beam_theory" rel="nofollow noreferrer">Euler-Bernoulli beam theory</a> is a <strong>lie</strong> since it ignores shear deformations. <a href="https://en.wikipedia.org/wiki/Timoshenko_beam_theory" rel="nofollow noreferrer">Timoshenko beam theory</a> (which does consider shear) is also a <strong>lie</strong>, since it considers perfectly elastic material behavior, which we saw above is a <strong>lie</strong>.</p>
<p>There is not a single "theory" (in engineering) which is not a <strong>lie</strong>, and that is simply because engineering doesn't care about providing a <em>perfect</em> description of reality. Engineers just want a theory that's <em>good enough</em>.</p>
<p><a href="https://engineering.stackexchange.com/questions/6020/why-do-we-even-use-engineering-stress">That's why we use engineering stress</a> (as opposed to true stress). That's why we use Euler-Bernoulli or Timoshenko beam theory. And here I'm talking about the fundamental laws governing structural engineering. If we can accept simplifying assumptions here, in the bedrock of our field, we can accept them on a case-by-case basis as well.</p>
<p>And that's why we usually consider supports to be rigid. Because it's a simplification which makes our lives infinitely easier and which is <em>good enough</em>.</p>
<p>Could you create a complex model, where you consider the stiffness of the columns? Sure. But the columns are resting on the foundations, so you'll need to consider the stiffness of those as well, and they are resting on the soil, so you'll need to consider the effect of that as well. There comes a time when you need to decide whether it's worth it. Is the deformation of the supports really going to make a difference?</p>
<p>If you have reason to believe that your supports may be significantly flexible (for example, your foundations are resting on a complex soil), then yes, you absolutely <strong>must</strong> consider the effect that will have on your structure. If, however, there is no reason to believe the support's deformations will be all that relevant, then, well, it's not relevant and there's no reason to consider it.</p>
<p>And, as a detail, in the simple case of isostatic structures (such as simply-supported beams), small support deformations don't cause any stress increase at all. Well, actually, that's a <strong>lie</strong> since it arises from Euler-Bernoulli theory. But if you pretend that Euler-Bernoulli is true (you really should), then yeah, no internal forces emerge.</p>
| 10450 | How can we assume that structural supports are rigid? |
2016-06-25T03:05:52.790 | <p>For those of us in the US, we're used to seeing roadways with yellow stripes separating traffic traveling against our direction of travel, and white stripes separating traffic traveling in the same direction. Of course, <a href="http://mutcd.fhwa.dot.gov/htm/2009r1r2/part3/part3b.htm" rel="nofollow">Chapter 3B</a> of the Manual on Uniform Traffic Control Devices requires this. However, I noticed that countries such as England, Japan, Russia, Australia, etc., don't distinguish these features by color. What particular rationales are there for using the color yellow to separate opposing directions of travel on roadway facilities?</p>
| |civil-engineering|highway-engineering| | <p>The most likely reason yellow is used as a color was probably the result of standardization efforts between the 1960s and 1970s (NCHRP Report 484, p8). Since the United States is a very influential country in North America, most other countries bordering them emulated its system to a degree.</p>
<p>Interestingly, the United States has considered moving to an all-white pavement marking system (<a href="http://onlinepubs.trb.org/onlinepubs/nchrp/nchrp_rpt_484.pdf" rel="nofollow">NHCRP Report 484</a>). The NCHRP (National Cooperative Highway Research Program) conducted a feasibility study for all-white pavement markings, documented as Report 484. At the time the report was conducted, another study was underway (Project 5-18) to determine the effectiveness of yellow pavement markings. The <a href="http://onlinepubs.trb.org/onlinepubs/nchrp/nchrp_rrd_328.pdf" rel="nofollow">summary</a> as well as the <a href="http://onlinepubs.trb.org/onlinepubs/nchrp/nchrp_w125.pdf" rel="nofollow">full report</a> can be found on the Transportation Research Board.</p>
| 10460 | Why do North American Roadways Use Yellow Instead of White to Separate Opposing Traffic? |
2016-06-25T08:14:36.540 | <p>We cast a bridge's concrete deck and kept samples for concrete strength tests at 7 and 28 days.</p>
<p>We took a test at 7 days, and the test result were over the design strength (designed for 30 MPa, result was 33 MPa). The control engineer insisted that we have to also do the 28-day test.</p>
<p>This seems pointless to me, as I believe concrete gains strength during curing, and doesn't lose it. Can the results of the 28-day test be worse than those of the 7-day test?</p>
<hr>
<p>I am asking this question because I want to get paid for the job. The control engineer (who signs the documents) is the one who is insisting that we have to wait until the 28-day test.</p>
| |concrete|strength| | <p>With the exception of anything extremely unusual: the concrete will continue to gain strength until 28 days. In fact, it will continue to gain strength for something like 50 years, but the main strength gain happens early on.</p>
<p>The motivation for waiting for the 28-day test is therefore nothing to do with strength. Far more likely: the later you are paid, the better it is for the main contractor. This helps the contractor manage cash flow, and/or allows them to make interest by having money in their bank account for longer. I have no idea what your contract says in terms of when you get paid, but it is completely normal for the control engineer to pay you as late as he is able.</p>
| 10463 | Can concrete lose strength during curing? |
2016-06-25T21:46:41.830 | <p>I am building a battery discharge system for LiPo batteries which accepts a charged battery and discharges it at a CONSTANT CURRENT regardless of the battery voltage, until a threshold is reached. The threshold is the LiPo storage voltage.</p>
<p>The purpose of the system is to measure actual delivered power under variable battery environmental conditions.</p>
<p>The question I have is regarding the load itself and the feedback system required to vary the load to meet the constant current requirement.</p>
<p>So what is the recommended load? e.g. motor or heater</p>
<p>How could the load resistance be varied in direct proportion to the battery voltage?</p>
<p>Any circuit references would be valuable.</p>
<p>(sorry I'm an ME not an EE)</p>
| |electrical-engineering| | <p>If you want a constant current load, design a constant current load. As you say, this is a circuit that draws the same current over some range of applied voltage. That's your spec right there.</p>
<p>There are various ways to achieve this. The conceptually simple is to get a measure of the current, then use feedback that adjusts a pass element to keep the current constant. Some current sinks do work this way, especially if a separate supply is available to power the control electronics. For example:</p>
<p><a href="https://i.stack.imgur.com/fXEfO.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/fXEfO.gif" alt=""></a></p>
<p>All current entering at SINK goes thru R1. The voltage across R1 is therefore proportional to the current being sunk. Due to how the feedback is arranged, the opamp adjusts the gate of Q1 to maintain the same voltage across R1 as Vref. Vref is therefore the control input that sets how much current to sink.</p>
<p>There are details of stability, bandwidth, and accuracy beyond the scope of a simple answer here. However, such circuits do work and are used in reality. In this case, you'd want a separate power supply for the opamp in most cases. </p>
<p>Choosing R1 is a tradeoff. You want it large for good signal to noise ratio, considering the opamp offset voltage as one source of noise. On the other hand, you want R1 low for lower minimum voltage the current sink can operate at.</p>
<p>Another less accurate but much simpler current sink is:</p>
<p><a href="https://i.stack.imgur.com/UwumW.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/UwumW.gif" alt=""></a></p>
<p>This exploits the property of a bipolar junction transistor where the collector current is largely independent of the collector voltage over a decent range of collector voltages. As before, all of the sink current flows thru R1. In this case, the base current also flows thru R1, but that's small compared to the collector current due to the gain of the transistor. For example, if the transistor has a gain of 50, then about 98% of the current thru R1 is the sink current and the other 2% is the base current.</p>
<p>The voltage on R1 will be Vref minus the B-E drop of the transistor. At first approximation, that can be considered constant, around 700 mV. This kind of current sink is also used in real life when a few percent accuracy and some load on Vref are OK. This is the case often enough to make this a useful circuit.</p>
| 10469 | Constant current load under varying input voltage |
2016-06-26T00:26:13.563 | <p>I'm asking this here because I really cannot find a better place to ask this.</p>
<p>Near my house (in the US), there is a pair of parallel train tracks operated by Union Pacific, primarily for cargo-carrying trains.</p>
<p>Along the track, there are various types of signals (for the trains): some are circular with three lights, some are linear with three lights (like a traffic light), and I think I've seen some with two lights. There's one pole on each side of the track, each with its own lights. Sometimes, on a pole, there are two sets of the lights - one above the other (I.e. Two traffic lights on the same vertical pole, and another identical one for the other set of tracks).</p>
<p>What do these signals mean? Why do they vary? How do I read them?</p>
| |rail| | <p>The signals mean exactly what you think they do: They tell the drivers of trains when it is safe to proceed, and often, how fast is safe. Everyone knows that trains are heavy and can take a long time to stop. Trains must be warned in advance long before they bump into one another so they can stop before that happens. </p>
<p>Railroad signals vary in many ways, even on the same railroad or the same route. There are different classes of signals which serve different purposes. The two main classes are <em>absolute</em> and <em>permissive</em>. Absolute signals require a train to stop and stay stopped when they are "red." Permissive signals allow a train to continue on at a slow, controlled speed. In the United States, these signals are more-or-less identical in form and appearance, except that permissive signals have a <em>number plate</em> indicating the track mile of the signal. Railroad signals, especially in the United States, can have many varied appearances, even of the same class. This is because signaling has evolved over the past 100+ years and different railroads and eras have held different views on how best to do it. Because of this, signals have formed a part of many railroads' identities because are unique to that railroad. Your description of the two different forms of signals near you sound just like you said, an implementation detail (of tradition). </p>
<p>I am the author of the second link about signaling posted by alephzero above. If you still have any more questions, feel free to ask.</p>
| 10470 | Train Signalling - What Do They Mean |
2016-06-27T11:58:14.507 | <p>What kind of resource provides summary data on how given kind of aluminum yields to machining? Are there norms, or such?</p>
<p>Explaining:</p>
<p>Different kinds of metals yield to machining differently; e.g. MO-59 brass has a small addition of lead, which makes it leave a clean, smooth cut; in contrast pure copper will pull, crease, drag into threads; where the milled surface of brass left a mirror shine, copper left a fleece-like mess of metal turned into myriads of tiny wires.</p>
<p>The data for a few materials I worked with, I got through word-on-mouth, or own experimentation. I <em>know</em> resources exist, from which professionals learn this in an organized manner, but I don't even know what keywords to look for.</p>
| |metallurgy| | <p>Usually data sheets will give a qualitative indication of maintainability along the lines of excellent/good/poor/not possible etc. </p>
<p>However you may need to seek out more subjective advice for specific alloys and processes as there are a huge variety of possible applications and there isn't a generally applicable metric of maintainability. For example lots of long stringy swarf may be a major problem for automated production but no more than a mild inconvenience for one-off manual operations. Equally some issues like work hardening may be reasonably easy to mitigate by correct procedures if you are aware of them. </p>
<p>In very general terms many grades of aluminium are at least reasonably machinable and you are much less likely to encounter the severe difficulties that you can get with some types of stainless or high alloy steels. </p>
<p>Similarly some alloys may be easy to rough out but very resistant to taking a fine finish. </p>
| 10488 | Where can I find how good given aluminum is for machining? |
2016-06-27T20:36:51.760 | <p>What is the most common material gas pipes made of?
I read in a unreliable website it's steel, is it true? </p>
<p>Also how good is steel at stopping or slowing down the leaking of gas? </p>
<p>The gas I am using or planning on using is Argon which is denser than air.</p>
| |gas|pipelines| | <p>In my admittedly limited experience working with plumbers, natural gas pipes are typically made from galvanized steel for interior work, or from black steel pipe for underground. Last-leg fittings and tubes which connect the stub-outs to various appliances are sometimes made from copper and brass. Other materials are possible, as noted <a href="http://www.engineeringtoolbox.com/approved-gas-pipes-d_1112.html" rel="noreferrer">here</a>. Some things to consider:</p>
<ul>
<li>When joining two pipes made from dissimilar metals (including steel and cast iron), always use a dielectric union to decrease the risk of galvanic corrosion between the two metals.</li>
<li>In US jurisdictions, exterior pipes must be painted to reduce the risk and rate of weathering.</li>
<li>In the US, natural gas typically contains a small amount of methyl or ethyl mercaptan. The additive has a foul smell which makes leaks readily detectible by most people. These compounds contain sulfur and will react with copper to form copper sulfide, eventually corroding through the pipe. Some copper pipes of types K and L are approved for use with gas. Only properly approved copper pipes should be used with natural gas, with appropriate fittings.</li>
</ul>
<p>For argon, the first point is still relevant, though the last is likely irrelevant if the argon is pure. The second point may be relevant if you need any pipe runs outdoors.</p>
<p>As for leaking, the pipes themselves should not leak, typically it is joints that leak. If you need high pressure, ensure that the schedule of your pipe is sufficiently high (i.e. wall thickness) and that it is rated for the required pressure. As with all pressure vessels, ensure you have appropriate mechanical safety valves and a pressure regulator to decrease the risk of explosion. If you are worried about diffusion through the pipe, unless you plan to heat the argon to many hundreds of degrees C (which you will need specialized pipe for), the diffusion rates will probably be trivial. Focus on the pipe joints first for leak reduction.</p>
<p>If you only need to transport small quantities of argon over short distances, as in a lab setting, copper and brass flare fittings are most likely sufficient if the pressures aren't too high. If you need to transport large quantities, you should consult an professional engineer.</p>
| 10496 | What material is used to make gas pipes? |
2016-06-27T23:09:25.320 | <p>I need to build an air tight system that can hold gas (Argon Gas). I see that most pipes use steel. Is steel tight enough to hold that gas? Argon is very dense, so I think it should have no problem. </p>
| |gas|pipelines| | <p>Just adding on to Mr.Johns's answer, do try to make your threaded fittings according to NPT (National Pipe Thread) or BSPT (British Standard Pipe Taper). Both have a tapered profile which will significantly reduce leakages. </p>
<p>Learned it the hard way here. All the best!</p>
| 10498 | Is steel air (argon) tight? |
2016-06-28T18:28:29.157 | <p>Fire broke out from wing of an airliner after emergency landing. All were safely evacuated.</p>
<p>In this video,
<a href="https://www.youtube.com/watch?v=cKLbjPmOIq4" rel="nofollow">video</a>, fire engines, standing by on run way as pilot have maydayed, rushed pass the airliner, positioned themselves in a 'formation' at up wind position, then, advanced toward the downwind airliners and started spraying foam. </p>
<p>In a documentary program, during a drill, the first fire engine started spraying foam as soon as it got within 'shooting range'. That is, saving extra time, 48 seconds in this real case, for all engines moving into upwind formation position. </p>
<p>What is the pro and con of these two approaches?</p>
| |fire-sprinkler| | <p>The first priority for any fire crew is to be part of the solution, and not become part of the problem. </p>
<p>Staying downwind of a kerosene fire would put them in the likely path of a lot of hot gas and smoke, even if the flames don't reach them. </p>
<p>Actually, there's an infamous story about a similar incident that happened in the company I work for, many years ago. They were carrying out a test that involved a decommissioned aircraft fuselage. They had a fire crew present as a safety precaution.</p>
<p>Something went wrong with the test, resulting in a fuel leak on to the ground followed by a fire. Unfortunately, the ground was sloping, and the burning fuel flowed gently downhill, heading straight towards the fire truck.</p>
<p>There is a video of the fire truck reversing away from the incident as fast as possible, which gets shown regularly to trainees as a classic example of "how <strong><em>not</em></strong> to do a risk assessment and make a safety plan"...</p>
| 10512 | Fighting airplane fire on runway |
2016-06-29T13:02:14.323 | <p>I would like to know with which existing types of power plants(thermal,wind,etc.) can compressed air energy storage(CAES) be used efficiently? And how can it aid power plants like thermal(a brief explanation is sufficient)?</p>
<p>As far as I know it requires multiple compressors and hence can be used with a gas turbine power plant. But why would a gas turbine need CAES since it can readily obtain air and compress it at any time?</p>
| |energy-storage| | <p>Classical CAES plants (all two of them currently in operation) need a heat source to generate power: The gas cools with expansion, thus loosing pressure, they need to heat it to realise all the potential energy. One way around his is to combine the CAES with a gas turbine, the air is mixed with gas and undergoes combustion in the turbine. So in effect the CAES provides the work otherwise neccessary for compression of the combustion air.</p>
<p>You probably could use this with other heat engines. But CAES are built for grid balancing where you want to adjust the output of your plant quickly, and quickest engines in this regard are gas turbines.</p>
<p>This is the system in place in the plants Huntdorf (Germany) and McIntosh (USA).</p>
<p>Contrast with adiabatic CAES, where the heat from compression is stored and reused when generating power. The furthest developed project along these lines, ADELE in Staßfurt (Germany), was canceled recently though.</p>
<p>Source: <a href="https://de.wikipedia.org/wiki/Druckluftspeicherkraftwerk#Druckluft-Gas-Kombikraftwerk" rel="nofollow">wikipedia Druckluftspeicherkraftwerk</a></p>
| 10519 | Compatibility Of CAES |
2016-06-29T14:03:33.373 | <p>Why do CAES systems have intercoolers after every compression stage?</p>
<p>Isn't it a waste of work, since the air will cool anyway in the storage cavern?</p>
| |energy-storage| | <p>It would be a waste of work, if the air did <em>not</em> cool in the storage cavern! </p>
<p>But as you say, it does cool, hence you'll have less pressure available when expanding the air than the pressure you needed to work against when compressing it. The excess pressure could be avoided by compressing in isothermal mode, and <a href="https://engineering.stackexchange.com/a/10525/5531">as Algo said</a> you approximate isothermal by putting an intercooler between two isentropic compressors.</p>
| 10523 | Why does compressed air energy storage use intercoolers? |
2016-06-30T23:05:17.100 | <p>Please note that the question is "Why", not "How".</p>
<p>For sure I'm getting this wrong, but I'm stuck and I don't know why! If a reverse thrust is applied during an aircraft landing wouldn't the total amount of work necessary to pull the air inside the turbine (hence pushing the aircraft forward) be canceled by the amount of work used to push the air forward (i.e. pushing the aircraft backwards)? I mean, why does it work? Why does the net work push the aircraft backwards and help to stop the plane?</p>
| |mechanical-engineering|fluid-mechanics| | <p>Imagine that the engine is a person throwing a 1 kilogram ball backward at a speed of 1 meter per second. The reverse thrust system can be modeled as a wall attached to the plane that bounces the ball so it goes forward.</p>
<p>The forward impulse caused by each ball is, obviously. 1 kg m/s. Then when it bounces, its velocity changes from 1 m/s backward to 1 m/s forward, for a net change of 2 m/s, applying a reverse impulse to the plane of 2 kg m/s. So the net total impulse is 1 kg m/s backwards.</p>
| 10537 | Why does aircraft reverse thrust actually work? |
2016-07-01T16:51:05.703 | <p>I was wondering, basicly every radio-transsmission works with electromagnetic waves which travels at the same speed whether I modulate the information with a low frequency or high frequency such as done with LI-FI.
What makes Li-fi faster? what makes any wireless transmission faster/slower if eventually I send my information with electromagnetic wave which travels at the same speed?
What makes lifi special?</p>
| |telecommunication|wireless-communication|wifi| | <p>The wave propagation speed is only relevant when you start approaching the theoretical limit of that type of wave's ability to transmit information.</p>
<p>For an analogy, consider people having a conversation in a room full of air. They're using sound waves to transmit information. The speed at which those waves propagate is the same for every speaker, in that particular medium. Does that imply that everyone transmits information at the same speed?</p>
<p>The answer is no, of course not. The information is overlaid on the sound wave in a particular way, and we call that speech. You can transmit the same information at the same speed across a wide range of frequencies. You can speak faster or slower without varying the pitch of your voice. (In fact, <a href="https://www.youtube.com/watch?v=4X4Fy8YqysY&t=130" rel="nofollow noreferrer">some people can transmit speech faster than you can receive it.</a>) You can use different languages that might allow you to communicate particular ideas at different rates.</p>
<p>The theoretical maximum speed at which information can be transmitted via audible speech is only loosely related to the speed of propagation of an audio wave. <a href="http://hyperphysics.phy-astr.gsu.edu/hbase/tables/soundv.html" rel="nofollow noreferrer">Put the same conversation underwater and the waves propagate four or five times faster</a> but you don't even notice because the time required for the wave to propagate is so many orders of magnitude less than the time required to encode and decode a single datum.</p>
<p>Digital signals using EM waves are a much more specialized technology, compared to the human systems that enable verbal communication. But we're still far from any theoretical limit on the rate at which EM waves could possibly transmit information. Each of the <a href="https://en.wikipedia.org/wiki/IEEE_802.11#Protocol" rel="nofollow noreferrer">802.11 protocols</a> used by your Wi-Fi device has a theoretical maximum bitrate, based on multiple elements of its design. At one point those were measured in Mb/s, now they're measured in Gb/s and there's plenty of room to keep going. <a href="https://physics.stackexchange.com/q/56240/82691">A question over on Physics SE asks about the upper limit for fiber optics</a> (which also use EM waves, of course) and you can see from the answers that, at Gb/s for Wi-Fi or Li-Fi, we're still many orders of magnitude away from hitting a ceiling on transfer speeds.</p>
<p>That's why the wave propagation speed isn't as meaningful as you expect. As far as what makes Li-Fi faster than Wi-Fi, I'm not even sure that's the case in isolation; practically speaking, what seems to be driving the development of the former is <a href="https://en.wikipedia.org/wiki/Bandwidth_allocation#Bandwidth_limitations" rel="nofollow noreferrer">over-saturation of the frequencies used by the latter</a>. That's more of an issue of congestion than anything else, which means that if Li-Fi had been developed first, the situation might be reversed, and popular science mags might be writing the same BS headlines about Wi-Fi instead.</p>
<p><a href="http://www.informationweek.com/mobile/lifi-wireless-at-light-speed-comes-with-drawbacks/a/d-id/1323344" rel="nofollow noreferrer">"Wireless at light speed."</a> Give me a break.</p>
| 10545 | Why is Li-Fi faster than regular wifi? |
2016-07-04T08:30:20.437 | <p>Are there any recommended clamps for holding samples of a viscoelastic material in place? My goal is to have the clamps hold the rectangular samples along lines drawn parallel to edges of the samples for elastic testing. </p>
<p>Hand screw clamps appear to be completely inappropriate as the material just squishes out of position. But, I think most clamps would suffer this weakness. </p>
| |mechanical-engineering| | <p>If practical rolling the strap sample a few times around a double cylinder mechanism which are designed to lock into each other with tension similar to seat belt or the old military uniform service belts would be the best setting.<br>
This clamp on elastic material would not perform evenly, it will allow more slip on the ends. So it has to be installed far enough to allow for redistribution of strain and has to be calibrated for establishment of the origin datum after loading.</p>
| 10561 | clamps for viscoelastic materials |
2016-07-05T21:10:10.850 | <h2>Background</h2>
<p>The Bailey Bridge panel bridge system has been around since WWII. It has been credited as being one of the top 3 technological advancements that allowed the allies to win the war. After the war, the surplus of Bailey Bridge parts were sold off and distributed around the world. While these bridges were intended to be for <em>temporary</em> use (military temporary use and civilian temporary use are two different things), some of these old bridges can still be found in use today.</p>
<p><a href="https://i.stack.imgur.com/yIZO3.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/yIZO3.jpg" alt="Bailey schematic" /></a>
<a href="https://i.stack.imgur.com/BflbI.png" rel="noreferrer"><img src="https://i.stack.imgur.com/BflbI.png" alt="Bailey Schematic with tapered transom" /></a></p>
<p>A quick search on google will return all sorts of beautiful modern bridges that are not Bailey Bridges. To me, a Bailey Bridge is a very specific type of panel bridge from WWII or post war sold by Bailey-Uniflote. However listening to colleagues, and even looking at the google search results, it appears that Bailey Bridge is just another word for Panel Bridge. Similar to Xerox and Photocopy, or Kleenex and tissues. So this has muddy some of my search efforts thus far.</p>
<h2>The Problem</h2>
<p>The problem I face is trying to determine which of the 4 (maybe more) types of panel bridge systems I may be looking at as they can be very similar in component layout at a quick glance and even some basic measurements.</p>
<p>So far my research has told me that some of the distinguishing factors for these panel type bridges are:</p>
<ul>
<li>Panel size</li>
<li>Panel diagonal members (I, C, HSS/TUBE)</li>
<li>Deck width</li>
<li>Transom location</li>
<li>Transom size</li>
</ul>
<h2>Panel Size</h2>
<p>The original Bailey Bridge had a panel length of 10'0" (3.048 mm) from centres of connecting pin holes and a height of 4'9" (1.448 mm) from centres of connecting pin holes. The overall height was 5'1". Anything outside of these dimension would indicate a newer panel system such as an Acrow 700xs or Mabey Compact 200 or the like.</p>
<h2>Diagonal Members</h2>
<p>In the Bailey Bridge panel there are two diamond shapes. The diagonals of these shapes were originally I sections. At some later time these sections changed to C-Channels and rumour has it HSS or tube sections were used by certain manufactures. Anyone have any confirmation on how to identify or rule out potential bridge type based diagonal section type? Maybe even some date ranges for the various section type?</p>
<h2>Deck width</h2>
<ul>
<li>The original Bailey Bridge has a roadway width of _____ and a clear width between trusses of ____.</li>
<li>The standard widened Bailey Bridge has a roadway width of 10'9" (3.28 m) and a clear width between trusses of 14'3" (4.34 m).</li>
<li>The Extra Wide Bailey Bridge has a roadway width of 13'9" (4.19 m) and a clear width between trusses of 15'8" (4.77m).</li>
</ul>
<p>I am not sure what the standard deck width is on an Acrow 300 or a Mabey 100.</p>
<h2>Transom Location</h2>
<p>On a Bailey Bridge, the transoms are located adjacent to the panel's end vertical and middle vertical. This can been seen in the following diagram and photo:</p>
<p><a href="https://i.stack.imgur.com/hfwaJ.png" rel="noreferrer"><img src="https://i.stack.imgur.com/hfwaJ.png" alt="Bailey Transom Location Diagram" /></a></p>
<p><a href="https://i.stack.imgur.com/gW1AC.png" rel="noreferrer"><img src="https://i.stack.imgur.com/gW1AC.png" alt="Bailey Single Transom Location Photo" /></a></p>
<p>If the transoms need to be double up to carry a heavier load then there are 4 transoms per bay. Two either side of the middle vertical and 1 inside of each end vertical as shown in the following photo:</p>
<p><a href="https://i.stack.imgur.com/JvA5E.png" rel="noreferrer"><img src="https://i.stack.imgur.com/JvA5E.png" alt="Bailey Double Transom Location Photo" /></a></p>
<p>For an Acrow 300 style panel bridge, the transom is located at the base of panel's diamond as seen in the following photo:</p>
<p><a href="https://i.stack.imgur.com/wstre.png" rel="noreferrer"><img src="https://i.stack.imgur.com/wstre.png" alt="Acrow 300 Bridge Transom Location" /></a></p>
<p>Mabey 100 transoms are located ?????</p>
<h2>Transom Size</h2>
<ul>
<li>A standard Bailey Bridge has a transom length of 18'0" (5.49 m) with a constant 10" depth. There will be 3 holes near the bottom flange to allow for 3 panel lines.</li>
<li>A Standard Widened Bailey Bridge has a transom length of 19'11" (6.1 m) with a 12 in depth between under the deck and a 10" depth at the panel lines. There will be 4 holes near the bottom flange to allow for 4 panel lines.</li>
</ul>
<p><a href="https://i.stack.imgur.com/Hv3b6.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/Hv3b6.jpg" alt="Wide DD Bailey Transom" /></a></p>
<ul>
<li>An Extra Wide Bailey Bridge has a transom length of 19'11" (6.1 m) with a 12 in depth between under the deck and a 10" depth at the panel lines. There will be 3 holes near the bottom flange to allow for 3 panel lines.</li>
<li>It also appears that transom with a tapered flange also exist while maintaining a uniform section depth</li>
<li>There is also the M2 and M3 Transoms identified in the US Bailey Bridge Field Manual</li>
</ul>
<p>Acrow 300 and Mabey 100 transoms are?????</p>
<h2>SUMMARY</h2>
<p>What are the key factors that distinguish Bailey Bridges?</p>
<p>So far I can only tell Acrow 300 by its transom position.</p>
<p>What about:</p>
<p>UK Bailey<br />
US Bailey<br />
Mabey 100</p>
<p>Just found this really <a href="http://www.thinkdefence.co.uk/2012/01/uk-military-bridging-equipment-the-bailey-bridge/" rel="noreferrer">nice write up on the Bailey Bridge</a> but it still does not completely answer my questions.</p>
| |bridges| | <p>This is well after your post but may answer some of your questions.<br>
I was an engineer in Papua New Guinea durng the late 1970's early 1980s when Mabey began supplying Compact Bailey (now Compact 100), and we had a fair bit of older bridging to the "standard" design as well. At the time we used the data from the Bailey Uniflote book, but even then we had issues, as tables of available live load were not transparent on true capacity - they assumed a lowish dynamic load factor and were calcualted using working stresses.
We tendered the supply (with Acrow and Mabey competing - well before the Chinese and Indians started supplying) based on the standard capacities. I don't recollect how much extra the Compact gave, if any, but they were only purchased in competitive tender against the standard Acrow product so we took no account of any extra capacity even if they had it. </p>
<p>Anyway, these days we still find ourselves occasionally load rating them and we still use the same approach - that is we use the data for the standard Bailey via the live load capacity tables, adjusted for dynamic factors and anything varying the dead loads. I believe, despite all the alloy and section changes, that the panels were designed to meet proof loading requirements, so in the end it still comes back to the table ratings.</p>
<p>We have only seen one compact among about a half dozen we have encountered - you can tell them easily by the absence of the transom clamp - the compact has a "permanent" bolted end. We treated it just the same.</p>
| 10582 | Determining the difference between US Bailey Bridge, UK Bailey Bridge, Mabey Compact 100 and Acrow 300 |
2016-07-06T06:44:36.443 | <p>I have a pressure regulator bringing an inlet pressure of 150 bar down to 10 bar which is then lead into a plenum followed by a nozzle of area 340 mm^2, exposed to atmospheric conditions.</p>
<p>Weirdly enough at both points the necessary condition of pressure ratio for choking is satisfied!</p>
<p><img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/7d4f2f098327a3e772615e3e2ba44bd4697d1ebc" alt="pressure ratio"></p>
<p>The critical exit pressure below which choking occurs (p*), is 5.8 bar at the nozzle exit and 79.7 bar at the regulator exit. How do I explain the fact that the flow chokes only at the nozzle? </p>
| |fluid-mechanics|thermodynamics|aerospace-engineering|compressed-gases|compressible-flow| | <blockquote>
<p>How do I explain the fact that the flow chokes only at the nozzle?</p>
</blockquote>
<p>To explain why the flow only chokes at a single point, it is important to remember that for a system at steady state, there can only be a single flow. All of the fluid entering the system must leave. Since there can only be a single flow rate through the system, you must then consider where that maximum flow can occur.</p>
<p><em>The following equations describe the flow through a frictionless nozzle where the expansion occurs adiabatically and isenthropically. They are from Perry's page 6-23. The actual flow through an orifice is usually handled by a flow coefficient since the flow through an orifice will be less than a frictionless nozzle.</em></p>
<p>$$
\begin{align}
\frac{p^*}{p_0} &= \left(\frac{2}{k+1}\right)^{k/(k-1)} \\
\frac{T^*}{T_0} &= \frac{2}{k+1} \\
\frac{\rho^*}{\rho_0} &= \left(\frac{2}{k+1}\right)^{1/(k-1)} \\
G^* &= p_0 \sqrt{\left(\frac{2}{k+1}\right)^{(k+1)/(k-1)}\left( \frac{kM_w}{RT_0} \right)} \\
w^* &= G^*A \\
V = V^* = c^* &= \sqrt{\frac{kRT^*}{M_w}}
\end{align}
$$</p>
<p>For the formulas above, the $^*$ represents the condition at choked flow and the $_0$ condition is inlet. The other variables are defined as:</p>
<ul>
<li>$p$ = pressure</li>
<li>$T$ = temperature</li>
<li>$\rho$ = density</li>
<li>$G$ = mass velocity (mass flow per unit area)</li>
<li>$w$ = mass flow</li>
<li>$A$ = nozzle exit area</li>
<li>$V, c$ = exit velocity</li>
<li>$R$ = gas constant</li>
<li>$M_w$ = molecular weight</li>
</ul>
<p>Choked flow occurs when the downstream pressure is less than the critical pressure or the pressure ratio is less than the critical ratio. This is shown in equation 1 and repeats your initial question. Once you know the flow will be choked, you can then use the remaining equations. Looking at the equation for the mass velocity, $G^*$, you can see that choked flow is a function of gas composition $(k,M_w)$ and inlet conditions $(T_0,p_0)$ and that changing downstream conditions has no effect on the mass velocity. To get to the mass flow rate $w^*$ you must also consider the orifice area $A$. With those variables known, you can determine which orifice will create the limiting flow rate. This can become an iterative process as changing upstream conditions may then limit downstream components.</p>
<blockquote>
<p>I have a pressure regulator bringing an inlet pressure of 150 bar down to 10 bar</p>
</blockquote>
<p>Since you have a pressure regulator, this tells you something about the dynamics of the system. The pressure regulator presents a variable sized orifice to the process until it is fully opened. Once it is fully open, it behaves like a fixed orifice size. Since the pressure regulator is able to adjust to maintain a downstream condition, it will not be the limiting component until it is wide open.</p>
<p>When the regulator is partially open, the system has established a steady state condition wherein:</p>
<ul>
<li>the flow out of the system (to atmosphere) is the maximum flow through the system</li>
<li>the position of the regulator (e.g. what % open) is such that its exposed flow area provides exactly the same flow as the outlet to atmosphere; this flow is a function of the temperature, upstream pressure, and composition of the gas at that choke point</li>
</ul>
| 10584 | How to determine the choke point in a compressed gas system? |
2016-07-06T14:02:42.530 | <p><strong>Disclaimer</strong></p>
<p>Designing the behavior of controllers is an important task for many applications. Every application needs its own requirements fulfilled. For example a CNC machine needs a controller with a strict asymptotic convergence to the set value. An overshooting behavior can’t be tolerated due to the fact that once removed material can’t be replaced by the same tool. Other examples for devastating results with an overshooting controller behavior are autopilots landing an aircraft or shuttles docking the ISS.</p>
<p><strong>Question</strong></p>
<p>Are there any real applications where a controller with an overshooting behavior can not only be tolerated but leads to an overall suitable performance for the system? I am looking for a specific example not just field of tasks where it might be the case.</p>
<p><strong>Thoughts of my own</strong></p>
<p>There are systems where an overshooting control behavior can be tolerated. Like if a robot needs to stay on a specific path on the way to its destination with a given tolerance. A controller that controls the distance to this given path can let the robot wiggle a little around the given trajectory as long as it keeps it inside the tolerance. But if you can design a controller with an asymptotic converge to its trajectory you would prefer it to get rid of this wiggle. I can’t think of any real application where you would prefer an overshooting controller over an asymptotic one if you had the choice.</p>
| |control-engineering|control-theory|pid-control| | <p>For some systems, the salient criterion is settling time to within some error band. Sometimes you can get faster settling by allowing earlier overshoot.</p>
<p>If you need a system to get to within some minimum error of before starting a process, you probably don't care what the system is doing before the process is started, only that it gets there as quickly as possible since you're waiting until then. It's not hard to come up with scenarios where settling time is important, and overshoot before settling largely irrelevant.</p>
<p>For example, let's say some process has to occur at a nominal 200 °C, ±10 °C. When you're powering on the oven, you care that it settles to the 190-210 °C range quickly. It's going to get there faster than a asymptotic controller if allowed to overshoot to 210 °C, then oscillate with ever-decreasing amplitude around 200 °C. The same holds true if you change the set point, and you can't do anything useful until it gets to within some tolerance of the new setpoint.</p>
| 10589 | When are overshooting controllers preferred over asymptotic ones? |
2016-07-06T14:16:18.777 | <p>What is the difference in temperature between a surface in the sun as supposed to one in the shade and how is it calculated?</p>
<p>For instance:
What will the temperature be if a metal sheet lies in the sun and then what will its temperature be if the same sheet lies in a shaded area?</p>
| |temperature|solar-energy|radiation| | <p>As Dr. Bad Science Ben Goldacre says, "I think you'll find it's a bit more complicated than that." It's not hard to dig up solar spectral power density data - which depends heavily on latitude and season, of course -- but how a "metal" sheet behaves depends on tons of parameters.<br>
Different metals (elemental or mixtures) have different spectral absorptivities as well as specific heat (how much energy per unit volume per unit temperature change). Further, for example, a roughened surface may have a higher net absorption.<br>
In summary, you'll need a lot more specific info, including sheet thickness.</p>
<p>In the shade, any material will reach thermal equilibrium with the local atmosphere.</p>
| 10590 | Surface in the sun vs. surface in the shade |
2016-07-07T05:31:12.577 | <p>I'm building a 3D scanning (photogrammetry) device with a rotating platform, a bunch of wheels and a stepper motor. What I want to know is how much weight that motor could rotate on the platform.</p>
<p>Currently the platform weights just over 2kg and is just under 40cm wide (see the image below). I estimate that the objects I will be scanning will weigh no more than 5kg (just to be safe).</p>
<p>So the link to that servo motor says it has a 'stall torque' of 15kg - cm at 6V. I have no idea what that means. Is that enough? Will it be able to rotate the my platform?</p>
<p><a href="https://i.stack.imgur.com/R1qu2.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/R1qu2.png" alt="enter image description here"></a></p>
| |stepper-motor| | <p>The limit on the stepper motor is the torque it can produce. This has nothing to do with the weight it is rotating, except that the weight may cause more friction, which the motor has to overcome.</p>
<p>The mass does matter somewhat, but only in a rotational sense, and only to determine how fast the motor's torque will speed up the rotation. The proper term for this "rotational mass" is <i>moment of inertia</i>. The torque divided by the moment of inertia about the rotating axis gives you the angular acceleration.</p>
<p>Another limit is what the motor bearings can handle. The motor datasheet should give you the maximum axial and radial forces the motor is rated for. In your case, the axial force is the weight of the platform and everything on it, minus the weight taken up by the wheels. That may be hard to determine, since it's not clear how much of the weight is supported by the wheels and how much by the motor shaft.</p>
<p>In the end in this case, the limiting factors will be the friction and therefore the torque required just to get the platform rotating at all, and how fast you want to get it up to speed. The former depends on too many things that are hard to know up front. Build it and see, then get a motor with enough torque to reliably overcome the friction.</p>
<p>You could also use gears or a belt drive to effectively turn a fast but low-torque motor into a slow but high-torque motor.</p>
| 10600 | How much weight can my stepper motor rotate? |
2016-07-07T17:12:18.673 | <p>In civil engineering calculation I have seen weight of IPE 300 profile expressed in kN/m<sup>2</sup> unit.</p>
<p>IPE 300 0.42 kN/m<sup>2</sup></p>
<p>I know IPE 300 is approximately 42.4 kg/m. How did they calculated that value expressed in kN/m<sup>2</sup>?</p>
| |mechanical-engineering|civil-engineering|beam| | <p>The density of structural steel is usually adopted as 7850 kg/m<sup>3</sup> (or thereabouts). According to <a href="http://www.b2bmetal.eu/i-sections-ipe-specification" rel="nofollow">this source</a>, the cross-sectional area of an IPE 300 section is 53.8 cm<sup>2</sup>. The linear mass of such a section is therefore 42.2 kg/m, which converts to a linear weight of 0.422 kN/m.</p>
<p>$$\begin{align}
m &= 42.2\ \text{kg/m} \\
w &= mg \\
&= 42.2 * 10 = 422\ \text{N/m} = 0.422\ \text{kN/m}
\end{align}$$</p>
<p>If you've seen anyone use 0.42 kN/m<sup>2</sup>, then you've seen someone doing it wrong and not paying attention to dimensional analysis.</p>
| 10609 | IPE Beam weight |
2016-07-07T18:19:33.907 | <p>I got the Ikea SKARSTA desk <a href="http://www.ikea.com/us/en/catalog/products/S49084965/" rel="nofollow noreferrer">http://www.ikea.com/us/en/catalog/products/S49084965/</a>, it's a sit stand desk that uses a manual shaft to rise and lower the desk.</p>
<p><a href="https://i.stack.imgur.com/BuzIv.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/BuzIv.jpg" alt="enter image description here"></a></p>
<p>So I plan to make the desk motorized using a satellite dish motor and a satellite positionner to control the motor (I removed the linear actuator part and kept only the motor).</p>
<p><a href="https://i.stack.imgur.com/LUelV.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/LUelV.jpg" alt="enter image description here"></a>
<a href="https://i.stack.imgur.com/QfWsf.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/QfWsf.jpg" alt="enter image description here"></a></p>
<p>The problem I'm facing is that the space between shaft and tabletop is about 1cm, so I'm not sure how can I transmit the rotation from the motor to the shaft (which is simply a long Allen key).
<a href="https://i.stack.imgur.com/bdE2j.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/bdE2j.png" alt="enter image description here"></a></p>
<p>I looked into pulley systems and cogs but couldn't find suitable pieces, please advise.</p>
<p>Many thanks.</p>
<p><strong>Update Jul 9:</strong></p>
<p><a href="https://i.stack.imgur.com/k2mDW.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/k2mDW.jpg" alt="enter image description here"></a></p>
<p>The problem is how to transmit the rotation from the motor (green) to shaft (orange), motor can be placed at any position relative to table top but shaft isn't.</p>
| |mechanical-engineering| | <p>You can couple to the hex shaft using a suitable socket wrench socket. (If you don't want to destroy the original shaft, get the correct size Allen wrench and cut off the short arm.) Then you can use standard socket wrench accessories to change the direction of the axis.</p>
<p>A quick search turned up options such as the MFE-6 flexible shaft or the M-140A universal joint in the <a href="http://www.snaponindustrialbrands.com/DSN/wwwsnaponindustrialbrandscom/Content/PDF/Snap-on%20Industrial%20Brands%20-%20CAT3%20227.pdf" rel="nofollow">Snap-On Tools catalog</a>. Other vendors have similar offerings.</p>
| 10612 | Transmit rotation in a tight space |
2016-07-07T18:47:41.170 | <p>I have a 2 mm piece of UV transparent acrylic sheet. The datasheet defines it as 80% transparent at 300 nm. However, when I use it with different temperatures (4-50°C), the transmittance characteristics are totally different. It is much more transparent at higher temperatures.</p>
<p>Does anyone know if this is true - that the light transmittance of acrylic varies with temperature? And what might be the reason?</p>
<p>To be more specific, I am interested in the 250-600 nm wavelength range.</p>
| |electrical-engineering|materials|optics| | <p>I'm tempted to say that you've measured this change, so what's the question here, but there are a couple things you can do to validate your tests. Most important: run your source & detector in exactly the same setup but without the acrylic sheet in place, and see if the response varies with temperature. You want to be sure that only the sheet is causing the observed change. </p>
<p>After that, there are less critical things to check: is your source polarized? Does the sheet move at all as the temperature changes? Acrylics often have stress lines which affect the polarization of light passing thru them.</p>
<p>And of course, do the deepest Google search you can for info or tech papers on thermal sensitivity of acrylics.</p>
| 10613 | Light transmittance of acrylic with temperature |
2016-07-08T14:42:04.260 | <p>I want to prototype a wah-wah guitar pedal using an arduino, a potentiometer, and some meccano I have lying around.
I thought of two ways to do this :</p>
<p>the first one is to put a pinion on the pot, and connect a rack to the pedal itself; however I can already see that it can't be as simple as that, because the rack to pinion angle will change as the pedal gets pushed more.</p>
<p>the second one is to use several gears, the first one on the axle of the pedal, the last one on the potentiometer, so that the full motion of the pedal results in a full potentiometer rotation; I don't like that one much, as I don't think I have the requisite gear ratios.</p>
<p>Having absolutely no background in mechanics, that's it for me; I don't even know the name of the concepts I might be needing. Can Stackexchange Engineering provide some pointers for making this at home, with no complex tools beyond a sander, a dremel, and such?</p>
<p>Thanks for your attention.</p>
| |gears| | <p>I know this is a bit late but an Arduino should be capable of interpreting pulse train or positional transducers. These are used in industrial machines so are really rugged. This would allow the software to map the signal onto all sorts of things.</p>
| 10624 | Building a wah-wah pedal |
2016-07-08T16:04:15.720 | <p>I'd like to prototype a kind of device to move a rope like a sinusoidal, or similar waveform, it doesn't have to be precise about the waveform, the goal is hide this mechanism in order to show a floating waveform in space for an exhibition. I think to use an electric motor to vibrate one side of the robe and to fix the other side, but I'm not an expert in this field. How to convert the rotatory movement to vibrate the rope? Any hints on how to do this? Thank you</p>
| |motors| | <p>Following on what Forward Ed said, a crank and connecting rod seems like a pretty straight-forward design. I think the motion of the rope will approach a sine wave as the connecting rod is made longer.</p>
<p>I think the free end of the rope might need some attention, too. If you want a travelling sine wave, I think it'll have to have a specific mass added to it (the free end) so that the end doesn't whip around (too little mass), or hang there near motionless (too much mass). Kind of like the kinetic equivalent of a terminating resistor.</p>
<p>If you want a standing sine wave, I think the free end could be held stationary (or maybe nearly-anchored, with just a little bit of stretch), but the relationship between the motor speed and length of rope becomes important. Only certain lengths of rope would produce a standing wave for a given motor speed.</p>
| 10626 | motor to move rope sinusoidally |
2016-07-11T14:57:05.657 | <p>I have general problem of understanding how forces are transfered from 3d members problem to plane problem.</p>
<p><a href="https://i.stack.imgur.com/R5EB9.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/R5EB9.jpg" alt="enter image description here"></a></p>
<p>For example what would be 2D representation of forces F1 and F2 acting on horizontal A beam, and what will be effect of beam A,B,C,D and column 1 and 2 on horizontal A beam.</p>
<p>How can I represent horizontal beam A in 2D so I can do calculations of bending moment, shear forces etc.</p>
<p>All joints can be considered as welded.</p>
<p><strong>EDIT</strong> (After Mr. Wasabis answer)</p>
<p>I can replace influence of force F1 with reactions of beam B</p>
<p><a href="https://i.stack.imgur.com/0nD2X.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/0nD2X.jpg" alt="enter image description here"></a></p>
<p>so reactions are influencing horizontal A (beam C and F2 are igonred for simplicity). Is this OK - With moment Ms1 in plane perpendicular to horizontal A causing torsion of horizonta A.</p>
<p><a href="https://i.stack.imgur.com/HoHRl.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/HoHRl.jpg" alt="enter image description here"></a></p>
| |mechanical-engineering|structural-engineering|civil-engineering|structures|structural-analysis| | <p>As a first estimate and assuming all the loads and stresses are below working stress limits and assuming the supports are simple pin connection:
Beam A support reactions are
$$R1=\dfrac{P_b(l-a) + P_c(b)}{l}$$
where $P_b$ is the reaction of beam B on beam A and $P_c$ is reaction of beam C on beam A and $a$ and $b$ are the distance of the beams $b$ and $c$ from the respective ends of beam $a$.</p>
<p>And the end reactions of beam $b$ or $c$ on beam $a$ are simply Pb = (F1 x distant of F1 to far end of b)/its length.</p>
<p>These are available in any engineering hand book or quick charts. However if we were to consider the welds to be fixed joints then it turns into an indeterminate structure! </p>
<p>there are approximate methods or computer programs available but the significant difference is that you have to have the member properties and loads as a starting point of analysis because then the entire structure comes into action and plays some role in supporting the applied load. i.e. If the beam across from beam a is too stiff and connected to two stiff columns with correct welded connections hypothetically in extreme case it can support the majority of loads and let beam a relieved of most of the load. Beam a can even be removed entirely and beams b and c act cantilever from the other beam! </p>
| 10650 | 3D problem to 2D plane problem |
2016-07-12T15:02:01.857 | <p>I have two parts that should be bolted together with four M6 bolts. The tolerances of the bottom part are quite big and the holes for this part are fixed to Ø6. The tolerances of both parts are fixed due to manufacturing and lowering of costs.
I need to determine the size of the holes of the top part in order for them to fit together. I have attached a photo showing the tolerances of both parts.</p>
<p>I am in doubt how to calculate the size of the holes. For instance for the left side, is this the correct way?
$$\begin{align}
\text{Diameter} &= \text{Delta}_{dim} + \text{Tol}_{dist} + D_{shaft} \\
&= (211.5-210)+3.5 + 6 = 1.5 + 3.5 + 6 \\
&= Ø11
\end{align}$$</p>
<p><a href="https://i.stack.imgur.com/RKbIO.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/RKbIO.jpg" alt="enter image description here"></a></p>
| |mechanical-engineering|design| | <p>The most straightforward solution to this problem is to drill the holes in the second plate at the time of assembly, using the first plate as a template. </p>
<p>Having said that it really does sound like there is a manufacturing issue which needs to be addressed if you have such large tolerance on a part this small. Even with the most basic manufacturing process I can't imagine that this can't be improved on without undue cost. Even just making a template for marking the hole centres with a marker pen should improve matters substantially, bearing in mind that having to drill 12 mm clearance holes for an M6 fastener and having to obtain and use oversized washers is going to add a significant amount of faff to the whole process. </p>
| 10661 | Dimensioning of hole from tolerances |
2016-07-12T18:05:06.050 | <p>I have to build a device to apply a pure normal force over a force plate. The requirement is that it <em>doesn't</em> apply any shearing (by means of friction) on the surface.</p>
<p>The problem is: for any given device structure, there will always appear some deformation, specially under load, and so I don't want to depend on geometric alignment, I would like to have my indenter to have an <em>appropriate joint</em> with the proper types of degrees of freedom.</p>
<p>I though about a ball bearing roller, but that would only decouple the force along one direction, since it could still transmit force by friction along the other direction.</p>
<p><a href="https://i.stack.imgur.com/neZAk.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/neZAk.png" alt="enter image description here"></a></p>
<p>So my question is:</p>
<blockquote>
<p>What is the recommended way to eliminate coplanar reaction forces, so that only normal force is transmitted and coplanar forces are neutralized?</p>
</blockquote>
<p>I think lubricant-based solutions don't fit well, since they are always quantitative in nature. I think a qualitative solution involving mechanical decoupling by a proper joint is better, similar to the ball-bearing one, but with an additional degree of freedom.</p>
| |mechanical-engineering| | <p>If you apply a force which can be decomposed to a non-zero horizontal component, that component has to be transferred onwards.</p>
<p>So, if you only want the vertical component to be transferred to your plate, you'll need to create a secondary structure to absorb the possible horizontal component.</p>
<p>One possible inspiration are <a href="https://en.wikipedia.org/wiki/Slide_plate" rel="nofollow noreferrer">sliding plate bearings</a>. These allow the free translation of an object along the horizontal plane, meaning no horizontal forces are transferred to the underlying structure (other than inevitable friction, which is low due to the adopted materials). Obviously, you'll want to limit these translations at some point, so you'll need to design a container which restrains the plate's movements. When the plate touches the container, a horizontal force will appear between them, so the container will have to rest on something other than the structure you're trying to protect.</p>
<p>If you'll forgive my horrible sketch, this is what I mean:</p>
<p><a href="https://i.stack.imgur.com/7Un76.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/7Un76.png" alt="enter image description here"></a></p>
<p>The low-friction pad allows the upper plate to move freely in both directions. If an inclined force is applied to the upper plate, it will move until it touches the bounding structure, at which point the horizontal component of the force will be absorbed by the wall (or whatever supports the bounding structure, which can't be the same as whatever it is you're trying to protect), while the vertical component is transferred directly to the underlying structure.</p>
| 10667 | Device to apply pure normal force over planar surface, without friction-based coplanar forces |
2016-07-13T07:36:24.107 | <p>I am making an aircraft in a simulator (using JSBSIM). Note that I am not a professional.</p>
<p>The manual states that it is limited below certain calibrated airspeed to a certain positive angle of attack.</p>
<p>I have tried different schemes to make this work, but I have not been successful.</p>
<p>For example the simplest was to just freeze the elevator when the limit was reached. But in for example a loop, the alpha will increase even with frozen elevators as the plane goes through the loop.</p>
<p>Another was to use a feedback of the exceeded alpha with a gain into reducing the elevator actuation. But that makes the plane pitch jerk. It should be smooth.</p>
<p>I hope someone can help me out. I prefer a solution that does not use PID controller as they are a pain to calibrate. Block diagrams or text help is very appreciated.</p>
| |control-engineering|aerospace-engineering|simulation| | <p>I solved this by converting the stick input to a Nz value. I then subtracted alpha in a certain range from that Nz input. Then converted that to elevator deflection using a integrator comparing actual Nz to the input.</p>
<p>I also found out that in the real aircraft I am simulating a certain stick input correspond to certain commanded G-force, so in all, its a good solution.</p>
| 10675 | How can I limit a simulated aircraft's angle of attack? |
2016-07-14T07:14:12.777 | <p>I want to power an Arduino project for several weeks on a remote location. At the location, data is gathered and sent over the internet. The project runs on 5V and draws 200mAh max. I need a power source that runs on its own, without cables.</p>
<p>To do so, I was looking into big batteries. My idea was to use a battery like <a href="http://www.produktinfo.conrad.com/datenblaetter/1300000-1399999/001372559-da-01-en-LATERNENBATTERIE_ALKALI_4LR25SZ__6V_50AH.pdf" rel="nofollow">this</a>. It has a capacity of 50Ah and an output of 6V which I can downstep with a voltage regulator.</p>
<p>I have several questions:</p>
<ol>
<li>How safe is this battery? Can directly touching the contacts cause harm?</li>
<li>Can my project draw 200mAh or will the battery push out more due to its high capacity?</li>
<li>And if it's dangerous, how can I protect myself and my system?</li>
</ol>
<p>Hope my information in this post is sufficient, thanks for any responses.</p>
| |power-electronics|battery| | <p>These batteries are designed for quite high current applications so can provide quite a punch. It won't hurt you but if you short circuit it, it will get very hot very quickly. Short-circuits aside, it should be fine. I used to set fire to steel wool using one of these...</p>
<p>I would not waste time or money on the DC DC converter as the arduino will handle 6V very effectively.
Also, if you look at the voltage curves on the datasheet, you will see that the voltage will rapidly decline to 5V anyway so a converter would just be draining power.</p>
<p>Have a look at <a href="http://gadgetmakersblog.com/arduino-power-consumption/" rel="nofollow">http://gadgetmakersblog.com/arduino-power-consumption/</a> which discusses arduino power consumption.</p>
<p>As others have suggested try to transmit your information infrequently to save power. Unless you really need real-time data, you could consider storing the data on an SD card and retrieving it when the project is over. Of course someone might run off with it and the internet is so much more fun!</p>
| 10690 | Low voltage, high ampere battery safety |
2016-07-15T08:17:18.590 | <p><a href="https://i.stack.imgur.com/cbxRR.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/cbxRR.jpg" alt="enter image description here"></a>I am sorry if this question has been asked before or if the answer is obvious, but I cannot seem to find the answer. A lot of the structural engineering problems we do involve calculating the elongation of certain types of beams/cables etc while under tensile stress. In particular there was a relatively straightforward question that asked for the total elongation of a structure comprised of two connected steel cables which have different cross sectional areas. Each of the two cables had consistent cross sectional area, and the structure was placed under an axial load of 800N.</p>
<p>Solving for the total elongation was simple with delta = (FL)/(EA) for each cable and adding the two deltas. My confusion however is with why this is considered the total elongation. </p>
<p>After the structure has undergone this elongation, it will become a 'new' structure, with a different length and different cross=sectional area for each cable. My question is, if the load continues to be applied after this 'initial' elongation (delta1+ delta1), then wouldn't the new structure still be under direct stress, and thus based on the new E of each of the now deformed cables, would each deformed cable undergo deformation again? And would this process not repeat until the object yields?</p>
<p>I understand that in tensile testing a specimen's stress-strain graph will be obtained by deforming the specimen until it has yielded. This is why I am confused because under load, would these cables not just deform until they yielded? And if so, wouldn't total elongation then be based on the final length just before the yield point?</p>
<p>Edit: The two steel cables in the questions are connected together, with the cross-sectional areas of the two cables being distinct, i.e. a smaller circle stuck onto a larger one</p>
| |steel|stresses| | <p>I got a -1 for my answer. I don't know what it means but it probably isn't good. So I will try again.
There have been what I thought were some very good answers but I’m not sure if the question has been totally resolved. The analogy to a spring seems to be a good one. For stresses below the proportional limit, a tensile force applied to a steel rod behaves just like a spring.</p>
<p>In the figure below, the linear elastic portion of the stress-strain curve is shown. The data for these curves is collected in the laboratory using a pull tester linked to a computer. The computer uses the data to plot the curves. Loading and unloading is applied to the specimen. When an applied force produces stress in a specimen below the proportional limit, no yielding will occur. If the force is removed from the laboratory specimen, the unloading curve will follow the linear-elastic portion all the way back to the origin. This behavior is consistent with the definition of elasticity. If the elastic limit is exceeded by a small amount prior to removal of the load, a permanent strain or "offset" will result. The unloading curve will follow the same slope as the elastic curve. However, the strain will not return to zero even though no loading is applied. In general, structures are designed to ensure that they remain elastic.
<a href="https://i.stack.imgur.com/87oFg.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/87oFg.png" alt="enter image description here" /></a></p>
| 10710 | How does a constant load affect the elongation of a steel cable? |
2016-07-16T00:58:48.913 | <p>To experienced supercar owners/engineers:</p>
<p>What is the best fuel for a turbocharged 6 liter V8, 7.6 liter V10, or 9.2 liter V12 (Audi R8 or Lamborghini Aventador type cars) in terms of speed/horsepower efficiency and minimization of wear on the engine? Mileage efficiency can be discarded (it can do 4 gallons per mile... not a problem :) ).</p>
| |energy-efficiency|fuel-economy| | <p>Use the fuel specified by the manufacturer in the car's manual and/or near the gas cap. Likely a High Octane Fuel based on your question.</p>
<p>If you're referring to Octane levels... The higher the Octane, the harder it is to ignite the fuel. In the case of High Compression Engines(typically in sports cars like you mention in your question) lower octane fuel can detonate on the compression stroke before the piston actually reaches the point where the spark plug fires. This is not good. Higher octane fuels will not pre-detonate under higher compression.</p>
<p><em>Newer</em> engines can handle lower octane, but performance will suffer.</p>
<p>You can read more here: <a href="http://www.contactmagazine.com/Issue54/EngineBasics.html" rel="nofollow">A bit about early detonation</a></p>
| 10722 | Best fuel for supercar engines |
2016-07-17T04:11:51.407 | <p>I used a very inexpensive circuit based on the TP4056 chip and built a Lithium Ion battery charger for 18650 batteries.</p>
<p>The problem is, the batteries I use have so much capacity that the circuit, which is limited to only charging at 1 amp, takes many hours to charge my batteries.</p>
<p>What I would like to do, is build a circuit that I can add to the one I purchased. This new circuit would (ideally) sample the current flow from the charger circuit to the battery, and multiply it by two. My batteries can handle a charge current of 3 amps safely. All I want to do is charge them at 2 amps.</p>
<p>I have attached a diagram that I made hopefully clearly illustrating what I am looking for. Obviously any and all of it is subject to modification based on new information, knowledge etc.</p>
<p><a href="https://i.stack.imgur.com/rkshO.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/rkshO.jpg" alt="Existing and desired circuit diagram"></a></p>
| |electrical-engineering|consumer-electronics| | <p>What you ask for won't work.</p>
<p>First, if you were to create a current amplifier as you describe, it would require a separate power source. You can't cheat physics. Twice the current out at the same voltage means twice the power. The additional power has to come from somewhere other than the input current.</p>
<p>If you have additional power available, it would be better to use it directly to create a higher current charger than trying to amplify the output of a low current charger.</p>
<p>Second, the charger is managing the battery by creating a particular current and voltage profile over time. The charger will get the wrong idea of what the battery is doing with a current amplifier between it and the battery.</p>
| 10739 | Looking for a current multiplier circuit - if possible |
2016-07-17T11:03:32.877 | <p>I want to build a small remote-controlled submarine, and have been thinking about how to handle the ballast tanks. The simplest option seems to be a piston shaft, which fundamentally needs some kind of (sealed) linear actuator with sufficient force to overcome water/air pressure at reasonable depths (say 0m to 50m).</p>
<p>The problem is I don't have an infinite supply of power available on-board and all the linear actuators I found use ludicrous amounts of it, such as over 1 amp at 12 volts... but I only need to move the piston every now and then, most of the time it just needs to sit wherever I tell it to and not move.</p>
<p>My question is, once a linear actuator has been "actuated" into place, can I de-power it and expect it to hold up to its maximum rated load? Does it depend on the design of the actuator? If not, what are some engineering approaches I can use to mechanically lock the piston into place until I need to unlock it, so that I can trade off precious electrical power for mechanical strength?</p>
| |mechanical-engineering|motors|actuator|linear-motors| | <p>Some linear actuators are self-locking — they can't be back-driven by force on the output connection. A leadscrew with a fine enough thread pitch is one example, and since you don't need a lot of speed for your application, wold probably be a good choice.</p>
| 10744 | Do linear actuators need to stay powered once in place? If not, how can I lock them? |
2016-07-17T22:49:27.983 | <p>The vast majority of wind turbines I have ever seen feature 3 blades, as shown here:</p>
<p><a href="https://i.stack.imgur.com/InIXy.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/InIXy.jpg" alt="3" /></a></p>
<p>Recently while driving through western new York state, however, I passed by several turbines featuring only two blades, as shown here:</p>
<p><a href="https://i.stack.imgur.com/G5FMx.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/G5FMx.jpg" alt="2" /></a></p>
<p>What advantages and disadvantages would a 2-bladed wind turbine have over 3-bladed versions?</p>
| |power-engineering|wind-power| | <p>A 2-bladed wind turbine is less stable mechanically than 3 (or more) blades. </p>
<p>Because the two blades are in line, it is much easier to twist the hub of the turbine in the direction along the line of the blades than to twist it at right angles to the blades. If the turbine is being rotated because the wind direction changes, this will cause an unbalanced twisting force on the hub (and the pole) at twice the rotation frequency of the blades.</p>
<p>This unwanted twisting of may feed back into the blades causing them to vibrate. If the speed of the twisting matches the natural vibration frequency of the blades, this could lead to catastrophic mechanical failure. As a consequence of this, 2-blade turbines usually have to be rotated slower than those with more blades, if the wind direction changes, but the wind blowing in an off-axis direction also produces the same unwanted twisting effects. Catch-22!</p>
<p>The hub mechanism and/or the pole itself may need to be designed stronger or stiffer to resist these unwanted twisting forces.</p>
<p>With 3 or more blades, the mechanical behaviour of the turbine is actually the same in <em>all</em> possible orientations of the blades. Even if there are even number of blades, for example 4, there is nothing "special" mechanically about the 90-degree angle between two adjacent blades but economically 4 blades are more expensive to make than three.</p>
<p>The mechanical issues with 2-blade turbines can be "designed out," but the optimum number of blades to generate the maximum <em>power</em> for a given size of turbine is typically 4, ignoring considerations like the extra weight of more blades. As such, a 3-blade design is often the best "minimum cost" compromise between a simple and reliable design, high power output, and low weight.</p>
| 10754 | Advantages/Disadvantages of two-bladed windmill? |
2016-07-18T14:27:39.313 | <p>I have come across various values for maximum flow velocity of water in pipes.
At university we were told that 2 m/s is the upper limit where flow becomes turbulent. </p>
<p>In various planning reports I have read recently, 1 m/s is stated as the upper limit. I could calculate the exact velocity at which water in a specific size and type of pipeline will no longer flow in a laminar way, but for infrastructure planning purposes, this exercise is not required. What would a reasonable value for potable water be? Alternatively, what would the value for raw water (river) be?</p>
| |civil-engineering|water-resources| | <p>Sometimes existing infrastructure limitations, such as chilled water pressure or pump capabilities, play a role in pipe sizing. In such cases, the available pressure drop may dictate the pipe size, regardless of general rules of thumb.</p>
| 10767 | Rule of thumb maximum flow velocity of water in pipes |
2016-07-18T18:23:18.170 | <p>I have an application that is drawing fluid from a reservoir via a standard dip tube. This application has a special requirement that once the fluid is depleted from the reservoir, the pump pulling liquid from the reservoir needs to shut off before drawing air into the system. </p>
<p>The function of this type of valve would be more or less the opposite of a floating shutoff valve used in toilet tanks where the floating mechanism closes the valve once the tank is filled. My application would do the opposite, where the float closes a valve on the dip tube or shuts of the pump once the tank is empty (or close to it).</p>
<p>Does anyone know if this type of valve exists on the market, or is it something I need to design on my own? Thanks!</p>
| |pumps|hydraulics|valves| | <p>Drop Valve, or Flapper Valve.</p>
<p>A Flapper valve is shown in this wikipedia diagram of a toilt flush cistern, it shuts off the outflow when the tank has emptied.</p>
<p><a href="https://en.wikipedia.org/wiki/Ballcock" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Ballcock</a>
<a href="https://en.wikipedia.org/wiki/Ballcock#/media/File:Gravity_toilet_valves_at_rest.svg" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Ballcock#/media/File:Gravity_toilet_valves_at_rest.svg</a></p>
| 10769 | Reverse float shutoff valve |
2016-07-20T03:40:14.553 | <p>In our dynamics class, it is highly recommend to us to use D'Alembert's approach rather than the traditional approach to solving dynamic problems. For example, consider the problem of a block resting on a surface (we know the cof): Rather than calculating the frictional force of the block on the slope and comparing that to the component of weight acting down the slope, we draw the free body diagram including the inertial force and directly solve for a, which will tell is if the block is moving or not. My question is why this method is recommended over the traditional approach? It seems the answer is usually with regards to making it easier to solve more complex problems, but I don't quite understand why this trivial manipulation of Newton's second law makes it much easier to solve problems? </p>
<p>I'm sorry if this has been asked before, but I couldn't quite find a satisfactory answer online!</p>
| |dynamics| | <p>I don't really understand how D'Alembert's principle "helps" in solving the problem in your OP, but I assume your teacher wants you to practise using it in simple situations before you try more complicated ones.</p>
<p>The real value of working in a <em>non-inertial coordinate system</em> is when such a system makes it easier to describe the problem.</p>
<p>For example, suppose you want to model the dynamics of an aircraft. Planes can move in any direction in 3-dimensional space, and the nose of the plane can also be <em>pointing</em> in any direction - not necessarily in the same direction that the plane is moving (see any video of an aerobatics display for examples!)</p>
<p>The easiest way to described the forces created by the plane's engines and control surfaces (rudder, tailplane, ailerons) is to use a coordinate system <em>fixed to the plane</em>. Since this coordinate system may be accelerating and/or rotating relative to the ground, it is not an inertial coordinate system and you have to include the d'Alembert forces depending on how it is moving.</p>
<p>Another example is the dynamics of rotating machinery. Suppose you want to model the vibrations of the blades of a wind turbine, while it is rotating at constant angular velocity. The easiest way to set up the model is in a coordinate system that rotates with the blades, including the d'Alembert forces. In general the blade vibration frequencies will <em>not</em> have any simple relation to the speed of rotation of the whole windmill, and they will depend on the rotation speed. In fact, because of the Coriolis forces on the system, even in the rotating system the blades don't vibrate back and forth in simple harmonic motion in a straight line, like the systems you study in a first dynamics course. Each point on the blade actually moves in around an elliptical path whose center is at the "non-vibrating" position. Trying to describe that motion, and set up the equations of motion, in an inertial coordinate system fixed relative to the ground would be much more complicated than using a coordinate system relative to the blade itself. And if you use a computer model to solve the dynamics of the system, the easiest way to visualise the results is to make an animation in the same rotating coordinate system - i.e. to display what you would see if you were "sitting on the blade" as it rotates.</p>
| 10781 | What is the advantage of using D'Alembert's principle when dealing with dynamic systems? |
2016-07-20T05:28:43.407 | <p>I have a 1.5 mm thick plastic sheet part which hast to be bent at a joint 0.3 mm thick by ~ 10 Deg. I know (found from design drawing) compressed and stretched lengths of the joint. How can I find the length of the joint in non-bent (flat) position so that when it is bent the compressed and stretched joint lengths have the initial known lengths ?</p>
<p><a href="https://i.stack.imgur.com/rH10e.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/rH10e.png" alt="enter image description here"></a></p>
| |mechanical-engineering|structural-engineering|materials|plastic|deformation| | <p>If we can assume that the material behaves in an elastic linear way and that the joint is only loaded by bending, then as you commented the neutral line lies in the middle and that deformation on the two extreme fibres is equal but in opposite directions. Then, original length is just the average of both deformed lengths.</p>
<p>Of course, to get this result we must assume too that the conditions for the Euler–Bernoulli beam theory holds, and according to your diagram we might approaching its limits (short span compared to width, not very small deformation...) but I think it's still a good approximation.</p>
<p>However, if the material is not linear elastic, location of neutral fibre depends on the constitutive equation of the material (the stress-strain diagram), and the original length may not be the average of both deformed lengths, specially if the constitutive equation is not symmetrical.</p>
| 10782 | Finding initial length of a bended plastic sheet (Bend Allowance) |
2016-07-20T06:27:43.213 | <p>I have a plate of aluminum 5 mm thick to which I would like to attach a hook. The plate runs roughly vertically, but only spans around 18 cm from anchor point to anchor point. The hook should be able to withstand upwards of 1000 kg.</p>
<p>I've drilled a 12 mm hole through the plate and my strategy is to insert the hook bolt in it and tighten it using a nut on each side.</p>
<p>However, in order to keep the aluminum from deforming, I'm going to need one or two steel washers.</p>
<p>My question is, whether it is necessary to use a washer on both sides or whether one side is sufficient?</p>
<p>If one is enough, does it make any difference, which side it rests on?</p>
<p>My reasoning goes like this: The job of the washer is to keep the hook from bending the aluminum by spreading the force over a large area. As long as a single washer is taut against the aluminum – held there by the nut – the hook won't be able to bend either way.</p>
| |mechanical-engineering| | <p>While your setup will probably work either way, it's advisable to use a washer on both sides.</p>
<p>A washer does help prevent local bending as you note, but that isn't really its main purpose. The biggest reason to use a washer is to prevent the nut from cutting into or shearing through the base material. In your case where the hook bolt and nut are steel, and the plate is softer aluminum, this is especially a concern. As you tighten the connection, without a washer the nuts are likely to dig into the aluminum and start to act like a drill. It also wouldn't be too hard for the nut without a washer to be pulled through the plate (causing a shear failure in the shape of the nut) if there isn't a washer.</p>
<p>Note that in a similar case, using a shouldered eye bolt where the bolt, the substrate, and the nut are the same material, it would be customary to only use the washer on the nut side, because the bolt is not designed to ever be loaded in compression, only in tension against the nut. Because of this the bolts are designed with a shoulder that isn't as big as a washer, but meets the needs of the bolt for the angles it is rated at. Washers may be added when required for thread length or orientation.</p>
<p>Also note that based on your description it sounds like you may be side-loading the hooks which is generally not how they are designed to be used. If you want to make a separate question with the plate, the hook selection, the geometry, and the load, you could make sure you are using the hardware correctly. 1000 kg is a very serious load and if it's going to be anywhere near people, the setup should be evaluated according to standard rigging practices in your country.</p>
| 10783 | Washers on both inside and outside? |
2016-07-21T04:29:35.613 | <p>I am referring to the answer <a href="https://engineering.stackexchange.com/a/8915/3353">here</a>:</p>
<p>It seems that on a simple structure </p>
<p><a href="https://i.stack.imgur.com/Lz5TT.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Lz5TT.png" alt="enter image description here"></a></p>
<p><em>(The diagrams on the left the images below have fully-fixed connections, while on the right the columns are pinned connections to the beams.)</em></p>
<p><strong>A: The deflection on beam is bigger in pinned model, but the deflection on beam/column joint is bigger in fixed model</strong> </p>
<p><a href="https://i.stack.imgur.com/TFSzW.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/TFSzW.png" alt="enter image description here"></a></p>
<p><strong>B: The moment on column in pinned model is 0</strong></p>
<p><a href="https://i.stack.imgur.com/np5DW.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/np5DW.png" alt="enter image description here"></a></p>
<p><strong>C: The axial load on column in pinned model is bigger, because the moment is being transformed into axial load (?? is this reasoning true??)</strong></p>
<p><a href="https://i.stack.imgur.com/yBlPd.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/yBlPd.png" alt="enter image description here"></a></p>
<p>My questions are:</p>
<ol>
<li>These conclusions are true for this particular simple model, but are these three conclusions always true on any general model?</li>
<li>And why? Can it be explained in terms of loading flow and statics? Or this is just FEM behavior that we can't explain more intuitively?</li>
</ol>
| |structural-engineering|structural-analysis|statics| | <p>Let me start by answering your second question: models such as this one, which involve only one-dimensional beam elements, are 100% analytical and can therefore always in theory be understood intuitively. There is no "FEM behavior" for such models. Sometimes the models may get complex with lots of bars and whatnot, which may make "intuitive explanations" more difficult, but the result will always be analytical.</p>
<hr>
<p>Let's start by looking at statement <strong>B</strong>:</p>
<p>Now, let's take a look at beam 4 in your model (the left-most beam). More specifically, it's bending moment diagram. As you've noticed, the pinned model displays zero moment at the left-most column. This is the very definition of a hinge and is expected behavior. The moment on the beam at the central column is non-zero because the beam itself is not hinged, but the central column is hinged and therefore displays zero moment at the node.</p>
<hr>
<p>Now, on to statement <strong>A</strong>, starting by looking at the beam's deflection:</p>
<p>Let's keep looking at the bending moment diagram. The <a href="https://en.wikipedia.org/wiki/Euler%E2%80%93Bernoulli_beam_theory#Static_beam_equation" rel="nofollow">beam equation</a> tells us that</p>
<p>$$\dfrac{\partial^2 }{\partial x^2}\left(EI\frac{\partial^2 w}{\partial x^2}\right) = q$$</p>
<p>which also tells us that</p>
<p>$$EI\frac{\partial^2 w}{\partial x^2} = M$$</p>
<p>that is: bending moment (divided by stiffness $EI$) is the second derivative of deflection. From calculus, we know that the second derivative of any function described the function's curvature. So bending moment describes deflection's curvature, which describes the "acceleration" with which the beam's tangent (the first derivative of deflection, and therefore bending moment's integral) changes.</p>
<p>So, the more balanced a bending moment diagram is between positive and negative bending moment, the more the total "acceleration" cancels itself out, implying in smaller tangent changes, and therefore smaller deflections. So yes, a fixed node will always lead to smaller deflections</p>
<hr>
<p>To answer the matter of the node's displacement, we first need to explain point statement <strong>C</strong>. For that we need to look at beam 4 in isolation. To do so, we need to replace the surrounding beams with elastic supports which describe their stiffness.</p>
<ul>
<li>The vertical supports' stiffness will be equal to the columns' axial stiffness (the node with the central column will also have a tiny addition due to the other beam's stiffness against imposed transversal displacements)</li>
<li>The horizontal supports' stiffness will be equal to the columns' stiffness against imposed transversal displacements</li>
<li>The rotational supports' stiffness will depend on the boundary conditions. If hinged, then the outer node will have zero stiffness and the central node will have a stiffness equal to the other beam's stiffness against imposed rotations. If fixed, then both nodes will have the columns' stiffness against imposed rotations, adding the other beam's stiffness as well for the central node.</li>
</ul>
<p>So, basically, the only difference between the hinged and fixed cases is in the rotational stiffness (as would intuitively be expected). This increased stiffness, however, causes the node to pull in a greater proportion of all forces, thereby increasing the axial forces in your outer column and reducing them in the central column in the fixed model.</p>
<hr>
<p>Returning to the issue of the node's deflections, they are now easy to explain. After all, in the fixed model the column suffers more axial forces, naturally increasing the vertical deflections. But it also suffers bending moment, which generates horizontal deflections as well as a tiny bit of additional vertical deflection.</p>
| 10789 | Is pin column-beam joint always have less moment, more deflection compared to fixed joint? |
2016-07-21T12:24:01.647 | <p>I'm spec'ing out a conveyor belt and want a control mechanism to make sure that it runs straight. I know there exists a product for this since I've seen the function in an older tender, but don't have a clear idea how the device looks like and what it does - googling *"Geradelaufkontrolle Förderband"' didn't yield anything helpful.</p>
<p>Is there a device that gives me an alarm when the belt runs off course, what is it called, and how does it work?</p>
| |mechanical-engineering|sensors| | <p><strong>Alignment Monitoring</strong></p>
<p>There are many switch systems that can be mounted to the frame that will detect when a conveyor is beyond the acceptable limits of alignment. A google of "Conveyor Belt Alignment Sensor" brings this within the first page.</p>
<p><a href="http://www.go4b.com/usa/technical-support/product-manuals/belt-misalignment/touchswitch-belt-alignment-sensor-manual.pdf" rel="nofollow">http://www.go4b.com/usa/technical-support/product-manuals/belt-misalignment/touchswitch-belt-alignment-sensor-manual.pdf</a></p>
<p>Sensors will assist with major changes in alignment and sudden failures but they do not compare with the ability of a trained belt manager who walks the belts at least every shift.</p>
<p><strong>Equipment Install</strong></p>
<p>While the travel rollers on a conveyor system can be set to correct for a minorly inconsistent alignment of the conveyor structure it adds additional wear to the bearings, rollers and belting. The additional wear will also reduce the efficiency of your conveyor if that is a concern to your operation. In all situations I recommend laser levels for the install of conveyor structure (even in semi-mobile installs) If this is a cross country type conveyor system that will deal with variable terrain crossing, vertical and horizontal curves or surface to elevated structure transitions using a knowledgeable surveyor is critical to alignment and the foundations should not be underestimated. </p>
<p><strong>Operating Alignment</strong></p>
<p>When major components on the conveyor are replaced or at the completion of the install the start up and training of the conveyor is important to insure that there isn't a failure due to premature wear. This site has a good set of basic knowledge about training and individual adjustment for travel rollers.
<a href="http://www.shippbelting.com/BeltTraining.aspx" rel="nofollow">http://www.shippbelting.com/BeltTraining.aspx</a></p>
| 10791 | How to determine / control if a conveyor belt runs straight? |
2016-07-21T22:56:32.703 | <p>Using WiFi or bluetooth frequencies 5.8GHz and 2.4-2.485GHz respectively, I want to place a hermetically sealed circuit at the bottom of a gold fish bowl, and want to know if the water will significantly attenuate WiFi or Bluetooth.</p>
<p>Gold fish bowl would be probably 6" to 1', or more.</p>
<p>So, I want to know rate of signal absorption per distance through intervening medium of water.</p>
| |electrical-engineering|materials|telecommunication|rf-electronics| | <p>There is information out there, but probably the best way to find out is to actually try it. Grab a small Bluetooth or Wifi device and seal it in a waterproof enclosure or bag, and submerge it to the bottom of the fish bowl under water. See what happens. This should provide more valuable insights than the simple theoretical RF attenuation of water per meter, since in the fish bowl you have an irregular geometry and also the device will have a short path for RF out the bottom of the bowl.</p>
<p>Some references that might be useful as well:</p>
<p>A forum thread reported that <a href="http://www.societyofrobots.com/robotforum/index.php?topic=8408.0" rel="nofollow">Bluetooth under water doesn't work</a>.</p>
<p>A white paper from Laird Technologies says, "The 5 GHz wave form is attenuated by common building materials to a greater degree than the 2.4 GHz wave form. <strong>On the other hand, the 2.4 GHz wave
form is optimally absorbed by water.</strong>" (<a href="http://www.digikey.com/Web%20Export/Supplier%20Content/Laird_776/PDF/laird-wireless-optimizing-operation-5ghz.pdf" rel="nofollow">Optimizing Operation at 5 GHz</a>, p. 4)</p>
| 10799 | Will the water in a gold fish bowl signficantly abosorb Bluetooth or WiFi if transciever is at bottom of bowl? |
2016-07-22T04:03:06.923 | <p>Following on from <a href="https://engineering.stackexchange.com/a/10793/255">this answer</a>, I am reading about the difference between hinge and pin connections <a href="http://www.tboake.com/SSEF1/pin.shtml" rel="nofollow noreferrer">here</a>, and I don't quite understand about it. I think I would need to see the mathematical representation ( the boundary condition) involving pin and hinge connection, in order to actually understand it.</p>
<p>How can I represent, in mathematical terms, the pin and hinge connection? And how can it be represented in the <a href="https://en.wikipedia.org/wiki/Direct_stiffness_method" rel="nofollow noreferrer">Stiffness matrix method</a>, as boundary conditions (maybe)?</p>
| |structural-engineering|structural-analysis|statics| | <p>I believe you have chosen a poor reference. Indeed, that link has absolutely nothing to do with structural analysis, but rather a specific implementation in steel structures. </p>
<p>A pinned support is a boundary condition which restricts all displacements but allows the structure to rotate.</p>
<p>A hinge (more formally, an internal hinge), on the other hand, is a description of the structure's behavior. Specifically, it removes the rotation compatibility between bars around a node (between all bars or a subset of them), thereby increasing the structure's degrees of freedom. </p>
<p>For a more visual demonstration of the difference, here's a structure with four pinned supports, along with its deformed configuration under the effect of a uniform load. You'll notice that the supports allow the structure to rotate around them, however one beam's inclination at a given node must be equal to the inclination of the neighboring beam on the same node, meaning there may be no discontinuity in the derivative of the deflection at the support (compatibility of rotations).</p>
<p><a href="https://i.stack.imgur.com/Sfa9R.png" rel="noreferrer"><img src="https://i.stack.imgur.com/Sfa9R.png" alt="enter image description here"></a></p>
<p>And now two variations on the same structure with the same loading, but with an internal hinge at different locations.</p>
<p><a href="https://i.stack.imgur.com/fQBHW.png" rel="noreferrer"><img src="https://i.stack.imgur.com/fQBHW.png" alt="enter image description here"></a></p>
<p>In this case, an internal hinge is placed on the same position as one of the supports. This removes the compatibility of rotations around that node, leading to a clear discontinuity in the derivative of the deflection around that support.</p>
<p><a href="https://i.stack.imgur.com/yPKKz.png" rel="noreferrer"><img src="https://i.stack.imgur.com/yPKKz.png" alt="enter image description here"></a></p>
<p>Now the hinge is placed in the middle of the central span, leading to a clear discontinuity in the derivative of the deflection at that point.</p>
<p>Hinges can also be placed to just one side of a node. For instance, look at the deflection of this frame under a uniform vertical load and a concentrated horizontal load.</p>
<p><a href="https://i.stack.imgur.com/NvOGv.png" rel="noreferrer"><img src="https://i.stack.imgur.com/NvOGv.png" alt="enter image description here"></a></p>
<p>The columns and beams all maintain their original perpendicularity. Specifically at the central column, notice that the beams maintain their angular compatibility (no discontinuity in the derivative of the deflection).</p>
<p>If, however, we add hinges to the tops of the columns:</p>
<p><a href="https://i.stack.imgur.com/FtWPe.png" rel="noreferrer"><img src="https://i.stack.imgur.com/FtWPe.png" alt="enter image description here"></a></p>
<p>The beams and columns no longer remain perpendicular in their deflected state. The beams over the central column, however, maintain their angular compatibility.</p>
<p>If, however, the nodes were fully hinged, then each span would behave independently like a simply-supported beam:</p>
<p><a href="https://i.stack.imgur.com/77nrI.png" rel="noreferrer"><img src="https://i.stack.imgur.com/77nrI.png" alt="enter image description here"></a></p>
<hr>
<p>In the direct stiffness method, the fundamental equation is</p>
<p>$$\{q\}=K\{d\}$$</p>
<p>A pinned support defines the contents of $\{d\}$, by setting the degrees of freedom of the given node to $(d_x, d_y, \theta) = (0, 0, \theta)$ (in the case of a 2D structure). An internal hinge, on the other hand, increases the dimensions of the calculation, since it increases the number of degrees of freedom. So, instead of a node having merely $(d_x, d_y, \theta)$, it will have $(d_x, d_y, \theta_1, \theta_2, \theta_3, ...)$, where $\theta_i$ represents the rotation on different sides of the hinge.</p>
| 10800 | Mathematical representation of hinge and pin connection |
2016-07-22T16:14:55.467 | <p>I have an application that requires torque limiting and I need to determine if VFD torque limiting will provide sufficient protection. I have used VFDs for torque limiting through gearboxes before with success, but I am hoping to find a way to calculate the result as opposed to just gut feeling.</p>
<p>10HP motor<br>
1775 rpm <br>
69.5:1 gearbox reduction (im assuming 50% efficient)<br>
25.5 rpm at driving sprocket<br></p>
<p>I have looked at some frequency drives to get an indication of their torque/amp sensitivity. Ive seen 0.1A, 0.01A, 0.1% of torque and 0.01% of torque as control resolutions, but have not found any literature specifically addressing if this resolution can be reasonably expected.</p>
<p>In this situation I am gearing down, so inertia, backlash and controller reaction time will not be an issue.</p>
| |control-engineering|motors|torque| | <p>Lots of assumptions. Some gearbox data and empirical motor current data would go a long way to predicting the performance.</p>
<p>      Sensorless vector mode <br>
      Startup torque will be ignored by the drive <br>
10     hp Powerflex 700 VFD <br>
10     hp motor <br>
1775     rpm <br>
29.6     ft lbs <br>
69.5     gear box ratio <br>
200     ft lbs assumed constant gearbox torque consumption, referenced on output <br>
50%     worst case planetary gearbox efficency (in additon to constant) <br>
26     rpm @ driving sprocket <br>
928     ft lbs @ driving sprocket <br>
95%     estimated chain efficency <br>
19     rpm @ driven sproket <br>
1176     ft lbs @ driven sprocket <br>
      <br>
      Full 10 hp has previously damaged mixing rotor <br>
1176     ft lbs Estimated emperical damaging torque <br>
700     ft lbs, desired torque limit <br>
553     ft lbs, pre gearbox, desired torque limit <br>
18.8     ft lbs seen by motor, desired torque limit <br>
500     ft lbs Estimated Normal opperational torque, (motor needs amp'd for emperical data here) <br>
395     ft lbs, operational torque <br>
14.2     ft lbs seen by motor, operational torque <br>
      <br>
15.4%     Torque as range of total <br>
      <br></p>
<p>0.1% percent of motor rated toque, torque resolution. It is not clear if this requires torque feedback or not.
<a href="http://literature.rockwellautomation.com/idc/groups/literature/documents/um/20b-um002_-en-p.pdf" rel="nofollow">Page 22 of powerflex700 programming manual</a> <br>
No closed loop torque sensor will be used as they are prohibitively expensive <br>
<br>
Worst published torque regulation on the drive is +/-5% in the <a href="http://literature.rockwellautomation.com/idc/groups/literature/documents/td/20b-td001_-en-p.pdf" rel="nofollow">powerflex700 technical data</a> <br>
<br>
15.4% > 5%, VFD torque limiting will be satisfactory in this application <br></p>
| 10807 | Torque sensitivity of a VFD induction motor through a gearbox? |
2016-07-23T19:13:19.997 | <h2>Ive been wondering about how small of a cooler you can build.</h2>
<p>Although this probably is just really low-level engineering, i have a rather particular question.</p>
<p>For my idea, the cooler has to be as small as possible, <em>rather efficient</em> and <em>not use a chemical compound or reaction that is not locally reversible</em>. It has to be an fully working circle that, theoretically, keeps working, solely relying on electricity for power.</p>
<p>While I understand how certain cooling systems work, I could not find a satisfying example, since mobile phones, tablets and other handheld devices rely on, <em>if any</em>, air cooling.</p>
<p>However, what I'm searching for woul'd be a small cooler that can reach slightly lower temperatures than the surrounding area, lets say about 5 °C, maybe less or even more, if it just gets <strong>below average surrounding temperatures</strong> (room-temperatures).</p>
<p>Would this be achievable in a <strong>watch-like size</strong>?
And <em>how efficient</em> woul'd that be, <strong>running on electrical reloading</strong> only?</p>
<p><em>EDIT</em>:
The heat would, as far as i understand thermodynamics, not be a big problem.
A short-range <em>"relocation"</em> (idk the term) <em>about a few centimeters</em> to a <em>small vent</em> would be enough.</p>
| |electrical-engineering|heat-transfer|cooling| | <p>A <a href="https://en.wikipedia.org/wiki/Thermoelectric_cooling" rel="nofollow">Peltier cooler</a> would fit the requirements you set out. They can be compact enough to fit the 'wristwatch sized' requirement and use electrical power to achieve the cooling effect. </p>
<p>You do still need some way of shedding heat from the hot side of the cooler and this may require either a passive heat exchange, heat pipe etc depending on the application. </p>
<p>The performance will depend not just on the temperature difference you want to achieve but also on the heat flux required. </p>
<p>The most common comparison would be with the coolers used in laptops which tend to be heat pipe based. Typically a modern CPU would generate up to about 100W of heat, depending on load and performance and look to limit internal temperatures on the hot side to about 40-50 deg C. This isn't exactly the same as the refrigeration you require but should give you some sense of what is possible. </p>
<p>If you have a bit more space to play with water cooling is now pretty mainstream for high performance PCs although most most units are substantially bigger than wristwatch size they will give substantially better performance than a laptop heat spreader. </p>
| 10819 | How small can coolers get? |
2016-07-24T16:31:49.647 | <p><a href="https://i.stack.imgur.com/Hshwjm.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Hshwjm.jpg" alt="this"></a> </p>
<p>From the picture above, in order to calculate the moment of load acting on the beam we take moment $M=\frac{W}{L}×L×\frac{L}{2}$.</p>
<p>Why can't we calculate the moment as $M=\frac{W}{L}×L×L$?</p>
<p>Why do we need to take the distance from center of gravity to a point as the distance?</p>
| |civil-engineering|beam|construction-management| | <p>A point force produces a moment with respect to a point equal to the product of the force by the distance. To compute the moment produced by a system of forces, we can compute the moment for each force and sum the individual moments, or we can just sum the forces and multiply them by the average distance, therefore getting the same moment. Finding the average distance is the same as finding the centre of gravity of the system of forces, since the distance to the center of gravity is the same as the average distance to the forces in the system.</p>
<p>If I'm understanding you, in your formula you are using the distance to the most distant point with distributed load, but most of the load is acting at a shorter distance and therefore producing less moment that the value you are accounting for. If we use the distance to the centre of gravity, some loaded points would be closes to reference points (producing less moment) and some will be further (producing more moment) but the global result will be right.</p>
<p>However, beware that the moment you gave in your question is valid for a cantilever beam with uniform load, and that's not what is in your drawing shows. Your drawing shows a simple supported beam, that is, supported at both ends. The maximum moment in a simple supported beam is $M=\frac{W}{L}\times\frac{L^2}{8}$. </p>
<p>And just a last note: beware that the definition of moment I've used is valid when the direction of the force is perpendicular to the distance to the point, as usually happens when computing moments in beams. If they are not perpendicular, definition of moment should be adjusted.</p>
| 10825 | Why do we need to take distance of center gravity of a distributed load to a point in order to calculate moment? |
2016-07-26T00:44:35.567 | <p>I have an old vertical and horizontal cutting machine that is working manually it is kind of useless because it is hard to make accurate measurements. I want to use two digital sensor (with output for any controller) to find the position with less than 1mm error on two dimensions, left-right (with rack and pinion mechanism) 0 to 2,5 m and up-down (with roller screw mechanism) 0 to 1m. What kind of sensor do I have to use under 300 euros?</p>
<p><a href="https://i.stack.imgur.com/4gLrh.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/4gLrh.jpg" alt="rack and pinion mechanism"></a>
<a href="https://i.stack.imgur.com/EyRiX.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/EyRiX.gif" alt="roller screw mechanism"></a></p>
| |control-engineering|sensors| | <p>Buy yourself a DRO (Digital Read Out) kit from eBay/aliexpress or somewhere similar. They are within or close to your budget and should give you significantly more precision and accuracy than 1mm.<br>
<a href="https://i.stack.imgur.com/UFakE.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/UFakE.jpg" alt="DRO kit"></a></p>
| 10838 | How to find the positioning of cutter machine with high accuracy with digital sensor? |
2016-07-26T20:48:06.300 | <p>I understand that poles on the left hand complex plane (LHP) make a continuous linear dynamic system stable. What's so significant about imaginary poles on the LHP that make a system stable? What does it mean to have poles and zeros on the LHP verses the right hand plane (RHP)?</p>
| |mechanical-engineering|electrical-engineering|control-engineering|control-theory| | <p>Briefly, poles in the left half plane (R < 0) symbolize ringing that dies down over time, at R = 0 ringing that stays the same, and in the right half plane (R > 0) ringing that increases in amplitude over time.</p>
| 10848 | In Control Systems Engineering, why do imaginary poles and zeros on the LHP indicate stability? |
2016-07-27T09:40:15.913 | <p>I'm looking for a device that breaks a circuit once the voltage of my battery gets too low. This to protect the battery.</p>
<p>Example: I have a 12V battery, once it reaches ~11V the huge voltage drop begins, but the device that is connected to the battery still runs and draws current. I need another device that can detect the voltage goes below 11V and then automatically breaks the circuit, preventing the circuit to draw more current from the battery. Basically, it shuts down the whole circuit.</p>
<p>Since English is not my native language and I'm not aware of any kinds of systems in my own language, I'm having a hard time finding something like this. Thanks in advance.</p>
<p>EDIT: or can I make my own?</p>
| |circuits|battery| | <p>The simple answer is that you can get off the shelf modules which does exactly what you want. </p>
<p>They are often called 'split charge modules' and are designed for installations like boats and motor-homes which have batteries charged off an engine alternator but prevent the starter battery form being fully discharged. </p>
<p>This is one example <a href="http://www.12voltplanet.co.uk/voltage-sensitive-relay-12v-140a.html" rel="nofollow noreferrer">http://www.12voltplanet.co.uk/voltage-sensitive-relay-12v-140a.html</a></p>
<p>They are not expensive and are easy to fit with basic knowledge of electronic systems. </p>
<p>There are also more sophisticated systems which allow for conditioning and management of auxiliary batteries </p>
<p>I have used this brand before : <a href="https://i.stack.imgur.com/EHlE2.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/EHlE2.jpg" alt="enter image description here"></a></p>
<p>I can't remember the exact specification but they would certainly be in the 1A sort of range for a 12V system. I suggest this one in particular as it is easy to disassemble so you could discard the case and screw terminals to make it a lot more compact to package with the rest of your circuit. </p>
| 10851 | Looking for a device that breaks the circuit if the voltage gets too low |
2016-07-27T13:26:01.567 | <p>What kind of simplified structural system would a Leonardo bridge use?</p>
<p>I've tried approximating the system to a simple statically determined beam (see below) but our calculations did not give any realistic results. I believe it is due to a negligence of the potential strain in each beam. Maybe using a Spring Load Constant combined with nodes would help?
<img src="https://i.stack.imgur.com/3zrdT.jpg" alt="">
A <a href="http://www.leonardodavincisinventions.com/leonardo-da-vinci-models/leonardo-da-vincis-self-supporting-bridge/" rel="nofollow noreferrer">Leonardo bridge</a> is a bridge constructed without any kind of binding, but just using uniform length pieces of Wood. It can be constructed in this manner:
<img src="https://i.stack.imgur.com/0V7RI.gif" alt=""></p>
<p>My own approximation is rough but each of the connections cannot be considered nodes because none of the pieces are loose or movable.
<img src="https://i.stack.imgur.com/oprqM.jpg" alt=""></p>
<p>P.S. It would greatly simplify my calculations if any answers would consider a statically determined system even if it is a simplification. </p>
| |structural-engineering|civil-engineering|structures|bridges| | <p>We will solve for n members with these assumptions.<br>
members are L ft long by 2H, (H=L/20) thick and assuming the notch depth at H on the center to accommodate next member linkage. </p>
<p>Loads are applied uniformly spread vertically over the exposed top of the bridge at m. ton/ meter.<br>
The structure as we add to members looks more and more like an arch. </p>
<p>The radius of which is deduced from similar triangles H/L = L/R<br>
$R=L^2/h$</p>
<p>From here we apply as an approximation the arch formulas with one caveat the moment will be converted to compression and tension on the joints by this formula<br>
$T=C=M/H$ and also shear converts to vertical joint loads.<br>
There are many ways to solve a pin end support arch but an starter is to assume one support on rollers and solve then superimpose the lateral load needed to bring the roller end back in. I have some hand books if you need, but here is a link.<br>
<a href="http://www.civilengineeringx.com/bdac/stresses-in-arches/" rel="nofollow">http://www.civilengineeringx.com/bdac/stresses-in-arches/</a></p>
| 10856 | What kind of simplified structural system would a Leonardo bridge use? |
2016-07-28T07:40:16.567 | <p>In <a href="https://en.wikipedia.org/wiki/Direct_stiffness_method" rel="nofollow noreferrer">direct stiffness method</a>, the beam and column connection are assumed to be done through center-line beam and column:</p>
<p><a href="https://i.stack.imgur.com/x9bAY.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/x9bAY.jpg" alt="enter image description here"></a></p>
<p>And hence we can represent them easily using frame elements ( 1D element) and do not have to model the surface contacts or using area elements.</p>
<p>However, in reality, beam and column may not always be connected via centerline: eg: such configuration is not uncommon</p>
<p><a href="https://i.stack.imgur.com/VnElg.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/VnElg.gif" alt="enter image description here"></a></p>
<p>It is easy to see that the offset of beam shall induce additional moment on the supporting column. </p>
<p>How should I modify my stiffness matrix in order to accommodate the offset of beam from the column center location? The frame representation of beam and column is just too convenient to give up. </p>
<p>For the purpose of this question, assume that I am writing a frame analysis software from first principle ( yes, from the equations on the wiki page that I link to), and hence can't make use of any existing software package </p>
| |structural-engineering|structural-analysis| | <p>Use a stiff element, <a href="https://engineering.stackexchange.com/a/10867/3353">as explained by AndyT</a>, to model the offset might create numerical issues for the finite element model. </p>
<p>Here are some alternatives:</p>
<h2>Alternative 1:</h2>
<p>Build <a href="https://engineering.stackexchange.com/questions/10866/how-can-i-model-beam-column-connection-that-doesnt-connect-at-centerline/10867?noredirect=1#comment19260_10867">on the comment by alephzero</a>, one can model the offset as Multi-Point Constraint (MPC). An explanation is provided <a href="http://what-when-how.com/the-finite-element-method/modelling-techniques-finite-element-method-part-3/" rel="nofollow noreferrer">here</a>:</p>
<blockquote>
<p>The basic idea of using MPC is to create a set of MPC equations that
gives the relation between the DOFs of the two separated nodes. It
assumes that the two corner nodes are connected by a rigid body. MPC
equations are then derived using the simple kinematic relations of the
DOFs on the rigid body.</p>
</blockquote>
<p><a href="https://i.stack.imgur.com/rJmIs.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/rJmIs.png" alt="enter image description here"></a>
<a href="https://i.stack.imgur.com/cjhBJ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/cjhBJ.png" alt="enter image description here"></a>
<a href="https://i.stack.imgur.com/Q2Aay.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Q2Aay.png" alt="enter image description here"></a>
<a href="https://i.stack.imgur.com/U88rT.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/U88rT.png" alt="enter image description here"></a></p>
<h2>Alternative 2:</h2>
<p>Another similar approach is to use the <a href="http://www.colorado.edu/engineering/CAS/courses.d/IFEM.d/IFEM.Ch20.d/IFEM.Ch20.pdf" rel="nofollow noreferrer">Direct Fabrication of offset beam stiffness ( section 20.4.3)</a>, where the stiffness matrix is explicitly modified to incorporate the offset.</p>
<p><a href="https://i.stack.imgur.com/GMDTd.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/GMDTd.png" alt="enter image description here"></a></p>
<p><a href="https://i.stack.imgur.com/lOkib.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/lOkib.png" alt="enter image description here"></a></p>
| 10866 | How can I model beam column connection that doesn't connect at centerline? |
2016-07-28T10:12:42.750 | <p>From what I know, the key to reducing losses due to eddy currents in a core is to reduce the path that they take, as is shown in this <a href="https://upload.wikimedia.org/wikipedia/commons/d/d1/Laminering_av_k%C3%A4rna.svg" rel="nofollow">diagram illustrating solid vs. laminated cores</a> (courtesy of Wikipedia).</p>
<p>Would taking this one step further and making the core a bundle of wires/rods instead of a stack of laminations reduce the losses further? If so, then why aren't there any commercial designs using it?</p>
| |electrical-engineering|electromagnetism| | <p>What stands between each lamination ? Isolator. If you switch from lamination to wires, you'll get much more isolator, and therefore less metal in the same volume of core. Doing so, you'll get a less efficient core, you'll have to increase its size, increasing cost, wire length on the coils, losses in the wires and that's often a bigger drawback than the benefit in metal losses reduction. I'm also doubtful about the mechanical resistance of such core.</p>
| 10869 | Inductor/Transformer core design - reducing losses from eddy currents |
2016-07-29T14:05:38.183 | <p>I have tested a gear box for many hours. Where the gears are made of a hardened steel alloy and the housing is 7000 series aluminum. Now I drained the gear box of its gear oil and I need to weight the metal shavings in that oil. So I can put in fresh oil and test the gear box for a couple more hours. And then do the test all over again. Can anyone give me a good idea as to how I can accurately separate the oil with out losing the tiny pieces of metal? I thought I could filter the oil through coffee filters then give it a thorough cleaning with hot water and soap, then drying it in the oven. Can anyone tell me if this is a good idea or can you come up with a better one?</p>
| |gears|fluid-filtration| | <p><a href="https://www.dartmouth.edu/~chemlab/techniques/vfiltration.html" rel="nofollow">Standard lab approach is vacuum filtration</a>: Get a Buchner funnel and filter paper sized to capture the smallest particles you care to measure. Now:</p>
<ol>
<li>Weigh filter.</li>
<li>Place filter in funnel and wet with solvent.</li>
<li>Connect funnel to vacuum filter flask.</li>
<li>Turn on vacuum.</li>
<li>Pour sample into funnel.</li>
<li>Continue to rinse sample with a volatile solvent until it is clean. For rinsing off oil you could use acetone.</li>
<li>Wait a few moments for solvent to evaporate.</li>
<li>Weigh filter with sample.</li>
</ol>
| 10881 | How can I separate gear oil from metal shavings to weigh worn metal? |
2016-07-30T10:06:42.380 | <p>Recently, I read an October 2006 IEEE JOURNAL OF LIGHTWAVE TECHNOLOGY, article titled, High-Sensitivity Detection of Narrowband Light in a More Intense Broadband Background Using Coherence Interferogram Phase written by the esteemed Ricardo C. Coutinho, Member, IEEE, David R. Selviah, Member, IEEE, and Hugh D. Griffiths, Fellow, IEEE from highly respected University College of London.</p>
<blockquote>
<p>In this excellent article which is available publicly, it describes
that the frame rate of the scan is 150 microns per seconds in steps
of 0.1 micron which is equivalent to 1500 seconds time duration for
detecting the presence of a coherent laser light in an incoherent
light background with averaging </p>
<p>Page 3656 "A 10-μm scan length, corresponding to a 20-μm path
difference range, was chosen, and this was scanned at a speed of 150
μm/s in steps of 0.1 μm."</p>
</blockquote>
<p>Please correct me if I am wrong about this calculation.</p>
<p>Then, tonight I read in the July 25 2016 Optics Express journal, an article titled "Optical digital Coherent detection technology enabled flexible and ultrafast quantitative phase imaging" by Yuan-Hua Feng et al. from Peking University Electrical Engineering Department. This article is currently open and availabe at the Optical Society Of America web portal.</p>
<p>Dr. Feng and his colleagues report that "we present a digital coherent time-stretch microscopy system that is capable of delivering ultrafast QPI of fixed cells in a scanning rate up to 100 MHz without averaging" . In addition, these researchers state that "bebefiting from a digital coherent receiver, we can retrieve intensity and phase information with high quality simultaneously. Because of the improved Signal to Noise (SNR) of the proposd imaging method, the amplitude and phase images were processed without averaging. So we can achieve an effective line-scan rate as high as 100 MHz."</p>
<p>Both of these papers are very important to use because we wish to protect pilot's vision from increasingly sophisticated laser pointers on the ground 150 meters away aimed at the Boeing 747-787 cockpit windows. If we can detect the presencen of coherent laser light in an incoherent background within 10 milliseconds of arrival , we can then activate an electro-optic shuttter to block the laser ray and then funnel the recovered incoherent light signal to an Planar(@TM) transparent display.</p>
<p>My question is which of the two papers is most effective for accomplishing the aforementioned purpose. In addition, how is the digital coherent receiver able to achieve line-scan rates as high as 100 MHz? Please use equations to explain this if necessary.</p>
<p>Any help is greatly appreciated. </p>
| |optics|lasers|digital-communication| | <p>You are wrong about that calculation. Pay attention to the unit analysis!</p>
<p>If you have 150 µm/s and 0.1 µm/step, then you divide the two to get the step rate:</p>
<p>$$\frac{150 \frac{\mu m}{s} }{ 0.1 \frac{\mu m}{step}} = 1500 \frac{step}{s}$$</p>
<p>Alternatively, the time taken per step is:</p>
<p>$$\frac{ 0.1 \frac{\mu m}{step}}{150 \frac{\mu m}{s} } = 0.667 \frac{ms}{step}$$</p>
| 10886 | How is an digital coherent receiver able to achieve 100 Mhz line-scan rate for detecting coherent light in the presence of incoherent light? |
2016-07-30T10:14:37.237 | <p>I recently flew on a A380 and I saw this interesting socket. </p>
<p><a href="https://i.stack.imgur.com/BznWK.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/BznWK.png" alt="enter image description here"></a></p>
<p>Could someone explain to me what is the purpose of these 4 golden connections?</p>
| |electrical-engineering|aircraft-design|audio-engineering| | <p>This jack allow planes to use the air based head sets or use the electrical based headsets. It is an upgraded seat and jack that meet either audio requirements. Older planes use air power headsets normally. Headsets on planes are odd, robust and almost fit nothing to deter theft.</p>
<p>The golden pieces are negative positive and right left speaker but I could be wrong and the magnets conduct electricity and the golden pieces are Left negative/positive and Right negative/positive.</p>
<p><a href="https://i.stack.imgur.com/T4fV6.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/T4fV6.jpg" alt="enter image description here"></a></p>
<p><a href="https://i.stack.imgur.com/4Ym42.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/4Ym42.png" alt="enter image description here"></a></p>
| 10887 | What is that airplane headphone socket? |
2016-07-30T13:44:05.873 | <p>I have trial version of STAAD.Pro software and I am trying to figure out why my calculation are not same as manual one. I have modeled simple structure. But I don't want to have beams 1,27,26,25,24. It is intended that all that beams are one beam and beams 6,11,16,21 are welded to that beam and central beam (there are also beams 3,5,10,15,20 and they are in reality one single beam). </p>
<p>Is that influencing results (Continues load is added to 8 middle beams). I have merged beams 1,27,26,25,24 into single beam , and beams 3,5,10,15,20 into single beam- and beams 2,7,12,17,22 into single beam also - but than it says that I have multiple structures. </p>
<p><strong>Top view of structure</strong></p>
<p><a href="https://i.stack.imgur.com/KVVsR.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/KVVsR.png" alt="enter image description here"></a></p>
<p>Can someone explain how to model intended? </p>
| |mechanical-engineering|structural-engineering|civil-engineering|software| | <p>I think it is imperative to first determine if the cross-beam to main-beam connection is a moment-(resisting-)connection. I am not sure if the cross beams are (or need to be) moment-connected to the main beams.</p>
<p>In the given case, beams 6,11,16,... 8,13,18... appear to be cross beams that carry local floor loads. It is not generally necessary to make them moment connected. Usually they are welded to a pair of angles, which will fit to some pre-drilled holes on the web of the main beams for ease of erection. Individual beams can be transported in different trucks. If this is the case, the connection is usually considered a pinned connection.
The following is what they look like: <a href="http://www.engr.mun.ca/~adluri/courses/steel/ppt%20files1/Topic%20-Connections%20-typical%20joints.pdf" rel="nofollow">Typical Steel Connections</a>
(scroll to about 1/3 down the page and look for beam-to-beam connections)</p>
<p>If they are "welded" to the main beams and are expected to transmit moments, they would likely look like these, and obviously cost more in fabrication: <a href="http://www.steelconstruction.info/Moment_resisting_connections#Welded_beam-to-column_connections" rel="nofollow">Moment resisting connections</a>
(scroll to the last fifth of the page to see beam-to-beam connections).</p>
<p>There is also a complication with perpendicular beam-to-beam moment connections in the sense that the beam-depths are usually different. In this case, the main beams are likely to be deeper than the cross beams. So again, moment connections are not usually called for. In any event, if beam 6 is moment connected to beams 1 and 27, the latter will provide very little torsional resistance, so I doubt if it is worthwhile to have them moment-connected.</p>
<p>One more reason that the cross beams are not moment-connected to the main ones would be if all these beams are welded in the shop, it would become a single piece that may not be transportable by trucks to the building site using highways. On the other hand, welding in the field would create extra quality assurance expenses, x-ray inspection, etc.</p>
| 10890 | Staad - structure design |
2016-07-30T17:37:20.110 | <p>Why don't cellphones use unidirectional antennas instead of omnidirectional (single dipole) antennas?</p>
<p>This could be accomplished by using a unidirectional array of dipole antennas (specifically, a broadside array) placed at different locations in the phone, where each antenna's phase could be changed in such a way to minimize radiation directed into the user's head and maximize radiation sent in the direction of the cell tower, producing a radiation pattern as is shown in the picture below.</p>
<p>But, with digital electronics, arbitrary phase-delays could be introduced to change the orientation of the major lobe.</p>
<p>Cellphone batteries would last longer and there would be less adverse health effects due to EMF exposure.
<a href="https://i.stack.imgur.com/QJDZR.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/QJDZR.png" alt="Unidirectional Array"></a></p>
| |electrical-engineering|wireless-communication| | <p>This is called <strong>beamforming</strong>, which exists in the 802.11 wireless standard.</p>
<p>Matthew S. Gast, <a href="https://isidore.co/calibre/#panel=book_details&book_id=7461" rel="nofollow noreferrer"><em>802.11ac: A Survival Guide</em></a>, glossary:</p>
<blockquote>
<p>Beamforming<br>
A method of using precise phase shifts on an antenna array that focuses the resulting transmission in a particular direction. Sending beamformed transmissions may require an exchange of control information to set up the antenna array.</p>
<p>Beamformee<br>
The receiver of a beamformed transmission. The beamformee may need to transmit some packets in a beamforming setup exchange, but the main purpose of the beamforming exchange is to receive a directional transmission.</p>
<p>Beamformer<br>
The sender of a beamformed transmission. The beamformer may need to receive some packets in a beamforming setup exchange, but the main purpose of such an exchange is to send a directional transmission.</p>
</blockquote>
<p><a href="https://isidore.co/calibre/#panel=book_details&book_id=7461" rel="nofollow noreferrer"><em>ibid.</em></a> Chapter 4. Beamforming in 802.11ac, § Beamforming Basics:</p>
<blockquote>
<p><a href="https://isidore.co/calibre/#panel=book_details&book_id=7461" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/FHbZH.png" width="320"/></a></p>
<p><em>Figure 4-4. Using multiple antennas to steer transmissions</em></p>
</blockquote>
<p>There is Single-User (SU) Beamforming and even Multi-User (MU) Beamforming.</p>
| 10894 | Why don't cellphones use unidirectional antennas? |
2016-07-31T21:49:08.563 | <p>In digital or discrete control, instead of continuous analog-type variables, a controller samples a process variable signal at regular discrete time intervals. So, effectively, the process variable does not change as far as the system is concerned between those samples, or if it does, the change won’t be detected until the next sample. And it will be abrupt if it changes at all.</p>
<p>One would think that the faster the sampling rate (the more samples in the same time interval, or shorter times between samples), the more closely the discrete control would emulate ideal continuous control. Apparently that is not the case. In fact, as I understand it, increasing the sampling rate past some critical point can actually make a discrete time or digital control system unstable, as it can push the transfer function poles out close to and even outside the unit circle of stability.</p>
<p>This seems to be counter intuitive, at least to me. Is there a simple explanation for this that I’m missing here?</p>
| |control-engineering| | <p>Based on the reference cited in the question comments, this particular situation occurs when you implement a controller designed in the continuous time domain on a discrete time system with measurement or model uncertainties.</p>
<p>Designing the controller in continuous time means that the sample rate is not taken into account/is assumed to be infinitely fast. Therefore the controller can behave very differently for different sample rates, even if the control gains remain unchanged. In particular, increasing the sample rate can destabilize the system by making it more sensitive to noise/uncertainty in the measurements (especially with PD and PID controllers).</p>
<h1>A simplified example</h1>
<p>You have a proportional-derivative controller, with a derivative gain <span class="math-container">$K_d = 1$</span> (the proportional part is not really important for this example). Your control system has a sensor which gives you a measurement of your error with a random uncertainty up to <span class="math-container">$w = 0.1$</span> units. At a certain moment in time the true error signal is changing at a constant rate of <span class="math-container">$\frac{de}{dt} = 10$</span> units/s.</p>
<p>First you implement your controller with a sample time of <span class="math-container">$T=0.1$</span> s. You estimate the current rate of change using a first order approximation, and it just so happens that you experience the maximum error in this timestep:</p>
<p><span class="math-container">$\frac{de}{dt}_{measured} = \frac{e(k)-e(k-1)}{T} = \frac{\frac{de}{dt}T + w}{T} = \frac{de}{dt} + \frac{w}{T} = 10 + \frac{0.1}{0.1} = 11$</span> units/s</p>
<p>Therefore the output of the derivative part of your controller is <span class="math-container">$11K_d=11$</span>, i.e. 10% higher than the ideal output of the continuous controller (<span class="math-container">$10K_d=10$</span>). Notice that the contribution of the error term to your controller output is inversely proportional to sample time, therefore smaller sample times (faster sample rates) increase its effect.</p>
<p>Then you decide, faster must be better, so you decrease the sample time to <span class="math-container">$T=0.01$</span> s. Now, in the same situation, the estimated rate of change of error is:</p>
<p><span class="math-container">$\frac{de}{dt}_{measured} = 10 + \frac{0.1}{0.01} = 20$</span> units/s</p>
<p>The output of the derivative part of the controller is now <span class="math-container">$20K_d = 20$</span> i.e. almost double what it was before, and 100% higher than the ideal output. Your controller is now behaving more aggressively even though the true rate of change of error has not increased.</p>
<p>The problem is exaggerated in this example by using a first-order approximation, but more complex approximations of the derivative will only mitigate, not eliminate, the problem. The fact is that changing the sample rate of the system can have a negative impact in the presence of any kind of measurement uncertainty. Now, whether your controller performs better or worse when you increase your sample rate depends on the relative magnitude of the uncertainty versus the rate of change of the error signal, the relative magnitudes of your derivative and proportional gains, and the type of approximation you use for your derivatives.</p>
<h1>Conclusion</h1>
<p>If you are designing a system where the controller cannot be approximated as continuous and there are measurement or other uncertainties present, increasing the sample rate will not necessarily increase performance because the influence of the uncertainty on the control input will increase.</p>
| 10901 | Discrete Control System Process Sampling Rate |
2016-08-01T19:54:58.853 | <p>I'm interested in designing a transmission to achieve a constant-speed, variable-torque output from variable-speed, variable-torque input -- something like a CVT, but applied backward. Is such a thing possible? Do any examples of a transmission like this exist commonly?</p>
<p>I'm happy to answer any clarifying questions, as I'm imagining this doesn't include all the needed information. Thank you in advance for your patience!</p>
<p>EDIT: I should mention that the specific effect that I'm looking for is that when the input speed increases, the output speed remains constant, but the output <em>torque</em> increases proportionally.</p>
| |mechanical-engineering| | <p>In an early solution to air conditioning of railroad cars the variable-speed motion of the axles was transmitted through such a coupling to provide constant speed drive for the A/C compressor. There are several mechanical and electromechanical CVT type mechanisms that could be used.</p>
<p>To find information about the various CVT technologies, you should also search "mechanical adjustable-speed drive" and "mechanical variable-speed drive." The term CVT is primarily used for vehicle components. The ASD and VSD terms are used with industrial equipment. In industry, electrical equipment is usually used, but there are mechanical alternatives. Note that descriptions of the equipment that you find will be be the direction opposite to what you want, but the reverse alternative can usually be achieved with the same or similar equipment.</p>
<blockquote>
<p>EDIT: I should mention that the specific effect that I'm looking for is that when the input speed increases, the output speed remains constant, but the output torque increases proportionally.</p>
</blockquote>
<p>You can regulate speed or you can regulate torque, but you can not regulate both simultaneously. The characteristics of the load must dictate one or the other. Perhaps you should explain the system more completely and add a diagram.</p>
| 10909 | Transmission for constant-speed, variable-torque output from variable-speed, variable-torque input? |
2016-08-02T07:21:55.107 | <p>I am looking to build a specific kind of lift for a project I'm working on but I don't know the name. I am willing to hire a mechanical engineer to help me design and build it (or buy one if it already exists) but I'd like to do as much research as possible before deciding on the next steps.</p>
<p>What I have in mind is like a ferris wheel, save that it is shaped more like a stretched rubber band than a circle (i.e. it has vertical sides and a short circular path at the top and bottom). Instead of carts that people sit in I'd just like a flat steel floor, say 4'x8', for loading cargo.</p>
<p>Using a switch I'd like the flat "carts" to move into a loading space, load from the front, then be sent up to make room for the next cart. Maximum load should be about 1,000lbs per cart.</p>
<p>What is this type of device called? I know similar things have been built for parking cars, but I don't want something quite so large and I'm interested in storing smaller things than cars.</p>
<p>I'm sorry if this isn't more descriptive, please ask if I can clarify anything and I'll be happy to.</p>
| |mechanical-engineering| | <p>At the scale you describe (8'x4') this would be similar to a <a href="https://en.wikipedia.org/wiki/Paternoster" rel="nofollow">Paternoster</a>, except that it would need to stop to load and unload the carts. Paternosters (and bucket elevators) move continuously.</p>
<p>Health and Safety regulations have closed down most of them (especially those that can be used by the general public) because of the risk of accidents when entering or leaving. YouTube has some videos, including what happens when you ride around the top or the bottom - which is designed to be safe.</p>
| 10915 | Lift-type identification - What class of lifts are ferris wheels in? |
2016-08-03T00:08:51.947 | <p>In historical or modern buildings, flat ceilings are typically made of concrete, wood, metal, solid bedrock (for carved tunnels), or generally a material that can withstand pushing forces (pressure) and pulling forces (tension).</p>
<p>A ceiling made from bricks / masonry would typically be vault-shaped, because otherwise it would crash. This is what we find in a lot of historic buildings.</p>
<p>I can so far only remember flat brick ceilings from imaginary buildings e.g. in a roleplaying context or in video games. But I wonder if there are cases where such a thing can be realistic.</p>
<p>In theory, I can imagine two ways to achieve a flat ceiling with masonry:</p>
<ul>
<li>By using very good cement, so the entire thing would behave like one solid piece of rock or concrete.</li>
<li>By imitating a vault, in a way similar to how a Fresnel lens imitates a regular concave lens:<br>
Use slightly V-shaped bricks, and stack them like this, more or less: `\\\|||///``.</li>
</ul>
<p>My question is:</p>
<ul>
<li>Can this ever work?</li>
<li>Does this technique have a name?</li>
<li>How far can you go with this, what are the limitations?</li>
<li>Are there historical examples?</li>
</ul>
| |civil-engineering|masonry| | <p>What you're looking for is called a "<a href="https://en.wikipedia.org/wiki/Jack_arch" rel="nofollow">jack arch</a>".</p>
<p>The top and bottom are flat, but the joints are angled in such a way that the the parabolic "path of compression" is entirely contained within the layer of masonry.</p>
<p>The loading (e.g., the outward force at the ends) is higher than with a thinner parabolic arch or vault, simply because the extra material is additional dead load.</p>
| 10925 | Can masoned ceilings be flat? |
2016-08-03T16:42:47.380 | <p>I understand that for a rectangular c-s the shear stress distribution is parabolic and the max shear stress occurs at the neutral axis and has a value of 1.5V/A. Where V is the 'applied shear force' and A is the cross-sectional area.</p>
<p>But this in turn then means that the shear force at this point is equal to 1.5V ( 1.5 times larger than the applied shear force) - which seems a tad strange physically.</p>
<p>Is this due to the fact the the average shear force (average shear stress x cs area) is equal to the applied shear force? This is the only way it makes sense to me.</p>
<p>Thanks!</p>
<p><a href="https://i.stack.imgur.com/fpDrt.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/fpDrt.jpg" alt="enter image description here"></a></p>
| |mechanical-engineering|structural-engineering|statics|stresses| | <p>If we need to calculate how much shear a rectangular beam can take this is the formula.<br>
V= 2/3 [A x tau(allowable)]. </p>
<p>We find the allowable tau in charts readily available but for lumber under normal humidity it is around 80-90 psi.</p>
| 10938 | Maximum Shear stress in Beams |
2016-08-01T17:03:23.443 | <p>first post so bear with me please!</p>
<p>Studying Mech Eng and in all notes provided and I have come across online the stress/strain distribution is shown as I have sketched belo (see pic). Surely this is only a representation of the right hand side of the beam (for hogging as shown) as the left would be as I have sketched below.</p>
<p>Am I correct in assuming this? Why do they always just show the first distribution? Why only show the 2 triangles and not the full square? What is the distribution in the middle of the beam (assuming that is where a single point load is applied)?</p>
<p>Hope this is clear enough to understand!</p>
<p>Thanks for you help!</p>
<p><a href="https://i.stack.imgur.com/40n9V.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/40n9V.jpg" alt="bending stress/strain"></a></p>
<p>EDIT: Additionally, I feel this is where my confusion orginated from:
<a href="https://i.stack.imgur.com/NQ6dY.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/NQ6dY.jpg" alt="enter image description here"></a></p>
<p>I now understand the stress distribution is constant throughout the length of the beam (although unsure as to why it's shown in that particular direction). But can someone please explain why the above image then makes sense. Is this only valid on the right hand side?</p>
| |structural-engineering|structural-analysis| | <ol>
<li>understand that the force diagram is just a representation and depends on many presuppositions, on the other hand if you consider it in an actual practice the results are different than that of standard diagrams, but we can say that the fluctuations are very minimal. Still, it has a different output every time you run it.</li>
<li>showing one diagram is enough to explain what is happening throughout the beam, more or less with a factor of safety.</li>
<li>now if you want to remember the simple concept of sagging and hogging and to clear the positive and negative confusions, you go through <a href="https://youtu.be/9CJDi0HTpL4" rel="nofollow">this video</a> this will sure take exactly 2 min 5sec of your life, but trust me you will remember this sign conventions for the rest of your life.</li>
</ol>
| 10941 | Beams in Bending: Bending stress / strain distrution |
2016-08-04T16:01:08.197 | <p>I'm building a set of monkey bars for my kids and want to use aluminum tubing (6061 T6) for the rungs. The tubes will be 3 feet long and 1 inch outside diameter. Supported on both ends with a circular metal pipe bracket. I have a choice of various wall thicknesses, all the way up to a full solid round bar. How do I calculate the minimum wall thickness I need to safely support (without any permanent deformation) something like 200 pounds at the middle point of the rung?</p>
| |beam|statics|aluminum| | <p>We use this equation for a simply supported beam loaded at its center with 200lbs. M= P x L/4 =200 x 3 /4 = 150 lbs.ft<br>
for calculating the tube stress we convert this to 150 x 12 = 1800 lbs.inch </p>
<p>For a hollow round tube we calculate section module which is </p>
<p>$Z = 0.78 (R_o^4 -R_i^4)/R_o$ assuming you have 16 gauge tube with 0.065 inch walls, Z=0.1838 . </p>
<p>Now we have $sigma = M/Z so 42000 (allowable aluminum stress) = M/Z $ </p>
<p>$M = 42000 x 0,1838/4= 1929 lbs$<br>
Usually in applications like this which is dynamic loading they use a safety factor of 3 so you can expect this tubing to be able to support approximately<br>
$ 600 lbs$. </p>
<p>Then again there are many standards of aluminum. For this illustration I assume 2024-T3 which has hi strength, you need to find out what you have.<br>
This answer is just an illustration and you should check with the manufacturer about their tests on strength and ductility. Also connections and wall construction has to be checked.</p>
| 10957 | How much weight can an aluminum tube beam support? |
2016-08-05T08:09:19.887 | <p>I have a continuous beam across a center column (not sure whether I've draw it correctly or not)</p>
<p><a href="https://i.stack.imgur.com/yvCEb.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/yvCEb.png" alt="enter image description here"></a></p>
<p><a href="https://engineering.stackexchange.com/q/9066/3353">The Displacement and force diagram is as shown:</a></p>
<p><a href="https://i.stack.imgur.com/3ic7e.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/3ic7e.png" alt="enter image description here"></a></p>
<p>The first diagram is the bending moment, second the axial force, the third is the displacement.</p>
<p>Now, what is the boundary condition at $R_A$, $R_B$ and $R_C$? </p>
<p>From what I can infer, it seems to be</p>
<p>$w(0)=w(L)=w(2L)=0$ (corresponds to the deflection at the three support)</p>
<p>$M(0)=M(2L)=0$, or $\frac{d^2w(0)}{dx^2}=\frac{d^2w(2L)}{dx^2}=0$ (corresponds to the moment).</p>
<p>But I suspect that I am still missing some boundary conditions in order to derive the complete displacement/force diagram for the continuous beam. Are there any boundary conditions that I've missed?</p>
| |structural-engineering|structural-analysis|beam|statics| | <p>Well you are missing the compatibility of slopes at the middle support:</p>
<p>$$\frac{dw(L^-)}{dx} = \frac{dw(L^+)}{dx}$$</p>
<p>In the case of symmetric geometry & loading, the slope of the beam at the middle support will be zero.</p>
<p>Since the bending moment has no derivative at x=L, you will need to derive the deflections of the two halves separately and 'join' them with compatibility.</p>
<p><strong>Update:</strong> derivation of beam deflection formula:</p>
<p>Starting from the <a href="https://en.wikipedia.org/wiki/Euler%E2%80%93Bernoulli_beam_theory" rel="nofollow">Euler-Bernoulli beam equation</a> (assuming constant EI), and taking x from the outer supports towards the center:
$$ q = EI \frac{dw^4}{dx^4} $$</p>
<p>Integrating four times:</p>
<p>$$ EI \frac{d^3 w}{dx^3} = q x + A $$
$$ EI \frac{d^2 w}{dx^2} = q \frac{x^2}{2} + Ax +B $$
$$ EI \frac{d w}{dx} = q \frac{x^3}{6} + A\frac{x^2}{2} + Bx + C $$
$$ EI w = q \frac{x^4}{24} + A\frac{x^3}{6} + B\frac{x^2}{2} +Cx + D$$</p>
<p>Noticing the problem is symmetric the boundary conditions are:
$$ w(0)=w(L) = 0$$
$$ \frac{d w(L)}{dx} = 0 $$
$$ \frac{d^2 w(0)}{dx^2} = 0 $$ </p>
<p>Therefore we can immediately see that:
$ B = D = 0$</p>
<p>We now have two equations with two unknowns (A,C). Solving we find:
$$ A = -\frac{3 L q}{8}$$
$$ C = \frac{L^3 q}{48}$$</p>
<p>We can now substitue all the constants back into the equation for w. Simplifying results in:</p>
<p>$$ w = \frac{q x}{48EI} (L-x)^2 (L+2 x) $$</p>
<p>Which is identical to the result referenced <a href="http://www.engineersedge.com/beam_bending/beam_bending22.htm" rel="nofollow">here</a> (note their coordinate system has x=0 at the centre). Also notice how this is exactly the same result as a <a href="http://www.engineersedge.com/beam_bending/beam_bending11.htm" rel="nofollow">propped cantilever</a>. This is due to symmetry, meaning that the beam slope at the center is zero (which is the same boundary condition as a cantilever support).</p>
<p>You can also substitute into the bending moment equation:</p>
<p>$$ M = EI \frac{d^2 w}{dx^2} = \frac{1}{8} q x (4 x-3 L) $$</p>
| 10963 | The boundary condition(s) for a continuous beam |
2016-08-05T17:11:14.880 | <p>I am trying to determine if it is possible to "physically" determine the volume of liquid (such as water) in a non-solid container like a fluid bladder like <a href="http://www.interstateproducts.com/pillow_tanks.htm" rel="noreferrer">this</a> or <a href="http://atlinc.com/pillow.html" rel="noreferrer">this</a>?</p>
<p>Assume that it is not possible to keep accurate track of the difference between how much fluid is placed in the bladder and how much is taken out, and that a "physical" measurement must be made (where "physical" can include using a sensor, or using some sort of actual measurement of the dimensions of the bag to compute the volume).</p>
| |fluid-mechanics|measurements| | <p>Here are some options:</p>
<ol>
<li><p><strong>Weight</strong>- Reliable if the density is known. Bladder can be hung from a load cell, or placed on a surface supported by load cells, and volume can be derived continuously.</p></li>
<li><p><strong>Bag in tank</strong> - One easy way to calculate volume is to place an awkward shape in a tank of water and measure the increase in volume. Simply inflate the bladder submersed in water in a straight wall tank and an ultrasonic (or other) level transmitter will be able to accurately give you an continuous volume measurement. <a href="https://en.wikipedia.org/wiki/Archimedes%27_principle" rel="nofollow">Archimedes' principle</a></p></li>
<li><p><strong>Open Loop Calibration Curve</strong> - Reliable if the bladder is designed such that it always grows in a predictable fashion (not one side then the other). A proximity sensor or distance sensor (possibly ultrasonic or laser) would be used to target a location on the bladder that changes the most during fill and empty (center of a whoopie cushion for example). You would then take an empty measurement, record the distance, fill with 10ml, record the distance, fill with 10 more ml, record the distance, and so on... The cure you create can then be used in a computer or PLC to continuously derive your volume from the distance sensor input.</p></li>
<li><p><strong>3D Scanning</strong> - I would only recommend this option if the others were not feasible, the geometry was unpredictable, and a continuous measurement was not needed. Multi-photograph reverse rendering, laser scanning, and LIDAR(as Inquisitive mentioned) are all possible options. The computer scans the object over a period of seconds or minutes then analytically computes the volume.</p></li>
</ol>
| 10972 | Determine volume of liquid in a fluid bladder |
2016-08-05T23:58:18.947 | <p>I am trying to simulate a winch as a speed-regulated motor that works through a gearbox to lift a mass. The output of the gearbox is a drum, which rotates to accumulate cable. </p>
<p>I feel comfortable <a href="https://en.wikipedia.org/wiki/List_of_moments_of_inertia#Moments_of_inertia" rel="noreferrer">converting the mass to a moment of inertia</a> and I also feel comfortable with converting <em>that</em> moment of inertia (output-side) to the moment of inertia "seen" by the motor (input-side) with <a href="https://www.physicsforums.com/threads/inertia-change-through-a-gearbox-question.387779/" rel="noreferrer">the gearbox ratio</a>. With a simple simulation, I have no problem writing the equations of motion. </p>
<p>My complication comes when I want to model "stretch" in the cable. I thought I could do this by simply putting a spring of arbitrary stiffness between the winch drum and the mass, as pictured below. </p>
<p><a href="https://i.stack.imgur.com/Zza1q.png" rel="noreferrer"><img src="https://i.stack.imgur.com/Zza1q.png" alt="Spring Inertia"></a></p>
<p>With this model, for the sake of simulation, I'm assuming I know the "drum height", which would be how far the drum has turned multiplied by the drum radius, and the height of the load. The spring force would be $k(\phi r - y)$, but <em>how do I apply this to the motor</em>? </p>
<p>I have a motor model:</p>
<p>$$
\frac{\Theta}{V} = \frac{K_T}{R_a Js+K_T K_b}
$$
and a PI controller model:</p>
<p>$$
\frac{V}{\Theta_\mbox{error}} = \frac{k_p \left(s + \frac{k_i}{k_p} \right)}{s}\\
$$
where $\Theta$ is the motor speed, $V$ is terminal voltage, $J$ is the inertia of the load and machinery, and $R_a$, $K_T$, and $K_b$ are the motor armature resistance, torque constant, and back EMF constant, respectively. </p>
<p>The interaction I'm interested in studying occurs when the PI controller is tuned to the <em>anticipated</em> load inertia $J$, which would be found with the motor, gearbox, drum, and load mass, but the system actually "sees" the springy-mass. </p>
<p>Simplification is done by setting the $k_i/k_p$ ratio equal to $K_TK_b/R_aJ$, giving:</p>
<p>$$
\frac{\Theta}{\Theta_{\mbox{error}}} = \frac{V}{\Theta_{\mbox{error}}}\frac{\Theta}{V} =
\left(
\frac{k_p \left( s + \frac{K_TK_b}{R_aJ} \right) }{s}
\right) \left( \frac{\frac{K_T}{R_aJ}}{s+\frac{K_T K_b}{R_aJ}} \right)
$$</p>
<p>(Note I can leave $k_p$ as variable because the ratio $k_i/k_p$ can be set to whatever I want via $k_i$ as long as $k_p$ isn't zero.)</p>
<p>So, in an <em>ideal</em> world, where the value of "total" inertia $J$ is known in advance, the pole cancels, and the entire system reduces to:</p>
<p>$$
\frac{\Theta}{\Theta_{\mbox{error}}} =
\left( \frac{k_p}{s} \right) \left( \frac{\frac{K_T}{R_aJ}}{1} \right) \\
$$
$$
\frac{\Theta}{\Theta_{\mbox{error}}} =
\frac{1}{\frac{R_aJ}{k_pK_T}s}
$$</p>
<p>Finally, $\Theta_{\mbox{error}} = \Theta_{\mbox{ref}} - \Theta_{\mbox{out}}$, so, with algebra:</p>
<p>$$
\boxed{\frac{\Theta_{\mbox{out}}}{\Theta_{\mbox{ref}}} = \frac{1}{\frac{R_aJ}{k_pK_T}s + 1}}$$</p>
<p>So, kind of sorry to shotgun so much detail, but I wanted to impress on anyone reading that I feel confident with all of my steps so far and that I have spent <em>considerable</em> effort working on this problem. Now, again to my question - I want to simulate <em>stretch</em> in the cable between the drum and the load, but I'm not sure how to use the spring force to modulate load inertia. </p>
<p>One thought I had was to try to fake an "equivalent mass", by assuming:</p>
<p>$$
F = m_{\mbox{equivalent}}a \\
m_{\mbox{equivalent}} = \frac{F_{\mbox{spring}}}{a} \\
$$</p>
<p>but this doesn't feel right, and I'm not sure what I would use for acceleration $a$. </p>
<p>I'm frustrated to be this far along on the problem and getting stumped by what seems like should be an easy issue, but I really can't think of a way to approach this problem. I think if I could frame it correctly I could work out the mechanics, but it's the force-to-inertia conversion I feel like needs to be made that has me stumped.</p>
<p>Finally, for the record, I have also tried back-tracking my motor model to include the load torque. This gives seemingly-reasonable results, but in the end I subtract the load torque from the motor torque to get net torque, then <em>apply that net torque to the total inertia</em> to get motor acceleration. That feeds on down the line and, again, I'm not sure that I'm treating total inertia correctly. </p>
| |motors|dynamics|springs|transfer-function| | <p>I realize this is an old thread, and I am not sure how deep of a dive you finally took on this, but one thing I don't see accounted for in your equations is drum/cable friction. This will be small, and like the accumulated mass of the wound steel wire rope you did not include, it may not be on your list. The cable could be pre-stretched and pre-loaded, however any movement between the cable and drum due to cable stretch will also encounter friction. In my industry (theater rigging, stage machinery design), the groove contacts a greater area than a flat drum application, and we usually have additional friction along the redirect sheaves and mules in the lineset to account for especially in 2:1 or 4:1 mechanical advantage systems. </p>
| 10978 | How does load force impact load inertia? |
2016-08-06T02:18:54.640 | <p>Is there a formula or method to calculate the pressure of failure of a cylindrical container of known material, thickness, and size?</p>
| |pressure|statics| | <p>In this answer, I aim to extend the discussion to thicker walled cylinders too. If you want to skip the maths, the pressures are summarised at the bottom.</p>
<h2>DEFINED TERMS:</h2>
<p>$\sigma_{rr}$ - radial component of stress</p>
<p>$\sigma_{\theta\theta}$ - circumferential component of stress</p>
<p>$\sigma_{zz}$ - axial component of stress</p>
<p>In this answer, I will assume that failure occurs in the curved walls of the cylinder. Different failure modes, could be considered, but that is beyond the scope of my answer. It is in my intuition that the curved walls are the weak points of failure, but I will leave the discussion of that to more qualified answers.</p>
<h2>FIRST YIELD</h2>
<p>When it comes to determining the failure of thick walled cylinders (defining failure to occur the instance yield occurs anywhere in the cylinder), Lamé's Equations are useful. These equations give the stresses that lie in the plane of the circular cross section of the cylinder:</p>
<p><a href="https://i.stack.imgur.com/EYsTB.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/EYsTB.jpg" alt="enter image description here"></a></p>
<p>Note the rotational symmetry axis is moving out of the page, normal to the cross section.</p>
<p>Lamé's Equations are derived from the governing equilibrium equation for an axisymmetric solid:</p>
<p>$$\frac{\partial}{\partial r} \left( r \sigma_{rr}\right) = \sigma_{\theta\theta}$$</p>
<p>Where $r$ is the distance of the point where stresses are being considered from the axis of rotational symmetry.</p>
<p>Lamé's equations themselves are given as:</p>
<p>$$\sigma_{rr} = A - \frac{B}{r^2}$$
$$\sigma_{\theta\theta} = A + \frac{B}{r^2}$$</p>
<p>Where $A$ and $B$ are arbitrary constants.</p>
<p>For a cylinder with internal pressure $p$, no external pressure, an internal radius of $a$ and thickness $t$, these equations become:</p>
<p>$$\sigma_{rr} = \frac{p}{m^2-1}\left(1-\frac{m^2a^2}{r^2}\right)$$
$$\sigma_{\theta\theta} = \frac{p}{m^2-1}\left(1+\frac{m^2a^2}{r^2}\right)$$</p>
<p>Where $m=1 + \frac{t}{a}$ (i.e. ratio of outer to inner radii). Note that as $t$ gets smaller, $m$ approaches 1 and the stresses approach that </p>
<p>To determine the final stress component, it is reasonable to assume that, away from the edges of the cylinder, the axial stress is constant in the walls. This stress will oppose the pressure acting on the cylinder ends:</p>
<p>$$\sigma_{zz} = \frac{p}{m^2-1}$$</p>
<p>Here is a rough plot of the three stresses as function of $r$:</p>
<p><a href="https://i.stack.imgur.com/MSSWI.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/MSSWI.jpg" alt="enter image description here"></a></p>
<p>With the three components of stress known, a suitable <a href="https://en.m.wikipedia.org/wiki/Yield_surface" rel="nofollow noreferrer">yield criterion</a> should be used. Usually Tresca or von Mises are suitable for metals like steel. Here, I will assume Tresca, which states that yield first occurs at a point on the cylinder whenever the following is satisfied:</p>
<p>$$\max\left(|\sigma_{rr}-\sigma_{\theta\theta}|,|\sigma_{rr}-\sigma_{zz}|, |\sigma_{\theta\theta}-\sigma_{zz}|\right)=\sigma_y$$</p>
<p>Where $\sigma_y$ is the yield stress of the material.</p>
<p>By looking at the plot, it can be seen that yield will occur on the inside of the cylinder, so that at yield:</p>
<p>$$|\sigma_{rr} - \sigma_{\theta\theta}| = \frac{2m^2}{m^2-1} p=\sigma_y$$</p>
<p>Therefore, failure by first yield occurs when:</p>
<p>$$p=\frac{m^2-1}{2m^2}\sigma_y$$</p>
<p>The maximum pressure before first yield increases with thickness, up to a maximum of $\frac{\sigma_y}{2}$ at infinite thickness.</p>
<h2>COLLAPSE</h2>
<p>Even when yield first occurs, the walls of the cylinder won't burst open, i.e. collapse has yet to occur. The point at which collapse occurs should occur after first yield: would you ever expect an infinitely thick cylinder to burst? The collapse pressure is useful to determine since it is at this point that the cylinder fails to retain its pressurised contents.</p>
<p>When first yield occurs, there is still regions in the cylinder that have yet to yield, and so the internals are retained. As you increase the pressure beyond that of first yield, you get a plastic region of yielded material that forms at the inner face of the cylinder and creeps along the cylinder thickness towards the outer face:</p>
<p><a href="https://i.stack.imgur.com/KTliT.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/KTliT.jpg" alt="enter image description here"></a></p>
<p>Collapse occurs as soon as the plastic zone reaches the outer face of the cylinder. The issue in trying to calculate this load is that elasticity no longer applies: Lamé's equations, which are only valid for linear elastic cylinders, will not be valid. However, the governing equilibrium equation still applies: equilibrium must be upheld regardless whether the material law is linear or not. Since the plastic zone has extended through the entire cross section, less us assume that the stresses abide by Tresca's criterion throughout the cylinder (this assumption is valid according to the Lower Bound Theorem, which returns an estimate of the collapse pressure which must be less than the actual collapse pressure, giving a conservative estimate):</p>
<p>$$\sigma_{\theta\theta} - \sigma_{rr} = \sigma_y$$</p>
<p>We can now substitute this into the equilibrium equation to give:</p>
<p>$$r \frac{\partial \sigma_{rr}}{\partial r} = \sigma_y$$</p>
<p>Solving this and applying boundary conditions then gives a collapse pressure of:</p>
<p>$$p = \ln{(m)} \sigma_y$$</p>
<p>Interestingly, for all positive $m$, $\ln (m)$ is greater than or equal to $\frac{m^2-1}{2m^2}$, so first yield must always occur before collapse. Only when $m$ approaches 1 (thin-wall) do the pressures for first yield and collapse approach being equal.</p>
<h2>IN SUMMARY</h2>
<p>For failure by first yield, the pressure must be:</p>
<p>$$p=\frac{m^2-1}{2m^2}\sigma_y$$</p>
<p>For failure by collapse, the pressure must be:</p>
<p>$$p=\ln{(m)} \sigma_y$$</p>
<p>Where $m$ is the ratio of external to internal radii.</p>
<p>Here is a rough plot for both:</p>
<p><a href="https://i.stack.imgur.com/DEJUu.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/DEJUu.jpg" alt="enter image description here"></a></p>
<p>As $m$ increases, the first-yield limit increases to a maximum of $\frac{\sigma_y}{2}$, while the collapse limit increases to infinity. For the collapse limit to be double that of the first-yield limit, $m$ needs to be around 2.2.</p>
| 10980 | Method of Calculating pressure of failure in a cylindrical container? |
2016-08-06T14:04:41.607 | <p>This question is a chemical engineering question related to ecology.</p>
<p>What is the list of possible threats of human feces to the environment? What about toilet paper? How do engineers deal with this problem? I mean, sure, there are sewers, but sewers consume energy, and process more than sit and toilet paper, also soap and other detergents. I want to know, these specific cases:</p>
<ul>
<li>how would you treat dung</li>
<li>how would you treat dung and paper (toilet paper and paper towels)</li>
<li>how would you treat dung and paper (toilet paper and paper towels) and soap</li>
<li>how would you treat dung and paper (toilet paper and paper towels) and soap and detergents/washing products</li>
</ul>
<p>In this logical order, I would assume that treating these wastes increases in complexity as we go down the list. Treating something further down is sufficient for treating something further up, but treating something further up alone (perhaps in some situations the extras provided by entries further down the list are not needed).</p>
<p>Then there is the question of whether what you eat affects the ability of human feces to act as fertilizer, and why would house our core dung be better than human dung? How would it be chemically different.</p>
<p>Thank you for your insight.</p>
| |chemical-engineering|waste-water-treatment|waste-disposal| | <p>The main issue is the spread of pathogens, human faeces can harbour some nasty diseases, notably e-coli, hepatitis and cholera. Accumulation can also provide an environment where pathogens can survive for a long time and there is a significant risk of them leeching into water supplies. </p>
<p>There is also a problem with any animal waste that if it accumulates in water supplies it creates excessive nutrient levels which promote the growth of algae and bacteria which, as well as being a health hazard can over-grow and crowd out other organisms. </p>
<p>It has also been suggested that metabolic products from drugs and things like hormone based contraceptives can cause environmental problems. Although it is obviously more complex to establish causality with this. </p>
<p>For the above reasons much more care needs to be exercise when using human waste as fertiliser than with manure from other animals. However if properly handled, composting can be a simple and effective way to make it safe. This both allows the more accessible nutrients which tend to promote unwanted microorganisms to be broken down and, if done properly can reach temperatures where harmful organisms are killed off, at least to safe levels. </p>
<p>A more technical solution is something like an anaerobic digester, these are becoming increasingly popular, especially in industrial agriculture to create revenue from crop and animal waste and have the advantage that they make use of the energy in the waste as well as the minerals and nitrates. Here controlled conditions and selected microorganisms are used to generate methane from the waste slurry. It is even potentially possible that they could be engineered to produce plastics or other chemical feed-stocks. </p>
<p>Obviously when you are talking about human waste you also need to take social and cultural behaviour into account. Most people in the developed world are now deeply ingrained with the idea of getting rid of their bodily waste instantly and, ideally without even seeing it and any significant change in this behaviour may be a hard sell, at least at first, unless you can offer some very significant benefits. This is, of course a very new thing and it is not that long ago that there was thriving industry in human waste for all sorts of industrial purposes. and there are some parts of the world where it is actually a challenge to persuade people to use toilets, even when they are available. </p>
<p>Of course another major source of waste is that we tend to spend huge resources on making water safe to drink and then use most of it just for flushing toilets, irrigation etc and only drink a tiny percentage of it. </p>
| 10984 | Disposal of dung, related products, and cleaning products, prioritized and case by case |
2016-08-06T19:23:54.370 | <p>All mobile phones have bars or some other type of indicator displaying signal strength. As mobile devices move away from the base station the signal strength becomes weak and at some point becomes out of range or establishes connections with a different tower. </p>
<p><strong>Question:</strong> For a LTE base station is it possible to determine the relationship between the signal strength and distance?</p>
<p><strong>Assumption:</strong> LTE base station is located in a plain field. </p>
<p>Back story: A cellular service provider is planning on installing a LTE base station using equipment from <a href="http://www.jmawireless.com/index.php" rel="nofollow noreferrer">JMA wireless</a>. The model number of the equipment is <a href="http://www.jmawireless.com/images/uploads/pdfs/X7CQAP-FRO-645-V_DataSheet.pdf" rel="nofollow noreferrer">X7CQAP_FRO_645</a>. Below is a H-plane pattern plot for 750MHz frequency.</p>
<p><a href="https://i.stack.imgur.com/1jskY.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1jskY.png" alt="enter image description here"></a></p>
<p>Can anyone provide insight as how to interpret the above plot in relation to signal strength verses distance? Do I need more information?</p>
| |electrical-engineering|wireless-communication|lte| | <blockquote>
<p>For a LTE base station is it possible to determine the relationship
between the signal strength and distance?</p>
</blockquote>
<p>Ideally yes. Practically, probably not very reliably. Given the following pieces of information and assumptions, you can find the distance between any transmitter and receiver:</p>
<ul>
<li>The transmitting antenna's radiation pattern and orientation (gain)</li>
<li>The receiving antenna's radiation pattern and orientation (gain)</li>
<li>The transmitter's power</li>
<li>An accurate measure of received power</li>
<li>Assume no obstructions (as you have)</li>
<li>Frequency of signal</li>
</ul>
<p>These are the terms needed in <a href="https://en.wikipedia.org/wiki/Friis_transmission_equation" rel="noreferrer">the Friis Transmission equation</a> to solve for Distance. </p>
<p>The plot you show is likely a gain plot along the horizontal plane in dB relative to an <a href="https://en.wikipedia.org/wiki/Isotropic_radiator" rel="noreferrer">isotropic antenna</a> (equal power in every direction). Every 3 dB represents a doubling of gain (and hence energy) transmitted in that direction. </p>
<p>Additionally, you can find a 'rough' measure of your phones received power by entering field test mode (google for instructions). The rest of the information is going to be tough to come by.</p>
| 10987 | How to determine signal strength verses distance for LTE base stations? |
2016-08-08T14:41:56.830 | <p>I want to make a small air filter using standard axial pc fans in order to suck smoke like that have been produced in soldering process. </p>
<p>But I assume filtering will make it difficult. Am i wrong? The air have to be forced through the granulated carbon filter. </p>
<p>I researched about radial fans which produced spesifically for this job(suction), but they have disadvantages like price, vibration and noise. They are at least 60dba. May not suitable to be used in room. </p>
<p>My main purpose is increasing suction of air, not the air flow. I can go with minor flow. (if my expression is not true please correct)</p>
<p>The question is, can a group of fan in same direction, increase the suction of air? Can it force the air to pass through carbon filter about 1-2cm thickness?</p>
<p><a href="https://i.stack.imgur.com/j7at7.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/j7at7.jpg" alt="enter image description here"></a></p>
| |airflow|aerodynamics| | <p>The short answer is yes, but they will be much more effective if they turn it opposite directions, or have straightening vanes between to take the spin out of the air.</p>
| 11013 | Can air suction be increased by consecutive pc fans in small air tunnel? |
2016-08-10T00:05:22.390 | <p>What material is the viewing screen of a microwave made of? I have been researching and all I find is that the holes on the screen are too small to let the bulky microwave radiation to pass through.</p>
| |materials| | <p>The shielding in a microwave must be an electrically conductive material (usually thin steel), because it reflects microwaves.</p>
<p>The viewing window is also made of an electrically conductive material (thin steel), but has holes cut in it that are large enough to permit shorter wavelengths like light to pass through (so we can view it), but reflects long wavelengths like microwaves.</p>
<p><a href="https://en.wikipedia.org/wiki/Microwave_oven#Design" rel="nofollow">Microwave Oven Design on Wiki</a></p>
| 11024 | What is the screen of a microwave made off? |
2016-08-10T08:40:45.043 | <p>What is the highest pressure possible today?</p>
<p>I have a sketchy idea of using a weather balloon with the lightest possible pressure vessel that can pack as much helium as possible and would like some pointers on where to lookup to.</p>
<p>Is it possible to have a weather balloon liftoff with a composite pressure vessel dynamically controlling/adjusting the gas ?
What would be the ideal pressure vessel in terms of highest pressure possible?
Weight of the vessel plus pumping the gas back in the cylinder in flight? Is it doable?</p>
<p>some numbers please</p>
| |fluid-mechanics|gas|pressure-vessel| | <p>It sounds like what you really want to know is what is the best ratio of mass of gas to mass of container. Pressure is part of this but by no means the whole story. </p>
<p>Industrial gasses are typically stored in portable tanks at around 250 bar, although higher pressures of 300 bar or more are becoming more common. </p>
<p>These cylinders are typically mass produced in steel and are designed for long lives in under fairy adverse conditions and as such have quite generous factors of safety. </p>
<p>With this in mind you could certainly get better performance by using higher performance materials and trimming safety factors in favour of more stringent quality control. Also, while cylindrical tanks are the most convenient shape for industrial use they aren't the most efficient in terms of rated pressure to weight ratio. For example toroidal composite tanks are a lot lighter. </p>
<p>There is no overriding reason to go for <em>very</em> high pressures in this context as the mass of the gas is the same however much you compress it. </p>
| 11030 | Highest pressure helium(lifting gas)can be stored at? |
2016-08-10T18:06:16.120 | <p>Why do some variable speed motor switches have the <em>highest</em> speed setting next to the <em>off</em> setting instead of the <em>lowest</em> speed setting next to the off setting?</p>
<p>For example, most 3-speed household air-circulating / cooling fans will have 3 speeds with 3 being the highest. And the sequence will be: Off > 3 > 2 > 1. Why isn't it Off > 1 > 2 > 3? What design considerations are taken into account?</p>
<p>By contrast, most rheostat settings (e.g., light dimmer knobs/switches) have the dimmest setting closest to the off position.</p>
| |design|motors| | <p>An electric motor needs more current on startup to overcome inertia. While the low setting will likely start a brand new fan (example: unplugging it and plugging it back in on the low setting), it may not start an old fan that has more friction from worn bearings, etc. Even on a new fan, internal heat generated from this slow startup on the low setting may reduce the life of the motor.</p>
<p>If the friction was high and the low setting did not start the fan, the motor would burn up rather quickly because the current would stay high and there would be no cooling from rotation. If that same worn out fan was started on high it may have been fine because the current to maintain rpm on the low setting would be much less than startup and the cooling would be present.</p>
<p>So essentially the Off > 3 > 2 > 1 sequence is a very low cost way to ensure that 95% of the time the user toggles through the full power setting 3 to get to the lower power settings. This brief full power on startup helps get the motor to full speed quickly and increase the overall reliability of the product.</p>
| 11036 | Design decisions for variable speed motor switch settings (vs., say, rheostats) |
2016-08-10T21:30:05.160 | <p>The other day I was in the waiting room of a doctor's office in a new building as part of UCLA medical complex. A new construction as part of multi million dollar extension of existing medical complex in west Los Angeles. </p>
<p>I felt a small tremor and because of California's proximity to San Andreas fault immediately suspected an earth-quake! When the tremor did not go on as I expected and died out I realized it must have been caused by the impact of steps of another patient. I paid more attention and noticed the vibrations are very noticeable specially when the stride of the person resonates with natural frequency of the floor. </p>
<p>Here is my question re. undesired vibration of beams and floors under foot traffic. </p>
<p>Consider we have a beam of steel WF 12x120 with 20 feet span with I(xx) = 1070 inch^4. </p>
<p>What is the deflection under the foot of a man weighing 180 lbs.<br>
Lets assume he is wearing work-boots with negligible flexibility and when he lands on his foot he exert equivalent momentum as if he fell down 2 inches?<br>
And what is the vibration of this beam if we assume 5% damping?</p>
<p>As a comparison static deflection of this beam under same load is.
0.001614 inch = $Wl^3/48EI$</p>
| |beam|vibration|deflection| | <p>An exact theoretical analysis of this is rather involved so I will take an approximate approach.</p>
<p>We can idealise the beam as a single degree of freedom (SDOF) dynamic system. The theory behind this can be found in a variety of dynamics textbooks. For example: Biggs - Introduction to Structural Dynamics (1964).</p>
<p>Because the beam has uniform mass and stiffness we can obtain a Load-mass factor $K_{LM} = 0.49$ for a point load in the center of a simply supported beam. This converts the mass and load of a distributed beam into an equivalent SDOF spring-mass system by assuming a mode shape.</p>
<p>For the central point load it is straightforward to show that the stiffness $k = \frac{48EI}{L^3}$.</p>
<p>We can then calculate the circular natural frequency of the equivalent SDOF system:
$$\omega_n = \sqrt{\frac{k}{K_{LM}M}}$$
where $M$ is the total mass of the beam.</p>
<p>And the damped natural frequency:
$$\omega_d = \omega_n\sqrt{1-\zeta^2}$$
where, $\zeta$ is the damping ratio (in this case we take this as 0.05 since you specify 5% damping).</p>
<p>Taking the situation you describe, we can consider the system as having an initial velocity, which we will determine by conservation of momentum.</p>
<p>Velocity of the man when he impacts the beam:
$$v_f = \sqrt{2 a d} = \sqrt{2(32.2 \frac{ft}{s^2})(\frac{2}{12}ft)}=3.3\frac{ft}{s}$$</p>
<p>If he 'sticks' to the beam the initial velocity of the beam-man system is found by conservation of momentum:</p>
<p>$$ v_i (0.49 M+m)= v_f m \rightarrow v_i = v_f \frac{m}{m+0.49 M}$$</p>
<p>The displacement of a damped single degree of freedom system subject to an initial velocity is (see a dynamics text for derivation):</p>
<p>$$x(t)=\frac{v_i e^{-\zeta \omega_n t}}{\omega_d} \sin{\left(\omega_d t\right)}$$</p>
<p>so with
$$v_i = 3.3 \frac{ft}{s}\frac{180lb}{180lb+0.49 (120\frac{lb}{ft})(20ft)}=1.59 \frac{in}{s}$$
$$k = 48 \frac{EI}{L^3} = \frac{48(29000 ksi)(1070in^4)}{\left[(20ft)(12\frac{in}{ft})\right]^3}=108 \frac{kips}{in}$$
$$\omega_n = \sqrt{\frac{k}{K_{LM}M+m}}=\sqrt{\frac{108\frac{kips}{in}(386\frac{in}{s^2})}{0.49 (2400lb)+180lb}} = 175 \frac{1}{s}$$
$$\omega_d = \omega_n\sqrt{1-\zeta^2} = 175\sqrt{1-0.05^2}=174.8\frac{1}{s}$$</p>
<p>and we can now calculate the beam mid-span displacement x(t):
$$ x(t)=\frac{v_i e^{-\zeta \omega_n t}}{\omega_d} \sin{\left(\omega_d t\right)} = \frac{1.59 \frac{in}{s} e^{-0.05 (175\frac{1}{s}) t}}{174.8\frac{1}{s}} \sin{\left(174.8\frac{1}{s} t\right)}$$</p>
<p>The response is:</p>
<p><a href="https://i.stack.imgur.com/6PKPa.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/6PKPa.jpg" alt="enter image description here"></a></p>
<p>The frequency of the loading, for example due to walking, will have a large effect on the response. Especially if the frequency of loading is close to the natural frequency of the element (<a href="https://en.wikipedia.org/wiki/Resonance" rel="nofollow noreferrer">resonance</a>).</p>
<p>For floor vibrations you may find the <a href="http://www.steelconstruction.info/Floor_vibrations" rel="nofollow noreferrer">steelconstruction.info page on this topic</a> interesting.</p>
| 11037 | Undesired vibration of a beam under foot traffic |
2016-08-11T01:14:26.430 | <p>I have encountered a type of unit that I cannot understand. The unit describes the rate of permeation through a substance:(cm^3 . mm)/m^2 . d . atm </p>
<p>Could someone please explain how to interpret these units? Please attempt to leave the answer in simple - to - understand terms as chemistry, not engineering, is my primary field.</p>
| |pressure|measurements|gas| | <p>$$
\frac{\left(\mbox{amount of gas}\right)\left(\mbox{thickness of membrane}\right)}
{\left(\mbox{membrane area}\right)\left(\mbox{time}\right)\left(\mbox{differential pressure of gas}\right)}
$$</p>
<p><a href="http://onlinelibrary.wiley.com/doi/10.1002/app.1975.070190915/abstract" rel="nofollow">Units of gas permeability constants</a> (source of the formula)<br>
<a href="http://www.soarnol.com/eng/solution/solution040507.html" rel="nofollow">Gas Barrier Introduction</a><br>
<a href="http://en.labthink.com/en-us/literatures/detailed-introduction-to-material-permeability-indexes.html" rel="nofollow">Introduction to Material Permeability Indexes</a></p>
| 11038 | Understanding Permeation Units |
2016-08-11T07:27:17.320 | <p>Can someone please explain if the hypocycloid described below would degenerate to a line. If possible, can someone provide an actual plot of the entire question?</p>
<p>The scenario is as follows:</p>
<p>A circle starts rolling on a flat surface. After it completes one revolution, the path starts curving up in the arc of a circle till it becomes vertical and then the path becomes vertical. During the curved part of the path, the circle performs exactly half a revolution and then completes the other half of the revolution on the vertical path. Plot the locus of the point which was initially in contact the ground.</p>
<p>EDIT:
The problem is 2-D. Everything happens in a plane (I've copied the question exactly, except the original asks you to plot on paper). By vertical, it actually means upwards, parallel to the vertical edge of the paper and similarly for horizontal.</p>
| |drafting| | <p>I discovered that this indeed hoes happen. Since radius of the path is twice of the disc, the locus is a straight line there. Used GeoGebra for the purpose!
<a href="https://s10.postimg.org/f340jevft/S2_ED_6.gif" rel="nofollow">Animation</a></p>
| 11044 | Hypocycloid degenerates to straight line |
2016-08-11T16:07:40.253 | <p>Is it possible for a weather balloons filled with helium gas to lift the weight of itself and a helium tank(highest psi possible) which can fit the volume of the lifting gas in it?</p>
<p>Assuming a typical weather ballon material is used as the ascending vehicle would the balloon with the helium pumped back in the tank fall down to the earths surface?</p>
<p>P.S. In water the weight as well as the surface area for that weight determine its buoyancy - ships etc. A 5tons of dense block steel would sink in water but spread around over wide enough area it should float instead.</p>
<p>Wouldn't the same thing apply to the difference between a helium gas filling the larger volume of the balloon vs the high pressure compact volume storage vessel?</p>
| |aircraft-design|compressed-gases|pressure-vessel| | <p>As a reality check lets consider industrial gas tanks as the specifications are easily available. </p>
<p>A BOC size D hydrogen tank has a gross weight of 57kg and contains 23kg of hydrogen at 230bar. </p>
<p>At atmospheric pressure hydrogen has a density of 0.09 kg per cubic metre so you get about 255 cubic metres of gas at atmospheric pressure. </p>
<p>This will give lift of about 250 kg, so substantially more than the gross weight of the tank. You could also improve on this a bit by improving the performance of the cylinder eg a composite tank could significantly reduce the 'dead' weight of the system. </p>
<p>The problem is in compressing the gas back into the tank which would require a high pressure compressor plus a power supply to run it. This is probably not entirely beyond the realms of possibility but is going to eat substantially into the payload and probably isn't really worth it as the only benefit of this system compared to just venting the gas is saving the cost of the gas itself. </p>
<p>So the short answer is that yes it is probably possible (albeit likely very expensive) but there aren't any real advantages in doing it this way. </p>
<p>If you are looking at something a bit larger and more sophisticated like a passenger airship then, maybe as you already have an on-board power supply and the overheads of the compressors and tanks may be worthwhile, but it would take a fairly serious feasibility study of an actual design concept to say for sure. </p>
| 11047 | Will helium gas lift more weight in air when it is spread over large volume compared to equal weight heavily compressed small volume? |
2016-08-11T18:03:27.203 | <p>So the below question might sound like a homework question but in reality it is an oversimplified explanation of what I'm trying to measure.</p>
<p>Let's say we have a perfect machine that pushes out 600cfm into a box. On the other end of the box is another machine that only consumes 300cfm from the box.</p>
<p>What is the pressure in the box and how do I calculate that? I've looked into Boyle's Law and something just doesn't seem right about the results I'm getting so I'm trying to make sure that this is the right law I should be looking into.</p>
<p>If it's relevant, assume a volume of 360in<sup>3</sup> and temperature of 310 K.</p>
| |mechanical-engineering|fluid-mechanics|pressure| | <p>You can use the ideal gas law at first for a rough estimate.
Assuming we start at atmospheric pressure, and the flow rate is in standard cubic feet per minute:
$$P = 14.7 psia$$
$$V = 360 in^3 = 0.2083 ft^3$$
$$R = 10.731 \frac{ft^3*psi}{R*lbmol}$$
$$T = 310K = 558^\circ R$$
Starting point in the box at time = 0:
$$n=\frac{PV}{RT}=0.0005115 mol$$
Flow rate for 300 cfm converted to moles per second:
$$\frac{14.7 psi * (\frac{300 cfm}{60s/min})}{10.731}*558^\circ R = 0.01227 mol/s$$
After one second of pushing 0.01227 moles of air into a box that already contains 0.0005 moles of air:
$$n = 0.0128 mol$$
$$P = \frac{0.0128 mol * 10.731 * 558R}{0.208 ft^3} = 367.5 psi$$</p>
<p>In this situation, the box fills up so fast that the outlet flow would increase drastically to a certain point until you either deadhead the blower or get a box failure. Or as @Ben Welborn suggested, you may create nuclear fusion resulting in a wormhole between the engineering SE and physics SE</p>
| 11051 | Relationship between Pressure and volume with Air |
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