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2016-01-28T15:13:30.727 | <p>I have been reading a lot on <a href="https://en.wikipedia.org/wiki/Flexure_bearing" rel="nofollow">flexure bearings</a>, but it's still not really clear what they are. From everything I've seen, it seems like a flexure bearing is just a spring. </p>
<p>Can someone give me a simple explanation of what they are and their use?</p>
| |mechanical-engineering| | <p>From the Wikipedia article you linked,</p>
<blockquote>
<p>A flexure bearing is a bearing which allows motion by bending a load element.</p>
</blockquote>
<p>So it's something that provides support, but also allows motion by <em>bending</em> a component (as opposed to sliding).</p>
<p>There are several examples on the Wikipedia page, but I would add (in addition to those examples) that probably the most common flexure bearing would be the vibration mount.</p>
<p>It is typically a piece of rubber that is designed to provide support, and it allows some relative motion by bending. </p>
<p>I have vibration mounts holding the engine and transmission in my car, holding the ball joints, sway bars, and other suspension components. They're in the blender in my kitchen, in my washing machine, etc. </p>
<p>You could consider any rubber foot on a component designed to prevent rattle/hum (refrigerators, etc.) as vibration mounts and thus flexure bearings. They all provide support and allow motion by bending. </p>
| 7133 | What is a flexure bearing? |
2016-01-28T19:13:25.423 | <p>When building things like floors to walk on, we use materials like 2"x10" wood joists or steel I-beams. I'm primarily interested in knowing why a 2"x10" piece of wood resists bending downward when you have it "standing up" (ie, the 2" side facing the sky) as opposed to when you have it "laying down" (ie, the 10" side facing the sky).</p>
<p>It seems intuitive to me based on experience, but I can't actually explain why we orient floor joists the way we do even though it seems obvious that will lead to a stronger floor. I imagine whatever the reason is, it also explains why I-beams have such a thin webbing between the two sides.</p>
| |civil-engineering|structural-engineering| | <p>Why do we orient floor joists in with the longest side vertically? as in the shape of the letter "I" ?</p>
<p>AS mentioned in other posts, the name of the theory is "mass moment of inertia" of which there are whole textbooks written to explain.</p>
<p>However, quite simply, the formula for calculated this moment of inertia (for a rectangle like a 2x10 joist) is (bh^3)/12 or in English "B" x "H Cubed" over 12. </p>
<p>when the joist is oriented this way, its longer side (10") is oriented as the "H" (and the base is 2). Therefore when oriented this way, the wood resists loads much heavier than if it was "laid flat". </p>
<p>Try to bend a plastic ruler different ways. There is one way which will be very difficult. This is explained by moment of inertia!</p>
<p>The reason why I beams are so slim in the middle is that they are able to retain their strength in this orientation, but to save on weight, they are minimal in design (because the weight of the beam also adds to the forces that need to be calculated) So why would you make the beam any heavier than it needs to be if it is just as strong without the extra weight?</p>
| 7141 | Why does the orientation of a floor joist affect its strength? |
2016-01-29T01:46:30.760 | <p>I'm a beginner at autodesk inventor.<br>
For some instances when I click a region to extrude it doesn't come alone.</p>
<p>The screenshot below illustrates something even stranger. </p>
<p>What should I do to make this not to happen? </p>
<p><a href="https://i.stack.imgur.com/Yg1xp.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Yg1xp.png" alt="enter image description here"></a></p>
| |autodesk-inventor| | <p>In a sketch, you cannot have two lines that intersect themselves. You have to split them with the Split tool. </p>
<p>Your bottom line for example, must be divided into 10 parts.</p>
| 7148 | How do I extrude a region in autodesk inventor? |
2016-01-30T01:27:48.890 | <p>I am prototyping a nested waveguide made of four cylindrical sections. Each piece is turned on a lathe from Delrin (acetal) rod. Each piece fits inside the next larger one.</p>
<p>There are no moving parts. It is an antenna.</p>
<p>My question is: how much larger must the ID of a cylinder be, than the OD of the one which fits inside it?</p>
<p>In my drawing, I have indicated that they be 10 thousandths of an inch larger, but I do not know if this is correct.</p>
<p>I would like to be able to disassemble the cylinders during experiment, to take measurements, and then put them back together.</p>
<p>Each cylinder is to be electroformed with nickel on the outside, so the thickness of the metal layer will also be taken into account.</p>
<p><a href="https://i.stack.imgur.com/snB0y.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/snB0ym.png" alt="enter image description here"></a></p>
<p>Here is an updated drawing with tolerances added and revised diameters.</p>
<p><a href="https://i.stack.imgur.com/AWiiQ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/AWiiQ.png" alt="enter image description here"></a></p>
| |mechanical-engineering|machining| | <p>0.01" is probably a decent place to start as a slip fit, though you can certainly go tighter. I might even question if you actually want a slip fit for this application. You say it's for an antenna, and if you want it to extend and stay extended on its own like other collapsible antennae, a slip fit probably won't accomplish that. You'll need something that creates some friction but not so much that it can't be overcome by a person pulling on it.</p>
<p>In any case, what's potentially more important than the actual dimension is the tolerance. Those nominal dimensions will not be what you actually get, especially when you call out things down to the thousandth of an inch. For a slip fit, how I would actually dimension it is to make the nominal dimensions the same for the inner and outer, and then create a plus tolerance (e.g. +0.005"/-0) on the OD and a minus tolerance on the ID. Using 5 thousandths on each end will get you a 10 thousandths slip at most. </p>
<p>However, you need to know that the place you're sourcing these parts from can match that tolerance. 0.005" isn't that tough for a good machine shop, though I'm not sure how workable that plastic is, I'm more used to dealing with steels and cast irons. </p>
| 7179 | How much larger must the ID of a cylinder be than the OD for a Slip Fit? Material: Delrin (Acetal) |
2016-01-30T08:24:05.997 | <p>I have two bevel small (POM) gears that I'd like to test. I'm hoping to test their strength and their maximum torque. One is 15-tooth and is around 16MM wide and the other is 45-tooth and is around 46MM wide.</p>
<p>Does anyone know of a good, simple way for building a rig that hold these gears (and their 6MM shafts) in place in order for the test to take place? Would it be a good idea to just use 3 timbre boards and screw them together? Something like this</p>
<p><a href="https://i.stack.imgur.com/GoAJN.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/GoAJN.png" alt="enter image description here"></a></p>
<p>where the boards are held by screws. Would this suffice?</p>
<p>Thanks</p>
| |mechanical-engineering|gears|torque|product-testing| | <p>Torque ratings for bevel POM (also called acetal or Delrin) gears are at <a href="http://www.huco.com/products.asp?p=true&cat=285" rel="nofollow">http://www.huco.com/products.asp?p=true&cat=285</a></p>
<p>I didn't find timbre boards on google, I assume you meant timber wood boards. I don't think it is feasible with wood boards screwed together to get the necessary accuracy to holding the center lines of the gears to each other for valid test results. A precision machined gearbox would be ideal, but the cost would probably be prohibited. For practical purposes I think it is best to go with manufacture's torque ratings.</p>
| 7183 | Gear testing rig |
2016-01-30T13:59:48.527 | <p>I'm going to simulate an <strong>automotive radiator</strong> . The method called $\varepsilon$- NTU , suggests following steps to calculate the heat rejection,$Q$, of a heat exchanger:
<br/>1-calculate $U$(overall heat transfer coefficient)
<br/>2- calculate $NTU$
<br/>3- calculate $\varepsilon$ and finally calculate $Q(W)$.
<br/>My question is about calculation of $U$ in automotive radiators. We can neglect fouling and wall resistances and write:
$$
\frac{1}{UA} = \frac{1}{\eta_aA_ah_a}+\frac{1}{A_wh_w}
$$
$U$: Overall heat transfer coefficient. <br/>$A$ : Air side or waterside area. <br/>$A_a$ : Air side heat transfer area<br/>$A_w$: Waterside heat transfer Area <br/>$h$ : heat transfer coefficient<br/> $\eta_a$ air side surface efficiency.<br/>I want to use <em>Shanoun & Webb</em> formula to calculate the first term of the above equation(Air thermal resistance). They propose following equation for <strong>automotive radiators</strong>: $$ (\eta Ah)_a=(\eta_f Ah)_{l} +(\eta_f Ah)_{S1}+(\eta_f Ah)_{S2}+h_eA_e$$ You can find regions named <em>louvered(l)</em>,<em>S1</em>,<em>S2</em> and <em>e</em> in following image: <a href="https://i.stack.imgur.com/BbHPm.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/BbHPm.jpg" alt=""></a> This is image of a fin between <em>2</em> flat tubes.<br/><strong>Question</strong>: <em>Shanoun and Webb</em> just enter areas relating to one fin in their equation. So when we want to calculate $UA$, we account total surface of water$A_w$ but just a small surface of air side$A_a$. It makes the answer wrong. How can we use this formula to calculate $UA$ ?</p>
| |fluid-mechanics|automotive-engineering|simulation|cooling|heat-exchanger| | <p>I should multiply the No. of fins to the right-hand-side statement. So we have:
$$(\eta hA)_a=No.of fins \times( (\eta_fhA)_l+(\eta_fhA)_{S1}+(\eta_fhA)_{S2}+h_eA_e)$$</p>
| 7189 | Simulation of Automotive Radiators |
2016-02-01T07:25:25.200 | <p>I've read that it's not a good idea to rub two surfaces of the same material against each other, e.g., plastic against plastic, metal against metal. Plastic against metal is preferred. Why is this the case? Is it because that the coefficient of friction is higher for same materials?</p>
<p>According to <a href="http://www.tribology-abc.com/abc/cof.htm">this site</a>, steel-steel gives 0.5, plastic-plastic gives 0.3, but plastic-metal gives 0.25.</p>
<p>Suppose two rollers rotate and rub against each other. Will it be better for one to be made of metal and one of plastic?</p>
| |mechanical-engineering|friction| | <p>When material A of coefficient of friction $\mu_A$ and material B of $\mu_B$. </p>
<p>If material A is being rubbed with material B.
The requirement for abrasion and heat generated would affect both parts.
Hence if the application was for grinding, the tool itself would get affected.
So it would make sense to use a harder material to smooth the softer material.
For a steel part, you could use diamond, or even steel itself, but here the point of contact is different. </p>
<p>Also a lot of other factors come into play,
if you have rollers, and a part going in might be hot rolled, hence there is temperature difference or material property difference which helps roller material stay in its own geometry.
Also rollers have to be designed so as to take care of camber.</p>
| 7208 | Not a good idea to rub two of the same material against each other? |
2016-02-01T09:38:51.200 | <p>Can a screw such as the one below be used as an air engine in pressure range between 700 kPa and 2500 kPa <br>
on other words Can I reverse compression process with a screw to work as air engine</p>
<p><a href="https://i.stack.imgur.com/eC33J.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/eC33J.png" alt="enter image description here"></a></p>
| |mechanical-engineering|compressed-air| | <p>I can't imagine an Archimedes screw ( figure) being effective moving air/ gas. If you want to get work out of high pressure air , A reciprocal machine, ie a "steam" engine would be about the only effective mechanism. </p>
| 7209 | Using compression screw as air engine |
2016-02-01T18:08:36.813 | <p>I'm looking at something called at a Steadicam that stabilizes cameras. It looks like this</p>
<p><a href="https://i.stack.imgur.com/hj29C.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/hj29C.jpg" alt="enter image description here"></a></p>
<p>Clearly, the weights at the bottom are used for dampening/stabilizing. If the arm holding the weights were shorter/longer, would the stabilizing effect be affected? Would the centre of gravity change if the arm lengths were different (and thus affecting stability)?</p>
| |mechanical-engineering|structural-engineering| | <p>I think the most simple version would place the CG of the whole system right at the handle! this would eliminate any need for active electronic or mechanical damping.
So the range of the adjustable counter weight lever would allow for working with different shapes and masses of camcorders.
you would adjust to system by holding it and moving fast to the sides and back and forth and observing any rotation or sway. then by fine tuning the distance of the counter-weight you minimize the imbalance till you're satisfied with calibration. I guess the knobs are to adjust the grip for intentional tilt and sway to capture dynamic sports scenes!</p>
| 7218 | How does the location of a mass affect its dampening performance? |
2016-02-02T02:35:47.273 | <p>So I'm a high school student doing some competition (a wallet-sized one, unfortunately) and I'm supposed to make an electric car. Think an RC-car, (about that size) without the remote control. I want to make one from scratch, so I decided on using an arduino for a micro controller. Unfortunately, I don't really have an adviser that's particularly knowledgeable about this stuff, so I decided to ask here. (brace yourself for quite a few possibly dumb and vague questions. I do understand this question will likely be edited/closed for being too vague, but if, by any chance, someone has answers, I'd also like explanations -- mostly because I want to learn something.)</p>
<p>I've decided on using an aluminum chassis and soft wheels, since the competition wants a car that can go as fast as possible. I've chosen soft wheels because that should give me higher friction, and an aluminum chassis because I want to reduce weight. Now, here comes the first problem. The competition specifies a maximum of a 9v battery to power everything, including the arduino and the motors. Now, of course, we only need to go about 10 meters (I just want sub 3s times). My original design involved having 4 separate dc brushed/brushless motors, and then connecting each dc motor directly to the wheel. The dc motors could be held on with brackets, and then connected to an Arduino connected to a MOSFET, at which point I could then use a motor encoder to see how far to go.</p>
<p>I now have a couple of questions:</p>
<ul>
<li>Should I try having a 9v to 12v converter? or a 9v to 24v converter? That might cut the current significantly, but I'd also be able to use a much more powerful motor. The problem I have with this is that it might be adding a lot of potential weight -- as I understand it, I can't convert directly from DC 9v to something else.</li>
<li>I've decided on having a rear wheel drive rather than a 4wd -- but is it worth getting a gearbox and an axle and just having one motor?</li>
<li>Should I use a gearbox at all? Or should I try to change the speed based on the size of the wheel I use?</li>
<li>Would I try to maximize RPM or torque by getting a geared-down brushless?</li>
<li>If I have a 9v battery, can I have 2 9v motors in parallel? Or is that unsafe because 9v is only nominal?</li>
</ul>
<p>If anyone has tips on how to improve the question, I'd be glad to incorporate them into my edits.</p>
<p>some info, by the way:</p>
<ul>
<li>looking for sub3 second times</li>
<li>using banebots 2-7/8 inch diameter wheels</li>
<li>hard floor</li>
<li>will need to back up after 9m</li>
<li>only going straight, doesn't need to be able to turn</li>
<li>thinking of a PID loop with a physical rotary quadrature encoder if one axle. thoughts?</li>
<li>the car will be ~1kg and I won't be carrying anything else.</li>
</ul>
| |mechanical-engineering|motors|electric-vehicles| | <p>Answers to your specific questions:</p>
<ul>
<li>No, voltage converters won't help, at best they will waste 10% of your power. Think about this : they involve transformers. two different windings on a magnetic core to translate one voltage to another. That might help you understand that any given motor (which is just a winding on a magnetic core to translate voltage into motion) can be tuned to run off any voltage by changing the winding. You'll see the same RC car motors with 13,15,17 turns to tune performance.</li>
</ul>
<p>Instead, learn the relationship between voltage and speed (Kv, rpm/volt, the relationship is simply Speed = Kv * voltage) and current and torque (Kt, Newton-metres/amp).
A high speed motor (high Kv) gives little torque, and a low speed motor gives higher torque. So with direct driven wheels you want a LOW speed motor.</p>
<p>You have a mass (1kg), a distance (10m), and a time (3s). If you've done basic high school physics you know how to get speed and acceleration from distance and time. Work out 2 cases : linear acceleration from 0 to twice the average speed, and covering the distance at average speed. </p>
<p>I'll recommend you stick to metric, it makes some of the math easier. If your motor Kv is specified in rpm/volt, you have to convert that to radians/second/volt. In metric units, the torque relationship Kt = 1/Kv, so torque is Kt * current, or 1/Kv * current.</p>
<p>If you think about electric power, P = V * A and mechanical power = Nm/second = torque * rotation rate in radians/second, and the law of conservation of energy, you'll see why Kt = 1/Kv and understand how (high speed, low torque) and (low speed, high torque) can provide the same power.</p>
<hr>
<p>Now acceleration and mass give you force, and force * wheel radius gives you the torque required for acceleration. </p>
<p>Also multiply mass * gravity(9.8N/kg) * coefficient of friction to get the force required to push the car at constant speed, and convert that to a torque. (You don't know the coefficient of friction : try 0.2 for a perfectly smooth floor and 0.6 for carpet)</p>
<p>Now you know the torque required just for acceleration, and for travelling across the floor. The actual torque you need is the sum of these. (adding a safety factor like another 25% isn't a bad idea).</p>
<p>Given the speeds you need, in m/s, and a specific wheel radius, divide speed by radius to get rotational rate in radians/second. (Convert that to RPM if you want).</p>
<hr>
<p>Now that you know the speed (rad/sec or rpm) and torque (Nm or Newton metres) you can go back to selecting a motor. Can you find a motor with the right Kv, where 9V * Kv is anywhere close to your desired rotation speed? What current do you need to achieve the torque? If it's over 0.5A you'd better forget that little 9V battery, and that's being optimistic...</p>
<p>I'm guessing not, and that's where you need a gearbox to translate from (high speed, low torque) to (low speed, high torque). So if your motor speeds are 10x too high, use a 10:1 gearbox, multiply your required speed by 10, and divide the required torque by 10. Can you meet your speed and current requirements now?</p>
<p>All this may look daunting but each bit is relatively simple maths. If there are specific sticking points, think about them for a bit, and ask if you need to.</p>
<hr>
<p>Having laid that groundwork, the rest of the questions should be easy...</p>
<ul>
<li>If you have 4 motors, each delivers 1/4 the torque and consumes 1/4 the current, so there's not much to choose between 1 motor and 4 in overall efficiency, but if one small motor can't supply the torque you need, 2 or 4 might work.</li>
<li>The huge chunk of maths above will tell you if you need a gearbox. You probably do.</li>
<li>You don't want to maximise RPM OR torque, you want just enough of both to cover the requirements and a bit (50% or so) to spare if possible. (I haven't mentioned inefficiencies like friction that eat away at performance little by little. This is long enough without that!). Brushless motors aren't a whole lot better at delivering torque, where they win is by running much faster, so you'd need to gear them down more. Probably not worth it here.</li>
<li>You can run motors in parallel, but remember their currents add up, and your little battery can't supply much current. (<a href="http://www.farnell.com/datasheets/2006032.pdf" rel="nofollow">Good batteries have datasheets</a> which show how much current they can supply, for how long. In this case, 0.5A for much less than an hour, down to 4.8V)</li>
</ul>
| 7221 | A couple of questions about making an electric hobby car from scratch |
2016-02-02T05:52:46.400 | <p>I am faced with a problem where I have to design a reinforced masonry beam for biaxial bending. The governing code is ACI 530-11. I cannot find a provision in this Code for biaxial bending. The only part that addresses this issue is in Section 2.2.3.1 which states:</p>
<blockquote>
<p>The unity formula can be extended when biaxial bending is present by replacing the bending stress quotients with the quotients of the calculated bending stress over the allowable bending stress for both axes</p>
</blockquote>
<p>Unfortunately this section deals with unreinforced masonry. It is very strange to me that reinforced masonry is not addressed. Any ideas?</p>
| |structural-engineering|structures|building-design|masonry| | <p>I think that the MSJC (ACI530) actually contradicts itself, a little, with regard to biaxial bending of reinforced masonry.</p>
<p>As you pointed out, Section 2.2.3 (unreinforced masonry) points to the use of the Unity Equation. However, the commentary of Section 2.3.4.2.2 (reinforced masonry) explicitly says,</p>
<blockquote>
<p>The interaction equation used in Section 2.2.3 is not applicable for reinforced masonry and is therefore not included in Section 2.3.</p>
</blockquote>
<p>So it seems that the omission of the unity equation from Section 2.3 is intentional. Further, it isn't clear what that statement means, exactly. Is it implying that the equation explicitly written in Section 2.2.3 ( $\frac{f_a}{F_a} + \frac{f_b}{F_b} \leq 1$ ) is not applicable to reinforced masonry, or that the Unity Equation in general is not applicable?</p>
<p>However, I think that the language provided in Section 2.3.4.2.2 is fairly clear, or at least clear enough, to indicate how one might approach biaxial bending or (biaxial) bending and compression. It states,</p>
<blockquote>
<p>The compressive stress in masonry due to flexure or due to flexure in combination with axial load, shall not exceed 0.45$f'_m$ provided that the axial compressive stress due to axial load component, $f_a$, does not exceed the allowable stress, $F_a$, in Section 2.2.3.1.</p>
</blockquote>
<p>This seems to imply that, for axial force with bending, one would need to satisfy <strong>both</strong>, </p>
<p>$$\frac{f_a}{F_a} \leq 1 \tag{1}$$</p>
<p>and,</p>
<p>$$\frac{f_a + f_b}{0.45f'_m} \leq 1 \tag{2}$$.</p>
<p>This would then lead me to rationally assume that for pure biaxial bending we'd simply need to satisfy,</p>
<p>$$\frac{f_b^1 + f_b^2}{0.45f'_m} \leq 1 \tag{3}$$ </p>
<p>which is just a form of the Unity Equation. Hence, the code seems to contradict itself, kinda.</p>
<p>A few final things worth noting:</p>
<ul>
<li>I have a couple references, written to ACI380-11, that qualify masonry beams for biaxial bending using equation 3 (published by PPI),</li>
<li>I see all sorts of stuff online, from reputable schools and organizations, using equation 3 to qualify reinforced masonry beams for biaxial bending,</li>
<li>Don't forget that you need to check the combined stress in your steel as well. This conversation has largely focused on the stress in the masonry, but ensure that
$$\frac{f_s^1 + f_s^2}{F_s} \leq 1 \tag{4}$$</li>
</ul>
| 7222 | How is biaxial bending considered for reinforced masonry design? |
2016-02-02T20:33:49.490 | <p>I see that most worm gears are self-locking but I'm looking to make my own non-self-locking POM worm gear. It'll only be of module 1. Does anyone know what needs to be changed in order to make a self-locking worm gear non-locking? Is it just the angle of the threads?</p>
<p><strong>EDIT</strong></p>
<p>Many motorised locks tend to have a worm gear on the motor shaft though - like <a href="https://i.stack.imgur.com/DWkYT.png" rel="nofollow noreferrer">this one</a>. The lock is designed to be manually turned as well. Does that mean the friction caused by manual turning will be high?</p>
| |mechanical-engineering|gears| | <p>Like Chris Johns mentioned, locking is dependent on lead angle and a coefficient of friction; specifically a coefficient of static friction.</p>
<p>To get a worm gear design to back drive you can increase the lead angle or decrease the friction coefficient, both of which increases the ratio of torque to friction. It will not be efficient, and even though it is spinning it may not transfer much load before locking.</p>
<p>Many "locking" worm drives can back drive when they are already in motion when a force was applied. This is because the dynamic friction coefficient is less than the static friction coefficient. This is why many safety dependent systems also have brakes since the "locking" characteristic of the worm drive is not a guarantee.</p>
<p>If you want to run a worm drive the other direction with reasonable efficiency you may want to look at a <a href="https://www.youtube.com/watch?v=hvz9A0c_HeQ" rel="nofollow">ball wormdrive</a> similar to a <a href="https://en.wikipedia.org/wiki/Ball_screw" rel="nofollow">ball screw</a> design. The ball bearings have rolling resistance which is much lower than sliding friction. Probably too expensive for most projects though.</p>
| 7230 | Designing non-self-locking worm gear |
2016-02-03T13:24:48.743 | <p>When iron is melted, I guess it has to be transported and contained. I think the container in which it is has to be able to withstand higher temperatures than what you want to melt.</p>
<p>According to <a href="http://www.engineeringtoolbox.com/melting-temperature-metals-d_860.html">this webiste</a>, "Iron, Wrought" has a melting temperature of 1482 - 1593 °C. There are a couple of other metals which have higher melting points (e.g. Wolfram (tungsten) with over 3400 °C), but all I can think of are much more expensive. So what material is the oven / "bottle" / "basin" (or however you call it) made of?</p>
<p>(Side question: Iron has been melted for quite a while now. I guess this has changed over the years. Of which materials was it before?)</p>
| |metallurgy| | <p>Other recycled materials such as <a href="https://millscale.org" rel="nofollow noreferrer">mill scale</a> can be used in refractories as well.</p>
<p>Refractories</p>
<p>Refractory material is made by crushing dolomite and mixing it with a flux suspension liquid or paint. Mill scale can be used as flux material that is combined with the liquid binder and ultimate used to produce the refractory material. <a href="https://en.wikipedia.org/wiki/Mill_scale" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Mill_scale</a></p>
| 7245 | What material is used to hold molten iron in a furnace? |
2016-02-04T20:53:40.973 | <p>I am looking into making a strong electro magnet.
I want it to be able to hold around 100 pounds of weight or more. </p>
<p><a href="http://rads.stackoverflow.com/amzn/click/B00GXTPBRY" rel="nofollow">Would a magnetic core like this work at all?</a>
I'm planning on wrapping wire around one of these 'C' shaped ferrite cores.</p>
<p>How much current should I expect to be using?
How many turns or windings around the core should I expect?
Is 22 gauge copper solid core wire recommended? </p>
<p>A rough estimate would help very much.
Also If there is any formula or information you could add so that I could start to answer some of these questions I would greatly appreciate it. </p>
| |power-electronics|magnets| | <p>The basic principle is that the force exerted by an electromagnet (solenoid) is proportional to the number of turns in the coil and the current passing through the coil. </p>
<p>The gauge of wire required will be determined by the current. The size of the core doesn't affect the strength of the magnet directly but the maximum field strength per unit area is limited so, in practice it does need to be scaled with the force required. </p>
<p>The best thing to use as a core would be soft iron or steel, it's dimensions are determined by its cross sectional area and the length required to physically fit the core on it. The ferrite core you linked to probably isn't large enough and in any case ferrite is quite brittle and probably won't stand up to being used to lift significant weights. </p>
<p>This <a href="http://www.daycounter.com/Calculators/Magnets/Solenoid-Force-Calculator.phtml" rel="nofollow">calculator</a> should help </p>
<p>You should also note that large electromagnets are potentially dangerous as they can cause injury if you trap your fingers with them and you are going to be dealing with fairly significant currents not to mention the fact that electromagnetic lifting devices do not fail safe and lifting 100lbs needs to be treated with caution</p>
| 7272 | Making a strong electro magnet |
2016-02-05T01:08:42.373 | <p>I am trying to work out how a liquid inside a container will be affected by the back and forth sliding of a Teflon coated plate such as the one in the picture. </p>
<p><a href="https://i.stack.imgur.com/eBZT7.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/eBZT7.jpg" alt=""></a></p>
<p>I know how to work out the shear rate of a fluid between two plates moving at different velocities using the v/h equation, but it doesn't take into account things like the walls on my container, or the fact that the plates may be hydrophobic. </p>
<p>My goal is to move the plate (which is in contact with the liquid) back and forth without the liquid moving. So my question revolves around firstly whether it is even possible to move the screen back and forth without significantly displacing a certain liquid it is in contact with underneath, and secondly how I would go about calculating to what extent the liquid would be disturbed. </p>
| |fluid-mechanics|microfluidics| | <p>As an addition to @ChrisJohns answer: </p>
<p>if you want to limit the effect of the plate oscillations on the motion of the liquid as much as possible you need to make the time scale for viscous diffusion much larger than that of the plate oscillations.</p>
<p>If the plate oscillates with an angular frequency $\omega$, the liquid has a kinematic viscosity $\nu$ and the length scale is the height of the box $H$ then you want the ratio:
$$\epsilon=\frac{\nu}{\omega H^2}\ll1$$
here $\frac{H^2}{\nu}$ is the viscous time scale and $\frac{1}{\omega}$ is the oscillation time scale.</p>
<p>As mentioned by @ChrisJohns, this requires very low viscosities or large angular frequencies. However, there will alway be some pertubation of the liquid by the plate but $\epsilon\ll1$ should limit it. </p>
<p>Take note, that this assumes that the height of the box is the characteristic length scale; this assumes the absence of any interference by walls. I would redesign the box to have a height much smaller than any of the other two dimensions if possible but still such that $\epsilon\ll1$.</p>
| 7275 | Shear rate of a fluid in a container with a sliding PTFE top plate. Will the liquid move with the plate? |
2016-02-05T07:46:49.950 | <p>I'm hoping to design a simple gearbox for a 1-seater car for kids. So far I'm only using discs of paper I cut out and rotating them by hand. I wonder what the canonical approach is to prototyping/experimenting with gearboxes. I guess the prototype won't need to be big, but just 10CM or so?</p>
<p>Should it be done virtually in a CAD software instead?</p>
<p>I'm very new so any tip will be appreciated.</p>
| |mechanical-engineering|gears| | <p>In designing a gearbox the first step is to work out the ratios you need in the context of the torques and rotational speeds you need, bearing in mind that other parts of the drive train such as the differential and wheel diameter will also effect the final drive ratio. This will allow you to select the most appropriate overall design and specify parts which can meet these requirements. </p>
<p>These basic parameters are simple enough to calculate in themselves but need to be considered in the context of the whole design as they depend on the overall performance requirements for the car. In particular the range of gear ratios required will depend on the mass of the car and the performance of the engine/motor.</p>
<p>For something like this designing gears from scratch is probably not something you want to do so this will likely be a case of sourcing component eg. gears, shafts bearings etc. </p>
<p>The next step is to start thinking about how the whole thing will be packaged which also needs to consider how the gearbox casing relates to other parts of the system. Here a lot depends on what you want to achieve and there are many potential different solutions. I'm assuming that a car for kids will be relatively low speed and power and so potential solutions could include chain and pulley drives as well as gears. </p>
<p>It is also well worth considering whether there is some existing off the shelf component which can be adapted to meet your needs. </p>
<p>To expand on my reply to the first comment. A gearbox is a case where you really need to get quite a lot of the design parameters calculated on paper before you think about prototyping. They key parameters like the ratios and loads on individual parts are largely defined by external factors such as engine characteristics and desired vehicle performance so at the very least you need to give yourself a reasonable starting point. </p>
<p>These parameters are reasonably easy to calculate and don't require FEA or calculus, especially if you are using off the shelf components. It's also entirely reasonable to do most of the design of a gearbox 'on paper' and a working prototype would normally only be needed for things like reliability testing. </p>
<p>Prototyping is most useful for working out fine detail like how the gearbox housing is physically connected to other parts of the vehicle in terms of its size, shape and where mounting points need to go and whether it interferes with other parts. But you can't even begin to do this until you have worked out the basic parameters. </p>
<p>These should be based on your best estimate of things like vehicle weight and engine performance as well as desired top speed and acceleration. </p>
| 7280 | Canonical approach to experimenting with a gearbox design |
2016-02-05T12:34:16.900 | <p>I'm studying how industrial plastic film pallet wrappers work. The wrappers apply a certain amount of pre-stretch to the plastic film (by means of different rolls) before applying it to the object that has to be wrapped.</p>
<p>If the object to be wrapped isn't spherical, i.e. it has edges, then the plastic film is pulled more when it reaches an edge, and so there is an variation in how much the plastic stretches. I believe that this variation in stretching leads to the plastic film breaking when the pallet is rotating very quickly and subsequently has high pull on the film.</p>
<p>Are there techniques that make it possible to even out these variations and to make the tension the film is under during application more uniform?</p>
| |plastic| | <p>One way used to obtain more uniform loading when wrapping around square corners is to have the film run through a spring loaded dancer bar. When a corner pulls the film faster, the dancer bar rotates. Also, the dancer bar is connected to rotary potentiometer, which is wired to a DC SCR control board and speeds up the DC motor. This method provides control at a relatively lower control cost. A servo, encoder, and sensor system could provide very precise control for using high pre-stretch ratios, but may not be implemented in commercial systems as stretch wrappers are relatively lower cost machinery.</p>
| 7287 | Plastic wrapper stretch |
2016-02-06T08:06:01.243 | <p>I'm hoping to make a small mechanical demo (~10cm tall) that models an orbit that is similar to the Earth-Moon orbit - that is, only one side of the Moon faces Earth. I'm trying to keep it simple, so here's my idea so far:</p>
<p><a href="https://i.stack.imgur.com/QtN7S.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/QtN7S.png" alt="enter image description here"></a></p>
<p>where the blue and yellow represent Earth and Moon, respectively. These are meant to represent a different planet and its moon though.</p>
<p>The orbit of the Planet (blue) and its Moon (yellow) has the following differences:</p>
<ul>
<li>Planet is eccentric</li>
<li>Only one face of the Planet and see its Moon, i.e. the horizontal bar on each disc should remain parallel. The bars are for explanatory purposes only.</li>
<li>Planet's rotation axis may shift (vertically along the green bar), but the Moon's orbital axis is fixed, i.e. Moon should be always in contact with the outer ring housing</li>
</ul>
<p>My question is: how should the two discs be connected in order to to have the above properties? I've thought about using springs, but I think it'd be quite flimsy though.</p>
<p>I know about orreries, but typically they don't self-adjust orbits though. Also, I was hoping for a simpler mechanism given that there doesn't need to be empty space between the two discs. I'm only hoping to model the motion on just one plane.</p>
| |mechanical-engineering|gears| | <p>I started these as comments but they're way too long for comments. Basically, please clarify what you want. </p>
<ol>
<li><p>The orbit of the Planet (blue) and its Moon (yellow) has the following differences: Planet is eccentric." Why would the eccentricity of the planet's orbit affect its moon? You don't have a sun in that model, so really I don't understand what the planet's orbit has to do with any of this.</p></li>
<li><p>"Only one face of the Planet [can] see its Moon." This isn't how our moon works. With our moon, only one side of it can see the planet. The way you have it described, it would be like if only the Western hemisphere could see the moon, and nobody in the Eastern hemisphere would have ever seen it. Also, the way you describe the motion (the bars are always parallel) is like both scenarios put together - the moon can only be seen from one hemisphere and that moon only has one face that can be observed. </p></li>
<li><p>"Planet's rotation axis may shift" Here's where you really lost me. Maybe it's because <em>your drawing isn't labeled</em> (!), but you state you are, "only hoping to model the motion on just one plane." As I mention above, there is no sun, so I'm assuming the spinning disc that is your planet it spinning on its axis. If it is shown as spinning on its axis, then you can't move the planet to adjust the tilt of that axis and keep it planar. It looks like you might mean to do this by shifting the point of rotation, but all that would do is cause the entire planet to rotate eccentrically.</p></li>
</ol>
<p>Please clarify your question by:</p>
<ol>
<li>Labeling your drawing</li>
<li>Add several drawings showing "snapshots" of the motion you would like to demonstrate. </li>
</ol>
<p>Also bear in mind that, to show rotation in a plane, the axis of rotation has to be orthogonal to the plane. As soon as you change the angle of the axis of rotation you <em>cannot</em> have a planar display. You could have a <em>projection</em>, but I'm not sure that you could depict a 2D projection of an off-axis sphere rotating with a "simple" setup. </p>
| 7296 | How would you mechanically model an Earth-Moon like orbit? |
2016-02-06T11:50:53.497 | <p>I was studying the designation of names to refrigerants. </p>
<p>The following is the basic formula:</p>
<p>R - (m - 1) ( n + 1) ( o )</p>
<p>where:<br>
m = number of carbon atoms in the refrigerant<br>
n = number of hydrogen atoms in the refrigerant<br>
o = number of fluorine atoms in the refrigerant</p>
<p>So R-134a has:</p>
<p>1 + 1 = 2 carbon atoms<br>
3 - 1 = 2 hydrogen atoms<br>
4 fluorine atoms</p>
<p><strong>What does the 'a' at the end mean?</strong></p>
| |chemical-engineering|refrigeration| | <p>If a refrigerant contains 'a' at its end it is the isomer of that refrigerant.
Example 'R123' is 'C2H2F3Cl2' and 'R123a' is the isomer of the same molecule.</p>
| 7297 | What does "a" mean at the end of a refrigerant's designation (R-134a)? |
2016-02-06T15:00:27.497 | <p><strong>Background</strong>: I'm a statistician working with a design of experiments example regarding plasma etching of circuits, and unfortunately, the author does not define what is meant by the units used for the etch rate ($\mathring{A}$/min). A google search has proved fruitless due to the specialized symbol and the fact that other texts also assume you already know what the $\mathring{A}$ symbol means (which is probably true for those who work in this field, but not for the rest of us). </p>
<p><strong>Question</strong>: What does the $\mathring{A}$ unit mean in a plasma etching experiment, for "Etch rate ($\mathring{A}$/min)."</p>
| |circuits|metrology|nomenclature| | <p>Anisotropic etching is a linear process, like welding, so presumably the units would be distance per time. Unlike welding it is at a small scale. See <a href="https://www.slideshare.net/mobile/minh65/plasma-etching">this slideshare</a>. In this case units are <strong>Angstroms per minute</strong>. One Angstrom is <strong>$10^{-10}\ \textrm{m}$</strong>, indicating the process is on an atomic scale. Since it appears to be used for integrated circuit fabrication, those units are sensible depending on the scale.</p>
| 7301 | Units for plasma etching of circuits: A (with degree symbol)/min |
2016-02-07T06:47:32.460 | <p>The problem in my text (Thermodynamics An Engineering Approach, Cengel & Boles, 2nd Ed.) is stated as follows:</p>
<p>Balloons are often filled with helium gas because it weighs only about one-seventh of what air weighs under identical conditions. The buoyancy force, which can be expressed as $F_b=\rho_{air} g V_{balloon}$, will push the balloon upward. If the balloon has a diameter of $30ft$ and carries two people, $140 lb_m$ each, determine the acceleration of the balloon when it is first released. Assume the density of air is $\rho_{air}=0.0724 \frac{lb_m}{ft^3}$, and neglect the weight of the ropes and the cage.</p>
<p>Answer: $45.1 \frac{ft}{s^2}$</p>
<p>I can easily find a solution in S.I. units. It is the U.S. Customary units that I am having difficulty with.</p>
<p>Given:
$$
r = 15ft \\
m_{people} = 2 \cdot 140 lb_m \\
\rho_{air}=0.0724 \frac{lb_m}{ft^3} \\
$$
Let us begin by finding $V$ and $F_b$:
$$
V_{balloon} = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi (15ft)^3 = 14137 ft^3 \\
F_b=\rho_{air} g V_{balloon} = \frac{0.0724\frac{lb_m}{ft^3}}{32.174\frac{lb_m}{lb_f}\frac{ft}{s^2}}\cdot32.174\frac{ft}{s^2} \cdot 14137 ft^3 = 1023 lb_f\\
$$
Next let us find all masses:
$$
m_{people} = 2 \cdot 140lb_m = 280 lb_m = \frac{280lb_m}{32.174\frac{ft}{s^2}} = 8.703slugs \\ (Q1: \: how\: do\: I\: make\: the\: units\: work\: out?) \\
m_{He} = \frac{\rho_{air} V_{balloon}}{7} = \frac{0.0724\frac{lb_m}{ft^3}}{7 \cdot 32.174\frac{lb_m}{lb_f}\frac{ft}{s^2}} \cdot 14137 ft^3 = 4.545 slugs \\
(Q2: \: how\: do\: I\: make\: the\: units\: work\: out?) \\
m_{total} = m_{people} + m_{He} = (8.703 + 4.545) slugs = 13.25 slugs$$</p>
<p>Now let us determine the $F_{net}$ and finally the acceleration by Newton's Second Law:
$$
F_{total} = m_{total}g = 13.25 slugs \cdot 32.174\frac{ft}{s^2} = 426lb_f\\
(Q3: \: how\: do\: I\: make\: the\: units\: work\: out?) \\
F_{net} = F_b - F_{total} = (1023-426) lb_f = 596.7 lb_f \\
a = \frac{F_{net}}{m_{total}} = \frac{596.7 lb_f}{13.25 slugs} = 45.1 \frac{ft}{s^2} \\
(Q4: \: how\: do\: I\: make\: the\: units\: work\: out?) \\
$$
I was able to duplicate the solution but I had to fudge on the units because I am not sure how to convert between $lb_f$ to $lb_m$ to $slugs$. Please advise by using the very same calculations in this problem. Please select any of Q1 to Q4 to demonstrate how to use unit conversion factors in U.S. customary units.</p>
| |fluid-mechanics|thermodynamics|statics| | <p>If it helps, convert the input values to whatever units you are comfortable with, then convert the answer to whatever units the question is asking for. If the question doesn't specify units for the answer, and this isn't implied from discussions in class, then answer in any reasonable units you like.</p>
<p>I said <i>reasonable</i> units, since deliberately tweaking the professor's or TA's nose is not a good idea. He might be fine with meters/second instead of feet/second or miles/hour, but furlongs/fortnight is just deliberate provocation.</p>
<p>In any case, pounds-force, pounds-mass, and slugs are all very common units that are well defined out there, so answering that part of your question is pointless. Go look them up. Whether you like them or not, they are units you should at least know how to use, perhaps just looking up the conversion constants when you need them.</p>
| 7316 | Find Acceleration Of A Buoyancy Force Of A Balloon In U.S. Customary Units (1-49E) |
2016-02-08T00:04:47.807 | <p>I'm looking for a linear actuator that can lift a microphone of about 700 grams. It's for a goose-neck microphone (together with shock-mount) that will pop out of a small pulpit. Only the top part is bent so it will stick out a small part when at the lowest position.</p>
<p>The <a href="http://www.alibaba.com/trade/search?fsb=y&IndexArea=product_en&CatId=&SearchText=electric%20linear%20actuator" rel="nofollow">electrical actuators</a> I have found so far are made for heavy lifting and are far to slow (and noisy). I need to be able to adjust the height quickly when a different speaker comes.</p>
<p>What I'm looking for would be more like a printer head carriage in an ink-jet printer. It does not have to be that fast, 10cm/second would be good enough with a stroke of +/-45cm. It would be great if it could also send feedback on its position (or if it could be driven to a specific point like a servo).</p>
<p>Does anything like that exist off the shelf or would I have to custom-build the whole thing? I'm planning to use an <a href="http://www.netduino.com/netduinoplus2/specs.htm" rel="nofollow">Netduino+</a> (with a motor shield) to drive it.</p>
| |mechanical-engineering|electrical-engineering|motors| | <p>If you're not set on the motor shield, you could use a stepper motor with a pulley instead, plus some form of linear guide (eg. an Igus slide, if you don't want it just dangling). Small Nema 17 size stepper motors easily generate enough torque to lift 700g, and you can use <a href="http://reprap.org/wiki/StepStick" rel="nofollow">StepStick</a>-style drivers, which can be driven easily from an Arduino.</p>
<p>If you need it to be very quiet, the <a href="http://www.watterott.com/en/SilentStepStick" rel="nofollow">SilentStepStick</a> is a driver which is completely inaudible, while many other stepper drivers cause audible hum/buzz in the motors.</p>
| 7318 | Fast and quitet linear actuator for lifting less than a kilo |
2016-02-08T20:50:21.843 | <h3>Crushing Question</h3>
<p>While reading a material specification for crushed aggregate, the text mentions both "crushed stone" and "crushed gravel" are acceptable. These materials sound very similar, but they are mentioned individually. I assume that this means that they are not the same.</p>
<ul>
<li>What is the difference?</li>
<li>How can I tell the difference if I see them on-site?</li>
</ul>
<h3>Example</h3>
<p>This <a href="https://www.michigan.gov/documents/aero/P-209_282160_7.pdf" rel="noreferrer">specification</a> from Michigan is an example:</p>
<blockquote>
<p>Aggregates shall consist of clean, sound, durable particles of crushed stone, crushed gravel, or crushed slag...</p>
</blockquote>
| |civil-engineering|geotechnical-engineering| | <p>It's of importance to note that historically, "stone", "gravel", and many other related terms have great regional variability. It is always advised to make certain what is meant by such terms when they are being used. </p>
<p>Although the term "gravel" does have a specific engineering definition (aggregate of a certain size range), in design specifications the terms "crushed stone" and "crushed gravel" have come to have a slightly different and generally agreed upon meaning. Using these terms together is a way of specifying several things at once, including the <em>source</em> of the material, and its size, as well as its shape characteristics. </p>
<p>In this context, "stone" is rock that is sourced- usually quarried- from some parent rock (such as granite, limestone, and dolomite). Stone has generally not been naturally created by weathering. On the other hand, "gravel" is rock fragments sourced from an existing deposit of weathered rock, often from rivers and streams, but also gravel pits. As such, gravel tends to be more rounded in shape. </p>
<p>The modifier <em>crushed</em> specifies two things at once. First, that the aggregate be angular in shape and not rounded. Second, that the aggregate consist of a variety of sizes. When required, these very general terms are often fleshed out a bit in the design specification by requiring the aggregate meet some kind of gradation curve, or a durability test (such as the LA Abrasion Test). </p>
<p>When these terms are listed such as in a design specification like the one quoted above, the specification is allowing for just about <em>any</em> material <em>source</em>, so long as the material meets the size and shape requirements. </p>
<p>If the material is high quality and meets specification, telling the difference between the two on site should be difficult. Look for smaller rounded pieces in the stockpile that were missed by the crusher to identify a gravel source. Rounded pieces from a quarried stone should be very rare. </p>
| 7328 | What is the difference between "Crushed Stone" and "Crushed Gravel" aggregate? |
2016-02-09T16:13:55.363 | <p>I've read that Mach 0.3 is pretty much the upper limit for treating air as an incompressible fluid. The sources I've read seem to treat this as a given, without proof or justification. </p>
<p>Why is this the limit? Is there a mathematical justification for this? Also, does this limit only apply to air? If not, then what does the limit depend on?</p>
| |fluid-mechanics| | <p><a href="https://en.wikipedia.org/wiki/Compressible_flow" rel="nofollow noreferrer">Wikipedia gives the reason</a> for Mach 0.3 as due to the fact that this achieves ~5% change in density.</p>
<p>I found a <a href="https://www.grc.nasa.gov/www/k-12/airplane/machrole.html" rel="nofollow noreferrer">NASA page</a> that describes (analytically!) the relationship. I cited the source, but I'll reproduce the work here for posterity, in the event their links change. </p>
<p>Start with conservation of momentum:</p>
<p><span class="math-container">$$
(\rho V) dV = -dp \\
$$</span></p>
<p>where <span class="math-container">$\rho$</span> is the fluid density, <span class="math-container">$V$</span> is the velocity, and <span class="math-container">$p$</span> is the pressure. for isentropic flow:</p>
<p><span class="math-container">$$
\frac{dp}{p} = \gamma \frac{d\rho}{\rho} \\
dp = \left( \frac{\gamma p}{\rho} \right) d\rho \\
$$</span></p>
<p>where <span class="math-container">$\gamma$</span> is the specific heat ratio. The ideal gas law gives:</p>
<p><span class="math-container">$$
p = \rho R T \\
$$</span></p>
<p>where <span class="math-container">$R$</span> is the specific gas constant and <span class="math-container">$T$</span> is the absolute temperature. So, substituting:</p>
<p><span class="math-container">$$
dp = \gamma R T d\rho
$$</span></p>
<p>The speed of sound can be calculated by:</p>
<p><span class="math-container">$$
\gamma R T = a^2 \\
$$</span></p>
<p>where <span class="math-container">$a$</span> is the speed of sound, so:</p>
<p><span class="math-container">$$
dp = a^2 d\rho \\
$$</span></p>
<p>Substituting the expression above into the conservation of momentum equation gives:</p>
<p><span class="math-container">$$
(\rho V)dV = -a^2 d\rho \\
-\left(\frac{V^2}{a^2}\right)dV/V = d\rho/\rho \\
-M^2 dV/V = d\rho/\rho \\
$$</span></p>
<p>where <span class="math-container">$M$</span> is the Mach number. <em>This</em> gives a Mach number of 0.3 to be approximately a 5% change in density. </p>
<p>As a note, this is based on the Mach number, which in turn is dependent on the speed of sound in the gas, so it's automatically adjusted on a per-gas basis.</p>
| 7344 | Why is Mach 0.3 the threshold separating compressible and incompressible flow? |
2016-02-10T00:06:52.717 | <p>I get that you normalize stress to engineering stress, so that it is independent of the cross sectional area. This means you can calculate (roughly) the stress needed to elongate the specimen to the degree you want. But why normalize the strain aswell?</p>
<p>I would have thought that the orginal length of the specimen does not matter. Lets say you want to make the specimen 2 mm bigger, then I would have thought that you would need the same engineering stress no matter how long the specimen is. Thus that elongation is only dependant on cross sectional area and load.</p>
<p>For example you have a specimen, with orginal length, of 60 mm and 100 mm and then you want (l-lo) to be 2 mm. I would have thought that it would take same amount of engineering stress. And that a 2% strain would lead to different stresses needed, since in one this means an elongation of 1.2 mm and with the other an elongation of 2 mm.</p>
<p>Does this mean that the length of a specimen also determines how 'easily' a specimen elongates?
That it more difficult to elongate a 100 mm to 102 mm than a 60 to 62 mm. Because the tensile stress-strain curve implies that to elongate a 60 mm specimen to 61.2 takes same amount of stress as elongating a specimen to 62 mm.</p>
<p>If the length of the specimen relates to how easily you can elongate it, does this then also explain why the gauge length should be specified when giving the percent elongation (%EL) when talking about ductility? Because most of the plastic deformation is at the neck, so not uniform and thus independant of the length of the specimen orginally. </p>
<p>So before necking the elongation is dependant of orginal length and engineering stress and after necking is independant of orginal length?</p>
| |materials|metallurgy| | <p>From what I understand of the question, there are three interlinked concepts that are causing confusion, each of which could be a separate question.</p>
<h2>1. Is axial stress dependent on length of a specimen?</h2>
<p>Axial <strong>stress is independent</strong> of length. @Wasabi's answer goes into more detail, but the core concept is that a stress of $\sigma$ on a specimen of length $L$ has the same effect on a specimen <em>per unit length</em> as a stress of $\sigma$ on a specimen of length $2L$, or indeed of any length.</p>
<h2>2. Is axial strain dependent on length of a specimen?</h2>
<p>Axial <strong>strain is independent</strong> of length. This follows from the answer to (1.) above, and from Hooke's law, specifically that</p>
<p>$$
\sigma = E \varepsilon
$$</p>
<p>so that if $\sigma$ is independent of length, so must be $\varepsilon$ since they are linearly related.</p>
<h2>3. Is axial deformation dependent on length of a specimen?</h2>
<p>Axial <strong>deformation is dependent</strong> on length. It also depends on cross-sectional area (assuming constant cross-section) and tensile modulus. This follows from a quick and dirty derivation of the axial deformation formula. Consider a constant-cross-section tensile member of area $A$, length $L$, experiencing a uniform load $P$, with modulus $E$. The member will experience a deformation of $\delta$. Note that by definition $\varepsilon = \delta / L$ and $\sigma = P / A$.</p>
<p>From Hooke's law,</p>
<p>$$
\sigma = E \varepsilon
$$</p>
<p>and substituting our notes gives</p>
<p>$$
\frac{P}{A} = E \frac{\delta}{L} \\
$$</p>
<p>which after some rearranging yields</p>
<p>$$
\delta = \frac{PL}{AE}
$$</p>
<p>which is the equation for axial deformation of constant-cross-section tensile members. As should be obvious from the equation, axial deformation depends on length, as well as area.</p>
<h2>How Can I Think About This Intuitively?</h2>
<p>Elasticity is just like springs, and indeed Hooke's law is also used for linear springs. Now consider two springs such that one has twice the number of coils as the other, but the rest of their parameters are the same. Assume they are also weightless. If you held just the end of each spring, hooked identical weights to the other ends and let them uncoil toward the floor, the spring with twice the number of coils would extend twice as far. As a result, the change in length per per unit original length, equivalent to the strain, would be the same.</p>
<p>The same is true of two tension rods, one twice as long, experiencing the same stress with the same cross-sectional area. The one that is twice as long would have double the deformation, but they would have the same strain.</p>
<h2>How Can I Use This Information?</h2>
<p>As an example, consider a tensile rod. Suppose you must minimize axial deformation. Then you would choose a material with high tensile modulus so that stress has less effect on strain, and thus less effect on deformation.</p>
<p>Now suppose instead you have a family of tensile rods all made of the same material. If each must not deform more than 1 mm, then the longer rods must have a larger cross-sectional area to make up for the additional "slack" from their added length.</p>
<h2>What About Plastic Deformation?</h2>
<p>The above three questions only apply in the elastic regime. In the plastic regime, volume is practically conserved in most materials, which means that any instantaneous extension of a tensile specimen must cause an instantaneous reduction of area such that the volume gained by increasing length is equal to the volume lost by reducing the cross-section area.</p>
<p>However, there is an instability inherent to tensile plastic deformation called necking, which you noted. Namely, if one length-wise segment of a specimen is slightly narrower than the rest of the specimen, stresses will be concentrated at the narrow region and it will reach the elastic limit slightly before the rest of the specimen. Plastic deformation will thus begin at the slightly narrower section, causing a greater decrease in area than the rest of the specimen, further concentrating stresses, which causes further and greater deformation. The specimen will tend to fail wherever it is narrowest, and will do so unstably in a smooth, constant-cross-section specimen. That is, it is impossible to predict where failure will occur without intentionally notching the specimen.</p>
<p>Beware! Necking does not necessarily proceed the same way in polymers. In polymers, necking is associated with polymer chain alignment, which locally strengthens the material so much it prevents further deformation there, unlike metals. Instead, the neighboring regions begin to neck causing them to strengthen, and so on, until the entire specimen becomes aligned.</p>
| 7354 | Why is strain also normalized to the parameters of engineering strain? |
2016-02-10T14:46:07.643 | <p>When calculating the reactions of a simply supported beam under any type of loading we assume that the moment about the reaction points is <strong>zero</strong>. What if our assumption is wrong and in actual practice there is a rotational effect about that point? Why do we make such inappropriate assumption?</p>
| |mechanical-engineering|beam|statics| | <p>This is something of a nonsensical question, as zero moment is not an assumption here but part of the <strong>definition</strong> of this class of beam problem. A "simply supported" beam is any beam that can be represented as supported at one end by a hinge/pin and at the other by a roller (or, sometimes another hinge/pin). These are, again, <em>classes</em> of supports that <em>represent</em> real-world objects; in particular, they represent real-world objects that are not capable of transferring a moment to the member they support. <strong>By definition</strong>.</p>
<p>"Rotational effect" is not a meaningful term. The requirement that moment is zero means that the ends of the beam are free to rotate. However, if you meant <em>what if there is a nonzero moment</em> transferred at the support(s) of an actual beam, the answer is simply that the type of analysis used for the "simply supported" class of problem would no longer apply.</p>
<p>If you chose to represent a real beam as simply supported in your analysis and then found that there was actually a moment being transferred at one or both ends, the only bad assumption would be that this real beam could accurately be represented as a simply supported beam. You would have to classify it as something else. Not every real-world beam can be adequately analyzed using some common framework like "simply supported" or "fixed-fixed" or "cantilever"—some problems are more complex, and don't have a name. Some, we don't have an analytical solution for at all.</p>
| 7364 | Why do we assume the moment is zero when analyzing a simply supported beam? |
2016-02-10T16:28:45.660 | <p>In the diagram below, the compression member is $AC$. What is its unbraced length for determining buckling capacity? This also matters because certain codes put limits on the maximum value of $\frac{KL}{r}$ for bracing members.</p>
<p><a href="https://i.stack.imgur.com/ntDtA.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ntDtA.png" alt="enter image description here"></a></p>
<p>I know that the conservative answer is to consider the entire length, AC. This can be too conservative when lots of braces are used (total weight of steel increases). It can also be conservative if the actual compressive force is low enough that a member size is increased solely to meet the $\frac{KL}{r}$ requirements.</p>
<p>I can justify to myself that the tension member braces the compression member in one direction (in the plane of the page), but can the tension member be considered to provide bracing in the orthogonal direction (in and out of the page)?</p>
<p>There are two reasons why I think this might be possible:</p>
<ol>
<li>The tension brace provides resistance solely from being present. (i.e. something is better than nothing)</li>
<li>The other brace is in tension so it will provide additional restraint similar to a bow string.</li>
</ol>
<p>Note: Some codes have requirements that a brace be able to withstand 5% of the axial force in the member being braced. I would think that this would be the lower limit for restraint coming from the tension member.</p>
| |structural-engineering| | <p>The second edition of the AISC Sesimic Design Manual (SDM) contains many wonderfully worked out example problems. I highly recommend that you purchase this resource if you are designing braced frames in a professional setting. It can be purchased from the AISC, and if you are ever planning to sit and take the SE exam, you'll need to buy it anyway.</p>
<p>Example 5.2.1 in the SDM is the complete design of the diagonal brace in an ordinary concentric moment frame. In this example, while discussing the effective length of the brace, it states:</p>
<blockquote>
<p>By inspection the laterally braced length of the diagonal brace in the in-plane direction is half the overall length. For buckling out-of-plane, <strong>if both of the diagonals are continuous for their full length and are connected at the intersection point, then the effective length factor, K, is 0.5</strong> (El-Tayem and Goel, 1986; Picard and Beaulieu, 1987).</p>
</blockquote>
<p>Emphasis mine.</p>
<p>This results in an effective length for <em>both</em> the in-plane and out-of-plane bending of the brace to be 0.5 * L.</p>
| 7368 | What is the unbraced length of the compression portion of an X brace? |
2016-02-10T19:07:51.610 | <p>I'm trying to design a small metal telescopic track (6cm x 2cm when extended) where small objects may be arbitrarily positioned inside:</p>
<p><a href="https://i.stack.imgur.com/GNTMw.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/GNTMw.png" alt="enter image description here"></a></p>
<p>In the above example, the green block jams the tracks as the bottom cannot move upward. Does anyone know of a better design such that the tracks won't be jammed? I've thought about putting a sleeve on the inside, but if it was 6cm long and covers the entire extended length, it'd overhang when retracted.</p>
<p>Any help will be appreciated.</p>
| |mechanical-engineering| | <p>If you beveled the edges of the end of the inner track so that they form a ramp rather than a square shoulder that would help a lot. </p>
<p>Alternatively, if the two tracks are not a very close fit and the objects in question could jam in the gap some sort of flexible baffle may help. In both cases the idea is to create a smooth ramp transition between the two rail elements so there are no sharp corners for things to snag on. </p>
<p>Similarly having non-parallel sides (some approximation of a 'V' section) is also likely to help.</p>
| 7370 | Avoiding jams in telescopic tracks |
2016-02-11T03:05:28.320 | <p>I am a draftsman. I make project of mechanical pieces and machinery. Some pieces works like knifes or sleeve, so to reduce the friction we design the ideal material to manufacture this pieces. The steel Vc 130 or 131 are options of choices. What is the usage of steel grades VC-130 and VC-131? What the difference in applications between them, on tools and gears projects?</p>
| |mechanical-engineering|steel| | <p>VC-130 and VC-131 are names from the Brazilian standard. A comparison can be found at <a href="http://www.joinville.ifsc.edu.br/%7Epaulosergio/Ciencia_dos_Materiais/Classifica%C3%A7%C3%A3o%20dos%20a%C3%A7os.pdf" rel="nofollow noreferrer">Paulo Sergio's website</a>. Table 4 suggests that VC-130 is equivalent to AISI-D3 and VC-131 is equivalent to AISI-D6. However, these stainless steels are low sulfur (see the <a href="http://www.villaresmetals.com.br/pt/Produtos/Acos-Ferramenta/Trabalho-a-frio/VC131" rel="nofollow noreferrer">Villares VC-131 page</a>). I think the "Tool & Die Steels" site has the composition wrong.</p>
<p>The main difference between the two steels is the Tungsten content. Tungsten tends to reduce pitting corrosion (see the <a href="http://www.outokumpu.com/en/products-properties/more-stainless/the-effects-of-alloying-elements%E2%80%8B/Pages/default.aspx" rel="nofollow noreferrer">Outokompu page</a>).</p>
| 7373 | Difference between VC-130 and VC-131 steel |
2016-02-11T06:28:32.960 | <p>I need to import an assembly from CATIA into SimMechanics but in CATIA, the .stl file format is available only for part files and not for assembly files. How can I do this? Is it possible to convert between the .CATProduct and .stl file formats?</p>
| |matlab| | <p><strong>Short answer</strong>: you can't directly import CATIA assemblies in SimMechanics.</p>
<p><strong>Long answer</strong>: SimMechanics supports import of CAD models (assemblies or parts) from Pro/Engineer, SolidWorks or Autodesk Inventor via SimMechanics Link (see <a href="http://uk.mathworks.com/products/simmechanics/features.html#importing-cad-models" rel="nofollow">Importing CAD Models</a> on the MathWorks website). The <a href="http://uk.mathworks.com/products/simmechanics/download_smxmlschema.html" rel="nofollow">SimMechanics Import XML Schema</a> should allow import from any CAD system, but you need to be able to export to XML from your CAD system, and the XML must follow the SimMechanics Import XML Schema. For you, that probably means having to write a custom interface layer in CATIA to export to the correct XML file schema.</p>
| 7374 | How to import a CATIA assembly into SimMechanics |
2016-02-11T06:38:11.087 | <p>What is the down-stand beam called at an edge of a surface bed or raft foundation? It is not exactly a grade beam or stem wall. It is more like a toe.</p>
<p><a href="https://i.stack.imgur.com/DtqXP.png" rel="noreferrer"><img src="https://i.stack.imgur.com/DtqXP.png" alt="enter image description here"></a></p>
| |structural-engineering|foundations| | <p>It is an "edge beam", or "grade beam", in a raft foundation, and a "downturn" of a slab-on-grade. The former is a structural component, usually spanning between columns, thus more massive and heavily reinforced. The latter is a strip footing or purposed to stiffen the slab edge and prevent outward migration of the subgrade.</p>
<p><a href="https://i.stack.imgur.com/iA1rV.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/iA1rV.png" alt="enter image description here" /></a></p>
| 7375 | What is the down-stand beam called at an edge of a surface bed or raft foundation? |
2016-02-11T20:53:05.213 | <p>Consider two streams of air flowing through the nozzles to an unlimited volume with atmospheric pressure.</p>
<p>The flow is the same, but the cross section areas of the circular nozzles and the speeds differ, so the flow through the smaller nozzle is proportionally faster.</p>
<p>Intuitively, the stream which is faster at the nozzle would transfer the air farther.</p>
<p>Is there any empirical rule which can be used to measure air stream dissipation rates given the initial speed and nozzle cross section area?</p>
<p>I am aware of Lattice-Bolzmann numerical methods, and can run the experiments too, but as i am new to air dynamics, may be I am overlooking something obvious.</p>
| |airflow| | <p>In a gas flow into an open volume there will be a fairly well defined point (at some distance form the nozzle) at which the flow transitions from laminar to turbulent and this is usually where most of the energy dissipation occurs. </p>
<p>This can be seem in the real world in the smoke pattern from a recently extinguished candle wick. </p>
<p>The most common formal definition of this is the <a href="https://en.wikipedia.org/wiki/Reynolds_number" rel="nofollow">Reynolds Number</a> which gives a reasonable approximation of whether flow conditions are dominated by viscous or inertial forces. </p>
| 7387 | Air flow dissipation rate |
2016-02-12T04:03:54.420 | <p>I am considering a two wheeler petrol engine. How is the mass of air entering the intake manifold calculated? </p>
<p>Given that I have:</p>
<ul>
<li>Manifold Absolute Pressure sensor</li>
<li>Intake Air Tmperature sensor</li>
<li>Engine Temperature sensor</li>
<li>Throttle Position sensor</li>
<li>Engine rpm sensor </li>
</ul>
<p>Please help me understand how to estimate the amount of air entering the engine for combustion.</p>
| |mechanical-engineering|electrical-engineering| | <p>An IC engine is (for these purposes) a pump. It moves a certain volume of gas through on each cycle (one revolution for a 2-stroke, two for a 4-stroke). That amount is reported as the displacement of the engine (110 cc) in your case. Then you need to determine the mass of that air in that volume.You can arrive at that using the ideal gas law and the molecular weight of the air.</p>
<p>$$ PV=nRT $$</p>
<p>Values in order: Pressure($Pa$), Volume (displacement in your case$m^{3}$), number of moles, the ideal Gas constant (8.314 $J\cdot K^{-1} \cdot mol^{-1}$) and temperature (of the air $K$). Solve for $n$ and multiply by the molecular weight (~.0288 $kg \cdot mol^{-1}$ for air). That'll be the mass of air.</p>
<p>This is method ignores the volume taken up by the fuel itself, pressure loss between the intake manifold and cylinder etc.. It's a pretty good estimate, but won't be accurate enough to make the engine run acceptably well. You'll need to apply an empirical coefficient to tune it to the right value to control your fuel injection. That coefficient will not be constant over all RPM and other operating conditions. The good news is that the coefficient will be close to unity and won't change (significantly) over the lifetime of the engine.</p>
<p>That tuning process can be involved. Ideally, you'd have an air-fuel ($\lambda$ sensor on the exhaust to measure how far from stoich the mix is. Tuning with an exhaust gas temp sensor is also possible (never had to do that myself), and may be cheaper. The last option would be to tune the engine on a dynamometer (measures power), although tuning to produce maximum power will usually produce a slightly fuel rich mixture. The $\lambda$ sensors are the best way to go and can be integrated into your controller to make the system more robust, but you could borrow one temporarily to tune and run the engine without it.</p>
<p>Notes:</p>
<ul>
<li><p>I've specified everything in base SI units. Other units will work too, but make sure that you get the value of $R$ right and use an absolute temperature (Kelvin or Rankine, not Celsius or Fahrenheit)</p></li>
<li><p>Engine temp and TP are not strictly necessary for the calculation, but TP is often used to tweak fuel delivery during changes in speed/TP.</p></li>
</ul>
| 7388 | How is the mass of air in the intake manifold calculated in an electronically fuel injected vehicle? |
2016-02-12T06:25:46.467 | <p>Similarly, does there exist an index with the maximum building depths and heights in the United States?</p>
<p>I found this map of soil types in California by UC Davis, but I don't think this directly translates into maximal building depths: <a href="http://casoilresource.lawr.ucdavis.edu/gmap/" rel="nofollow">http://casoilresource.lawr.ucdavis.edu/gmap/</a></p>
| |civil-engineering|structural-engineering| | <p>The deepest human made excavation is the <a href="https://en.wikipedia.org/wiki/TauTona_Mine" rel="noreferrer">Tau Tona Mine</a> in South Africa, which is 3900 metres deep. This is accessed by two vertical shafts, each 2000 metres long. At the bottom of the first shaft there is a short tunnel, at the end of which is the top of the second shaft.</p>
<p>The depth limit on the shafts is due to the capabilities of the winding ropes: due to the weight of the ropes, the strength of the steel used, the unavoidable manufacturing flaws in the ropes and the fatigue induced in the ropes over time as the ropes are wound and unwound from the drum.</p>
<p>If we want to dig an excavation for a building we will need to get the unwanted material out of the excavation so we will need a winder and a kibble (the bucket used to hoist the material out of the hole). With current technology, this limit is 2000 metres.</p>
<p>Other important factors which limit the depth to which we can dig and construct anything underground are:</p>
<ul>
<li>Ground stresses - if they're too large the excavation for your construction will collapse or be squeezed so its cross-sectional area is reduced.</li>
<li>Geothermal heat - if the temperature is too hot, people can't work in the excavation for the construction</li>
<li>Competency of the material being dug - the strength of the material being dug will determine the maximum dimension of the walls of the excavation and of the overlying roof, if applicable.</li>
<li>The material being dug - dig in rock and you can go down to 2000
metres; dig in sand and you won't get too far, but if you freeze the sand so it can be dug like rock you can dig deeper into it.</li>
</ul>
<p>Other factors include:</p>
<ul>
<li>The proximity of geological structures, such as faults and shears;
how active such structures are and the extent of the fractured zone
of rock surrounding the structure.</li>
<li>The availability of engineering construction materials such as steel,
concrete, reinforcing bars.</li>
<li>How the walls of the excavation will be supported during and after
construction.</li>
</ul>
| 7389 | What practical geographical and engineering constraints limit the depth to which you can build underground? |
2016-02-12T06:39:03.733 | <p>I have been investigating pros and cons USB Type-C. Obviously their are many advantages in innovation, so are the disadvantages. On such disadvantage is <a href="http://money.cnn.com/2016/02/05/technology/usb-c-cords/" rel="nofollow noreferrer">"Those new USB-C cords can fry your laptop"</a>.</p>
<p><a href="https://i.stack.imgur.com/05CeZ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/05CeZ.png" alt="enter image description here"></a></p>
<p>Advantages:</p>
<ul>
<li>Good bye to proprietary laptop power brick.</li>
<li>Support both USB 2.0 and USB 3.0</li>
<li>Higher data rate of 5 Gbps with USB 3.0</li>
</ul>
<p>Disadvantages:</p>
<ul>
<li>Poor design could fry your expensive laptop.</li>
<li>Shorter cable length. USB 3.0 cable length is 10 feet, where is USB 2.0 is 16 feet.</li>
</ul>
<p><strong>Question</strong>: What are other disadvantages that need to be considered when recommending USB Type-C?</p>
<hr>
<p><strong>References:</strong> </p>
<ul>
<li><a href="https://en.wikipedia.org/wiki/USB#USB_Type-C" rel="nofollow noreferrer">USB</a></li>
<li><a href="https://en.wikipedia.org/wiki/USB_Type-C" rel="nofollow noreferrer">USB Type-C</a></li>
<li><a href="https://giuseppemelle.wordpress.com/2010/06/06/usb-3-0-the-pros-and-cons/" rel="nofollow noreferrer">USB 3.0 – The Pros and Cons</a></li>
<li><a href="http://www.laptopmag.com/articles/usb-type-c-faq" rel="nofollow noreferrer">USB Type-C FAQ: Everything You Need to Know</a></li>
<li><a href="http://www.usb.org/developers/presentations/pres1108/SuperSpeed_USB_DevCon_Cable_and_Connnector_Ling.pdf" rel="nofollow noreferrer">SuperSpeed USB Developers Conference - USB.org</a></li>
<li><a href="http://www.chassis-plans.com/should-usb-3-0-be-used-instead-of-usb-2-0-in-your-next-computer/" rel="nofollow noreferrer">Should USB 3.0 Be Used Instead of USB 2.0 in Your Next Computer?</a></li>
</ul>
| |electrical-engineering| | <p>No -- any cable built <strong>wrong</strong> can fry your laptop. The kerfluffle in the blogosphere is about off-brand cables which were incorrectly wired. The whole point of USB-C is that the connector itself is orientation-agnostic and it (the <em>connector</em>) cannot be inserted in a way that sends power to the wrong lead. Only a miswired cable can do that.</p>
| 7390 | What are the documented pros and cons of USB Type-C? |
2016-02-12T15:18:07.183 | <p>I have heard a couple of times that an operating nuclear power plant which was shut down (non-emergency; e.g. for a regular check) needs over 24 hours (up to 72 hours?) to get up running again.</p>
<p>Why does it take that long?</p>
| |power|nuclear-technology|electrical-grid| | <p>They are many reasons for the time it takes to startup or return to full power operation in commercial nuclear power plants. In the US there are two main types of plants, Boiling Water Reactors(BWRs) and Pressurized Water Reactors (PWRs). Answers will differ according to the type of reactor and even which version of the type. A common explanation that I did not see mentioned is that all commercial nuclear power plants avoid making >15% thermal power changes in any 4 hour period. This is to protect the integrity of the fuel cladding. I worked in the commercial nuclear power industry for almost 20 years - and have been out of it for over 20 years -so maybe they have improved the fuel cladding and this is no longer an issue - but it was a mandatory constraint in my day.</p>
| 7394 | Why does it take so long to restart a nuclear power plant? |
2016-02-13T00:19:47.997 | <p>I have an abrasive cutoff saw that I use to cut metal, it runs at 3800RPM. You cannot cut aluminum with these saws, so I bought a carbide blade rated for 3800RPM. </p>
<p>I went to install it, but after looking in the manual I don't think it is such a good idea. It says "No toothed blades". A regular saw runs around 1200RPM. </p>
<p>Why can't I put toothed blades in my cutoff saw? The blade is rated for that speed, would there be a problem?</p>
| |machining|tools| | <p>Imaging a strong hard blade snapping a chunk off on the 27th cut and bounces off the steel into your hand, arm, guy, teeth or eye. That will go right through eye protection and any leather. The point load at some portion of the blade was just we'll over 50,000psi to snap the steel and now it's all embedded in the chunk heading for you tendons and arteries. Us old guys can tell you stories of set ups like that killing and mutilating people for hours. The classic is when a buddy says look at this and starts pulling an x-ray out of a manilla envelope. We cringe and have to ask right away "Ring Shank?" 80percent of the time the answer is "Yep". The reason ER doctors had to start buying titanium pry bars and pliers is the improvement in construction steel. I was at a hospital Christmas party and walked up to the ER doc with a Guinness and said this is your beer. He looked at the two full beers I had and his almost empty one in his hand and said "this is my beer". "Nope this is your next beer because you glued face back together 2 months ago that glued my face back together. Everyone complemented his work since no on could see the scars. Then I had a life changing moment happen to me. The hospital was Denver Health that created the "level 1" triage system and the flight for life system for the world and is the best ER school in the world. 4 ER doctor relaxed anfer a few drinks smiled at me and one said what happened and when I said construction accident they all flinched and look away from me. A lot of groans and muttering "construction is the worse damage we see, and motorcycle accidents" turns out motorcycle get their head spun around and a massive amount of bone breaks. That's treatable if you survive. Construction accident a have no standard injury. It's just mutilated flesh of some random sort. The top ER doctors in the world wear uncomfortable with me after that and I could see the images flashing in their minds when I said anything so I excused myself. That pushed home just how dangerous it is and safety was easy after that. Been stitched up 3 times in the last year and had an ER trip from a pinhole inclusion on my foot. I knew it was bad and was on the phone giving the ER direction on treatments to prep as I drove toward the ER. I was getting prepped for surgery when a motorcycle accidents came in and bu.ped me. I was on IV antibiotics as soon as I got there and 4 hours after the accident I already had a fever over a 100 and rising. Literally a 1 millimeter round hole at the edge the steel toe in my boot. I wouldn't have a half my foot if I wasn't wearing steel toes. But I would want to trade places with the motorcycle rider that bumped me. (It was a commercial pressure washer, water 'disects tissue" and runs random through tissue. Pinhole in the top of my foot and it went forward and split 3 ways, around the side of my foot and between three toes, curved back toward my heel on the bottom of my foot spraying out farther back than it went in on the top. Over two months until I could walk and work.) Be safe. No amount of construction money can overcome a mutilated body when you want to hook up. But I still recommend doing practice prep with each partner explaining what to do when you say we have to go to the ER, NOW.</p>
| 7406 | What prevents using a toothed blade with a cutoff saw designed for abrasive discs? |
2016-02-13T23:35:37.203 | <p>I was looking at hydrogen as alternative fuel and came across <a href="http://www.autoblog.com/2009/08/20/greenlings-why-choose-a-fuel-cell-or-an-internal-combustion-eng/" rel="nofollow">an article</a> stating that although $\textrm{H}_2$ produces more energy than conventional fuel, it produces less horsepower in internal combustion engines. </p>
<p>I thought that more energy ($143\ \textrm{MJ/kg}$ for $\textrm{H}_2$ vs $46.4\ \textrm{MJ/kg}$ for gasoline) would mean more heat, more temperature, more pressure and thus more force and horsepower. Why, then, does $\textrm{H}_2$ produce less horsepower than gasoline in internal combustion engines?</p>
| |mechanical-engineering|combustion| | <p>One consideration is the relative densities of hydrocarbon fuels compared to hydrogen. If we assume that the fuels are completely vapourised in the combustion chamber then the mass of fuel depends on the volume, temperature and pressure. So energy density per kg doesn't necessarily tell you much about how many molecules of fuel you can get in one charge or indeed the chemical energy available per molecule. </p>
<p>There are also issues in IC engines relating to the combustion properties of hydrogen. One particular problem is that hydrogen will detonate at a broad range of air/fuel ratios which has knock on effects for ignition timing and compression ratios which in turn affect how well the chemical energy available from combustion can be converted into useful mechanical power. Premature detonation under compression (especially in hot cylinders) is a particular problem with hydrogen. </p>
<p>Hydrogen is often associated with low compression ratio engines like Wankel (and other rotary) engines for exactly this reason. </p>
<p>Hydrogen also burns hot compared to other fuel gases, so even when overall energies are comparable hydrogen engines can produce localised hot spots in the combustion chamber. </p>
<p>Gasoline is a mixture of fairly short chain hydrocarbons which can be approximated to C8 but the specific mixture of different molecules has a significant effect on its combustion behavior and there is quite a lot of scope for adjusting a particular fuel formulation to get the best performance. This is much more difficult with hydrogen. </p>
| 7425 | Why does hydrogen fuel produce less horsepower than gasoline in internal combustion engines? |
2016-02-14T09:14:21.797 | <p>There seems to be no noiseless thermic motor for cars or motorcycles on the market.</p>
<p>Is there a technical obstacle at making such a product ?</p>
<p>Is there a way to isolate/eliminate noises from the explosion process and from the moving parts ?</p>
| |motors| | <p>Considering the nature of what an explosion is, I doubt we can just suppress the sound at it's source. The only way is to use a different motor technology.</p>
<p>However, car manufacturers tend to isolate the motor compartment from the outside with basic noise-suppressing technologies which are for example : covering the compartment with absorbing materials like foams or using bushed fixations (<a href="https://en.wikipedia.org/wiki/Bushing_%28isolator%29" rel="nofollow noreferrer">here</a>). Most highly priced cars are well isolated, for example I know that you barely hear the sound of the motor in recent BMW cars.</p>
<p>Here is a picture of some bmw compartment. You can see that when the cover is closed, a plastic joint isolates the cover from the chassis.
<a href="https://i.stack.imgur.com/bFyPA.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/bFyPA.jpg" alt="enter image description here"></a></p>
| 7429 | Technical obstacles in making a noiseless thermic motor? |
2016-02-15T09:32:36.263 | <p>An example is <a href="http://www.omega.com/pptst/PX409-485_PXM409-485.html" rel="nofollow">this pressure transmitter</a> which has a BSL accuracy of 0.08%. I know the term "Best fit straight line" - BFSL. Is BSL just a short hand for this ("Best straight line")?</p>
| |sensors|metrology| | <p>BFSL (Best Fit Straight Line) is the same as BSL (Best Straight Line). It is calculated from the calibration's extracted data, where the manufacturer test the sensor for several known input values (in this case, pressures). Once these calibration points are measured, they are approximated with a straight line (least-square method) and the BSL value is an error value that indicates how close the measured points are to this straight line.</p>
| 7441 | What does "BSL Accuracy" mean? |
2016-02-16T00:33:18.647 | <p>I'm unsure if this is the correct community, so feel free to move it if you feel it's in the wrong area.</p>
<p>I'm trying to develop a water-tight/sealed box for a backup flash drive. The idea is to make something that is weatherproof enough so I can keep the box outside or in my basement for years, and if I ever need to retrieve the data, I can cut the box open and retrieve the USB stick. This would need to completely resist water/moisture and withstand temperature changes between say -40 C to 40 C without it breaking down.</p>
<p>Here is the concept I have so far: </p>
<ul>
<li><p>First layer: Put USB stick into air-tight plastic container. Could be something cheap like a sealable zip-lock bag to protect from:</p></li>
<li><p>Second layer: fill a small box - not much larger than the stick - with a polymer sealant/caulk that will eventually solidify, fitting the form of the box.</p></li>
</ul>
<p>My question is, what is the best material can I use for layer #2?</p>
| |materials|seals| | <p>I'd like to mention something else that will improve the retention of your data. If you want your data to be really safe - make a few copies. To be safe even if parts of the data are corrupted.</p>
<p>If you have multiple physical copies of your data (say: 3 memory cards holding the same files) you can use a tool like <a href="https://www.gnu.org/software/ddrescue/" rel="nofollow noreferrer">ddrescue</a> to heal problems in one copy with the other one.</p>
<p>This way even if some of the data gets corrupted - you have a very high chance of recovering it all as random errors will most likely not corrupt the same bytes in all of your copies. The more copies you have - the more recoverable your data is.</p>
<p>If you still have some holes in your ddrecovered LZIP archive - you can use <code>lziprecover</code> to fix any residual errors.</p>
<p>Another approach could be to create a ZFS pool on multiple USB drives or memory cards and set them to work in mirror mode. ZFS is a filesystem/volume manager designed to prevent data loss used in data centers and NAS deployments around the world.</p>
<p>You'd need to use Linux or Solaris for that though.</p>
| 7451 | Best way to make a sealed container for a long term storage of a USB drive? |
2016-02-16T00:38:33.303 | <p>I'd like to make my own structure that will hold roughly 50-100lbs of equipment. Since I need this to be portable, I feel like aluminum tubing would be the most practical. </p>
<p>My structure will basically be a 2.5'-10' cube. I want to get 2.5' tube sections that I can connect very sleekly (word?) and make a cube that is 2.5' at its smallerst and 10' at its largest.</p>
<p>Taking a cue from a photography backdrop, they have 1" diameter tubing that sort of fits into itself. The tube is not flared, but it's pinched so it's the same size as the inner diameter. </p>
<p>Is there a name for this "pinch" I'm talking about? And if so, is there a tool I can buy to do this? </p>
<p><img src="https://i.stack.imgur.com/Ww71m.jpg" alt="enter image description here"></p>
| |aluminum| | <p>For a one-off structure swaging or flaring tube is probably a bit of a long way round to do it. There are actually quite a few off the shelf solutions for joining tube, <a href="http://www.rosscastors.co.uk/tube-inserts-fittings/connect-a-tube.html" rel="nofollow">an example at random</a></p>
<p>Something else to consider is that, for something which is disassembled and moved about a lot, if your system relies on one tube fitting inside another then and damage to the ends of the tube will mean that they no longer fit properly and the method of fit will have significant effect on the stiffness of the whole structure. With this in mind any system which either puts a permanent rigid insert in the end of the tube or clamps on the outside is likely to be preferable in terms of both durability and stiffness. </p>
<p>It is also worth saying that any operation which involves ductile forming of metals tends to be a bit fiddly as the parameters depend a lot on the <em>precise</em> material properties of the raw materials which can vary a bit from batch to batch. </p>
| 7452 | What is the opposite of flaring a tube (pinching?) and is there a tool to do this with aluminum? |
2016-02-16T08:47:22.857 | <p>I recently got a gizmo installed at home that samples my electricity usage once every 10 seconds. I've become a little transfixed watching the graph update as various appliances in my home turn on and off.</p>
<p>For instance, turning on lights results in a small and simple bump up a couple of hundred watts. Turning on the microwave a huge 1.2KW jump. The washing machine usage moves up and down depending on where it is in the cycle.</p>
<p>I started wondering whether there'd be some ways to automatically work out what devices are running by what you might call their 'power signature'. </p>
<p>Essentially what I want is an algorithm that watches what is effectively a 0.1Hz signal representing usage in watts, and can say 'Looks like your washing machine was on between 09:00 and 11:00 today.)</p>
<p>The complication of course is that multiple appliances can be running. As a human you can sometimes resolve that (E.g. This shape looks like the washing machine was on, but someone also turned on the microwave in the middle).</p>
| |power|energy-efficiency|signal-processing| | <p>The word that will unlock your searching is <strong>disaggregation</strong> - there's quite a lot of research going on right now into energy disaggregation.</p>
<p>These generally involve some kind of learning algorithm, and the energy-disaggregation field is quite young, so you won't find a reliable off-the-shelf package.</p>
<p>But you will find current research which will help you design your own.</p>
<p>Daniel A Kelly (aka Jack Kelly, and currently of Imperial College London) has worked in this field - check out his papers; and Clym Stock-Williams of Uniper (formerly part of E.On) recently reported at a Data Science for Energy conference organised by the Alan Turing Institute (Edinburgh, Jan 2016) that data at 0.1 Hz was as good as data at 1 Hz for disaggregation - the more frequent 1-second data did not lead to a more skillful model (YMMV).</p>
<p>In short, do a literature search (Google Scholar or similar) for energy disaggregation or electricity disaggregation, and you'll find new papers every month or so, in this fast-evolving field.</p>
<p>But for a single dwelling, where you know what's going on at any one moment, the best way to do this would be to train your own model, as you'll be able to get very good data on what is being used, when: even as far as listening for when the fridge and freezer compressors turn on and off.</p>
| 7457 | Are there any methods to work out what appliances are running in my home from an aggregate power consumption sampled 6 times a minute? |
2016-02-18T15:54:37.907 | <p>I have made Image which is self explanatory. It has the question</p>
<p><a href="https://i.stack.imgur.com/KzIPd.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/KzIPd.png" alt="enter image description here" /></a></p>
<p>Suppose I have a metal plate that is travelling at V m/s.<br />
A stationary body acts as a shearing body. When the metal plate contacts the still body, it starts to shear off. The plane in which it will be sheared is shown in diagram (red line). So I want to ask how much force will get applied for the shearing.</p>
<p>This is just an idea that I'm working through. I haven't seen this in any book that I have read (I haven't yet seen various books yet). This problem came in my mind yesterday while I was dreaming.</p>
<p>Also I am assuming that the acceleration of the moving plate will be equal to change in velocity (after shearing the plate stops), so a = (0 - V) / 1 (unit time)</p>
<p>So question is</p>
<blockquote>
<p>how much mass should be considered for the equation F = ma ?</p>
<p>basically I want to calculate <strong>force required</strong>.</p>
</blockquote>
<p>If am wrong in some calculation then correct me.</p>
| |design| | <p>It seems like what you're really after is the force, not the mass. Now, if you're assuming that the plate will shear off, then the easiest way of calculating the force isn't through Newton's laws, but by calculating the force necessary to shear off the plate.</p>
<p>Simplistically, that can be found by</p>
<p>$$ F = A\tau $$
where $A$ is the transversal cross-section of the plate along the shearing plane (assuming thin-walled theory applies here) and $\tau$ is the shear strength of the chosen material.</p>
<p>This doesn't consider dynamic effects (buckling, warping, possibility of the plate simply bouncing off instead of shearing), but that's impossible to do with the information presented.</p>
| 7499 | shear force calculation with velocity and mass |
2016-02-18T16:43:51.280 | <p>I'm following <a href="http://openmicros.org/index.php/articles/94-ciseco-product-documentation/raspberry-pi/217-getting-started-with-raspberry-pi-gpio-and-python" rel="nofollow">this tutorial</a> on how to control Raspberry Pi GPIO pins with Python scripts. Instead of connecting the test circuit (LED + resistor) between the ground and the GPIO pin however, they connect it between the +3v pin and GPIO (see section 4, last paragraph). The author says this is to have the current come from the 3v pin rather than the GPIO pin. However, from my limited understanding of electricity, the magnitude of the current through the GPIO pin should be the same regardless of how the polarity has been set, the only difference would be its direction. </p>
<p>Am I wrong about the current? Is it actually better to draw power from the 3v pin instead of the GPIO pin?</p>
| |electrical-engineering|embedded-systems| | <p>Early TTL outputs (those typical of 74xx00 series logic gates) were capable of sinking (creating a current path to ground) more current than they could source (creating a current path to VCC). If you wanted a predictable current flow which you could set by your choice of external current limiting resistors then the output transistor path to ground did a better job.</p>
<p>As far as I know this is still true.</p>
| 7500 | What is the advantage of using GPIO pins as sinks instead of sources? |
2016-02-18T22:03:55.527 | <p>I am trying to decipher some as-built drawings for an old wood frame building. The purpose is to generate a preliminary load takeoff. I cannot figure out what the following is saying: <a href="https://i.stack.imgur.com/JRjSv.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/JRjSv.jpg" alt="enter image description here"></a></p>
<p>It looks like "3 Ply Built up roof " but the rest makes no sense. Any seasoned structural engineers can tell me what this says?</p>
| |structural-engineering|drafting|architecture| | <p>I am a practicing structural engineer with plenty of wood experience. That said, this is a bit of a tricky note because I believe it was probably written by an architect on a set of architectural drawings (as opposed to by a structural engineer on a set of structural drawings). Therefore, I think the note includes information about both the structural makeup of the roof diaphragm as well as the architectural components of the roofing system. That said, I can at least get you started. I am assuming that you are in the US, but if you're in another country, the abbreviations may be a bit different. I am less sure about the second line, but am confident about the meaning of the other 4 lines.</p>
<ol>
<li><p>The first line reads, <code>3 PLY BU ROOF</code></p>
<p>I agree that it is communicating something related to the (mostly) architectural composition of a built up roof. It likely means that the roof is a 3 ply <a href="http://construction.about.com/od/Finishes/a/Types-Of-Built-Up-Roofing-And-Its-Benefits.htm" rel="noreferrer">built up roof</a>. Depending on your intent here, if it's structural for example, this doesn't have too much importance to you.</p>
</li>
<li><p>The second line, <code>o/ 3/4" O.F. PLYWOOD INO.</code></p>
<p>This line is a description, of some sort, of the plywood. <code>o/</code> is, I think, an abbreviation for the word "over." So the objects on this line are under the 3 ply built up roof. <code>3/4"</code> is describing the plywood thickness, <code>OF</code> is an old way to describe "oriented fiber" material, this is similar to oriented strand board (OSB) that you see at the hardware store today. This is a little confusing, since plywood is not oriented fiber board. It could have another meaning. <code>OF</code> could also mean "outside face" and this may be an instruction about how to apply a coating related to the built up roof to the outside face of the plywood. Also, since the plywood is given a span rating (see next point), no plywood thickness note is required so the <code>3/4"</code> may not be referring to the plywood at all.</p>
</li>
<li><p>The third and fourth line, <code>32/16 (UNBLKD) W/ 8d @ 6" EN & RO</code></p>
<p>These lines are decribing the strucutral requirements of the plywood and associated fasteners.</p>
<ul>
<li><p><code>32/16</code> is a "span rating" for the plywood. From the <a href="http://www.apawood.org/apa-trademark" rel="noreferrer">APA</a>,</p>
<blockquote>
<p>Two numbers separated by a slash. The left-hand number is the maximum recommended center-to-center spacing for supports in inches when the panel is used for roof sheathing with long dimensions across supports. The right-hand number is the maximum center-to-center spacing of supports in inches when the panel is used for subflooring with the long dimension across supports.</p>
</blockquote>
</li>
<li><p><code>(UNBLKD)</code> is instructing that the edges of the plywood do not need to be <a href="https://en.wikipedia.org/wiki/Blocking_(construction)" rel="noreferrer">blocked</a>.</p>
</li>
<li><p><code>W/</code> is an abbreviation for the word "with"</p>
</li>
<li><p><code>8d @ 6"</code> is instructing the use of 8d nail at a spacing of 6" on center.</p>
</li>
<li><p><code>EN & RO</code> is instructing that the previously mentioned nailing is applied as <strong>E</strong>dge <strong>N</strong>ailing and at <strong>R</strong>ough <strong>O</strong>penings. This means that the 6" oc nailing is applicable at the endges of the roof and around any openings.</p>
</li>
</ul>
</li>
<li><p>The fifth line, <code>12" FN</code> is instructing a 12" on center "field nailing." This means that the plywood should be nailed to all the rafters/joists in areas that are not diaphragm edges at a 12 oc spacing.</p>
</li>
</ol>
<p>Good luck and please comment when you figure it out. I've passed your image around the office and it's generated quite the lively discussion!</p>
| 7502 | Wood roof specs on engineering drawing |
2016-02-19T12:31:09.313 | <p>I have to write a scientific paper where I refer to a CAD drawing. Now, I'm a bit confused, because I have to refer to the length "parameters" on that drawing (as opposed to the angle "parameters"), e.g. the segment lengths <em>l1</em>, <em>l2</em>, ... Should I refer to these length parameters as "length measurements" or as "length measures"? I used both as a synonym so far, but I'm not sure if that is accurate.</p>
| |mechanical-engineering|computer-aided-design| | <p>The conventional term would be a <em>linear dimension</em> if you just want to specify that your measurement is of a length. There are also <em>angular dimensions</em> for angles, and more complicated dimensional features like tolerances.</p>
<p>If you're discussing the CAD drawing in terms of the CAD program (not just to describe the object it shows) and you want to identify your parameters you would want to distinguish between a <em>driving dimension</em> and a <em>driven dimension</em>. A driving dimension is a length, angle or other number that you enter which is driving your model. A driven dimension is a length that you are showing in your drawing, but which is automatically determined based on the driving dimensions you provide.</p>
<p>It's worth noting that specific CAD packages have more complicated names to describe how a dimension is created (eg. Autocad distinguishes between linear and aligned dimensions to describe how it sets the axis of the dimension) but in the context of the final drawing, most dimensions still fit into these two categories.</p>
| 7510 | 'Length measure' vs. 'length measurement' |
2016-02-20T02:19:39.323 | <p>In my stem class we are going to have a competition where we have to build a toothpick bridge that can support as must weight as possible, while using only toothpicks and glue. These are the rules:</p>
<ul>
<li>Each toothpick costs \$10,000.</li>
<li>Each end of a toothpick glued to another end of a toothpick costs \$10,000.</li>
<li>The bridge must span at least 1 foot. </li>
<li>We can't build beams down to the floor for support.</li>
<li>The base of the bridge, the part cars would drive on if it were a real bridge, can't have any large gaps in it, although I'm not entirely how much this matters.</li>
<li>We will have a time limit when building it, but I don't know what it will be. We can bring a hairdryer to help the glue dry faster.</li>
<li>It must hold a lot of weight, and be cheap. I don't know how much each of those factors matter, but I'd imagine the weight it can hold is much more important.</li>
</ul>
<p>My teacher wants us to do lots of research before the project and have a clear plan, so I have been looking into what kind of bridge to build. It seems like the obvious choice is a beam bridge, but I could be wrong. I found <a href="http://www.warwickallen.com/bridges/BeamBridges.htm" rel="nofollow">this site</a> talking about how a truss can help a beam bridge hold more weight, but I don't quite understand how it helps.</p>
<p>If a beam bridge is my best option, would a truss significantly help at this scale? And if so, which kind of truss? The warren looks more efficient as far as costs are concerned, but I don't know if the pratt or any other kind has a major advantage. Any advice on this would be greatly appreciated, thanks!</p>
| |structural-engineering|civil-engineering|bridges| | <p>You didn't specify cost of glue. If the glue is of no cost (other than end to end joints as you mentioned) then my inclination is to think outside the box here. </p>
<p>You could shred the toothpicks into fibers (pull them apart, grind them up, whatever) and then use the glue to create a composite matrix using the fibers. </p>
<p>You could get a lot of stringy fibers from very few toothpicks. </p>
<p>I don't know all the rules, but If you can bring stuff in other than just a hairdryer, then I would bring in a food saver type vacuum bagger and some 10 or 20 lbs foam, and some Vaseline. </p>
<p>Carve a negative form for the bridge into the foam and coat it with vaseline (as a mold release agent). Mix up the wood fibers and glue and spread this mixture out into the form. Slip the whole thing into a food saver bag and turn on the vacuum pump to apply pressure while it cures. </p>
<p>Use the hair dryer to apply heat over the whole thing to help cure while it's in the bag. </p>
<p>Finally, once your WFRG (wood-fiber-reinforced-glue) bridge is cured you would remove the whole thing from the bag, and pop the bridge out of the mold. Done. </p>
<hr>
<p>I didn't see time limit listed in your question, but I would imagine all of this would be <em>way</em> faster than building a more traditional bridge gluing individual toothpicks together, so I assume the time allotment would be sufficient. </p>
<p>Again, depending on the rules about bringing stuff in, you might be able to make the mold ahead of time and bring it with you, so all you have to do during the construction phase is shred wood, mix with glue, apply, and cure. </p>
<p>Something I learned at an early age is it's better to ask forgiveness than ask permission. If the rules don't specifically ban you from bringing in this outside stuff then I would say it's fair game. </p>
<hr>
<p>I believe this fiber-reinforced-matrix construction could work well. However I've never done it myself, so I'm not positive. It's similar in concept to fiberglass or carbon fiber, GRFC, or FRP, and conceptually would work the same way and offer the same benefits. </p>
<p>Traditionally, fiber-reinforced products work because the fibers are suitably strong (overall, not individually) to resist great tensile loads, while the matrix binder (in this case the glue) provides adequate support to hold the fibers in place. </p>
<p>As it's your homework assignment, I would do your research on FRP's (fiber reinforced plastics) and other composite materials to determine whether or not this will indeed work for you. </p>
<p>An H-Beam or I-Beam form I think would prove suitable.<br>
You could also look at adding gussets and ribbing within your foam mold to increase strength and rigidity at areas of stress concentration.
Looking at the rules you've written, I don't see how this would violate any of them. </p>
<p>If you have access to 3d modeling software and FEA software at your school, you could do an optimization study to give you a good idea of how to build the bridge and where to add gussets to assure structural integrity. Many FEA suites offer the ability to apply composite materials. </p>
<p>Also you can look at the way many plastic playsets and slides are constructed. These are typically roto-molded plastic, not composites, but the beam construction and ribbing could give you some hints as to the approach to take. </p>
<hr>
<p>Finally, </p>
<p>You're mileage may vary, but I'll tell you this - if I was a teacher, who did this project every year and watched countless students create countless iterations of the same basic bridge design using girder construction, I would be pretty impressed if someone brought in a completely fresh approach to the problem. </p>
<p>This attitude of thinking differently has served me well in my career, it may or may not work for you. Just a thought. </p>
| 7524 | Most efficient way to build a toothpick bridge |
2016-02-20T03:42:11.050 | <p>First of all, I'm interested in train <strong>track</strong> only, not the rolling stock.</p>
<p>How are tracks built to cope with really cold weather? An example might be some place in Canada or Siberia. Ice would accumulate all the time, changing the shape of the rail.</p>
<p>How does the ballast or foundation cope with the thawing tundra in summer? What about the really nasty stuff like blizzards dumping snow over the track, or sleet, or freezing rain?</p>
| |materials|temperature|rail|transportation| | <p>I was a hoghead for 25yrs here in the NW and we had a few real cold but not very many. Our main fear was water. If the tracks got under mined and you were the train to go over it when it decides to go, the rest of your trip isn't much fun. Whereas if the rail breaks during cold, the dispatcher will see that almost instantly. The undermined rail won't break until a car going over that spot. Whether it's the engine or a trailing car. Hopefully it goes on a trailing car. That way your train would separate and go into "emergency". And if the engineer "hoghead" knows what he's doing, you come to a smooth stop.</p>
| 7526 | How do train tracks handle really cold weather? |
2016-02-20T09:00:15.823 | <p>I just watched this video: <a href="https://www.facebook.com/HigherPerspective/videos/1162007547164896/" rel="nofollow noreferrer">https://www.facebook.com/HigherPerspective/videos/1162007547164896/</a></p>
<p>I don't understand what I see. If the spinner's magnetic, the bowl is made of iron and that the wood piece is magnetic too. Shouldn't the iron bowl mess the magnetic field? Can someone explain how this works?</p>
<p>Another similar image from Wikipedia:</p>
<p><a href="https://i.stack.imgur.com/vatCu.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vatCu.jpg" alt="enter image description here"></a></p>
| |magnets| | <p>This is an example of spin stabilised magnetic levitation. </p>
<p>It's really just two magnets repelling each other, one in the wooden base, one in the spinning top; with opposite poles facing each other. This system of two magnets isn't stable on its own but the fact that the top is spinning provides gyroscopic stabilization. The bowl, bubbles, smoke and bits of wood are just cosmetic and have nothing to do with how it actually works.</p>
| 7528 | How does a Levitron work? |
2016-02-20T19:44:45.290 | <p>On my laptop, I set a variable that makes it so the line tool defaults to Cartesian coordinates. On my desktop, I can't seem to find the variable name. I have been searching for hours.</p>
| |autocad| | <p>The variable is DYNPIFORMAT.</p>
<blockquote>
<p>Controls whether pointer input uses polar or Cartesian format for coordinates.</p>
<p>This setting applies only to a second or next point.</p>
<p>0 Polar</p>
<p>1 Cartesian</p>
</blockquote>
<p>(From: <a href="https://knowledge.autodesk.com/support/autocad-lt-for-mac/learn-explore/caas/CloudHelp/cloudhelp/2015/ENU/AutoCAD-LT-MAC/files/GUID-03D36D36-D734-43C3-BA4D-EA68C573F73C-htm.html" rel="nofollow noreferrer">https://knowledge.autodesk.com/support/autocad-lt-for-mac/learn-explore/caas/CloudHelp/cloudhelp/2015/ENU/AutoCAD-LT-MAC/files/GUID-03D36D36-D734-43C3-BA4D-EA68C573F73C-htm.html</a>)</p>
| 7532 | How do I set the default coordinate system for the line tool in Autocad? |
2016-02-21T00:46:55.283 | <p>I'm referring to the process shown in this video:</p>
<p><a href="https://youtu.be/Quyr5R1Rbfw?t=20" rel="noreferrer">https://youtu.be/Quyr5R1Rbfw?t=20</a></p>
<p>Or this image from Wikipedia:</p>
<p><a href="https://i.stack.imgur.com/U2vdC.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/U2vdC.jpg" alt="enter image description here"></a></p>
<p>In it, a large warship is launched into the water by essentially dropping it sideways down some ramps and off of a pier. The ship rolls hard to one side, and then oscillates back to the other, making the process seem like a fairly risky one. For instance, if it rolls back towards the pier too aggressively it might strike the structure and cause damage both to the pier and the shiny new warship. Or if it rolls too far on the initial drop, the ship might capsize. </p>
<p>So my question is, what are the advantages of launching a large ship sideways like this, as opposed to, say, dropping or gently lowering it vertically into the water?</p>
| |mechanical-engineering|structural-engineering|safety|naval-engineering|ships| | <p>End to end, it would bend and snap in half under its own weight.
I know it's just a movie, but watch "Titanic" for a visual.
The support required otherwise would be huge and expensive.
A boat/ship will naturally rock side to side without breaking.</p>
<p>Impact with the water will act as a brake and keep it from rocking too far.
The weight of the ship will cause it to sink more and more, adding to friction with the water. The water <strong>does not</strong> want to compress downward so it surges <strong>upward</strong> into a wall. Gravity pulls the wall back down, even though only part of it is still pressing backwards on the ship.</p>
<p>Physics will also fight it rocking back against the dock, inertia and gravity.
Think about a bolt in your ceiling attached to a weight, like a bowling ball with a rope. Pull it back from the center where it hangs, to your face, and let it go. It swings out away from you, and then swings back towards your face.
It will not swing back far enough to hit you because gravity is pulling it down, and friction with the air slows it down. (Water has even more friction than air!) Internal friction of the individual strands of the rope against each other also comes into play, but does not illustrate the point enough to mention in great detail.</p>
| 7536 | Why would you launch a large ship by dropping it sideways? |
2016-02-22T08:52:02.870 | <p>Recently I have been researching torque and horsepower. I know that the lower the car gear, the more the torque. However, whenever I see specifications of a car, it says : </p>
<blockquote>
<p><em>X lb-ft of torque at Y rpm.</em></p>
</blockquote>
<p>Now, shouldn't they specify the gear that was measured in? Or is this the torque the engine makes, measured before the gearbox? </p>
<p>I am a little confused, I would be happy if you could help me with some explanations.</p>
| |torque|engines| | <p>The torque indicated by the car manufacturer is usually measured at the engine's output, without use of the gearbox.
However it is still a good indicator of the car's global performance, but only if you pay attention to which rpm provides the higher torque.</p>
<p>Example : 100 Nm @3000 rpm is a better performance than 100 Nm @7000 rpm since the first one is available in "normal" driving while the second is available pushing the engine very high in rpm which only occurs in automobile sports or "unlawful" driving.</p>
| 7555 | Torque in different car gears? |
2016-02-22T11:57:04.473 | <p>I'm trying to find the canonical method for transferring heat from a 70–80 °C source to heat up some air in a box to 45–50 °C. My understanding of heat pipes is that the liquid inside has to evaporate in order for heat transfer to occur. Many heat pipes use water, but water doesn't boil at 50 °C.</p>
<p>What is the canonical/best way for transferring heat at 70 °C?</p>
<p>Here's a simple sketch:</p>
<p><a href="https://i.stack.imgur.com/Phuab.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Phuabt.png" alt=""></a></p>
<p>where the black rod is the proposed heat pipe insulated with some rubber (green).</p>
| |thermodynamics|heat-transfer| | <p>The advantage of a heat pipe is that it is a passive way of getting a high rate of heat transfer between two points using a single, compact component. </p>
<p>However a heat pipe is only one part of a heat transfer system and you also need heat exchangers of some sort at the source and sink to effectively transfer heat from one medium to another. </p>
<p>The most common way to transfer heat to air is to use a radiator (they actually work by convection not radiation bu that's what they are called). Essentially you pump hot water through a network of pipes with a large surface area exposed to the air. Air has a low coefficient of thermal conductivity so the surface area has a big impact on its performance.</p>
<p>The situation you describe is broadly similar to a domestic central heating system or water cooling in an engine. </p>
<p>Whether you use a heat pipe or pumped coolant is largely down to what is most convenient for your particular installation. Heat pipes tend to be more attractive for small compact systems whereas pumps will tend to give more performance in large scale and more demanding applications and also give more flexibility in the relative positions of the source and sink as you can make the pipe runs as long as you want as long as it is well insulated. </p>
<p>Pumps also have the advantage that you can use the flow-rate to control the rate of heat transfer. </p>
| 7558 | Transferring heat using a pipe at temperatures below the boiling point of water |
2016-02-22T13:20:31.930 | <p>I am trying to solve the problem below without using moments. To find the resultant force is simply a matter of adding the force components together, but I cannot find any method of finding a location of the resultant without using moments. Is there any other way?</p>
<p><a href="https://i.stack.imgur.com/z8PKg.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/z8PKg.jpg" alt="enter image description here"></a></p>
<p>The problems is from Hartog's <em>Mechanics</em>.</p>
<p>Edit: This page explains how to solve it using moments: <a href="http://www.leancrew.com/mechanics/problem006.html" rel="nofollow noreferrer">http://www.leancrew.com/mechanics/problem006.html</a></p>
| |mechanical-engineering| | <h2>Method to Solve the Problem</h2>
<p>This problem requires only 2 things: </p>
<ol>
<li>one must remember that a force applied to a rigid body may be considered to act on the body anywhere along its <a href="https://en.wikipedia.org/wiki/Line_of_action" rel="nofollow noreferrer">line of action</a>, and </li>
<li>the application of the <a href="https://en.wikipedia.org/wiki/Parallelogram_of_force" rel="nofollow noreferrer">parallelogram of forces (PoF) method</a>.</li>
</ol>
<p>To solve the problem, simply find the resultant force of two of the forces by applying the parallelogram of forces method. Proceed by finding the point of intersection of this resultant force's line of action and the line of action of one of the remaining original forces. Again apply the parallelogram of forces method. You will now have the resultant force from 3 of the original forces as well as its location. Proceed in this manner until you are left with a single resultant force and no original forces.</p>
<h2>Solving the Problem</h2>
<p>Let's go ahead and perform this analysis on the problem given.</p>
<p>Start by considering the 500 unit force applied to the upper left corner of the plate and the 500 unit force acting at the midpoint of the left edge of the plate. By inspection we see that the intersection point of their lines of action is at the midpoint of the left edge of the plate. Using the <a href="https://en.wikipedia.org/wiki/Law_of_cosines" rel="nofollow noreferrer">Law of Cosines (LoC)</a> we can find the length (and angle) of the resultant of these two forces,</p>
<p>$$
c^2 = a^2 + b^2 -2ab \cdot cos (\gamma)
$$
$$
c^2 = 500^2 + 500^2 - 2 \cdot 500\cdot500\cdot cos(150^\circ)
$$
$$
c = 966
$$</p>
<p><a href="https://i.stack.imgur.com/3LyLlm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/3LyLlm.png" alt="enter image description here"></a></p>
<p>We proceed in the same manner. I did this manually (mathematically via LoC) but I'm too lazy to type it all out. I did, however, check my work by doing it graphically in AutoCAD. Here are some pictures of each step.</p>
<p>Find intersection of lines of action for resultant and top 500 unit load:
<a href="https://i.stack.imgur.com/7KELBm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/7KELBm.png" alt="Find intersection of lines of action for resultant and top 500 unit load"></a> </p>
<p>Apply PoF method:</p>
<p><a href="https://i.stack.imgur.com/THG6pm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/THG6pm.png" alt="Apply PoF method"></a></p>
<p>Find intersection of lines of action for resultant and bottom 300 unit load:</p>
<p><a href="https://i.stack.imgur.com/sFykHm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/sFykHm.png" alt="Find intersection of lines of action for resultant and bottom 300 unit load"></a></p>
<p>Apply PoF method:</p>
<p><a href="https://i.stack.imgur.com/TsY9Vm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/TsY9Vm.png" alt="Apply PoF method"></a></p>
<p>Find intersection of lines of action for resultant and right 300 unit load:</p>
<p><a href="https://i.stack.imgur.com/d2OtPm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/d2OtPm.png" alt="Find intersection of lines of action for resultant and right 300 unit load"></a></p>
<p>Apply PoF method:</p>
<p><a href="https://i.stack.imgur.com/BXcBom.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/BXcBom.png" alt="Apply PoF method"></a></p>
<p>Final Result:</p>
<p><a href="https://i.stack.imgur.com/U07AEm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/U07AEm.png" alt="Final Result"></a></p>
<p>So there you are, how to find the magnitude, location, and angle of the resultant resulting from multiple force vectors applied to a rigid body.</p>
<h2>Graphic Statics</h2>
<p>The above are a few of the basic principles of a pretty cool but obscure method of static analysis called <em>Graphic Statics</em>.</p>
<p>If you'd like to know more, you should check out the book <em>Shaping Structures | Statics</em>, by Zalewski & Allen. As, alpehzero alluded to, graphic statics used to be an extraordinarily handy tool to solve otherwise tricky analysis problems prior to the advent and subsequent proliferation of computers. Note that research into graphic statics methods, particularly with respect to optimizing truss deisgn, <a href="https://architecture.mit.edu/sites/architecture.mit.edu/files/attachments/lecture/StructuralOptUsingGraphicStatics_r.pdf" rel="nofollow noreferrer">is still being done today</a></p>
| 7560 | Find location of resultant of forces without using moments |
2016-02-22T15:20:08.283 | <p>I am thinking of playing around with food in partial vacuum conditions (ie water boiling at room temperature). But, as it is more a curiosity, I'd rather not spend too much money right now. That inspired me to be a little creative with stuff I already have.</p>
<p>I have a pressure canner (heavy duty aluminum), an AC vacuum and sheets of 3/16" acrylic. The acrylic would be nice to use as I can keep an eye on what is happening.</p>
<p>I don't think the acrylic is thick enough to withstand the pressure, so I was thinking of ways to combine them somehow.</p>
<p>Here is a proposed solution. For the first sheet on the low pressure side, drill some holes in the sheet to reduce the pressure, but make the holes small enough that it is not able to come to an equilibrium while the pump is in operation. Then another sheet is placed on top of it with a gasket to keep them separate. This next sheet would have smaller/fewer holes in order to .. In theory, this could continue indefinitely, then an unperforated sheet at the top.</p>
<p>Here is an illustration (note, I missed gaskets around the vacuum outlet):<a href="https://i.stack.imgur.com/ZfzMS.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ZfzMS.png" alt="vacuum semiperforated seal"></a></p>
<ul>
<li>a=pressure canner </li>
<li>b=sheets of acrylic </li>
<li>c=gaskets </li>
<li>d=vacuum outlet</li>
<li>e=larger holes</li>
<li>f=smaller holes</li>
</ul>
<p>I am not particularly sure I want to do this, but I am curious if the general concept is sound.</p>
<p>One possibility is that it could work, but the holes would be smaller than I could reasonably do in my home shop. Acrylic isn't great for drilling.</p>
| |vacuum| | <p>No, this idea won't work. Except for the start-up transient, there will be no flow of air through all of those holes, and the top (unperforated) plate will still be experiencing the full force of the pressure difference between outside and inside.</p>
<p>Better to just stack up N sheets with no holes and no space between them.</p>
| 7564 | Spreading vacuum load using stacked perforated sheets |
2016-02-23T01:03:06.067 | <p>The power generated by a wind turbine is given by:</p>
<p>$$\mathrm{Power} = \frac{1}{2}C\rho AV^3$$</p>
<p>Where:</p>
<ul>
<li>$\rho = \text{Air density}$</li>
<li>$C = \text{Coefficient of performance}$</li>
<li>$A = \text{Frontal area}$</li>
<li>$V = \text{Velocity of the wind}$</li>
</ul>
<p>In other words, the power is proportional to the square of the length of the blades and the cube of the velocity of the wind. As we know, the velocities of winds are high at high altitudes. So instead of building many smaller wind turbines, why can't we just build a giant wind turbine that is, say, 1000 m tall? That may be an engineering challenge at first, but won't that be more economical in the end? After all, the Burj Khalifa in Dubai has been 828 m tall.</p>
<p>Why can't we build three instead of one pillars to support such a structure? Why can't we build one at the sea?</p>
<blockquote>
<p><a href="https://en.wikipedia.org/wiki/Wind_turbine#Records" rel="noreferrer">The Vestas V164 has a rated capacity of 8.0 MW, has an overall height
of 220 m (722 ft), a diameter of 164 m (538 ft), is for offshore use,
and is the world's largest-capacity wind turbine since its
introduction in 2014.</a></p>
</blockquote>
| |electrical-engineering|civil-engineering|structural-engineering|renewable-energy|wind-power| | <p>Weight. As turbines get larger, it's the point of diminishing returns it takes a steady and constant wind velocity in order to accomplish the same goal. The bigger the machines get...........the efficiency rate declines. It's all based on the Cube square law. <em>When an object undergoes a proportional increase in size, its new surface area is proportional to the square of the multiplier and its new volume is proportional to the cube of the multiplier.</em></p>
<p>The turbine blade get's bigger, it's surface area improves. BUT it's weight increases and structural members to increase it's rigidity.
<a href="https://i.stack.imgur.com/zP4TV.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/zP4TV.jpg" alt="enter image description here" /></a></p>
<p>And does nothing to change overall energy market. Global energy use' percentage devoted by hydrocarbon was 87% in early 1990, and 84% roughly now. This otherwise small percentage point decline took nearly 30 years and 2 Trillion in global capital investment. And the Betz limits maximum theoretical turbine power output, making them inefficient as they get heavier, the solution is to install them at he sites with highest potential wind loads.</p>
| 7567 | Why don't we just build a giant wind turbine? |
2016-02-23T01:51:29.803 | <p>How can large bridges, with spans on the order of 1 km, be made resistant to earthquakes?</p>
<p>I'm no expert on quakes, but there are at least two kinds of shaking: lateral and vertical. The vertical shaking in particular really worries me. I don't know how any kind of shock absorbtion can be built into a massive, tall structure like a building or bridge tower.</p>
<p>(I was originally going to ask about suspension bridges, but then I read that suspension bridges are not great for heavy trains. The background of this question lies in my exploring the idea of a Bering Bridge (from Alaska to Siberia, crossing the Bering Strait), which would be primarily a train bridge. And it would need to handle freight trains, the heaviest of all sometimes exceeding 100 tons per car.)</p>
<p>So I will just ask the question about large bridges in general.</p>
<p>I think the largest quake in Alaska was a 9.4 Richter Scale in 1964, hitting Fairbanks. (I don't know how to convert to Moment-Magnitude scale). Is it possible to build large bridges that won't collapse from that? Ideally, we want the bridge to not collapse during a quake even if it's under full load.</p>
<p>P.S. I know it's not cost effective to build such a bridge. For one thing, far east Siberia has no rail network (or much civilization in general). There was a tunnel project proposed which I think they said was cheaper than a bridge, which I couldn't understand because digging through tough rock seems a lot harder than driving a pile in 50-m deep water. The tunnel project was suspended IIRC, no surprise.</p>
<p>Anyway, I'm just exploring if such a bridge is technically possible and can be quake-resistant. If you need something closer to reality, I suppose we can look at the Golden Gate Bridge in San Francisco. I saw <a href="http://kalw.org/post/what-could-collapse-golden-gate-bridge#stream/0">here</a> they were working to make it safe in quakes up to 8.3, but it didn't go into the details. And keep in mind the Golden Gate Bridge does not handle freight trains.</p>
<p>Anyway, is it possible to quake-proof or seriously resist quakes in large bridges loaded with a freight train? The bridge would not necessarily have to remain completely undamaged. I just don't want it to snap and drop the train in the ocean.</p>
<p>Is this possible?</p>
| |structural-engineering|bridges|seismic| | <p>Actually very long bridges (and super-tall buildings) often have less issues with earthquakes than their smaller brethren. This is due to them generally being much more flexible and therefore having lower fundamental periods which makes them less susceptible to resonance in their fundamental modes. The fundamental modes are the swaying patterns that include the majority of the structural mass. An extreme simplification would be that the main structure sways so slowly that it barely notices the quick movements of an earthquake. A bit like a large ship in small waves.</p>
<p>In general "mid-size"-structures, with fundamental frequencies between say 1Hz and 10Hz, are usually much more harshly affected as there is a much larger risk of fundamental resonance leading to very large load effects. For very large and slender structures the wind engineering is generally a bigger challenge than the earthquake engineering.</p>
<p>However the piers and abutments and their connections to the main bridge deck are critical as they usually are much stiffer than the bridge as a whole. And given the amount invested and the potentially horrific consequences of a failure of a large structure a lot of effort will of course be put into performing and checking (and triple checking) the earthquake engineering of every part of the structure. I'm just pointing out that the problems aren't simply proportional to scale, larger structures aren't necessarily harder to "earthquake-proof" than smaller ones.</p>
| 7568 | How can large bridges be made resistant to earthquakes? |
2016-02-24T16:50:51.693 | <p>I am modelling a railway track in ANSYS where the different soil layers have different material properties (including different damping ratios).</p>
<p>Since I know what frequency of vibrations to expect in the track, I can model the damping ratio using beta-damping in the soil elements only. I do not want to use alpha-damping, since it is defined globally in ANSYS and therefore would apply to all other materials for which I do not want to include damping.</p>
<p>I have no problems in the results, but I need to justify why it is okay not to include the alpha-coefficient. I have found a <a href="http://designspace.com/staticassets/ANSYS/staticassets/resourcelibrary/confpaper/2002-Int-ANSYS-Conf-197.PDF" rel="nofollow">report online</a> that says "In many practical structural problems, the $\alpha$ damping (or mass damping) which represents friction damping may be ignored ($\alpha=0$)."</p>
<p>However, I need either a published paper or relevant textbook, or a detailed justification why beta-damping is a good approach, and I have been unable to find it so far.</p>
| |ansys|finite-element-method| | <p>The report in your link explains it briefly in the sentence after your quote. The beta term models "structural" or "hysteretic" damping, which is described earlier in the report.</p>
<p>The key fact about hysteretic damping is that, for a fixed amplitude of motion, it dissipates the same amount of <em>energy</em> in each cycle of the motion, <em>independent of the frequency</em>. That is often a good model for "internal energy dissipation" within a structure. I'm not a civil engineer but I could imagine it is a plausible model for the ballast or earth under a rail track squidging about as it dissipates energy. It is certainly a good model for the (small) amount of energy dissipated by the metal rails themselves as they vibrate.</p>
<p>The alpha term doesn't have an obvious physical interpretation, but it can be mathematically useful. It has the same mathematical property as beta, namely that the mode shapes of the damped system are identical to the undamped. (In general that is not the case, and the damped modes can have different phases in different parts of the structure, i.e. they can look more like "travelling waves" than the undamped vibration modes). Using a combination of alpha and beta, you can get a fairly constant level of equivalent modal damping factors within a narrow band of frequencies, while the damping factors for modes with both higher and lower frequencies are more heavily damped. (That can be nice if you are modelling a system which is not fixed to anything, because the alpha term damps out any rigid body motion which you are probably not interested in, but the beta term does not).</p>
<p>Sorry, I don't have a reference for this (I've known it for too many decades already!) but a search for "hysteretic damping" should turn up something.</p>
<p>Bear in mind that modelling damping is rarely an "exact science". The best source of advice would be "best practice" in your particular application - either specialist textbooks, or relevant papers in journals. You are probably not the first person who wants to model this!</p>
| 7593 | Rayleigh damping in Finite Element Models using beta-coefficient only |
2016-02-25T00:53:13.350 | <p>Individual-head (i.e. not deluge) sprinkler systems come in three flavors:</p>
<ul>
<li>Wet-pipe systems have water pressure at the head at all times -- they are the quickest to respond, but can't be subject to freezing conditions and are too risky for water-sensitive spaces</li>
<li>Dry-pipe systems have compressed air in the pipe between the head and the dry-pipe valve -- when the head fuses, the dry-pipe valve is opened by the pressure change and delivers water to the system. This can be used when a freeze risk exists, but is slower to respond.</li>
<li>Preaction systems are used to provide external confirmation (from say an alarm system) before the sprinklers activate -- the alarm system must open the preaction valve before water can flow onto the fire. This is used when accidental discharges must be controlled at all costs.</li>
</ul>
<p>This raises the question: under NFPA 13/13R/13D, can individual fire zones be protected by a dry-pipe or preaction system while the remainder of the building has wet-pipe sprinkler protection? For instance, the attic, garage, and unfinished basement in a home could be dry-pipe protected to avoid freeze-ups while the finished spaces are protected by a conventional wet-pipe system.</p>
| |safety|building-design|fire-sprinkler| | <h2>Yes, this is permissible</h2>
<p>NFPA 13 7.1.3 explicitly permits this:</p>
<blockquote>
<p><strong>7.1.3 Auxiliary Systems.</strong> A wet pipe system shall be permitted
to supply an auxiliary dry pipe, preaction, or deluge system,
provided the water supply is adequate.</p>
</blockquote>
| 7598 | Can a dry-pipe or preaction valve be used to create a branch off a wet pipe sprinkler system? |
2016-02-25T03:36:22.737 | <p>See here are the shapes which are made on the suction head of compressor, <a href="https://i.stack.imgur.com/s9dUq.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/s9dUqm.jpg" alt="enter image description here"></a></p>
<p>What are these, and why are they on the suction head pipe? </p>
| |mechanical-engineering|fluid-mechanics|piping|compressors| | <p>You can see that the elbow is immediately upstream of the diameter reducer, which means it's immediately upstream of the pump. That's a problem, because the elbow should be a large-radius curve, and at least 5 diameters away from the pump.
The reason the elbow should be away from the pump is the same reason that the pipe is oversized and reduced at the pump inlet: to eliminate unbalanced flow at the pump inlet.</p>
<p>Since the elbow is sharp and adjacent to the pump, of course it has inline flow conditioning. As shown, there's not even room to insert a simpler straight flow-conditioning section (which would be cheaper)</p>
<p>Inline flow conditioning inserts don't always show up on outside of the pipe, but since we know that they must be there (in a properly designed system), our immediate observation is that those odd shapes correspond to internal flow conditioning vanes.
<a href="https://www.researchgate.net/figure/Flow-pattern-a-90-rectangular-elbow-b-90-rectangular-elbow-with-turning-vanes-c_fig1_351405638" rel="nofollow noreferrer">figure</a> <a href="https://www.researchgate.net/publication/351405638_An_adaptive_approach_to_duct_optimization_of_an_industrial_boiler_air_supply_system_using_airfoils" rel="nofollow noreferrer">source</a></p>
| 7600 | What are these bent shapes on this elbow in a compressor suction pipe? |
2016-02-25T07:38:41.467 | <p>I open the door for 30s of a 140L capacity 80W (240VAC, 0.8A) barfridge with a 20L freezer compartment (partition open to main compartment at one end) at a temperature of -5℃ with an internal main compartment temperature at 2℃, with initially 30L of 50-50 glass and plastic containers container mainly water inside, in a room at an ambient temperature of 41℃ and 85% humidity (Sydney today), to <strong>take out</strong> a 6-pack of medicinal beer.</p>
<p>Presumably, the cold, denser air from inside the fridge flows out,
and is replaced by the hot, lighter air from the room.</p>
<ul>
<li><p><strong>How much of the fridge and freezer compartment air is exchanged in that time period?</strong></p></li>
<li><p><strong>How long does it take the exchanged air in the fridge to equilibrate
to its initial temperature of 2℃? What equation describes this? Is there an initial temperature rise?</strong></p></li>
<li><p><strong>How many grams of ice will precipitate from the exchanged air into
the freezer compartment?</strong></p></li>
</ul>
<p><strong>EDIT UPDATE</strong></p>
<p>After @Chris Johns' answer, I rolled the barfridge out and actually looked at the manufacturer's plate, and then updated my question from my original eyeball estimates.</p>
<ol>
<li><p>Changed 40L to <strong>140L</strong></p></li>
<li><p>Added power info: <strong>"80W (240VAC, 0.8A)"</strong>.</p>
<p>From my calculations, $ P = V_{rms} * I_{rms} $ = 240VAC * 0.8A = 192 W,</p>
<p>so efficiency, $ \eta = P_{out} / P_{in} \%% $ $ = {80W \over 192W} = 41.7\%% $</p></li>
<li><p>Changed freezer compartment capacity eyeball estimate from 10L to <strong>20L</strong> </p></li>
<li><p>The refrigerant is $CCl_2F_2-R12$, 85g.</p></li>
<li><p>Clarified the internal contents of the fridge as containing a total of <strong>30L</strong> of plastic and glass containers (50-50) containing mostly water.</p></li>
<li><p>Clarified the temperature of the freezer compartment at <strong>-5℃</strong>. </p></li>
</ol>
<p>I presume you can calculate the coefficient of performance (CoP) from this data?</p>
| |thermodynamics|heat-transfer|airflow|cfd|gas| | <p>If you open the door of a small upright fridge for 30 seconds it's reasonable to assume that virtually all of the cold air inside will be replaced with warm air from outside. </p>
<p>The missing information in this question is the power and coefficient of performance of the fridge, this tells you the rate at which it is able to move heat from inside to outside. </p>
<p>As a first approximation you are looking at the heat exchange required to cool 40L of air from 41 deg C to 2 deg C. This is simply the specific heat capacity of air multiplied by the mass of air multiplied by the temperature difference. And works out at around 40 Joules so basically very little. </p>
<p>A more detailed approach would consider the change in enthalpy of condensing out and freezing the water vapour present and also the heat transfer from the fridge body and contents during the time that the door is open, but again we are talking about quite small numbers.</p>
<p>You will put much more demand on the fridge when you put in cans of beer which are at ambient room temperature as the heat capacity of the liquid is much greater than that of the air that it replaces. </p>
<p>Eg the heat capacity of 1 litre of water is around 4200 J/C so a 40 deg C temperature difference would require 168 KJ of heat transfer to achieve. </p>
<p>Say the fridge has a power consumption of 100W and a CoP of 3 then is can transfer 300W of heat and so would take at least 10 minutes to cool 1 litre of water, although this figure will be further limited by the heat transfer rate between the water container and the air. </p>
<p>So in summary : </p>
<ul>
<li>Just opening the door for a short and taking something out is fairly minor in terms of load on the fridge.
<ul>
<li>The minimum time (and energy requirement) to cool an object from its current temperature to fridge temp and return to equilibrium depends on the performance of the fridge, the initial temperature difference and heat capacity of the object being cooled.</li>
<li>The actual time taken will depend on the rate of heat transfer between the cold air in the fridge and the object. </li>
</ul></li>
</ul>
| 7601 | Refrigerator temperature - compartment gas volume exchange, time to equilibrate? |
2016-02-25T08:32:59.877 | <p><a href="https://i.stack.imgur.com/6bpGs.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/6bpGs.jpg" alt="determine the length of b of the triangular load and its position a on the beam such that the equivalent resultant force is zero and the resultant couple moment is 8 kN*m clockwise."></a></p>
<p>This is the question, but with me 6 kN/m is 4 kN/m and 2 kN/m is 2.5 kN/m. And the length is not 4 but 9 metres.</p>
<p>I am not yet at the chapter of equations of equilibrium, plus it says the couple moment is not 0, so I assume it just about the loadings and not including the reactive forces and reactive moment.</p>
<p>So if I call the top one F1 and the bottom loading F2 I got:</p>
<p>$$F_1+F_2=0= -4b\dfrac{1}{2}+2.5(b+a)\dfrac{1}{2}$$
solving this i got $a=0.6b$</p>
<p>Moment around the free end of the bar (not $A$, but opposite side) due to the loadings is (counter clockwise positive):</p>
<p>$$M= -F_1\cdot\text{center of triangle} + F_2\cdot\text{center of triangle}$$</p>
<p>Because you can replace the loading with force $F_1$ and force $F_2$ and its line of action is through the center of the triangle area ($\dfrac{1}{3}\text{base}$)</p>
<p>$$M= -8= -4b\dfrac{1}{2}\cdot\dfrac{1}{3}b+2.5(b+a)\dfrac{1}{2}\cdot\dfrac{1}{3}(b+a)$$</p>
<p>So what am I doing wrong?
Because $M$ can never be negative with me, plus the answer should be $b=5.625$ and a= $1.539$.
But this to me makes no sense, because then $F_1+F_2\neq0$.
And if I should take the reactive forces into account at $A$ then you can never have still a moment, because then it is not static anymore.</p>
| |statics| | <p>Defining $F_1$ as the triangular load, we have</p>
<p>$$F_1 = -\dfrac{4b}{2} = -2b$$</p>
<p>Defining $F_2$ as the uniform load, we have</p>
<p>$$F_2 = 2.5(a+b)$$</p>
<p>As you stated, $F_1+F_2 = 0 \therefore b = -5a$. This result may seem odd, given that it's negative, but it's correct. After all, the resultant force of a triangular load is half of what it would be if the load were uniform and equal to the maximum value. So, for a triangular load to have the same resultant force as a uniform load while both occupy the same length ($a=0$), the triangular load must have a maximum value which is double that of the uniform load. If the triangular load's maximum value is greater than double, then it can occupy a shorter length while having the same resultant force ($a>0$, that's the case displayed in the original question with 6 and 2 kN/m). However, if the triangular load's maximum value is less than double the uniform load (as is your case), then the triangular load must occupy a greater length to have the same resultant force ($a<0$).</p>
<p>Now, the moment due to a force couple is $M = F \times D$, where $D$ is the distance between the forces in the couple. Now, the centers of action of $F_1$ and $F_2$ are (from the free end):</p>
<p>$$\begin{align}
D_{F_1} &= \dfrac{b}{3} \\
D_{F_2} &= \dfrac{a+b}{2}
\end{align}$$
Therefore $D = \left|\dfrac{a+b}{2} - \dfrac{b}{3}\right| = \left|\dfrac{3a+b}{6}\right| = \left|\dfrac{-2a}{6}\right| = \dfrac{|a|}{3}$. Thus, $M = -2b \times \dfrac{|a|}{3} = 10a \times \dfrac{|a|}{3} = -8 \therefore a = -\sqrt{2.4} \therefore b = 5\sqrt{2.4}$.</p>
<p>Checking our work:
$$\begin{align}
F_1 &= -10\sqrt{2.4} \\
F_2 &= 2.5(-\sqrt{2.4}+5\sqrt{2.4}) = 10\sqrt{2.4} \\
&\therefore F_1 + F_2 = 0 \text{ OK!}\\
M_A &= -10\sqrt{2.4}\left(4-\dfrac{5\sqrt{2.4}}{3}\right) + 10\sqrt{2.4}\left(4-\dfrac{4\sqrt{2.4}}{2}\right) = 10\sqrt{2.4}\left(\dfrac{5\sqrt{2.4}}{3}-\dfrac{4\sqrt{2.4}}{2}\right) \\
&= 24\left(\dfrac{5}{3}-2\right) = -\dfrac{24}{3} = -8\text{ OK!}
\end{align}$$</p>
| 7604 | Reduction of a simple distributed loading |
2016-02-25T08:36:38.920 | <p>I am working on modeling of air-heater component. A simplified representation of system is as follows:
<a href="https://i.stack.imgur.com/NIKAS.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/NIKAS.png" alt="Heater duct block diagram"></a></p>
<p>Cold air flow enters steel duct at one end. Inside duct is Calrod heating element shaped as ellipse. Follwing heat transfer occur simulteneously:-</p>
<ol>
<li>Air gains heat from surface of heated calord. </li>
<li>Air also exchanges heat with duct inner surface. </li>
<li>Calrod element radiates heat to duct inner surface. </li>
<li>Duct loses heat by radiation to surroundings as well as convection.</li>
</ol>
<p>System is solved by taking energy balance equations on Air mass, Duct mass and Heater element mass.</p>
<p>My question is in regards to convection between air and duct. In model, I have considered whole assembly to be made of 5 parts along duct length. In each part respective thermal balance on air, duct and heated element mass is taken. In each part, I consider air to be lump-sum mass. Entering air gains heat from calrod and gains some temperature ( which would in real world be average temperature). Problem occurs when that temperature is used for convection transfer with duct. In actual process, air flow is fast enough (0.013 $\frac{m^3}{sec}$) so that air that actually is in contact with heater element doesn't reach duct surface. So air that exchanges heat with duct inner surface is at lower temperature than average air's temperature. This gives incorrect duct temperature predictions. How can I properly set $\Delta T$ for convection heat exchange with duct? </p>
| |heat-transfer|modeling|convection|radiation| | <p>There are several issues going on that I believe are making the model not very accurate. They all in some way relate to your question about the convection:</p>
<ol>
<li><p>Breaking up the flow into 5 discrete chunks is a good start, but probably not a sufficient number. Keep increasing the number of nodes in your model until the answer stops changing. </p></li>
<li><p>The next thing to check is the Reynold number of your flow and the length it will take to become fully developed. If your flow is fully developed for most of the length, the heat transfer coefficient will do a good job modeling the problem you are describing with the air not contacting the wall. If the flow is not fully developed (which is my gut guess) then the heat transfer coefficient won't do a good job. </p></li>
<li><p>In general, you should consider the high level thermal resistances between the coil and the duct wall. Since air is such a poor conductor and has such a low density, it is a poor heat transfer medium. Since the heat transfer coefficient on the surface of the coil to the air will be similar to the heat transfer coefficient from the air to duct, the extreme difference in dT between the coil and air and the air and wall will result in a very large Q to the air and a very small Q to the duct wall from the air. It is not impossible that the air actually cools the duct wall instead of heating. If you work out the air-to-duct heat transfer coefficient you may find that it is so trivial that it is likely not the source of error in your model.</p></li>
<li><p>Without numbers I can't really say for sure, but my guess is that the air in your model is not actually the primary means of heat from the coil to the wall. Instead, the radiation in dominant. Thus, I would look to improve the accuracy of that calculation first if you want to predict wall temperature. For example, what emissivity are you using for the duct wall? How are you calculating the view factor between the coil's interior edge and the wall? Etc etc. </p></li>
</ol>
<p>Hope this helps and that this is still even remotely relevant 1 year late. </p>
| 7605 | Heat Transfer coefficient calculation for hot air flow inside rectangular duct |
2016-02-26T05:35:26.377 | <p>A textbook is attempting to convince me that the following figure:</p>
<p><a href="https://i.stack.imgur.com/ShjuW.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ShjuW.jpg" alt="enter image description here"></a></p>
<p>Represents the generic zones for a Case B wind loading including roof overhangs based on the ASCE 7-10 Chapter 28, Part 2 (Enclosed Simple Diaphragm Low-Rise Buildings) procedure. This procedure addresses roof overhangs via footnote 8 on Figure 28.6-1:</p>
<p>"Where zone E or G falls on a roof overhang on the windward side of the building, use EOH and GOH f`or the pressure on the horizontal projection of the overhang. Overhangs on the leeward and side edges shall have the basic zone pressure applied."</p>
<p>I absolutely do not agree that the figure is a correct application of the Chapter 28, Part 2 wind loading provisions for Case B wind direction. Please see below for ASCE 7-10 Figure 28.6-1 Case B zoning. You may observe that zones H, E , A and C correspond, while zones G and F have been flipped.</p>
<p><a href="https://i.stack.imgur.com/1CC2J.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1CC2J.jpg" alt="enter image description here"></a></p>
<p>I would like to know if the textbook is not applying the wind provisions properly, or if I am missing something in the code? Thanks.</p>
<p>Assuming ASCE 7-10 Figure 28.6-1 is correct and footnote 8 applies, I believe the following (edited) figure shows the correct distribution.
<a href="https://i.stack.imgur.com/J9AGB.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/J9AGB.jpg" alt="enter image description here"></a></p>
| |civil-engineering|structural-engineering| | <p>I believe you are correct, the diagram in your book is incorrect in the way it has labeled Zones F and G. This is a bit of a stretch, but I think this may be related to/as a result of an error in Figure G6-5 of a publication released by ASCE called, <em>Wind Loads, Guide to the Wind Loads Provisions of ASCE 7-10</em> by Mehta & Coulbourne and published by ASCE Press that I came across last month. This publication is an ASCE officially endorsed "How To" guide for ASCE7-10 wind provisions.</p>
<p>Figure G6-5 in this publication (below) is an attempt to replicate the image shown in Figure 28.6-1 of ASCE7-10 but was pretty horribly botched. You'll notice the same mistake with Zones F and G for Case B (called "Longitudinal in my book's figure) as well as some general drafting errors with the linework and arrow hatches.</p>
<p><a href="https://i.stack.imgur.com/Q526P.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Q526P.png" alt="Figure G6-5 from Guide to the Wind Loads Provisions of ASCE 7-10 by Mehta & Coulbourne"></a></p>
<p>There doesn't seem to be a published errata for this publication and when I attempted to contact ASCE about the figure last month I didn't get a response. </p>
<p>In addition to that point, I also think that there are at least three other errors in your figure as well. </p>
<ol>
<li><p>Note 8 of Figure 28.6-1 of ASCE7-10 states (emphasis mine),</p>
<blockquote>
<p>Where zone E or G falls on a roof overhang on the windward side of the building, use E<sub>OH</sub> and G<sub>OH</sub> for the pressure on the horizontal projection of the overhang. <strong>Overhangs on the leeward and side edges shall have the basic zones pressure applied.</strong></p>
</blockquote>
<p>Therefore, I think that the pressures on the overhangs on the side edges should be E and G instead of E<sub>OH</sub> and G<sub>OH</sub> as shown in your figure. The application of E<sub>OH</sub> and G<sub>OH</sub> on the windward side in your figure is correct.</p></li>
<li><p>There is also an error with the boundary line between E<sub>OH</sub> and G<sub>OH</sub> on the windward side. The boundary between E<sub>OH</sub> and G<sub>OH</sub> should project parallel to the axis of the building.</p></li>
</ol>
<p><a href="https://i.stack.imgur.com/NttQem.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/NttQem.png" alt="Correct E<sub>OH</sub> and G<sub>OH</sub> boundary"></a></p>
<ol start="3">
<li>There is nothing in Figure 28.6-1 of ASCE7-10 that suggests the length of Zone E on the gable side of a structure is equal to $ 2a$ as implied in your figure. I believe that Zone E (and G) should extend all the way to the ridge as shown in Figure 28.6-1 of ASCE7-10. The width of Zone E is equal to $2a$ <strong>only on the eave side</strong> of the structure.</li>
</ol>
| 7633 | ASCE 7-10 Wind Loading with roof overhangs |
2016-02-26T12:32:43.340 | <p>What is the meaning of <em>clutch fill</em>?
See for example <a href="http://www.dcsc.tudelft.nl/~bdeschutter/private_20100705_acc_2010/data/papers/0918.pdf" rel="nofollow">this paper</a>.</p>
| |mechanical-engineering|fluid-mechanics|automotive-engineering|terminology| | <p>The paper you referenced defines what the term means. </p>
<blockquote>
<p>To ensure precise synchronization [when changing the drive path from one clutch to another one], before clutch engagement, it is
necessary to actuate the oncoming clutch to a position where the
clutch packs are in contact. This process is called clutch fill...</p>
</blockquote>
| 7636 | What is the meaning of 'clutch fill'? |
2016-02-26T12:51:41.597 | <p>Please take a look at following figure:
<a href="https://i.stack.imgur.com/S2cBw.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/S2cBw.png" alt="Rectangular narrow channel with heat transfer by convection"></a></p>
<p>System considered is a narrow rectangular channel of 2.5 mm depth (W x L is 60 mm x 150 mm). Since channel has low depth, it can be considered as mini-channel. Bottom surface is hot and at $~ 100\,^o \text{C}$, top surface is initially at ambient $20\,^o \text{C}$. Air flow enters channel at $20\,^o \text{C}$.
I want to know how for such narrow channel heat transfer is defined for <strong>air</strong> as working fluid. Specifically:
$$ Nu = f(Re, Pr)$$
a correlation of Nusselt number as function of Reynolds and Prandtl number. Re range is in 500 - 2000.
Any suggestion of literature reference are welcome.</p>
| |heat-transfer|convection| | <p>From "Heat Transfer, 2nd Edition" by A.F. Mills, equation 4.51 gives a formula for the average Nusselt number for flow between two parallel plates. This might be appropriate for your duct because you can probably ignore the sides and just treat it like two parallel plates:</p>
<p>$$
\overline{Nu} = 7.54 + \frac{0.03\frac{D_H}{L}Re\cdot Pr}{1+0.016 \left[ \frac{D_H}{L}Re\cdot Pr\right]^\frac{2}{3}}
$$</p>
<p>Here, $D_H$ is the hydraulic diameter, which is just twice the plate spacing for your case. $L$ is the plate length, and $Nu$, $Re$, and $Pr$ are the Nusselt, Reynolds, and Prandtl numbers.</p>
| 7637 | Heat transfer characteristics in narrow rectangular channel |
2016-02-26T20:12:05.647 | <p>I'm attempting to find the equations of motion (and eventually transfer functions) for a mass-spring-damper system, but one that is slightly different from your generic damped system example.</p>
<p>Below I've given a picture of essentially what the system looks like.</p>
<p>There is a large box with mass m and spring k inside of it. This box is damped by damper c to the left-hand wall of the room. I apply a force F(t) to the side of the box, causing both the box and perhaps the mass inside to move. x is measured from the left-hand side of the room to the center of mass, so x will increase <em>both</em> when the box moves and when the mass inside the box moves.</p>
<p>What I imagine happens is that for low frequency forcing functions, the spring is essentially nonexistent and the equation of motion looks like:</p>
<p>$$F(t) = m\ddot{x} + c\dot{x}$$</p>
<p>When the forcing function frequency approaches the resonance frequency of the system, not only will the box move to the right, but the mass will vigorously experience motion <em>inside</em> of the box. I'm not quite sure what the equation of motion looks like for this case.</p>
<p>I'm looking for one differential equation that captures both of these - low frequency and high frequency forcing functions. At the end of the day I really want a transfer function to use in Simulink.</p>
<p>$$\frac{w_n^2}{s^2 + 2{w_n}{\zeta}s + {w_n^2}}$$</p>
<p>This transfer function doesn't seem to model what I want, even though it's the generic second order system transfer function.</p>
<p>Any help would be greatly appreciated. ^^</p>
<p><a href="https://i.stack.imgur.com/WmVnu.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/WmVnu.png" alt="enter image description here"></a></p>
<p><a href="https://i.stack.imgur.com/Mq14A.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Mq14A.png" alt="enter image description here"></a></p>
| |mechanical-engineering|electrical-engineering|dynamics| | <p>Call the displacement of the box (where it is attached to the damper and spring) y(t).</p>
<p>The force of the damper is $$-c\cdot \dot{y}$$</p>
<p>The force of the spring is $$k\cdot(x - y)$$</p>
<p>y(t) is the point at which the three forces on it balance to zero:</p>
<p>$$F - c\cdot \dot{y} + k\cdot(x - y) = 0$$</p>
<p>x(t) depends only on the spring force and the mass:</p>
<p>$$k\cdot(x - y) + m\cdot \ddot{x} = 0$$</p>
<p>Combine and solve!</p>
| 7644 | Mass-Spring-Damper Dynamic System |
2016-02-27T05:32:40.703 | <p>I am working on a flywheel currently, and I was wondering what is the force needed to recenter the flywheel when it is spinning and it is nudged slightly off axis? Is it related to gyroscopic precession?</p>
| |mechanical-engineering|flywheels| | <p>For this answer I'll assume "off-center" means the axis of rotation differs from the desired axis of rotation by angle <span class="math-container">$\theta$</span></p>
<p>The goal is to realign the axis of rotation, which can be accomplished by changing the rotational momentum vector.</p>
<p>This goal can be achieved by applying forces to the axle.</p>
<p><a href="https://i.stack.imgur.com/nmKnm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/nmKnm.png" alt="Fly Wheel force diagram" /></a></p>
<p>Applying symmetric forces as shown of magnitude <span class="math-container">$F$</span> a distance of <span class="math-container">$R$</span> from the center will add rotational inertia at a rate of <span class="math-container">$2FR$</span> Which will rotate the axis of the Gyroscope at angular rotation rate of <span class="math-container">$\frac{2FR}{I\omega}$</span> the axis of which will be perpendicular to the axis of the flywheel and parallel to the force being applied.</p>
<p>This process will not change the magnitude of the angular velocity of the flywheel.</p>
| 7648 | Calculate Forces on a Flywheel |
2016-02-28T16:28:45.943 | <p>We call air which is pressurised as "compressed air". But we don't call pressurised liquid as compressed. I know that liquids are incompressible.</p>
<p>So,</p>
<ol>
<li><p>why particular word "compressed" is the created, we could call pressurised air instead of compressed air?</p></li>
<li><p>Is there any significant difference between them (aside from the words), if yes then explain?</p></li>
</ol>
| |pressure|compressed-air| | <p>As your question implies 'pressurise' can apply to increasing the pressure of any fluid whereas 'compression' indicates a change in volume. </p>
<p>To illustrate; a typical industrial gas cylinder will be filled to around 200 bar and might have a volume of 100l at 1 bar that gas would occupy 20000l. Often this is as much about storing and transporting produced gasses in a convenient package (eg argon for welding) as it is using the stored pressure to do work. </p>
<p>An incompressible fluid on the other hand occupies almost exactly the same volume (disregarding changes of state) regardless of pressure so 1kg of water has a volume of 1l at 1 bar or 100 bar. </p>
<p>Similarly using the term 'compressor' for a category of machine distinguishes it within the more general category of pumps. </p>
<p>It is not unusual to use terms like 'pressurised gas' but, in very general terms, we talk about compressed air simply because it is produced by a compressor. The term also implies that it is available at (more or less) ambient temperature as opposed to air which has been pressurised by heating. </p>
<p>Similarly without any other context 'compressed air' tends to imply high pressure air used to operate air tools or similar pneumatic machinery or plant. </p>
| 7656 | Basic difference between "compress" and "pressurise" |
2016-02-29T04:40:15.863 | <p>Building codes such as the Uniform Building Code (UBC) or International Building Code (IBC) require the determination of <a href="http://timber.ce.wsu.edu/supplements/seismic/eslf.htm" rel="nofollow">Equivalent Static Lateral Forces</a> in order to compute $P-\Delta$ effect and to perform checks on the structural design.</p>
<p>However, all these overturning moment computations and structural checks are done (in the building code) assuming a <a href="http://www.scc.kit.edu/scc/sw/msc/Nas120/appendix_f_final.pdf" rel="nofollow">lumped mass model</a> (a cantilever column). But what if the structure is irregular? How can we compute the Equivalent Static Lateral Forces in an irregular building/structure discretized by finite element? </p>
<p>These quantities have clear physical meaning in a cantilever column, but may not even necessarily have clear meaning/clear equivalent in an irregular building.</p>
<p>I am looking for the FEM way of computing equivalent static forces that act on global level. Any ideas?</p>
<hr>
<p>To illustrate why I need FEM version of Equivalent Static Load: <strike>consider that the overturning moment is defined as $M_o=F_xh$, where $F_x$ is the lateral load at <em>floor</em> level, and $h$ is the distance between the top floor to the point where overturning moment is calculated.</strike> </p>
<p><a href="http://site.iugaza.edu.ps/sshihada/files/2012/02/Handout-11-12.pdf" rel="nofollow">consider $P-\Delta$ checking ratio</a>:</p>
<p>$\frac{P_x\Delta}{V_xh_s}$</p>
<p>where </p>
<p>$P_x$ is the total unfactored gravity load at and above level $x$</p>
<p>$\Delta$ is the seismic story drift by seismic forces</p>
<p>$V_x$ seismic shear between level $x$ and $x-1$</p>
<p>$h_s$ sotry height below level $x$</p>
<hr>
<p>Consider also accidental torsional effects ( <a href="https://drive.google.com/file/d/1XZcPZpa_ACp-Fw9ROSnKY2iJDgHe-FBY9L8rrGCTqgPe_Cy65Evs40bMHFoPR40mHK6n4mN4_qEsYFhz/view" rel="nofollow">EC8 example, section 2.5.3, pp 39</a>)</p>
<p>$M_t=F_xe_y$</p>
<p>Where </p>
<p>$F_x$ is the equivalent lateral static force in $x$ direction</p>
<p>$e_y$ is the accidental eccentricities</p>
<p>Take note that all these involved quantities are evaluated <strong><em>on floor basis</em></strong>. </p>
<p>But if we do FEM seismic dynamic analysis, then all I get is the forces for each node, not at each floor. So we need to be able to resolve FEM seismic nodal forces into floor forces. </p>
| |structural-engineering|finite-element-method|structural-analysis|seismic| | <p>The short answer to your question,</p>
<blockquote>
<p>How can we compute the Equivalent Static Lateral Forces, the overturning moment and other quantities in an irregular building/structural discretized by finite element?</p>
</blockquote>
<p>is, simply, you can't. ASCE7-10 speaks directly to this in Section 12.6 when it describes the conditions under which Equivalent Lateral Force (ELF) can and cannot be used. If your structure is located in an area with moderate to high seismic activity (SDC C-F), you cannot use ELF if your structure has "structural irregularities." See Table 12.6-1 in ASCE7-10 for details, I've simplified a bit. There are certain structural irregularities that do not preclude ELF.</p>
<p>That said, there <em>are</em> a couple different ways you can go about calculating equivalent static seismic forces for highly irregular structures using FEM. These forces are not directly analogous to the forces you'd get using ELF method from ASCE 7, but they are forces obtained via rational analysis and commonly used/accepted in order to simplify the seismic analysis of irregular structures. They are:</p>
<ol>
<li>Perform a (fixed base) time history analysis on your structure. Use the results to generate acceleration response spectra at each level of your structure. Use the accelerations from all of your RS to design the structure. (Most conservative) </li>
<li>Perform a (fixed base) time history analysis on your structure. Pull the peak acceleration in all 3 directions across all your time steps for each node. Use these accelerations to design the structure. </li>
<li>Perform a (fixed base) time history analysis on your structure. Use the accelerations at each time step to design your structure (at each time step). The most conservative design controls. (Least conservative)</li>
</ol>
<h2>Time History to RS Method (#1)</h2>
<p>This method is the most conservative of the three. The general procedure would be to apply the peak acceleration from your response spectra to the structural elements at each level. In my industry, when using this method, we even multiply the peak acceleration by 1.5 to ensure that we envelope the actual acceleration including any multi-modal effects. This is generally only done for designing components within the structure or modifications to an old structure when time & budget do not allow a more accurate (\$$$$) analysis. It is considered less conservative for component design because you, theoretically, know the natural frequency of the piece of equipment you are working with, and therefore can pull a more accurate acceleration off the response spectra.</p>
<h2>Max Time History Acceleration Method (#2)</h2>
<p>This method is fairly straightforward. After performing your time history analysis, you can extract acceleration time histories for every node in your FE model. For each node in your model, pull the highest acceleration in the X, Y, and Z directions. Multiply these accelerations with your tributary nodal mass to obtain an equivalent static seismic nodal force in all three directions. This is still considered (highly) conservative because you are pulling accelerations from different time steps and combining them.</p>
<p>There are several variations of this method related to exactly which accelerations/time steps you design to. The tricky part is in convincing yourself and/or the body having jurisdiction over your design that you are enveloping reality.</p>
<h2>Time step Time History Acceleration Method (#3)</h2>
<p>This method is similar to #2 except you will design your structure/run your analysis for the seismic accelerations at each time step. At each time step, for each node in your model, pull the acceleration in the X, Y, and Z directions. Multiply these accelerations with your tributary nodal mass to obtain the equivalent static seismic nodal force in all three directions.</p>
<p>This method is the least conservative but, by far, most computationally expensive approach and is mostly unrealistic. Your time histories will likely have upwards of 10,000-100,000 time steps. It is incredibly expensive to chug through the analysis.</p>
<p>I've personally used all 3 methods at one time or another in my career. #2 seems to hit the sweet spot with respect to the accuracy:expense ratio. </p>
| 7662 | How to derive Equivalent Static Load for irregular structure modeled with FEM? |
2016-02-29T09:19:07.290 | <p>I am interested in structural stability of earthen bunds. They are susceptible to being eroded by water. Let us assume I have a rice field that is surrounded on all four sides by earthen bunds. I have seen in my travels two geometries (shape of bunds). Some are rectangular and others are curved. Is there an objective way to analyze(mathematically I mean) which geometry would be more resistant to flooding i.e. rectangular(or square) vs. curved ? Water can hit the bund vertically from above(during precipitation) and/or horizontally (water from surrounding fields can flow towards a bund)</p>
<p>Here is an example of a rice field surrounded by bunds - <a href="http://www.rkmp.co.in/content/bund-preparation-and-maintenance-for-water-management" rel="nofollow">Earthen bunds</a>.
This one is a bit curved but I have seen rectangular bunds as well. </p>
<p>Let us assume I have flat surface area. </p>
| |structural-engineering|geotechnical-engineering|structural-analysis| | <p>Bund stability will be affected by a number a factors:</p>
<ul>
<li>materials used to construct the bund</li>
<li>compaction of the materials used</li>
<li>the height of the bund</li>
<li>the width of the bund</li>
<li>shear strength of the material used to construct the bund</li>
<li>the degree water saturation and depth of saturation</li>
<li>whether any part of the bund is allowed to dry out completely</li>
<li>the slope angle of the walls of the bund</li>
<li>whether the bund has been protected by facing materials like hard
stones that will absorb the energy of water hitting the bund</li>
<li>the force of the water hitting the bund</li>
<li>water drainage through the bund, if applicable</li>
</ul>
<p>Unfortunately, there are no simple formulae that can be used to analyze slope or bund <a href="https://en.wikipedia.org/wiki/Slope_stability_analysis" rel="nofollow">stability</a>. The Swedish slip circle method or similar slicing methods would be the most appropriate techniques as these methods are used for slopes composed of soils and I'm assuming the bunds you are referring to are made of soil or a combination of soil and stones.</p>
| 7664 | Is a particular bund/embankment geometry more stable than others? |
2016-02-29T15:48:24.223 | <p>Is there a good way to estimate the power consumption of an oven/heating device given the volume of the oven? I'm trying to do really, really dirty/quick math - I know that the average oven uses probably between 2-4000 Watts, but I'd be looking for a similar estimate with a volume of about 1.3 cubic <em>inches,</em> so it would be far, far less than a normal oven. </p>
<p>I'm under the assumptions that heating up and insulating a smaller space would be significantly easier (better than just the linear ratio in size) than a regular oven. Is there a way to figure out how much power input would be needed to heat up that much volume in ~20 minutes (or any time)? I'm not sure of all of the factors that would need to come into play (I'm sure insulation/heat retention is important, but I don't know a typical number for that). I'm not sure if convection or conduction would be better, but presumably the one that would use less power? The maximum temperature would be ~400F.</p>
<p>This is for a project I just thought of, which is why everything is somewhat vague. I'm basically just looking for the worst case ceiling, and the best case ideal to have something to start from. Thank you!</p>
| |mechanical-engineering|heat-transfer|chemical-engineering|power| | <p>If you can make a guess at the interior and exterior surface temperatures, then you can use the following simplified steady-state equation for conduction through an arbitrary shape. It is from A.F. Mills, "Heat Transfer 2nd Edition", equation 3.32:</p>
<p>$$
\dot{Q} = k \Delta T \sum_i S_i
$$</p>
<p>where $\dot Q$ is the rate of heat transfer, $k$ is the thermal conductivity of the material, $\Delta T$ is the temperature difference across the wall of the material, and $S_i$ is a "shape factor" for surface $i$, which is a function of the size and shape of the material through which the heat is flowing. If your oven will be a cube with interior width $W$ and uniform insulation thickness, $L$, on all sides, then it can be decomposed into two basic shapes; 6 walls and 12 edges. I'm ignoring the corners because the amount of heat lost there will be small. According to table 3.2 from the same book, the shape factors for these surfaces will be:</p>
<p>$$
\sum_i S_i = 6S_{wall} + 12S_{edge} = 6\frac{W^2}{L} + 12(0.54W)
$$</p>
<p>For example, I'll assume the interior of your oven will be a cube, ~1 inch (0.0254 meters) on all interior dimensions, with insulation thickness of ~1 inch. So, $W=L=0.0254m$. The total shape factor will be</p>
<p>$$
S_{tot} = 0.152+0.165 = 0.317m
$$</p>
<p>If the interior temperature is 400F (477K) and the outer surface is at 120F (322K), then the temperature difference will be $477K-322K = 155K$</p>
<p>Assuming fiberglass insulation, like the kind on a common kitchen stove, the thermal conductivity will be $\approx 0.11 \mathrm{\frac{W}{m\cdot K}}$.</p>
<p>Putting it all together:</p>
<p>$$
\dot{Q} = (0.11\mathrm{\frac{W}{m\cdot K}})(155K)(0.317m) = 5.4 W
$$</p>
<p>So, after the oven is heated up and running at steady state, you'll need about 6 watts to keep it at operating temperature. Like I said, this solution depends on knowledge of the surface temperature, so if it's way off from my guess of 120F, you'll need to re-do the calculation.</p>
| 7672 | How to estimate power consumption for an oven given volume? |
2016-03-01T10:48:56.533 | <p>On January 1991, the esteemed Swedish professor of optics filed a patent titled "<a href="http://www.google.com/patents/WO1996036893A1?cl=en" rel="nofollow">Coherence filter</a>", that blocks coherent light regardless of laser light wavelength using destructive interference and lets through ordinary incoherent light. Sometime in 1999, several University College of London professors published "Detection of coherent light in an incoherent background"which uses both optical and digital signal processing with an interferometer to measure the self-coherence function of the incident radiation to produce an interferogram which is subsequently narrow band filtered to yield a sinc envelope.</p>
<p>I was wondering if it is possible to make a tunable coherence filter with 1.8 nanometer bandwidth for visible light which lets through ordinary incoherent light with less than 5db loss. The reason I chose a bandwidth of 1.8 nanometer is because Edmund Optics manufactures part 87293 which is a spectrophotometer with that wavelength resolution.</p>
<p>In certain applications, sensing , filtering and protection from laser radiation is required for protection of eyes.</p>
| |optics|lasers|signal-processing| | <p>Since you mentioned "pilot," I'm going to guess you're looking for a way to block laser pointers from distracting airline pilots. It really should be sufficient to use the same thing used in laboratories, i.e. goggles which physically attenuate at and near the laser wavelength. Despite that patent's claims, I'm skeptical that their grating system will operate over the angular range (field of view) needed. </p>
<p>If you can post the actual requirements you're trying to meet, we might be able to suggest other alternatives.</p>
| 7694 | Is it possible to make a tunable coherent light filter with 1.8 nanometer bandwidth for visible light? |
2016-03-01T20:26:43.833 | <p><a href="https://engineering.stackexchange.com/questions/7680/unexpected-results-from-my-transfer-function">I recently asked a question about odd results from a transfer function</a>, and while looking over my work, I conjectured that maybe I'm having a more fundamental problem. Since this is a design problem I don't expect anyone to give a full-fledged answer, but any hints or prodding in the right direction would be appreciated. Here goes.</p>
<hr>
<p>I'm looking to model a one-axis gimbal in Simulink, which requires a transfer function to plug into my simulation. The gimbal has the following properties that I need to capture in a transfer function:</p>
<ul>
<li><p>Pushing the gimbal (giving it an impulse $\delta$) will move it some <em>finite distance</em> $x$. This is equivalent to having an impulse response that converges to a finite value.</p></li>
<li><p>The gimbal has a resonant frequency of $\omega$. Since I will be using active PID control on it in Simulink, if I attempt to drive the gimbal with a forcing frequency near $\omega$, I should expect high amplitude oscillations in the position $x$.</p></li>
</ul>
<p>These two responses combined will produce a very realistic expectation: pushing on the gimbal will move it (though damping will slow it down to a halt if the force is removed), but if you get near the resonance frequency of the structure, you'll excite the first elastic mode and have large-amplitude oscillations.</p>
<p>Below is a picture of what I <em>thought</em> the system should look like, approximated with springs and dampers (because finding the transfer functions of these elements is easy to compute):</p>
<p><a href="https://i.stack.imgur.com/wojVY.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/wojVY.png" alt="enter image description here"></a></p>
<p>My rationale is as follows: imagine just the mass $m$ and the damper $c$. This will produce the first response required, having the impulse response converge to a finite non-zero value. To force the resonancy condition, I should attach to the mass a spring-and-dashpot so that I can use the simple equation of:</p>
<p>$$\ddot{x} + 2\zeta\omega\dot{x} + \omega^2x$$</p>
<p>However, if you look at the link I've provided, I'm having a whole host of problems reproducing the two required conditions in my model. I think something is wrong with how I've modeled the system, but I don't quite know what's going wrong.</p>
<p>Any help would be greatly appreciated. Further, if anyone knows of papers that describe modeling of one-axis gimbal systems (I've found <em>plenty</em> on two-axis gimbal systems, but those are way too complicated for what I'm trying to accomplish here), that would also be helpful.</p>
| |mechanical-engineering|dynamics|systems-design| | <p>In my "day job" I do a lot of (mostly nonlinear!) dynamics modelling and simulation of mechanical systems. One thing that always raises a "red flag" for me is degrees of freedom with no mass, especially if they are at places where somebody is trying to apply a load to the model, as you have done.</p>
<p>I'm assuming that in the physical structure there is a rod of shaft connecting the "payload" (i.e. the mass $m$) to the "motor" (which applies the force $F$ ). The shaft will have a torsional stiffness (modelled by your $k$) but it also has some mass. The rotating part of the motor also has mass - possibly more than the connecting shaft. </p>
<p>The model should be capable of representing the basic behaviour of the system if you remove all the damping. </p>
<p>With no damping, your model has no "resonant frequency" at all. It is just a disconnected mass, with a massless spring attached to it. Trying to create a non-zero-frequency vibration mode from nowhere by adding the dampers isn't likely to work.</p>
<p>But if you add a second mass at the point where the force is applied, representing the real mass of the shaft and the rotating part of the motor, then (again ignoring the dampers) you now have a two-degree-of-freedom system with two modes. One mode is a zero-frequency mode corresponding to free rotation at constant angular velocity, the other is your "resonance" with a non-zero frequency $\omega$. </p>
<p>The two modes now represent the two "interesting" things that physical structure does, and two dampers will act to control those two modes. It's a reasonable guess that $c$ will mainly damp out the rigid body rotation, and $d$ will mainly damp out the unwanted vibration in the drive shaft.</p>
<p>Note: The $m$ and $k$ in your model would create a vibration mode if you applied a fixed-amplitude <strong>displacement</strong> to the end of the spring, instead of a <strong>force</strong>. That would be equivalent to mounting $m$ and $k$ on a vibration testing table and shaking the end of the spring. But if you try to use the model in that way, the damper $c$ is irrelevant, because it has a prescribed displacement at both ends (one end fixed, the other end moving). But the way I interpret your diagram and description, $F$ is meant to represent a force, not a displacement.</p>
| 7709 | Approximating a Gimbal using Springs and Dampers |
2016-03-02T19:11:53.217 | <p>I was wondering about the main difference between using heat pipes and a fluid-based cooling system. In both cases the heat is transported away from the heat-generating area to a place where the heat is dissipated. What is now the advantage of using a fluid instead of heat pipes? Is it the lack of heat transfer points (a fluid has no heat exchange points, except of the radiator and thermal pad, while heat pipes have to be connected)?<br>
Thus, when is it more beneficial to use a fluid cooling system instead of heat pipes?</p>
| |heat-transfer| | <p>It really depends on the application.</p>
<p>Heat Pipes are sort of a middle-ground between heat sink and liquid cooling technology. They take advantage of phase-changing a working gas/liquid to transfer thermal energy from a higher density of thermal energy (greater temperature) to a lower density. At any rate, no matter the method, you still are ultimately going to be transferring thermal energy to an ambient medium (namely air). Think about cars and motorcycles. Motorcycles are typically passively air-cooled and have engines with heat sinks built into the cylinders. There is little reason to employ a liquid cooling system here because they have plenty of exposure to air due to their smaller, open designs. Cars on the other hand, virtually always have a liquid cooling system with radiators mounted at the front of the vehicle, evacuating thermal energy from the closely packed engine cylinders to the more open and exposed front.</p>
<p>Because air (at 1 atmosphere) is such a poor conductor of thermal energy, how effectively you can get rid of heat is going to ultimately boil down to how much surface area is exposed to air, how much air you can move across the exposed surfaces, and the ambient temperature of the air.</p>
<p><strong>When to use a Heat Sink:</strong>
Most straight-forward option. Use a heat sink, add a fan and you're all set.</p>
<ul>
<li>Simplest, generally most cost-effective Solution</li>
<li>Smallest Possible Footprint</li>
<li>High thermal capacity and tolerance</li>
<li>The greater the difference in temperatures (object to be cooled and ambient temperature), the more effective the heat sink becomes</li>
<li>Air flow is either available or can be readily controlled</li>
</ul>
<p><strong>When to use Heat Pipes:</strong>
A little more planning may be involved, and space may get to be a concern, but all in all, heat pipes are a very effective means of cooling components within a specified range of temperatures.</p>
<ul>
<li>Minimal Hassle and Maintenance</li>
<li><p>Reduced Weight/Portability (for same capacity heat sink or liquid cooling)</p></li>
<li><p>Component to be cooled has relatively low surface area by which to transfer heat (such as a computer cpu)</p></li>
<li><p>Vertical Mounting Available: Many heat pipes MUST be mounted so that gravity can properly transfer the liquid after it condenses at the top of the unit.</p></li>
<li>Specific temperature range (to work properly, working medium must phase change). If component and ambient temperatures are too low, the heat pipes will only function as a mediocre heat sink. If component and/or ambient temperatures are too high, the gas won't be able to condense and thermal handling will be reduced.</li>
<li>Use when space isn't an issue or heat pipes can be custom designed to fit --- for example, the Acer Predator 21 X uses heat pipes produced specifically to fit inside of the laptop. Stock heat pipe solutions otherwise tend to be somewhat bulky to accommodate the working medium (liquid and gas) and transfer heat through a radiator.</li>
</ul>
<p><strong>When to use Liquid Cooling:</strong>
If you need to cool a component which is tightly packed amongst other components and otherwise has little access to air flow, liquid cooling is the way to go, so long as you have space elsewhere for a radiator and fan setup. Essentially, you can use this method to transfer heat from a space-restricted location to a more open location. Also, depending on the working liquid, liquid cooling will function optimally within a greater range of temperatures than heat pipes, but is also prone to more serious failures if temperatures become to low or high.</p>
<ul>
<li>When Space is a serious issue --- a heat exchanger can take up very little space and plumbing can be flexible</li>
<li>When Air flow is a serious issue --- components to be cooled do not have ready access to air</li>
<li>Heat needs to be transferred directly away from components without building up</li>
<li>Excellent thermal capacity and heat exchange rate</li>
<li>When properly used, can maintain temperature closest to ambient</li>
<li>May be combined with other cooling options to lower temperature below ambient</li>
</ul>
| 7718 | Heat pipes vs fluid cooling |
2016-03-03T03:14:46.340 | <p>I have the following to wheels that rotate on the same axis:</p>
<p><a href="https://i.stack.imgur.com/fscJy.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/fscJy.png" alt="enter image description here"></a></p>
<p>Assuming that they are stuck together, if I rotate the green one the red one rotates at the same speed and when I stop the green one the red one stops. </p>
<p><strong>I'm looking for the name of the mechanism that makes it so that when I stop rotating the green wheel and I hold it in place, the red one keeps turning.</strong></p>
<p>PS: I'm building a flywheel (represented here by the red wheel) and the green wheel here is connected with a belt to a crank. I want to make it so that the flywheel keeps turning even when the crank is fully stopped.</p>
| |mechanical-engineering|terminology|flywheels| | <p><a href="https://www.youtube.com/watch?v=nO5nOZVhx74" rel="nofollow">Another solution which is based on spur gears+springs rather than a clutch</a></p>
<p><a href="https://www.youtube.com/watch?v=-p8zrvU-cYE" rel="nofollow">Another solution which is based on spur gears+differentials rather than a clutch</a></p>
<p>Found using the keyword "rotation rectifier" (since it's the mechanical equivalent of an electrical rectifier in which alternating current is converted to direct current).</p>
| 7726 | What do you call a mechanism where rotation is transfered in one direction only? |
2016-02-25T14:35:57.193 | <h3>Bulding</h3>
<p>I live in a new (~10 years old) 2 story duplex home in the Eastern United States.</p>
<p>We have a completely interior bathroom on the first floor. It is between 2 other (exterior) rooms, the garage, and the firewall. It has a shower (plastic covering, not tile) but no tub.</p>
<p>We also have an unfinished basement. The foundation is concrete slab with CMU walls, which extend about 18 inches above ground level. There is a small (I think it's called a "port") window in one corner. At the opposite corner is a steel door that opens to a poured concrete stairwell topped with Bilco doors. It's an open space, I'd estimate a little under 1500ft<sup>2</sup>. We use about 2/3 of the space for general storage (decorations, unused furniture, food, the usual stuff) and 1/3 for my woodshop (hand tools, power tools, lumber, etc).</p>
<hr>
<h3>Risk</h3>
<p>According to <a href="http://www.tornadohistoryproject.com" rel="nofollow">http://www.tornadohistoryproject.com</a>, my County has seen 31 recorded tornadoes since 1950, with the following breakdown by scale:</p>
<pre><code>| Fujita Scale | Number |
| ------------ | ------ |
| 3 | 1 |
| 2 | 11 |
| 1 | 13 |
| 0 | 6 |
</code></pre>
<p>(the Tornado History Project gets it's data directly from NOAA but uses the term "Fujita Scale" consistently - I don't know whether that means all their data after 2007 is actually scaled on EF or not - they don't seem to specify).</p>
<hr>
<h3>Advice</h3>
<p>The general advice is always "get to a basement or an internal room on the lowest floor."</p>
<p>This makes sense. Roofs and exterior walls are for obvious reasons the first surfaces to be damaged and destroyed. Upper floors are more susceptible to structural damage. Direct wind and debris from tornadoes also is much less likely to extend below ground level, making basements in general a good place to be. In a lot of cases (as in mine) the basement is used as storage for food and/or provisions - helpful in the event of being trapped.</p>
<p>It's also clear why fully interior rooms are an acceptable alternative - they put more structure between you and the storm. Since wind-induced structural failures happen from the outside in, they're also the rooms most likely to be left standing.</p>
<hr>
<h3>Questions</h3>
<ol>
<li><p><strong>Are there any other factors to consider?</strong> What are some other risk/benefits that I've not considered?</p></li>
<li><p>Speaking in the most general sense, if you have both options, <strong>does one tend to be ubiquitously safer</strong>, overall?</p></li>
<li><p>Speaking specifically to my home, given that A) the historical risk of major structural failure is low and B) the basement is wide open and filled with sharp would-be missiles, <strong>which would you recommend for my family</strong>, and why?</p></li>
</ol>
| |structural-engineering|safety| | <p>The Wikipedia page on <a href="https://en.wikipedia.org/wiki/Tornado_intensity" rel="nofollow noreferrer">Tornado Intensity</a> has example pictures of damage.</p>
<p>Your location has mostly EF1 and EF2 tornados historically (I'm assuming EF scale).</p>
<h3>EF1</h3>
<p><a href="https://i.stack.imgur.com/XipYam.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/XipYam.jpg" alt="enter image description here" /></a></p>
<h3>EF2</h3>
<p><a href="https://i.stack.imgur.com/VyuIAm.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/VyuIAm.jpg" alt="enter image description here" /></a></p>
<h3>EF3</h3>
<p><a href="https://i.stack.imgur.com/14cIHm.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/14cIHm.jpg" alt="enter image description here" /></a></p>
<h3>Result</h3>
<p>Just looking at those pictures, I wouldn't want to be above ground if the basement is an option. You don't want to make a decision based on how strong you think the tornado is. Just go to the basement if it is possible.</p>
<p>It isn't likely that the wind will be strong enough to throw your tools around in the basement. And if it is, I certainly wouldn't want to be above ground.</p>
| 7728 | Tornado Safety: Basement or first floor interior room? |
2016-03-03T06:17:12.893 | <p>How strong is the adhesion between Ultra High Molecular Weight Polyethylene(UHMWPE) Fiber and Epoxy Resin normally? If we take the UHMWPE fiber sheet as it is without subjecting it to any chemical or plasma treatment (for improved surface bonding between UHMWPE fiber and Epoxy resin), can a laminate be formed through compression molding? Will there be any issues or loss of mechanical properties like impact resistance for UHMWPE?</p>
| |materials|composite|composite-resin|adhesive|epoxy| | <p>I am not a chemist or material scientist, but it is my experience that reliable adhesion to a smooth Polyethylene surface is nearly impossible even with "plastic" rated epoxy.</p>
<p>However where it is a fiber sheet, the epoxy, even with zero surface adhesion, will mechanically hold each fiber it encapsulates. I can not speak to what mechanical properties will result as it would be highly dependent on the fiber geometry. Some online research and testing is probably your best route.</p>
| 7732 | How strong is the adhesion between UHMWPE Fiber and Epoxy Resin? |
2016-03-03T06:21:07.887 | <p><a href="https://i.stack.imgur.com/34p8p.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/34p8p.png" alt="enter image description here"></a></p>
<p>Calculate the horizontal and vertical components of the reactions at the supports.</p>
<p>I found the horizontal force for the pin by adding the forces to 0 and I know I can find the vertical reaction of the pin by summing the forces in the y direction. But I get stuck on what do do to find the vertical reaction at the roller support which I need to calculate the vertical force at the pin. If I did a moment calculation at the pin do I count the 10kN force at the top?, don't you only count the stuff that's only in a straight line from the place you're calculating the moment from?.</p>
| |civil-engineering|statics| | <p>You don't always have to use moments to calculate reactions. In this situation it's possible just by balancing horizontal and vertical forces.</p>
<p>You have one force at an angle. With all our reactions at right angles, you should convert this force into it's two resultants (7.1kN in both the vertical and horizontal direction). </p>
<p>There is only a single horizontal reaction so this must solely resist the 7.1kN horizontal load. Therefore this reaction is 7.1kN. </p>
<p>The other force is 7.1kN upwards. You can proportionally divide this between the two supports, as we know that the closer support will take more of the load. The proportion would simply be 3/4 to the pin and 1/4 to the roller as we know the distances. </p>
<p>Dividing 7.1kN into these proportions means the rollers vertical reaction is 1.76kN downwards and the pin is 5.33kN downwards. </p>
<p>As a final check, taking moments around any point will give a result of 0 which is correct. </p>
<p><a href="https://i.stack.imgur.com/MBza9.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/MBza9.png" alt="enter image description here"></a></p>
| 7733 | Horizontal and vertical forces on supports |
2016-03-03T10:07:57.973 | <p>Is there some sort of equivalent standardized schematical (topological) diagram -such as Piping and Instrumentation Diagram, Electrical Schematics, Process Flow Diagrams, etc.- for the realm of mechanical engineering? I just need to draw the overall architecture that would describe the main components of the system and how they are connected, without going into too much detail. I have seen some diagrams but they seem to be ad hoc, as they have no fixed resemblance to each other. </p>
<p><a href="https://i.stack.imgur.com/O93J7.png" rel="noreferrer"><img src="https://i.stack.imgur.com/O93J7.png" alt="enter image description here"></a></p>
<p><strong>Image 1</strong>: I'm looking for something in this complexity. </p>
<p>Maybe there is more than one way to do this but it would be nice to be more universally understood. Looked into OMG SysML but that didt at first glance seem to have what i was looking for.</p>
| |mechanical-engineering|drafting|standards|diagram| | <p>The diagram most closely related to the original question seems to be a <strong>kinematic diagram</strong>. Kinematic diagrams are standardized in <strong>ISO 3952</strong>, although it seems that most often people are just copying what they have seen without reading the standard (which is quite old, but then this is common anyway).</p>
<p>The standard is in 4 parts and covers symbols, and alternate pictograms for such things as: </p>
<ul>
<li>Simple kinematic pairs such as used in bar linkages (part 1),</li>
<li>gears and cams (part 2),</li>
<li>geneva drives, ratchets, clutches and couplings (part 3),</li>
<li>belt/chain drives, lead screws, flywheels, bearings (part 4).</li>
</ul>
<p>If we re-draw the image in the original question:</p>
<p><a href="https://i.stack.imgur.com/lTJLH.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/lTJLH.png" alt="enter image description here"></a></p>
<p><strong>Image 1</strong>: Kinematic diagram, uses the same thin thick line style as normal engineering drawings.</p>
<p>ISO 3952 standard does not have a motor symbol. That is because they are concerned with motions not dynamics. Either way it is a human readable format so you can always add your own clarifications. </p>
| 7736 | Schematic Diagrams for Mechanical Systems |
2016-03-03T10:23:55.467 | <p>I'm trying to analyze a simple shaft supported by two bearings and subject to a single point load. But I keep coming across assumptions that only consider the vertical reactions and neglect the moment reactions of the bearings.</p>
<p><a href="http://faculty.ccri.edu/rpreliasco/website/engr%202050_files/hw9.pdf" rel="nofollow">This article</a> states this assumption in examples: 5-73 and 5-86. I've tried to find an explanation for this but keep finding examples that just state the assumption.</p>
<p>I'm at a loss not having my textbooks with me at the moment but from looking at examples I've noticed in some cases that the bearings take the form of a simple pin support hence the bending moment is zero at those points. See <a href="http://www.roymech.co.uk/Useful_Tables/Drive/Shaft_design.html" rel="nofollow">here</a> for an example (although no assumption stated).</p>
| |mechanical-engineering|applied-mechanics| | <p>Many bearings (single row ball bearings for example) are meant to allow slight misalignment of angle without significant counteracting torque. This allows them to be used in situations where there are two bearings supporting a shaft without having to have the alignment between the two bearing supports be extremely critical.</p>
<p>This means that unless there is significant bending in the shaft, the torque applied by the bearing would be negligible compared to the side or axial loading.</p>
<p>Many bearings that are meant to handle torque (double row angled ball bearings for example) are in reality composed of two (or more) bearings that are really using radial loads to apply the torque, so even these bearings could be modeled without using bending moments.</p>
| 7738 | Bearing Alignment Moment Reactions |
2016-03-03T15:42:45.900 | <p>I have this figure for my engineering drawing I am currently taking this semester I am quite baffled about this $1\dfrac{1}{4}$ radius circle with $1\dfrac{1}{4}$ Drill hole. How do I do that? as a reference I have this picture.</p>
<p><a href="https://i.stack.imgur.com/qeP7L.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/qeP7L.jpg" alt="enter image description here"></a></p>
| |drafting|autocad| | <p>The answer with 5/8 radius is irrelevant. Well not completely. Could help find answer.</p>
<p>The leg has a 1 1/4 in hole in it. Therefore the leg with a 1 1/4 radius will be a 2.5 inch dia leg half way rounded. </p>
<p>Now. How long and wide is your actual part?</p>
| 7746 | What is the Radius of 1 and 1/4 inch drill on a 1 and 1/4 Circle |
2016-03-04T01:43:38.007 | <p>Imagine a structure that is held above the water by pylons that are grounded on the ocean floor. In between these pylons is a pontoon that, when tides are rising holds and lifts a heavy weight. When the tide is falling the weight is supported by the structure, which can gradually let the weight down, and in doing so create electricity. Would this be a practical or even viable way to create electricity? Why or why not?</p>
| |mechanical-engineering| | <p>I agree with the comments about this question being a poor fit for the site, but one can nevertheless point out the common problem of tidal systems: small amount of vertical movement.</p>
<p>For example, say that you wanted to build a relatively small, 1 megawatt power plant based on tidal power. You have tidal movements of, say, 2 meters every 6 hours. Thus the weight you would need is:</p>
<p>$$ \frac{1 \mathrm{\,MW} \cdot 6 \mathrm{\,h}}{9.8 \mathrm{\frac{m}{s^2}} \cdot 2\mathrm{\,m}} = 10^9 \mathrm{\,kg}$$</p>
<p>If you made it out of steel, it would cost approximately 500 million dollars. And at 10 cents per kWh, this gives a repayment time on investment of 5000 years.</p>
| 7754 | Would this be a practical way to convert tidal energy into electrical energy? |
2016-03-03T05:15:29.027 | <p>In my mechanics of materials class we are required to be able to solve problems with bolts in shear. Unfortunately our textbook doesn't cover this subject we only had a little bit of information given on slides and I am having trouble finding other resources that expand on this concept. </p>
<p>Here's an example question for context:</p>
<p><a href="https://i.stack.imgur.com/THEGu.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/THEGu.png" alt="Eccentric loaded bolts"></a></p>
<p>So far these problems are straight forward. The total shear on a bolt is equal to the direct shear plus the torsional shear. First you get the direct shear component by distributing the force evenly on each bolt so $\displaystyle\tau_D=\frac{P}{4A}$ where $A$ is the cross sectional area of the bolt parallel to the force. Then you calculate the torsional shear by finding the torque $T$ of the force $P$ about the centroid of the bolts and $\displaystyle\tau_T=\frac{Tc}{J}$ where $c$ is the distance of the bolt from the centroid and $J$ is the polar moment of inertia of the bolts. The direction of the torsional shear is found perpendicular to a line from the centroid to the bolt in the sense of the rotation of the torque.</p>
<hr>
<p>I have no trouble doing these types of problems but I'm curious what happens if the bolts are not all the same diameter. </p>
<p>What I'm guessing is that the torsional shear will change because the centroid and polar moment of inertia will change but the torque stays the same. Is that correct? Is the force in the direct shear component still evenly distributed on the bolts? </p>
<p>In the above example if one bolt has a larger diameter than the others would the direct shear on the bolts still be $\displaystyle\tau_D=\frac{P}{4A}$? (obviously $A$ is now different in the larger bolt) or would the force in the direct shear be proportional to the relative size of the bolts?</p>
<p>Can anyone explain this to me to verify if my line of thinking is correct?</p>
| |mechanical-engineering|stresses|bolting| | <p>If the bolt diameters vary (rare in civil engineering constructions, we don't want the workers to get bolt sizes mixed up on the tenth floor from ground), then the centroid of the bolt areas will change, hence the value of J (all relative to the centroid). Therefore the torque component of shear will change since the value of J and c (distance from centroid to bolt) will vary.
The direct shear component will change as well, since the total direct shear (P) must be shared prorata according to the areas.
Do not forget that the eccentricity e must be modified according to the new location of the centroid.</p>
<p>Just as a side note, this is the classical method based on superposition of the two components of an eccentric force into a force applied through the centroid, as well as a moment (torque). It is considered conservative,... too conservative.
Kulak has done a good review of this method, as well as the ICR (instantaneous centre of rotation) method, which is considered realistic. However, the ICR method requires iteration to find the centre of rotation, and hence is not suited for a first course. The ICR method is realistic because it takes into account of a certain degree of slipping, and local failures in shear.</p>
<p>If and when you have time, here are some detailed readings:
<a href="http://www.boltcouncil.org/files/2ndEditionGuide.pdf" rel="nofollow">http://www.boltcouncil.org/files/2ndEditionGuide.pdf</a>
<a href="http://personal.stevens.edu/~shassiot/SteelDesign/AISC-DesignExamples/pdf/RP/CrawfordKulak1968.pdf" rel="nofollow">http://personal.stevens.edu/~shassiot/SteelDesign/AISC-DesignExamples/pdf/RP/CrawfordKulak1968.pdf</a></p>
| 7770 | Effect of varying diameter bolts on shear stress under eccentric loading |
2016-03-05T09:19:33.870 | <p><a href="https://i.stack.imgur.com/PzcDr.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/PzcDr.jpg" alt="Dislocation motion"></a></p>
<p>In this picture do we speak of (microscopic) plastic deformation if the edge dislocation has moved all the way to the right and emerges out of the surface? Or if you apply shear stress and stop when the edge dislocation has moved from A to B is it then plastically deformed aswell?</p>
<p><a href="https://i.stack.imgur.com/Nldvu.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Nldvu.gif" alt="Screw dislocation"></a></p>
<p>How about with this situation? Is it only plastically deformed when the dislocation line has moved all the way to the left and thus the screw dislocation emerging from the crystal?</p>
<p>Ow and one last small question:
'before and after the movement of a dislocation through some particular region of the crystal, the atomic arrangement is ordered and perfect; it is only during the passage of the extra half plane that the lattice structure is disrupted'</p>
<p>This particular region is that the region B to D, when the dislocation starts at A and moves untill emerges? And if the dislocation motion stops while the edge dislocation is at position B or C then this region isn't ordered and perfect? Because there is a linear defect in it?</p>
| |materials|metallurgy| | <p>Plastic deformation is defined as non-recoverable deformation. When a stress is applied to the material, it will first deform elastically. When the yield limit is reached, it will deform plastically which for metals means by dislocation motion. If during dislocation motion the stress is removed, the dislocation does not vanish, and the atoms stay in their new locations. Therefore, the answer to your questions is that any dislocation motion of any distance constitutes plastic deformation.</p>
<p>For your last question, when the dislocation has caused any strain, the lattice is imperfect. So the first two subfigures of the top image are imperfect, while the last is "perfect" in the sense of no lattice strain.</p>
| 7781 | Plastic deformation due to the motion of dislocations |
2016-03-05T14:00:00.670 | <p>I think that some of you know what is rolling resistance. It is force that works in opposite direction to roll and is caused due to deformation of tire. It also affects car wheels.
There is one problem with it. I have found 2 equations around:</p>
<ul>
<li><p>$F_\mathrm{roll} = -V_\mathrm{car} * C_\mathrm{rr}$</p></li>
<li><p>$F_\mathrm{roll} = W_\mathrm{load} * \sin\alpha * C_\mathrm{rr}$</p></li>
</ul>
<p>However, none of them seems to be right one. First one seems to be incomplete, inaccurate, but second one doesn't depend on speed making car shake when being in standstill.<br>
I need to know what is accurate(but still game-compatible, not too heavy) equation for rolling resistance.</p>
| |automotive-engineering|deformation|wheels| | <p>to calculate fulcrum based rears wheels
my formula : f²(w²divide by unit decimal digits)
derive : V = mass gravitational force x psf vol
: f²(w² x w¾)
as value to minimal point to sling aerodinamic idea</p>
<p>time by ff (force G break)
tq</p>
| 7784 | Wheel rolling resistance |
2016-03-07T10:33:32.003 | <p>I have a question on my work which I cannot figure out. I have three chains which support a beam. On the beam acts a distributed load. I want to calculate how much load will go to each chain.</p>
<p>I know the sum of each force/moment is equal to zero. The magnitude of the distributed load ($w$ Newton/meter). The length of the beam $L$ (Meter) and the sub-distances ($x_a$, $x_b$, $x_c$, $x_d$ Meter). $F_{vt}$ (newton) represents the concentrated force of the distributed load ($F_v = w*l*0.5$)
and $F_{av}$ is the concentrated load in section $x_a$.</p>
<p>So if you go through all the point you get four formulas. The first one for point $a$ is: </p>
<p>$$\begin{gather}
\sum M_a = 0\\
F_b(x_b) * F_c(x_b+x_c) - (w *L)* \dfrac{L}{2} = 0
\end{gather}$$</p>
<p>But in $x_a$ section forces are also action, which should a moment around point $a$. But how can I calculate the resultant force ($F_{av}$) for section $x_a$? </p>
<p>I have tried the following but I get a net force in this section of 0 N ($F_{av} = w\cdot\dfrac{0}{L} = 0$). So there is no force acting in section $x_a$?</p>
<p>Can somebody help me out with this part?
<a href="https://i.stack.imgur.com/VXRiK.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/VXRiK.png" alt="enter image description here"></a></p>
| |mechanical-engineering| | <p>I have looked at the other solution that you mentioned. That solution would be ok provided that the two assumptions hold. That is, the beam is not flexible, and the chains are adjusted before any load is applied, to be of equal length and carrying equal almost-zero loads. Depending on the Load/Extension value of the chains, this may involve very small adjustments to their initial lengths, and resulting tensions. Without the length and elastic modulus of the chains, and knowledge of the temperature stability of the support structure, it is all guesswork, but I expect that an adjustment in length of the centre chain of about 0.1 mm would be enough for it to go from carrying no load, to carrying the whole load.</p>
<p>The loads are so low, is an accurate answer really necessary? Could you change it to only two chains? Could you hinge the beam where the middle chain is attached?</p>
<p>Have a look at strand7.com, and their demo download. This is a fully working Finite Element package, limited to 20 beams, so ok for your problem. In my personal opinion it is the best FE package for user-friendliness. (Or - it is dead easy to understand and use.)</p>
<p>Sorry it took so long to come back. I am currently struggling to understand how emails work, and how ubuntu works.</p>
| 7801 | Distributed load with an offset? |
2016-03-07T16:45:06.763 | <p>Wikipedia says that <a href="https://en.wikipedia.org/wiki/Anti-roll_bar" rel="nofollow">an anti-roll bar is a device that balances the weight of car between the wheels, preventing it from rolling</a>. I want to simulate this in a game, but I don't know the equations behind it. I need to know how much weight the anti-roll bar transfers to each wheel. That means that I need to obtain a value that I can add to the load of wheel to manage it.</p>
<p>$W_{load} = W_{static} + W_{transfer} + W_{anti-roll}$</p>
<p>The static load is dependant on weight distribution. I do it like so:</p>
<p>$W_{static} = L_{0} / L * m * g / 2$</p>
<p>Where $L_0$ is the longitudinal distance from the opposite (e.g., if checking the rear wheel, than this is the front wheel's distance) wheel joint axle to the centre of gravity, $L$ is wheelbase and $m$ is mass of the car.</p>
<p>The $W_{transfer}$ is acceleration-dependent.</p>
<p>How can I obtain the $W_{anti-roll}$ value?</p>
| |automotive-engineering|applied-mechanics| | <p>I found my answer <a href="http://www.edy.es/dev/2011/10/the-stabilizer-bars-creating-physically-realistic-stable-vehicles/#comment-30477" rel="nofollow">here</a>. So I basically need to get the difference between compression in the two opposite wheels and multiply this difference by the bar's stiffness to calculate changes of suspension force on each side. Pretty simple. It is even simpler to do in my setup than what's described in the link, since I actually add this value to the suspension force.</p>
| 7810 | Find the load applied to each wheel by the anti-roll bar |
2016-03-08T15:12:43.860 | <p>I recently went on a tour of a glass manufacturing facility who produce bottles for the prestige alcohol market. I was informed on the tour that red glass cannot be directly produced - any bottles that I have seen had made by applying a spray to a plain finished bottle as opposed to adding the colour into the molten glass. </p>
<p>Why is this?</p>
| |materials|manufacturing-engineering| | <p>It seems likely that costs associated with the material prevents economical production of red glass. Rather than a matter of possibility, it may be a matter of viability. Without knowing more about how the glass is produced or the specific materials used for their bottles, it is difficult to say the exact reason. The best answer I can think of is that the only glass I am aware of which selectively transmits red light requires gold nanoparticles suspended in the silica glass network. Such glass is referred to as Cranberry Glass or Gold Ruby Glass (<a href="https://en.wikipedia.org/wiki/Cranberry_glass" rel="nofollow">Wikipedia</a>).</p>
<p>It isn't clear how much gold is required, though the abstract of <a href="http://www.sciencedirect.com/science/article/pii/S1296207408001040" rel="nofollow">this paper</a>, which uses an alternate production process, suggests about 0.2% by weight. The price of gold is about 40,000 USD/kg per the current spot price from <a href="http://www.apmex.com/spotprices/gold-price" rel="nofollow">apmex.com</a> as of 08 March 2016 at ~2:45 PM CST. Thus a glass bottle weighing 1 kg would have an added cost of about 80 USD from the gold alone. Expanding the process to mass-produce such bottles would require storing and shipping significant quantities of gold, which may add logistical and overhead costs for security. There would also be equipment requirements for producing colloidal aqua regia in the proper concentration and drying, handling, and mixing the nanoparticles into the liquid glass.</p>
<p>As a side note, flint glass (<a href="https://en.wikipedia.org/wiki/Flint_glass" rel="nofollow">Wikipedia</a>) contains zirconium and/or titanium. I can't think of any reason that either element would interfere with the cohesion of the gold particles, but I also can't strictly rule it out. If such a thing did happen, then Flint glass specifically might be impossible to make red by the use of gold nanoparticles.</p>
| 7827 | Why can't red flint glass be produced? |
2016-03-08T21:57:18.477 | <p><a href="https://i.stack.imgur.com/3XnKn.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/3XnKn.png" alt="enter image description here"></a></p>
<p>So above I have a memory design including a 74LS138 3-to-8 multiplexer which enables 8 memory banks comprised of 2732s which are 4Kx8 in size.</p>
<p>My question is why the memory addressing starts F8000 instead of F0000?</p>
| |electrical-engineering| | <p>Addresses are: A19 18 17 16 15 .... </p>
<p>This expands on Dave Tweed's correct but brief comment:</p>
<p>F8000 = 1111 <strong>1</strong>000 0000 0000 0000 <- A15 = 1<br>
F0000 = 1111 <strong>0</strong>000 0000 0000 0000 <- A15 = 0 </p>
<p>For F8000, A15 is high<br>
For F0000, A15 is low</p>
<p>As Dave Tweed notes, the decoder is only selected when all inputs to the '30 NAND gate are high.<br>
As A15 is one of the inputs it must be high for the decoder to be selected.<br>
As A19 18 17 16 15 must be high the applicable address is F8xxx as above .</p>
| 7834 | Why do these memory spaces start at F8000 instead of F0000? |
2016-03-08T22:12:29.320 | <p>I have a small assembly with several parts misaligned. By turning the grey wheel, the grey block is to rotate about its axis (Axis 1). The problem I have is that the grey wheel can't be turned further anti-clockwise given the current position. <a href="https://youtu.be/ozMX3gkA3W8" rel="nofollow noreferrer">It seems to be jammed</a>.</p>
<p><a href="https://i.stack.imgur.com/G9IDI.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/G9IDIt.png" alt="enter image description here"></a>
<a href="https://i.stack.imgur.com/Hbk1L.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Hbk1Lt.png" alt="enter image description here"></a>
<a href="https://i.stack.imgur.com/cEXqc.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/cEXqct.png" alt="enter image description here"></a></p>
<p>The red connecting rod can freely slide in the yellow bore, which itself can rotate with the rod in the grey bore. I'd expect the red rod to extend further and turn the yellow connector. </p>
<p>Does anyone know why the wheel might be jammed or how it could be modified to allow continuous rotation?</p>
| |mechanical-engineering| | <p>It isn't clear what is constraining the motion of the gray ring (blue in the video), but it seems to be rotating on its own axis. If so, the mechanism jams because the pin at the end of the red rod hits the end of the slot in the orange L-bracket. Regardless of the position of the yellow piece WRT the gray bar, there is a fixed maximum distance between the yellow pin and the red pin.</p>
| 7835 | Unexpected jamming of a misaligned assembly |
2016-03-08T23:25:51.293 | <p>I want to perform a structural analysis on a portal frame using a finite element solver to get a displacements and reactions for a load. The software library only accepts loads applied to node, not to the member. The node loads have 6 parameters: 3 translation (x,y,z) and 3 rotation (xx,yy,zz). My load is a uniform load of 1 kN/m along one rafter. </p>
<p><a href="https://i.stack.imgur.com/lhqwy.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/lhqwy.png" alt=""></a></p>
<p>Is it possible to convert the uniform member load to point loads that I can apply at the nodes? </p>
<p>Portal frame dimensions:</p>
<ul>
<li>span: 10 m</li>
<li>eave height: 5 m</li>
<li>apex height: 6.34 m</li>
<li>rafter length: 5.18 m</li>
<li>roof pitch: 15°</li>
</ul>
<p>EDIT: fixed apex height </p>
| |beam|structural-analysis| | <p>If your FEM S/W can't handle distributed loads, throw it away and find something which does. It's a pain having to convert DL to NL all the time. Your software should make your life easier, not harder.</p>
| 7837 | Is it possible to convert uniform loads to point loads? |
2016-03-09T20:09:46.613 | <p>I'm planning to build a head-mounted camera stabilizer for pro GoPro camera. The idea is that when I tilt my head the camera would stay at a user-set angle. </p>
<p>I'm using an accelerometer and a gyro to measure camera angle $\theta$ and angular velocity $\dot{\theta}$, and some motor to control the camera angle. I derived the following differential equation for the system (which is basically an inverted pendulum) where $l$ is length of the rod, $\theta$ is the angle of the camera due to gravity, $m$ is the mass, $g$ is the gravitational acceleration, $b$ is damping coefficient (due to friction) and $\tau$ is torque. $\tau_{motor}$ will be the input of the system. </p>
<p>$\tau_{total} = \tau_{motor} - \tau_{gravity} - \tau_{friction}$</p>
<p>$I\ddot{\theta} = \tau_{motor} - mgl\sin\theta - b\dot{\theta}$</p>
<p>$ml^2\ddot{\theta} = \tau_{motor} - mgl\sin\theta - b\dot{\theta}$</p>
<p>$\ddot{\theta} = \frac{1}{ml^2}\tau_{motor} - \frac{g}{l}\sin\theta - \frac{b}{ml^2}\dot{\theta}$</p>
<p>From here I'll continue designing the control algorithm using state-space methods such as state-feedback control (I choose camera angle $\theta$ and angular velocity $\dot{\theta}$ as states and $\tau_{motor}$ as input). </p>
<p><a href="https://i.stack.imgur.com/uKxCa.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/uKxCa.jpg" alt="enter image description here"></a></p>
<p>Now here's the question:</p>
<p>I want to test how my control system behaves when a person tilts his head (I'll be using Matlab and Simulink). How should I incorporate the tilting of the head (or rotating of the body of the motor since the motor will be attached to head) into my control design? I guess it can be thought as a disturbance to system (or another input)? How should it be incorporated to the system model (meaning the differential equation above)? Should there be maybe more differential equations? I thought it would be nice to give the system the the angle of the body of the motor (meaning the angle of the head) as function of time and observe how the control system works. In other words I would like to see how the controller works if I tilt my head for example back and forth.</p>
<p>On the other hand, is this a good approach or should I just use another torque $\tau_{disturbance}$ to test the controller? I'm pretty sure that it isn't the same as tilting the motor body, but would perhaps lead to good enough model for the controller?</p>
| |control-engineering|modeling|control-theory| | <p>There's a couple questions you can ask:</p>
<ol>
<li>Can you control the signal? If not, then it's either a noise or disturbance input. </li>
<li>Is it affecting your system (plant), or is it affecting your measurements? If it's affecting the system, it's a disturbance input; if it's affecting the measurements, it's noise. Note that noise can be colored, biased, periodic, etc. That is, pretty much any signal can be noise or a disturbance. </li>
<li>Does the disturbance have enough power that is effectively a position regulator?</li>
</ol>
<p>Regarding point 3, if the power output of the human neck is comparable with the power output of your stabilization system, then its position will vary with reaction forces applied by your stabilization system. If the power output of the neck is significantly higher, then it becomes effectively rigid and is better modeled as a position regulator.</p>
<p>Based on these questions, I would guess that the power of your stabilizer is much less than the power of the neck, so I would treat the neck as a positioner. You can't control it, but it affects the plant, so it's a disturbance input. </p>
<p>I would include it with the definition of angular acceleration $\ddot{\theta}$, such that:
$$
\ddot{\theta} = \frac{1}{ml^2}\tau_{motor} - \frac{g}{l}\sin\theta - \frac{b}{ml^2}\dot{\theta} + \ddot{\theta}_{\mbox{head}} \\
$$</p>
| 7846 | System model for motor controlled inverted pendulum with rotating motor body (used in camera stabilizer application) |
2016-03-10T02:33:49.343 | <p>So there's a permutation of rotation and translation each being either fixed or free for a column. What do each mean?</p>
<p>If the column is toe nailed to a plywood / beam underneath which is it?</p>
<p>If it's in a steel bracket nailed to a column below what is it?</p>
<p>If it has a 2x6 plate of a wall on top nailed to it what is it?</p>
<p>Does bracing in the middle of the column for either affect it?</p>
<p>My case is a 8x8 wood post</p>
| |structural-engineering|structural-analysis|wood|columns| | <blockquote>
<p>So there's a permutation of rotation and translation each being either fixed or free for a column. What do each mean?</p>
</blockquote>
<p>To say that a specific degree of freedom of a support is "fixed" is to say that the support cannot translate along, or rotate about, that axis. For example, if it is stated that the translation of a beam is fixed in the X axis at one end, that means that the beam cannot translate, or move, along the X axis (in the direction of +/- X) at that end. And similarly for rotation, if it is stated that the rotation of a beam is fixed about the X axis at one end, that means that the beam cannot rotate about the X axis at that end.</p>
<p>When we design things, we often assume an ideal, or perfect, boundary condition. For example, idealizing a beam as being "simply supported," which means we assume it can rotate about either end and translate along its longitudinal axis at one end, usually enveloped the beam design forces of bending moment and shear force.</p>
<p>In reality nothing we design is ever actually perfectly rigid (fixed) or perfectly free to translate or rotate, we make those assumptions to simplify the design process. In practice, we usually model support conditions that are a combination of fixed translations and free rotations, or vice versa. A very common boundary condition is called "pinned." A pinned support is a support that is not allowed to translate (fixed for translation) but is allowed to rotate freely about any axis (free for rotation).</p>
<p>To answer you specific questions, for an 8x8 post,</p>
<blockquote>
<p>If the column is toe nailed to a plywood / beam underneath which is it?</p>
</blockquote>
<p>Pinned (free to rotate, not allowed to translate).</p>
<blockquote>
<p>If it's in a steel bracket nailed to a column below what is it?</p>
</blockquote>
<p>Depends on the bracket and attachment to the bracket, but treating as pinned (free to rotate, not allowed to translate) will usually be conservative.</p>
<blockquote>
<p>If it has a 2x6 plate of a wall on top nailed to it what is it?</p>
</blockquote>
<p>Pinned (free to rotate, not allowed to translate) .</p>
<blockquote>
<p>Does bracing in the middle of the column for either affect it?</p>
</blockquote>
<p>No</p>
| 7850 | For a column, what does "rotation fixed and translation fixed" mean |
2016-03-10T07:36:36.050 | <p>In structural dynamic mode analysis, one would have to carry out modal analysis in order to get different modes that contribute to the response of the seismic ( or wind) force. <a href="http://www.vibrationdata.com/tutorials2/ModalMass.pdf" rel="nofollow">Here</a> is a nice summary. </p>
<p>In a large structural model, there are so many modes, and we can't take them all. So we can only take the first few prominent modes ( with low frequencies), assuming that the modes will grow less and less important as the frequency becomes higher and higher. </p>
<p><strong>Then the issue arises</strong>: how to judge the relative importance of different modes ( whatever the term 'importance' may mean?) I find that the often-cited <a href="https://en.wikipedia.org/wiki/Structural_dynamics#Modal_participation_factor" rel="nofollow">mode participation factor</a> </p>
<p>$\Gamma=\frac{\sum M \phi}{\sum\phi^T M \phi}$</p>
<p>doesn't work, because it depends on how we normalize the eigenvector, as one can readily see from the above formula. </p>
<p>Is there any other factor that allows us to judge whether a mode is more important or not?</p>
| |structural-analysis|seismic| | <p>You should be normalizing your eigenvector so that the generalized mass matrix (defined by $ \hat{m} = \phi^T M \phi$) is the identity matrix, and therefore, the generalized mass of the <em>r</em>th mode (defined by $ M_r = \phi^T_r M \phi_r$) has the value of 1. This should give you consistent modal participation factors.</p>
<p>The modal mass participation ratio is widely used as the metric to determine the relative significance of modes in a modal response spectrum analysis. It has even been codified in many of the codes of record. For example, ASCE 7-05 (7-10 is at home, sorry), Section 12.9.1 states that,</p>
<blockquote>
<p>The analysis shall include a sufficient number of nodes to obtain a combined modal mass of at least 90 percent of the actual mass in each of the orthogonal horizontal directions of the response considered in the model.</p>
</blockquote>
<p>And to elaborate on what I think may be an underlying question in your query (and the code stipulation), you calculate total effective mass as, $$ M_{eff} = \Sigma\Bigg[\frac{\Gamma_i^2}{\hat{m}_{ii}} \Bigg] $$</p>
<p>If $ M_{eff} $ is greater than 90% of your overall system mass, you've considered an appropriate number of modes.</p>
<p>Likewise, the effective mass participation of each mode may be used as a guide to determine the relative "importance" of each node, $$ m_{eff,i} = \frac{\Gamma_i^2}{\hat{m}_{ii}} $$</p>
<p>Dynamics of Structures by Chopra is a really good reference if you are attempting to do this all by hand.</p>
<hr>
<p><strike>Edit: Check out this <a href="http://www.vibrationdata.com/tutorials2/ModalMass.pdf" rel="nofollow">pretty good discussion and examples</a> written by Tom Irvine.this</strike> <em>Oops, this was your original link!</em></p>
| 7853 | How to calculate the relative contribution of each mode in seismic dynamic mode analysis? |
2016-03-10T10:44:40.557 | <p>How can I find out the function equations in a diagram with 5 curves? Measuring by hand is very inaccurate and takes a lot of time. Any other possibilities?<a href="https://i.stack.imgur.com/sZIQo.png" rel="noreferrer"><img src="https://i.stack.imgur.com/sZIQo.png" alt="Example: See uploaded Picture!"></a></p>
| |mathematics| | <p>You have a single-valued function dependent of two variables. There are many ways to model that.</p>
<p>If you know something about what this function represents, then going back to the physics might yield a useable equation with only a few coefficients. Digitize a bunch of points, throw them at a least-squares error minimizer and see how close the result is.</p>
<p>If you know nothing more about the physics behind the function, then the standard answer is a polynomial. From the general look of the curves, I'd say you need at least a 3rd order polynomial (a cubic). With two independent variables and the various crossover terms, that comes out to 10 coefficients to adjust to get these curves.</p>
<p>Either way, I'd digitize 10 values for each curve, one where they hit each marked X division. It really wouldn't take long to just sit down and do it. That gives you 50 points, which should do a reasonable job of allowing a least-squares error solver to divine the coefficients for you. For the polynomial case it would be trying to solve for 10 value. In the physics modeling case, hopefully fewer.</p>
| 7855 | How can I determine a function equation from a graph image? |
2016-03-10T19:26:09.497 | <p>I am bit confused. The slip direction is the direction of movement of a dislocation which is denoted by the burgers vector.</p>
<p>This makes sense in an edge dislocation, because the stress is applied perpendicular to the dislocation line and the movement of the line is in the direction of the applied stress. The burger vector is also perpendicular to the dislocation line for edge dislocation. So it makes sense that the direction of motion corresponds to the direction denoted by the burgers vector. </p>
<p>But for screw dislocation, the motion of the dislocation is perpendicular to the applied stress and since the burger vector points in the direction of the dislocation line; the burger vector cannot point in same direction as the direction of movement?</p>
<p>The professor even said that the direction of motion for a screw dislocation is perpendicular to the burgers vector. So how can a burger vector then correspond to the direction of slip when this direction of slip is the direction of movement?</p>
<p>So the points that confuse me:</p>
<ul>
<li><p>slip direction: specific direction along which dislocation motion occurs</p></li>
<li><p>burger vector: direction corresponds to a dislocation's slip direction</p></li>
<li><p>screw dislocation: direction of motion is perpendicular to applied stress and the motion is perpendicular to the burger vector</p></li>
</ul>
<p><a href="https://i.stack.imgur.com/qHRls.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/qHRls.gif" alt="screw dislocation motion"></a></p>
<p>So these 3 things conflict with each other:
the motion of the dislocation is here perpendicular to the burger vector (and i have heard is always perpendicular for screw dislocations), so how can a burger vector denote the slip direction if the slip direction is the direction of the motion of the dislocation? For it to denote the motion of the dislocation shouldn't it align/be parrallel?
Or is that the mistake I am making, that a burger vector tells about direction of the dislocation motion, but that it doesn't mean that the motion is in the same direction?
That there is just always a fixed relation between the two depending on the type of dislocation, but that the relation isn't always parrallel.
It is 90 degrees for screw and 0 degrees for edge related to the dislocation motion. So that indeed the burger vector says something about the direction of the motion but just that it doesn't say that they always in same direction?</p>
<p>Because again clearly here the dislocation motion is not in the same direction as the direction of the burger vector;
The motion is from front to back, while the burger vector points in direction of the shear stress in this picture</p>
| |materials| | <p><strong>Summary: For an edge dislocation, the Burgers vector is parallel to dislocation motion. For a screw dislocation, the Burgers vector is parallel to the dislocation. The Burgers vector is always parallel to slip.</strong></p>
<p>The diagram below shows both edge and screw dislocations in an indealized cubic lattice. The edge dislocation is on the front face and the screw dislocation is on the right face. Burgers vectors are denoted with white-tipped arrows, and point in the same direction for the two dislocations shown. I have added red arrows that indicate applied shear. Slip occurs along the resolved shear direction, which in this case happens to also be the applied shear direction. Note that slip is the same for both dislocations, and the Burgers vector points in the same direction as slip in both cases. Note that the edge dislocation itself is perpendicular to slip while the screw dislocation is aligned with slip.</p>
<p>If slip continued, the edge dislocation would proceed in the same direction as slip. The screw dislocation would recede away from the front plane, toward the back plane. It must move that direction in this case because otherwise the Burgers vector would have to get longer as strain proceeds and new edge dislocations are introduced, which is impossible. An analogy would be tearing a piece of paper. As you pull the ends of the tear further apart, the tip of the tear recedes along the length of the paper.</p>
<p><a href="https://i.stack.imgur.com/3VRRZ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/3VRRZ.png" alt="Dislocation diagram showing shear in red."></a></p>
<p>The image is a modified version of an image found at <a href="http://www.geology.um.maine.edu/geodynamics/AnalogWebsite/UndergradProjects2010/PatrickRyan/Pages/Background.html" rel="nofollow noreferrer">www.geology.um.maine.edu</a>. Original credit: Passchier and Trouw, pg 33 (2005).</p>
| 7864 | Relation between burgers vector and direction of motion for dislocations? |
2016-03-10T19:42:15.843 | <p>For my racing game I am using a simplified Pacejka Magical Formula:</p>
<p>$\mu_{lateral} = \mu_{0} * \mu_{1} * D * \sin(C * \arctan(B * x - E * (B * x - \arctan(B * x))))$</p>
<p>Where:</p>
<ul>
<li>$\mu_{lateral}$ is sideways coefficient of friction</li>
<li>$\mu_{0}$ is wheel friction coefficient</li>
<li>$\mu_{1}$ is surface coefficient friction</li>
<li>$B$ is wheel lateral stiffness</li>
<li>$C$ is wheel lateral peak factor</li>
<li>$D$ is wheel lateral shape factor</li>
<li>$E$ is wheel lateral curvature</li>
</ul>
<p>However, I don't know what are realistic values for $B, C, D, E$. I know well the first ones and I also assume that $C$ may be around <code>1.1</code>. However, what's realistic stiffness, shape factor and curvature in sideways directions for different types of real-life tires?</p>
| |automotive-engineering|friction|wheels| | <p>The actual parameters for real tyres tend to be heavily protected intellectual property, you won't find those anywhere. However, have a look at these various pages, they give indicative values of the various parameters:</p>
<ul>
<li><a href="http://www.edy.es/dev/docs/pacejka-94-parameters-explained-a-comprehensive-guide/" rel="nofollow">Pacejka ’94 parameters explained – a comprehensive guide</a></li>
<li><a href="http://www.racer.nl/reference/pacejka.htm" rel="nofollow">Pacejka's Magic Formula</a></li>
<li><a href="http://www.biblioteca.uma.es/bbldoc/articulos/16512881.pdf" rel="nofollow">An Alternative Method to Determine the Magic Tyre Model Parameters Using Genetic Algorithms</a> (uses a Michelin XZA 11R22.5)</li>
</ul>
| 7866 | Pacejka Formula - natural values |
2016-03-11T01:51:25.283 | <p>Say I have a spring in the shape of the backbone of DNA. If I unwind the two parts which would have more potential energy when pushed all the way down, the double helical or both of the individual springs combined?</p>
| |design|springs| | <p>I think two equal springs would test very close to a spring with double the spring constant. See <a href="https://en.wikipedia.org/wiki/Series_and_parallel_springs" rel="nofollow noreferrer">adding springs in parallel equation</a>.</p>
<p>Two springs may be somewhat less material efficient than a single spring; because the cross section of a spring wire has an <a href="https://en.wikipedia.org/wiki/List_of_area_moments_of_inertia" rel="nofollow noreferrer">area moment of inertial</a> (just like a beam) that provides the restoring force and the equation is not linear. Think of it like the strength of a cantilever wood 2x4 vs the lower cantilever strength of an equivalent mass of four wood 1x2 members.</p>
<p>A reason you may want to do it would be to achieve a desired spring constant with springs you already have or within what is available. However, most of the time this is done with nested springs because the ends are typically flattened and would not permit another spring without modification to the ends. Another issue would be spring travel and fill. Notice that the green and brown die springs below would not be able to even fit a duplicate.</p>
<p><a href="https://i.stack.imgur.com/xy81I.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/xy81I.jpg" alt="enter image description here"></a>
<br><br><br>
One spring design that does have space advantages over a standard coil is a <a href="https://en.wikipedia.org/wiki/Wave_spring" rel="nofollow noreferrer">wave spring</a>. It has a shorter height than a coil spring of equal spring constant. Here are some <a href="https://www.rotorclip.com/wave_spring_advantage3.php" rel="nofollow noreferrer">wave spring applications</a>.
<a href="https://i.stack.imgur.com/Oa23L.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Oa23L.png" alt="enter image description here"></a></p>
| 7872 | Would a double helix spring be more or less effective than two springs? |