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2017-09-06T17:12:13.427 | <p>I'm currently writing a simulation environment in python/c for heat networks and got to the point where I'm implementing the controls for the environment.<br>
The PID controller is already implemented and should be working, but now I stumbled upon a question where I need some advice which way is the best or most common way to go.</p>
<p>Let's say I've got a heat generator with the power $P=40\,kW$, through which water with a heat capacity $c_p \approx 4000\,\frac{J}{kgK}$ is flowing. The inlet flow temperature is $T_{in} = 50\,^°C$, current outlet flow temperature at timestep $n$ is $T^{n}_{out} = 70\,^°C$ and the massflow is $\dot m^{n} = 0.5\,\frac{kg}{s}$. The desired outlet flow temperature $T^{*}_{out} = 90\,^°C$.<br>
For a constant $c_p$ I need to tell the pump to reduce the massflow to $\dot m^{n+1} = 0.25\,\frac{kg}{s}$ to get $T^{n+1}_{out} = 90\,^°C$ (assuming infinitely long timesteps to get a steady state condition).</p>
<p>To make the example fast and easy, I assume to only have a P-controller with a proportional factor of $K_p = 1.2$ and a <em>pump control algorithm</em>, which is an equation to correlate massflow and temperature, but has no P/I/D-parts!
The <em>pump control algorithm</em> equation is:
$$\dot m = \frac{P}{c_p \left(T_{out} - T_{in}\right)}$$</p>
<p>Now I've got two possibilites which values to pass as <em>setpoint</em>-values to the P(ID)-controller:</p>
<ul>
<li>I can pass the desired outlet temperature $T^{*}_{out} = 90\,^°C$ as setpoint-value and the current outlet temperature $T^{n}_{out} = 70\,^°C$ as process-value to the controller.<br>
This means that the control variable output of the controller will be $C = \left(90\,^°C - 70\,^°C\right)K_p = 24\,^°C$.<br>
This will be fed into my <em>pump-control-algorithm</em> equation with $T_{out} = T^{n}_{out} + C$ which then yields $\dot m^{n+1} = \approx 0.2272 \frac{kg}{s}$
and will then be set as the massflow.</li>
<li>Or I can pass the desired outlet temperature $T^{*}_{out} = 90\,^°C$ to the <em>pump-control algorithm</em> to get the desired massflow <strong>and then pass the desired massflow as setpoint variable</strong> to the P(ID)-controller.<br>
Thus the desired massflow will be according to the <em>pump control algorithm</em> equation: $\dot m^{*} = 0.25\,\frac{kg}{s}$. This will be passed to the P(ID)-controller as setpoint-value and the current massflow $\dot m^{n}_{out} = 0.5\,\frac{kg}{s}$ as process-value, which yieds the control variable: $C = \left(0.25\,\frac{kg}{s} - 0.5\,\frac{kg}{s}\right)K_p = -0.3\,\frac{kg}{s}$.<br>
So the massflow set by the pump will be: $\dot m^{n+1} = \dot m^{n} + C = 0.2\,\frac{kg}{s}$</li>
</ul>
<p>Considering both possibilities, the second path with passing the massflow to the controller seems alot more aggressive. Since I have little to no experience with controlling water flows with a pump, I don't know which is the common way to go. Or is there anything I forgot to consider?<br>
What also could be interesting is the region/nation in which it is made in one or the other way and for which applications. My application would be on the one hand a district heating network and on the other hand potable hot water stations, both situated in germany.</p>
<p>Thanks for your help in advance!</p>
<hr>
<p><strong>Edit</strong><br>
As several people keep telling me, that a system like this is unlikely to be built: It is common and very much likely in central europe, where energy efficiency is rated quite highly!<br>
In fact systems like this <strong>are being built</strong> often! (I definetely know this, as I am working at operation optimization for systems like this. Thus I'm working with alot of systems like this.)</p>
<p>So this question is <strong>not</strong> about if systems like this are likely to exist. It is if the control flow usually first goes to the controller and then to some algorithm which calculates a value to set for an actuator, <strong>or vice versa</strong>.</p>
<p>I know that usually a voltage or PWM is passed to the actuator and <em>not</em> a massflow which has to be set. This is the only point where I agree that it is unlikely it is going to be done in reality. But this is how it works best for my simulation environment and it <strong>does not affect</strong> my question which is about the order of the control flow!<br>
So... Help is <em>really</em> appreciated, but I don't want to discuss if this system exists or not.</p>
<p>Thanks for help in advance!</p>
| |control-engineering|pid-control|optimal-control|flow-control| | <p>Control problems are usually easier if you specify</p>
<ol>
<li>The measured variables, I assume $T_{in}$ and $T_{out}$</li>
<li>The manipulated variables, I assume the pump speed according to your description</li>
<li>The objectives of the controller. Do you need to be exactly at 90 degC or would and outlet with a certain offset be acceptable? Or is the temperature very important and you would vary the pump outlet a lot if needed in order to be exactly at 90 degC?</li>
</ol>
<p>Your approaches are almost feedforward: you rely on a basic model of how water gets heated in order to control your system. Although that can be a sensible option, it can fail if there are heat losses, changes in $T_{in}$ that you are not measuring, etc. For such a system good old feedback should work.</p>
<p>If you can accept a certain offset, the easiest by far is to use just a proportional controller using $T_{out}$ as a controlled variable (with setpoint 90 degC or slightly higher) and the pump speed (or voltage) as the manipulated variable. </p>
<p>If you need to be exactly at 90 degC, then you would require integral action. The tricky part is that you can make your controller unstable if you specify a very tight integral action or have very long oscillations if the integral action is too slow. However, for a system like yours, it should not be a problem. </p>
| 17000 | Layout of a controller for controlling a water pump |
2017-09-07T08:05:55.277 | <p>Let's say I'm designing a table surface which has to be stiff, rigid, but at the same time, not too heavy. It is rectangular in shape and ratio of length:width is quite large. It is only supported on both ends of its length. If I use a 50 mm slab of gray cast iron for the surface of the table, during maximum loading the table will bend and the Z-axis displacement at the center would be 0.04 mm. This table will be used as a machining table, and the maximum tolerance of the workpiece will be 0.005 in (0.127 mm), so I think the 0.04 mm deflection of the table is already good because it will give room for tolerance from other component of the machine, such as cutting tools, spindle, etc.
Problem is, with 50 mm steel slab, the maximum weight criterion is exceeded. The whole table is designed to be able to tilt on A-axis (it's a trunnion table), driven by a motor. The heavier the table is, the bigger and exponentially more expensive the motor would be. I have to reduce the weight of this structure, but I have to keep the stiffness.</p>
<p>What if instead a simple flat 50 mm slab I make it like 20 mm slab with extra stiffener profile on its underside? Like honeycomb pattern, or any other stiffener configuration. which is possible because it's a cast iron anyway, just have to create the mould.. by doing that I could reduce the weight by almost 40kg</p>
<p>The question is: Is it theoretically possible? 50 mm slab gives 0.04 mm bending displacement, now I want to make it like 20 mm slab plus some stiffener to reduce weight, but I would like to keep same performance with 0.04 mm maximum displacement. The maximum load is 1200 N, concentrated in small area in the middle of the table.</p>
<p>EDIT: previously I wrote the requirement as 0.005 mm deflection, but it is actually 0.005 in (0.127 mm). However, this +- 0.127 mm figure is basically the machining tolerance of the workpiece that will be made on top of this table, so I guess the tolerance of table itself should be much lower than that, I don't know the exact number yet but it just won't go as low as 0.005 mm that's probably too extreme. I'm still in the early stage of the design so a lot of design requirements is still open to revision..</p>
| |mechanical-engineering|structural-analysis|machining|deformation|stiffness| | <p>The formula for central displacement shows that the displacement is inversely proportional to both the elastic modulus, E, and the second moment of area, I.</p>
<p>You need to increase either E or I, eight times. Increasing E is not allowed because it is an intrinsic property of the material that you are using. </p>
<p>I is proportional to the cube of height, so doubling height would directly solve your problem. However, the increase in mass would not be allowed.</p>
<p>What you can do is make a hollow rectangular section and thus increase the height of the cross section as much as possible. This will take some calculations ( by hand or by using a spreadsheet program). </p>
<p>It seems quite doable.</p>
| 17006 | Increasing the stiffnes of a structure by adding stiffener, but at the same time reducing main surface thickness |
2017-09-08T09:40:03.983 | <p>Isoparametric triangular finite elements can be formulated using "triangular coordinates" $(\zeta_1, \zeta_2, \zeta_3)$, also sometimes called area/areal/barycentric coords, or using "Cartesian coordinates" $(r,s)$ defined on a master element. These two systems can be related to each other without difficulty, and seem to be functionally equivalent.</p>
<p>It seems that several texts and journal papers use triangular coords $(\zeta_i)$ exclusively, particularly those that are more recent and/or mathematically oriented. On the other hand, I have older textbooks and use certain commercial FEA packages that consistently present element formulations with $(r,s)$ coordinates. I have a feeling that this might be due to historical reasons, but perhaps there is more to it.</p>
<p>Provided that the element geometry is restricted (straight sides, and any side nodes are uniformly spaced), I understand that triangular coords may be useful because they lend themselves to analytical integration, and thus avoid potentially costly numerical integration procedures. However, when these conditions are not met, numerical integration must be used. In that case, is there any benefit from using triangular coordinates, conceptual or computational?</p>
| |finite-element-method|mathematics| | <p>Triangular coordinate systems (or tetragonal systems, in 3D) have the advantage of giving all the vertices of the element equal status. That becomes clear in numerical integration procedures, when you look at the positions and weights of the integration points.</p>
<p>The three coordinates are not independent - they should sum to 1 at every point in the element. But usually, preserving the symmetry in the formulas is more useful than losing the independence of the coordinates.</p>
<p>Early papers often picked one vertex (arbitrarily) as "special", and mapped that vertex to the origin of a local system and the adjacent sides along the X and Y axes. For an arbitrary shaped triangular element, this will give <em>different results</em> depending which vertex you put at the origin, which can be confusing when trying to understand why two identical-looking models don't give the same answers!</p>
<p>Before the era of automatic mesh generation and refinement, triangular meshes often had a systematic bias anyway (e.g. start with a rectangular mesh and draw all the diagonals pointing in the same direction) so treating one vertex of the triangle as "special" might have been a <em>good</em> thing to do, in conjunction with that type of mesh. But that's all history now.</p>
<p>Of course you <em>could</em> re-map a "symmetrical" numerical integration procedure onto an "old-school" local coordinate system, if you really wanted to - but it's easier not to bother doing that.</p>
<p>Hardly anyone uses analytic integration for "real world" problems any more - even if the geometry is suitable for it, properties like Youngs Modulus often vary (e.g. with temperature) within a single element. In any case, you can create numerical integration rules that give <em>exact</em> results for the special cases that can be integrated analytically, so why bother to use two different integration methods when one method will handle both situations?</p>
| 17020 | Do triangular/area coordinates offer computational advantages for numerically integrated triangles? |
2017-09-08T16:15:59.243 | <p>What are the tolerance specs on the major diameter of imperial bolts? </p>
<p>Specifically, a 1/4-20 imperial bolt has a major diameter of 0.25in, but what is the tolerance on that dimension. I (and everyone I've asked) have always taken that value to be exact, but that's obviously incorrect. </p>
| |standards|tolerance| | <p>Here is a link to a screw thread chart. </p>
<p><a href="http://www.engineersedge.com/screw_threads_chart.htm" rel="nofollow noreferrer">http://www.engineersedge.com/screw_threads_chart.htm</a></p>
<p>Note there are different tolerance classes, and also note the max is never over the nominal size.</p>
<p>For the example 1/4-20</p>
<pre><code>class 2: 0.2408 < d < 0.2489
class 3: 0.2419 < d < 0.25
</code></pre>
| 17024 | Tolerance on bolt major diameters |
2017-09-09T03:39:16.520 | <p>If a 4x4 (true 4x4 not AWD) has 3 wheels slipping, does the one tire with traction receive all the torque available from the transmission? Is there typically a reduction in power transmission in 4x4 systems to the wheels that are not normally powered (i.e. front wheels on a pickup)?</p>
| |transmission|car|traction| | <p>It depends on what type of differential it has. </p>
<p>The nature of an open differential (as found on normal road cars) is that the spinning wheel(s) will get <em>all</em> of the torque.</p>
<p>Proper off-road vehicles will usually have a diff-lock selector which works on some of all of the differentials which means that both driveshafts the diff is connected to get the same torque, regardless of slip. This should only be used at low speeds on slippery surfaces as the locked differential won't be doing its normal job anymore. </p>
<p>More sophisticated vehicles may have limited slip differentials which are a compromise between an locked and open differential and in some cases allow adjustment of the degree of slip either by manual selection of computer control. Depending on the degree of slip selected the differential will always send some fraction of torque to both wheels regardless of their relative speeds. </p>
<p>Limited slip differentials are particular useful on vehicles which are intended to drive at speed on loose surfaces eg in rallying and for permanent AWD cars as they can also be used to adjust turn-in characteristics. </p>
<p>Another option is traction control which monitors the relative speed of the driven wheels and applies the brakes to lock the spinning wheel which transfers torque to the opposite wheel. The are also manual versions of this known as 'fiddle brakes' where the driver has separate brake levers for each side of the vehicle (or front/rear) this system can also be used for 'skid steer' a bit like the way that most tracked vehicles are steered. </p>
| 17033 | Equal power transmission in 4x4 vehicles? |
2017-09-09T22:54:46.257 | <p>Has anyone designed a pond that extends beyond its surface perimeter?</p>
<p>Example cross section:
<a href="https://i.stack.imgur.com/ywPlj.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ywPlj.png" alt="sub surface example"></a></p>
<p>brown - existing surface (or brought to grade)
black - designed pond
blue - capacity</p>
| |water-resources| | <p>I would describe this negative incline portion of the pond as a "shelf". They are common and need to be self-supporting with a material that has compression and tensile strength; concrete and rebar, rock, fiberglass, etc.</p>
<p>There are lots of examples of this:</p>
<ul>
<li>covered irrigation canals</li>
<li>dam water turbine intakes</li>
<li>water diversion intakes</li>
<li>zoo habitats</li>
<li>natural shelf under waterfalls</li>
<li>naturally in water caves</li>
</ul>
| 17040 | Is it possible to create a sub-surface pond? |
2017-09-10T00:11:34.397 | <p>QUESTION(see image attached): <a href="https://i.stack.imgur.com/S14iQ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/S14iQ.png" alt="enter image description here"></a></p>
<p>Really need help here can someone help me show how we get the solution as shown in the image file attached above.</p>
<p>okay since I was asked by one of the users commenting asking me to show what I had done; please see attached image file below</p>
<p>Thank You.</p>
<p><a href="https://i.stack.imgur.com/fFKwB.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/fFKwB.jpg" alt="enter image description here"></a></p>
| |mechanical-engineering|statics|moments| | <p>First you have to notice that the contact at A is a roll support, so the reaction force there is perpendicular to the surface. Therefore Fa has 60 degree against horizontal axis. When you draw free body diagram, decompose the Fa reaction force into its horizontal and vertical components:</p>
<p>Fax = Fa Cos60</p>
<p>Fay = Fa Sin60</p>
<p>Then you solve the balance equations of X-axis, Y-axis, and Moments. I would start from the sum of Moments at C equal zero. You will get the value of Fa.
Then with the balance equations of X-axis and Y-axis with Fa already known you will get the value of Fcx and Fcy.</p>
<p>I get a slightly different result, probably due to rounding:</p>
<p>Fa = 365.24 N</p>
<p>Fcx = 782.62 N</p>
<p>Fcy = 316.3 N</p>
| 17041 | Determine the reactions at A and C, knowing that α = 30º |
2017-09-10T07:49:23.540 | <p>Let's assume I want to build a subway system. I dig two subway stations and now I start digging a tunnel from both of these endpoints simultaneously.</p>
<p>How can I ensure that the two tunnels dug from the different endpoints meet in the middle? I don't believe I can use GPS underground. What methods do professionals use to ensure that the two tunnels meet? Are there some positioning technologies that work underground?</p>
| |civil-engineering|tunnels|positions-measurement| | <p>This is more for a historical aspect (not about state of the art)</p>
<p>If you search for Eupalinian aqueduct on Wikipedia, you will find a reference for the first tunnel excavated from both ends through the use of geometry (and it is actually the second ever excavated in recorded history).</p>
<p>The tunnel was approximately 1 km long and it was build in 6th century BC (approximately 2600 years ago) - long before GPS or any fancy instrumentation.</p>
<p>In the Wikipedia article, there is a fairly detailed description of how Eupalinos used a "mountain line" to create a guide which then compared to the direction of the tunnel.</p>
| 17045 | How are tunnels dug from two endpoints joined? |
2017-09-10T14:47:09.877 | <p>The other day I was sawing some wood on my mitre saw. I have a single light bulb in that room and it was on. The moment I press the button for the saw to turn on, the lights dim a little and go back to full brightness while the saw is running. The saw was connected to the same room. Why does this happen only when the saw is started? The saw has supposedly slightly less than 2 kW power and uses a universal electric motor.</p>
| |electrical-engineering|motors| | <p>The wires in your home have resistance. Current going through those wires causes a voltage drop. The startup (inrush) current of electric motors can be several times their normal operating current. It's impossible to know exactly what the startup current is for your motor without knowing specifics supplied by its manufacturer. The high startup current occurs because the wires in the armature windings appear almost as a short until they set up a magnetic field (become energized). The momentary high current doesn't last long enough to cause the breaker to trip or fuse to blow. There's a pretty good explanation <a href="http://sciencing.com/calculate-motor-inrush-current-7791777.html" rel="noreferrer">here</a>. <br>So...the momentary high current causes a like momentarily high voltage drop which in turn causes the light to dim.</p>
| 17051 | Why do the lights dim when I turn on a mitre saw? |
2017-09-10T16:21:50.790 | <p>Most if not all of the off-the-shelf epoxies I've used have generated a fair amount of heat while they are hardening. This seems to indicate to me that these are exothermic reaction, i.e. they generate heat from the reaction. </p>
<p>The common practice at my last company was to apply heat (heatgun or hairdryer) to epoxied joints to get them to cure faster.</p>
<p>As I understand the chemistry though, adding heat to an exothermic reaction should slow the reaction and effectively take longer for the joint to cure.</p>
<p>Was my company's 'standard practice' wrong to be adding heat to epoxy joints and extending the curing time, or is epoxy some type of anomalous material that cures faster when adding heat to the exothermic reaction. OR, is epoxy actually an endothermic reaction and the heat perceived during curing is something else? </p>
| |epoxy|chemistry| | <p>Per the <a href="https://en.wikipedia.org/wiki/Arrhenius_equation" rel="nofollow noreferrer">Arrhenius Equation Wiki</a>; "the reaction rate doubles for every 10 degree Celsius increase in temperature."</p>
<p>The fact that temperature increases the reaction speed is independent of endothermic or exothermic. The reaction rate increases because the motion of particles is greater when the temperature is higher, which increases the likeliness (and speed) that two reactant molecules will meet one another.</p>
<p>I have used a microwave (very briefly) to get a head start on fiberglass epoxy in cold weather, but heating your work space is a much better idea. Since it is exothermic like you mentioned, it is easy for it to have a run-away reaction where the heat it generates further speeds it up. Too fast of cure time will result in a much weaker material; too slow (like days in the cold) and it is possible for the hardener wick into a surface more than the resin and the epoxy will never fully harden.</p>
| 17052 | Epoxy curing, heat or no heat? |
2017-09-10T17:20:25.727 | <p>I'm doing a project on Theo Jansen's Mechanism, where I have to modify the mechanism. What is the most efficient way to design these mechanisms? Is there any good software where I can just draw the links and input crank and test out the mechanism? I have used GeoGebra, but it doesn't simulate the mechanism physically, and while using Creo I found that the pin joints often get stuck - like they suddenly fold up and do not move. Also, creating the links one by one takes quite a lot of time.</p>
| |mechanical-engineering|mechanisms|linkage| | <p>3D CAD software has a hard time with lots of pin connections, because without a range limitation the solution is indeterminate. I have not used Creo, but if you can limit the range of motion with interference or other, the model may work.</p>
<p><a href="http://www.design-simulation.com/wm2d/" rel="nofollow noreferrer">Working Model 2D</a> is nice for designing mechanisms. I used it as a student; not sure how much it costs commercially.</p>
| 17053 | How to quickly design and simulate crank based mechanisms? |
2017-09-11T09:21:51.083 | <p>What is the relationship between nozzle size and sound produced? If there are references on it,kindly put it the comment space.</p>
| |mechanical-engineering| | <p>Imho the nozzle diameter should only affect the impedance, but not the frequency. The frequency is affected by the length of the nozzle. Everything else is really easy to look up using some google skills and keywords like "acoustic resonance".</p>
| 17062 | relationship between nozzle radius and sound produced |
2017-09-11T11:24:39.927 | <p>Below is a stack of reinforced concrete slabs with hollows (camera is positioned along the hollows' axis). Such slabs are used for floors - you build walls of the next floor, then lay such slabs such that they rest onto walls with the end shown on the picture and the opposite end. The dark round dots are steel rebars covered in rust.</p>
<p><a href="https://i.stack.imgur.com/QqSej.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/QqSej.jpg" alt="stack of slabs with hollows"></a></p>
<p>I've seen slabs with round hollows and with oval hollows but this shape is new to me. It's something like a rounded hexagon with an extra cavity on the bottom.</p>
<p>What's the name of this shape and why is it used for hollows in a reinforced concrete slab?</p>
| |structural-engineering|civil-engineering|concrete|reinforced-concrete| | <p>It looks like a typical precast concrete "hollow-core" or "voided" slab. A quick Google image search reveals the variety of void shapes employed by manufacturers.</p>
<p><a href="https://i.stack.imgur.com/XAXZfl.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/XAXZfl.jpg" alt="Voided Slab Examples"></a></p>
<p>The image on the left (from Oldcastle Precast, incidentally) nicely shows the variation of void shape with increasing slab thickness.</p>
<p>As noted in the other answers, we care about the concrete profile more than the void profile. The voids are simply introduced to minimize concrete area where it is not structurally necessary.</p>
<p>In theory, a rectangular void would minimize concrete. However, the corners are better off rounded to reduce the stress concentrations (and therefore minimize cracking). Hence, the oblong voids. In thinner slabs, the voids may indeed approach a circular shape. One possible explanation for the irregular void shape in your posted question, is that the irregular void can set proper concrete cover for the rebar.</p>
<p><a href="https://i.stack.imgur.com/jd6rRl.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/jd6rRl.jpg" alt="Thoughts on Void Shape"></a></p>
| 17065 | What's the name and purpose of this specific shape of hollows in a reinforced concrete slab? |
2017-09-12T13:01:12.643 | <p>The CNC spindle is supposed to operate its single shaft in two different modes:</p>
<ul>
<li>high RPM (30,000-80,000 RPM), variable, possibly quite high torque, a straightforward PID to maintain the RPM as torque varies ("work mode", RPM control).</li>
<li>very low RPM (exact value doesn't matter), nearly no torque (just friction of bearings), rotation of shaft to precise angle. ("optical scan of the tool, from all sides", angle control). </li>
</ul>
<p>The first is achievable through a good AC or DC motor. The latter would be achievable through a stepper motor with microstepping. I can cut off any of the motors (leave it idle) and drive the other to achieve either of the modes.</p>
<p>But I don't think typical stepper motors are capable of surviving 80,000 RPM applied externally - and I'd prefer to avoid a mechanical clutch decoupling the stepper. </p>
<p>Are there steppers that can just survive these kinds of speeds? Or other neat solutions that don't overly increase the mechanical complexity?</p>
| |motors|stepper-motor| | <p>Couple the stepper motor to the high speed spindle motor only indirectly by means of magnets rotating around the spindle motor shaft which, at low speeds, will align themselves with specially designed ferrous spots fixed to the high speed spindle shaft and sync with it. The magnets should create a detent force strong enough to move the spindle at slow speeds, but would break free at higher speeds, particularly if the stepper was held in place electrically.</p>
| 17083 | Stepper on high-RPM shaft |
2017-09-13T15:23:15.637 | <p>Historically the longest lasting building materials seem to be stone. But I was wondering what material modern technology can create that would last longer under various conditions.</p>
<p>The various things the material would need to be tolerant to would include extreme heat, cold, water, wind, plants, animals, sunlight, earthquakes, as well as a determined effort by people to deface, damage or destroy.</p>
<p>I had considered steel blocks with a coating of an alloy or metal that would not react to air or water, or maybe an advanced form of composite glass or plastic such as bullet and bomb resistant glass that would be multiple feet thick and reinforced with metal as well.</p>
<p>I was planning to ask this question in an architecture stack, but all I could find was home improvement and I thought engineering may have more people versed about materials science than do it yourselfers.</p>
<p>So, if a modern team was going to build a pyramid and wanted it to last much longer than those of Egypt what materials would likely be used?</p>
| |materials| | <p>Steel buildings are a more long-lasting and modern building material compared to wood or any other material because durable steel better resists general causes of deterioration such as termites, moisture, and mold. <a href="https://www.qualitysteelbuildings.com/" rel="nofollow noreferrer">Building made with Steel is long-lasting</a> because galvanized steel is used, and it is affordable as well as can withstand harsh weather conditions for years to come.</p>
| 17096 | What is the longest lasting modern building material? |
2017-09-13T19:12:36.927 | <p>I have a trajectory which is a function of time that returns tuples of position, velocity, acceleration and jerk for each moment. So at every control loop iteration I would like the motion of my actuator to match these values. However it is not clear how to build a control system which will be able to follow position, velocity, acceleration and jerk commands altogether. </p>
<p>The actual question is that of control system input. In the vast majority of cases input is just a force which in context of PWM is just a number. Therefore natural question arises: should I anyhow map my 4 trajectory parameters to this one input of control system(if yes then how?) or is there any other way?</p>
| |control-engineering|control-theory| | <p>If you want to track a trajectory and its derivatives is equivalent to just tracking that trajectory. However depending on your controller and reference signal you might get a steady state error.</p>
<p>In order to know how to choose an appropriate controller we have to look at how a potential reference could look like. In the worst case the highest derivative of the reference we want to track is a constant, so of the form $t^n$. The Laplace transform of this is $n!\,s^{-1-n}$. So in order for the steady state error to be zero the <a href="https://en.wikipedia.org/wiki/Sensitivity_(control_systems)" rel="nofollow noreferrer">sensitivity transfer function</a> should have bigger positive slope at low frequencies than the negative slope of the Laplace transform of the reference. This implies that in general you want your openloop (plant times controller transfer function) to have at least $n+1$ integrators.</p>
<p>But I should also mention that you can also get a huge performance improvement by using feed forward.</p>
| 17104 | How to execute trajectory? |
2017-09-14T09:51:31.930 | <p>I have to model the behavior of an electric car. For this, I use <a href="https://books.google.fr/books?id=6aWSaJUZuOEC&pg=SA7-PA47&lpg=SA7-PA47&dq=equation%20electric%20vehicle&source=bl&ots=oHIOhDcJaq&sig=o9-kql0qtUwJb--clA0VRwFspVA&hl=fr&sa=X&ved=0ahUKEwjKhffDsqTWAhVH1RQKHdAvATUQ6AEIPTAB" rel="noreferrer">these equations</a> and I can observe by "playing" with parameters that, when going downhill at constant speed, the car has a negative consumption (i.e. recovers energy). I was wondering if this is realistic?</p>
| |mechanical-engineering|electric-vehicles|regenerative-braking| | <p>The basic equations governing an electric car are reversible, so a simple analysis would say that the battery is getting charged. In the real world, there are more complications. In the case of an electric bike, there is a ratchet that disengages the engine from the wheel when the wheel is going faster from the engine (just like how on a regular bike -- where "regular" doesn't include track bikes -- if you stop pedaling, the wheel keeps moving even though the chain isn't). In an electric car, the motor can be disengaged from wheels either mechanically or electrically. Electric motors generally can't handle as much power when being driven backwards as when going forward, so electric cars are built to limit how much power can be transferred from the wheels to the battery. Absent those modifications, any time the wheels are going faster than the motor, there's going to be electricity flowing "backwards". Whether that energy will be charging the battery or frying the motor is a different matter.</p>
<p>And if you include heat, all cars "recover" energy in some sense when going downhill, either through increased kinetic energy, or heating of brake pads.</p>
| 17118 | Does an electric vehicle going downhill recover energy? |
2017-09-15T08:45:35.660 | <p>I have been trying to determine the response coefficients for seismic analysis of Non building structures in <strong>ASCE 7-10</strong> and found that the value of <strong><em>R, the response coefficient</em></strong>, is larger in case steel connections are detailed using <strong>AISC-341</strong> than <strong>AISC-360</strong>, this would imply a larger design load in case of AISC 341 than AISC 360. </p>
<p>So basically,</p>
<pre><code>AISC 341 -> Large seismic design load
AISC 360 -> Smaller seismic design load
</code></pre>
<p>Does that imply AISC 341 design results in a more conservative steel connection than AISC 360 or is there a difference in steel design philosophy?</p>
| |civil-engineering|steel|aisc| | <p>I think that it should be like this:</p>
<p>AISC-341 = more ductile = lower R = higher seismic design force</p>
<p>AISC-360 = less ductile = higher R = lower seismic design force</p>
| 17131 | AISC 360 vs AISC 341 |
2017-09-15T19:15:35.050 | <p>The book "Instrument Engineers' Handbook vol. 1 - Liptak" said that for the first-order system forced by a step or an impulse, the time constant is the time required to complete 63.2% of the total rise or decay; at any instant during the process, the time constant is the quotient of the instantaneous rate of change divided into the change still to be completed.</p>
<p>How can I express the last sentence mathematically?</p>
<p>Thank you very much.</p>
| |control-theory|education| | <p>Does the book not give you the mathematics? The underlying expression is:</p>
<p>$$parameter = 1-e^{ -t/ \tau }$$</p>
<p>so you see at $t = \tau$</p>
<p>$$parameter = 1 - e^{-1} = 0.63$$</p>
<p>Now the second sentence says divide "change to be completed" which is $1 -parameter$ , or $e^{-t/\tau}$, by the first derivative:</p>
<p>$$\frac{d parameter}{dt} = \frac{e^{-t/\tau}}{\tau}$$</p>
<p>so</p>
<p>$$\frac{e^{-t/\tau}}{\frac{e^{-t/\tau}}{\tau}} = \tau$$</p>
<p>pretty simple..?</p>
<p>I will say the wording of that sentence is pretty awkward. I would reverse it as "the time constant is the quotient of the change still to be completed <em>divided by</em> the instantaneous rate of change."</p>
| 17134 | Time constant - formula |
2017-09-16T00:31:36.987 | <p>I am currently into an estimating job for a small construction company. In my previous experiences, in calculating the labor cost for a certain job, </p>
<p>Labor cost = material cost * 0.45</p>
<p>I did this to simplify the computation. Are there any other specific labor factor for different job classification for the following class:</p>
<ol>
<li>concreting </li>
<li>masonry</li>
<li>carpentry</li>
<li>painting</li>
<li>steel</li>
<li>rebars, etc. </li>
</ol>
<p>that I can adopt? </p>
<p>Note: This will not involve equipment in the process.</p>
| |civil-engineering|construction-management| | <p>In the Philippines, there are 2 types of labor rates that we consider.
We have provincial and non-provincial rates.</p>
<p>For non-provincial rates, we can go as far as 45% material cost, while on provincial rates, we only go up to 35% of the material cost. This is due to our present labor rules and regulations (which i hope should be changed soon).</p>
<p>But the above percentage varies according to the difficulty of work, e.g. painting and carpentry requires a lot of skills than concreting that's why I sometimes go as much as 60% for these type of jobs.</p>
<p>Hope this helps.</p>
| 17140 | Relationship between Material cost and Labor cost |
2017-09-16T01:32:53.053 | <p>Most of resources I've been reading don't explain the scientific reason for the existence of shear force. Most resources just teaches how to solve problems and about properties. Nothing goes deep (<strong>not even a little bit deep</strong>, sorry if I'm being unfair, for me it was just <strong>drawing diagrams and solving them</strong>).</p>
| |structural-engineering|civil-engineering|strength| | <p>This is from <em>Wikipedia</em>, <a href="https://en.wikipedia.org/wiki/Shear_force" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Shear_force</a> :</p>
<p>"Shearing forces are unaligned forces pushing one part of a body in one direction, and another part of the body in the opposite direction. When the forces are aligned into each other, they are called compression forces."</p>
<p>In engineering terms, it is the force that forces the body to be <strong>sliced</strong> and to be <strong>split</strong> with regards to the direction of the force. It is like running a boiled potato into your hands, the skin of that potato will be removed. Same goes with the materials. All materials are assumed to have infinite <em>layers</em> which can be removed like the potato skin when a force parallel to that layer is applied. </p>
<p>See images below for the failure caused by shear.</p>
<p><a href="https://i.stack.imgur.com/i7L3W.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/i7L3W.jpg" alt="enter image description here"></a></p>
<p><a href="https://i.stack.imgur.com/O0MUp.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/O0MUp.jpg" alt="enter image description here"></a></p>
<p><a href="https://i.stack.imgur.com/rdWyV.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/rdWyV.jpg" alt="enter image description here"></a></p>
<hr>
<p>As an answer to OP's comment, take a look at the pictures:</p>
<p>Assume that a block is loaded by two forces acting in the opposite directions, the red layer will be the plane that will carry the shear force.
<a href="https://i.stack.imgur.com/2l2GA.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/2l2GA.png" alt="Original Block"></a></p>
<p>What we don't see is that the block is composed of infinite number of differential planes, see picture below, that also counteracts the applied force.
<a href="https://i.stack.imgur.com/vP33T.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vP33T.png" alt="Next Block"></a></p>
<p>As we stay by the rule of statics, the force will then be distributed over the other layers giving the <strong>same amount of force</strong> for each layer. See picture below. <a href="https://i.stack.imgur.com/b24Sn.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/b24Sn.png" alt="Block 3"></a></p>
<p>Therefore, shear force <em>can</em> make the any of the differential plain fail.</p>
| 17144 | What causes shear force? |
2017-09-16T13:44:42.530 | <p>I'm studying vibrations in a gearhead, for which I have a general dynamic model of 12 degrees of freedom (only translational coordinates) and I have calculated its natural frequencies. </p>
<p>Now I want to try and correlate my model with a physical system. I did some impact testing with a hammer and measured the response with one accelerometer placed on the horizontal direction of the gearheads case.</p>
<p>My theory on mechanical vibrations and modal testing is quite basic and I don't know how to proceed and interpret the results of the readings.</p>
<p>I tried simulating the forced response in order to correlate the model I'm using to the measurements. For example, in one of the experiments, I hit the gearhead horizontally and measured the response on the opposite side of the gearhead with the accelerometer, in the same axis. In the forced response simulation, I applied a unit force on what would be that equivalent degree of freedom.</p>
<p>In the magnitude response of the test, I have a few peaks ranging from 2khz to 24khz (its also the width of my window res). But in my force response simulation I get resonance peaks much higher, from 9khz up to 80khz. The natural frequencies range from 1khz to 16khz.
I have a few questions regarding what I'm trying to do:</p>
<ol>
<li><p>Should the natural frequencies I calculated all show in the forced response simulation within a frequency range of up to 20khz, or only those close to the degree of freedom of the applied force in the force vector? In my results so far, none of the peaks in the forced response are close to one of the natural frequencies.</p></li>
<li><p>How can I correlate the test measurements with the forced response simulation? none of the peaks in the magnitude response of the tests seem to be close to the simulation.</p></li>
<li><p>From the modal test results, I measured some of the peaks and about 3 of them are very close to some of the natural frequencies, does this mean that this frequency were excited in this specific test? or can it be just coincidence? as I explained before, in the forced response simulation I couldn't find any relation.</p></li>
</ol>
<p>I should also note that my model is undamped, and my equipment was quite limited (only one sensor and small hammer(22khz)). </p>
<p>I read some theory on varius text books on vibrations and modal testing, they all show how to interpret a general FRF of course, but I can't find practical examples relating a theoretical model and modal testing like I'm doing, not explained in detail at least, most just skip to results, so the general methodology is what I'm lacking. </p>
<p>Thanks.</p>
| |mechanical-engineering|dynamics|vibration|modal-analysis|frequency-response| | <p>You can't make much use of just one FRF, especially if none of the peak responses line up between the model and the test. </p>
<p>Fix the accelerometer at the position of one of your model DOFs and then impact the model at all 12 DOF positions (in the correct directions, of course). Move the accelerometer to the other DOF positions in turn, and repeat. </p>
<p>You should be measuring the hammer force as well as the accelerometer response, so you can make the FRFs constisent with each other.</p>
<p>You also need to do some analog filtering of the measured signals <strong>before</strong> you digitize them, otherwise the results will be a mess to interpret because of artefacts caused by aliasing. For example if you digitize at 22k samples/sec but don't do any filtering, you can't tell the difference between resonances at 11+1 = 12kHz and 11-1 = 10kHz in the digitized data and the FRFs - or even worse, there is no difference between resonances at 21kHz and 1kHz! </p>
<p>You will then have 144 FRFs in total. You can cross-correlate them to see which peaks are "real" and which are just random artefacts. More important, you can also find the <em>mode shape</em> corresponding to each peak, and try to identify that in the modes from your model.</p>
<p>You need some software to do this, of course. Commercial modal analysis software isn't cheap, but there are some free packages - for example <a href="http://hosting.umons.ac.be/html/mecara/EasyMod/index.html" rel="nofollow noreferrer">http://hosting.umons.ac.be/html/mecara/EasyMod/index.html</a> (which also needs Matlab). That is not a personal recommendation for EasyMod - I've never used it - but modal analysis and testing seems to be a speciality in Belgian universities so it "comes from a good home" IMO.</p>
<p>A good book to read is Ewins, D.J. “Modal Testing; theory, practice and application”. You can probably find the first edition free on the internet - that will tell you more than you ever wanted to know about the subject. Note the words <em>theory, practice and application</em> in the title. It's a good book because it describes all three aspects. Just having a mathematical understanding of FRFs (which is about as much as you will learn in a typical B-Eng degree) is only a part of the total picture!</p>
<p>Note, it is not at all unusual for a first attempt at modelling to give modes that are all "way off" from anything you measured - but unless you can identify some corresponding <em>pairs</em> of modes in the test and model data, you won't get very far trying to improve the situation except by pure good luck.</p>
| 17152 | Modal Testing and Frequency Response Interpretation |
2017-09-18T09:23:21.940 | <p>I read that trains in pratice instead of acceleration to a certain speed level and holding the speed, they accelerate and then cost, accelerate again, und keep this pattern. This was shown by experiments. What is the explanation why this way of driving is more energy-efficient?</p>
| |energy-efficiency| | <p>Just keeping a train at a fixed speed over level terrain takes a small fraction of the engine's power output. As a result, the engine runs very inefficiently.</p>
<p>In pulse and glide, the short bursts from the engine during the pulse phase are much higher power. These allow the engine to run at higher efficiency.</p>
<p>Roughly the same overall average power is produced by the engine either way, but with pulse and glide, that power is produced at a more efficient operating point of the engine.</p>
| 17167 | What is the explanation why Pulse and Glide is more energy-efficient for trains |
2017-09-18T17:18:54.027 | <p>How can I create a gear system which keeps, say, 60 rpm stable on output wheel, when I don't care about force? I am using a hand-turned crank as an input wheel.</p>
<p>I know a flywheel could be used, but it is weak to small accelerations and slowdowns over time.</p>
| |mechanical-engineering|gears| | <p>The keyword you are looking for is a <a href="https://en.wikipedia.org/wiki/Governor_(device)" rel="nofollow noreferrer">governor</a></p>
<p>The most simple type is a parted circle that are hinged to be able to move outward into a ring by centrifugal force and the friction counteracts the increased speed.</p>
| 17170 | Keeping rpm stable with input of floating rpm |
2017-09-19T12:09:05.050 | <p>A door of volume l x b x h and weight w is hanging vertically on a horizontal hinge. How can I determine the force/torque to open or close this door?</p>
| |mechanical-engineering| | <p>Assuming that:</p>
<ul>
<li>the door all one material and the weight is evenly distributed</li>
<li>the door must be opened all the way to horizontal</li>
<li>you open it by pushing on the edge furthest from the hinge</li>
</ul>
<p>then the force required is half the weight. In other words if the mass is 10kg, the weight $w = mg = 10kg*9.8ms^{-2} = 98$N, and 49N will be required to open it. If you want a formula for $F$ then it's probably:</p>
<p>$F = mg/2$</p>
<p>If you can't push on the edge, then more force is needed. If you can't push directly upwards, then more force is needed. If there is any friction in the system, slightly more force is needed. If the door doesn't need to go horizontal, slightly less force is needed. If the door has heavy bits at the outside edge, more force is needed. All in all, it would be wise to design the system with a good amount of extra force.</p>
| 17178 | Force required to open a door of volume lxbxh and weight w |
2017-09-19T16:43:32.853 | <p>I was solving problems from the Meriam and Kraige Engineering Mechanics Statics vol-1 and I couldn't get the correct way through this problem.</p>
<p>The question is to find out the force in member $DK$ of the loaded overhead sign truss.</p>
<p><img src="https://i.stack.imgur.com/j5ckD.jpg" alt="enter image description here"></p>
<p>My attempt</p>
<p>I tried eliminating member NM,FG,GT,HI.i tried using the method of section but any section passing through DK is of more than 3 members.</p>
<p>Please suggest how to approach this problem.Thanks.</p>
| |mechanical-engineering| | <p>At first glance it’s a fairly daunting problem, but it’s actually easily solved using K-truss analysis.</p>
<ol>
<li><p>Solve for the external reactions by summing the moments about Point U, and summing vertical forces.</p></li>
<li><p>Use K-truss analysis to determine the net vertical force in diagonals QD/QJ and RD/RK. This is done by making vertical section cuts and then summing the vertical forces. We need only consider the vertical component of the force in the diagonals.</p></li>
<li><p>Distribute that equivalent vertical force we calculated in step 2 between the diagonals. (Basically we look at another free body diagram and consider vertical equilibrium again.) Because the diagonals are at the same angle, the equivalent vertical force distributes equally.</p></li>
<li><p>Use method of joints at Joint K to solve for the force in element DK. (5kN in tension)</p></li>
</ol>
<p><a href="https://i.stack.imgur.com/GXe3Jl.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/GXe3Jl.jpg" alt="Solution Method"></a></p>
| 17182 | Force in a member of the loaded overhead sign truss |
2017-09-20T07:10:05.767 | <p><a href="https://i.stack.imgur.com/NjES8.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/NjES8.png" alt="The different support reactions at B and C"></a></p>
<p>I know that pins prevent horizontal and vertical translation and that rollers prevent only the vertical one.
Therefore, what I do not understand here is that while the support at B reacts with two forces C support has only one, even though both look like pins to me.Where am I going wrong here?</p>
<p>(I could not find anything on the site about whether I can upload images this way. I hope there are no copyright issues related to my post because
I do not own the images. If there is, please notify me and I will immediately take them.)</p>
| |statics|connections| | <p>The support at C looks like a roller.</p>
<p>The reaction from roller is always perpendicular to the surface on which it rests. Hence it will be from left to right, since the surface is vertical here.
So the figure shows the correct direction of the reaction, and yes there is only one reaction at C.</p>
| 17187 | Isn't a pin support supposed to prevent translation both horizontally and vertically? |
2017-09-20T13:30:03.870 | <p>I've been working on implementing a new company standard for our file naming convention (super fun), and it has been mentioned to me that there is a "standard" that says that you shouldn't use spaces in file and folder names when saving documents.</p>
<p>I've done some looking and I haven't found a published standard that says one way or the other whether this is acceptable. Does anyone know of a standard relating to this?</p>
<p>example</p>
<p>"Released Drawings" as opposed to "ReleasedDrawings" as a folder name</p>
<p>or </p>
<p>"123 - Rev 0 - Important part" as opposed to "123-Rev0-Importantpart"</p>
<p>I'm hoping to find a a specific standard, not just a rule of thumb, or best practice.</p>
| |standards| | <p>If you are likely to share your files with other organizations that use different operating systems, you need to research what works best on all the main options - i.e. Windows, MacOS, and the various flavors of Unix. The "If it works on Windows 10, that's OK because it's the only system we use" is probably digging a hole to fall into in future!</p>
<p>Spaces in file names generally cause pointless aggravation, when a system takes only the first "word" of the name as the complete name, and first complains that the file doesn't exist, and then complains that it can't understand what follows it. You can work round that by putting the whole name in quotes, but why make people do that? If you don't like ReleasedDrawings, use Released-Drawings or Released_Drawings instead.</p>
<p>There are other invalid characters - the most obvious being / which is the separator between <em>parts</em> of the file name (e.g. folders and directories) and the equivalent \ on Windows. Others are quote marks, ">", "<", "?", "*", etc. These characters have special meanings on command-line instructions in most operating systems.</p>
<p>Some operating systems have case sensitive file names (ReleasedDrawings and Releaseddrawings are different files!) but others do not. Windows is a compromise. </p>
<p>Some systems have "reserved" file names - for example COM and NUL on Windows.</p>
<p>There may be a limit to the total length of a file name (including all the folder names that precede it).</p>
<p>A Microsoft reference (obviously Windows-specific) is <a href="https://support.office.com/en-us/article/Invalid-file-names-and-file-types-in-OneDrive-OneDrive-for-Business-and-SharePoint-64883a5d-228e-48f5-b3d2-eb39e07630fa" rel="nofollow noreferrer">https://support.office.com/en-us/article/Invalid-file-names-and-file-types-in-OneDrive-OneDrive-for-Business-and-SharePoint-64883a5d-228e-48f5-b3d2-eb39e07630fa</a></p>
<p>You could do worse than stick with the POSIX specification for "fully portable filenames" which permits only characters A-Z, a-z, 0-9, "_", "-", and "." ("-" must not be the <em>first</em> character of the name) and a maximum length of 14 characters, or ISO 9660 (used on CDs and similar devices) which is case-insensitive and only allows A-Z, 0-9, "_", and "." with a maximum length of about 180 characters.</p>
| 17193 | Filename standards and spacing |
2017-09-20T18:34:43.420 | <p>I would like to build a clock where the face rotates. Imagine a standard clock, but instead of hands pointing to the hour, there would be a rotating face plate with a small window, as it were. The face plate would obscure everything except the current hour, which would be visible through the window. If it makes any difference, I need it to be a 24-hour, not a 12-hour, clock.</p>
<p>I have considered buying a 24-hour high-torque clock movement (because I'm not interested in building the actual gears, etc.), but I get the impression that it would be hard to attach anything to it other than standardized clock hands. I could try to glue the face plate onto the hour hand, but I don't know if that would hold effectively.</p>
<p>Any ideas?</p>
<p>Thank you. (Also, if this isn't the right forum, please send me in the right direction.)</p>
| |mechanical-engineering| | <p>With the disks light enough, mounting (gluing) them to the clock hands (trimmed short enough not to obscure the disks) shouldn't pose much of a problem. </p>
<p>Print the disks on transparency (transparent foil for overhead projectors; printable with inkjet or laser, be sure to pick the type that matches the printer you have), obscuring (printing over) everything that's not supposed to show, leaving the windows unprinted (transparent) plus a ring on the bottom disk to let the top disk's window "see through" all the way down to the fixed face plate with the digits (you can print that one too). Carefully cut the transparencies to shape, detach the clock hands (they are usually push-on, plain friction / tight fit mount), cut short not to obscure the windows and paste them to the reverse side of the transparencies, then mount on the clock. With each disk weighing about a gram, and being a centered weight, they won't affect performance of the clock in the least.</p>
<p>If you want the windows to be immobile and the number dials to turn instead, you need to print three disks: bottom with hour numbers, middle with minutes and a transparent ring to reveal the hours beneath, and topmost which would be larger than the two, affixed to the clock case, with the windows.</p>
| 17196 | building a clock with a rotating face |
2017-09-21T13:40:07.983 | <p>A colleague of mine sent me this image of some racking and I've been trying to figure out the design intent of some of the features.</p>
<p>Image one (key holes)</p>
<p><a href="https://i.stack.imgur.com/Y4rfh.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Y4rfh.jpg" alt="enter image description here"></a></p>
<p>Looking at the keyholes on this racking the one side of the lobe is curved while the other side is straight, sloping away from where the skids are going to be placed. </p>
<p>Image 2 (top view paint sketch of the racking profile)</p>
<p><a href="https://i.stack.imgur.com/5qnkQ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/5qnkQ.png" alt="enter image description here"></a></p>
<p>This image is a quick and dirty sketch of the shape of the racking profile,, sort of a W shape.</p>
<p>When I looked at these images I was thinking that the two shapes were made to work together, the straight side of the keyhole being made to slightly pull the shelves of the racking apart when placed in them. I.e. not fully resting on the bottom of the keyholes under their own weight, and then the 'W' shape of the racking uprights being used to act as a bit of a spring/dampener feature. My thinking/reasoning for this is so that when a forklift drops/lowers a skid onto the shelves, the racking uprights will have room to flex slightly and absorb some of the impact (forklifts not being particularly precision instruments). This way the racking pins don't get impact loaded every time a skid gets loaded, and also the flexing of the uprights would allow for a greater tolerance in the levelness of the shelving. (Shelving/racking safety requirements being remarkably strict)</p>
<p>My question is, does my reasoning sound accurate, or is there another reason for these somewhat 'complicated*' shapes? </p>
<p>note- by complicated I don't mean intricate, these are very simple shapes to make, but they appear to have more design intent because they go beyond simple holes and extrusion processes. I come from industry where all geometry requires justification and calculations not just "because it looks cool".</p>
| |mechanical-engineering|structural-engineering|design| | <p>The main reason for the shape (W-shape) of one side of the rack is to improve structural strength.</p>
<p><a href="https://i.stack.imgur.com/4EIHu.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/4EIHu.png" alt="See picture as reference."></a></p>
<p>As depicted in picture $a$, the system is to resist forces that will tend to bend the vertical pole in its local x-axis, see picture below.</p>
<p><a href="https://i.stack.imgur.com/DVKiF.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/DVKiF.png" alt="Deflected shape of the pole."></a></p>
<p>Thus the moment of inertia of the profile shall be designed to be the maximum considering also the part of contact between the rack and the pole. This will be proven when you compare the moment of inertia of the two parts. Also, if you fold a piece of paper or cardboard into V-shape, it is harder to bend then when it is flat.</p>
<p>Also, if the profile is to be constructed like in picture $a$, the side in which the racks are connected will easily bulge out the frame. See image below.
<a href="https://i.stack.imgur.com/sqK87.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/sqK87.png" alt="enter image description here"></a></p>
<p>Therefore, this design gives room for the deflection to occur without compromising both the strength and the aesthetics of the racking system. See also <a href="https://qph.ec.quoracdn.net/main-qimg-19ea8f8ed5515c7f13406fa7c347efce" rel="nofollow noreferrer">this image.</a></p>
| 17207 | Racking design question |
2017-09-22T13:19:19.083 | <p>Is it possible to simulate a layer of dust deposited on top of a PV cell in a controlled manner?</p>
<p>One potential method I had was to put a given weight of a particulate/powder (chalk powder, talcum powder, etc.) into water, and pour the water on top of the solar panel, letting it dry and thus leaving the powder on top of the panel. Would this be viable, and if so, what would be the best powder to use?</p>
| |simulation|experimental-physics|photovoltaics| | <p>More to the "even-distribution" question, I'd recommend your idea of water distribution. I'd use spray nozzles - they're designed for relatively even distribution. To that end, I'd recommend <a href="http://www.bete.com/products/is.html" rel="nofollow noreferrer">IS</a> nozzles (lower budget) or <a href="http://www.bete.com/products/nc.html" rel="nofollow noreferrer">NC</a> nozzles (higher budget), with a distribution manifold. These will provide an easily repeatable test method. Simply dial down to the minimum levels for consistent spray (1 psi for IS, 3.5 psi for NC). Then adjust the concentration of dust in your spray solution. Use a pressure regulator to ensure consistent pressure across the spray system, and a timer to ensure the exact amount of solution sprays across the panels. Finally, as a test to the system, weigh the dried off powder at the end of the experiment on each panel to verify how much dust was deposited on each panel.</p>
<p>For the powder, if you have access to a very low humidity area I'd recommend powdered milk or creamer. Talcum powder is also used readily. But that's mostly a personal opinion.</p>
| 17212 | Simulate Dust Layer on PV Cell? |
2017-09-22T14:51:22.127 | <p>I would like to plug a 25A welder on a 10A house plug. I understand that it will not accept full power or the circuit breaker would switch.
However, my question is if I may run under some lower power setting: for instance setting 70A out of 200A on the welder.</p>
<p>From a very simplistic point of view:
A welding machine with a maximum of 200A for 25A@220V mean, the welder at 60A would require about 10A@220V </p>
<p>However, a welding machine is not such a simplistic situation and uses several variants of waves, AC or DC, several frequencies, etc. which make all more complex.</p>
<p><strong>Question</strong>
May I plug this 25A Welder on a 10A-protected house plug?</p>
| |electrical-engineering|welding|welds| | <p>Yes, you can plug a higher rated load into a lower rated breaker if the load is monitored to minimize trips. If you are using a different plug, make sure that the other plug is removed entirely there are no exposed wires or contacts. I assume you are in europe and the standard wall voltage is indeed 220V (If someone in the US is reading this in the future, a standard wall outlet is 110V and will not work for a 220V device).</p>
<p>The breaker is sized to protect the wiring, not your device. It will allow greater than 10A for a small period of time (say 15A for 30 seconds) for start up loads. If you exceed 10A for a length of time the circuit breaker will trip to protect the wiring. This will not harm your device, it is just like unplugging it or the power going off. You can increase the load and trip the breaker a couple times to test how high you can run your welder, but like agenpt mentioned in the comments, you should not run it in a way where you are tripping the breaker frequently. Breakers are only rated for so many trips (say 100) before they risk not operating correctly and need to be replaced.</p>
| 17216 | Plugging a 25A device (Welder) on 10A house plug |
2017-09-23T06:06:40.240 | <p>I am solving problems from the book: Meriam Kraige Engineering Mechanics statics vol 1. The problem is as follows:</p>
<p>The sliding door rolls on the two small lower wheels $A$ and $B$. Under normal conditions, the upper wheels don't touch their horizontal guide. The question is to find out the force $P$ required to slide the door at a steady speed if wheel $A$ becomes "frozen" and doesn't turn in its bearing. The coefficient of kinetic friction between a frozen wheel and the supporting surface is $0.3$ and the center of mass of the 64 kg door is at its geometric center. We have to neglect the small diameter of the wheels.</p>
<p><img src="https://i.stack.imgur.com/VmuAN.jpg" alt="enter image description here"></p>
<p>My attempt</p>
<p>I calculated the normal reactions at the two wheels to be equal to $32g$ each. I then applied $P=0.3N$ which gave me the incorrect value of $P$ .I couldn't get where I am wrong.Any help shall be highly appreciated.Thanks.</p>
| |civil-engineering|statics| | <p>See picture below for reference. The thing is, the forces are now applied at a body. This is quite different when you are analyzing a particle because the position of each force are significant enough to change the effects of other forces. </p>
<p><a href="https://i.stack.imgur.com/rYv7I.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/rYv7I.png" alt="Reference drawing."></a></p>
<p>Notes:</p>
<ol>
<li>The inertia is neglected because the velocity is to be constant (steady speed)</li>
<li>The friction force, $F_f$, depends linearly on the normal force, $R_a$. the relationship is given by: $F_f=\mu\ N$</li>
<li>Since the point of contact is at point $A$, the friction force acts at point $A$ as well.</li>
<li>The friction in the rotating wheel, $B$ is neglected.</li>
</ol>
<p>So to sum up, you <strong>need</strong> to consider the reduction of reaction force at A due to turning moment of force $P$. Such that:</p>
<p>Summing up forces at $B$, gives:</p>
<p>$$P\ (1)\ -\ 64\ (0.35)\ +\ R_a\ (0.7)\ =\ 0$$</p>
<p>and summing up forces along x-axis gives:</p>
<p>$$P\ =\ F_f\ =\ \mu_k\ R_a$$</p>
<p>$$P\ = 0.3\ (R_a)$$</p>
<p>Solving the two equations simultaneously gives:</p>
<p>$$P\ = \ 6.72\ kg$$</p>
<p>$$R_a\ = \ 22.4\ kg$$</p>
| 17224 | Force P required to slide the door if one of the wheel becomes frozen |
2017-09-23T10:32:42.323 | <p>Two automobiles both of which have the mass center located midway between the front and rear axles,are identical except that one is front-wheel-drive and the other is rear-wheel-drive.The cars are driven at constant speed at ramps of various inclinations.The question is to judge which car could climb the ramp of higher inclination angle from a theoretical point of view.</p>
<p><img src="https://i.stack.imgur.com/YCPa6.jpg" alt="enter image description here"></p>
<p>My idea was that the front wheel drive would be more effective if more weight acts on the front wheel.But in this case the center of gravity is midway between the rear and front wheel axles .I couldn't see how I can approach this problem.Any ideas?Thanks.</p>
| |statics| | <p>The rear-driven car is able to climb higher inclination.</p>
<p>Rational:</p>
<ul>
<li>Where is the engine and the weight has for sure an impact, but it is not related with rear or front driven car: You have rear-driven cars with engine in the front and reverse. Additionally, you specify that the center of gravity is in the same place (center), so the place of the engine has no impact.</li>
<li>However, if you consider the forces acting on the car, the gravity will lower the weight component on the front and increase the weight component to the rear. This increase the capacity of a rear-driven car to climb. </li>
</ul>
<p>Note: Consider however, that rear-driven cars will have lower direction control while climbing, and for this reason, they might have big issues in climbing icy roads. </p>
<p>Note 2: Driving a front-wheel-driven backward allows to climb higher inclinations than forward. But probably not something to try on the highway.</p>
<p>Note 3: The ideal situation for climbing strong inclinations is the rear-wheel-driven car with rear direction (you do not have the problem of control). You may find several construction machines designed in this way, but it is not practical (difficult to drive) at high speed, and thus not used for cars.</p>
<p><a href="https://i.stack.imgur.com/MJzfn.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/MJzfn.png" alt="enter image description here"></a></p>
| 17228 | Car which can climb the ramp of higher inclination |
2017-09-23T13:05:10.520 | <p>This is a problem from the book Engineering Mechanics Statics vol-1 by Meriam Kraige</p>
<p>The rectangular steel yoke is used to prevent slippage between the two boards under tensile loads $P$ .If the coefficients of static friction between the yoke and the board surfaces and between the boards are all $0.30$ the question is to determine the maximum value of $h$ for which there is no slipping and to find out the corresponding normal force $N$ between the two boards for $P=800N$ if impending motion occurs at all surfaces.
<img src="https://i.stack.imgur.com/X076V.jpg" alt="enter image description here"></p>
<p>Here is what I tried</p>
<p>For the first part if $\theta$ denotes the angle between the top part of yoke and horizontal then $\cos \theta = \frac{200}{210}=\frac{150}{h}$ which gives $h=157.5 mm$ </p>
<p>For the second part I tried to draw the FBD of the top block.</p>
<p><img src="https://i.stack.imgur.com/9NVfo.jpg" alt="enter image description here"></p>
<p>From FBD $$P=f_1+f_2+f_3+f_4$$ and $$mg=N_3-N_1$$ Also by taking moments about COM I got $P=f_3-f_1$.However I am not able to proceed after this.Any help shall be highly appreciated.Thanks</p>
| |statics|friction| | <p>I will try to answer it. </p>
<p>First, draw the forces acting upon it. (I considered only the top plank/board as it is symmetric with the bottom board). Refer to the figure below. </p>
<p><a href="https://i.stack.imgur.com/tlbKs.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/tlbKs.png" alt="Reference drawing"></a></p>
<p>Separating the upper plank and analyzing the forces upon it, you will get the figure below.</p>
<p><a href="https://i.stack.imgur.com/q13Hn.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/q13Hn.png" alt="Figure 2"></a></p>
<p>Note that $F_{\text f1}=P_y (0.3)$ and $F_{\text f2}=N(0.3)$, therefore:</p>
<p>Summing up forces in the x-axis gives:</p>
<p>$$0.3(P_y)+0.3(N)=800$$</p>
<p>and Summing up forces in the y-axis gives:</p>
<p>$$N-P_y=0$$</p>
<p>Solving both, </p>
<p>$$P_y=N=1,333.33\ N$$</p>
<p>Solving also for $F_{\text f1}$ gives $F_{\text f1}=400\ N.$</p>
<p>Solving for the angle of inclination of the yoke gives</p>
<p>$$\theta = arctan (400/1333.33)=16.70\ degrees$$</p>
<p>therefore, $h = 150/\cos(16.70)=156.60\ mm$</p>
| 17232 | Normal force between two boards |
2017-09-24T05:22:31.613 | <p>This is a problem from the book Engineering Mechanics Statics vol-1 by Meriam Kraige.</p>
<p>The elements of a unidirectional mechanical clutch are shown in the figure.A torque $M$ applied to the outer ring is transmitted to the output shaft through frictional interaction between the outer ring and the balls and between the balls and the inner driven member.If the direction of rotation of the outer ring is reversed,the wedging action of the balls is absent,and no torque can be transmitted to the output shaft.For given values of $r,r_o$ and coefficient of friction $\mu$ which is applicable for both pairs of interacting surfaces,the question is to find the minimum dimension $b$ of inner member which will permit the transmission of torque without slipping.</p>
<p><img src="https://i.stack.imgur.com/EfxH8.jpg" alt="enter image description here"></p>
<p>I couldn't get how to approach this problem.i tried drawing FBD of the balls but could get nothing out of it.Any help shall be highly appreciated.Thanks.</p>
| |statics|friction| | <p>Let me give you a headstart.</p>
<p>$T=F*d$, therefore, $F_{f1}=(T/2)/r$. The torque is divided by two because there are two balls.</p>
<p>From there, you can compute $N_1$. Sum up moments at the center of the ball, you will get:</p>
<p>$F_{f1}=F_{f2}$</p>
<p>Now that you have $F_{f2}$, you will get $N_2$. </p>
<p>Looking at the output shaft now, you will get the position of the force as shown in the right figure. You will need to sum up moment at the center of the output shaft. The resulting moment needs to be equal to $T/2$ as we need to transmit the torque to the output shaft. </p>
<p>This will require hard mathematics but I think this will do it. I know there is a simpler solution so you we will have to find out.</p>
<p><a href="https://i.stack.imgur.com/sKp5y.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/sKp5y.png" alt="See image below."></a></p>
| 17241 | Dimension of the unidirectional mechanical clutch |
2017-09-24T07:29:27.510 | <p><a href="https://i.stack.imgur.com/7xNfU.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/7xNfU.png" alt="enter image description here"></a></p>
<p>Can someone please explain why the support at C does not have a vertical reaction force while the support at E has?
Doesn't the support at C have the same kind of pin connection as the support at E?</p>
| |statics|beam|connections| | <p>The rod BC is a <strong>two force member</strong>. Therefore, the forces on either end must be along the axis of the member. i.e. since the rod is horizontal, the forces on either end can only be horizontal. </p>
<p>The beam AGDE is not a two force member. Therefore, it can have forces on the ends in both directions.</p>
<p>In other words, the reason that the reactions are different does not have to do with the pinned boundary, but one of the members being two force and the other not.</p>
<p>If this doesn't make sense, I would suggest you look up "two force member" in your textbook.</p>
| 17242 | Why are the support reactions of the two pin-connected members different? |
2017-09-25T02:09:50.273 | <p>I have this question assigned as part of my homework, and I am having trouble figuring out the solution. Below is the question and my attempts at solving it. I need to have the homework ready tomorrow morning, so I would really appreciate any help or direction in finding the solution.</p>
<hr>
<p><strong>Question:</strong></p>
<p>Suppose that a composite solid consists of alternating materials $A$ and $B$, with layer thickness $L_A$ and $L_B$, respectively. Both materials are isotropic, with thermal conductivities $k_A$ and $k_B$. It is desired to predict the average, steady-state heat flux in a sample that contains many layers. The flux in the composite is written as
$$\boldsymbol{ Q=-k\cdot G}$$
where $\boldsymbol{Q}$ and $\boldsymbol{G}$ are the average heat flux and temperature gradient, respectively, and $\boldsymbol{k}$ is the effective thermal conductivity. With coordinate axis of material interfaces aligned with normal along the $y$-axis, the effective conductivity will be of the form
$$\boldsymbol{k}=\left[\begin{array}{ccc} k_{xx} & 0 &0\\0&k_{yy}&0\\0&0&k_{zz}\end{array}\right].$$
(a) Evaluate $k_{xx}$, $k_{yy}$, and $k_{zz}$ in terms of $k_A$, $k_B$, and $\phi$, where $\phi=L_A/(L_A+L_B)$ is the volume fraction of material $A$ ($L_A$ is the length of material $A$, and $L_B$ is the length of material $B$). (<em>Hint</em>: The steady heat flux $q$ through a slab of thickness $L$ and thermal conductivity $k$ is $q=k\Delta T/L$, where $\Delta T$ is the temperature difference between surfaces.)</p>
<p>(b) Show that, in general, the vectors $\boldsymbol{Q}$ and $\boldsymbol{-G}$ will not be parallel to one another. That is, the directions of the heat flux and temperature gradient will tend to differ.</p>
<hr>
<p><strong>Work:</strong></p>
<p>We can consider the orientation of axes such that the positive $x$-axis comes out of the page towards us, the positive $y$-axis goes to the right of the page, and the positive $z$-axis goes to the top of the page. We can also consider a segment of the composite solid with material $A$ on the left and material $B$ on the right along the $y$-axis, i.e., material $A$ is from $0$ to $L_A$ on the $y$-axis, and material $B$ is situated from $L_A$ to $L_B$ along the $y$-axis. </p>
<p>The vector equation $\boldsymbol{Q=-k\cdot G}$ can be broken down into 3 equations (using the definition $\boldsymbol{G}=\nabla T$):
$$q_x=-k_{xx}\frac{\partial T}{\partial x}\\q_y=-k_{yy}\frac{\partial T}{\partial y}\\q_z=-k_{zz}\frac{\partial T}{\partial z}$$</p>
<p>Regarding $k_{xx}$ and $k_{zz}$, since the difference in material is only along the $y$-axis, I would think that $k_{xx}$ and $k_{zz}$ would not depend on $\phi$, $L_A$, or $L_B$, rather only on the thermal conductivity of the particular material $A$ or $B$, depending on the location along the $y$-axis. However, if the materials are assumed to be infinite along the $x$-axis and $z$-axis (this may be a logical inference, since the height and width of the composite solid are not given), then the flux along the $x$ and $z$ directions should be $0$.
If this assumption is correct, then $\frac{\partial T}{\partial x}=\frac{\partial T}{\partial z}=0$, so this tells me nothing about $k_{xx}$ or $k_{zz}$. Is this assumption correct? Did I miss something regarding the evaluation of $k_{xx}$ and $k_{zz}$?</p>
<p>To figure out $k_{yy}$, we can construct equations for q for each material along the $y$-axis:
$$q_A=k_A\Delta T_A /L_A\\q_B=k_B\Delta T_B/L_B$$
Then, using the given hint, we can consider temperature to the left of material $A$ to be $T_0$, the temperature at the interface between materials $A$ and $B$ to be $T_1$, and the temperature to the right of material $B$ to be $T_2$. Accordingly, $\Delta T_A=T_0-T_1$ and $\Delta T_B=T_1-T_2$, so the equations become the following:
$$q_A=k_A\left(T_0-T_1\right)/L_A\\q_B=k_B\left(T_1-T_2\right)/L_B$$
The Equations can be rearranged:
$$T_0-T_1=q_A L_A/k_A\\T_1-T_2=q_B L_B/k_B$$
Then the two equations can be added together to form $\Delta T=T_0-T_2$:
$$\Delta T=\frac{q_A L_A k_B +q_B L_B k_A}{k_A+k_B}$$
From here, however, I am not sure how I would proceed; $\phi$ does not explicitly appear, and $q_A$ and $q_B$ do appear but the question implies they should not appear.</p>
<p>An alternate approach would be to add the two equations together:
$$q_A+q_B=\frac{k_A L_B \left(T_0-T_1\right)+k_B L_A \left(T_1-T_2\right)}{L_A+L_B}$$
From the definition of $\phi$, we also know that $1-\phi=L_B/(L_A+L_B)$. Using this fact, the above equation can be rearranged as follows:
$$q_A+q_B=k_A(T_0-T_1)+\phi\left[k_A(T_1-T_0)+k_B(T_1-T_2)\right]$$
However, here again I do not know how to proceed; I still have $T_0$, $T_1$, and $T_2$ separate, and this equation is only useful if we assume $q_y=q_A+q_B$. Is this last assumption correct? Is there a simplification of the equation that I missed?</p>
<p>Regarding (b), I presume that once I find $\boldsymbol{Q}$ and $\boldsymbol{-G}$, I can show $\boldsymbol{Q\cdot -G}\neq1$, but I do not yet have $\boldsymbol{Q}$ or $\boldsymbol{-G}$ (or even $\nabla T$, from which I can find $\boldsymbol{-G}$) to figure this out.</p>
<p>Any help or direction on any part of this problem would be much appreciated! As I mentioned, I need to have the homework ready tomorrow morning, so I am kind of in a rush. Thank you in advance!</p>
| |fluid-mechanics|thermodynamics|heat-transfer| | <p>The thought process is not entirely correct in the work section. The heat fluxes are not additive, rather they should be equal due to conservation of energy. </p>
<p>Thus, in the $y$ direction, $q_y=q_{yA}=q_{yB}$.
Since $q=k\frac{\Delta T}{L}$,
\begin{align}
q_{yA}&=k_A\frac{\Delta T_A}{L_A}\\
q_{yB}&=k_B\frac{\Delta T_B}{L_B}.
\end{align}
Therefore,
\begin{align}
q_{y}&=\frac{k_A}{L_A}(T_0-T_1)\\
q_{y}&=\frac{k_B}{L_B}(T_1-T_2).
\end{align}
Rearranging the equations and replacing in for $T_1$,
$$
T_0-\frac{L_A}{k_A}q_y=T_2+\frac{L_B}{k_B}q_y,
$$
from which
$$
T_0-T_2=q_y\left(\frac{L_A}{k_A}+\frac{L_B}{k_B}\right).
$$
Defining $\Delta T=T_0-T_2$,
$$
q_y=\frac{1}{\frac{L_A}{k_A}+\frac{L_B}{k_B}}\Delta T.
$$
Since $G_y=-\frac{\Delta T}{L_A+L_B}$,
$$
q_y=\frac{L_A+L_B}{\frac{L_A}{k_A}+\frac{L_B}{k_B}}\frac{\Delta T}{L_A+L_B}=-\frac{L_A+L_B}{\frac{L_A}{k_A}+\frac{L_B}{k_B}}G_y.
$$
Rearranging and using the definition $\phi=\frac{L_A}{L_A+L_B}$,
$$
q_y=-\frac{k_Ak_B}{(1-\phi)k_A+\phi k_B}G_y.
$$
Since $q_y=-k_{yy}G_y$,
$$
k_{yy}=\frac{k_Ak_B}{(1-\phi)k_A+\phi k_B}.
$$
For $k_{xx}$ and $k_{zz}$,
the paths through $A$ and $B$ are parallel, so $\frac{\Delta T}{L}$ is the same for $A$ and $B$, and $G_x=G_z=-\frac{\Delta T}{L}$ is the same for $A$ and $B$.
Since the paths are parallel,
\begin{align}
q_x&=\frac{L_A}{L_A+L_B}q_{xA}+\frac{L_B}{L_A+L_B}q_{xB}\\
q_z&=\frac{L_A}{L_A+L_B}q_{zA}+\frac{L_B}{L_A+L_B}q_{zB}.
\end{align}
Now,
\begin{align}
q_{xA}&=q_{zA}=k_A\left(\frac{\Delta T}{L}\right)_A\\
q_{xB}&=q_{zB}=k_B\left(\frac{\Delta T}{L}\right)_B,
\end{align}
but
$$
\left(\frac{\Delta T}{L}\right)_A=\left(\frac{\Delta T}{L}\right)_B=-G_x=-G_z,
$$
so
\begin{align}
q_x&=\phi k_A(-G_x)+(1-\phi)k_B(-G_x)=-\left[\phi k_A+(1-\phi)k_B\right]G_x\\
q_z&=\phi k_A(-G_z)+(1-\phi)k_B(-G_z)=-\left[\phi k_A+(1-\phi)k_B\right]G_z.
\end{align}
Since
\begin{align}
q_x&=-k_{xx}G_x\\
q_z&=-k_{zz}G_z,
\end{align}
$$
k_{xx}=k_{zz}=\phi k_A+(1-\phi)k_B.
$$
Therefore,
$$
\mathbf{k}=\left(
\begin{array}{ccc}
k_{zz}=\phi k_A+(1-\phi)k_B&0&0\\
0&\frac{k_Ak_B}{(1-\phi)k_A+\phi k_B}&0\\
0&0&k_{zz}=\phi k_A+(1-\phi)k_B
\end{array}
\right).
$$</p>
<p>For part b, in order for two vectors $\mathbf{a}$ and $\mathbf{b}$ to be parallel, the ratio $a_i/b_i$ must be constant for all $i$.
Here,
\begin{align}
\mathbf{Q}&=-k_{xx}G_x\hat{x}-k_{yy}G_y\hat{y}-k_{zz}G_z\hat{z}\\
-\mathbf{G}&=-G_x\hat{x}-G_y\hat{y}-G_z\hat{z}.
\end{align}
\begin{align}
\frac{-k_{xx}G_x}{-G_x}&=k_{xx}=k_{zz}=\frac{-k_{zz}G_z}{-G_z}\\
\frac{-k_{yy}G_y}{-G_y}&=k_{yy},
\end{align}
but unless $k_A=k_B$, $k_{yy}\neq k_{xx}$, so the ratio does not hold and the vectors are not parallel.</p>
| 17252 | Thermal Conductivity of a Layered Composite |
2017-09-25T07:46:57.117 | <p>I am currently rating two motorcycles to decide which one to buy. The criteria which is most important to me is the power-to-weight ratio. Now, I want to calculate the power of the motorcycle by looking at its displacement. The engine displacement is the amount of air-fuel mix that an engine can burn at a time. Looking at google, I cannot find a concrete formula to convert the engine displacement (cubic centimeter) to power (horsepower). </p>
<p>Currently, I am looking at two models:</p>
<ol>
<li>Suzuki Raider J 115 Fi (fuel injection)</li>
<li>Honda XRS 125 (Carburetor)</li>
</ol>
<p>I know that the Honda model will provide more power but I want to look at the other model's fuel efficiency. I want to convert to see whether the power difference is not that significant. For the sake of comparison/formula, assume that you feed both with the same amount of fuel.</p>
<p>Can you please advise?</p>
| |mechanical-engineering|automotive-engineering| | <p>There's no simple relationship. When manufacturers quote a figure for power it's usually a peak value, which doesn't mean much without knowing the engine speed (rpm) at which this is developed. Power is time based, so an engine capable of spinning twice as fast should be able to burn twice as much air/fuel in the same time for a given displacement.</p>
<p>If you can find power and torque curves, these will tell you more than a single value. You'll get different power and torque curves depending on how the displacement is achieved (broader piston or longer stroke).</p>
<p>Power to weight ratio is a factor, but if you're looking at low speed acceleration you'll probably find that torque is more relevant.</p>
| 17257 | Engineering formula to relate engine displacement and power |
2017-09-25T20:26:50.753 | <p>The problem I have at hand is the following: I have a container C with a given volume V. I need to fill it with water, fast, up to a given pressure P. I have a number of high pressure hydropneumatic pumps to choose from, but they are expensive and take a while to get here. I don't even have such pump readily at hand to test the equations. The flow rate for these pumps is easily calculated as:</p>
<p>$$
Q = Q_{max} - K_{b} P
$$
$K_{b}$ is a pump parameter per datasheet. Obviously pressure $P$ won't go higher than $Q_{max}/K_{b}$, the pump stops working. Now I start simplifying things: I consider no thermal factors, no sudden pipe diameter change, no turbulence, no back flow, constant ambient pressure, always the same fluid; basically no hydrodynamics. And the longest shot, I consider the pressure per volume rate of the container constant:
$$
\frac{P}{V} = K_{t}
$$
I treat $K_{t}$ as some sort of elasticity constant which in reality comprises the actual container elasticity,water compressibility, container's volume, etc. It's the only empiric data I have: how many liters of water I stuffed in the container to get the target pressure. Taking differentials, and considering t=0 the moment the container is full with P = 0, I get:
$$
P(t) = \frac{Q_{max}}{K_{b}}(1-e^{-K_{t}K_{b}t})
$$
Which makes sense for a single pump. But I can't find a way to adjust it to multiple pumps in parallel with different parameters. I need an equation to later iterate and get the best combination of pumps possible. Trying to get an "equivalent" pump, the flow rate is the sum of flow rates as in: $Q(t) = Q_{1}(t) + Q_{2}(t) ...$, easily resolving to an equivalent pump having parameters $[\sum{Q_{{max}_i}},\sum{K_{{b}_{i}}}] $, but that does not make sense with the equation above. Keep in mind this is for pressures up to 2000 Bar.</p>
| |fluid-mechanics|pumps| | <p>Well, I did a numerical approach on this using a backward euler form, and it turns out the equation above is actually correct. My mistake was that I was not considering anywhere in the formulation that indeed the low pressure pumps shut off.</p>
<p>The optimization algorithm must take this into consideration and recalculate the equivalent pump ruling out the lowest pressure pump as soon as that pressure is reached (or at least considered so, at about 70% of the pressure).</p>
| 17269 | High pressure water pumps filling a container. How long does it take to obtain P pressure? |
2017-09-26T00:19:42.780 | <p>This is a simple question to follow up the question from the thread below:</p>
<p><a href="https://engineering.stackexchange.com/questions/17257/engineering-formula-to-relate-engine-displacement-and-power">Engineering formula to relate engine displacement and power</a></p>
<p>Motorcycle manufacturers tend to show the displacement value in mass production motorcycles incorporated in their names such as Honda Wave 125i, Suzuki Raider 150, Kawasaki Z1000, etc. Most of these are below 200cc displacement. (Exceptions are Yamaha YZF-R1M, Kawasaki Ninja H2, etc).</p>
<p>The question is: If power is not determined by the displacement alone based on answers and comments on the thread above, does this imply that the displacement is the most important factor in determining the engine's overall performance?</p>
| |mechanical-engineering|automotive-engineering| | <p>If you take "performance" as a mix of how fast the bike will accelerate and travel along its intended terrain, you can roughly approximate it as a ratio of thrust to mass.</p>
<p>As mentioned by others, thrust can be increased by having a larger capacity engine with a corresponding mass penalty. Depending on the terrain, this penalty may be more or less important - there are plenty of 1000cc street bikes but very few off road bikes.</p>
<p>There are other reasons for numbers in the name though, one being marketing (I don't remember Mazda being big on publishing the engine capacity of their rotary cars) the other being licensing. In the UK, I can ride a 125cc bike on my current license. Having numbers in the name makes it incredibly clear what I need to be looking at. </p>
| 17271 | Is displacement the most important factor in determining a motorcycle engine's overall performance? |
2017-09-26T03:51:22.323 | <p>I often seen printers <a href="http://www.office.xerox.com/multifunction-printer/color-multifunction/xerox-color-c60-c70/enus.html" rel="nofollow noreferrer">like this</a> in offices and schools. Considering how we can have printers small enough to fit on a desk, why are these printers taking up half a room? What are they capable of doing, and why do they need to be that large?</p>
| |mechanical-engineering|computer-hardware| | <p>One requirement is serviceability and time between refills (of either ink or paper. </p>
<p>When printing 15 copies of a 100 page report you don't want to have to stand next to it and refill the paper and ink 10 times before it's done. Instead you want to push the button and then pick up the stack for binding later after you had your coffee.</p>
<p>Having components be serviceable means that you need to be able to access them or remove them. This means adding space for an arm to reach in (and possibly a head) or adding bulk for the interface of the module and the rest of the machine (or both).</p>
| 17273 | Why are corporate printers so large? |
2017-09-26T19:23:55.173 | <p>I'm new to mechanical engineering, gears, and motors. Coming from a software/hardware background, I am trying to understand how to leverage gear reduction to maximize the torque output of a stepper motor for an equatorial mount. The speed can be very slow so I would like to focus on maximizing torque for a heavy payload.</p>
<p>By my understanding, the output torque is the product of the input torque and the gear ratio. Is this the same for the "holding torque" of a gear system? Is the output holding torque the product of the input holding torque and the gear ratio?</p>
| |mechanical-engineering|gears|torque|stepper-motor| | <p>Yes, the holding torque goes up proportionally with the gear ratio. Plus the friction of the gear train, and any other seals, bearings along the gear train that add friction. </p>
| 17289 | How does gear reduction affect a stepper motor's holding torque? |
2017-09-27T11:16:20.450 | <p>If we have an object on a frictionless inclined bench, why does the minimum force applied to keep it steady have to be parallel to the incline (or perpendicular to the reaction force)?</p>
<p><a href="https://i.stack.imgur.com/eBEbX.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/eBEbX.jpg" alt="Object on an inclined bench"></a></p>
<p>In other words, why does the angle θ in the image have to be 0 in order for the force to be minimum?</p>
<p>Consider we know the weight of the object and the reaction of the incline.</p>
| |statics| | <p>Since I posed the question in comment, here is how to treat the friction case, and of course at the end it reduces to the expected friction free result.</p>
<p>Assume the applied force has components $f_x$ parallel (up) the plane and a normal component $f_y$ downward into the plane:</p>
<p>The normal force between block and plane is:</p>
<p>$$f_n = W\cos(t) + f_y$$</p>
<p>The force $f_x$ required to hold against sliding, assuming static coulomb friction with coefficient $u$ is </p>
<p>$$f_x = W\sin(t) - u f_n = W\sin(t) - u (W\cos(t) + f_y)$$</p>
<p>The magnitude of the applied force is then</p>
<p>$$f_{mag} = \sqrt{f_x^2+f_y^2}$$</p>
<p>without showing all the messy steps, setting $d f_{mag} / d f_y == 0$ we can solve for $f_y$:</p>
<p>$$f_y = u W\sin(t)\cdot\dfrac{1 - u /\tan(t)}{1 + u^2}$$</p>
<p>and</p>
<p>$$f_x = W\sin(t) ( 1 - u\cdot\dfrac{1/\tan(t) + u}{1 + u^2})$$</p>
<p>For the friction free case ($u=0$) this results in the expected $f_x=W\sin(t),f_y=0$ (ie parallel to the plane) but with friction angling the force toward the plane reduces the required force magnitude.</p>
| 17297 | Why does the minimum force have to be parallel to the incline? |
2017-09-28T04:04:06.247 | <p>What are the Principle directions? Are they unit normals of three planes perpendicular to each of the principle stresses?</p>
| |materials| | <p>I think that "Principle directions" can be used both in terms of stress (the direction of principal stresses) and in terms of inertia (the 3 direction where the inertia matrix is diagonal).</p>
<p>I mainly use "principal direction" in terms of inertia. </p>
<p>I think that you can ganeralize: principal directions are the unit vector where something 3D distribution (inertia, stresses...maybe other things) can be represented as a diagonal matrix. </p>
| 17316 | Principle Directions of a Stress Tensor |
2017-09-28T14:09:23.037 | <p>I am working on split type AC compressor. I actually want to know, if i changed the frequency of the compressor to run on different power level, will it affect the life of the compressor. My actual work to run the Compressor with different power level. I will use VFD ( variable frequency drive) to change the frequency of compressor and it will take different power with respect to the frequency. can you tell me that all rotary compressor is designed for a specific frequency. ?</p>
| |compressors|compressed-air| | <p>The answer is, we don't know what compressor you have so we can't answer definitively. MANY compressors are indeed designed to operate at a very narrowly specific speed because the lubrication of it is dependent on that speed. Also, are you working with a compressor that would use a 3 phase motor? Because for the most part VFDs only work on 3 phase motors. There are a few that work on SOME types of single phase motors, but the type of motors that will work are not the types that are typically used on compressors.</p>
| 17323 | can we run rotary compressor of Air conditioner with different frequency level? |
2017-09-28T18:28:09.293 | <p>Steel prefabricated buildings cost less, require less time and provide a reasonable quality compared to concrete buildings.</p>
<p>They are common for warehouses, offices and other types of buildings. Also, there are prefabricated steel houses for residential use.</p>
<p>However, I have noticed that steel buildings/apartments are rare/non-existent. Is there a reason for this.</p>
<p>I am specifically asking regarding the usage of prefabricated steel in apartment and/or multistory buildings for regular use. </p>
| |steel|building-design| | <p>It is actually quite common in Scandinavia in my experience, using steel beams and columns with concrete hollowcore slabs. <a href="http://www.stalforbund.no/uploads/source/files/Materialet/Hulldekker/Veileder_Hulldekker_2008_2.pdf" rel="nofollow noreferrer">This PDF</a> describes one example of such a system.</p>
| 17328 | Why aren't Steel Prefab Buildings common for residential use? |
2017-09-30T09:58:25.223 | <p>This is an above view of a system in which tin cans arrive via conveyor (1) to a "feeding star" ("<strong>подающая звезда</strong>" in Russian; No. 5) and are placed firmly in slot 4. (<a href="http://www.freepm.ru/Models/105263.html" rel="nofollow noreferrer">source in Russian</a>)</p>
<p><a href="https://i.stack.imgur.com/OG0oo.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/OG0oo.gif" alt="enter image description here" /></a></p>
<p>What does one call this "feeding star" in English?</p>
<p>I'm actually translating a text about a pharmaceutical filling-and-capping line, in which there is a "feeding star" and a "stoppering star". My text comes with no drawings, but I googled and found out that these "stars" are basically star-like cogwheels that take something from a moving conveyor system and position it precisely for further use. In my text, this "feeding star" probably passes drug-filled vials to the "stoppering star" which is part of the vacuum-operated automatic stoppering unit.</p>
| |terminology| | <p>Looks like it may be called <strong>starwheel feeder</strong> or <strong>starwheel infeed</strong> - here's an excerpt from <em><a href="https://goo.gl/A2FRG3" rel="noreferrer">Design of Automatic Machinery</a>:</em></p>
<blockquote>
<p><a href="https://i.stack.imgur.com/6mlAx.png" rel="noreferrer"><img src="https://i.stack.imgur.com/6mlAx.png" alt="enter image description here"></a></p>
</blockquote>
| 17350 | What do I call this "cogwheel" that feeds tin cans to a specific position? |
2017-09-30T16:11:52.777 | <p>I read that when you want to double the max speed of e.g. a boat you would have to increase the power eightfold! I could understand that you have to square it (= fourfold) because you are working against "a wall" of water which makes it an area (= power of two)... but why to the power of three?</p>
<p>Could you please give a physical/mathematical explanation but also an intuition on why this is the case. </p>
| |power| | <p>It's a pretty rough approximation, and only holds where skin friction is the dominant source of drag throughout the speed range considered. Skin friction increases about as speed ^2. So for double speed, the resistance force is quadrupled. Plus the rate is doubled. Power is the rate of doing work, so 4 X force and 2 X rate equals 8 X power. This doesn't work if wave making is significant, and it isn't very good in the planing regime where surface area is changing with speed.</p>
| 17354 | Intuitive explanation for why doubling max speed is proportional to eightfold power |
2017-10-01T06:44:40.173 | <p>I was browsing for Gibraltar strait tunnel where due to greater depth it is not being build.
Question popped up as to why tunnel building is difficult at lets say 1000m. we have coal mines which go deeper than that, than why not tunnel?</p>
| |civil-engineering|tunnels| | <p><strong>Tunnel Access</strong></p>
<p>To begin, to construct and use a tunnel, it needs to be accessed. Access will be via either a vertical or sub-vertical shaft or a ramp that wheeled vehicles can use.</p>
<p>Shafts may be practical for mines, but for a traffic tunnel access via a shaft is impractical because of the rate of transfer of vehicles from one level to the other. A small number of vehicles will be able to fit into a cage in a shaft & then the cage will be lowered or raised, transporting the vehicles. This is OK for mines but not for a large number of vehicles carrying trade goods or cars with families.</p>
<p>This means that the only practical way of accessing such a tunnel is via a ramp. Irrespective of whether the ramp is straight or spiraled, its overall length will be dictated by the gradient of the ramp.</p>
<p>Assuming a uniform gradient of 1 in 10 (10 percent), for a tunnel 1000 m (1 km) deep, the horizontal length of the ramp will be 10 000 m and its slope length will be 10 050 m (10.05 km). Such a ramp will need to be constructed at both the entry and exit of the tunnel. Thus, the overall length of the tunnel will need to be an additional 20.1 km. This would add significant cost to the construction of the tunnel.</p>
<p>For vehicular traffic, will one tunnel, divided in half length-ways be enough, or will two tunnels be required: one allowing for traffic to flow in one direction and the other tunnel for traffic to flow in the other direction. If two tunnels are needed, then double the cost of construction.</p>
<p><strong>Tunnel Ventilation</strong></p>
<p>Irrespective of the type of vehicles that use the tunnel, electrically powered vehicles or combustion engine powered vehicles, ventilation will need to be provided to the tunnel.</p>
<p>Ventilation can be provided via ventilation shafts or another tunnel dedicated to ventilation and servicing the vehicular transportation tunnel. If the tunnel passes under water, such as the Straits of Gibraltar, then ventilation shafts cannot be used along the route of the tunnel - a dedicated ventilation tunnel must be used.</p>
<p>Then there is the cost of powering the ventilation fans and how the energy will be sourced to do so - solar, wind, gas, nuclear, coal power stations.</p>
<p><strong>Ground Control</strong></p>
<p>Another issue that will need to be considered is that at 1000 m depth ground stresses may be large which will dictate the form of ground control measures and the subsequent cost.</p>
<p><strong>Tunnel Rock Temperature</strong></p>
<p>Additionally, at 1000 m depth the rock temperature will be high. Depending on location, temperatures of 40 C to 60 C are not unusual. To cater for this, will the vehicular transportation tunnel need to be cooled via an air cooling system. This will cost a lot of money to install and operate. Again, the issue of a power source for such a system will need to be addressed.</p>
| 17360 | Tunnel at depth problem |
2017-10-03T18:58:23.867 | <p>Suppose we have the following truss-like structure, with a load of 1N on the middle joint and two reactions on the bottom joints.</p>
<p><a href="https://i.stack.imgur.com/JOVqB.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/JOVqB.png" alt="Truss"></a></p>
<p>If I try to analyze this using a Cremona diagram, I run into the problem that I do not know how to handle a load on a joint inside the structure. Is it possible to analyze this with a Cremoda diagram, and if so, how does the naming of the different faces and panels work? Because we can only depict forces on joints between faces, right?</p>
<p>Is there even a solution?</p>
| |structural-engineering|structural-analysis| | <p>The vertical reactions at A and B are clearly 0.5N each, by symmetry.</p>
<p>Since there are reactions at A and B, those points must be connected to something external. If the connections are fixed points, there is no force in member AB because the length of AB can't change.</p>
<p>Interpreting the diagram literally, there are no horizontal components of the reaction at points A and B.</p>
<p>Removing AB makes the truss a statically determinate structure and you can solve for the other member forces in the usual way - for example, start at joint A and find the forces in AC and AD.</p>
<p>If you don't accept the argument that there is no force in AB, there is no unique answer to the problem, because the structure is statically indeterminate.</p>
| 17387 | Internal load in cremona diagram |
2017-10-05T15:41:13.167 | <p>I need to cut a 7mm solid Aluminium plug (Non-hollow and non-threaded) to replace a part in my car. </p>
<p>Is there a simple drill adapter tool to cut a solid metal cylinder (Plug) from an Aluminium mass ?</p>
<p>(Hole saw cutters are not in small sizes like 7mm)</p>
<p>Edit:</p>
<p>Commercially these plugs are called "Solid Aluminium Rivets". But 7mm size is not available commercially (Available maximum size is 1/4" or 6.4mm). Actually head part of the rivet is not necessary for the function. It is needed for remove the plug for rebuild. But practically, that head got damaged when I try to remove it by a plug remover. So I had to drill-out the plug. So there is no photo to show. So, since the head is useless, I just need to make a plug to close the Idle port of my old carburetor.</p>
| |metals|drilling|cutting| | <p>It depends a but what you mean by M7 size. Specifying M7 rather than 7 mm diameter implies that the hole is threaded. If this is the case you are better off with a bit of M7 threaded rod cut to length. The best material will depend on the application aluminium male threads are rare but do exist. In most cases stainless should be ok. If you do use a metal other than aluminium galavnic corrosion may be an issue but using thread locking or anti-seize compound should mitigate that pretty well unless it is constantly exposed to water or coolant (in which case a standard thread may not provide adequate sealing anyway). </p>
<p>If you do mean a plain 7mm diameter hole then you can just buy 7mm diameter rod and cut it to length, although getting it to stay in the hole may not be trivial and you might be better off tapping the hole anyway. </p>
<p>Edit : based on the additional information in the question I would be tempted to tap the hole with an M8 thread and use a grub screw with some thread locking compound as the plug. If sealing is an issue you could also use a button head set screw with a soft washer. </p>
<p>Actually if you set a 6.4 mm solid rivet properly it the shank should expand to fill a 7mm hole but it sounds like this will be difficult in this case as I assume you have limited access to the back side of the hole. </p>
| 17403 | Is there a drill adaptor or attachment tool for cut a metal solid cylinder |
2017-10-05T17:57:39.807 | <p>I was reading a text talking about Joule's experiment(mechanical equivalent of heat)<br>
The mechanical equivalent of heat was a concept that had an important part in the development and acceptance of the conservation of energy and the establishment of the science of thermodynamics.. The apparatus used in the experiment is shown below:
<img src="https://i.stack.imgur.com/O1Tcq.gif" alt="enter image description here"> </p>
<p>I have noticed that there is a spring balance in that apparatus. Why is there a spring balance in the mechanical equivalent of heat apparatus? What will happen if it is removed? </p>
| |mechanical-engineering|thermodynamics|heat-transfer| | <p>The spring balance is used to load a <a href="https://en.wikipedia.org/wiki/Prony_brake" rel="nofollow noreferrer">Prony brake</a>.</p>
| 17407 | Why is there a spring balance in the mechanical equivalent of heat apparatus |
2017-10-05T21:46:49.833 | <p>I am doing a project looking into the advantages of radio/microwave heating of oil reservoirs. I've seen research indicating that RF/Microwave heating can be used for environments such as shallow, tight, high permeability, fractured, etc zones. What are the disadvantages of applying techniques such as SAGD in these zones?</p>
| |geotechnical-engineering|rf-electronics|petroleum-engineering| | <p>Heating oil makes it less viscous which makes it easier to flow and thus easier to pump. RF/microwave heating is an easy and relatively safe way of reducing the viscosity of the oil a reservoir.</p>
<p>Additionally, the greater the permeability of a rock mass the more easily a fluid will flow through it. If a rock mass has large pore spaces, such as sandstone, the more permeable it will be. For rock which has small/tight pore spaces the permeability can be increased by fracturing the rock mass.</p>
<p>SAGD (steam assisted gravity drainage) is another way to reduce the viscosity of the oil. The effectiveness of SAGD would rely on how precise the horizontal steam holes were drilled relative to each other.</p>
<p>Rock mass have varying properties and there may be softer and harder portions within a rock mass. This can affect the path of the drill bit and steel when the holes are drilled. Consequently, the holes may not always be exactly where they are required for efficient SAGD recovery of the oil. Some parts of the reservoir may be cooler or hotter than needed. Applying additional heat via RF/microwave heating will increase the likelihood of more of the oil being heated and having its viscosity reduced and thus potentially increasing the recovery of oil from the reservoir.</p>
| 17410 | What types of oil reservoirs are applicable to radio/microwave heating and advantageous to SAGD? |
2017-10-06T15:10:48.693 | <p>I'm currently trying to extend the shock-absorbing capabilities of a welded steel structure that is shown in the following image. It is a structure that supports a slat conveyor. A load is to fall approximately in the middle half of the structure. The base of the structure rests in a concrete foundation.</p>
<p>I need to expand the shock-absorbing capabilities because I'm worried about the structure failing due to pure impact forces. I'm not as worried by vibrations as by the maximum force exerted on it due to the impact. The load impact is difficult to quantify, even for a worst case scenario, but it is safe to say that the structure has resisted an impact force that is 25% lower than the one that is to fall now. </p>
<p><a href="https://i.stack.imgur.com/HuQwD.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/HuQwD.png" alt="Welded steel structure"></a></p>
<p>I would prefer not to change all the beam sizes, since the slat conveyor that rests on top of it is already assembled, but I could add some supports if needed. </p>
<p>According to <em>Design of Welded Structures</em>, however, making the beam span between supports shorter raises the impact stresses suffered by the structure, even if it lowers the static stresses (once the load is in place).Is there anything I can do to minimize these impact stresses?</p>
| |structural-engineering|civil-engineering|structures|dynamics| | <p>Large displacement shock absorbers used to dampen the effects of seismic events may be suitable for this application. These could be used to isolate the part of the structure that is grounded and the part that experiences the impact.</p>
<p>Seismic dampers sit between the ground slab and the building and can effectively reduce the shock loadings in all directions. Hydraulic dampers are often used to attenuate forces in a specific direction.</p>
<p>The dampers are available in many sizes and formats, from simple steel springs to sophisticated viscous piston systems.</p>
<p><a href="https://i.stack.imgur.com/3NfDL.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/3NfDL.jpg" alt="Steel spring used in a building footing"></a></p>
<p><a href="https://i.stack.imgur.com/aChhs.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/aChhs.jpg" alt="Hydraulic dampers"></a></p>
| 17421 | How can I absorb shock on an existing steel structure? |
2017-10-08T16:59:11.033 | <p>Viscosity is resistance to the motion of fluid layers sliding over one another. Density is a measure of forces of attraction between atoms or molecules of same species. Is there any relation between the two? </p>
| |fluid-mechanics| | <p>No, mercury is very dense but pours easily
Honey is lighter but is very viscous - (changes with temperature though)</p>
<p>Density and viscosity are two different characteristics of a fluid.</p>
<p>Viscosity has the units Poise for dynamic viscosity and Stokes for kinematic viscosity.</p>
| 17452 | Relation between density and viscosity |
2017-10-08T18:18:39.300 | <p>I have a homework which should solve by me. My problem is questions are really simple or should I think outside of the box? Like, bode diagram, nyquist or etc.? And, are my answers correct?</p>
<p>Thanks.</p>
<h3>Question-1</h3>
<blockquote>
<p>$$G(s) = K\dfrac{As+1}{Bs+1}$$</p>
</blockquote>
<p>For which values $K, A$ and $B$ is the system always stable? Should I look directly to the pole of the system?</p>
<ul>
<li>$Bs+1=0$</li>
<li>$s=-1/B \implies \text{So, must } B>0$</li>
</ul>
<p>Is it enough? Or, anything else? What about K, A?</p>
<h3>Question-2</h3>
<blockquote>
<p>$$G(s) = K\dfrac{As+1}{(Bs+1)(Cs+1)}$$</p>
</blockquote>
<p>For which values $K, A, B$ and $C$ is the system always stable? Should I look directly to the pole of the system or anything else?</p>
<p>$$Bs+1=0 \wedge Cs+1=0$$
$$s=-1/B \wedge s=-1/C$$
$$\implies B>0 \wedge C>0$$</p>
<p>Is it enough? Or, anything else? What about $K$ and $A$?</p>
| |control-engineering|transfer-function|stability| | <p>I would like to extent the already given answer by MrYouMath.</p>
<p>So question 1 is pretty straight forward and you already got it right. If there's no right half plane (RHP) pole then it doesn't matter what gain you chose. Even for $A = B$, $G(s) = K$ yields a finite response. </p>
<p>For Question 2 have a look at the Routh Hurwitz Array</p>
<p>\begin{array} {|r|r|}
\hline
s^2 & B \cdot C & 1 \\
\hline
s^1 & B+C & 0\\
\hline
s^0 & 1 & \\
\hline
\end{array}</p>
<p>In order for the system to be stable there must not be any sign changes in the first column, hence</p>
<p>$$BC > 0 \quad \land \quad B+C > 0$$</p>
<p>From $BC > 0$ we derive that B and C must have the same sign. $B+C > 0$ yields that the sign has to be positive.</p>
<p>As you see neither $A$ nor $K$ are involved in that. </p>
<p>If you want to explore other methods like root locus, bode, ... Keep in mind that you have variables ($A$,$B$,$C$) in there. I know that you can see the gain margins for root loci in Python, Matlab, etc. but I think that's it. I don't think (but I stand to be corrected) that you can derive the values for $A$,$B$,$C$ that way. I think with Bode plots this may work, however as you've seen it's much easier to solve with Hurwitz or by just looking at the poles.</p>
| 17454 | How to solve system's general stability from transfer function? |
2017-10-08T19:09:01.987 | <p><a href="https://i.stack.imgur.com/3FPvG.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/3FPvG.jpg" alt="enter image description here"></a></p>
<p>I’m following Anderson’s Fundamentals Of Aerodynamics. However, the proof for the above theorem is not provided. Every time I try it, things get really lengthy and cumbersome. Is there some nice and short method to getting around to prove the time rate of change of circulation over a chose closed loop 0 in an inviscid, incompressible flow ? </p>
| |aerodynamics| | <p>The circulation is given by</p>
<p>$$\Gamma = \oint_{C(t)}\boldsymbol{u}d\boldsymbol{r}.$$</p>
<p>Now, take the material derivative of this expression (remember to use the product rule because our contour is time-dependent) </p>
<p>$$\dfrac{D \Gamma}{Dt} =\oint_{C(t)}\dfrac{D \boldsymbol{u}}{Dt}d\boldsymbol{r}+\oint_{C(t)}\boldsymbol{u}d\left[\dfrac{D\boldsymbol{r}}{Dt}\right].$$</p>
<p>Now, we assume that the flow in inviscid and all volume forces are conservative (have a potential $\Psi$). Such that we can use Euler's equation for the first integral </p>
<p>$$\dfrac{D \Gamma}{Dt} =\oint_{C(t)}\left[-\dfrac{\nabla p}{\rho}-\nabla \Psi \right]d\boldsymbol{r}+\oint_{C(t)}\boldsymbol{u}d\boldsymbol{u}.$$</p>
<p>The term on the right-hand side is integrable and has an anti-derivative of $\boldsymbol{u}^2$. As we are integrating on a circle (starting point = endpoint) it must be zero. If we assume that $\rho(p)$, which is called barotropic, then we can conclude that the closed contour integral on the left-side is also zero. Note, that the closed contour integral for conservative fields $-\nabla \Psi$ is zero, also called path independence. Hence, we conclude</p>
<p>$$\dfrac{D \Gamma}{Dt} =0.$$</p>
<p>This means that for inviscid, barotropic flow's and conservative volume forces the circulation $\Gamma$ is conserved.</p>
| 17455 | Proving Kelvin’s Circulation theorem |
2017-10-08T21:34:32.057 | <p>If the force in member OA is 0.25F and inclined at $\alpha$ degree from the horizontal. Calculate the angle $\beta$ to keep the system in equilibrium and calculate the angle $\omega$. $\alpha$ = 20°. F = 100 N.</p>
<p><a href="https://i.stack.imgur.com/zONCH.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/zONCH.png" alt="enter image description here"></a></p>
<p>Attempt:</p>
<p>By alternate interior angles:</p>
<p><a href="https://i.stack.imgur.com/e8Hra.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/e8Hra.jpg" alt="poorly drawn alterior angles"></a></p>
<p>$\beta - \alpha$ is the angle between OB line and horizontal axis x.</p>
<p>Sum of forces in X direction should be equal to 0:</p>
<p>$$\sum F_{x} = -0.25F \cdot \cos(\alpha) -F_{OB} \cdot \cos(\beta - \alpha) + 100N$$</p>
<p>$$\sum F_{y} = 0.25F \cdot \sin(\alpha) - F_{OB} \cdot \sin(\beta - \alpha)= 0 $$</p>
<p>is this correct? I'm not sure what $F_{OB}$, it's not given. So I have two equations, and two unknowns: $F_{OB}$ and $\beta$. I tried to solve via <a href="http://www.wolframalpha.com/widgets/view.jsp?id=ae438682ce61743f90d4693c497621b7" rel="nofollow noreferrer">wolframalpha</a> but it didn't find any solutions. What do?</p>
| |civil-engineering| | <p><span class="math-container">$ 100- 25cos (20) = 76.5076 $</span></p>
<p>Force in member ob.</p>
<p>We divide the tributory of 75 N by this to get the cos(b-a)
75/ 76.5076 = 0.98029</p>
<p>Arccos (0.98029)= 11.3934</p>
<p>This added to 20 is angle beta.</p>
<p>Beta = 20 + 11.3934 = 31.3934</p>
<p>And Omega = 180- 31.3934= 148.606</p>
<p>Sum of angle in a triangle =180°
Omega=180-(70+31.394)</p>
| 17457 | Equilibrium of forces in 2d plane, find angle alpha and w |
2017-10-10T17:07:24.540 | <p>Just to leave some space here I link to an introduction of linear transformation of state. How is it useful? Can nayone provide an example where it is indeed useful and has to be done?
<a href="http://www.staff.ul.ie/burkem/Teaching/st-tr.pdf" rel="nofollow noreferrer">http://www.staff.ul.ie/burkem/Teaching/st-tr.pdf</a></p>
| |control-engineering|control-theory| | <p>The examples in which it is useful are already given in your PDF file. </p>
<p>The first reason is modal decomposition (use $x=Vz$, in which $V$ is the matrix containing the eigenvectors of the matrix $A$). Assuming we have n distinct eigenvalues, then modal decomposition helps us to turn a difficult system into a diagonal system, in which we can solve each line of the ODE by a simple integration.</p>
<p>The second application is the controllable canonical form. This system also has a very simple structure. Additionally, it is used to simplify the pole placement method when doing it by hand, without using the Ackermann formula.</p>
<p>A third application is the <a href="https://en.wikipedia.org/wiki/Kalman_decomposition" rel="nofollow noreferrer">Kalman decomposition</a> of a state space system into controllable & observable, controllable & not observable, not controllable & observable and not controllable & not observable subsystems.</p>
<p>In general, you can conclude that a linear transformation might be used to transform the system into a form which is easier to handle hand the original system or it gives you additional information that you couldn't simply see without the transformation.</p>
| 17483 | Why do we use linear transformation of states in linear control technique |
2017-10-11T01:43:43.130 | <p>Can a Plasma Cutter use for melt metals like Iron, Brass, Stainless Steel ect, ?. Particularly for molding purposes.</p>
<p>If can't, is there any other electric item that doesn't use gases (Acetylene, Butane, Oxygen) for melt metals ?</p>
<p>Edit: </p>
<p>I want to heat a mass of metal up to the melting point and pour that melted metal to a Plaster Parris impression cast. I want to find a electric solution to melt the metal instead of gas.</p>
| |metals|molding| | <p>No.</p>
<p>Plasma cutters use a jet of air along with the arc to cut they pretty thoroughly oxidise the metal, plus the jet would just spray molten metal everywhere if you tried to contain it. </p>
<p>In terms of electrical only heating an induction furnace is efficient and scalable, for low melting point metals like pewter and possible some brasses a straightforward restive heating kiln should work but will be less efficient. </p>
<p>Also plaster isn't suitable for casting ferrous metals, the temperatures involved are way too high. Sand or ceramic shell moulds are required for iron and steel. </p>
| 17486 | Can a Plasma Cutter use for melt metals? |
2017-10-11T09:52:48.413 | <p>In my application the liquid would be tapped from a fast-varying flowrate, and I want to measure the static pressure change during this by measuring the air pressure above it. Will this work?</p>
<p><a href="https://i.stack.imgur.com/dVFa0.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/dVFa0.jpg" alt="If the image doesn't work, all it shows is just a fixed volume of gas connected to the top of a liquid filled tube OK"></a></p>
| |mechanical-engineering|pressure|sensors|compressible-flow| | <p>Yes, the pressure will be identical (with some qualifications). If you think about it, the gas will compress until the force on the boundary between the gas and liquid are equal (i.e., the pressure). Even with a compressible liquid, this should also be the case. You will want to have a reasonable initial pressure for your gas, as well - it may not work as well if your gas volume shrinks to a very small size in response to a very high fluid pressure.</p>
<p>You will get some amount of gas dissolving into the liquid, according to Henry's law. I'm not sure whether that will be an issue with the fluid you are working with. The gas will evaporate out of the liquid / dissolve in response to the pressure increasing and decreasing.</p>
<p>How quickly the gas responds to the increase or decrease in pressure will be another question. It will depend on the gas you are using (based on whether it is monatomic, diatomic, molecular, etc - based on its capacity to store and release energy in various modes). It will also depend on the degree of fluctuation of pressure, as well as the temperature of the gas and the size of the volume pictured here. If either the gas or fluid are supercritical, or if the initial gas pressure is at a low vacuum, things will get weird, but as long as you stay within the regimes where ideal gas assumptions are relevant you should be fine.</p>
<p>One thing to note is that a big tap combined with a small chamber should reach equilibrium pretty quickly. You may want to be careful about getting into very high pressures though if you are designing the seals / shell for this pressure vessel. Using inert gases is obviously preferable. Additionally, if you're using a pressure transducer, you may want to fit it with a snubber to prevent damage from fluid hammer.</p>
| 17489 | Is the pressure of the gas identical to the static pressure of the liquid in the diagram? |
2017-10-11T14:09:25.460 | <p>Suspension often requires different ranges of behavior. Small amplitude and high sensitivity for vibrations and small sharp bumps but also high amplitude and low sensitivity for larger smoother road deformation.
So my question is whether or not it's possible to use two suspension system on top of the other.
I was thinking of a small fast shock (strong spring with weak damper) for small amplitude/high frequency and then a long slow shock (weaker spring with a stronger damper) for high amplitude/low frequencies.</p>
<p>My main concern is of course interactions between them.</p>
<p>Do you have any example of this setup?</p>
<p>Thanks </p>
| |mechanical-engineering| | <p>My answer will probably sound simplistic, but examples of this are more common than people think - the spring/damper arrangement of most vehicle suspensions, coupled with a pneumatic tyre, does exactly what you're describing (tyre = low amplitude high frequency, spring/damper = high amplitude, low frequency).</p>
<p>[I think joojaa was thinking along similar lines, and any wheel with a damped elastic response will have this effect.]</p>
<p>There are much smarter ways of doing it, as Mike and Carl have indicated.</p>
| 17494 | Is it possible to "add" suspension systems? |
2017-10-14T05:09:48.997 | <p>I know that Formula One and some sports cars have KERS systems to recover kinetic energy, but is there an expense/practicality barrier to installing them in consumer vehicles? As the technology clearly serves efficiency purposes beyond just providing extra horsepower, and as KERS can capture energy electrically or mechanically, it seems that it would be a useful step towards increased consumer vehicle efficiency.</p>
<hr>
<p><strong>Note:</strong> I recognize that there's a significant cost associated with thermoelectric capture, and that <a href="https://engineering.stackexchange.com/questions/389/thermoelectric-technology-to-harvest-energy-from-internal-combustions-engines">this suggested duplicate</a> raises the concern of how it's impractical to slow heat dissipation. However, if we limit the energy capture to brakes (typically cooled by natural convection and therefore not subject to any air/liquid cooling), could we see a more practical energy capture solution than trying to capture kinetic energy straight from an internal combustion engine? </p>
| |mechanical-engineering|thermodynamics|regenerative-braking| | <p>Actually, a lot of consumer cars <i>do</i> use kinetic energy recovery. Every hybrid and electric car that I know of does this, and there are significant numbers of hybrids and electric cars out there now.</p>
<p>Since these cars already have a electric motor connected to the wheels, it only takes a little addition to the control electrics to run the motor as a generator when you want to slow down. The cost of the additional electronics is minimal compared to the cost of the motor and the electrical system as a whole, so this is a no-brainer in most cases.</p>
<p>The same system is used to charge the battery when going down a hill.</p>
<p>My Honda Civic hybrid works this way, for example. Stepping on the brake lightly while moving first runs the motor in reverse. I can see the battery charge/discharge indicator going to charge. Stepping harder on the brake causes more charging, and therefore more drag. Eventually the friction brakes are also engaged. This is all done quite smoothly and seamlessly.</p>
| 17523 | Why don't most consumer vehicles use kinetic energy recovery systems? |
2017-10-16T02:21:09.867 | <p>There is a national level competition <strong>Ecokart</strong> (Electric kart design & manufacturing) has been held every year in Delhi, India. there are some rules for making the kart which are specified in the rule book.</p>
<p><a href="http://www.ecokart.in/index.html" rel="nofollow noreferrer">ECO KART SERIES</a> </p>
| |mechanical-engineering|electrical-engineering| | <p>There is no simple calculation which will give you a single correct answer as gear ratios are essentially an optimisation problem. A higher final drive ratio will give a higher top speed (up to a point at least) while a lower ratio will give you more torque (although usable torque is also limited by the traction of the drive wheels. </p>
<p>The overall approach for designing something like this is to start with fairly simple calculations based on reasonable approximations to work out rough orders of magnitude and where the limits are and then progressively refine your numbers as you get closer to what a sensible design should look like. </p>
<p>It is also helpful to look at existing real world examples to get a sense of what is likely to be possible. </p>
<p>The diameter of the wheels and tyres will affect the final drive ratio too so you may want to look at what wheels you are using first. </p>
<p>Th basic concept of <a href="https://en.wikipedia.org/wiki/Gear_train" rel="nofollow noreferrer">gear ratios</a> is simple enough and similarly for every turn of the road wheels you move forwards by the circumference of the wheel. </p>
<p>eg. if you have a wheel with circumference of 0.6m turning @ 600 rpm (ie 10 revolutions per second the road speed is 0.6m * 10/s = 6m/s or about 13.4mph </p>
| 17538 | What are the calculations required to design transmission system for an Electric Kart ( i.e Go Kart)? |
2017-10-16T11:04:25.543 | <p>I'm working on a design where I need a ball bearing with an inner diameter of 8mm. The ball bearings will support small 3D printed wheels, looking kind of like a skateboard wheels. In order to fasten these wheels, I'm planning to use 8mm screws. The screws that I have found are threaded all the way, and I'm wondering if it is possible/recommended to use threaded screws to support ball bearings? Ideally I would use screws that are not threaded all the way, so the ball bearing can sit on the part that is not threaded. However, I couldn't find such screws. So what if I would use threaded screws? Would it work at all? Is there some workaround I can do using washers/extra bolts?</p>
<p>Edit:</p>
<p>Here is the design, plus a new design I came up with...</p>
<p><a href="https://www.dropbox.com/s/c86cxw1wyc9zi7v/IMG_3628.JPG" rel="nofollow noreferrer">https://www.dropbox.com/s/c86cxw1wyc9zi7v/IMG_3628.JPG</a></p>
<p>What I have learned so far... There is such a thing as a shaft collar, and that's what I should be using in this case. Here is one that is 3D printed:</p>
<p><a href="https://www.thingiverse.com/thing:400921" rel="nofollow noreferrer">https://www.thingiverse.com/thing:400921</a></p>
<p>And on there there is also an example of how it is used together with ball bearings. That's what I should be inspired by...</p>
| |mechanical-engineering| | <p><strong>So yes, you can - but</strong></p>
<p>I recently made a device using skateboard bearings (608zz) and mounted them onto a M8 bolt. It had noticeable slop despite using a bolt with a smooth shank where the bearings were located. This is because a M8 bolt is designed to fit through an 8mm hole, so the shaft will be smaller than 8mm.</p>
<p>According to my calipers, my M8 bolt's shank was was 7.84mm near the head and 7.65mm near the start of the thread. The bearing inner diameter was 8.03mm. In my device this meant one bearing had a slop of ~0.2mm and the other closer to 0.4mm.</p>
<p>With an offset of 0.2mm:</p>
<ul>
<li>Two bearings 60mm apart mean the shaft can have an angular deviation of 0.5 degrees.</li>
<li>A 1kg load will create up to 45 grams of torque.</li>
<li>A 1kg load spinning at 1000 RPM offset by 0.4mm will induce 200g of linear force that turns into vibration.</li>
</ul>
<p>So for low speed, low precision devices - sure, go ahead.
But if you are trying to build a precise device or one with high shaft speeds, you'll need to look for other solutions.</p>
<hr />
<p>In my case it turns out that angle grinder disks are not precision flywheels and electrical tape is a poor flexible shaft coupling, so bearing slop is not a problem yet....</p>
| 17542 | Can I use a threaded bolt to support a ball bearing? |
2017-10-16T11:49:28.763 | <p>I am trying to ascertain the maximum line pull for an old winch I have in my workshop and I am arriving at values that are considerably higher than what is in the original manufacturer's manual. I am not sure if I am doing this right and want to validate my calculations.</p>
<p>The winch is rotated hydraulically with a motor. The motor shaft is connected to a wet brake which in turn is connected to a planetary gear hub. This hub is then connected directly to the drum.</p>
<p>Some specs:</p>
<p>If the motor builds up 3000 PSI and has 38 LPM of flow it produces a torque of 295Nm with 316RPM. (or 30.08163KGm)</p>
<p>The wet brake disengages and produces negligible friction.</p>
<p>The ratio in the gear hub is +53.58:1</p>
<p>The diameter of the drum on the winch is 0.81m</p>
<p>The diameter of the cable used to pull is 0.0202m</p>
<p>From what I understand, to find linear force for a winch I must use this formula: Torque = Force X Arm</p>
<p>Where Arm = radius of drum + radius of cable radius drum = 0.81m/2 = 0.405m radium cable = 0.0202m/2 = 0.0101m Arm = 0.405m + 0.0101m = 0.4151m</p>
<p>Where Torque = Torque of motor X ratio of gear hub Torque = 30.08163KGm X 53.58:1 = 1611.7737354KGm</p>
<p>Therefore Force = Torque / Arm Force = 1611.7737354KGm / 0.4151m = 3882.856505420381KG</p>
<p>Am I correct to therefore say that the maximum line pull before the winch stalls is 3882.85KG? Therefore We can say it can lift anything bellow this weight?</p>
<p><a href="https://i.stack.imgur.com/8uVG3.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/8uVG3.png" alt="enter image description here"></a></p>
| |mechanical-engineering|torque| | <p>The main constraint would be the cable braking strength. Per Engineering Toolbox, this size cable has a maximum safe load of 42.3 kN. Your device produces 38 kN at stall point, so the cable is under it's safe load. This implies that the device will indeed handle this level of load without a problem. The system may have derated the pump as approaching the stall point of most motors, including hydraulic motors, can wear excessively and cause early breaking, so it is best to stay within nameplate parameters for day-to-day operations.</p>
<p>Of course, due to angles, we could have additional loading beyond the weight. See this handy diagram and table from Engineering Toolbox:</p>
<p><a href="https://i.stack.imgur.com/l3iRr.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/l3iRr.png" alt="enter image description here"></a></p>
<pre><code>α β Increased Force
0 90 1.00
5 85 1.00
10 80 1.02
15 75 1.04
20 70 1.07
25 65 1.10
30 60 1.16
35 55 1.22
40 50 1.31
45 45 1.41
50 40 1.56
55 35 1.74
60 30 2.00
65 25 2.37
70 20 2.92
75 15 3.86
80 10 5.76
85 5 11.5
</code></pre>
| 17543 | Computing winch maximum line pull (torque vs lift capacity) |
2017-10-17T06:11:58.843 | <p>The aircraft carrier <a href="https://en.wikipedia.org/wiki/USS_Nimitz" rel="nofollow noreferrer">USS Nimitz</a> cost \$4.5bn. The cruise ship <a href="https://en.wikipedia.org/wiki/Allure_of_the_Seas" rel="nofollow noreferrer">Allure of the Seas</a> cost \$1.2bn. Why is the difference so huge?</p>
| |ships|naval-engineering| | <p>The USS Nimitz is nuclear powered, whereas the Allure of the Seas is not. Nuclear reactors are not cheap.</p>
<p>Also, the USS Nimitz, like all aircraft carriers has a catapult to help launch aircraft off the decks. The Allure of the Seas does not have a catapult. Catapults are expensive.</p>
<p>The USS Nimitz has kevlar armor over vital spaces, but civil cruise ships don't</p>
<p>Add to this the radar systems and armaments that aircraft carriers must have but are not required for civil cruise ships.</p>
<p>Aircraft carriers are build to handle severe weather where some cruise ships may not be. They must also have the propulsion systems and structural integrity to travel long distances at high speed. Civil cruise ship don't.</p>
| 17555 | Why are aircraft carriers so expensive to build? |
2017-10-17T10:44:14.957 | <p>I am looking for something like this to make my robot ideas as prototype. But I don't know it's name! Also if there are better options I will happy to know.</p>
<p><a href="https://i.stack.imgur.com/BpW14.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/BpW14.jpg" alt="" /></a><br />
<sub>(source: <a href="https://cdn.instructables.com/FQC/7ENO/IW5JB2UM/FQC7ENOIW5JB2UM.MEDIUM.jpg" rel="nofollow noreferrer">instructables.com</a>)</sub></p>
<p><img src="https://cdn.instructables.com/F3J/BE2K/IW5JB58G/F3JBE2KIW5JB58G.MEDIUM.jpg" alt="" /></p>
| |metals|robotics|building-design|prototyping| | <p>In addition to 80/20 you could use "unistrut" or "channel strut" framing which may be more readily available at a big box store. </p>
<p><a href="https://i.stack.imgur.com/Gh9NF.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Gh9NF.jpg" alt="unistrut"></a></p>
| 17562 | I am looking for some basic mechanical parts to build structures(prototypes)? |
2017-10-17T15:37:50.817 | <p>Is it only possible to write a User Defined Function for Ansys Fluent in C? All of the tutorials I could access online do so. I was wondering if this could be done using Java, or any other language for that matter?</p>
| |fluid-mechanics|ansys|fluent| | <p>You can use only C language for writing UDF. As UDF is an add on customization of Fluent, you have to use C language. Because all APIs are exposed in C language only.
For automating ANSYS Fluent you can use scheme language.
In future ANSYS may provide multi language support, but as of now (ANSYS 19.2) there is support of only C language for writing UDFs.</p>
| 17570 | Writing a User Defined Function for Ansys Fluent |
2017-10-17T19:30:38.700 | <p>I hope this is the right place to ask this question; if not I'd really appreciate some tips as to where else I can go for answers.</p>
<p><a href="https://i.stack.imgur.com/pC24m.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/pC24m.jpg" alt="enter image description here"></a></p>
<p>The symbol above (apologies for the blurry quality) is on a Chinese/Taiwanese-manufactured piece of testing equipment. It's located on the manufacturer's sticker. I can find absolutely no trace of it elsewhere on the Internet (I've absolutely, positively searched).</p>
<p>Can anybody shed some light as to what this symbol might mean? E.g. an electrical safety certification, a manufacturing standard...?</p>
<p>Thank you very much!</p>
| |mechanical-engineering|electrical-engineering|manufacturing-engineering| | <p>This appears to be related to GB/T 50430-2007, the Chinese "Code for Qualtity management of Engingeering Construction Enterprises". Unfortunately I can't find the full text translated into english, but found reference to the CMC mark <a href="https://books.google.co.uk/books?id=nPumDgAAQBAJ&lpg=PA63&ots=d_alubKM3b&dq=CMC%20certification%20test%20equipment%20china&pg=PA63#v=onepage&q=CMC%20marking&f=false" rel="nofollow noreferrer">in section 11.5.1</a></p>
| 17575 | What's this mystery symbol on a manufactured machine? |
2017-10-19T08:50:35.357 | <p>I have a small device emitting up to ~90 Watts thermal power over a surface of 2x2 square centimeters. What is the best way (both commercially and DiY) for keeping the temperature of the device constant, maybe +- 1 degree at room temperature? </p>
<p>I have a thermoelectric element I could use, which has a surface area of 18x10 square centimeters for the cold plate. It is capable of removing up to ~75 Watts thermal power at 0 degrees temperature difference. How efficient will it be when I put the electric device on top of it? How much will the size difference effect the efficiency of the TEC? I know that it is rated for lower thermal powers, but it can be used as intermediary solution while keeping the device at lower powers.</p>
<p>The device itself is a solid state laser diode, which means that even small changes in the temperature (i.e. degrees) will shift the wavelength, something I do not want (I prefer having it at ~20 degrees). The diode can emit up to 90 Watts, depending on how much output power I request, but the upper limit is 90 Watts.</p>
| |cooling| | <p>A CPU cooler is the first thing that springs to mind DIY - Heatsinks coupled with forced convection are a tried-and-tested method of heat removal.</p>
<p>Provided the heat output from the diode is constant, and the room doesn't change, then this should remain at a constant temperature, albeit above the ambient temperature of the room.</p>
<p>In the application you describe, it seems that some form of thermostatic control would be required to keep the diode temperature constant as you vary the output power. DIY, This means hooking up a thermocouple to an arduino or similar, to control the power delivered to the thermoelectric element that you described previously. Adding a Heat-Sink/Fan to the hot side of the element will improve efficiency, and give you more control. Varying the fan speed will have an effect on the cold-side temperature, and this can be used in addition to simply varying the voltage into the element.</p>
| 17598 | Cooling a small high-power device |
2017-10-19T10:34:07.320 | <p>I am Comp-sci student doing a software engineering paper about designing a program, planning the software like you were actually planning a real-life software project, short of writing actual pseudo-code or code. For my project, I picked doing an app that can detect a physical man in the middle attack. Like monitoring for a tap on the line.</p>
<p>It got me thinking, can you monitor for someone catching wireless communication? Like the latest WEP2 hack. If someone were monitoring your wireless traffic passively to capture your secure handshake, could you detect them snooping wirelessly without them transmitting packets?</p>
<p>My knowledge of EE stops at MOSFETs and timing of electrical currents through logic gates and metastable currents propagating through circuits because the timing of logic gate was off. My knowledge of EE certainly falls short of radio broadcasting. So I am not even sure if such a feat is possible, I suspect it isn't but it's out of my ken.</p>
<p>I suspect if you were close enough a receiver would leave behind some kind of detectable EM footprint.</p>
| |electrical-engineering|sensors|wireless-communication| | <p>many radio receivers contain a circuit called an IF oscillator. the output of this oscillator is added to the radio frequency signal that the radio is receiving to downconvert the incoming high frequency to a lower frequency that is easier to process. Old radios used IF oscillators that were powerful enough to actually transmit signals on their IF frequency which other receivers could then pick up and home in on. during the second world war, this technique was used to find enemy ships at night, until they learned how to shield their receivers so they would not inadvertently transmit on their IF frequency and thereby reveal their location. I do not know how well-shielded radios of recent manufacture are against "spurious transmission", which is what this effect is called.</p>
| 17601 | Can you detect the operation of a wireless radio (or other wavelength) receiver? |
2017-10-21T13:39:37.783 | <p>I was wondering about the ethics of an interviewer asking to see your design notebooks in an interview.</p>
<p>First off, I'm not sure if it's even legal for someone to ask that, since many of the designs I have are covered by (Non disclosure Agreements) NDA'S with various companies which the interviewer is not a part of. </p>
<p>Secondly, I'm not sure if the request to review my notes is even ethical? Your design notebook is basically a screen shot of EVERYTHING you've worked on good, bad, right and wrong asking to see it is almost like asking to see your work email account from another company, in my opinion.</p>
<p>Guidance on this would be appreciated.</p>
| |ethics| | <p>The simplest way to handle is if the work is propitiatory to your previous employer then indicated up front. Also remember looks like you are interviewing for a new position, so take the opportunity to ask if there anything that is of particular interest to the interviewer that you use alternative methods to answer the interviews questions.</p>
<p>Also, sometime interviewer ask this questions to validate how you protect propitiatory information. So if you offer up propitiatory information from previous employers, chances are you will also offer up their propitiatory information too. </p>
| 17637 | Ethics in an interview |
2017-10-21T20:15:00.317 | <p>Can polycarbonate be used to store a cryogenic liquid for a short duration, let's say 15-20 minutes.</p>
<p>I've heard that PVC will be really brittle at low temperatures.</p>
<p>What I'm trying to do is a rocket and I need a tank to store my LOX. I need it to be something like PVC or polycarbonate because they are light and cheap enough for my purposes. Also, they are often sold for tubing, the shape I need.</p>
<p>Any other suggestions of material will be greatly appreciated.</p>
| |materials|cryogenics| | <p>in designing the flask to hold your LOX, please note that the LOX will slowly warm up as it sits in the flask, and will then pressurize the flask. In addition to preventing heat loss and not being brittle at low temperatures, the flask will also have to be strong enough to withstand pressure buildup inside it without exploding. </p>
| 17642 | Polycarbonate as a short-term cryo-tank? |
2017-10-21T21:53:54.827 | <p>Question: <a href="https://i.stack.imgur.com/BSWNf.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/BSWNf.png" alt="enter image description here"></a></p>
<p>what i thought; see below.</p>
<p><a href="https://i.stack.imgur.com/mJ5Iz.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/mJ5Iz.jpg" alt="enter image description here"></a></p>
<p>we do not know density if i where to use that formula on the image file .</p>
<p>and i a m wondering why i was given the Reynolds number .</p>
<p>need help.</p>
| |fluid-mechanics|airflow|drag| | <p>So the formula you need is Density = P /(R*T) and the value of R you have been given is for air.</p>
<p>Once you have re-worked your answer then you should calculate the Reynolds number and then you can check your result here :
<a href="https://www.grc.nasa.gov/www/k-12/airplane/dragsphere.html" rel="nofollow noreferrer">Drag on a sphere by NASA</a></p>
| 17643 | Coefficient of Drag (Cd) Question |
2017-10-22T12:43:33.683 | <p>I have to determine the centre of mass for a 3D object (bounding box: approximately $0.30 \text{ m} \times 0.20 \text{ m} \times 0.10 \text { m}$) that is not regular and does not have a uniform density. I know how to determine the centre of mass for a flat object. I thought that I could try to use a similar approach for the 3D object, but the problem that I have is that I cannot determine the intersection of the planes as it is somewhere inside the object.</p>
<p>Then I thought I could use a balance for three directions and determine the centre of mass by finding the point of balance for all three "main directions".</p>
<blockquote>
<p>Is the last method a good method? What is the best way to determine
the centre of mass by measurement?</p>
</blockquote>
| |measurements| | <p>Another method is to pendulum the thing and calculate the pendulum arm from the swing period that corresponds to the CG. Note, there are two options here, a single pendulum where the object also rotates with the swing, or multiple suspension points of the same length where the object stays flat as it swings. As in the other methods, to get the exact cg, you need to do this at least 3 times in noncoplanar orientations and calculate the intersection. This is actually an approved method for computing the VCG of smallish boats. </p>
| 17651 | Experimental method to determine the centre of mass for 3D object |
2017-10-23T11:33:16.233 | <p>I'm designing a probe to provide a stable and continuous stream of exhaust gas sample to an analyser. The probe has to be inserted into the exhaust for about 20–30 cm to keep outside air from entering the probe. The static pressure there is about 1000 Pa above atmosphere. The dynamic pressure is probably way higher. The material of the probe is brass.</p>
<p>The problem here is that the fluctations in the environment inside the exhaust may have significant influence on my sampling. The analyser precisely dilutes the exhaust gas to match the sensor's specs, and environmental changes may alter the dilution ratio, resulting in incorrect readings. The probe has to offer a sample at a pressure as constant as possible.</p>
<p>The pressure inside the exhaust will force exhaust gas into the probe, and with changing engine load, the temperature, static, and dynamic pressure will change, forcing a different amount of exhaust gas into the probe. I want to have that pressure as stable as possible, so that I can correct for it in my calculations.</p>
<p>I have studied the pitot tube and my design is now as illustrated below:
<a href="https://i.stack.imgur.com/pDv4n.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/pDv4n.png" alt="probeShape"></a></p>
<p>I have tried to eliminate the influence of dynamic pressure, but I'm not sure if this is the best way to create a sampling of constant pressure.</p>
<p>So <strong>tl;dr</strong>, consider image above, is there a better way to offer samples of constant pressure, in an environment that has fluctuating static and dynamic pressure, and temperature?</p>
| |pressure|airflow|gas| | <p>It sounds like you are unable to affect the analyser end of things. My first suggestion would be to adjust the analyser software to undersample the probe.</p>
<p>Since this does not sound possible and without knowing more about your design constraints, I would suggest adding an accumulator. With only 1K pa over Atmos. you should be able to calculate the volume of a small accumulator that could smooth pressure spikes and allow the gas to cool slightly for a better read. I've seen designs in applications similar to what you are describing which have a two port accumulator (more of a settling chamber) where one is intake from the probe front end and the second contains an orifice calculated to provide a set flow for a given pressure range to the end sensor.</p>
| 17662 | Can I improve this exhaust gas probe to be immune to pressure changes? |
2017-10-23T12:02:11.097 | <p>I have a question about the <em>significant figures</em> topic:</p>
<p><strong>Problem:</strong>
Find the volume of a cylinder having a diameter measured as 2.3 ft and length measured as 8.25 ft.</p>
<p><strong>Solution:</strong>
V = π(D^2 /4)h = π((2.3)^2)(8.25)/4 = 34.276739 cu ft = 34 cu ft</p>
<p><strong>Question:</strong><br>
In my workbook is explained that the final answer is rounded to two significant figures because the quantity 2.3 has two significant figures.<br/>
Why will I have to round to two significant figures if the quantity 8.25 have three significant figures? </p>
| |civil-engineering| | <p>Selter's answer follows the book's approach. A proper analysis, however, should go a little deeper.<br>
First, we stick with the convention that the uncertainty in any provided value is +/- half the least sig digit, i.e. 2.3 means 2.3 +/- 0.05 . Next, we notice that the diameter enters into the formula squared and the height linearly. The standard approach is to look at the partial derivatives and do variational analysis. Following the mechanisms presented, e.g., in <a href="https://smile.amazon.com/Reduction-Error-Analysis-Physical-Sciences/dp/9339221206/" rel="nofollow noreferrer">Bevington</a> Chapter 3, we will get ("Var" is variance, or sigma^2) </p>
<p>$ \frac{Var(volume)}{Vol^2} = \frac{\pi}{4} *( 2*\frac{Var(diameter)}{Diameter^2} + \frac{Var(height) }{height^2} ) $</p>
| 17663 | Significant figures |
2017-10-24T13:04:39.607 | <p>I’m pretty sure it’s something to do with the fitting, possibly specifying to use the minimum circumscribed or maximum inscribed circle to tolerance the diameter but I can’t remember.</p>
<p>The tolerance would be in the form</p>
<blockquote>
<p>⌀123 ± 0.01 Ⓔ</p>
</blockquote>
<p>In case the Unicode doesn’t work correctly that’s</p>
<blockquote>
<p>[Diameter Symbol] [value] [plus minus symbol] [value] [E in a circle]</p>
</blockquote>
| |drawings|technical-drawing| | <p>Like promised, I am now looking at <em>EN ISO 14405-1:2016</em> where this E symbol (Ⓔ) is defined. More accurately is defined in section "3.8 envelope requirement". </p>
<ol>
<li><p>The upper value of a external feature defines a virtual cylinder that the entire feature has to fit inside. Likewise a internal features minimum value defines a cylinder that has to fit inside the hole. </p></li>
<li><p>Each slice of needs to be smaller or bigger (for internal features) than the other tolerance value measure as a two point size (imagine measuring with a caliper)</p></li>
</ol>
<p>Note: Nowehere does it say the cylinder is aligned to the centerline</p>
<p><a href="https://i.stack.imgur.com/h4NmH.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/h4NmH.png" alt="enter image description here"></a></p>
<p><strong>Image 1</strong>: Cylinder needs to be inside the envelope and not point measured smaller than the minimum size.</p>
<p>See relevant standard for more robust explanation.</p>
| 17684 | What does an E in a circle after a diameter mean? |
2017-10-25T19:09:14.373 | <h1>Some background for context only</h1>
<p>I am in the process of designing a pressure system where my pressurized fluid needs to be kept incredibly clean, with typical contaminants at the ppt or below level. The fluid in question is more compressible than a typical hydraulic fluid, so I need to build in an expansion volume for when I cycle to lower pressure. This expansion volume is going to look very much like a hydraulic accumulator. </p>
<p>The most common hydraulic accumulators are pneumatic, and the gas charge is separated from the fluid by a bladder, diaphragm, piston, or metal bellows. All of these are problems for my cleanliness requirements, as the moving parts will tend to shed metal shavings, etc. </p>
<h1>The question</h1>
<p>My question is, why is the separation of the two volumes necessary? In my case, this is for a fixed piece of equipment, so I can guarantee that the accumulator will always be vertical. Can I just have a nitrogen-filled cylinder with the fluid connected at the bottom and rely on the fluid to act as my piston? If this does work, why isn't it more common? </p>
<p><strong>Edit to address some of the comments:</strong></p>
<p>I'm not asking for advice on how to design a clean accumulator; I only want to know the fairly general question of whether it is possible to not separate the charge gas from the fluid. </p>
| |design|hydraulics| | <p>gases are soluble in oils. this means that as your oil enters the gas-filled accumulator and the pressure in it builds up, gas will probably dissolve into the oil. Then, when you release the pressure on the system, the gas will boil out of solution and cause the oil to foam up vigorously. the foamy oil then enters your plumbing and creates big problems, especially if it is ingested by a pump.</p>
| 17713 | possibility for hydraulic accumulator with no moving parts? |
2017-10-26T09:54:18.667 | <p>I have come across a problem, which has troubled me for some time now. What needs to be done is the following:</p>
<p>A mass on a rod 0.6m (mass less) has a mass of 1 kg attached at the end of it. The rod needs to be rotated 60°, within t=120 sec (see image). What I would like to do is size a rotational spring (located at the pivot point) and a damping system, such that it that will damp the spring force. Thus the rotation happens within the specified amount of time.</p>
<p>I have written the generic differential equation of the system:</p>
<p>$$J\theta'' + C\theta' + K\theta = 0$$</p>
<p>(typical differential equation of damped spring system)</p>
<p>and for a critical damped system, and for $t=0$, $\theta=0$ I have the solution:</p>
<p>$$\theta(t)=A t \exp(-bt)$$</p>
<p>where $A$ is a constant, and $b$ is the damping coefficient. </p>
<p>My question is how can I continue, such that I can size the damping coefficient and the spring constant?</p>
<p>And if I can continue from here, how should I then proceed to size my system?</p>
<p>Please note $g=0$, no gravity.</p>
<p><a href="https://i.stack.imgur.com/CYLbM.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/CYLbM.png" alt="problem decription, PLEASE NOTE g=0, no gravity"></a></p>
| |mechanical-engineering|vibration|linear-motion| | <p>The equation should be</p>
<p>$$ J \, \ddot{\theta} + K \, (\theta-\theta_T) + C \, \dot{\theta} =0 $$</p>
<p>where $\theta_T$ is the target angle and the rest position for the spring.</p>
<p>Things simplify wit the following substitution
$$ \begin{cases}
K = J \Omega^2 \\
C = 2 \zeta J \Omega
\end{cases} $$
where $\Omega$ is a parameter the relates to the stiffness, and $\zeta$ a parameter that relates to the damping.</p>
<p>The general solution of the equation $\ddot{\theta} + \Omega^2 (\theta-\theta_T) + 2 \zeta \Omega \dot{\theta} =0$ is</p>
<p>$$ \theta = \theta_T + C_1 \exp\left( -\Omega t \left(\sqrt{(\zeta^2-1)}+\zeta \right) \right)+ C_2 \exp\left( -\Omega t \left(\sqrt{(\zeta^2-1)}-\zeta \right) \right) $$</p>
<p>where the coefficients $C_1$ and $C_2$ are found based on the boundary conditions. In this case, the rod is at rest when $t=0$ and thus</p>
<p>$$\begin{cases}
C_1 =\frac{ \theta_T}{2} \left( \frac{\zeta}{\sqrt{(\zeta^2-1)}}-1 \right) \\
C_2 =-\frac{ \theta_T}{2} \left( \frac{\zeta}{\sqrt{(\zeta^2-1)}}+1 \right)
\end{cases} $$</p>
<p>The fastest response is when $\zeta \rightarrow 1$ since that minimizes the value of the first exponent. With optimum damping the solution becomes</p>
<p>$$ \theta = \theta_T \left( 1 - {\rm e}^{-\Omega t}(1+\Omega t) \right) $$</p>
<p>and $C = 2 J \Omega$.</p>
<p>But as you stated, you cannot find the correct $\Omega$ to reach the target angle at the specified time because $\Omega$ cannot be isolated from the solution. </p>
<p>But it can be isolated from the general solution when $\zeta>1$. From the two terms, the first one approaches zero much faster than the second one. We can find the coefficient $\varphi = \Omega t$ which makes the second exponent near zero by a value $\epsilon$ (in degrees).</p>
<p>$$ \left. \frac{ \theta_T}{2} \left( \frac{\zeta}{\sqrt{(\zeta^2-1)}}+1 \right) \exp\left( -\varphi \left(\sqrt{(\zeta^2-1)}-\zeta \right) \right) = \epsilon \right\} $$ $$ \varphi = \frac{ \ln\left( 2 \frac{\epsilon}{\theta_T} ( \zeta \sqrt{(\zeta^2-1)}-\zeta^2+1)\right)}{\sqrt{(\zeta^2-1)}-\zeta} = \Omega\, t_T $$</p>
<p>where $t_T$ is the target time. A good compromize occurs when $\varphi = \frac{1}{\sqrt{(\zeta^2-1)}}$ which makes $\frac{{\rm d}\varphi}{{\rm d}\zeta}=0$ and hence minimizes $\varphi = \Omega \, t_T$.</p>
<p>This produces the solution</p>
<p>$$\begin{cases}
K = \frac{J}{t_T^2 \gamma (\gamma+2)} \\
\zeta = \frac{ \sqrt{(J+K t_T^2)}}{t\,\sqrt{K} } \\
C = \frac{2}{t_T} \sqrt{J (J+K t_T^2)}
\end{cases} $$</p>
<p>where $\gamma \ll 1$ is a small positive value (defined as $\zeta = 1+\gamma$).</p>
| 17748 | Simple mass spring damping system sizing K, and C factors |
2017-10-26T20:45:11.510 | <p>It is easy to found stepper motors with torque over 2NM for a few coins, however, only expensive industrial servomotors achieve this torque.</p>
<p>Examples:</p>
<ul>
<li>Stepper motor: 10usd for 1.3nm with 2.8V at 3.5A rpm 0.6kg</li>
<li>Servo AC motor: 278usd for 2.39nm with 220V~ at 3A 3000rpm 6kg</li>
<li>Servo DC motor: 67usd for about 0.4nm 24V at 7.8A 3000rpm 1Kg</li>
</ul>
<p>Torque is not power, and closed loop has many benefits, but I am still confused:I believe many industrial purpose require quite high torque (CNCs, robotic, etc.) and it would be preferable (to my understanding) a lower rpm with higher torque. Additionally, it seem to me that any purpose for a servomotor require more than a few 0.4nm</p>
<p>Why is this torque that different? </p>
<p>What is the benefit / purpose of those non-hybrid servo-motors?</p>
<p>What am I missing?</p>
| |motors|stepper-motor|servo| | <p>Operating principle is different. In essence if you buy just a motor it will be just that. You can now add gearing that will reduce speed and increase torque. Since torque is directly proportional to gear ratio and mechanical advantage;</p>
<p>$$
MA = \frac{v_{input}}{v_{output}}.
$$</p>
<p>You get a hundred times more torque if you gear 3000 rpm down to 30 rpm. So now that DC motor you listed would be 40Nm and that AC servo a 239Nm monster. So indeed those are way more powerful.</p>
<p>Further there may be gearing inside your stepper motor, its quite common. You can get geared servos. It is just more uncommon as designing the drive train needed is kind of the point of using normal motors. As the drive train is the part where you optimize things for your application. That or manufacture/design the motor yourself.</p>
| 17757 | Why servomotor has such low torque compared with stepper motors? |
2017-10-27T15:55:00.083 | <p>In the event of an explosion through a high pressure gas pipe, is there a way to vent the pressure wave, in order to protect downstream objects? The mechanism must be faster than sound; pressure relief valves and even rupture discs are actually too slow and will not respond to such a "fast" pressure wave.</p>
| |mechanical-engineering|fluid-mechanics|pressure|mechanisms| | <p>They are called blast valves. You need a delay loop in the pipe so the trigger can get ahead of the shock wave. Our interest was in closing off ventilation air conduits before the nuclear weapon's overpressure could penetrate the site.</p>
<p>See the pipe loops over the Launch Control Equipment Building? Those are the delay loops for ventilation air.</p>
<p><a href="https://i.stack.imgur.com/l3MFI.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/l3MFI.png" alt="enter image description here"></a> </p>
<p><a href="https://i.stack.imgur.com/RVSPk.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/RVSPk.jpg" alt="enter image description here"></a></p>
<p><a href="http://www.dtic.mil/get-tr-doc/pdf?AD=AD0636321" rel="nofollow noreferrer">BLAST-CLOSURE VALVES</a></p>
| 17771 | Is there a pressure relief mechanism that operates faster than sound in a material? |
2017-10-27T20:48:52.903 | <p>When a reservoir is used to store water for electricity, it's pipes are usually laid in several segments on the ground to reach its source, such as a lake. It seems that if the pipes went directly to the lake by being suspended, less piping could be used and less energy lost due to friction would occur. Why are systems not built like this?</p>
| |water-resources|energy|renewable-energy|energy-storage| | <p>Suspended pipes are heavy and need support - any change of direction also creates a force - and there is the aerial impediment to consider : Birds, parachutes planes etc.</p>
<p>Also some "pipes" are cut directly into the rock saving on material costs and consider that some of these pipes have a water flow rate of 45m3/s so the inertial forces are massive.</p>
| 17780 | Why do pumped storage hydroelectric facilities use bent pipes? |
2017-10-28T10:41:43.673 | <p>I would like to know what industrial engineers use for planning and developing a process:</p>
<ul>
<li>What is the most widely <code>business analysis</code> modeling language used (Think of it as the stage
consisting of 'How the product is intended to be?')? <strong>(Before the process)</strong></li>
<li>What is the type of <code>software</code> and what are the <code>methods</code> used for <code>managing</code> the process? <strong>(Within the process)</strong></li>
</ul>
| |process-engineering|industrial-engineering|project-management| | <p>SysML is a good choice when it comes to modeling use cases and understanding systematic approach (business modeling), in a generic industrial context.</p>
<p>SQL is used to mine business data. R is great for analysis and simulation. VBA has good portability when coupled with Excel.</p>
<p>PDM software is good for setting up a document control process. Checking files into a database, monitoring revisions, and controlling the approval process, of BOM's (bill of materials) (Product Lifecycle Management (PLM)).</p>
| 17791 | How does an industrial engineer develop a process? |
2017-10-28T19:52:13.167 | <p>I was asked to solve the following problem on my test: </p>
<p>A tank is filled up to height H with a liquid having specific gravity 1.6. The rest of the tank is filled with air. A pressure gauge attached on top of the tank measures the pressure to be 11.5 m of water and another pressure gauge attached at the bottom of the tank reads 16.8 m of water. Now, what is the value of H? </p>
<p>I don't know where to start. I have solved U tube manometer problems but never faced something like this one. I don't have a single clue. Even the total height of the tank is also not given. Is it really possible to solve the problem with only single data (the SG of the liquid) ? What do you think?</p>
| |fluid-mechanics| | <p>I'll only outline the steps :
1) SG to density : Sg of 1.6 gives density as 1600kg/m3</p>
<p>2) pressure difference is given from the two gauges 5.3m of WATER...</p>
<p>3) calculate the pressure created by 5.3m of water</p>
<p>4) re-calculate for Height as Pressure / (density * gravity) but this time use 1600...</p>
<p>Neat that you have to spot that the gauges are in metres of water....</p>
<p>I assumed gravity as 9.81 m/s2,
and ignored gauge pressure as it's the difference...</p>
| 17794 | Pressure measurement problem |
2017-10-29T08:55:25.683 | <p>While focusing on Quality and strength of welded joints produced but both methods..</p>
| |mechanical-engineering|welding|welds| | <p>It depends a bit on which arc process you compare it to. Broadly speaking there is no fundamental reason why one should be any better or worse than the other. </p>
<p>Perhaps the biggest limitation of gas welding is shielding. A correctly adjusted gas flame will protect the joint from oxidation but is certainly not entirely inert which can be a problem with more active metals (especially aluminium) and alloys which are particularly susceptible to contamination. Certainly for welding stainless steel and aluminium MIG and TIG are almost universally preferred. </p>
<p>Also gas welding is now much less common in commercial and industrial applications so you are more likely to find up to date standards. </p>
<p>Compared to an electric arc (especially TIG) a gas flame is cooler and broader and so the heat affected zone tends to be larger and you end up putting more heat into the work overall. This is more of an issue for some materials than others </p>
| 17802 | How does Electric arc welding compared to gas welding? |
2017-10-30T20:31:01.867 | <p>I bought recently 10 meters of Kanthal A1 AWG18 wire, which is a type of heating element (FeCrAl alloy). I intend to make a coil out of it, but it's no use doing that right now, I need more of it for my project (electric furnace), have ordered more of it online and am waiting a couple weeks till it arrives.</p>
<p>My question is this, if I use the heating element a few times in a different project (assume here that the wattage is the maximum possible), in a way that I won't twist or wind it in any way, will this affect the physical properties of the wire in ways that might make it harder for me to turn it into a coil later on, for my initial project? (I'll be winding the coil manually, it will be between 5 and 10 mm of diameter)</p>
<p>I ask that because heating elements tend to get more brittle with time, but I don't know how long it might take for such changes to happen.</p>
| |materials|thermodynamics|heat-transfer|heating-systems|thermal-conduction| | <p>Kanthal and nichrome/ chromel are rather different; iron base vs nickel base. Neither has a "martensite" transformation so can not be hardened by fast cooling. Little change would occur on use , possibly grain growth at high temperatures could cause limited embrittlement. You could maximize ductility by annealing: Anneal by heating to red or orange heat and quench into water. </p>
| 17823 | Winding a wire heating element after a few uses |
2017-10-31T06:29:26.127 | <p>Is there a purely mechanical device that takes in clockwise or counter-clockwise rotation 'indiscriminately' and outputs in both cases clockwise rotation?</p>
| |mechanical-engineering|gears|transmission| | <p>How about something like this, uses two one-way clutches :
<a href="https://i.stack.imgur.com/ZznI1.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ZznI1.png" alt="enter image description here"></a></p>
<p>Output is clockwise from the front...</p>
| 17827 | Automatic translation of either-sided rotation to fixed directions? |
2017-10-31T16:08:03.483 | <p>I'm trying to design a device that can stretch a tissue sample of set length, right now about 2 mm, to desired stretch ratios. This will be controlled using a interface where the user can input their desired ratio, and the controller will tell a 6 in stroke WindyNation linear actuator to move to a position based on the input ratio. The controller also needs to communicate with a force sensor and display the force that is being exerted on the sample. </p>
<p>For this type of problem, is it better to use an Arduino, Raspberry Pi, or both to power the controller? I've found that the Pi won't be very good due to issues with timing. Will this be a serious issue?</p>
| |design|biomedical-engineering| | <p>All depends on the sampling rate you require:</p>
<p>RaspberryPi will run an OS, which has multi-tasking scheduler, drivers/IO interruptions, etc. (mostly like any computer). In general computing, this is great, but for sampling a sensor, it make difficult to get a precise frequency: you will have an average of -+25us jitter and up to nearly +500us in some cases.</p>
<p>On the other hand, Arduino is a MCU, it has no complex operating system. It cannot run a desktop application, but will have no issue in sambling precisely a sensor.</p>
<p>In general, if it is fine to sample at 1000Hz or slower, then either systems are fine. The RaspberryPi would give more processing power and many tools for showing in screen, sending by network or anything else with the data.</p>
<p>If it is required to get high sampling rate (e.g. 50KHz), then the Arduino will do a better job.</p>
<p>Ideally, you could sample with the Arduino, send the data to a RaspberryPi which could do complex stuff with the data.</p>
| 17836 | Designing of a tissue stretching device using a linear actuator |
2017-10-31T20:23:24.980 | <p>Can someone explain the how/why Nylons brittleness is to some degree a function of it's moisture content. I don't understand HOW the water content changes the elongation at break of the material. </p>
<p>Water doesn't bond or change the chemistry of the nylon when absorbed, my understanding is that it just fills some of the voids within the nylon. so then how is this changing the slope of the stress/strain curve?</p>
| |mechanical-engineering|materials|chemical-engineering|chemistry| | <p>Nylons are a class of polymers formed from di-acid and di-amine monomers. The amine groups bond with the acid groups to form amide groups, end to end. The resulting amide groups have a positive charge at the amine hydrogen, and negative charge at the acid double-bond oxygen. The result is a polymer with periodic polar moments across the chain. The polar moments align between neighboring chains, creating a strong van der Waals force between them, in the form of hydrogen bonding. This is part of why nylon fibers are so strong: they resist chain sliding due to the strong inter-chain bonding.</p>
<p>Water is also a polar molecule and also exhibits hydrogen bonding, giving it some interesting properties compared with similar molecules. Because of its polar nature, water readily infiltrates nylon materials and is attracted to the polar amide groups of nylons. Water creates space between chains, increasing the ability for chains to move around one another more easily, improving ductility. As Chris Johns noted, it acts as a plasticizer.</p>
<p>It is worth noting that nylon and other amide polymers are somewhat special in that they are hydrophilic and oleophobic, unlike other common engineering polymers.</p>
| 17844 | Nylon brittleness as function of moisture content |
2017-11-01T14:29:21.613 | <p>There is ice build in my Samsung RF266 refrigerator. I have posted the following questions in <a href="http://diy.stackexchange.com">Home Improvement Stack Exchange</a> hoping to find a solution. The questions are as follows:</p>
<ul>
<li><a href="https://diy.stackexchange.com/questions/125403/samsung-rf266-refrigerator-not-cooling-fan-making-noise">Samsung RF266 refrigerator not cooling - Fan making noise</a></li>
<li><a href="https://diy.stackexchange.com/questions/126094/ice-buildup-inside-samsung-rf266-refrigerator">Ice buildup inside Samsung RF266 refrigerator</a></li>
</ul>
<p>Bel
below is an image of the ice build up</p>
<p><a href="https://i.stack.imgur.com/F4xx8.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/F4xx8.jpg" alt="enter image description here"></a></p>
<p>I believe the purpose of the fan is to circulated the cold air and coils are what generates the cold air. I don't believe the ice build up is normal.</p>
<hr>
<p><strong>Question:</strong> From an engineering standpoint, how is this suppose to work? What would cause the ice build up? What design feature are present in a refrigerator to prevent ice buildup? </p>
| |refrigeration| | <p>the presence of severe ice buildup in a refrigeration system strongly suggests a door gasket leak. Check to see if the gasket is in good shape and also if the door is closing properly against the body of the refrigerator. In this connection, note that most refrigerators have reversible door hinges which let the owner remount the door so it opens left-handed instead of right-handed. If the hinge attachment screws are loose or if the top and bottom hinges are skewed, then the door won't seal properly when closed. A simple way to check the gasket seal integrity is to place a lit flashlight inside the refrigerator and carefully observe the entire gasket seal for light escaping- indicating a bad seal. </p>
| 17858 | How to prevent ice build up in a refrigerator? |
2017-11-01T18:07:46.737 | <p>I am trying to configure a force sensor (a piezoresistive force sensor) in a device to measure small forces (less than 0.5 mN). This is difficult due to noise of the system. One of the ways I have considered to address this problem is to mechanically amplify the force applied to the force sensor with a gearbox. The small force applied would be transferred to the gearbox with a gear on a drive bar, and the amplified force from the increased torque would be transferred to the sensor with another gear and drive bar (see schematic). However, I have no experience with gearboxes and would like some opinion on feasibility or if this system even makes sense. I am particularly worried that static friction forces in the gears would cause inaccuracies, especially because gearboxes are designed for a completely different purpose.</p>
<p><a href="https://i.stack.imgur.com/biLdg.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/biLdg.jpg" alt="Schematic"></a></p>
| |mechanical-engineering|gears| | <p>Any gearbox or lever linkage is going to have backlash to go with your other concerns. By the time you get to a system using this kind of mechanical amplifier to get the sensitivity you need along with the low noise, accuracy and precision you want, you're into making a what's essentially a precision watch... Can it be done? Sure. It's going to be time consuming and expensive though.</p>
<p>It will be faster, cheaper and more efficient to simply purchase a <a href="https://www.novatechloadcells.co.uk/ds/f329.htm" rel="noreferrer">load cell designed to measure micro Newton level Forces</a>. If you're a student, doing a project on a student budget with no corporate sponsor, you should call the company anyway, they might help you anyway.</p>
<p>If you really need to build the measurement device yourself, I suggest looking into a <a href="https://www.britannica.com/technology/torsion-balance" rel="noreferrer">Torsion Balance</a>. They're typically used to measure things like the gravitational attraction force between two objects, so the concept is in line with your measurements. You could also build a strain gage based sensor, perhaps some kind of canteliever beam. </p>
| 17861 | Can a hobby gearbox be used to mechanically amplify small forces for better force sensor resolution? |
2017-11-02T02:22:04.797 | <p>I'm building a product prototype for a client. We are trying to figure out the best way to build a hollow helix capable of transporting liquid at home. Something similar to this: </p>
<p><a href="https://i.stack.imgur.com/ftFPo.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ftFPo.jpg" alt="enter image description here"></a></p>
<p>So far our best idea is to use standard tubing and try to use rings mounted on the side of a support cylinder which the helix loops through but this seems almost impossible to pull off at a small scale, and wouldn't look too good.</p>
<p>We have also considered heating sand and using that to bend acrylic tubing, but this certainly still doesn't feel ideal. </p>
| |plastic|molding| | <p>I don't know if you finished your project,but if you didn't have to see the liquid, I would use copper tubing. I would carefully wrap two tubes on a dowel of the right diameter (it will reduce a little in diameter when you stretch them) and after removing the dowel stretch the coils appropriately. It should be self supporting but you could wrap solid bare wire 16 - 10 AWG around each coil (assuming 3/16" - 3/8" diameter tubing) from the bottom to the top of the helix in 3 or more evenly spaced supports. If sagging is still a problem you can work springs over the support wires, but be aware there may be galvanic corrosion from the dissimilar metals, and you may want to dip the springs in liquid electrical tape or spray them in insulator paint MIL-E-22118.
If you need to see the liquid I would use clear vinyl tubing supported with self adhesive cable clips sized appropriately with 4 or more evenly spaced wood or plastic supports.</p>
| 17867 | How to build a transparent double hollow helix at home |
2017-11-02T10:47:51.403 | <p>I'm solving this problem in mechanics, </p>
<p>Two cylinders A and B, weighing 100 lb and 200 lb respectively, are connected by a rigid rod curved parallel to the smooth cylindrical surface shown in Fig. P-329. Determine the angles α and β that define the position of equilibrium.</p>
<p><a href="https://i.stack.imgur.com/ST0bl.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ST0bl.gif" alt="Fig. P-329"></a> </p>
<p>I already understood that the relation between $\beta$ and $\alpha$ is $\cos \alpha = 2 \cos \beta$ by doing summation of moment with respect to O and equating to 0. What I don't get is how come the angle from this picture becomes 90 degrees? </p>
<p><a href="https://i.stack.imgur.com/zPpws.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/zPpws.jpg" alt="enter image description here"></a></p>
| |statics| | <p>It's given by the question. If you pay attention to the original image, you'll see that the internal angle ($180−\alpha−\beta$) has the classic right-angle symbol.</p>
<p><a href="https://i.stack.imgur.com/07NUKs.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/07NUKs.png" alt="enter image description here"></a>
<sup><a href="https://upload.wikimedia.org/wikipedia/commons/thumb/6/6c/Right_angle.svg/1200px-Right_angle.svg.png" rel="nofollow noreferrer">Source</a></sup></p>
| 17872 | Included angle in a semicircle |
2017-11-03T14:34:05.923 | <p>Can you please suggest what kind (type) of hinge / pins I should use to make the mechanism in the attached file. I was considering to use the pins shown in the second image. But I am not sure if it will work.</p>
<p><a href="https://i.stack.imgur.com/7XlU2.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/7XlU2.png" alt="enter image description here"></a> </p>
<p><a href="https://i.stack.imgur.com/v5zWym.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/v5zWym.jpg" alt="enter image description here"></a></p>
| |mechanical-engineering|applied-mechanics|mechanisms| | <p>What is shown in the picture is what people usually refer to as a pin connection. There are multiple ways you can design a pin connection as suggested in the comments above and there is no one right design for it. Disregarding practical considerations, such as materials, load cycles and etc, and in pure design engineering terms, you will need to find the shear strength of the items you have selected for your pin, and make sure it is reasonably stronger than what you have calculated for the shear load on the pin based on the configuration you have. The shear load on the pin can be determined from the normal force at the pin connection, and ut can be something less if you have double shear or triple shear mechanisms in your pin connection in question.</p>
| 17888 | Type of hinge I should use |
2017-11-04T04:16:16.027 | <p>I'm solving this problem, <br>
<a href="https://i.stack.imgur.com/aLweG.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/aLweG.jpg" alt="Problem"></a>
<br>
I used the formula $Q_p = mc_p\Delta T$. Since it is a mixture, I did this:
$Q_{Total} = 0.3 × 0.867 × (50-150) + 0.7 × 1.043 × (50-150 ) $ <br>
$ = -99.02 kJ/kg $<br>
Is my solution and final answer correct? If not, where am I doing wrong? Thankss. </p>
| |mechanical-engineering|thermodynamics| | <p>You're assuming that both gases have the same mass, which isn't true.</p>
<p>$$\text{mass} = \text{density}\cdot\text{volume}$$</p>
<p>$V_{\text{CO}_2}=0.3 V_\text{total}$ and $V_{\text{N}_2}=0.7 V_\text{total}$</p>
<p>Use the calculated mass to find $Q_p$ unit is in $\text{KJ}/\text{m}^3$.</p>
| 17899 | Heat released by gas mixtures that undergo constant pressure cooling |
2017-11-05T02:01:35.780 | <p>I'm used to calculate the minimum radius required for bending operations on both aluminum and steel sheets, given the following parameters: <a href="https://i.stack.imgur.com/yxEBc.jpg" rel="nofollow noreferrer">https://i.stack.imgur.com/yxEBc.jpg</a></p>
<p><strong>Problem statement:</strong></p>
<blockquote>
<p>For a sheet metal stock with inch thickness, determine the minimum
tool radius for both the steel and aluminum alloys that will not tear
the material. Assume the sheet of material is in pure bending (i.e.,
no additional tension is applied during forming).</p>
</blockquote>
<p>I have already learned about the following formulas:</p>
<ul>
<li>R = 1/k</li>
<li>k(y) = e(y) / (h/2)</li>
<li>e(y) = k.y</li>
</ul>
<p>Given that R = 1/k, the next step is to find the maximum curvature k for each sheet.</p>
<p>The problem is that I don't know how to find the maximum curvature, and I'm still stuck.</p>
<p>Any thoughts?</p>
| |mechanical-engineering|manufacturing-engineering|metals|moments|yield-point| | <p>I hope the following helps. This is from a book I used in college. Manufacturing Engineering and Technology, 5th Ed. by Kalpakjian and Schmid</p>
<hr>
<p><a href="https://i.stack.imgur.com/R1aHB.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/R1aHB.png" alt="enter image description here"></a></p>
<p><a href="https://i.stack.imgur.com/7ApdV.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/7ApdV.png" alt="enter image description here"></a></p>
<p><a href="https://i.stack.imgur.com/iUmbu.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/iUmbu.png" alt="enter image description here"></a></p>
<p><a href="https://i.stack.imgur.com/6T4Tq.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/6T4Tq.png" alt="enter image description here"></a></p>
<p><a href="https://i.stack.imgur.com/HFfS1.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/HFfS1.png" alt="enter image description here"></a></p>
<p><a href="https://i.stack.imgur.com/nHkmW.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/nHkmW.png" alt="enter image description here"></a></p>
<hr>
<h1>Minimum Bend Radius</h1>
<p><code>Where **r** is the tensile reduction of area.</code></p>
<p><a href="https://i.stack.imgur.com/QQghx.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/QQghx.png" alt="enter image description here"></a></p>
<hr>
| 17907 | Sheet Metal Bending: stuck on calculating the minimum radius required |
2017-11-05T21:41:05.130 | <p>I'm trying to build a step stool that has a rotating chair back that will rotate from top position (e.g. chair) into bottom position (e.g. step stool). I do not want a free rotation of the chair back so I want to avoid it slamming on the floor. I've been looking for a bearing that acts as a mechanical stepper to click every 15 degrees with enough force requirement to stop the chair back from falling to the ground without force being applied (other than gravity).</p>
<p>I've been searching google for a while now and cannot figure out what this part would be called.</p>
| |mechanical-engineering|bearings|bolting| | <p>It sounbds like you are describing a <a href="https://en.wikipedia.org/wiki/Detent" rel="nofollow noreferrer">detent</a> mechanism </p>
<p>Depending on the design this can be a positive lock like pin engaging with a hole which needs to be manually disengaged with each step or a 'soft' latch for example a ball bearing engaging with a conical pit which can be overcome just by applying a bit more force. </p>
<p>You can buy the sprung balls as a complete assembly which could then be mated to a disk with a shallow hole every 15 degrees. </p>
<p>Also a pawl and ratchet can be modified so that the teeth work like cams against a sprung pawl rather than providing a positive stop. </p>
| 17917 | What is the hardware called that can step a lever N degrees upon manual force? |
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