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2019-04-28T20:13:28.620 | <p>I have this circuit:</p>
<p><a href="https://i.stack.imgur.com/tIgot.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/tIgot.png" alt="enter image description here"></a></p>
<p>If D2 was a resistance, I could just combine it with R2 to get some composite resistance that I could put in series with R1. That way, I could calculate the resistance of the entire circuit and, thus, the total current.</p>
<p>Now, this is not the case here but I know a few things here:</p>
<ul>
<li>whatever voltage drops at D2, also drops at R2</li>
<li>the current through R1 is the same as through R2+D2</li>
</ul>
<p>But I do not know the current from the beginning, so I cannot calculate the voltage drop at R1. And the diode is non-ohmic, so I cannot find out anything about it.</p>
<p>In this way, I do not see how I could apply Kirchhoff's laws or any other methods because I seem to be missing information... where do I start here?</p>
| |electrical-engineering|circuits| | <p>One good place is to start is <a href="https://media.digikey.com/pdf/Data%20Sheets/Lite-On%20PDFs/LTL-307EE.pdf" rel="nofollow noreferrer">LTL-307EE</a> datasheet. From the datasheet take look at current profile for the device. Below is the forward current and voltage characteristics.</p>
<p><a href="https://i.stack.imgur.com/51RzZ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/51RzZ.png" alt="LED forward current and voltage graphs"></a> </p>
<p>Also take look at absolute maximum rating for the device. </p>
<p><a href="https://i.stack.imgur.com/iIL9b.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/iIL9b.png" alt="Absolute Max ratings"></a></p>
<p>Now you have to size <em>R1</em> resistor such that no more than in this case 30mA of current flows through the LED. <em>R1</em> is called a current limiting resistor. I would size the R1 to make ensure 25mA of current, with will provide Relative Luminous Intensity of around 3.0. Like wise <em>R2</em> can be used to manage current LED current flow. </p>
<p>I hope this help you get started. </p>
| 28137 | How do I calculate voltage at an LED in parallel? |
2019-04-28T21:34:24.237 | <p>Is there a mathematical analysis of the <a href="https://en.wikipedia.org/wiki/Hydraulic_jumps_in_rectangular_channels#Jump_location" rel="nofollow noreferrer">submerged hydraulic jump</a>, such as under what condition it is formed, the speed and trajectory of the circular flow therein, etc.? Is there a good mathematically sophisticated reference on this topic?</p>
| |fluid-mechanics|hydraulics|open-channel-flow| | <p>In open channels when the location of supercritical flow and hydraulic jump are very close, or restriction in down stream encourages higher elevation, submerged or partially submerged hydraulic jump happen.</p>
<p>For example, when they use flumes to measure the flow rate of an irrigation open channel because of the downstream conditions or obstacles the hydraulic jump may build up and partially or totally submerge the supercritical flow up-stream.</p>
<p>In closed pipe flow such as sewer system, the incidental blockage before an elbow or other constrain will cause submerged jump leading to deposit of more solids, that's why they design those locations with access openings, clean-outs.</p>
<p>because of wide range of conditions leading to this situation, I think the only practical way of detailing and calculation the submerged jump is using FEM and simulation programs.</p>
| 28142 | The mechanism of submerged hydraulic jump |
2019-04-29T21:42:05.840 | <p>So, I have a problem with fatigue cracking originating from hot punched holes. From failure forensic reviews, it seems like surface conditions contributed to some of the failures, especially where there are leftover shear marks and laps from punching operations. I did a quick research but it doesn't seem like there are well established standards regarding the finishes of holes generally in terms of hole surface quality. Am I over thinking and that nobody had the same problem previously? Or are there some standard that I am not already aware of that exists? </p>
<p>One of my thoughts is to establish Ra Rp Rz acceptable ranges that can be applied for the holes. But I have doubts about measurement consistency as those measurements are usually more geared towards flat surfaces. What are your thoughts on this? </p>
<p>Thanks for all your help!!</p>
| |metallurgy|surface-preparation| | <p>When talking about R values and holes, what you really should be measuring is the S value instead that bases on area. There are generally two techniques you can measure interior surface roughness of holes. One of them is to fill the hole with something and remove it, then measure that piece directly. Alternatively, you can also scan it optically to create a highly accurate 3D impression of the hole and measure Sz Sa Sp that way. R and S values are not equivalents, so care should be taken to review relevant literature on the subject when you are specifying a criteria to meet certain performance requirements. </p>
<p>Good luck!~</p>
| 28156 | Engineering standards or testing standards for hot punched metal holes for rolled steel |
2019-04-30T05:04:32.567 | <p>I have designed a system using a minimum realization.</p>
<p>I have the A, B, C matrices and I have assigned closed loop poles for the system.</p>
<p>I used the place command to find my K matrix. </p>
<p>I am wondering if it is possible to use the lsim matlab command to simulate my closed loop system (A-B*K)?</p>
<p><span class="math-container">$$x' = Ax+Bu$$</span>
<span class="math-container">$$u=-Kx$$</span>
<span class="math-container">$$x' = (A-BK)x$$</span>
<span class="math-container">$$y=Cx$$</span></p>
<p>All examples I have found for lsim use the system transfer function or the open loop system.</p>
<p>I have seen examples such as this:</p>
<pre><code>t = 0:0.01:10; % simulation time = 10 seconds
u = zeros(size(T)); % no input
x0 = [0.1 0.1 0.1]; % initial conditions of the three states
sys= ss(A,B,C,D);
lsim(sys,u,t,x0)
</code></pre>
<p>But I am not sure how to use this with the closed loop (A-B*K) system.</p>
<p>Any help would be appreciated! Thank you.</p>
| |control-engineering|control-theory|matlab| | <p>Yes, you can using the <a href="https://uk.mathworks.com/help/control/ref/feedback.html" rel="nofollow noreferrer"><code>feedback</code></a> command. However, if your open-loop system is defined by your state-space system <code>sys= ss(A,B,C,D)</code>, then your closed-loop system would be defined as follows (and not by <code>A-B*K</code> as you suggest):</p>
<pre><code>CL_loop = feedback(sys,K); % K is your controller/compensator, this needs to be an LTI object
lsim(CL_loop,u,t,x0); % you might have additional states due to the feedback loop, and so would need additional IC
</code></pre>
<p>Note that you may also need to change your input vector, as it's now representing your desired trajectory.</p>
<p><strong>Update after comment & question edit</strong></p>
<p>I think it's just a matter of creating a new state-space representation with the appropriate matrices, e.g.:</p>
<pre><code>CL_loop = sys(A-B*K,zeros(size(B)),C,zeros(size(D))); % implement x'=(A-B*K)*x & y = C*x
lsim(CL_loop,u,t,x0);
</code></pre>
| 28161 | How to use lsim in matlab for a closed loop system |
2019-05-01T08:01:17.103 | <p>I am looking for non-destructible way which can be performed in home conditions to tell if the material is titanium (in my case it is a mug). Advices like "grind it" or "use electronic microscope" could be useful but I cannot apply them as stated above. I am targeting primarily aluminium and steel alloy because of their competitive prices and availability. In pure form detection would be easy (weight) but what about the alloy?</p>
<p><em>I am also asking for non-expert level ways -- I can weight something, boil water, measure temperature or resistance, those kinds of stuff. I am not a chemist for sure.</em></p>
| |materials| | <p>In adition to measuring the density of your object as described by Fred there is a few other tests you can easily make. You can try to measure:</p>
<ul>
<li><p>Heat capacity of your object. By heating the object in hot water measure tempertaure. Then move it over to another cooler water container where you know the volume and measure the temperture of the water after this.Then compare to table values</p></li>
<li><p>Specific electrical resistance is also very quick to do if you happen to have a multimeter. This would very quickly rule out aluminium. Requires very minimal setup time to do this measurement.</p></li>
</ul>
<p>Now density is best but adding these to the mix can give you a better confidence in your results. Resistance measurement also works if you suspect there is a hollow cavity in your object.</p>
| 28170 | How to make sure the material is titanium? |
2019-05-02T10:12:19.463 | <p>I have a self balancing robot control loop that uses the PID controller. I feed the robot's angle (rad) to sum block that generates the error between the set point and the angle.</p>
<p>The error then goes to the PID and goes to DC motor block that needs the input to be in volts and generate an output in N.m. Then the signal goes to the self balancing robot block that need the input to be in N.m and outputs a rad angle.
<a href="https://i.stack.imgur.com/oS7OU.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/oS7OU.png" alt="enter image description here"></a>
As you see I don't know how to make the PID output a voltage (V). How i can do that?</p>
| |control-engineering|control-theory|pid-control| | <p>The PID simulation you've built does not care about the units of numbers being passed around your control loop. However, you may want to ensure that the units are correct at different points in the loop to make it easier for you to interpret and debug. To do that, you can simply add gain blocks (no frequency dependence) with the appropriate conversion factors. If you do not add separate gain blocks, then the unit conversion will be implicitly included in the gain of each transfer function. </p>
| 28188 | Do variables dimensions matter in PID simulation + I want the PID output to be volts? |
2019-05-02T10:48:11.617 | <p>Given the following situation:
<a href="https://i.stack.imgur.com/m5cgk.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/m5cgk.png" alt="in"></a></p>
<p>Pipe A and B have a known and identical length and diameter, and are in parallel. I'm assuming the flow trough pipes A and B are also identical, as is the roughness of the pipes.
Pipe C also has a known length and diameter (which is different from the diameter of pipe A/B).</p>
<p>I'm interested in the total head loss at the end of Pipe C. I am aware of the following Darcy-Weisbach equations that are relevant in this case (parallel and series):</p>
<p><a href="https://i.stack.imgur.com/zZ8D4.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/zZ8D4.png" alt="enter image description here"></a></p>
<p>Here the total headloss is the sum of the headloss per pipe, placed in series. </p>
<p><a href="https://i.stack.imgur.com/udlw6.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/udlw6.png" alt="enter image description here"></a></p>
<p>Here the total headloss over set of 2 parallel pipes is equal to the headloss of one of them.</p>
<p>But, how do I combine these two equations to retrieve the total headloss over pipes A, B and C?</p>
| |fluid-mechanics|hydraulics|pipelines| | <p>The total headloss through pipes A+C is equal to the headloss through pipes B+C. Therefore, you only need to calculate one of the paths, and your first equation gives you the desired total headloss (where 1 and Q in the first term are for pipe A, and 2 and Q in the second term are for pipe C).</p>
<p>Of course, the total headloss is equal to the pressure difference that is driving the flow. For example, if the tank is filled with a liquid to depth H, and if the end of the pipe C is opened to the atmosphere, the pressure driving the flow due to the elevation difference is known. In other words, you usually know the total headloss because you know the pressure difference, and the calculation is to find the flow rate. You can determine that from the first equation because you are assuming that <span class="math-container">$Q_A=0.5Q_C$</span>.</p>
<p>In addition to the Darcy-Weisbach wall losses, there are also losses due to the inlets into pipes A+B, the tee where A+B combine into C, and the exit loss at the end of C. These may be significant depending on the flow rate and length of pipes.</p>
| 28189 | Determining the headloss according to Darcy-Weisbach with pipes in parallel and series |
2019-05-03T08:00:18.267 | <p>I'm having some work done in the UK (Eurocode) which involves a steel column to support a beam at 1st floor (US: 2nd floor?) level. The column bears a very-close-to-vertical total load of 47.3kN (including any relevant safety factors I guess, as that's the final figure in his calcs).</p>
<p>The engineer has specified a 152x152 23kg UC to be used for the column, with suitable end+base plates and fixings.</p>
<p>My issue is that the engineer has offered a low rate as its for a "charity job" (house belongs to an elderly frail relative). I've gone back to him with a set of 3 queries and got replies that are professional but curt and I don't want to go back again to ask if anything else is possible, unless I'm already privately very sure it is and can back it with calcs, as I don't want to lose his goodwill. </p>
<p>The only time he's said yes to any change, was when I also attached calcs to justify why I thought it should be OK. Unhelpful and not my job, I know!</p>
<p>In this case, I really need a slimmer column, to run flush with the 100mm partition wall. Unfortunately I don't understand column slenderness calculations well enough to check if there's scope to replace the 3m 152x152 23kg UC by, say, a 100x100x10 25kg RHS, or even a 100x180x5 21kg RHS. My tentative figures suggest that because of the RHS vs UC profile, both of these would be stiffer in any direction, than the UC and they'd fit nicely flush to the wall, so maybe they'd work.</p>
<p>I don't need an absolute calculation <em>("Does this profile work for that length and load")</em>, but only a relative one <em>("Would this profile be at least as satisfactory as the one he's already said is good enough")</em>?</p>
<p>What calculations are needed, to confirm if a more slender (in one dimension at least!) RHS profile would be at least as satisfactory/rigid in performance as a column, compared to the currently-proposed UC?</p>
| |steel|columns|eurocodes| | <p>The root problem of your question is that you're asking us to make some engineering judgement calls and calculations for a building where we don't know all the relevant details. You could however calculate the slenderness of the existing column and then propose a section that has this slenderness or less.</p>
<p>The slenderness parameter that is commonly used here in the US is <span class="math-container">$ \sqrt{\frac{KL}{r}} $</span>
where <span class="math-container">$r$</span> is the radius of gyration <span class="math-container">$L$</span> is the column length, and <span class="math-container">$K$</span> is the effective length factor you can assume for comparison that this is <span class="math-container">$K=1$</span>. Be sure and use consistent length units.</p>
<p>You can find tabulated section properties that should give you the <span class="math-container">$r_{x}$</span> and the <span class="math-container">$r_{y}$</span> value (radius of gyration) for different shapes (including the one originally specified). In general you will want to take the smaller of the <span class="math-container">$r$</span> values as that direction buckling will control.</p>
| 28199 | RHS column profile able to replace 3m UC 152x23 column? |
2019-05-03T12:29:52.543 | <p>What causes the spider gear to spin about it's own axis while the vehicle is taking a turn?</p>
| |automotive-engineering|power-transmission| | <p>The reason it turns about its center is the difference in rotation of the two wheels, which are connected through the sun-gears to planet-pinion gears. By rotating it allows the two sun-gears and consequently, the two wheels go at deferent speeds without either losing power, or having to skid.</p>
<p>And the reason for the difference in the wheels speed is because in a turn the inside wheel has less speed then the outside wheel, having to go around a smaller radius.</p>
<p><a href="https://i.stack.imgur.com/bE5fE.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/bE5fE.gif" alt="diagram of the differential"></a></p>
| 28202 | Working of spider gear in differential axle wheel mechanism |
2019-05-04T05:56:39.310 | <p>It might be a little off topic but I'm trying to find source for why is that model called magic formula? I read that it's a simple function of commonplace functions, a data fitting. Is it simply an empirical model with an unusual name? </p>
<p>The model (or rather series of models) I'm referring to:
<a href="https://en.wikipedia.org/wiki/Hans_B._Pacejka" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Hans_B._Pacejka</a></p>
<p>So far only this lecture mentions the history of the name:
<a href="http://textofvideo.nptel.ac.in/107106080/lec15.pdf" rel="nofollow noreferrer">http://textofvideo.nptel.ac.in/107106080/lec15.pdf</a></p>
| |automotive-engineering|simulation|friction|wheels| | <p>"Magic X" is a common expression for things which aren't adequately explained but that "just work".</p>
<p>For instance, in programming, it is considered poor form to use "magic numbers". These are numbers which simply appear in a given code without any explanation of what they are. For example, if code has <code>x * 0.707</code>, you immediately start wondering what that 0.707 comes from. You'll need to look at the code and think about it and remember that the cosine of 45 degrees is approximately equal to that.</p>
<p>Instead, one should probably define a named variable (for example <code>var cos45 = 0.707</code>) with that number as its value. Then you just do <code>x * cos45</code> and it's clear as day what the code does, no thinking required.</p>
<p>That being said, if you do come across a program which does <code>x * 0.707</code>, you might not really understand where that number comes from, but you can tell it "just works".</p>
<p>In a similar vein, "magic formula" simply means it's a formula which, for reasons unknown, "just works". That much is stated in the <a href="https://en.wikipedia.org/wiki/Hans_B._Pacejka#The_Pacejka_%22Magic_Formula%22_tire_models" rel="nofollow noreferrer">Wikipedia article</a> you linked to (emphasis added):</p>
<blockquote>
<p>Pacejka has developed a series of tire design models over the last 20 years. <strong>They were named the 'magic formula' because there is no particular physical basis for the structure of the equations chosen, but they fit a wide variety of tire constructions and operating conditions.</strong></p>
</blockquote>
<p>One of the <a href="http://phors.locost7.info/phors21.htm" rel="nofollow noreferrer">references</a> to the Wikipedia article even keeps the symbolism going (emphasis added):</p>
<blockquote>
<p>The next step is to compute the product of a new helper, <span class="math-container">$B$</span>, times <span class="math-container">$b_0$</span> and the aforecomputed <span class="math-container">$D$</span>. The <strong>magicians</strong> who created the formula tell us that <span class="math-container">$B b_0 D = (b_3 F_z^2 + b_4 F_z)\exp(-b_5F_z)$</span>. This slurps up a few more of the <strong>magical</strong> eleven empirical numbers, and a pattern emerges [...]</p>
</blockquote>
<p>Basically, the "magic formula" is so-called because it is totally empirical. There are many other empirical formulas in life, though, so you might be asking why "only" this one is called thus. The answer is that this one is <strong><em>really</em></strong> empirical. Other empirical equations are mostly intuitive: they may have empirically-determined coefficients, but the overall form of the equation makes some sense. This one is just a bunch of coefficients added, multiplied and exponentiated and somehow it all works out.</p>
<p>Or, as that same reference I linked to above puts it:</p>
<blockquote>
<p>Once again, don't try to find any physics in here: it's just a convenient formula that fits the data reasonably well.</p>
</blockquote>
| 28219 | Pacejka "Magic Formula" tire models origin of the name |
2019-05-06T13:03:46.977 | <p>I'm recently interested in print-heads, especially thermal inkjet print heads.
I would like to design a simple print head with nozzles in a row with no chambers, but only a straight linear pipe-like structure, and the heating resistors stuck on the side.</p>
<p>I'm quite lost here. I don't know what kind of material I should use, or what should be the diameter of the nozzle, and etc.</p>
<p>I want to look at the earliest version of thermal inkjet print heads but all that I can get in google patent is the latest, and high tech version of print heads that I can barely understand.</p>
<p>Can anyone tell me what to do, or where to look at?</p>
| |mechanical-engineering|electrical-engineering|fluid-mechanics| | <p>For starters, read the May 1985 issue of the <em>HP Journal</em> magazine; it explains the construction of the very first thermal inkjet printhead that HP brought to market in March of 1984, which I helped design. </p>
<p>That printhead was called the <em>Thinkjet</em>; you can try googling on that. </p>
<p>I wrote extensively about printhead device physics and design during my 34 years in the inkjet field and would be happy to discuss your project with you. You can contact me directly through my website, www.nielsenkillowatt.com.</p>
| 28236 | How to design and make thermal inkjet print head / nozzles |
2019-05-06T14:50:37.853 | <p>ME Design uses hourly climate data files (HCD) as input into the pavement design algorithm. As a user I can enter an arbitrary number of years of climate data. If I enter say 30 years of climate data and then set an analysis period of 20 years, what data does the program use? Does it use the first value in the list or the last 20 years of the file? What if my analysis period is greater than the amount of climate data entered?</p>
| |pavement| | <p>Pavement ME Design checks to see if the pavement construction date is within the entered climatic data range. If it is, the analysis uses that date.</p>
<p>If the pavement construction date is not included in the entered climate data range, the analysis always uses the earliest month/day in the list that matches the pavement construction date. It does this so that the answer is the same at years 1,2,5,10 whether an analysis is run for 10, 20 or 30 years.</p>
<p>If the analysis period is longer than the available climate data, the data is reused starting from the earliest applicable date.</p>
| 28237 | How does Pavement ME Design handle climatic data? |
2019-05-07T21:08:06.780 | <p>The following images are taken from Mechanics of Materials, Ferdinard Beer et al. 6th ed (2012)
Just wondering that why the effective length of the column is taken 0.7 L (highlighted in the sample problem) ? Souldn't it be 0.5 L since in the xy plane we have two plates supporting the column with a reaction force in the y direction and a moment (which is the same as if the column were fixed at both ends).
Any thoughts would be appreciated.</p>
<p><img src="https://i.stack.imgur.com/MuIhF.png" alt="enter image description here">
<img src="https://i.stack.imgur.com/h1D8J.png" alt="enter image description here"></p>
| |solid-mechanics| | <p>In real life cases these diagrams are all highly exaggerated to convey the idea that effective length has to do with the inflection point and or, it starts at a pin support.</p>
<p>Otherwise for all intents and purposes the smooth rounded edge of the two plates are close enough to a pin support at that joint. </p>
<p>Now if we want to be too fussy, even if the two plates are too tight and fit to the column, under the load they yield just a smidge to make it a pin connection. And we would see the yield line as shiny horizontal score lines after. </p>
| 28247 | Buckling in Columns |
2019-05-08T04:06:53.857 | <p>So I am working on a project. I need a drone (quadcopter) to ascend for 5 seconds, then cut off power to the drone, then let it stay in the air for as long as possible without power. My idea for allowing it to glide is to have some rubber bands attached to some parachute like material, and the rubber bands will be stretched and held in place by some pins. After 4 to 5 seconds, the pins will disengage, allowing the rubber bands to compress, expanding the parachute-like material. An image is included to illustrate what I mean. </p>
<p>The red lines illustrate the rubber bands. The black are the quadcopter arms. The blue circles are where the pins would be located. After the timer stops, the image on the right demonstrates the rubber band straightened, and the purple is the parachute area. There will be one parachute per rubber band, and each rubber band will have its own locking mechanism. </p>
<p><a href="https://i.stack.imgur.com/ApvpG.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ApvpG.png" alt="enter image description here"></a></p>
<p>So what I want to figure out is how to design a mechanism that will retract a piston/pin after 4 to 5 seconds. I did some research on pneumatic cylinders, and I figured I could have pressurized air entering the cylinders with air leaving the system slowly until the piston disengages the rubber band. There would be some measuring to get the required air pressure needed to have the pistons last 5 seconds, but these cylinders are too expensive and too long for this application. Then, I read the post linked at the bottom of this post, and I think the mag-locks would be great, but I have only found them in rectangular forms; none of them have the geometry needed for this, so I figured I could have the mag-lock attached to the bottom of the drone, and the mag-lock would pull pins that were attached to springs that are attached to the top of the drone; when the mag-lock deactivates, the pins would retract, allowing the rubber band to disengage. </p>
<p><a href="https://i.stack.imgur.com/l4s1m.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/l4s1m.png" alt="enter image description here"></a></p>
<p>However, I am having issues with finding timers that are suitable for my task. Kitchen timers don't work well when it comes down to setting them for 5-seconds. I want to know if you guys have any ideas regarding getting a timer that would be well suited for this task and if you guys have any suggestions as to how I could improve the design of the parachute mechanism I have planned. </p>
<p>Post mentioning mag-locks: <a href="https://engineering.stackexchange.com/questions/10288/is-there-a-device-like-this">Is there a device like this?</a></p>
| |design| | <p>Have you worked with Arduino at all? A single servo motor looped with an Arduino code could do that at a fraction of the weight. Have an input that determines current to the motors and when that input is zero, output a half-twist on the servo.</p>
| 28252 | Use 4 to 5-second timer to disengage/engage mechanical system |
2019-05-08T10:11:53.450 | <p>As I understand, a steam machine needs a pressurised gas to work. This can be <a href="https://en.wikipedia.org/wiki/Pneumatic_motor" rel="noreferrer">compressed air</a> but it used to be <a href="https://en.wikipedia.org/wiki/Steam_engine" rel="noreferrer">steam</a>. Energy was provided to the steam engine by heating up water. I don't understand why it used water as the pressurised gas could have been generated only by heating air in a closed tank.</p>
<p>Is the design of heating air in a tank instead of heating water to generate steam feasible? If so, why do steam engine work with steam instead of hot air?</p>
| |steam| | <p>It's important for mechanical engineering students to get what we old timers describe as an “intuitive feel” for their subject. Luckily such is possible in this field, but not always in other fields, such as in Quantum Mechanics. Here's my suggestion for such an intuitive feel.</p>
<p>Many of the answers give “correct” suggestions, but I don't see in any of them a way presented for the student to understand the physics that occurs as steam is giving up its internal energy in exchange for useful work.</p>
<p>Yes, it's due to the fact that latent heat is involved and that latent heat is orders of magnitude larger than sensible heat, or potential energy of compression, as you would have with using air as the working fluid. With compression of air, there's the added difficulty to hold onto all the energy put into compression. Compression increases the temperature of the air, and if the air storage tank isn't insulated, you will lose much of the compression energy in the form of heat to the environment.</p>
<p>Here's the crux. Steam is usually introduced into piston engines and turbomachinery in the supersaturated state. In this state, it contains energy in a way similar to compressed air: there's no phase change. Thus, in the first moments in its travel through the machinery, it gives up its enthalpy (the best measure of energy transferred) just like compressed air does, until it begins to condense, which will occur when its pressure and temperature allow it, as it flows through its change of state within the machinery. With condensation, latent energy is released, and that “extra” energy can be intuitively visualized as a way for the liquid-to-vapor transition to add to its volume, with a tendency to increase its pressure above the pressure that it would experience without the phase change. As the steam flow continues down to lower pressures, more latent heat is given up, effectively keeping it's pressure at higher levels than it would be without the change of phase.</p>
<p>There's a very complicated relationship among the actual pressure experienced, the actual temperature experienced, and the amount of liquid water formed. There can be no latent energy release without the formation of liquid water, and the liquid water can cause damage to turbomachinery. Thus the process is controlled and by the time the steam exits the machine, very little actual liquid water is present. It's a bit magical.</p>
<p>As some others point out here, that exiting steam is introduced to the condenser, which causes a partial vacuum in the condenser, increasing the overall pressure difference across the machine. When the steam enters the condenser, it's “quality” is relatively low. Quality is the mass fraction of vapor to liquid in a saturated fluid. Zero quality is all liquid. 100% quality is all vapor. And also as pointed out, the liquid water leaving the condenser requires much less energy to pump up to the boiler pressure, than if it were in gaseous form. That's because water is nearly incompressible, and work done is force times distance. With very little distance, there's little work done.</p>
<p>From this view, one can see very easily that latent heat is a very important aspect to the use of steam as a working fluid. It takes much energy to vaporize liquid, but you get much of that energy back in the expansion process. Latent heat can be viewed as a large bucket, enabling you to carry a large amount of energy to the workings of the machine. With air, there's no such mechanism and you can't carry nearly as much energy to the machine.</p>
| 28254 | Why use steam instead of just hot air? |
2019-05-08T14:05:43.033 | <p>I have designed a flow resistance element and I would like to be able to keep the total pressure drop across the element to < 5.0 inches of water (from 5cc/min to 5,000 cc/min). I have run some bench tests on the current design and the pressure drops are a bit too high. I would like to avoid changing the diameter of the bore and would prefer to shorten the length of the flow path. What would be the best way to calculate the ideal flow path length to achieve the pressure drops I am looking for?</p>
| |mechanical-engineering|fluid-mechanics|pressure|airflow| | <p>Use the Darcy-Weisbach equation and the Swanee-Jain correlation in order to calculate the pressure drop and the friction factor:</p>
<p><span class="math-container">$pressure drop = rho*f*(L/D)*v^2/2 $</span></p>
<p>And this formula (the first one): <a href="https://es.m.wikipedia.org/wiki/Ecuaci%C3%B3n_de_Swamee-Jain" rel="nofollow noreferrer">https://es.m.wikipedia.org/wiki/Ecuación_de_Swamee-Jain</a></p>
<p>Where rho is the density, f the friction factor, L the ideal length, D the diameter and v the speed. </p>
<p>Finally, K is the absolute roughness and are the Reynolds number. </p>
<p>If your element isn't exactly a pipe, you can model it as a pipe by using the equivalent length: </p>
<p><a href="https://www.cedengineering.com/userfiles/Pneumatic%20Conveying%20Systems.pdf" rel="nofollow noreferrer">https://www.cedengineering.com/userfiles/Pneumatic%20Conveying%20Systems.pdf</a></p>
<p>Go to page 57.</p>
| 28259 | How to determine ideal flow path length for specified pressure drop |
2019-05-08T17:27:02.043 | <p><strong>Disclaimer</strong>: I'm not a (device) engineer. I don't know the terminology. So I suppose some of my statements, questions or ideas could sound either naive or vague.</p>
<hr>
<p>As far as I undestand, the most advanced (and publicly known) omnidirectional treadmill is the Infinadeck. But with current state of the progress on this treadmill, it's clearly not ready for a mass market. One of my biggest concerns is the shape of the treadmill: why don't the Infinadeck's engineers use a disk instead of a square? </p>
<p>I was thinking about possible implementation. I'm not sure how easy it would be to implement. Because, of course, the biggest issues are often related to implementation details. I would like to hear your thoughts about it. As well as some comments on Infinadeck's shape.</p>
<p>This description isn't touching the vector generation or movement initiation, it's just about directionality of the treadmill.</p>
<hr>
<p><a href="https://i.stack.imgur.com/iu5LA.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/iu5LA.png" alt=""></a></p>
<p>Figure 1.</p>
<hr>
<p><a href="https://i.stack.imgur.com/Z3JEc.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Z3JEc.png" alt=""></a></p>
<p>Figure 2. Single engine: the sphere and the conveyor parts. This figure doesn't show the spikes, though.</p>
<hr>
<p>So the shape of the device is cylindrical. There are three major elements: shell, engines and platform. </p>
<ol>
<li>Shell is just a container for the engines. </li>
<li>Platform is a thick metal disk wrapped in a thin elastic and seamless case. So it's like in regular one-directional threadmills, but radial. The case, of course, is smooth on the inside to lower the friction. But on the outside it has numerous tiny sockets that are used for pulling the case in the desired direction. And of course it's not fixed by some kind of joints or anything. It lies on top of the engines. There also may be some supporting columns beside the engines. On the other hand it needs to be restricred by the shell, so that it stays inside.</li>
<li>There are multiple engines, spread inside the shell. Each one of them consists of the two major parts:
<ol>
<li>The sphere that translates its movement into movement of the case above it. The sphere isn't a smooth ball. It's covered with numerous spikes all around it which precisely fit the sockets of the platform's case. So when the sphere rotates, it pulls the case in the same direction.</li>
<li>The same spikes are used to rotate the sphere. There is a pair of conveyor belts(see Figure 2) that translate their movement into the sphere's. I suppose the easiest way is to make the spikes penetrate the belts. And this pair of conveyors can move around the sphere horizontally depending on the required direction but independently of the vertical movement of the belts itself.</li>
</ol></li>
</ol>
<hr>
<p><strong>UPD</strong>: The platform should be able to move in <em>any</em> direction at <em>any</em> moment (with reasonable latency, of course). And the vector is generated in real time by Neural Network watching the user of the treadmill. So the person just walks in random direction and the neural network instantly generates the vector, which is then translated into engines movement.</p>
<hr>
<p>How far this concept from reality?</p>
| |mechanical-engineering|mechanisms|mechanical| | <p>A disk shaped threadmill seems somewhat problematic from a technical perspective. Placement of actuators and moving of the surface material seems much more problematic in a disk than other shapes. Speed of actuators need to be nonuniform for one.</p>
<p>Tension on a disk seems like a unknown problem. On a disk the corners make up more of the area while in a square its uniform. Wether you can even roll such a surface at all without elevating the user quite much is a question i cant answer. In anycase a disk shaped structure needs to withstand much more two directional bending and nonuniform stretching.</p>
<p>But really many of the technical problems could be solved by adequate funding.</p>
| 28268 | Why is there still no truly omnidirectional treadmill? (technical reasons) |
2019-05-08T21:49:45.923 | <p>I am currently for the past few years creating a Brushless Motor on and off. The stator I have built has 8 Stators and I cannot seem to find out how to wire this. I see most Brushless Motors with 9 Stators. Is this possible and if so how would I be able to get this working? If you do know how to wire this can you please provide a small diagram on wiring this? Thanks!</p>
| |motors| | <p>I believe that you mean you have built a motor with eight <strong>poles</strong> on one stator. The reason that brushless motors generally have some multiple of three poles on the stator is because they are three phase. If you want to carry on with your 8-pole stator, you need to wind it as a 2- 4- or 8-phase motor -- and then you need to design and build a driver that can handle that.</p>
| 28276 | How to Wire a Brushless Motor with 8 stators? |
2019-05-09T00:35:20.313 | <p>I have a motor with a 4mm shaft. I've purchased a propeller that's designed to fit a 1/2 inch shaft. How can I securely couple the two?</p>
| |mechanical-engineering|motors| | <p>Following on from Solar Mike's answer, I'm not sure you'll be able to find something off-the-shelf with the diameters you have (mix of imperial and metric units), so you may need to get something custom made instead. I have sketched the following (not knowing what shaft lengths you had, I made a few assumptions - adjust to suit your needs):</p>
<p><a href="https://i.stack.imgur.com/63brf.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/63brf.png" alt="enter image description here"></a></p>
<p>The outer diameter is 1/2 inch, while the ID is 4mm. I have toleranced it as H7 hole so that it's a snug fit on the 4mm shaft. The adapter is secured onto the 4mm shaft with 2 M3 grub screws at 90 degrees, you can also use Loctite 638 if you want to be sure. I have added a 1mm chamfer on the end of the shaft adapter.</p>
| 28278 | How to connect a motor with a 4mm shaft to a propeller that takes 1/2 inch shaft? |
2019-05-09T07:19:14.920 | <p>Is it possible to build a brushed DC motor with a ring magnet stator, instead of a stator with gaps as shown in the GIF below?</p>
<p><a href="https://i.stack.imgur.com/0LzJx.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/0LzJx.gif" alt=""></a></p>
| |electromagnetism| | <p>Yes, but most available ring magnets will not work because they are made magnetised with opposite magnetic poles on opposite faces. This is called "axially magnetised". What you will need to find is a ring magnet with opposite magnetic poles on opposite halves of the magnet. This is called "diametrically magnetised".</p>
<p>See illustration.
<a href="https://i.stack.imgur.com/g1T2d.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/g1T2d.png" alt=""axially magnetised" compared with "diametrically magnetised"" /></a></p>
| 28287 | Can two-pole brushed DC motors have a ring magnet stator? |
2019-05-09T13:57:21.200 | <p>In an electric motor like this one, are the stator magnets radially magnetised, as in, each of the two magnets has opposite poles on the inside and outside diameter?</p>
<p><a href="https://i.stack.imgur.com/Uvr4M.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Uvr4M.jpg" alt="enter image description here"></a></p>
| |motors|electromagnetism| | <p>You should be able to find a picture on the web, but -- yes. One is north on the inside and south on the outside, the other is south on the inside and north on the outside (so pay attention if you have a pile of taken-apart motors that you're reassembling!). The steel can of the motor helps to convey the flux between the outside poles of the magnets, which increases the motor's torque -- this is why some of the larger "can" motors have an extra ring around the outer case right where the magnets are located.</p>
| 28291 | Are stator magnets in brushed DC motors radially magnetised? |
2019-05-09T18:05:33.553 | <p>I've been going down the rabbit hole as far for pipe vs tube and have ran into the same statement from co-workers and on the internet. That statement is</p>
<p>"TUBE IS STRONGER THAN PIPE"
<a href="https://www.metalsupermarkets.com/difference-between-pipe-and-tube/" rel="nofollow noreferrer">https://www.metalsupermarkets.com/difference-between-pipe-and-tube/</a></p>
<p>Which makes no sense to me, shouldn't how strong they are be a function of wall thickness and pipe ID? It seems like it's just comparing apples and oranges. Anybody got some insight on this ?</p>
| |materials|piping| | <p>Given pipe and tube of same material/strength properties, the manufacturing tolerances also matter. HSS has higher standard to meet thickness properties, whereas pipe can vary up to something like 7% of the nominal thickness. so, yes your wall thickness can be 0.45" on a 1/2" wall pipe.</p>
| 28297 | Tube Strength vs Pipe Strength |
2019-05-09T19:51:27.083 | <p>I'm currently working on a project which involves a small fridge encased in a wooden box. It's designed for hotels and the boxes vary slightly, but one common trait is that the box has an open vent towards the back which is meant for ventilation. Unfortunately, the hotels like to jam the box up against a wall, which prevents the vents from working properly. We have no control over how the wooden box is manufactured (in fact, the fridge was made to fit inside this box) and the vent hole varies in height from hotel to hotel. </p>
<p>The point is, the hotels demand that we be able to cool down the temperature inside the box, without any added electricity. I have already looked at Thermoelectric Couples and the temperature difference is about 18°C between the ambient interior of the box and the compressor which isn't enough to run a small fan. I have attached some passive heat sinks to the compressor itself, which did reduce the ambient temperature, but increased the temperature inside the fridge. I'm stuck, and I am thinking of using aluminum foil extending from the fridge to outside of the box to help conduct the heat away. Can anyone suggest any other ideas?</p>
| |thermodynamics|heat-transfer|cooling|temperature| | <p>All the refrigerators have specific clearance behind the fridge radiators.</p>
<p>Many hotels have louvered doors for the cabinet enclosing the fridge, for the exact reasons you mentioned.</p>
<p>I am not sure wrapping the aluminum foil around the radiator will help.
It will impede the convection circulation around the fridge.</p>
| 28300 | Cooling Down Fridge Encased in wooden box |
2019-05-10T17:42:50.200 | <p>I am wondering if a pneumatic-based, paddle wheel-type lifting apparatus would have any practical application within the aeronautical or aerospace fields.</p>
<p>Please reference the conceptual drawing below.</p>
<p>It is showing radial bladed impellers ('paddle wheels') embedded halfway within an enclosure. The impellers are rotated by electric motors (although they can be any kind of motor). This enclosure has lower ambient pressure within it due to the suction of the intake pipe of the attached centrifugal blower. The centrifugal blower is being rotated by its own electric motor.</p>
<p>Since the impeller blades that are outside of the enclosure encounter higher air resistance/drag as compared to the blades within the enclosure, the paddle wheels should generate a net force in the direction indicated by the arrows on the impellers.</p>
<p>Although I do not believe that this lifting apparatus would be a replacement for aircraft propellers, I believe that it may have a practical application somewhere within the aeronautical/aerospace field, yet at this time I don't know what that would be. Perhaps one application would be that this lifting apparatus could be used on a STOL/VTOL aircraft or a drone to provide vertical lift/hovering capabilities. </p>
<p>Would this pneumatic-based, paddle wheel-type lifting apparatus have any practical application within the aeronautical or aerospace fields?</p>
<p><strong>EDIT</strong></p>
<p>Edited the Title text and made wording edits to the Body text. Also, revised the original drawing and reposted it.</p>
<p><a href="https://i.stack.imgur.com/7VYF5.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/7VYF5.png" alt="enter image description here"></a></p>
| |mechanical-engineering|fluid-mechanics|design|aerospace-engineering| | <p>Simply put it will not work. And the reason is very simple. It's the fact that lift is created when the wind is almost tangent to the wing or the blade of a windmill, or else it will stall.
Not when it is pushing it. This is what made the Wright brothers machine fly as opposed to those who made flapping wings and crashed.</p>
<p>Many people who start learning sailing make the same mistake. Best wind is when it is blowing almost parallel to the sail, not when is blowing into the sail and stagnates. Best you get from 10 miles back wind is probably 7 miles boat speed, because of the friction. But same wind when it blows 90 degrees will give the sail lift at 50 miles speed.</p>
<p>Impeller has to let the air wash around it to give it lift, not to shovel the air.</p>
| 28326 | Pneumatic-based, paddle wheel-type lifting apparatus for aircrafts |
2019-05-10T22:30:23.227 | <p>I am an undergraduate researcher and I am working at enhancing a calibrating system for a hydrographic surveying system that my supervisor uses on his hydrography course. The rope is used to lower a bar check system, and it comes out of a coil. I want to find a reliable way to measure how much rope has come out of the coil.</p>
<p>Currently the rope has markings on it at every meter so that the person deploying it can know how much rope has been used by counting the number of markings that passed.</p>
<p>I was wondering if there is a more reliable way to measure how much rope has come out of the coil, preferable by some sort of measuring device. The most obvious idea would be some sort of pulley system that records the number of rotations of the pulley, but that is prone to slipping and the error could add up over 30 something meters. Do you know of such a device or idea that could help with that?</p>
<p>(p.s.: this is a very minor aspect of the whole design, and doesn't affect at all my work, it would just be a "nice to have" thing in the design)</p>
| |mechanical-engineering|measurements|mechanisms| | <p>Depends on how much accuracy and at what cost you're looking for.</p>
<p>There are several manufacturers which make devices like what you are asking.
Here is one. <a href="https://www.safewareinc.com/itemdetail/CMC%20267030?ads_cmpid=1774051606&ads_adid=75849742504&ads_matchtype=&ads_network=g&ads_creative=342995678301&utm_term=&ads_targetid=pla-737801560713&utm_campaign=&utm_source=adwords&utm_medium=ppc&ttv=2&gclid=CjwKCAjwwtTmBRBqEiwA-b6c_zisxVy5lBK-HOOeMktWypeKYksA5AintRAr3syNy4q8DthBPzd8JhoC3-gQAvD_BwE" rel="nofollow noreferrer">Cord meter</a></p>
<p>Also you can use a basic computer optical mouse and an LED counter attached to an Arduino sensor for around $20-30. </p>
| 28330 | Whats a reliable way to measure the amount of rope going out of a coil? |
2019-05-12T08:55:46.720 | <p>I am learning the method of joints, but there's no explanation on how to figure out the direction of internal forces in my course.</p>
<p>How do you know that BA is tension and BC is compression?</p>
<p><a href="https://i.stack.imgur.com/aCKFg.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/aCKFg.png" alt="enter image description here"></a></p>
| |mechanical-engineering|structural-engineering|civil-engineering|structural-analysis| | <p>There are a few ways of doing it.</p>
<h2>Just math it out</h2>
<p>This basically means doing all the calculations paying attention to your sign convention and seeing what the final result is. In complex cases, with a large truss dealing with many loads in multiple directions, this is the only real option.</p>
<p>For a simple truss such as this one, though, there are two easier ways of doing it via inspection:</p>
<h2>So flexible!</h2>
<p>This method is basically just looking at the deformed configuration under load. Feel free to exaggerate as much as you need.</p>
<p>In this case, you have a horizontal load at node B. This force is going to try to move B rightward. Let it. In fact, lets pretend that force moves B a full meter to the right (remember, exaggerate!), such that the truss becomes an <a href="https://en.wikipedia.org/wiki/Isosceles_triangle" rel="nofollow noreferrer">isosceles triangle</a>.<sup>1</sup> Now consider what that does to beams AB and BC:</p>
<ul>
<li>AB went from being 2 m long to <span class="math-container">$\sqrt{2^2 + 1^2} = 2.2\text{ m}$</span> long.</li>
<li>BC went from being <span class="math-container">$\sqrt{2^2 + 2^2} = 2.8\text{ m}$</span> long to <span class="math-container">$\sqrt{2^2 + 1^2} = 2.2\text{ m}$</span> long.</li>
</ul>
<p>The numbers themselves are irrelevant (after all, they're based on a made-up displacement of B). All that matters is that AB got longer and BC got shorter. No matter what the real displacement of B is, this will always be the case: AB gets slightly longer, BC gets shorter. Which tells us that AB is under tension and BC, under compression.</p>
<p>That's the basic gist of this method: look at the expected deformed configuration and see which elements get stretched (tension) and which get squeezed (compression).</p>
<h2>Node equilibrium</h2>
<p>The last way is to look at the individual nodes. The equilibrium diagram in your question (just the arrows, ignoring the values) is a cheat-sheet: all arrows pointing towards the node (such as the one representing BC) are compression, those pointing away (such as AB) are tension.</p>
<p>The way to see this is to think in terms of <a href="https://en.wikipedia.org/wiki/Newton%27s_laws_of_motion#Newton's_third_law" rel="nofollow noreferrer">Newton's Third Law</a>: the node is applying a force on the beam, and the beam is applying an equal and opposite one on the node.</p>
<p>If a beam is under compression, the external force it feels is pointing inwards (think of a column with a downwards force above and an upwards reaction force below). Therefore, the column's reaction points outwards from the column... which is "towards" the node. </p>
<p>Likewise, a beam under tension feels outward-facing external forces, so its reaction points towards the column, and away from the node.</p>
<p>Therefore you just need to figure out the direction for each of the arrows in the node diagram.</p>
<p>In this case, you have a rightwards external force at B. Only BC has a horizontal component, so it has to be solely responsible for countering that force. That happens by making BC's horizontal component point towards the left. That being so, BC's vertical component must point up, since BC's resultant force must be purely axial and therefore parallel to BC's inclination. Since that resultant will point left-and-up, it's pointing towards the node and is therefore compression.</p>
<p>Given that, the node now has an upwards force which must be countered by AB, which must point down. Since that makes it point away from the node, it must be tension.</p>
<hr>
<p><sub><sup>1</sup> In reality, there'd also be some vertical movement, but it's much smaller than the horizontal and can be safely ignored for this thought experiment.</sub></p>
| 28343 | Determining the direction of internal forces in a truss |
2019-05-12T11:20:24.540 | <p>Given <a href="https://engineering.stackexchange.com/a/28258/2817">this answer</a>, I understand the importance of phase change (liquid --> gas) for the fluid used in a steam engine. Thus I understand it is important to use a fluid that is liquid at pressure and temperature this engine may work in (between 600 and 1000hPa, between -20°C and 50°C (I may take large margin for extreme climate condition including high altitude mountains)).
Thus, we may use any fluid that is liquid in this condition and easy to boil (I'm thinking of ethanol that has a lower boiling point but other fluids may be suited).</p>
<p>My question is: why water and not other fluids such as ethanol?</p>
<p><strong>Edit</strong>: I cite ethanol but I don't want the answer to be focused on this fluid in particular. It can be any fluid that is liquid in normal temperature and pressure condition and easy to boil. I have no idea what fluid can meet this requirement.</p>
| |steam| | <p>Water has a very high specific heat and heat of vaporization, so it takes more energy to raise it's temperature and to boil it than almost any other substance.</p>
<p>In the same way that it takes more energy to melt aluminum than copper, even though aluminum has a lower melting point than copper because aluminum has a higher specific heat and heat of fusion than copper.</p>
<p>Also, other substances could have a better vapor pressure curve than water, which means there could be a greater increase in pressure for the same increase in temperature.</p>
<p>If there was an advantage to using another substance than water it would have to be enough to still be advantageous even after the expenditure of the energy required to recover and recycle the expansion medium, also the engine would have to be stationary, as in an electrical power generation plant where an increase in efficiency of a few percent would be enough to justify implementing the modifications.</p>
<p>So, preferably we are looking for a slightly polar molecule with a low molecular weight and preferably nonflammable.</p>
<p>Also, it might be possible to use lower boiling liquids under higher pressure because there could be a very good vapor pressure curve with a large increase in pressure for the same increase in temperature.</p>
| 28344 | Why use steam water and not other fuild in steam engines? |
2019-05-13T18:33:49.793 | <p><a href="https://i.stack.imgur.com/Fk3zH.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Fk3zH.png" alt="enter image description here"></a></p>
<p>Where does the T-beam come from? There's just a straight beam resting on two supports. And where do 10 and 80 come from? I am so confused.</p>
| |mechanical-engineering|structural-engineering|civil-engineering|structural-analysis| | <p>To find the neutral axis of any section including the T section in your question, ( called T beam because its cross section is a T):</p>
<ul>
<li><p>one picks a horizontal axis parallel to x axis (or the axis they need to find the beam's neutral axis).</p></li>
<li><p>break the section into simple geometric shapes.</p></li>
<li><p>find the area moment of each section by multiplying its area by the distance from its CG to the axis picked at the first step. (in your case they assumed the top surface's, Line Z, as that random reference axis, and 10 is the distances of the horizontal 80 by 20 flange and 60+20 is the vertical web CG from the top axis)</p></li>
<li><p>sum all the area moments and dived by total area of all the part, you get the neutral axis, which they did and got 52 mm from the top surface as the neutral axis of the T beam.</p></li>
</ul>
| 28359 | Maximum tensile and compressive stress in a T-beam |
2019-05-14T19:27:41.663 | <p>So there is a lot of information out there about designing an external Geneva drive, but I can't seem to find anything about designing an internal one. I'm trying to design something similar to this (but with a different number of positions):</p>
<p><a href="https://i.stack.imgur.com/WQu5K.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/WQu5K.png" alt="Internal Geneva Drive"></a></p>
<p>I know the internal drive is based off a calculations from a triangle, and I'd assume the the internal one is too, but it's just not clicking for me. Can anyone point me in the right direction? Thanks!</p>
| |gears| | <p>So i finally stumbled upon something last night that had had some good tables and figures. See the link below</p>
<p><a href="http://uotechnology.edu.iq/tec_magaz/2016/volum342016/No.09.A.2016/[11Text.pdf" rel="nofollow noreferrer">Kinematics Characteristics of The Internal & External Geneva
Mechanism Under The Effect of Multi Design Parameters</a></p>
<p>Specifically it had this image which helped to visualize things better. Turns out it is using the same triangle that the external gear uses for figuring out the center point of the driver and driven gears.</p>
<p><a href="https://i.stack.imgur.com/Tm6kR.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Tm6kR.png" alt="enter image description here"></a></p>
| 28379 | Internal Geneva drive calulations/design |
2019-05-15T15:19:44.337 | <p>I have an application wherein I need a hermetic seal at cable exit points from a plastic housing. The cables are not removable. I'd like to achieve this seal using the plastic housing itself by creating a friction fit between the hole and PVC cable jacket (since there's only about 5mm of vertical space to work with). A couple questions:</p>
<ul>
<li><p>Is this possible? How can one prevent the seal from breaking as the cable is moved into different positions?</p></li>
<li><p>What are the best practices for such designs? </p></li>
</ul>
<p>Thanks for your help!</p>
| |product-engineering|waterproofing|seals| | <p>The solution you have suggested will not be waterproof for long. The reason is that, if you pull on the cable it will stretch and thin very slightly. This is enough to break the seal.</p>
<p>If you can make the space, a cable gland is the best option.</p>
<p>If you can't, and O-ring seal is probably the next best thing. Make sure to use a thick o-ring with some compression so the seal will be maintained even if the cable stretches slightly.</p>
<p>Another option is to use a rubber "plug" where the cable passes through the center. This can be water tight, but you will have to take great care to make sure the plug is retained and remains under compression with regards to the seal.</p>
<p>You can do a similar thing using potting compound, but it's not trivial. A rigid epoxy will have the same issues as the press fit you have suggested, so you need something semi-flexible. I don't know enough to recommend a product unfortunately.</p>
| 28389 | Hermetic Cable Seal With Space Limitations |
2019-05-15T17:14:36.203 | <p>Imagine a cantilevered beam, fixed on one end and with a tension line connecting the other end to ground (e.g. perpendicular to the length of the beam).</p>
<p>If I increase the tension, does the modal frequencies of the beam change?</p>
<ul>
<li>On the one hand, tension affects the frequency of a string. So, would pulling on the beam (in tension) cause it to have a higher frequency?</li>
<li>On the other hand, the bending stiffness of a beam is not affected by a lateral force. So, if the stiffness is not affected, then neither should the frequency.</li>
</ul>
<p>What can help me reconcile these two comments?</p>
<p>Bonus aspects of the question:</p>
<ol>
<li>If the stress is high enough to affect the slope of the stress-strain curve, then the stiffness changes because the modulus changes.</li>
<li>In Nastran, you cannot apply a load while performing an eigenmode / eigenvalue normal modes analysis. Does that mean it is not a linear phenomena?</li>
</ol>
| |beam|vibration|frequency-response|modal-analysis|nastran| | <p>The statement of the question, using "internal/external stresses" is confusing. I don't understand what "external stress" can be here. You can have external forces but not external stresses. From the question elaboration, it appears to ask whether the internal state of stress affects vibration modes, yet the questioner's example of a "tension line" cannot change the internal state of static stress without altering the boundary conditions on the beam. Altering the boundary conditions will certainly affect vibration modes. In addition, an "internal state of stress" always exists in a material, whether undergoing vibration or not. It seems that the questioner is asking whether the "static" state of stress affects the modes, and static stress are internal stresses of motionless material. I suggest the questioner clear up these issues, and I'll give the answer to the simple question which I believe he really wants to ask: "Does the static internal state of stress affect vibration modes of a beam"? </p>
<p>The answer to that question is No, as long as the elastic limit of stress in the material is not exceeded. A different static state of stress can result in total stresses during vibration that exceed the elastic limit, which may affect the vibration modes. But the answer to that would be too complicated to discuss here. For purposes here, the simple answer is "No."</p>
<p>Explanation is that, first of all, vibration of a taught string is a different phenomenon, where the restoring (spring) force is the tension itself. With the beam, called a "bar" in vibration science, the restoring force is the elastic force, and even more, it's the change of elastic force, not the total (change + static), caused by the vibration. Static forces don't enter into the equations of motion. If you change the static situation, the dynamic changes of internal stress only occur about a changed mean value, and as long as the response is elastic (obeys Hook's law), the dynamic solution will not change. </p>
<p>In fact, we know very well that the static state of internal stress normally has no effect on vibration modes Most all hardened metals have some nonzero internal static stresses, and we are not concerned with them when we calculate vibration modes in such materials. </p>
<p>For instance, with 1095 Swedish spring steel, there can be residual internal stresses in high temperature quenching during the hardening process that are not eliminated by sufficient uniform annealing (tempering), and that doesn't affect the vibration of a cantilever made of such material. Such a device is used as sound source in free reed musical instruments, and makers aren't directly concerned with the internal state of stress in the reeds they make. Now, there are complications when you consider other practical effects, such as metal fatigue and breakage. Fatigue can occur over many cycles when stress levels are even below the elastic limit, a static concept. There is also an "endurance limit," which is below the elastic limit, above which fatigue lifetime is greatly reduced.</p>
| 28391 | Do internal / external stresses affect modal frequency? |
2019-05-16T08:54:41.140 | <p>I'm designing a control system for a balloon-borne telescope experiment. As the balloon is constantly moving and the gondola is rotating, I want to stabilize the telescope which can be controlled in 3 axis of rotation. I have encoders for each axis motors and gyroscope located on the electronics box around 20cm apart from the telescope's center of rotation.</p>
<p>Is there a possibility of performing a sensor fusion, like Kalman filter, for these two groups of sensors?</p>
<p>I have a prepared model of the control loop, but have then realized that the sensors are in different places and while gyroscopes measure gondola movement, the encoders measure the telescope rotation.</p>
<p><a href="https://i.stack.imgur.com/e9zNg.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/e9zNg.png" alt="enter image description here"></a></p>
| |control-engineering|control-theory|matlab| | <p>I think you're getting confused because you're trying to combine the gyroscopes and encoders into one data source.</p>
<p>An easier way would be to calculate an angular deflection in each axis using your gyros. Now lets call wherever the telescope is pointing "zero". The gyroscope provides a target angle that is necessary to maintain "zero", and the motors and encoders will use PID to try to stay at that angle.</p>
<p>Also, the stars are far away, so you can probably just pretend your gyro and telescope exist at a single point.</p>
| 28396 | Stabilization of telescope on gimbal using data from encoders and gyroscope |
2019-05-16T17:17:40.910 | <p>I'm looking for a lightweight material that will hold its shape when stretched out. Something like a rubber band that doesn't snap back. So it starts out small, but once it's stretched it stays large. (Apologies if there's a name for such a property that I don't know.)</p>
<p>Thanks!</p>
| |materials| | <p><strong>Ductile and Plastic Material for Making Parts? Suggest you try Polycaprolactone.</strong></p>
<p>If you're looking for a highly ductile/purely plastic material to form into a shape -- one that's easy to work with at home, you might try: <a href="https://en.wikipedia.org/wiki/Polycaprolactone#Hobbyist_and_prototyping" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Polycaprolactone#Hobbyist_and_prototyping</a>
granules, after warming in hot water. There are a host of brand-names all selling essentially the same granules: Polymorph, Thermomorph, and Friendly Plastic are just a few. Quite safe; medicos use this stuff to splint fingers sometimes.</p>
| 28401 | Material that holds shape when stretched |
2019-05-16T22:06:10.890 | <p>As I've progress through my undergraduate engineering program, I've noticed a shift away from physical intuition - understanding why certain phenomena or equations make sense because of basic patterns we are already familiar with - to mathematical intuition - showing why equations and their phenomena should occur by mathematical derivation.</p>
<p>An example of this I came across today in aerodynamics:</p>
<p>For Isentropic Flow: <span class="math-container">$\frac{p_0}{p} = (1 + \frac{\gamma-1}{2}M^2)^{\gamma/(\gamma-1)}$</span></p>
<p>I can usually follow the derivation of the equation, identify the variables in it, and apply it to appropriate cases. But, I cannot describe why this equation is justified by simply looking at it - I must follow the mathematical derivation. There is no way I'd be able to describe this equation to a layperson in a way that could satisfy them. I can't come up with an explanation that satisfies myself; I want to know why <span class="math-container">$1+$</span> is there, why <span class="math-container">$\gamma- 1$</span> should be divided by <span class="math-container">$2$</span>, and why the Mach number should be squared, without solely relying on the mathematical derivation. I have this problem with many formulas I have learned.</p>
<p>I know a fundamental part of engineering is understanding the equations and phenomena we use to to describe physical systems. Now, it seems like I should just trust the equations and make sure I'm not using them improperly.</p>
<p>Should I, as an engineer-in-training hoping to complete research, focus on trying to understand equations to my satisfaction, or should I instead just become well acquainted the equations, their use cases, and their general behavior?</p>
<p>Note: I acknowledge questions on this site should have answers and not just be discussed. I hope this question has an answer that has some consensus among experienced engineers. </p>
| |aerospace-engineering|mathematics|aerodynamics| | <blockquote>
<p>But, I cannot describe why this equation is justified by simply looking at it - I must follow the mathematical derivation. There is no way I'd be able to describe this equation to a layperson in a way that could satisfy them.</p>
</blockquote>
<p>Every student starts with equations in their math classes, at school. With a thorough understanding of the various interrelated concepts in math, she is able to explain these equations to laymen. By the time we start college, we have the ability to explain why some variables are present and why it is present specifically in that exact form (atleast for the concepts we have been very familiar with).</p>
<p>But can we do that outside math? Is it possible in engineeering? I have viewed a lot of videos where a very comprehensive understanding of concepts have been presented, which can be presented to a general audience. But now i (have come to) believe that some equations are only empirical. Intuition can be developed, but i am not sure as to what extend.</p>
<p>Maybe simulations at the micro-scale so that our analysis becomes perfect? Quantum mechanics, molecular/DFT simulations could provide more insight.</p>
<p>P.S. I am not an experienced engineer, but an engineering student. I have gone through a futile search to explain some equations i have learned. It involves a lot more work. The deeper i go into theory, the point is lost, because an engineer provides real-life solutions, and i feel going deeper into theory is science rather than engineering; are we supposed to do that as engineering students (i await an answer to that). Developing such intuition has been easier for me in math, rather than engineering.</p>
<p>This is not an answer to the question, this is more intended as a comment, but i do not have the ability to comment at the moment. Perhaps people with similar experience could share their learning here.</p>
| 28404 | Should Engineers Understand Equations without Relying on Derivations? |
2019-05-17T17:13:47.563 | <p>I have a second order system and I need to control it using a PI controller. I need to tune the gains of the controller in order for the system to satisfy the below specifications:</p>
<p><span class="math-container">$$ OS\% < 10\% \rightarrow Overshoot $$</span>
<span class="math-container">$$ t_r < 1.2 \ seconds \rightarrow Rise \ Time $$</span></p>
<p>I have derived the third order transfer function of the closed loop system with the controller and I am not able to understand which characteristic polynomial I have to use in order to achieve the specified requirements.</p>
<p><strong>EDIT:</strong> </p>
<p>Transfer function of the plant is: </p>
<p><span class="math-container">$$ G(s) = \frac{10}{(s+1)(s+9)} $$</span> </p>
<p>Transfer function of PI controller is: </p>
<p><span class="math-container">$$ C(s)=\frac{K_p(s+c)}{s} $$</span> where <span class="math-container">$\ c = \frac{K_i}{K_p} $</span> </p>
<p>The closed loop transfer function (linear unity feedback) is:</p>
<p><span class="math-container">$$ T(s) = \frac{10K_p(s+c)}{s(s+1)(s+9) + 10K_p(s+c)} $$</span></p>
<p>So now how should I proceed in order to compute the gains <span class="math-container">$\ K_p $</span> and <span class="math-container">$\ K_i $</span> of the controller in order to satisfy the given specifications ? </p>
| |control-engineering|control-theory|pid-control|transfer-function| | <p><strong>Step 1:</strong> Draw the root locus of the system.</p>
<p><a href="https://i.stack.imgur.com/lCrwe.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/lCrwe.png" alt="Root locus of the system"></a>
<em>Here you can see the two poles of your plant <span class="math-container">$G(s)$</span> (marked with an x), at <span class="math-container">$p_1=-9$</span> and <span class="math-container">$p_2=-1$</span>, the pole of your controller <span class="math-container">$C(s)$</span> at <span class="math-container">$p_c = 0$</span> and the zero (marked with an o) at <span class="math-container">$z_c = -c$</span> (for now just at a random location).
The purple squares indicate the poles of the closed loop system for a certain gain <span class="math-container">$K_p$</span> and can only move over the blue lines (the root locus).</em></p>
<p><strong>Step 2:</strong> Add the design requirements to your root locus.</p>
<ul>
<li><p><strong>Less than 10% overshoot</strong> can be indicated by adding diagonal lines through the origin with a slope of <span class="math-container">$\frac{\Re}{\Im} = \pm\left|\frac{\ln(10\%)}{\pi}\right|\approx \pm 0.73$</span>, where <span class="math-container">$\Re$</span> and <span class="math-container">$\Im$</span> represents the real and imaginary axis respectively. All the closed loop poles of your system have to be between these two lines in the left open half plane.</p></li>
<li><p><strong>Rise time less than 1.2 seconds</strong> does not have an analytic expression so cannot be translated directly to the root locus. My suggestion is to rewrite this requirement as a more conservative requirement on the settling time, such that the system is at least at 10% of it's final value within 1.2 seconds. The settling time requirement can be represented as a vertical line at <span class="math-container">$\Re = \frac{-\ln(0,1)}{t_s(10\%)} \approx -1.9$</span>.</p></li>
</ul>
<p><a href="https://i.stack.imgur.com/7xdKN.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/7xdKN.png" alt="Root locus with design requirements"></a>
Here you can see the design requirements, the diagonal lines indicate the overshoot at 10% and the vertical line indicates the settling time of 1.2 seconds. To fulfill all the design requirements, all the poles of the closed loop system (purple squares) have to be in the white area. Hence for the current configuration were do not satisfy the design requirements.</p>
<p><strong>Step 3:</strong> is to tune the system. Since the purple block (pole) at the right cannot pass the zero and thus not fulfill the design requirements, we start with replacing the zero (e.g. at -1.5, thus <span class="math-container">$c = 1.5$</span>). This will change the root locus significantly.
<a href="https://i.stack.imgur.com/p6H7Y.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/p6H7Y.png" alt="Root locus with replaced integrator zero."></a></p>
<p>Now we can start changing the gain <span class="math-container">$K_p$</span> until all blocks are in the white area (e.g. <span class="math-container">$K_p = 2.4$</span>)
<a href="https://i.stack.imgur.com/PEYjo.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/PEYjo.png" alt="Root locus with updated gain"></a></p>
<p>Hence, the controller <span class="math-container">$C(s) = \frac{2.4(s+1.5)}{s}$</span>.</p>
<p><strong>Step 4:</strong> Check if the designed controller fulfills the design requirements by taking the step response.
<a href="https://i.stack.imgur.com/47DsO.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/47DsO.png" alt="Step response of closed-loop system"></a></p>
<p>It can be seen that the overshoot <span class="math-container">$D = 8\%$</span> and the rise time <span class="math-container">$T_r = 0.593 -0.106 = 0.487$</span> seconds thus all requirements are fulfilled with the current controller.</p>
| 28421 | PI controller for second order system |
2019-05-18T14:38:13.463 | <p>I'm wondering how to compare the efficiencies of vertical and horizontal ground loops for heating. [Assumptions follow, please correct if any are wrong!]</p>
<p>The biggest differences seem to be the higher cost of vertical loops vs the land required for horizontal loops, so therefore I'm having trouble finding any information on how to compare the energy input/output of similarly sized systems.</p>
<p>I'm interested comparing the two systems, so at the risk of getting off the question: in my area, average winter temp is a low of -6F/-20C, and a high of 12F/-11C, with cold snaps of -40deg. Frost gets to 70in/ 1.7m in the colder winters. If comparisons want to use that info, I would be very curious. </p>
<p>Thanks! </p>
| |mechanical-engineering|heat-transfer| | <p>Geothermal heating requires detailed gathering of several kind of data for planning and then you can judge which system is better for your particular case.</p>
<p>Some of Those factors are</p>
<ul>
<li><p>Outdoor design temperature (coldest night of the winter usually published by the building authority), </p></li>
<li><p>Soils and geology data (frost line and aggregate of the layers of soil at 10-20- 50 feet) </p></li>
<li><p>Tonnage of the heat pump, related to the size of your home and its thermal leakiness, etc.</p></li>
<li><p>The availability of land.</p></li>
</ul>
<p>Vertical ground loops cost more but don't need a large area of land so they lend themselves more to urban areas, they may need high pressure plumbing and bigger pumps. </p>
<p>Here are schematic diagrams of , horizontal loop
<a href="https://i.stack.imgur.com/vE1Rk.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vE1Rk.jpg" alt="horizontal loop"></a> </p>
<p>and vertical loop, --- <a href="https://i.stack.imgur.com/RRvPO.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/RRvPO.jpg" alt="vertical loop"></a></p>
<p>There are courses that teach how to estimate and design an efficient systems, International Ground Source Heat Pump Association has a course called, IGSHPA training. Here is the <a href="https://igshpa.org/training/" rel="nofollow noreferrer">course</a></p>
<p>There are people who have installed their system after taking similar courses, but one needs to check with the manufacturer of the pumps to see what kind of a warranty restriction the may have on installer qualification.</p>
| 28438 | Vertical vs Horizontal Ground Loop Heating |
2019-05-19T16:35:29.067 | <p>how to know if a deflection in a supersonic isentropic flow causes compression (oblique shock) or expansion (Prandtl-Meyer expansion)
Thanks in advance.</p>
<p><a href="https://www.chegg.com/homework-help/air-flows-supersonically-surface-changes-direction-twice-fig-chapter-9-problem-146p-solution-9780073398273-exc" rel="nofollow noreferrer">an example of a problem that is related to my problem</a></p>
<p><a href="https://i.stack.imgur.com/wR1my.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/wR1my.jpg" alt="enter image description here"></a></p>
| |fluid-mechanics| | <p>Really easy: Calculate the deflection angle beta. If the angle is positive, then it's an oblique shockwave, but if it's negative, then you have an expansion. If the angle beta is zero, then there's neither compression nor compression and if beta is 90º (π/2) you have a normalmente shockwave. </p>
<p>An expression that may help you (I did a report on this and demonstrated the formulæ):</p>
<p><a href="https://i.stack.imgur.com/3O8G2.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/3O8G2.png" alt="enter image description here"></a></p>
| 28446 | Supersonic isentropic flow |
2019-05-19T23:59:55.497 | <p>This <a href="http://wentec.com/unipower/calculators/power_torque.asp" rel="nofollow noreferrer">link</a> shows a motor torque, power, and speed relationship, as shown below: </p>
<p><a href="https://i.stack.imgur.com/FbMrJ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/FbMrJ.png" alt="enter image description here"></a></p>
<p>With regards to its units, how does kW and RPM become equivalent to N.m?</p>
| |motors|torque|power| | <blockquote>
<p>With regards to its units, how does kW and RPM become equivalent to N.m?</p>
</blockquote>
<p>Power = force * speed. When you're dealing with rotary motion, power = torque * rotation rate. Their calculation of torque from HP and RPM is just a matter of getting the units consistent, and solving for torque (noting that for the power as a function of torque and rotation rate, the rotation rate must be in radians/second).</p>
| 28452 | Motor torque calculation, about the units |
2019-05-20T02:22:28.450 | <p>What's an authoritative definition of whether the idealized pinned support and pin joint can transfer moments in 3D?</p>
<p>Most sources say both of them cannot transfer moments but are apparently only considering 2D. A real pin joint obviously can transfer moments about all axes normal to the pin's axis. So is an implementation of the ideal pin joint really more like a ball joint than a pin joint?</p>
<p>I assume a pinned support is the same as a pin joint but with one end fixed to ground. But maybe they have different numbers of degrees of rotational freedom?</p>
<p>For example, if this support can carry moments about two perpendicular axes (red arrows), would it be correctly called a pinned support?</p>
<p><a href="https://i.stack.imgur.com/UlygO.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/UlygO.png" alt="Pinned support"></a></p>
<p>In comparison, this one has all 3 rotational degrees of freedom, so is it still a pinned support?</p>
<p><a href="https://i.stack.imgur.com/jrNHn.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/jrNHn.jpg" alt="enter image description here"></a></p>
| |mechanical-engineering|structures|connections| | <p>Yes it can, and in many designs it should be a pin support along one axis and fixed support along other. In many trusses and bridges that is the case.</p>
<p>The design software when defining a joint have the joint degrees of freedom choices. Among those are the option to define the joint a hinge along one axis but fixed along the others.</p>
<p>You can even have joints or supports that can have, rigid, pin, or spring or even predefined varying stiffness and ductility restriction on rotation, settlement, lateral displacement, moment transfer. This is becoming more advantageous in seismic design. Where one expects different behavior of a structure under different spectra of earthquake.</p>
<p>Between the two extremes of a pin joint and fixed join one can define the semi fixed joint or fixed but after a certain stress behaving like a hinge. </p>
<p>In real construction of a joint there is always joints that are legally considered pin or fixed but they behave a certain mix of both, but you design them depending on what the code defines that type ofjoint. </p>
| 28457 | Can pinned support / pin joint transfer moments in 3D? |
2019-05-20T07:10:31.147 | <h3>TL;DR</h3>
<ul>
<li>Is it possible to calculate the mean diameter of a collar?</li>
<li>Can I ignore the collar's component when calculating the raising torque?</li>
</ul>
<hr />
<p>I have been trying to <a href="https://3dprinting.stackexchange.com/questions/7695/z-axis-stepper-motor-and-lead-screw-torque-calculation/9827#9827">answer</a> this question, <a href="https://3dprinting.stackexchange.com/questions/7695/z-axis-stepper-motor-and-lead-screw-torque-calculation">Z axis: Stepper motor and lead screw torque calculation</a>, over on the <a href="https://3dprinting.stackexchange.com">3D Printing</a> site.</p>
<p>I obtained an equation for the raising torque from
<a href="https://ncalculators.com/mechanical/torque-required-raise-load-calculator.htm" rel="nofollow noreferrer">Torque Required to Raise Load (TR) Calculation</a> and the <em>full</em> <a href="https://i.stack.imgur.com/3G23E.jpg" rel="nofollow noreferrer">equation</a>, for <em>both</em> thread and collar, is given by:</p>
<p><span class="math-container">$$T_R = \frac{Fd_m}{2}\left(\frac{l+\pi\mu{d_m}}{\pi{d_m}-\mu{l}}\right)+\left(\frac{F\mu_cd_c}{2}\right)$$</span></p>
<p>Neither of the mean diameters for the square thread, nor the collar were given.</p>
<p>However, I found that the mean diameter of the square thread can be determined from the pitch, <span class="math-container">$l$</span>, and nominal diameter, <span class="math-container">$d_n$</span>, by</p>
<p><span class="math-container">$$d_m = d_n-\frac{l}{2}$$</span></p>
<p>However, I do not know how to calculate the mean diameter of the collar. I was hoping that it would simply be</p>
<p><span class="math-container">$$d_c = d_n+\frac{l}{2}$$</span></p>
<p>but, I think that might be incorrect, although I am not sure - as this would seem to be the formula for calculating the major diameter of the screw, when given its mean diameter.</p>
<p>Looking at a <a href="http://users.encs.concordia.ca/%7Enrskumar/Index_files/Mech344/Solutions/Problem%20Set%206-Chapter%2010_Power%20Screws%20and%20Fasteners-Selected%20Problems.pdf" rel="nofollow noreferrer">series of tutorials</a>, I see that the mean diameter of the collar may need to be provided. For example, from Solution 10.3</p>
<p><a href="https://i.stack.imgur.com/uS0Wy.png" rel="nofollow noreferrer" title="Diagram of load, screw and collar"><img src="https://i.stack.imgur.com/uS0Wy.png" alt="Diagram of load, screw and collar" title="Diagram of load, screw and collar" /></a></p>
<p>After reading around the topic, I see that often the component relating to the collar's friction is discarded (or, as the tutorials state, considered to be eliminated). Why is that? Is that a common simplification? The OP had this simplified version of the above formula in their question,</p>
<p><span class="math-container">$$T_R = \frac{Fd_m}{2}\left(\frac{l+\pi\mu{d_m}}{\pi{d_m}-\mu{l}}\right)$$</span></p>
<p>So, do I actually need to worry about the collar's mean diameter?</p>
| |torque| | <p>It's not possible to calculate the collar diameter from the thread dimensions. The collar can be made any size greater than the thread diameter. The bigger it is, the bigger the torque needed to overcome its friction, so the diameter can be important. If you can show that collar friction is negligible, then discard it. Perhaps if it's on roller bearings and the thread isn't, then you can do a rough estimate of the friction to show that it's negligible for any reasonable collar diameter.</p>
| 28460 | Mean diameter of the collar when calculating the raising torque |
2019-05-23T08:06:33.983 | <p>I made a simulation of this cylinder that is being suspended mid air and held up by 8 rods, 4 on top and 4 on the bottom. The entire model is cooled to 15K</p>
<p>I managed to find the forces required for 6 of the rods to have a safety factor of above 1 as pictured below (green being above 1):
<a href="https://i.stack.imgur.com/gWVGL.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/gWVGL.png" alt="enter image description here"></a></p>
<p>The two rods in red have a safety factor of less than 1. All rods have a parametrised force so I would like to know how to create a design study to get all of the rods to be above 1 or at least all have the same safety factor under 1. Basically, I am only interested in changing the forces that are applied to the rods.</p>
<p>I've tried an Adaptive Multiple Objective with the goals of all rods having a safety factor above 1 but it didn't produce any meaningful results.</p>
| |mechanical-engineering|ansys|optimal-control|thermal-expansion| | <p>Found out about Response Surfaces and using all the points that I already created, created a Response Surface Optimization which let me easily find the answer without a significant amount more computation.</p>
| 28505 | Direct Optimisation - How to Create Time Efficient Study |
2019-05-24T04:53:05.207 | <p>I have a flat roof (dead load = 20 psf, live load = 30 psf) spanning 11 feet with a 2 ft. overhang. I'm trying to calculate the total load in plf on the header marked as X.
<a href="https://i.stack.imgur.com/FFKdT.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/FFKdT.png" alt="building profile"></a></p>
<p>Is this calculation correct:
TL = (DL + LL) × (11/2 + 2)
= (20 + 30) × 7.5
= 375 plf</p>
| |structural-engineering| | <p>No it is not correct.</p>
<p>If you just want to calculate the load on the header with service loads and no load factors, you find the load on the header by getting the moments about an arbitrary point.</p>
<p>In this case the easy choice is the point at the left end of the rafters. let's denote it A and the point right on top of the wall with the header B.</p>
<p><span class="math-container">$$ \Sigma M_{A}=0 \quad M_{A} = w*l*l/2 -R_{B}*11\ ft=0 \\ (20+30)*(11+2)*13/2 =R_{B}*11 $$</span></p>
<p><span class="math-container">$$ R_{B}= 4225\ lbs/11 = 384,09\ lbs.ft $$</span></p>
<p>So each lineal foot of the header is supporting 384.09 lbs.</p>
| 28513 | Window header sizing: total load on header |
2019-05-26T17:18:26.667 | <p>I wonder if there exist a formal guideline to help engineers choosing the right type of bearing. </p>
<p>Except in some few cases, this is not very evident to me.
As far as i know, the plain bearing systems are really simple, so why bothering to design the rolling element bearings? I guess there is a reason, but i don't see it. </p>
<p>I've done some research on the web, besides the economic factors and some other vague statements i couldn't find anything useful. </p>
| |mechanical-engineering|bearings| | <p>To add to the list of criteria in other answers:</p>
<ul>
<li>Cooling. If a bearing supporting a rotor transmitting say 10MW of power dissipates only 0.01% of that power in friction, that is still 1kW of heat being generated in the bearing. Rolling element bearings are relatively easy to cool via the lubricating oil since there an unobstructed flow path between the rolling elements. Sliding bearings are not.</li>
</ul>
<p>At the other end of the design spectrum:</p>
<ul>
<li>Size. Manufacturing a 2mm diameter journal bearing can be as trivial as drilling a 2mm diameter hole in something. Manufacturing a 2mm diameter rolling element bearing is slightly more difficult - though it has been done: <a href="https://www.nskmicro.co.jp/english/products/products04.html" rel="nofollow noreferrer">https://www.nskmicro.co.jp/english/products/products04.html</a></li>
</ul>
| 28531 | How to make a choice between rolling element bearing and journal (plain) bearing? |
2019-05-27T11:34:01.003 | <p>I was passing by a row of motorbikes and mopeds yesterday, and noticed they all have really small radius wheels compared to bicycles. Not all motorbikes are equal (some motorbikes are literally bicycles with engines), but generally it seems that motorbikes have smaller wheels.</p>
<p>I assume both vehicle types have been optimised and it's not just a fashion choice on the part of the manufacturers. What are they optimising for which makes them different? And why do mopeds/scooters have the broadest range of wheel radiuses when they're the lowest power type of motorbike?</p>
| |automotive-engineering|bicycles| | <p>It's all to do with the torque output and rpm of the "motor". A fit cyclist puts out about the same torque as a schoolbus diesel engine. But they do so at much lower rpm and therefor much less power. So a bicycle needs to gear the 80 rpm source up by a factor of 5 or so to hit 40 mph. Five is about as high as you want to go with sprocket ratios. The rest has to be done with wheel diameter. </p>
<p>My old Katana spun 9,600 rpm, and needed to convert that to about 160mph. That would be 2400 rpm for 40 mph. That's 30 times the gearing, so instead of gearing up by a factor of 5 as on the bicycle, it would have to gear down by a factor of 6 if it had the same sized tires as a bicycle. It's better to make the wheel diameter smaller and go with less gearing.</p>
| 28541 | Bike wheel sizes |
2019-05-27T14:19:56.370 | <p><a href="https://i.stack.imgur.com/sHebp.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/sHebp.png" alt="Diagram"></a></p>
<p>In this case, L, theta and P are given, there will be 4 reactions at the wall? (2 horizontal and 2 vertical). </p>
<p><span class="math-container">$$\begin{align}
\sum F_v &= 0 \therefore F_{v,B} + F_{v,A} = P \\
\sum F_h &= 0 \therefore F_{h,B} = F_{h,A}
\end{align}$$</span></p>
<p>What will be the tensile stresses along AB and AC?</p>
<p>However, I do not know how to obtain the other 2 equations to solve all the reaction forces at the wall and the tensile stresses? Can anyone please help? One of those equations are from the moment?</p>
| |mechanical-engineering|civil-engineering|solid-mechanics| | <p>@AndyT and @kamran explained how to technically solve this question.
I would like to add a short comment about the thing seemed to confuse you the most.</p>
<p>Indeed, there are 4 reactions in this problem. If the triangle was made of a solid body, you would deal with a statically indeterminate problem and find it quite hard to resolve those reactions. However, since we are dealing with a truss, you should examine carefully whether its statically determinate or not:</p>
<ul>
<li>The number of unknowns is 6 - the above mentioned 4 reactions + axial
forces along the truss members</li>
<li>The number of equation is also 6 - 3 pin joints, each has 2
equilibrium equations (in X and Y directions)</li>
<li>Since the number of unknown equals the number of equations - you deal
with statically determinate problem and can resolve all the
reactions without dealing with the truss stiffness</li>
</ul>
| 28548 | Truss Analysis Problem |
2019-05-28T05:49:47.407 | <p>Is it in the same direction of the applied load or in the co-planar perpendicular direction to the load applied?</p>
| |mechanical-engineering|civil-engineering|beam|stresses|strength| | <p>Bending moment of inertia or second moment of inertia is taken on the surface of the beam parallel to applied loads' vertical component and perpendicular to the long axis of the beam, about a horizontal line on that surface, called neutral axis.</p>
<p>When loads are applied to a beam they want to bend the beam into a deflected down curve. They create bending moment in the beam which is countered by tension stresses on the lower part and compression stresses on the upper part of the beam.</p>
<p>Moment of inertia is a characteristic property of the beam section which defines how that cross section bends and how much stress both tension and compression it takes to bend, and how this stress is distributed on that section of the beam.</p>
<p>It is a engineering representation of the stiffness of the beam at that point.
In prismatic beams, extruded shapes, moment of inertia or I is constant along the length of the beam. But if the Beam's cross section varies along its length, then the I varies along the length of the beam.</p>
| 28560 | Along which direction is the area moment of inertia taken in bending moment equation? |
2019-05-29T01:57:14.743 | <p><span class="math-container">$ P_m=P_n*{G\frac1{G_{STC}}}*[1-\gamma*(T_c-25)]$</span> </p>
<p>This equation is applied on sollar cells. <span class="math-container">$ P_m$</span> is the mean power produced in 15 minutes, <span class="math-container">$P_n$</span> is the nominal power [W], G is the insolation [W/m2], <span class="math-container">$G_{STC}$</span> (the subscript still in the wrong place, I don't know why) is the radiation [W/m2] in standard testing conditions <a href="https://sinovoltaics.com/learning-center/quality/normal-operating-cell-temperature-noct-definition-noct-definition/" rel="nofollow noreferrer">reference here</a>. Last, but nost least, <span class="math-container">$T_c$</span> is the cell temperature. </p>
<p>What is the parameter gamma (<span class="math-container">$\gamma$</span>)? I have a value for gamma, 0.005 [adimensional], but I don't know what it means or its definition. </p>
| |solar-energy|photovoltaics| | <p>It's the <a href="https://sinovoltaics.com/solar-basics/measuring-the-temperature-coefficients-of-a-pv-module/" rel="nofollow noreferrer">temperature coefficient</a> of the panel. It varies by panel, but 0.005 (0.5%) is a fairly normal value for it. Its unit is 1/degree C. It's the rate at which cell efficiency drops off, as the cell temperature rises above 25 degrees C.</p>
| 28573 | What's the name of this parameter? |
2019-05-30T15:22:16.273 | <p>After doing some research, I have come to the realization that 3 blade design horizontal turbines have the greatest maximum coefficient of performance. Most sources seem to agree a Cp of ~0.45 at TSR=7 is optimal for 3 blade designs.</p>
<p>The power output of a turbine is P=0.5*Cp<em>rho</em>A*V^3. So what I cannot wrap my head around is why, all other things being equal, more blades (of course up to a limit) does not equal a greater coefficient of performance and thus more power. How is only 3 blades this limit on something 200m in diameter?</p>
<p>It seems there would be a significant portion of the air passing freely undisturbed through the rotor.</p>
<p>I welcome any answers and explanations, but preferably backed by formulas.</p>
| |power|turbines|energy-efficiency|wind-power| | <p>Because in the real world, all else is never equal, but let's start there analytically. </p>
<h2>What does "all else being equal" actually mean?</h2>
<p>With respect to a flat swept disk perpendicular to an ideal fluid with a uniform free stream flow field, from Momentum Theory, the induced flow field associated with optimum performance has slowly been specified in ever more realistic and convenient forms. Prandl, Glauert, Betz, and a host of others have have developed the theory to the point where we now have a closed solution describing the induced flow that is associated with maximum power extraction. (see, for example, section 3.5 in <a href="https://pdfs.semanticscholar.org/9ecb/d17ae8d276d0395927e6604d94ca9d97ffde.pdf" rel="nofollow noreferrer">https://pdfs.semanticscholar.org/9ecb/d17ae8d276d0395927e6604d94ca9d97ffde.pdf</a> )
The relevant parameter is the ratio of the blade speed at a given radius to the free stream wind velocity. This is r/R * TSR. This determines the magnitude and direction of the ideal induced flow at this radial position. The actual rotor lift is proportional to, and opposed to, the momentum in the induced flow. Now all you have to do is design a blade set that creates this precise induced flow across it's entire span. It doesn't matter how you do it - whether with one blade or a thousand, you just need to match the theoretical induced flow field. One note - Differing numbers of blade will have significantly different properties in terms of <em>individual</em> span loading distributions in order for the overall rotor assembly to have the required span loading distribution. This is due to the interaction of the shed vorticity on the trailing blades. But according to momentum theory <strong>there exists a blade design for each differing number of blades which will produce the same optimum performance given the same TSR</strong>. So all else being equal means the TSR is the same, and the blades are optimized for the actual blade count so as to create the optimal induced flow vector at every radial distance. Blade count does affect the design of the individual blades, but not the theoretical performance limit of the rotor.</p>
<p>When we add in an actual vortex model, the <em>potential jumps</em> in the flow change from a single jump at the disk to spiral jumps in the wake trailing behind each blade. These two models are highly compatible. Using a complex Fourier expansion of the vortex flow, you can separate the induced flow into a constant induced flow and a cyclic induced flow. Interestingly, the proportion of these two are constant radially, and they always have the same vector direction. This is why the actuator disk model appears to overperform it's rather pathetic assumptions, as does lifting line theory applied to modest AR wings, and Michell's integral applied to the wakes of not-so-thin ships. <a href="https://apps.dtic.mil/dtic/tr/fulltext/u2/433976.pdf" rel="nofollow noreferrer">DTIC vortex rotor model evaluated with complex Fourier Analysis, 1964</a> The takeaway is that TSR needs to be matched to the blade number, and it has to be reduced for greater blade numbers. <strong>This combined with the result from the above paragraph puts a downward trend on best possible performance as the blade number increases.</strong> This performance limit vs TSR curve is called the Schmitz whirlpool power coefficient. See page 7 here - <a href="https://mragheb.com/NPRE%20475%20Wind%20Power%20Systems/Optimal%20Rotor%20Tip%20Speed%20Ratio.pdf" rel="nofollow noreferrer">https://mragheb.com/NPRE%20475%20Wind%20Power%20Systems/Optimal%20Rotor%20Tip%20Speed%20Ratio.pdf</a> (This result probably isn't the same as the power coefficient formula in the first link, but I'm not certain.)</p>
<h2>So what is missing from Momentum theory?</h2>
<ol>
<li><p>Aerodynamic drag</p></li>
<li><p>Tip losses due to the real vorticies not behaving like the highly
concentrated ideal tip vortices.</p></li>
<li><p>Structural constraints related to performance.</p></li>
<li><p>Economic constraints and "rest of system" interfaces. </p></li>
<li><p>Real wind behaves badly.</p></li>
</ol>
<p>These all influence blade count.</p>
<p>The aerodynamic drag limits HAWTs to TSRs best suited to at least three blades. The best two-blade designs don't quite match the best 3 blade designs because they can't quite run at their optimum speed due to excessive drag. Here I'm taking about skin friction and profile drag. These are the components of basic 2D airfoil L/D ratios. Vortex drag is handled next.</p>
<p>Very early on researchers realized that wing tips didn't work as well as the simple models predicted they should. Glauert and Prandl each had tip correction factor formulae. These favor more blades, not fewer, in an all else being equal scenario. But for high aspect ratio designs, the differences are quite small.</p>
<p>The biggest problem is structural. If you change from a three bladed to a four bladed rotor, you have to run a bit lower TSR to maximize performance. This means a higher torque at the same power. So the rotor ends up using more material with more blades. Also, the profile drag total is higher because you have to support that extra torque with thickness. And the blade area total has to be larger because with lower TSR, the apparent wind speed is less. So you don't save much on aero drag even though it is running slower (but this part can get tricky due to the complicated details of the geometry differences).</p>
<p>Rest of system issues - for rotors of any respectable size, you don't want to have to deal with slower. You already have gearboxes with 30 ton gears in two stage gearboxes. There are huge losses in gearing the rotor rpm up to a suitable generator rpm. This is one of the reasons why three-blade rotors are designed to run a bit quicker than the calculated aerodynamic optimum. It's a bigger problem for the large machines, but still an issue for kilowatt sized units. To save on gearbox cost and losses, you want to run at the fast end of the flattish Cp vs TSR area. So you can trade a bit of aero Cp off for a more manageable and efficient gearbox. Gearbox losses are typically proportional to the maximum torque rating of the gearbox, so looking at this a different way, you'd need a lower cutout speed with the higher blade count and the same gearbox torque rating.</p>
<p>And lastly, real wind behaves badly. It isn't the uniform flow that the models assume. There is usually a substantial vertical wind gradient, and some transverse wind gradient, wind speed variations (gusts), and wind direction variations. These each produce excursions from the ideal induced flow field that we are chasing. And these excursions are not symmetric around the rotor, so imbalances occur that produce large fluctuating forces and moments on the rotor shaft. The largest imbalance is normally the flap moment of the blade. The load imbalance isn't a statics problem. The shed vorticity from the blades "remembers" the load at the time it was created and carries that information into the swirling wake. Here, the wake influences the induced flow of the following blades based on earlier conditions seen at a different location on the disk. The result is often described as a form of hysteresis, but that isn't the right term. Hysteresis is statically stable, what's going on here is strictly dynamical outside of actual blade stall conditions. The physical response of the individual blades and the net imbalances seen at the hub are a massive engineering concern. We just know more about these issues in the three-bladed rotor than any of the others. The two-bladed rotor has a particular imbalance problem that can be molified with the three bladed design. And more than three blades make all but the smallest machines heavier and more costly to produce without any performance improvements.</p>
| 28596 | Do 3 blade wind turbines generate more power than a 9 blade, all else being equal? |
2019-05-30T16:34:10.963 | <p>Does anyone know what these three symbols are? I am working to identify them but they do not seem to appear in the site: <a href="https://www.edrawsoft.com/pid-legend.php" rel="nofollow noreferrer">https://www.edrawsoft.com/pid-legend.php</a></p>
<p>Does anyone know a better reference for P&ID diagram symbols?</p>
<p><a href="https://i.stack.imgur.com/xwgJi.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/xwgJi.png" alt="Sample from a project I am working on"></a></p>
| |piping|technical-drawing|instrumentation| | <p>Two of those are not standard usage symbols in that way. You may to have to use context and/or a physical inspection to determine what they are. <strong>Where is that diagram from?</strong></p>
<p>Various organizations define symbols for piping and/or control. The <em>International Society of Automation</em> uses <em>ANSI/ISA-5.1-2009</em>, which you can officially <a href="https://www.isa.org/store/products/product-detail/?productId=116630" rel="nofollow noreferrer">find on their website</a>.</p>
<p><br></p>
<ol>
<li>The <strong>top symbol</strong> is probably meant to be a shut-off valve, probably ball type. The usual symbol is something like:<br>
<a href="https://i.stack.imgur.com/tOJRs.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/tOJRs.png" alt="ball valve"></a><br>
<br></li>
<li><p>The <strong>middle symbol</strong> is a fairly standard <strong>Solenoid-controlled valve</strong>.<br>
<br></p></li>
<li><p>The <strong>bottom symbol</strong> could be meant to be a butterfly valve (for flow control), or a "blowout" valve / rupture disk (for over-pressure safety).<br>
The standard symbols for those are like:<br>
<a href="https://i.stack.imgur.com/3Q3p7.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/3Q3p7.png" alt="butterfly valve"></a> <strong>and</strong> <a href="https://i.stack.imgur.com/zLMDy.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/zLMDy.jpg" alt="pressure"></a> <strong>or</strong> <a href="https://i.stack.imgur.com/ZqlBF.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ZqlBF.jpg" alt="enter image description here"></a></p></li>
</ol>
| 28597 | P&ID Symbol Identification Question |
2019-05-31T08:07:57.110 | <p>I'm a programmer working on a project where we will be controlling a crane's boom for motion control. We are currently a year out from having the actual crane in house, but are doing the software ahead of time as we can 3d simulate a lot of it. I thought it would be a fun idea to make a little miniature version to play with (and it would be a fun toy for the office). Our crane boom consists of 4 segments and to scale it down somewhat proportionally each square tube will be 1.5 feet long and the first extension will be 2 inch outer diameter, with each inner tube getting 1/4 smaller. Making a little mechanism to pan and tilt the boom is simple, but I'm not sure how to go about controlling the extension with a motor.</p>
<p><a href="https://i.stack.imgur.com/Pco3T.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Pco3T.jpg" alt="enter image description here"></a></p>
<p>Ideally I'd like to control the extension with just one motor I can mount at the base of the boom. My question is, is there a mechanism that can create this collapsible column with just one motor? </p>
<p>I realize the easy solution is probably 4 separate motors that use a rack and pinion to extend each segment, but I don't think Ill have the space for the motors at this scale.</p>
<p>Another option I've pondered is using a spring and then having the motor just wind up a string that pulls to compress the spring but I feel acquiring a spring of the right diameter, length and strength would prove challenging as at full extension this will be about 4 feet, though I'd like to make it as small as possible. </p>
<p>Hydraulics and pneumatics are completely out of my ability.</p>
<p>In this, each segment should move the same amount as the boom extends. I feel like there has to be some sort of mechanism for this but I can't think of anything else. I think our actual crane uses a bunch of pulleys but I'm not sure how that works. My boss is designing the crane, but I'd like to do this as a surprise for him so I don't want to ask him for his input.</p>
| |mechanisms| | <p>Many moons ago, I was involved in a process which required a transportable antenna tower. The assembly was similar to your requirements. Tilt the tower module vertical and crank a winch handle to cause each of the "telescoping" segments to elevate. One crank, many pulleys and a strong wire cable.</p>
<p><a href="https://i.stack.imgur.com/T9tTE.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/T9tTE.png" alt="extensible crank assembly"></a></p>
<p>The drawing above is not to scale and placement of the pulleys are not critical. The individual segments to be extended would require guiding mechanisms as the segments are offset from the centerline. In the case of the antenna tower, the guiding mechanisms were a combination of low-friction plastic guide blocks and boat trailer rollers.</p>
<p>A series of nylon or teflon buttons would likely serve your requirements for guidance.</p>
| 28607 | Motorized 4 part extension column mechanism? |
2019-05-31T10:24:05.113 | <p>According to many sources, one of the purposes of building's foundations is "To distribute the weight of the structure over large area so as to avoid over-loading of the soil beneath." <a href="https://en.wikipedia.org/wiki/Foundation_(engineering)" rel="noreferrer">(Wikipedia)</a></p>
<p>On the Wikipedia page on foundations, there is the following picture:</p>
<p><a href="https://i.stack.imgur.com/FnVb1.png" rel="noreferrer"><img src="https://i.stack.imgur.com/FnVb1.png" alt="enter image description here"></a> </p>
<p>So the foundation seems to consist of some type of "legs", with a larger surface area on the bottom. But if the point is to distribute the weight of the building over an area, why not put the bottom of the building in contact with ground? Isn't the bottom of the building itself a much larger contact area than the area the foundation can provide? In the picture, the bottom of the building does seem to rest at ground, so why do we require the foundation to have these large surface area rectangles at the bottom if the whole weight of the building is already resting on ground?</p>
<p>Here is another picture from the same page:</p>
<p><a href="https://i.stack.imgur.com/9QElx.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/9QElx.jpg" alt="enter image description here"></a></p>
<p>This seems to be some sort of lodge or a cabin, with stones used as foundation. Here it is evident that the building itself is raised above ground with the stones used as a contact to ground. But now the small stones suffer all the weight of the structure. Wouldn't they be under quite a heavy load over small contact area and possibly sink into the ground? Why not simply have the building's floor, a much larger contact area, rest on the ground?</p>
| |structural-engineering|foundations| | <p>Building foundations perform many tasks.</p>
<p>They provide a strong base to take the weight of the building, penetrate through incompetent layers of soil to reach a layer strong enough, they resist the heaving forces of expansive clay soils, they anchor the building down against toppling in earthquakes and getting blown by the wind or washed away by the floods.</p>
<p>Add to this many other tasks such as insulating the floors from moisture, freezing, animals, enemies, etc.</p>
<p>In many parts of the word they act as a passive thermal climate control. Because the soil strata under ground have more moderate temperature and do not change much during winter and summer, and because the foundations are massive they have large capacity to act as a thermal battery and moderate the temperature of the building, which has been done in hot arid areas of middle east, or in basement and cellars in many urban areas.</p>
<p>In some areas of my old home country, Iran, specially in rugged mountainous villages the foundation of the entire village is integrated to resist the strong winds, heavy snows and bridge the soft patches of ground soils, span melted snow streams, and provide the support to the covered protected alleys.</p>
| 28610 | What is the purpose of building foundations? |
2019-05-31T19:39:47.157 | <p>Suppose that the engine cranking has been successful but I keep holding the ignition switch at the START position instead of releasing the switch to go at the RUN position. Does this mean that the starter motor is still spinning or has it somehow disengaged (probably through the starter solenoid ) from the battery?</p>
| |automotive-engineering| | <p>The flywheel will accelerate faster than the starter motor and the ring gear will accelerate the starter dog since there is a bendix drive or over-run clutch to allow this to happen so the starter motor itself will not overspeed. This does create noise which most notice...</p>
| 28618 | Is starter motor running if I hold the ignition switch at the START position? |
2019-06-01T09:20:39.607 | <p>Considering the fact that the starter solenoid needs a maximum of 50 Amps initially (in order for the pinion gear to engage to the flywheel through the plunger and lever) and then some 8 Amps during cranking, do we need a relay to supply power to the solenoid? Is this relay of electromechanical type or a solid state one?</p>
| |automotive-engineering| | <p>Earlier cars did not use a relay, as the ignition switches were built sufficiently well to carry the 50A current for the pull-in winding current.</p>
<p>Newer cars have components like the ignition switch that are lighter and have more functions and need a relay to drive the starter solenoid. This relay is usually in the fuse box in the engine compartment.</p>
<p>We had a Commer van (petrol) that the ignition switch had the following positions : off, ignition, start. Another van (diesel) had off, on, heat, start...</p>
<p>Most relays are the electro-mechanical type as they are cheap and easy to get hold of, but as cars get more computer well more micro-processor control then some things are being replaced with solid-state components.</p>
<p>Edit: here is a link kindly provided by a comment :</p>
<p><a href="https://autosystempro.com/starter-control-circuit-components/" rel="nofollow noreferrer">example starting circuits</a></p>
<p>It does not include all possibilities, early Landrovers just had a big spring-loaded push button that directly operated the starter - no relays, solenoids at all...</p>
| 28626 | Is a starter relay needed to supply the current to the starter motor solenoid? |
2019-06-02T11:57:44.513 | <p>I'm designing a steering mechanism following the Ackerman principle. I've met the condition that my steering arms' normal projections must cross in the middle of the rear axle however when I do a turn my wheels' axles do not cross in a single point on the rear axle. I'm applying some visuals</p>
<p><a href="https://i.stack.imgur.com/aU4dB.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/aU4dB.png" alt="enter image description here"></a></p>
<p>where the bottom most line is the x axis of my drawing, so the wheel base is 1210mm. I've set the length of my steering arms at 150mm for demonstration purposes, where 756.82mm seems to be the distance between their joint points. The white lines that go off to the right are the projections of my wheels' axles. </p>
<p>Here's a visual of when turning at 15 degrees.</p>
<p><a href="https://i.stack.imgur.com/jW6Hi.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/jW6Hi.png" alt="enter image description here"></a></p>
<p>The lines should cross, I can't figure out what I'm doing wrong.</p>
| |mechanical-engineering|automotive-engineering|mechanisms| | <p>If you just first turn the king pin on the interior wheel by say 25 degrees and draw a line perpendicular to the same wheel going up till it intersect the rear axel continuation, then from that point draw a line to the center of the other wheel and make that wheel perpendicular to the line, there is only one straight tie-bar that will connect the ends of wheels' bars so that they mach.</p>
<p>Now try for 15 degrees, 30 degrees and compare your length of rods and tie-bar. This gives you a pretty good idea of the geometry.
Here is a link with more info. <a href="http://datagenetics.com/blog/december12016/index.html" rel="nofollow noreferrer">Steering geometry</a> </p>
<p>just so you know the car manufacturers take advantage of three dimensional toe, caster, camber, drop arc, and lateral expansion of suspension and the lateral fricshion in road puddles as to give the car its road handling and comfort characteristics and sharing the handling of bumps with the shocks. basically the suspension twists and deform in a way that breaks the sudden shakes or jerks and vertical moves. Also they make the car surefooted in sudden jerks and even they design them to self destroy the whole suspension and wheels as sacrifice mass in an accident to break the crash and absorb a lot of energy. </p>
| 28640 | Ackerman steering |
2019-06-02T15:51:47.973 | <p>Sorry, it seems like the answer to this question is just a quick google search away; but I have not found a page defining what the axis is.</p>
<p>My guess was that the axis of a piple is the imaginary line running down through the middle of the pipe. However, this doesn't make sense based on a question I am looking at in a calculus book. The question is "write the vector field describing the flow if the axis of the pipe is the unit circle in the <span class="math-container">$xy$</span> plane." This makes me think that the axis of a pipe is cross section of a pipe, but that doesn't really make sense either because usually an axis is a line, not a circle.</p>
| |mechanical-engineering|fluid-mechanics|civil-engineering| | <p>I'm guessing that the pipe in the problem is a torus. </p>
<p>I wonder if "centerline" would be a better word than "axis" in this case?</p>
| 28641 | What is the axis of a pipe? |
2019-06-03T20:25:04.613 | <p>Since copper is excellent at dissipating heat and does not rust, I am surprised that brake manufacturers do not electroplate cast iron/steel brake rotors with a thin layer of copper (~.5mm). There are brake manufacturers who offer brake rotors with cadmium or zinc-plated coatings at an additional cost but these coatings usually do not last long especially in areas that use a lot of rock salt. </p>
<p>I think copper plating would be the most ideal coating for ventilated brake rotors since it should keep the cooling vanes from rusting and becoming clogged with rust. Clogged cooling vanes reduce the rotor's cooling capacity resulting in higher temperatures, faster rotor wear, premature warping, and a shortened lifespan. </p>
<p>Moreover, since copper is very good at dissipating heat, air flowing through copper-plated cooling vanes should draw away more heat from the brake rotor, resulting in a cooler ventilataed brake rotor, maximizing or perhaps even extending its intended lifespan. </p>
<p>Also, since there should be almost no rust on a copper-plated brake rotor, removing an worn out, copper-plated brake rotor should be no problem. Anyone who has worked on brakes knows how time-consuming it is to remove a rusted-on brake rotor or drum.</p>
<p>The brake manufacturer should not electroplate the friction areas (as shown in the picture below) since the high pressure and friction from contact with the brake pads would most likely warp the copper coating, but all the other surfaces would have a copper coating, most importantly the surfaces of the cooling vanes.</p>
<p>Will electroplating a ventilated brake rotor with copper improve its cooling capacity and maximize its lifespan?</p>
<p><a href="https://i.stack.imgur.com/zBI28.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/zBI28.jpg" alt="enter image description here"></a></p>
| |mechanical-engineering|materials|automotive-engineering| | <p>A thin layer of copper plating will have no effect on <em>convective</em> heat transfer to rapidly flowing air, which is what gets the heat out of a cored/ventilated steel brake disc. In addition, a few taps of the brakes would scrape most of the plating off the disc. </p>
<p>To have any effect on heat rejection rate in a predominantly steel disc, the copper cladding would need to be thick enough to conduct heat laterally outwards from under the brake puck's contact patch against the disc before the heat has a chance to soak into the steel underneath it. This requires a significantly thick copper layer relative to the thickness of the steel disc, as in the case of a copper-clad stainless steel frying pan bottom.</p>
| 28659 | Will electroplating a ventilated brake rotor with copper improve its cooling capacity and maximum its lifespan? |
2019-06-04T02:17:50.630 | <p>This question is a bit general but hopefully someone can point me in the right direction.</p>
<p>If you have a metal container, welded shut that's fitted into a hole that is say 10 meters wide and 30 meters deep.</p>
<p>This has the volume of about 21,205</p>
<p>If you pump pure oxygen into the tank until all the water was pushed out of that cylinder, what would be the pressure of the oxygen in the tank. How much oxygen would be in the tank?</p>
| |mechanical-engineering|pressure|pressure-vessel|compressed-gases| | <p>If the hole that the water drains out from is a hole with a check valve located at the height H from the top of the tank the pressure of the gas is equal to the pressure of water at the depth H. And the bottom part below that level will be water only. </p>
<p>However if you want to drain the entire tank and replace the water with the gas you should have the hole on the bottom of the tank, otherwise there is always water remaining below the discharge level.
In that case the pressure of the gas will be 3 atmosphere. Therefore you can use ideal gas low to calculate the weight of the oxygen:</p>
<p><span class="math-container">$$ PV=nRT $$</span>
-P is pressure</p>
<p>-V is volume, liters</p>
<p>-n is number of gas molecules</p>
<p>-R is the ideal gas constant= 0.08206 L atm mol–1 K–1</p>
<p>-T is temperature in kelvin.</p>
| 28663 | hydrostatic storage of gasses |
2019-06-04T15:47:15.823 | <p>I am interested in knowing how many newton's of force is needed to keep 1 kg of mass hovering above the moon's surface.</p>
<p>And whether the current most powerful ion thrusters would be up to the task? I believe ESA tested an ion thruster recently which achieved 100 kilo newton's. (I massively mis-remembered this part. I was thinking of the X3 thruster, and the figure is 5.4 Newtons.)</p>
<p>Thank you in advance! </p>
| |aerospace-engineering|propulsion| | <p>You don't need any energy or thrust, if your 1kg mass is already in orbit.</p>
<p>Basically an object orbiting the moon relatively near the surface, say one hundred miles, stays on orbit with no energy needed and rotates with a speed of v = SQRT(G*M/R). </p>
<p>In this case approximately 1.022 km/s.</p>
<p>Lifting up some mass and placing it on mons orbit is a different story!
you need to calculate the difference between the potential energy of the mass on the moon and in orbit altitude</p>
<p><span class="math-container">$E1-E2 = -GMm*((1/R)-(1/(R+d))$</span></p>
<p>R= moon radius and d is the orbital altitude.</p>
| 28672 | Amount of thrust needed to Hover on the moon? |
2019-06-06T18:08:03.937 | <p>Stress = Force / Area </p>
<p>Lets see with static force... what happens to stress output when decrease the area. </p>
<pre><code># Stress = Force / Area
F = 50000
A = [15.0,10.0,5.0]
S = zero(A) # Initialize the output
for i in 1:length(A)
S[i] = F / A[i]
end
m = DataFrame(hcat(A,S))
names = ["area","stress"]
names!(m, Symbol.(names))
print(m)
3×2 DataFrame
│ Row │ area │ stress │
│ │ Float64 │ Float64 │
├─────┼─────────┼─────────┤
│ 1 │ 15.0 │ 3333.33 │
│ 2 │ 10.0 │ 5000.0 │
│ 3 │ 5.0 │ 10000.0 │
</code></pre>
<p>With decreasing area we increase the stress. </p>
<p>So in thinking about hoop stress then how the minimum required thickness is obtained. Using the ASME B31.3 pressure tmin to find t </p>
<pre><code>304.1.2 Straight Pipe Under Internal Pressure
t = PD / 2(SEW + PY)
</code></pre>
<p>The numerator has P * D so anything in the denominator will dilute the numerator... </p>
<p>So to see how P and D affect the t output we can iterate through a range of static pressure and changing Diameter... so lets see what this looks like: </p>
<pre><code># Define numerator inputs
P = 200 # static P
D = collect(1.0:1.0:24.0) # varying D
# define the denominator static inputs
S = 20000
E = 1.0
W = 1.0
Y = 0.4 # temp coefficient
# initialize output
out = zero(D)
for i in 1:length(D)
out[i] = P * D[i] / (2*(S*E*W) + (P*Y))
end
# plot the output
plot(D,out,seriestype = :bar, title="t = static P in relation to Diameter")
</code></pre>
<p><a href="https://i.stack.imgur.com/GjK34.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/GjK34.png" alt="enter image description here"></a></p>
<p>So a big factor in the minimum required thickness for pressure containment is the P vs diameter.. at say static P increase the D and we need a thicker wall for pressure containment. </p>
<p>So in thinking about S = F / A </p>
<p>How does this relate to the pressure design thickness calculation.... S in the tmin calculation is in the denominator and S is static in this equation governed by the 3:1 ASME safety factor in relation to materials tensile stress. </p>
<p>Intuitively i can see how at a static F over varying Area - how the stress INCREASES... (as we saw top of the post) because there is less area to distribute the stress over... so stress would increase pounds per sq inch... </p>
<p>But doesn't that mean for smaller D in a pipe, we have less A so that the stress would increase... but running pressure design thickness calculation.. at smaller D we need a thinner wall for pressure containment. </p>
<p>For sure im missing something here! (then again i see no area in the pressure tmin calculation) </p>
<p>Anyone can help me connect the dots? </p>
<p>Thanks</p>
| |stresses|piping| | <p>The "area" terms in the equations all cancel out. Think about a section of pipe of internal diameter <span class="math-container">$d$</span>, length <span class="math-container">$l$</span>, thickness <span class="math-container">$t$</span> with internal pressure <span class="math-container">$P$</span>.</p>
<p>Now imagine you slice the section of pipe (and the fluid causing the internal pressure) into two semicircular parts, and think about the forces acting one of the parts.</p>
<p>The "area" of the cut through the fluid is d wide by <span class="math-container">$l$</span> long, and the fluid pressure of <span class="math-container">$P$</span>. So the force exerted by the fluid is pressure <span class="math-container">$\times$</span> area = <span class="math-container">$P\,d\,l$</span>.</p>
<p>This is balanced by the hoop stress in the pipe wall. The area of the cut through the walls is <span class="math-container">$2\,t\,l$</span> (2, because there are two cuts, on either side of the pipe).</p>
<p>If the average hoop stress is <span class="math-container">$S$</span>, the force on the cut surfaces of the pipe is stress <span class="math-container">$\times$</span> area = <span class="math-container">$2S\,t\,l$</span>.</p>
<p>So <span class="math-container">$2Stl = Pdl$</span> or <span class="math-container">$$S = \frac{Pd}{2t}.$$</span></p>
<p>The actual hoop stress varies between the outside and inside of the pipe wall, and this formula doesn't include any safety factors, so in the ASME formula the <span class="math-container">$2t$</span> term is replaced by something more realistic for design purposes.</p>
| 28701 | Pressure design thickness and how relates to hoop stress |
2019-06-07T18:20:00.987 | <p>Small deformation shear strain <span class="math-container">$\gamma$</span> in engineering <a href="https://en.wikipedia.org/wiki/Deformation_(mechanics)#Shear_strain" rel="noreferrer">can be expressed</a> as </p>
<p><span class="math-container">$$\gamma_{xy}=\alpha +\beta ={\frac {\partial u_{y}}{\partial x}}+{\frac {\partial u_{x}}{\partial y}}$$</span></p>
<p>where <span class="math-container">$u$</span> represents the displacement of the edges of the infinetesimal element. But in the strain tensor we have defined</p>
<p><span class="math-container">$$\epsilon_{xy} = 1/2 \gamma_{xy}$$</span></p>
<p>The shear strain represents the change in the angle between the lines of the infinitesimal element. So why do we define the tensorial component as half of this angle? Why not just use the angle itself?</p>
| |structural-engineering|deformation| | <p>Consider a displacement map <span class="math-container">$u : A \subset\mathbb{R}^3 \rightarrow \mathbb{R}^3$</span>. This map displaces a reference material point <span class="math-container">$x \in A$</span> by the action <span class="math-container">$x \mapsto x + u(x)$</span>. We have <span class="math-container">$\gamma_{xy} = \frac{du}{dx} + \frac{du}{dy}$</span> as you stated in your question when small angle approximation applies.</p>
<p>Now we will consider a separate, seemingly unrelated problem. Consider some reference point <span class="math-container">$x_0 \in A$</span>, and choose some other point <span class="math-container">$x$</span> very close by. We may call <span class="math-container">$\Delta a \equiv x - x_0$</span>. The displacement map <span class="math-container">$u$</span> sends <span class="math-container">$x$</span> to <span class="math-container">$u(x) + x$</span> and <span class="math-container">$x_0$</span> to <span class="math-container">$u(x_0) + x_0$</span> and so gives rise to a <span class="math-container">$\Delta b = (u(x) + x) - (u(x_0) - x_0)$</span>. We seek to characterize the difference in squared length, after distorting through the map <span class="math-container">$u$</span>, i.e. approximate <span class="math-container">$|\Delta b|^2 - |\Delta a|^2$</span> for all <span class="math-container">$x$</span> very close to <span class="math-container">$x_0$</span>.</p>
<p><span class="math-container">$\Delta b = (u(x) - u(x_0)) + (x - x_0)$</span><br />
<span class="math-container">$\Delta b = (u(x) - u(x_0)) + \Delta a$</span></p>
<p>Since <span class="math-container">$x$</span> is very close to <span class="math-container">$x_0$</span>, we may Taylor expand <span class="math-container">$u$</span> about <span class="math-container">$x_0$</span>.<br />
<span class="math-container">$\Delta b \approx \nabla u|_{x_0}\cdot (x - x_0) + \Delta a$</span><br />
<span class="math-container">$\Delta b \approx \nabla u|_{x_0} \cdot \Delta a + \Delta a$</span><br />
<span class="math-container">$\Delta b \approx (\nabla u|_{x_0} + I)\cdot \Delta a$</span></p>
<p>And since <span class="math-container">$|\Delta b|^2 = \Delta b^T \Delta b$</span>,<br />
<span class="math-container">$|\Delta b|^2 \approx \Delta a^T(\nabla u|_{x_0} +I)^T(\nabla u|_{x_0} + I)\cdot \Delta a$</span><br />
<span class="math-container">$|\Delta b|^2 \approx \Delta a^T(\nabla u|_{x_0}^T\nabla u|_{x_0} + \nabla u|_{x_0} + \nabla u|_{x_0}^T + I)\cdot \Delta a$</span></p>
<p>For infinitesimal strains, <span class="math-container">$\nabla u|_{x_0}^T\nabla u|_{x_0}$</span> is quadratic in the entries of <span class="math-container">$\nabla u|_{x_0}$</span> and therefore we can approximate it as <span class="math-container">$0$</span>.</p>
<p><span class="math-container">$|\Delta b|^2 \approx \Delta a^T(\nabla u|_{x_0} + \nabla u|_{x_0}^T + I)\cdot \Delta a$</span><br />
<span class="math-container">$|\Delta b|^2 \approx \Delta a^T(\nabla u|_{x_0} + \nabla u|_{x_0}^T) \Delta a + |\Delta a|^2$</span></p>
<p><span class="math-container">$|\Delta b|^2 - |\Delta a|^2 \approx \Delta a^T(\nabla u|_{x_0} + \nabla u|_{x_0}^T) \Delta a$</span></p>
<p>This is the sought-after approximation. Call the tensor <span class="math-container">$\nabla u|_{x_0} + \nabla u|_{x_0}^T \equiv T_{x_0}$</span>. You can see that <span class="math-container">$T_{x_0}$</span> captures completely how infinitesimal segments (with one endpoint attached to <span class="math-container">$x_0$</span>) change lengths under <span class="math-container">$u$</span>. Therefore we cannot do better than <span class="math-container">$T_{x_0}$</span> when trying to formulate the concept of a strain tensor. And since the derivation does not depend on coordinate system, it is invariant to choice of <span class="math-container">$x$</span>, <span class="math-container">$y$</span>, and <span class="math-container">$z$</span> directions, which makes it practical to manipulate mathematically, as @alephzero's answer indicated.</p>
<p>You can see that any multiple of <span class="math-container">$T_{x_0}$</span>, say <span class="math-container">$kT_{x_0}$</span>, will provide exactly the same information, but will introduce a scale factor when using it to compute difference of squared lengths. <span class="math-container">$\Delta a^T \cdot k T_{x_0} \cdot \Delta a = k|b|^2 - k|a|^2$</span>. This scaled <span class="math-container">$kT$</span> is less natural for computing the difference of squared lengths, but may be more natural for some other purpose.</p>
<p>We may want to seek the <strong>most</strong> natural and intuitive scaling parameter <span class="math-container">$k$</span>, by somehow examining the entries of <span class="math-container">$T$</span> and comparing them to simpler engineering notions of strain. This would completely resolve the motivation of your post. However, look at the entries of <span class="math-container">$T_{x_0}$</span> in some coordinate system.</p>
<p><span class="math-container">$[T_{x_0}]_{ij} = \frac{du_j}{dx_i}(x_0) + \frac{du_i}{dx_j}(x_0)$</span>.</p>
<p>For <span class="math-container">$i \neq j$</span>, the entries are <span class="math-container">$\gamma_{ij}(x_0)$</span>, the engineering shear strains at reference point <span class="math-container">$x_0$</span>.<br />
For <span class="math-container">$i = j$</span> the entries are <span class="math-container">$2 \epsilon_i(x_0)$</span>, twice the engineering direct strains at reference point <span class="math-container">$x_0$</span>.</p>
<p>So we cannot at once scale the entries of <span class="math-container">$T_{x_0}$</span> to seem natural for both direct strains and shear strains. One of them will end up getting "unnaturalled" by a factor of 2. But now remember that engineering shear strain is formulated somewhat arbitrarily. You might even argue that it's <strong>more</strong> arbitrary than how direct engineering strain is formulated. (It's defined by taking the outer angles formed by one corner of a deformed quadrilateral. Why not take the opposite corner as well into the definition?)</p>
<p>Therefore, choose the scaling <span class="math-container">$k$</span> to be natural in terms of the direct engineering strain, i.e. set <span class="math-container">$k = \frac{1}{2}$</span>.</p>
| 28714 | Why is tensorial shear strain half of engineering shear strain? |
2019-06-07T20:51:04.020 | <p>At any point inside a structural element under deformation, we can always find three planes such that when we define a coordinate system with basis vectors lying on those planes, the shear stresses at that point are zero and the stress tensor is diagonal.</p>
<p>The principal stresses are the eigenvalues of the stress tensor, and are <strong>invariant of the rotation of the coordinate system.</strong></p>
<p>This last feature on bold is recited everywhere.But my question is, so what? Why is it so useful to find these stresses whose components don't change with rotation of the coordinate system? And why would I even want to rotate my coordinate system?</p>
| |structural-engineering|stresses| | <p>I'll take your word for it that your quote is "recited everywhere". The point of it is not that you can "rotate your coordinate system" just for fun, but that the principal stresses are <em>independent</em> of whatever coordinate system you choose to work in.</p>
<p>The physical structure doesn't know or care what coordinate systems you are using, so any conclusions you draw from your analysis about failure criteria, crack propagation directions, the onset of plasticity, etc, etc, <em>must be independent of your choice of coordinate system</em> - otherwise they are guaranteed to be wrong!</p>
<p>The principal stress values, and the direction of the coordinate system corresponding to them, at every point in the structure, <em>are</em> a complete description of the state of stress which is independent of the analyst's coordinate system(s), and also very easy to work with since the stress tensor is diagonal. </p>
<p>To give a couple of examples, in brittle materials cracks propagate in the direction of the largest (tensile) principal stress. In ductile materials, a useful failure criterion is the maximum shear stress acting in any direction at a point, and that is simply the difference between the maximum and minimum principal stresses.</p>
<p>Of course it is possible to devise formulas that reproduce those results in a different coordinate system, but they are much more complicated, and don't give much <em>insight</em> into what is going on in the structure.</p>
| 28715 | What really is so special about principal stresses or stress invariants? |
2019-06-08T15:33:23.283 | <p><a href="https://i.stack.imgur.com/0mtA3.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/0mtA3.jpg" alt="opamp"></a></p>
<p>In this case, can I say that <span class="math-container">$V_E$</span> correspond to -0.7? if yes, why can this be?</p>
| |electrical-engineering|electrical| | <p>In simplistic explanation Bipolar Junction Transistor is made of two diodes back to back. </p>
<p><a href="https://i.stack.imgur.com/BSh1f.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/BSh1f.png" alt="enter image description here"></a></p>
<p>The minimum voltage required switch on the diode between Base and Emitter is 0.7V. Since <span class="math-container">$V_B$</span> = 0V, in order for the diode to be on the voltage across <span class="math-container">$V_{BE}$</span> has to be -0.7V. </p>
<p>I hope this helps. </p>
<p><strong>References:</strong></p>
<ul>
<li><a href="https://www.electronics-tutorials.ws/transistor/tran_1.html" rel="nofollow noreferrer">Bipolar Transistor</a></li>
</ul>
| 28722 | Operational Amplifier |
2019-06-08T16:45:05.210 | <p>I need to find the effective length of a column (kl) from this chart which can be found in the "design of concrete structure" By Dr. Arthur Nilson.<a href="https://i.stack.imgur.com/uwFOm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/uwFOm.png" alt="chart"></a></p>
<p>the question states "if Psi top is 1.5 and the bottom joint is a firm (fixed) joint what is k? if the column is braced" what I don't know is if the joint is fixed should I take Psi as 0 or infinity? this is a diagram of the column <a href="https://i.stack.imgur.com/oYw4b.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/oYw4b.png" alt=""></a></p>
| |civil-engineering|design|reinforced-concrete|columns| | <p>You should take <span class="math-container">$\psi=0$</span>, typically for fixed braced concrete column, design aids set the bottom as zero flexibility.</p>
<p>Common sense then tells us that if we have a column with top and bottom fixed; it has 2 inflection points at 1/4 length of the column from the bottom and top. Hence we get K= 1/2, which is as shown on this chart.</p>
| 28723 | How do i find the effective length factor (k) for a column from this chart? |
2019-06-09T19:32:10.643 | <p>Cooling towers for nuclear power plants often have the shape of a hyperboloid. At first glance, I thought that an obvious reason for this would be the smaller sectional area, meaning the tower would be less exposed to wind.</p>
<p>However, I did a quick calculation, based on a cooling tower nearby. I modeled the cooling tower as a hyperboloid and calculated its volume and its sectional area. I then calculated the radius needed for a cylindrical tower with the same volume. Its sectional area was just 1.5% larger. By reducing its height from 150 to 145 meters (and increasing the radius, of course), the sectional areas would have been the same.</p>
<p>I know that hyperboloids can be built using only straight pieces of reinforcement. However, those towers are quite big and the circular shape could be very well approximated by some sort of polygon with e.g. 1000 vertices making circularly shaped reinforcement unnecessary.</p>
<p>I can imagine other reasons:</p>
<ul>
<li><p>The waisted shape could probably lead to a certain compression of the hot vapor and thus help/accelerate the condensation. Is this true? How big could that effect be?</p></li>
<li><p>Some documents I have read say one needs less material to build a hyperboloid cooling tower. Is there really a big difference in terms of concrete? After a rough calculation, assuming a thickness of 1 meter of concrete all around, I'd estimate that a cylindrical shape would only need approximately 1.5% more concrete.</p></li>
</ul>
<p>So, why do almost all cooling towers have this particular shape?</p>
| |structural-engineering|civil-engineering|cooling|building-physics| | <p>The hyperbolic shape of cooling tower is preferred as this shape is stronger against wind pressure. Consequently, the concrete wall thickness can be reduced and optimised. from 1.5m at the base of the viel, the shell thickness rapidly fall to ~30cm.</p>
<p>The venturi effect resulting in the usage of this shape is, contrary to popular belief, not improving the performance of the cooling tower. Indeed, the Venturi effect increases the speed at the throat which creates a very slight increase in the pressure drop by friction on the walls. This reduces the draft and therefore the air flow available for cooling. This very marginal loss of performance is largely compensated by the gain in resistance of the shell to external winds.</p>
<p>The height of the tower is determined by :</p>
<ul>
<li>Local permit for maximal construction height.</li>
<li>A balance between performance / cost / erection time.</li>
</ul>
<p>If the tower is high, the draft will be stronger, thus the airflow will be higher thus the cooling effect will be better.
However, if the tower is high, the cost increase rapidly with the difficulty to bring concrete at such height. But also due to the increased thickness of the shell (due to self weight and the wind load).
Sometimes, construction time is a limiting factor, rule of thumb is 1m of shell per day. If the tower is 200m, it might be too long for a project.</p>
<p>Reference : 20 years as a hyperbolic cooling tower design engineer</p>
| 28727 | Cooling tower for nuclear power plants |
2019-06-09T23:43:51.520 | <p>My understanding of swash plates led me to believe that the pistons in a swash plate pump rotate around with the driving shaft along with the chambers in which they aspirate (e.g. <a href="https://www.youtube.com/watch?v=1FpT9Z4at-w" rel="nofollow noreferrer">M&S Hydraulic Youtube video on swash plate plumps</a>). As the axial position of the pistons change, the piston chambers draw and expel fluid.</p>
<p>However, I just watched a training video from "Our Virtual Academy" (which costs money to view unfortunately) where the pistons do not rotate with the shaft; instead, the swash plate rotates causing the pistons to move axially. The angle of the swash plate is varied by forcing pressurized fluid into the control chamber (here is an <a href="https://www.toyota-industries.com/company/business/automobile/compressor/kind_4/index.html" rel="nofollow noreferrer">example</a>).</p>
<p>I can't imagine why one would use this latter form of swash plate pump. What are the reasons for the different approaches? (What is the correct terminology to distinguish the two so I can carry on my own research)</p>
| |pumps| | <p>Following up on Sam's categories of pumps -</p>
<ol>
<li><p>There are industry examples of your first category - rotating swash plate + stationary barrel pumps. Here is one animated example of a pump that I've used personally - <a href="https://www.youtube.com/watch?v=sP3gc4b8Z5k" rel="nofollow noreferrer">https://www.youtube.com/watch?v=sP3gc4b8Z5k</a>. The relevant part of the video starts at 0:48. Pardon the marketing material. The pump's barrel <em>must</em> remain stationary because each piston chamber is connected directly to an outlet check valve. As they show at 1:30 in the video, the concept of a stationary barrel allows for separate flows from a specialized cover plate, which isolates 1 or more piston chambers from the others. A similar "isolation" concept is used by Bosch-Rexroth, Hydac, and HAWE for multi-chambered radial-piston pumps.</p></li>
<li><p>This is the most common variety of axial piston pump. The barrel / housing (valve plate) clearance does indeed allow leakage for pump cooling as you describe. Note that this is only 1 of 4 typical leakage sources in a rotating-barrel pump...
<a href="https://i.stack.imgur.com/ArIHk.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ArIHk.png" alt="enter image description here"></a>
(<span class="math-container">$Q_{lp}$</span>) piston / bore clearances, (<span class="math-container">$Q_{ls}$</span>) slipper / swashplate clearances, (<span class="math-container">$Q_{lv}$</span>) barrel / valve plate clearance, and (not pictured) swash plate control system return oil. </p></li>
<li><p>Good description on the bent-axis category. </p></li>
</ol>
<p>Overall your post categorizes axial-piston pumps correctly into the 3 general groups. However some of the descriptions are only true in specific situations. </p>
<ul>
<li><p>For pump category 1, you mention 120C as an upper operational limit. 120C is an extreme upper limiting point for petroleum-based hydraulic oil used in <em>any</em> type of axial piston pump. However there is no single number for temperature limit. It completely depends on the working fluid used with any specific pump design. Each fluid has a unique relationship between temperature and relevant fluid properties - viscosity, density, compressibility. Also, each pump may have interfaces with unique clearances optimized for a particular fluid. Pump failure often results from viscosity falling below some functional limit (1.0-10 cSt) due to temperature rise. Also pump failure can be caused by temperature alone in the case of elastomer seal degradation. Lastly note that bent-axis type pumps (category 3) may survive longest at elevated temperatures because they are designed without piston slippers for axial load support. Less sliding components that can fail.. </p></li>
<li><p>Category 1 is not inherently simpler or more efficient than the other categories. It's true that the barrel / valveplate sliding interface is eliminated in Category 1. However multiple new valves are introduced for periodic sealing of each piston chamber. Additional bearings are needed to support the side-load on the rotating swashplate. Also <em>far</em> more expense is involved to achieve variable-displacement operation. Lastly the weight of the rotating group has negligible influence on power consumption. Component weight only affects the accel/decel time needed to change speeds. Steady-state power consumption is a function of viscous fluid shear on surfaces, along with any power diverted for displacement control purposes. </p></li>
</ul>
| 28732 | Swash Plate Pump Design |
2019-06-10T19:03:28.043 | <p>I'm doing an experiment where 2 electrodes are inserted into 2 openings in a cylinder holding electrolytes to measure its impedance. We want the openings to be tightly sealed because the cylinder holding the electrolyte will be pressurized. Our idea is to insert the electrodes through the holes of tube fittings. The tube fitting will screw onto the threaded cylinder and it will lock into the electrodes. Will this work, and what type and size of tube fitting should I use for this? Please give any other suggestions. </p>
<p><a href="https://i.stack.imgur.com/K4X8U.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/K4X8U.jpg" alt=""></a></p>
<p>Here's a picture of what it looks like. The cylinder will be remade with only 2 holes. That's the electrode with a rod attached. We're hoping the tube fitting will dig into the rod. The diameter of the rod is about 1/4 in. </p>
| |pressure|pressure-vessel|seals| | <p>Further to my comment about connectors or fittings with olives, consider something like this as the olive will compress around the electrode and usually seal nicely.</p>
<p>These can be had with tapered or metric threads, you will need to tap the cylinder as needed. PTFE tape can be used on the thread into the cylinder and can also be put around the olive to aid the seal.</p>
<p><a href="https://i.stack.imgur.com/bDWSX.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/bDWSX.png" alt="enter image description here"></a></p>
<p>Another possibility is to use chamfered holes and a fixing plate so that those rubber seals you show get compressed onto the electrode.</p>
| 28747 | Are tube fittings a feasible way to create tight seal? |
2019-06-11T01:08:58.003 | <p>I am working on a personal project and wanted to use a solid oxide fuel cell to power up a load (I have not decided what yet), so I am trying to find a way to estimate the power I would need to heat up my fuel cell to a working temperature, so I was considering calculating the heat capacity of the fuel cell. But as it turns out the fuel cell is composed of different materials. Should it be enough to obtain the heat capacity from each material and then add each value to compute the overall energy needed?
Thanks!</p>
| |thermodynamics| | <p>This is a case where you want to determine a property in a composite using some form of a rule of mixtures.</p>
<p><span class="math-container">$$ P^n = \sum f_j P^n $$</span></p>
<p>Here, <span class="math-container">$P$</span> is the composite property, <span class="math-container">$f_j$</span> is typically the volume fraction, and <span class="math-container">$P_j$</span> is the property of the component <span class="math-container">$j$</span> in the mixture. When <span class="math-container">$n = 1$</span>, this is the simple rule of mixtures. When <span class="math-container">$n = -1$</span>, this is the inverse rule of mixtures.</p>
<p>An example of this rule applied to fiber composites is given at the link below.</p>
<p><a href="https://ipfs.io/ipfs/QmXoypizjW3WknFiJnKLwHCnL72vedxjQkDDP1mXWo6uco/wiki/Rule_of_mixtures.html" rel="nofollow noreferrer">https://ipfs.io/ipfs/QmXoypizjW3WknFiJnKLwHCnL72vedxjQkDDP1mXWo6uco/wiki/Rule_of_mixtures.html</a></p>
<p>For heat capacity, the formulation is written using mass specific heat capacity and mass fraction. The link below gives an explanation and a calculator.</p>
<p><a href="https://thermtest.com/rule-of-mixtures-calculator" rel="nofollow noreferrer">https://thermtest.com/rule-of-mixtures-calculator</a></p>
| 28749 | Is there a way to calculate the heat capacity of a solid or volumetric heat capacity of a solid oxide fuel cell? |
2019-06-14T03:53:00.317 | <p><a href="https://i.stack.imgur.com/CdcaR.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/CdcaR.jpg" alt="image"></a></p>
<p>What is the minimum value of <span class="math-container">$R_2$</span> (in <span class="math-container">$R_1$</span>) to minimize loading effect?</p>
<p>Please explain the concept.</p>
| |electrical-engineering|electrical| | <p>As described in <a href="https://www.electronics-tutorials.ws/filter/filter_2.html" rel="nofollow noreferrer">Passive Low Pass Filter</a> "<em>to reduce the loading effect, 10x from the previous stage implement the impedance such that R2 = 10 x R1 and C2 = 1/10th C1</em>".</p>
<p>Conceptually the target 2nd-Order Filter Corner Frequency and 2nd-Order Low Pass Filter -3dB Frequency will dictate the R and C component values. </p>
<p>In practice the ideal R and C valued don't exist in the open market, and engineers make compromises to design the best circuit. The calculation may require a resistor value of 11.6 Kohm and capacitor value of 157 micro farads. These components don't physically exists in the market. Therefore designers are required to make compromises. </p>
<p>Below are some equations and graphs to complete this discussion. </p>
<p>2nd-Order Filter Corner Frequency</p>
<p><a href="https://i.stack.imgur.com/GUOsS.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/GUOsS.png" alt="enter image description here"></a></p>
<p>2nd-Order Low Pass Filter -3dB Frequency</p>
<p><a href="https://i.stack.imgur.com/o4WEe.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/o4WEe.png" alt="enter image description here"></a></p>
<p>n is the order number, which 2 for this discussion. </p>
<p>Frequency Response of a 2nd-order Low Pass Filter</p>
<p><a href="https://i.stack.imgur.com/t2xj6.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/t2xj6.png" alt="*enter image description here*"></a></p>
| 28792 | LPF Second Order use cascade |
2019-06-14T10:46:06.467 | <p>I have in mind a device somewhere in between a firearms trigger and a clock escapement. One way it could be constructed is as follows:</p>
<ol>
<li>A crank would displace a linearly moving slider backwards, so that it compresses a spring.</li>
<li>Eventually the crank would let the slider slip, so that the spring shoots it forward, where it could again be caught by the crank on its next turn.</li>
</ol>
<p>However, this description is not essential to my question. I simply wish to deliver the idea to the reader, all because I do not have a name for it.</p>
<p>It is this name that I am seeking, so that I can further look it up, to hopefully find, study and re-use previously developed constructions of this kind.</p>
| |mechanical-engineering|kinematics| | <p>I remembered the concept but not the name so I searched a few terms and found that Wikipedia calls it a <a href="https://en.wikipedia.org/wiki/Quick_return_mechanism" rel="nofollow noreferrer">quick return mechanism</a>. Many such mechanisms are <a href="https://en.wikipedia.org/wiki/Linkage_(mechanical)" rel="nofollow noreferrer">linkages</a> (requiring no springs and catches, unlike your idea) and therefore may be called more specifically quick return linkages.</p>
| 28795 | How do I call a kinematic assembly that converts rotary motion to sawtooth linear? |
2019-06-14T15:24:14.323 | <p>I have thin, variable diameter PVC wires that run along a metal housing. Problem being, they can get pulled on and disconnect. I want to bond them down to the metal housing as a sort of strain relief but have not had success. There are holes in the metal so a glue would have to be pretty viscous. Taping over (tried VHB, teflon) still allows them to move around under the tape and instant adhesives (tried hot melt, acrylic) peel. Accepting all suggestions, and thanks in advance!</p>
| |mechanical-engineering|design|adhesive| | <p>If you don't want to use a mechanical fastener, I would suggest <em>encapsulating</em> them, by wrapping a thick adhesive around them and sticking it to the base material. You might try Sugru moldable silicone putty, or just a very thick (and non-corrosive!) RTV silicone. </p>
<p>Make sure to prepare surfaces well! Even slight fingerprints will trash most adhesive bond strengths!</p>
| 28800 | Bonding PVC Cables To Metal Housing |
2019-06-14T18:32:36.300 | <p>I have a quick question related to the topic of mechanics of materials.</p>
<p>If I were to have a beam that is fixed at both ends and has a central point force, do moments develop at the fixed supports?</p>
<p>e.g.</p>
<p><a href="https://i.stack.imgur.com/tVPCm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/tVPCm.png" alt="enter image description here"></a></p>
<p>I was thinking, wouldn't there be an absence of moments at the fixed supports since the balanced vertical reaction forces at both ends would prevent the development of a moment?</p>
| |mechanical-engineering|applied-mechanics|mechanical|solid-mechanics| | <p>When trying to figure out whether or not a given reaction will exist at a given support, it's worth remembering what a reaction actually is.</p>
<p>A reaction is the means by which the support resists the movement of the beam at that point. Force reactions resist the beam's attempts to deflect up-down or left-right at the support. Moment reactions resist the beam's attempts to rotate at the support.</p>
<p>So, let's assume there's no bending reaction at the beam's extremities. In that case, we'll be simply dealing with a simply-supported beam with pinned supports at the ends. These pinned supports will generate vertical forces which resist the beam's deflection at those points.</p>
<p>But how would the beam deflect in this case? Well, it'd obviously be shaped like a parabola (well, it'd actually be a cubic function, but it looks parabola-ish).</p>
<p><a href="https://i.stack.imgur.com/P3Dwv.png" rel="noreferrer"><img src="https://i.stack.imgur.com/P3Dwv.png" alt="enter image description here"></a></p>
<p>But that means there was rotation at the extremities: the beam was previously horizontal at the supports, but now it's tilted.</p>
<p>But the definition of a fixed support is that it doesn't allow for rotations at that point.</p>
<p>So what does the fixed support do to stop that rotation? Well, it applies a concentrated bending moment to the beam, rotating it back to horizontal:</p>
<p><a href="https://i.stack.imgur.com/HgWlb.png" rel="noreferrer"><img src="https://i.stack.imgur.com/HgWlb.png" alt="enter image description here"></a></p>
<hr>
<p><sub>Figures created with <a href="https://www.ftool.com.br" rel="noreferrer">Ftool</a>, a free structural analysis program.</sub></p>
| 28802 | Moments of Fixed-Fixed Beam |
2019-06-14T22:06:19.037 | <p>Would it be considered safe to use super glue, which contains ethyl cyanoacrylate, to attach a PVDF film to a metal alloy?</p>
<p>It says super glue is not recommended for polyethylene or polypropylene. However, I am not sure about the hazards for polyvinylidene fluoride. </p>
<p>Thanks! </p>
| |mechanical-engineering|materials|plastic|building-design|adhesive| | <p>The issue is not one of safety; it is just that PE, PP and PVDF are almost impossible to get anything to stick to, including glue. It is doubtful that crazy glue would work on it. you can try latex-based glue, this is what's used in stickytape. There are industrial-strength versions which fall under the category of pressure-sensitive adhesives. 3M makes these. </p>
| 28808 | Adhesion of Super Glue (containing ethyl cyanoacrylate) and PVDF (polyvinylidene fluoride) film |
2019-06-15T07:43:24.957 | <p>I am a language editor of academic papers in civil engineering (typically written in English by Poles). </p>
<p>I frequently encounter the term 'assumption' being used to describe a value that is adopted for the sake of calculations. For example, an author might not precisely know the original thickness of a corroded steel shell of a silo, but makes a decision about what it might have been in order to perform calculations/modelling. This would typically be written as something like: "assumption for thickness of steel shell = 6 mm".</p>
<p>I really feel uncomfortable with words like 'assumed' and 'assumption' on the basis that they are, almost by definition, unscientific. What would be the correct way to describe the above example?</p>
| |terminology| | <p>There are a lot of assumptions in science. Thing is most of us dont notice these assumptions because people leave them out. Does not mean assumptions are not there. </p>
<p>I can guarantee that a lot of the physics formulas you were given at any level of school had some simplifications associated with them, you just missed that section in the text.</p>
<p>Now if you introduce a new formulation of something it is prudent to tell what simplifications of reality you made. So using a sentence with a assumption is a perfectly normal way to do this. This way the reader knows wether or not the simplification is suitable for them. In fact leaving this out, except for cases where the assumptions are grandfathered from known formulas, is not very engineringly or scientifical for that matter.</p>
<p>Both engineering and science is to a great extent about honesty to yourself and others. Not documenting it as such is dishonest. There is nothing unscientific about simplifying reality, in fact that is what science is.</p>
| 28814 | Describing values as being 'assumptions' or 'assumed' - what is the correct approach in academic papers for civil engineering? |
2019-06-16T12:42:42.713 | <p>I am trying to understand this problem:</p>
<p><a href="https://i.stack.imgur.com/AaHM5.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/AaHM5.jpg" alt="enter image description here"></a></p>
<p>If we break down the 10T, they behave like this:</p>
<p><a href="https://i.stack.imgur.com/R4TRl.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/R4TRl.jpg" alt="enter image description here"></a></p>
<p>The right end of the beam can move as much as it wants, but the left end is pinned.</p>
<p>Why is there no horizontal reaction at A, if there is a horizontal part of the forces being applied on the beam axis?</p>
<p>I imagine, the force will push one section, and each section will push the next one, until the end, where it will create a reaction, just like the vertical forces are doing?</p>
<p>Thank you a lot.</p>
| |structures| | <p>You are more than overthinking this. Keep it simple. For an object at rest:</p>
<ul>
<li>ΣF(x) = 0 [sum of vertical forces = zero]</li>
<li>ΣF(y) = 0 [sum of horizontal forces = zero]</li>
<li><p>ΣMoments() = 0 [sum of moments about any point = zero]</p>
<p>Free body diagram of beam. </p></li>
</ul>
<p><a href="https://i.stack.imgur.com/2CmYj.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/2CmYj.jpg" alt="Free Body Diagram"></a></p>
<p>Again, ΣF(x) = 0. With the diagram drawn this way, what do you thing Force A (in the x direction) is?</p>
| 28828 | Pin support, why is there no horizontal reaction force? |
2019-06-18T08:51:55.563 | <p>When rubber (I don't know the specifics, but I assume it's butyl rubber) gets old, it turns into one of two states usually:</p>
<ul>
<li>It becomes sticky and gummy.</li>
<li>It becomes hard and brittle.</li>
</ul>
<p>I'd like to know if there's a way to change hardened or sticky rubber back into its original condition. The only thing I tried so far, is heating it up gently with a hot air gun to ~60°C which makes hard rubber more malleable.</p>
<p>I understand there are different kinds of synthetic rubbers, made of different polymers, and henceforth, there might be a range of chemicals that might be used to treat those various kinds. I couldn't find any information on such "pairings", for instance I was searching specifically to treat butyl rubber, but I couldn't find anything.</p>
<p>The air ducting in cars is made of some sort of hard rubber (I'm not sure what kind of polymer), and I've heard that Ballistol can be used to soften it, but it's hearsay, essentially, I'm not sure about that.</p>
<p>I'm not looking for a way to completely "rejuvenate" rubber parts like that, just to have them restored a bit, so they are easier to work with.</p>
<p>Same thing with sticky rubber. Is there a way to make it less sticky?</p>
| |chemistry| | <p>The embrittlement and goop-forming reactions are fundamentally different, and different measures would be needed to deal with each. </p>
<p>Rubber embrittlement occurs when the rubber oxidizes, and stops being rubber anymore because the rubber molecule chains get cut and the chemical bonds between adjacent molecules get broken. This degradation process is made worse by exposure to heat, ozone, and ultraviolet. Once it happens, you cannot repair the rubber part simply because it's no longer rubber. </p>
<p>The goop reaction occurs because to make rubber soft and flexible, it has mineral oil or other oils called <em>plasticizers</em> milled into it during its manufacture. The oil molecules form tiny globs of oil within the rubber matrix and interfere with the bonding between adjacent rubber molecules, letting them slip past one another instead of getting caught and pulling back against each other. With the passage of time, the oil sometimes tends to diffuse out of the bulk and accumulate at the surface, which renders the surface sticky. Sometimes the oil globs slowly dissolve into the bulk rubber and disconnect most of the molecules from one another, and the once-solid rubber object turns into something like tar and dribbles apart. </p>
<p>There is a substance called <em>ARMOR ALL</em> which can be used to rejuvenate decaying rubber. It contains a water emulsion of silicone oil that soaks into the rubber surface, but the improvement is just cosmetic since it can't re-connect broken chemical bonds.</p>
| 28847 | What chemical can be used to restore synthetic rubber? |
2019-06-18T12:49:14.670 | <p>Following situation: I need to mount a console holding a load to the edge of a 25cm concrete plate. I want you to check my reasoning. This is the situation:
<a href="https://i.stack.imgur.com/oZ364.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/oZ364.png" alt="enter image description here"></a></p>
<p>My thinking is: The load is 170kg, which amounts to 1.67 kN. The length of the cantilever is 600 mm. So the tension load on the upper two fasterners (of four, installed in square patterns) is a total of 1.67 kN * 600 mm / 150 mm = 6.69 kN.</p>
<p>Looking the <a href="https://www.hilti.de/medias/sys_master/documents/h89/h84/9342591827998/Technical-data-sheet-for-Hilti-HIT-1-and-Hilti-HIT-1-CE-Technical-information-ASSET-DOC-7567931.pdf" rel="nofollow noreferrer">specifications of a likely system</a>, I see that 2 M10 fasteners at embedment depth 100mm offer me 2 * 8.7 kN tensile load, which contains decent margin of safety. The shear load per fastener is in the same range, there's 4 fasteners, so even considering 10 or 20 kg for the console (likely too much) this is ok.</p>
<p>I'm out of school too long. Is the above correct?</p>
| |structural-engineering|fasteners| | <p>You'd have to check for these factors.:</p>
<ul>
<li><p>The fastener must penetrate into and fully develop its strength inside the steel cage tie bar or L shaped rebar inside the concrete which is a continuous part of its reinforcement. Concrete alone is susceptible to cracking under tension and is not reliable.</p></li>
<li><p>Dynamic load factor. A load of P, even if lowered gently onto a support has 2 to 3 times its weight impact factor. Because it needs to deforms the member to its deflected shape and beyond and back to equilibrium. So the load factor of 3 is advisable.</p></li>
</ul>
| 28849 | Correct sizing of fastener for console |
2019-06-18T22:07:38.527 | <p>Why do some guyed broadcast towers come to a point at the base. It seems that it would be more stable if each vertical bar was anchored separately rather than being balanced at one point. </p>
<p><a href="https://i.stack.imgur.com/xBs0Y.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/xBs0Y.jpg" alt="enter image description here"></a></p>
| |structural-engineering|structures| | <p>If the base of the tower is "rigidly" fixed to the ground, the foundations have to withstand the bending moments from the side loads (e.g. wind) on the tower. Otherwise, the foundations will work loose and not support the tower at all. Depending on the local geology, that may mean deep foundations to support a tall tower. Foundations are expensive, because unlike the tower itself they can't be prefabricated to a standard design off site and delivered by truck!</p>
<p>By using guy ropes and a tower base as shown, the only force on the foundation is downwards, and if the foundation gradually sinks into soft ground over time, that can be compensated for cheaply by adjusting the guy ropes.</p>
| 28856 | Why do some guyed towers come to a point at the bottom? |
2019-06-19T15:21:18.067 | <p>As per Bryton Cycles, most of TurboJet engines' Thermodynamic efficiency should be around 40%. Rest of energy gets rejected as waste heat in exit gases of after the turbine. </p>
<p>But if these hot gases from the turbine are passed through Nozzle of 95% efficiency (either Convergent or CD), then overall efficiency should be 95% Right? Because Nozzle is scavenging all the waste heat and pressure, converting it into Kinetic energy of gases. </p>
<p>As per my knowledge <strong>nozzles are extremely efficient devices with efficiencies above 95%</strong> (Please do comment if that's not the case). And for the sake of discussion please assume Combustion efficiency as 100% and no mechanical loss. And here I'm considering whole purpose of turbojet engine it propell the gases as fast as possible. So overall efficiency here is kinetic energy of exit gases vs chemical energy released by fuel.</p>
| |mechanical-engineering|fluid-mechanics|thermodynamics|jet| | <p>The complete Brayton cycle of an aircraft gas turbine engine has a couple of sources of energy loss - loss from the propulsive thrust power that is. The nozzle you mention is only part of the whole cycle, and yes can be very efficient if shaped properly.</p>
<p><a href="https://i.stack.imgur.com/opHF3.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/opHF3.png" alt="from an old uni book, English terms are my own translations"></a></p>
<p>Above image quantifies the energy streams for a bypass engine with a high bypass ratio. Every stage in energy transformation comes with a loss, quantified in the stage efficiency. The stages in the energy transformation are:</p>
<ol>
<li>Chemical energy => heat. Has a very high efficiency, heat radiation to the environment is minimal, as is CO and soot formation.</li>
<li>Heat energy => gas power. We're converting heat into useful power, which can never be fully accomplished: after full expansion, gas temperature has increased.</li>
<li>Gas power => propulsive power. Either into shaft power for turboprops/turboshafts, or into kinetic power for jet engines. Cannot occur loss free (isentropically) due to heat transfer, friction losses etc.</li>
<li>Propulsive power (of the medium = air, combustion gasses) into thrust power (of the aircraft). The medium is propelled with a higher velocity than that of the aircraft, which is an efficiency loss. A necessary one of course.</li>
</ol>
<p>End efficiency of the whole cycle is about 34%, probably a bit higher nowadays since the depiction is several decades old. Notice that the jet nozzle (exhaust) mentioned in the OP is only one aspect, with a high efficiency indeed. </p>
| 28867 | Efficiency of TurboJet engines with nozzles |
2019-06-21T08:38:02.440 | <p>I'm struggling to identify which equation is used within the first red square. </p>
<p>Could someone please help identify which equation was used as well as why?</p>
<p><a href="https://i.stack.imgur.com/SXlm7.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/SXlm7.png" alt="Reference"></a></p>
<p>Thankyou!</p>
| |mechanical-engineering|structural-engineering|civil-engineering| | <p>The calculation for <span class="math-container">$e$</span> that you've boxed is essentially arrived at through geometry. We simply want to find the distance from the centroid where total stress is zero. Drawing the axial, bending, and total stress along the cross-section makes this clear.</p>
<p><a href="https://i.stack.imgur.com/gd9Co.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/gd9Co.jpg" alt="stress distribution"></a></p>
<p><span class="math-container">$$\begin{align}
\sigma_{axial} &= \frac{P}{A} \\
\sigma_{flexure} &= \frac{My}{I} \\
\sigma_{total} &= \sigma_{axial} + \sigma_{flexure}
\end{align}$$</span></p>
<p>So, we can readily calculate <span class="math-container">$e$</span> based on the slope of the flexural stress. I tend to phrase it in my head as "at what distance <span class="math-container">$e$</span> from the centroid does the flexural stress perfectly counteract the axial stress?"</p>
<p>Find <span class="math-container">$e$</span> such that </p>
<p><span class="math-container">$$\sigma_{axial} = \frac{2\sigma_{flexure}}{d}e$$</span></p>
<p>Where <span class="math-container">$d$</span> is the depth of the section and <span class="math-container">$\sigma_{flexure}$</span> refers to the stress at the extreme fiber (distance <span class="math-container">$d/2$</span> from the centroid). Rearranging that expression, and substituting, we see the equation used in the given solution.</p>
<p><span class="math-container">$$e = \frac{\sigma_{axial} \cdot \frac{d}{2}}{\sigma_{flexure}} = \frac{\sigma_{axial} \cdot I}{M}$$</span></p>
<p>It's worth emphasizing, that from the standpoint of developing understanding, memorizing their particular expression for <span class="math-container">$e$</span> is not all that useful. It's more important to understand what's happening - that we're simply adding axial and flexural stress, and finding the location where total stress is zero.</p>
| 28891 | Max Tensile & Compressive Stress Question - Identify the equation |
2019-06-21T15:10:34.193 | <p>I am trying to get an idea of the amount of hydrogen that I can store in some pressurized steel tanks underground.</p>
<p>Based on the ideal gas law PV=nRT</p>
<p>I did a simple calculation
pressure = 200bar
volume = 2000Liters
temp = 25C</p>
<p>My results were about
16136.7 moles of hydrogen</p>
<p>1 mole of any gas is about 22.4 liters but that's at standard temperature and pressure.</p>
<p>At these elevated pressure it should be about 361,671 liters?</p>
| |thermodynamics|pressure|gas|pressure-vessel|compressed-gases| | <p>Theory: </p>
<p>1 mole at those conditions takes up 0.123 l. Why? If 16,136.7 moles of hydrogen take up 2,000 l, then 1 mole will take up 0,123 l. You can also apply the ideal gas equation to check it: </p>
<p><span class="math-container">$V=n*R*T/p=1 [mole]*0.082 [atm*l/mol*K]*298.15 [K]/[(200*10^5/101,325) [atm]]=0.123 [l]$</span></p>
<p>And what about reality? Well, ideal gas equation fails at such high pressures because it doesn't regard the space occupied by the mollecules of the gas (important at those high pressures becuase collisions are more frequent and repulsion forces more intense, therefore there's an increase in volume). Note: This is true for hydrogen, but not for all real gases.</p>
<p>Let's correct the ideal gas equation with the compressibility factor <span class="math-container">$z$</span>:</p>
<p>Now we'll have <span class="math-container">$p*V=z*n*R*T$</span> </p>
<p>At 20 MPa (200 bar) and 25 ºC we have <span class="math-container">$z=1.124$</span> (mean value between 1.0601 and 1.1879, highlighted down here):</p>
<p><a href="https://i.stack.imgur.com/EdQNr.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/EdQNr.jpg" alt="z of hydrogen for different temperatures and pressures. Source: https://h2tools.org/hyarc/hydrogen-properties"></a></p>
<p>Source:<a href="https://h2tools.org/hyarc/hydrogen-properties" rel="nofollow noreferrer">Hydrogen compressibility at different temperatures and pressures</a></p>
<p>So let's do the math:</p>
<p><span class="math-container">$V=z*n*R*T/p=1.124*1 [mole]*0.082 [atm*l/mol*K]*298.15 [K]/[(200*10^5/101,325) [atm]]=0.139 [l]$</span></p>
<p>And 2,246.7 l if you consider the 16,137.7 moles. So prepare a bigger tank for your hydrogen!</p>
| 28894 | Ideal gas law, theory vs reality |
2019-06-21T17:32:21.510 | <p>I'm seeking an explanation in reference to the photo below. I'm unsure how the solution within the red box calculates the max tension of the Aluminium beam and the max compression off the Aluminium beam.</p>
<p>How are you able to differentiate whether the Aluminium is in tension or compression within this scenario?</p>
<p>As well as how they are able to make the statement that <em>"the steel is entirely compressive bending stress and the max bending stress is at the top of the flange"</em>
<a href="https://i.stack.imgur.com/rahAI.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/rahAI.png" alt="enter image description here"></a></p>
<p>Thankyou!</p>
| |mechanical-engineering|structural-engineering|civil-engineering| | <p>It has to deal with the location of the Neutral Axis (NA). That whole calculation for Ῡ =37 mm is key here. Bending occurs about the NA. That big yellow arrow represents the moment in the beam, or in other words which direction it is bending. Based on the moment arrow everything above the NA will be compression, and everything below it will be tension.</p>
<p>The calculations show the NA at a distance of 37 mm from the bottom and from the cross section dimensions we see the aluminum section is 40 mm deep. This means all the steel is above the NA which in turns means all the steel for this sections is in compression. The portion of the aluminum between the NA and the steel is also in compression. The maximum stress occurs at the furthest distance from the neutral axis.</p>
| 28897 | Max compression & Tension |
2019-06-22T01:02:43.343 | <p>Im looking for clarification on how the solution identified the moment to be hogging, as well as the axial stress to be in compression?</p>
<p>Also I'm confused as to how the solution came to a conclusion that the bottom of the cross section is compressive stress, whilst the top of the cross section is tensile stress.</p>
<p>Please reference the photo below with the red box's</p>
<p><a href="https://i.stack.imgur.com/RwzG1.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/RwzG1.png" alt="enter image description here"></a></p>
<p>Thankyou!</p>
| |mechanical-engineering|structural-engineering|civil-engineering| | <p>Once again, the <a href="https://en.wikipedia.org/wiki/Superposition_principle" rel="nofollow noreferrer">superposition principle</a> is our friend. It basically means we can look at each load in isolation, find the individual results, and then add them all together to get the final result.</p>
<p>In this case, we are actually dealing with two loads:</p>
<ul>
<li>There's obviously the force <span class="math-container">$P$</span>;</li>
<li>And then there's the bending moment <span class="math-container">$P\cdot e$</span>, where <span class="math-container">$e$</span> is the eccentricity of the load <span class="math-container">$P$</span> in relation to the beam's neutral axis. Since the beam is rectangular with height 200 mm, we trivially know the neutral axis is at 100 mm. And since the load is only 30 mm from the bottom, we know <span class="math-container">$e=100-30=70\text{ mm}$</span>.</li>
</ul>
<p>Now let's look at these loads in isolation:</p>
<h2>The force <span class="math-container">$P$</span></h2>
<p>This one is pretty obvious. The force alone (remember, we're not yet considering the bending moment effect, so you can think of the force as acting at the neutral axis) obviously causes uniform compression throughout the beam. This means the beam remains perfectly straight, just a bit shorter. Not much to discuss here, I believe.</p>
<h2>The bending moment <span class="math-container">$P\cdot e$</span></h2>
<p>Herein lies the rub.</p>
<p>The bending moment is counter-clockwise in this case. This seems pretty intuitive to me, so I won't spend any time demonstrating this.</p>
<p>The issue is how the beam deflects given this bending moment.</p>
<p>Looking at the free end, it can rotate in one of two ways: clockwise or counter-clockwise. It once again seems pretty intuitive to me that the beam will rotate counter-clockwise, but here's a demonstration, just in case it isn't to you:</p>
<p>Let's transform the bending moment into a <a href="https://en.wikipedia.org/wiki/Couple_(mechanics)" rel="nofollow noreferrer">force couple</a>: two opposing forces which cancel each other out other than generating a bending moment. A classic example is a <a href="https://i.stack.imgur.com/jEAOz.png" rel="nofollow noreferrer">crank valve</a>.</p>
<p><a href="https://i.stack.imgur.com/lc04I.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/lc04I.png" alt="enter image description here"></a></p>
<p>So, if you had a crank valve at the end of a beam, and you applied the forces shown above, how would you expect it to rotate? It seems clear the beam would rotate counter-clockwise. By this I mean that the free end (and indeed any other slice along the beam) would rotate counter-clockwise, such that the top fiber moves a bit to the left and the bottom fiber moves to the right.</p>
<p>And that tells us all we need to know: if the top fiber moves to the left, then the total length of that fiber has increased (from 700 mm to... 701 mm or whatever). And we know that if an element is lengthened, it must be under tension<sup>1</sup>. Likewise, if the bottom fiber moved to the right, that fiber's length shortened and it is therefore under compression.</p>
<h2>Total result</h2>
<p>So, summing both loads' results, we get that the top fiber is under compression due to the force and under tension due to the bending moment. Whether the final result is compression or tension will depend on which stress is greater.</p>
<p>The bottom fiber is under compression due to both loads, so that's simple enough.</p>
<h2>Hogging</h2>
<p>As for why the moment is considered hogging, well, that's just the technical definition of hogging moment.</p>
<blockquote>
<p>When the beam bends in such a way that it forms concavity downwards (cup-shaped) it is called as sagging. Whereas the bending which results in convexity upwards (like a hump); is known as hogging. <a href="https://civilengineer.webinfolist.com/mech/bm.htm" rel="nofollow noreferrer">[source]</a></p>
</blockquote>
<p>Basically, positive moment is sagging, negative moment is hogging.</p>
<p>And if you remember your sign conventions:</p>
<p><a href="https://i.stack.imgur.com/OXLrT.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/OXLrT.png" alt="enter image description here"></a></p>
<p>You can see that counter-clockwise moments on the left side of the beam are negative, and therefore hogging.</p>
<hr>
<p><sup>1</sup> Or possibly under a positive thermal load, if you want to be a pedant.</p>
| 28904 | Identifying Compression and Tension as well as Hogging |
2019-06-23T01:23:06.900 | <p>I am trying to calculate <em>mass</em> fractions of a Hydrogen Oxygen gas mixture as a function of equivalence ratio before ignition. My work is the following:</p>
<p>Equivalence Ratio: <span class="math-container">$$\phi=\frac{FA}{FA_s}=\frac{\frac{m_f}{m_{air}}}{\frac{m_{fs}}{m_{air,s}}}=\frac{\frac{n_f}{n_{air}}}{\frac{n_{fs}}{n_{air,s}}}$$</span></p>
<p>Where <span class="math-container">$FA$</span> is the fuel to air ratio and a subscript <span class="math-container">$s$</span> stands for the stoichiometric reaction.</p>
<p>The balanced combustion reaction equation (assuming a perfectly stoichiometric mixture) is:</p>
<p><span class="math-container">$$2H_2+1O_2\rightarrow2H_2O$$</span></p>
<p>Thus, <span class="math-container">$n_{fs} = 2$</span> and <span class="math-container">$n_{air,s}=1$</span>. Now, when I write the mass fractions as a function of equivalence ratio, my work is:</p>
<p><span class="math-container">$$y_{O_2}=\frac{m_{air}}{m_f+m_{air}}=\frac{1}{FA+1}=\frac{1}{\phi\frac{n_{fs}}{n_{air,s}}+1}=\frac{n_{air,s}}{\phi n_{fs}+n_{air,s}}=\frac{1}{2\phi+1}$$</span></p>
<p>By a similar calculation:</p>
<p><span class="math-container">$$y_{H_2}=\frac{n_{fs}\phi}{n_{fs}\phi+n_{air,s}}=\frac{2\phi}{2\phi+1}$$</span></p>
<p>A quick check that should be true reveals that <span class="math-container">$y_{H_2}+y_{O_2}=1$</span> </p>
<p>The reason I am asking about this calculation is that I did this calculation for a previous project and the results the class said was correct was the reverse (i.e. the expression with <span class="math-container">$\phi$</span> in the numerator was for <span class="math-container">$y_{O_2}$</span> not <span class="math-container">$y_{H_2}$</span>. Is there something I did wrong/did I calculate them the wrong way round? The code I am basing my previous work off of (written in MATLAB) is: </p>
<pre><code> yO = 32/(4*phi+32);
yH = 4*phi/(4*phi+32);
</code></pre>
<p>The 32 crops up because the molar mass of <span class="math-container">$O_2$</span> was in the expression for some reason and I have not been able to figure out exactly how the above code expressions were derived. I just know they gave results that were marked correct for the class I did this for.</p>
<p>EDIT: I have figured out the source of my confusion. The statement</p>
<p><span class="math-container">$$\phi=\frac{FA}{FA_s}=\frac{\frac{m_f}{m_{air}}}{\frac{m_{fs}}{m_{air,s}}}=\frac{\frac{n_f}{n_{air}}}{\frac{n_{fs}}{n_{air,s}}}$$</span></p>
<p>is true by itself, but this does not mean that <span class="math-container">$FA_s=\frac{n_{fs}}{n_{air,s}}$</span>! I made this assumption in my head after I canceled out the molar masses used to transform the masses into moles.</p>
<p>The correct expressions are: <span class="math-container">$$y_{O_2} = \frac{n_{o,s}M_O}{\phi n_{f,s}{M_f}+n_{o,s}M_O}$$</span></p>
<p>and </p>
<p><span class="math-container">$$y_{H_2} = \frac{\phi n_{f,s}M_f}{\phi n_{f,s}{M_f}+n_{O,s}M_O}$$</span></p>
| |combustion| | <p>The derivation has a mistake in setting <span class="math-container">$FA = \phi n_{fs}/n_{O2s}$</span>. Here is the correction. For simplicity, I use <span class="math-container">$F$</span> rather than <span class="math-container">$FA$</span>, <span class="math-container">$m_O$</span> and <span class="math-container">$n_O$</span> as the mass and moles, respectively, for O<span class="math-container">$_2$</span>, and a <span class="math-container">$\star$</span> for stoichiometric. I use the term <span class="math-container">$\omega$</span> for mass fraction and <span class="math-container">$M$</span> (capital M) for molar mass.</p>
<p><span class="math-container">$$ \phi \equiv \frac{F}{F_s} = \frac{m_f/m_O}{m^\star_f/m^\star_O} $$</span></p>
<p><span class="math-container">$$ \omega_O \equiv \frac{m_O}{m_f + m_O} = \frac{1}{F + 1} $$</span></p>
<p><span class="math-container">$$ F = \phi\ F_s = \phi\ \frac{m^\star_f}{m^\star_O} = \phi\ \frac{n^\star_f\ M_f}{n^\star_O\ M_O}$$</span></p>
<p>For the given reaction, <span class="math-container">$n^\star_f = 2$</span> and <span class="math-container">$n^\star_O = 1$</span>. Therefore</p>
<p><span class="math-container">$$ \omega_O = \frac{1}{2\phi M_{f/O} + 1} $$</span></p>
<p>where <span class="math-container">$M_{f/O}$</span> is the ratio of molar mass of fuel (H<span class="math-container">$_2$</span>) to oxygen O<span class="math-container">$_2$</span>.</p>
<p>The expression for <span class="math-container">$\omega_f$</span> is found from a mass balance <span class="math-container">$\omega_O + \omega_f = 1$</span>.</p>
| 28921 | Calculating Mass Fractions of a Hydrogen and Oxygen Gas Mixture Pre Ignition |
2019-06-24T09:25:43.303 | <p>sorry for my english. I was looking to buy a house. I saw some in my hometown, which is very affordable.
but on a picture of the house they are selling I found on the net troubled me.
no pillar/column, just hollowblocks and some steel bars.<a href="https://i.stack.imgur.com/Rl8eC.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Rl8eC.jpg" alt="enter image description here"></a>
kindly look at the pic. </p>
<p>I found it in this site
<a href="https://fiestacommunitiesblog.wordpress.com/tag/fiesta-communities/" rel="nofollow noreferrer">https://fiestacommunitiesblog.wordpress.com/tag/fiesta-communities/</a></p>
| |civil-engineering|building-design| | <p>As @SamFarjamirad states it's best to consult a knowledgeable construction engineer where you live, or in your region, because such people would know the building codes & construction practices for your region. A column or pillar for a two storey house may not be necessary if the load bearing walls are designed & built properly. This may include internal load bearing wall, inside the house. The other thing to be wary of is the quality of the foundations for the load bearing walls. Also, if the house to be renovated later <strong>all load bearing walls must not be removed or altered</strong>.</p>
<p>It looks like the steel is being threaded through the hollow block in the external walls. This would be done to increase the flexibility of the house during earthquakes and should a typhoon/cyclone/hurricane affect the house. Internal load bearing walls should also have such steel reinforcement. </p>
<p>If designed and built correctly such steel reinforced walls <strong>should not</strong> collapse. They might crack during extreme natural events but they <strong>should not</strong> fall and collapse the house.</p>
<p>Looking at the picture you included, some things that I found strange were:</p>
<ul>
<li>Looking at the bricks surrounding the doorway, the bricks on the left
side of the doorway vary in width and they are not aligned vertically
very well, so there are gaps of varying sizes between the bricks and
the door frame. It would be better if the bricks were uniform in size
and vertically aligned.</li>
<li>It might be the way the picture has been taken but the bricks at the
top of the window holes in the left of picture look like they may be
sagging. They should be horizontal and supported by a solid lintel.</li>
<li>I'm used to seeing bricks laid in an interlocking pattern where the
bricks that are laid above one layer are slightly offset. This may
not be possible if the walls are threaded with steel reinforcing. Ask
a local engineer about this. In the picture, the bricks are laid vertically on top of one another.</li>
<li>I looked at the website you provided a link to and the second picture
showed what appear to be how the roofing iron was attached to the
house. Just laying the roofing iron on the walls as shown I consider
to be a primitive form of construction.</li>
</ul>
| 28933 | Is it OK for a 2 story house not have a column or pillar |
2019-06-24T09:47:22.757 | <p>I'm looking at designing a small gadget (think something like a Fitbit in size) where the case has a couple of buttons on it. For various reasons the buttons have to be around the narrow side (edge) of the case. I am fairly lost as to what the standard button mechanical design for this type of application would be. I mean, I know about (much bigger) buttons with a cantilever or a linear spring or spring and pivot, or rubber dome switches, but here... the whole thing it tiny! The largest buttons that can fit are roughly 12x5mm, and the "mechanism" has to fit in a few mm of thickness.</p>
<p>The prototype has oval plastic buttons which are ~12x5mm and 4mm deep, and C&K PTS840 tactile switches on a PCB behind them. These switches have a tiny little plunger 1.7x1mm with 0.2mm of travel that the buttons can hit. In general the buttons can push the plunger just fine. However there isn't really anything keeping the buttons in place or restricting them to move in a straight line, they kinda rattle around in holes in the case and can both tilt / jam at an angle and fall out.</p>
<p>I think this is probably a solved problem for cell phones / smartwatches / whatever, but I don't know what the standard solutions are. Could you please let me know what the major different of approaches are and what their pros and cons are?</p>
<p>P.S. This would be designed for injection molding in the future; PC/ABS material. Wall thickness is around 1.2-1.4mm.</p>
<p><strong>UPDATE</strong> I took apart some sports watches; they all have buttons with a metal spring loaded plunger that goes through a O-ring seal and hits a tactile switch. The total height of the mechanism is about 5mm (from the outside surface of the button to the switch on the PCB). That's a good mechanism (and waterproof) but I don't have room to fit something like that; and it probably can't be easily made out of just plastic parts. I'm still looking for a good solution.</p>
| |mechanical-engineering|mechanisms|plastic|molding| | <p>As you mentioned in your update you're going to want to start with a PCB edge mounted tactile switch. These can be quite small. So now what you've got to do is add a button.</p>
<ul>
<li><p>A common solution would be to use a small plastic button with a
retaining flange. You've said you're worried about this rattling or
jamming, but 1. the travel is very short, so it won't jam. And 2. The
button will be so small that the rattling will be inaudible. The buttons in your prototype sound like they were designed with too much linear travel.</p></li>
<li><p>Another option is to make the button part of the case, and retain it
using a living hinge. For a cheap product this is probably the way to
go, because it eliminates a part.</p></li>
<li><p>A 3rd, weirder option is to make the case flexible enough that the
button can be pressed <em>through it</em>. Unfortunately this is not well suited to an edge button.</p></li>
</ul>
<p>A flexible sticker or silicone coating can be added over the preceding assemblies to add some water/dust resistance.</p>
| 28934 | Push button design for small gadget |
2019-06-24T12:11:55.853 | <p>So I am learning for thermodynamics now and there is one thing that is bugging me.
In the book it shows a laval nozzle (very simplified) and a diffuser. </p>
<p><a href="https://i.stack.imgur.com/OljO2.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/OljO2.jpg" alt="laval nozzle"></a>
<a href="https://i.stack.imgur.com/aOp0T.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/aOp0T.jpg" alt="diffuser"></a></p>
<p>with the formulas:
<a href="https://i.stack.imgur.com/TkWUi.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/TkWUi.jpg" alt="formulas"></a></p>
<p>(a is the speed of sound)</p>
<p>According to those images and its explanation that would mean, if I make a paper cone with a hole in it and blow into it, at the hole speed of sound would be reached. That is obviously not the case. So what am I missing?</p>
<p>At the beginning it says "Isentropic flows" but how does one actually know that the flow is isentropic? Or is it because of turbulence?</p>
| |thermodynamics|airflow| | <blockquote>
<p>According to those images and its explanation that would mean, if I make a paper cone with a hole in it and blow into it, at the hole speed of sound would be reached. That is obviously not the case. So what am I missing?</p>
</blockquote>
<p>You are not blowing hard enough. There needs to be enough pressure drop from the big part of the cone to atmospheric to accelerate the air to the speed of sound. Do the math to figure out how hard to blow, and blow that hard. (You may want to figure out the bursting strength of your cone. And your lungs).</p>
| 28935 | What happens in a cone-shaped nozzle? |
2019-06-26T17:14:35.287 | <p>I am trying to create a program which estimates the final temperature of a magnet made of copper wire. I have different configurations in mind, but I will start off with a straight square wire with a circular cooling channel inside of it. Inside runs water which cools it off. The coil is wrapped in some type of electrical insulator. The wire is exposed to air. I am trying to model this using unsteady-state conditions and run it up to a steady state position to calculate the final temperature of the inside (middle if it makes it easy) of the wire, the temperature of the outside of the wire as well as the final temperature of the water coolant. </p>
<p>I have been at this, and coming up with some equations, but I have not taken heat transfer before and am struggling somewhat on this problem. I know that for unsteady state heat transfer </p>
<blockquote>
<p>∂T/∂t = α ∇^2(T)</p>
</blockquote>
<p>where the partial of temperature versus time is the laplacian of temperature times a factor alpha. There are some numerical methods to go about doing this and I am able to include a convective heat transfer coefficient, but I am not sure how to add in the rest of the geometries nor how to include a heat generation term. It just seems like there re so many things that have to be simultaneously solved and I am not sure where to go. I am not sure if the textbook I am reading is not sufficient or if I am not seeing where to go in general. Thanks in advance. </p>
<p><img src="https://i.stack.imgur.com/wJSoT.jpg" alt="general sketch"></p>
<p>The wire is square with an inner cooling channel. A layer of insulation wraps around the wire which may be neglected for now. I'm trying to reconcile both the convection from inside the wire, the convection outside the wire and the conduction inside the wire. No temperature is fixed which makes this very difficult for me to try to design around.</p>
| |mechanical-engineering|fluid-mechanics|heat-transfer|thermal-conduction|convection| | <h1>Foundations</h1>
<p>The heat transfer problem has three considerations. First is steady state versus unsteady state. The former case is far easier and is all that I will consider here. Second is the temperature profiles axial in <span class="math-container">$z$</span> and radial in <span class="math-container">$r$</span>. The best approach is to presume that separation of variables will work. Finally, the last consideration is the influence of heat transfer internally and externally to the cylinder while generating uniform power in the shell <span class="math-container">$P = i^2R$</span>.</p>
<p>Also, for simplicity, neglect the (electrical) insulation layer on the outside of the wire.</p>
<p>The starting point is to build each of the cases one step at a time.</p>
<h1>Axial versus Radial Gradients - Biot Number Estimates</h1>
<p>In the ideal case, we allow for separation of variables to make <span class="math-container">$T(r,z) = Tr(r)Tz(z)$</span>. A starting point for this is to determine the effective Biot numbers for radial heat transfer.</p>
<p><span class="math-container">$$Bi_{rw} = h_w \Delta R/k \hspace{1cm} Bi_{ra} = h_a \Delta R/k $$</span></p>
<p>In these, <span class="math-container">$h_w, h_a$</span> are the convection coefficients of the flowing water or surrounding air, <span class="math-container">$\Delta R$</span> is the outer minus inner radius <span class="math-container">$\Delta R = r_o - r_i$</span>, and <span class="math-container">$k$</span> is the thermal conductivity of the wire. When <span class="math-container">$Bi < 0.1$</span>, the wire functions in a lumped analysis. The radial gradients can be neglected.</p>
<h1>Axial Gradients Only</h1>
<h2>Water Cooling Only</h2>
<p>Consider the system for the case that <span class="math-container">$Bi_{rw} < 0.1$</span> with perfect thermal insulation around the wire. This will give the maximum temperature that can be expected in the wire. The energy balance along the flow says that total power input <span class="math-container">$P$</span> must be taken out by the flowing fluid (water) with mass flow <span class="math-container">$\dot{m}$</span> and specific heat <span class="math-container">$\tilde{C}_p$</span>. The fluid temperature changes from input <span class="math-container">$T_{wi}$</span> to output <span class="math-container">$T_{wf}$</span>.</p>
<p><span class="math-container">$$ \dot{m}\tilde{C}_p(T_{wf,max} - T_{wi}) = P$$</span></p>
<p>The subscript "max" indicates the case of maximum output temperature on the fluid. You can also use this to establish that, in this case, the temperature profile in the fluid is linear along the axis.</p>
<p><span class="math-container">$$ \dot{m}\tilde{C}_p(T_w(z) - T_{wi}) = P \frac{z}{L}$$</span></p>
<p>or</p>
<p><span class="math-container">$$ T_w(z) - T_{wi} = P\left(\frac{z}{L}\right)\left(\frac{1}{\dot{m}\tilde{C}_p}\right)$$</span></p>
<p>The differential energy balance on the wire with no radial gradients and with the external wall as a perfect thermal insulator is as below with <span class="math-container">$r_i$</span> as the radius of the channel.</p>
<p><span class="math-container">$$ h_w 2\pi r_i\left(T_z(z) - T_w(z)\right) dz = P \frac{dz}{L}$$</span></p>
<p>This gives the temperature in the wire as below with <span class="math-container">$A_i$</span> as the area of the water channel.</p>
<p><span class="math-container">$$ T_w(z) - T_{wi} = P \left(\frac{z}{L}\right)\left(\frac{1}{\dot{m}\tilde{C}_p} + \frac{1}{h_w A_i}\right)$$</span></p>
<h2>Including External Convection to Air</h2>
<p>When we add convection to air, the energy balance to the internal fluid (water) written in differential form is as below with <span class="math-container">$r_o$</span> as the external radius of the wire and <span class="math-container">$T_a$</span> as the air temperature.</p>
<p><span class="math-container">$$ \dot{m}\tilde{C}_p dT_w + h_a 2\pi r_o (T_z(z) - T_a) dz = P \frac{dz}{L}$$</span></p>
<p>The differential energy balance on the wire at any axial position <span class="math-container">$z$</span> is the equation below with <span class="math-container">$r_i$</span> as the channel radius.</p>
<p><span class="math-container">$$ h_w 2\pi r_i(T_z(z) - T_w(z))dz + h_a 2\pi r_o(T_z(z) - T_a)dz = P \frac{dz}{L}$$</span></p>
<p>These are two coupled differential equations to solve for <span class="math-container">$T_w(z)$</span> and <span class="math-container">$T_z(z)$</span>. In all cases, the water and wire temperatures will be lower than the case for water cooling only because a portion of the power generation is now removed by the air.</p>
<h1>Radial Gradients</h1>
<p>The radial gradient problem is easily solved for a hollow cylinder with fixed internal and external wall temperatures <span class="math-container">$T_{ci}$</span> and <span class="math-container">$T_{co}$</span>. The steady state profiles are well-documented in textbooks on heat transfer (see Chapter 17 in <a href="https://www.wiley.com/en-us/Fundamentals+of+Momentum%2C+Heat%2C+and+Mass+Transfer%2C+Revised+6th+Edition-p-9781118947463" rel="nofollow noreferrer">Welty, et.al</a>). At any position <span class="math-container">$z$</span>, the radial temperature profile will have the form below.</p>
<p><span class="math-container">$$ \frac{T_r - T_{co}}{T_{ci} - T_{co}} = \frac{\ln(r/r_o)}{\ln(r_i/r_o)} $$</span></p>
<p>Following the format of separation of variables, the radial gradient is superimposed on the axial profile from above.</p>
<p>The link between the wall temperatures <span class="math-container">$T_{ci}, T_{co}$</span> and the water or air temperatures occurs as convection. This is an entirely separate issue to tackle. It can only be done once a proof is established that separation of variables is valid.</p>
<h1>Unsteady State</h1>
<p>This level of analysis is best done step-wise. Include time on the axial profile with insulated boundary, then include time on the axial profile including air convection, and finally consider time only on the radial boundary.</p>
| 28958 | Heat Transfer in Different Wire Configurations |
2019-06-29T04:55:39.220 | <p>Is there any material that will absorb very low frequency radio light? I know that stealth aircraft use radar absorbant materials in their design that absorb radio light best at certain frequencies, but I cannot find any information on materials that are explicitly stated to absorb very low frequency radio light.</p>
<p>I am not looking for any aspect of geometry such as faraday cages - only materials.</p>
| |materials|telecommunication|radio| | <p>a chunk of material will absorb electromagnetic radiation if 1) its characteristic impedance is close to that of a vacuum (to minimize reflections) and 2) if its impedance contains the right amount of ohmic resistance, so as to dissipate the radiation into heat and not re-radiate it.</p>
<p>that said, to interact significantly with EM radiation of any wavelength requires (ideally) that the size of the interacting object be a significant fraction of the wavelength of that radiation. This means that to absorb centimeter-range radar requires something like a centimeter of thickness to the absorbing layer. </p>
<p>Different rules apply if you are interested in just reflecting away the incident EM waves. In this case, a faraday cage will work.</p>
| 28990 | Material to absorb very low frequency radio |
2019-06-29T23:44:12.790 | <p>I'm working on a balancing robot. I'm familiar with the simple PID control loop but i can't understand this PID control loop.</p>
<p><a href="https://i.stack.imgur.com/pIB0b.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/pIB0b.png" alt="pdpi"></a></p>
<ol>
<li>Is this actually a cascade control loop? The programming states that the angle control PD loop is executed every 5ms, the speed control PI loop is executed every 40ms and the steering PD loop every 20ms. From my understanding, based on the loop/function execution timing, the innermost loop is the angle control loop, then the steering loop and the outermost loop is the speed control loop. Correct?</li>
<li>Why is this complex control loop better than a classic control loop with P, I and D in one single controller?</li>
<li>How many control loops can be nested within each other? I have only seen one inner loop and one outer loop. Is it possible to have 2 inner loops and one outer loop? Or 3 loops, each nested within the other?</li>
</ol>
| |control-engineering|pid-control| | <p>To address your second and 3rd question it's important to remember the purpose of cascading PID loops.</p>
<p>Lets say you want to control a ships heading, so you create a rudder position signal using a PID loop. But the rudder doesn't take a position signal, it's just got actuators that take a "force" signal. So you run <em>those</em> on a PID loop to match the position setpoint. And now you've got a cascaded loop. </p>
<p>It may be possible to combine these into one loop, but you can see that logically it wouldn't make sense, they're each separate control problems with their own loops.</p>
<p>Now lets go deeper. Lets say the actuators are self contained devices that take a force signal, but internally they're just a dumb cylinder and valve. So you add a PID loop to that: it takes the force signal, and adjusts the valve using PID to match the force setpoint. That's 3 cascaded loops, and you can see how this could go on and on.</p>
| 28993 | Is this a PID cascade control loop? |
2019-06-30T03:46:09.383 | <p>So, I was trying to make something like this </p>
<p><a href="https://i.stack.imgur.com/HJpst.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/HJpst.jpg" alt="enter image description here"></a>. </p>
<p>However, While I have the motor, the motor has a vertical shaft and I need something to connect a rod to it in a T shape</p>
<p>Is this even possible? And if yes, then what should I get?</p>
| |mechanical-engineering| | <p>Yes, you could design a "top hat" that sits on the motor shaft and is secured with a couple of grub screws at 90°, with a hole for the rod to go through (the rod can also be secured with a couple of grub screws or if the rod is threaded, use a threaded hole instead).</p>
<p>Here's an example with a 6mm diameter motor shaft and a 6mm diameter rod:</p>
<p><a href="https://i.stack.imgur.com/8vXV0.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/8vXV0.png" alt="enter image description here"></a></p>
<p>A detailed drawing of the part in question would look something like this:</p>
<p><a href="https://i.stack.imgur.com/iA0Gl.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/iA0Gl.png" alt="enter image description here"></a></p>
| 28995 | Can a rod be attached perpendicular to the motor shaft |
2019-06-30T20:54:41.780 | <p>Some tight-coiled springs are made so that they are already under tension - how are they manufactured?</p>
| |springs| | <p>I believe you are referring to Extension Springs, which have a certain degree of pre-tension that holds the coils against each other until a certain force has been applied.</p>
<p>As the spring is wound, the wire is under tension. Imagine wrapping a some elastic around your finger a few times while it is stretched, there will be an inwards force, the diameter wants to be smaller, it wants to crush your finger. So too with the spring - the wire wants to relax, and become shorter, but it can't - the other spring coils are in the way.</p>
<p>Take a spring, hold it at opposite ends, and twist - you will see that it naturally wants to get longer and fatter, or shorter and narrower depending on which way you twist. With an extension spring, it can't get any shorter, so the initial tension from winding remains.</p>
| 29000 | How are tight coiled springs made |
2019-07-01T10:44:12.113 | <p>In the case of a static calculation of a structural model, one can verify the model by checking the applied and reacting forces to be equal zero. </p>
<p>I would like to adapt the same principle for a structural model with a response spectrum as an input.</p>
<p>Is it correct to do the following:
1) perform a frequency extraction procedure to obtain 90% of the eigenmodes contributing to the mass and rotational inertia.
2) calculate the mass of structure being excited and multiply it by the peak acceleration amplitude given in the response spectrum, eg 20g.
3) its not clear what this step should be... is there something I can do with the modal participation factors to help me calculate the total reaction load?</p>
<p>Any suggestions will be welcome.</p>
<p>Best Regards</p>
| |frequency-response| | <p>In response spectrum analysis of MDOF structures, the most common methods of modal combination are: </p>
<ul>
<li>absolute sum (assume the peak response of all modes occurs simultaneously)</li>
<li>SRSS (square root of the sum of the squares)</li>
<li>CQC (complete quadratic combination) </li>
</ul>
<p>If what you're aiming for is a pencil-and-paper check of the software output, the first two methods are most appropriate, and the absolute sum method should be conservative (sometimes overly conservative). CQC is more rigorous, and can give better solutions for structures with closely spaced modes, but it's best suited to a computer. It's worth noting that many off-the-shelf analysis packages will allow the user to choose a modal combination method when setting up the response spectrum analysis.</p>
<p>So, if eigenanalysis has supplied the frequency and mass participation for each mode, we can pull the peak acceleration for each mode from the acceleration response spectrum curve. Then, mass * acceleration gives the the base shear for each mode, and we find total base shear with our chosen modal combination method.</p>
<p>As for how many modes to consider, the standard of practice I'm familiar with (United States - bridges) is to include sufficient modes to capture at least 90% mass participation (not 90% of the modes).</p>
<p>Note that for seismic design purposes, you may need to take one final step and consider some combination of the longitudinal and transverse responses. The pertinent design code should specify.</p>
| 29002 | How do I verify the reaction forces of an FE model due to a response spectrum |
2019-07-01T17:38:58.450 | <p>Consider a four-wheeled cart that has swiveling casters for the two front wheels, and has wheels that don't swivel in back. For example, a common shopping cart, or trolley if you prefer. If such a cart is pushed and let go, it will continue to roll in the same general direction, and so I'll call it stable.</p>
<p>On the other hand, if the same cart is turned around, pushed, and let go, then the cart won't continue in the same direction for long; it has a tendency to spin around, so I'll call it unstable.</p>
<p>Why is a cart with swiveling casters in front stable, but is unstable when the same cart is turned around?</p>
| |mechanical-engineering|wheels|stability| | <p>Wikipedia mentions shopping carts, explicitly:<sub><a href="https://en.wikipedia.org/wiki/Caster_angle" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Caster_angle</a></sub></p>
<blockquote>
<p>In the extreme case, such as the caster wheel on a shopping cart, the system is undamped but stable, as the wheel oscillates around the 'correct' path.</p>
</blockquote>
<p>The shopping cart wheels are not normally/ever casted, so the beginning of @ray-butterworth's effort is wasted in this case. He also mentions driving wheel a lot, which is unrelated to the question at hand.</p>
<p>Where I think he hits the nail is these two paragraphs (with regard that the cart's swiveling wheels have trail):</p>
<blockquote>
<ul>
<li><p>On a vehicle with front-wheel steering, the centrifugal force also
acts on the center of gravity and pushes all the wheels to the left.
The rear wheels are fixed and so continue following the vehicle. The
front wheels can turn, so as they push to the left against the ground,
the ground pushes them to the right. With caster, or trail, this force
provides negative feedback that makes the wheels return to their
normal straight-forward position unless a steering wheel is forcing
them to stay in a turned position.</p></li>
<li><p>On a vehicle with rear-wheel steering, it is the front wheels that are
relatively unaffected by the sideways force, and the rear wheels that
react to it. The force of the road on the wheels makes them want to
turn to the left. But the vehicle is turning because the rear wheels
were already turned to the left, so in this case the centrifugal force
provides positive feedback. The more the vehicle turns, the more the
wheels tend to make it turn. Rather than forcing the vehicle to turn,
the steering wheel must be used to prevent the vehicle from turning.</p></li>
</ul>
</blockquote>
<p>I wish he had made those two paragraphs the answer. But I also appreciate the deeper tour I got about how steering works with automobiles, fork lifts, trains, boats and airplanes.</p>
<hr>
<p>Meta note: In the eyes of the <a href="https://engineering.stackexchange.com/tour">Engineering SE tour</a>, the shorter and precise answers - maybe with links to blogs or articles that provide the full story - would be preferred.</p>
| 29005 | Why is a cart with swiveling casters in front stable, but not if the swiveling casters are in back? |
2019-07-03T12:10:21.420 | <p>I am rebuilding a car engine. With some of the fasteners I am unscrewing, I notice that a lot of torque is required to undo the fastener. It feels like a snap when undoing a fastener, rather than a smooth twist. If I tighten the fastener to a set torque again and loosen it, it requires the same torque as it took to fasten it.</p>
<p>Why would this be? My guess would be something to do with heat cycling.</p>
<p>Note: There is no threadlocker on these bolts.</p>
| |bolting|fasteners| | <p>A standard bolt will stretch when tightened. This stretching increases force on the threads of both the bolt and hole and thus increases the force required to overcome static friction and loosen the bolt. If we exclude other materials and chemical changes to the surfaces then that snap that you experience is the moment you overcome that static friction between the two parts.</p>
| 29027 | What causes a fastener to lock? |
2019-07-03T18:28:23.030 | <p>Given an Archimedes screw pump (shown below) how would you go about calculating the torque required to operate it? I'm assuming a situation where you have to design for a given flow rate, Q (m^3/s) and the diameter, angle of incline and head are known. The fluid to be pumped is water.</p>
<p>I believe the power required to lift water in such a pump is just the hydraulic power, which we have all the variables for:</p>
<pre><code> P(hyd) = rho * g * Q * H
(Power = density * gravity * flow rate * head )
</code></pre>
<p>But I'm struggling to conceptualise how this translates into a value for torque.</p>
<p>Any help would be appreciated!</p>
<p><a href="https://i.stack.imgur.com/g7fRx.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/g7fRx.jpg" alt="Archimedes screw pump"></a></p>
| |torque|pumps|power| | <p>Hydraulic power of a pump is <span class="math-container">$P_h=QgH\rho$</span>, with flow <span class="math-container">$Q$</span>, gravitational constant <span class="math-container">$g$</span>, hydraulic height <span class="math-container">$H$</span> and fluid density <span class="math-container">$\rho$</span>. Note that <span class="math-container">$H$</span> is a bit higher than the actual useful hydraulic height:</p>
<p><a href="https://i.stack.imgur.com/os1PB.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/os1PB.png" alt="enter image description here"></a></p>
<p>Shaft power (The power delivered at the shaft of the pump) is <span class="math-container">$P_s=P_h/\eta$</span>, with hydraulic effiency <span class="math-container">$\eta$</span> - for screw pumps, this is usually around 75%. The strength of screw pumps is that this effiency is achieved over a wide range of flow rates.</p>
<p>Now you know the shaft power an can plug this into the formula provided by Tim: <span class="math-container">$T=\frac{P_s}{2\pi\omega}$</span>.</p>
<p>Now you need to know <span class="math-container">$\omega$</span>. RPM is <span class="math-container">$n=\frac{50}{\sqrt[3]{D^2}}$</span> with outer Diameter <span class="math-container">$D$</span> given in meters, then <span class="math-container">$n$</span> is given in rpm. This is the formula of Muysken, quoted after Nagel and Radlik. The actual rpm may be lower, at higher speeds the effiency suffers because more water flows back over the "shovels".</p>
<p>Wether your sizing a motor and gear or building the pump house, you should work closely with the manufacturer of the pump.</p>
<p>Source: Nagel / Radlik: Wasserförderschnecken - Planung, Bau und Betrieb von Wasserhebeanlagen. </p>
| 29035 | Calculating the torque required by an Archimedes screw pump |
2019-07-06T02:46:27.793 | <p>A question about performance of a turbojet.
Why are compression ratio and thrust are not in linear but increase exponentially with RPM?</p>
| |aerospace-engineering|turbomachinery| | <p>Because the compression in a turbojet is an aerodynamic effect caused by the rotating compressor blades.</p>
<p>Lift and drag forces on an object moving through a fluid are proportional to the square of its relative velocity:</p>
<p><span class="math-container">$$F \propto v^2$$</span></p>
<p>and that velocity is linear with the rpm of the shaft because:</p>
<p><span class="math-container">$$v = \omega \cdot r$$</span></p>
<p>hence, since work done is also linear with the applied force:</p>
<p><span class="math-container">$$W \propto F \propto \omega^2$$</span></p>
<hr>
<p>Thrust is not linear with the compression ratio, however, and also depends on an array of other factors. Thrust is the net effect of the Brayton cycle in the engine, and that cycle develops along an <a href="https://en.wikipedia.org/wiki/Adiabatic_process#Ideal_gas_(reversible_process)" rel="nofollow noreferrer">adiabatic</a> line during the compression step.</p>
| 29062 | Relationships among turbo jet rpm and compression ratio and thrust |
2019-07-06T10:22:15.880 | <p>I'm trying to understand how telephone wires work.<br>
I've stumbled upon this kind of identification method with colours.<br>
I can't manage to understand the reasons why is used and if it's still common to use it.</p>
| |telecommunication| | <p>At the most basic level, wires are color coded to simplify things during installation. And that's all wiring, not just telephone wires. </p>
<p>Think of the challenges involved if you're running a new telephone jack into a room that's a reasonable distance away from the phone block. If the wires weren't color coded, you would have to set up some sort of ringer on each of the lines in order to determine which of the wires to use and where. </p>
<p>On a basic phone outlet (two wire) it's not horribly complicated. On a standard outlet (four wire) it starts to get annoying. With standard telephone wire (8 wire) then it gets really annoying. Color coding simplifies all that. </p>
<p>In case you haven't wired any of those types of connections, <a href="http://www.westernet.net/Help/RJ45.htm" rel="nofollow noreferrer">this webpage</a> shows the standard color coding for telephone and ethernet wiring.</p>
<p>An additional aspect of color coding is standardized use of the wires. When someone new comes in to perform maintenance on an existing installation, they'll have a reasonable degree of assurance that specific color wires are used for certain tasks. Not following the standard can actually impact the data transfer rate on a cat-5 ethernet cable.</p>
<p>When it comes to higher voltages, there is a safety element as well. Incorrectly wiring a 240V outlet's multiple wires can potentially cause a fire or damage the equipment. </p>
| 29064 | What is the purpose of colour code for telephone wires? |
2019-07-06T13:33:20.623 | <p>Before I describe the problem I would just like to say that I'm no engineer this is my hobby, all I know is from google or if someone told me or from books. Something obvious to you may not be obvious to me. Thanks</p>
<p>I want to move a gear with the shaft like they were one in one moment and in the other I want to stop the gear using clutch brake while continuing to rotate the shaft.</p>
<p>One way to achieve this is using air bearings which would be the perfect solution if they weren't so impractical because of the required compressor and other parts.</p>
<p>Second solution would be to use two clutches. One to connect the gear to shaft and another clutch to brake the gear when the first one is not connected but I don't want to use more than one clutch pack because they are expensive.</p>
<p>Third solution is to use sync cone but that is much slower than clutch and that is why I don't want to use it.</p>
<p>Fourth solution which I found is using a shaft mounted brake to lock the gear with the shaft but I'm not sure if this would count as a clutch brake and if it would lower the efficiency of the system as whole.</p>
<p>So my question would be is there a way to achieve the problem described without using more breaks or clutches (if shaft mounted brake wouldn't count as a break then that is also OK)? The perfect solution would be a bearing which I could "turn off" and let it rotate with the shaft but I'm not sure if something like that exists.</p>
| |mechanical-engineering|automotive-engineering| | <p>Yes use a electromagnetic clutch, often used on car a/c compressors to disconnect them when not needed.</p>
<p>When engaged they drive and freewheel when not energised.</p>
| 29065 | Is it possible to move gear on a shaft in one moment and in the other allow it to rotate freely? |
2019-07-07T18:43:34.067 | <p>Determine the resultant internal loadings acting on the cross section at B of
the pipe shown in Fig. 1–7a. End A is subjected to a vertical force of 50 N, a
horizontal force of 30 N, and a couple moment of 70 N m. Neglect the
pipe’s mass.</p>
<p><a href="https://i.stack.imgur.com/BOCvQ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/BOCvQ.png" alt="fig 1-7a"></a>
<a href="https://i.stack.imgur.com/me0iw.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/me0iw.png" alt="solution of problem in book"></a></p>
<p>Well, I can't feel the solution. I would add upward reaction force (70/0.5=140 N) at B due to moment. But I wouldn't add 70 N m to moment equation at point B. can you explain the reason?</p>
| |mechanical-engineering|statics| | <p>The counterclockwise moment applied at A will be transferred as torque to the 90 elbow and from there it will impart a moment of 70N.m all the way along the length of that section of the pipe, in similar way as when you apply this moment to the free end of a cantilever beam.</p>
<p>Therefore the moment of 70N.m is the constant along the length of that section of the pipe starting from the elbow to and including the point B.</p>
<p>The solution is correct and you must include 70N.m in the sum of the moments.</p>
| 29083 | In static equilibrium body, how can an applied moment about an axis contribute to reaction moment to another point away from the axis? |
2019-07-08T16:46:00.300 | <p>I am curious as to how pump salad spinners work: not in terms of how they spin and use centripetal forces to get rid of water. </p>
<p>I mean in terms of how the pump itself works. How does the 2D motion of pushing the pump get converted into the rotation and spinning of the salad spinner itself? Are there gears involved? I would like to know the underlying mechanism of it.</p>
| |mechanical-engineering|gears| | <p>Inside the pump-like mechanism, there is a vertical shaft with helical grooves cut into its length, which wrap around the shaft. the bottom of this shaft is connected to the salad basket and supported by a bearing at its base, so the shaft and the basket can revolve. </p>
<p>The "pump" handle is a hollow tube which slips over the basket shaft. Inside it are a couple of protrusions which fit into the spiral groove. When you press down on the handle, those protrusions engage the groove and as the protrusions descend, they urge the shaft to revolve, which sets the basket into rotation.</p>
<p>At the bottom of the handle's stroke, the sprial groove blends into a circumferential cut or "runout groove" in the shaft so the shaft and basket can continue to revolve without any further motion on the part of the handle. </p>
<p>When you release pressure on the handle, a spring urges the handle back up again while the protrusions retract. At the top of the handle's travel, there is another runout groove that the protrusions fall into, and when you push down a second time on the handle, the protrusions find the spiral groove and the downward force applied by your hand on the handle forces the shaft to turn faster, and the process repeats. </p>
| 29095 | How do Salad Spinners work? |
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