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engineering.stackexchange.com

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2016-04-27T20:22:54.610

<p>I have been working with an old code for modeling incompressible, 2D viscous flow out of a tank to understand a chemical process. There isn't much documentation and I'm not a fluid dynamicist, so I have been trying to check that things are working as they should by comparing results with those from another code. The comparison is great when I use a velocity inlet condition. However, there is a major discrepancy when I use a pressure inlet condition. In both cases, the outlet pressure is zero. </p> <p>I have spent several weeks trying to figure out what the problem is with the pressure inlet condition (and have an active question about it) because I have pressure inlet values that I want to use in my problem. I don't know velocity inlet values. But I have to acknowledge I've hit a wall with figuring out my pressure inlet boundary condition problem. </p> <p>This may be a dumb question, but I would really appreciate it if someone would help me confirm if I can transform my pressure values into an inlet velocity condition. If I can, maybe I can simply use an inlet velocity and my problems will be over! </p> <p>I understand that: $$P = \dfrac{1}{2}\rho U^2 + \rho gz$$</p> <p>where $\rho$ is fluid density, $U$ is velocity, $g$ is the acceleration of gravity, and $z$ is depth of the tank inlet below the tank outlet at 0. </p> <p><strong>So if I simply plug in my inlet pressure value and rearrange, will I get a valid inlet velocity that I can use for a velocity boundary condition?</strong> My concern is that I'm dealing with viscous flow and I think this expression is related to the Bernoulli equation, which does not account for viscous flow. </p> <p>If I'm right and I can't use this expression to calculate an inlet velocity, does anyone know if there's an alternative? </p> <p>The reason I say I probably can't is that I've tried it and the results from the two codes don't match. I'm just trying to figure out where my problem is--if it's likely to be a bug in one of the codes, which code is the problem, or if I'm making some mistake when inputting my boundary condition values.</p>

|fluid-mechanics|statics|mathematics|

<p>It's difficult to tell how your problem should be approached with the information that you provide, for example in what sense you find a discrepancy. I assume your system is as you described in the question "<a href="https://engineering.stackexchange.com/questions/8566/incompressible-2d-pressure-driven-flow-for-navier-stokes-equations-in-nondimens">Incompressible 2D pressure-driven flow: for Navier-Stokes equations in nondimensional form, how should I express pressure boundary conditions?</a>", with the picture that I copy at the bottom of this post.</p> <p>The flow is upwards. Normally, the way to replace a velocity boundary condition by a pressure boundary condition is by trial and error until you have the pressure that will generate the same average velocity over your inlet plane. If that is what you did, but you found a different pressure than with the other calculation method, then something is wrong with at least one of the calculation methods and you have to a systematic comparison between the two models, checking all parameters that you can plot and figure out which fundamental parameter is wrong. For example, you could have forgotten to switch on the effect of gravity in one of the codes.</p> <p>If your question is: how do I calculate the pressure to set at the inlet from just the viscosity and a given inlet velocity, then you are using the wrong approach. The pressure at the inlet will be the result of two contributions: gravity ($P_g=\rho g h$) and flow resistance at the outlet pipe. For that, you need to take the <a href="https://en.wikipedia.org/wiki/Hydraulic_diameter" rel="nofollow noreferrer">hydraulic diameter</a> $D_H=2a$ of the outlet pipe, get the average outlet velocity $U_o$ and calculate the Reynolds number, $\mathit{Re}=\rho U_o D_H/\eta$. The pressure drop at the outlet will generally have several contributions:</p> <ul> <li><p>Bernoulli: $\Delta p=b\rho U_o^2/2$, where $b$ is a factor to account for the velocity profile of the fluid at the outlet; $b=1$ for a uniform velocity and $b=2$ (I think) for a parabolic velocity profile. This is sometimes called: "exit loss".</p></li> <li><p>Friction along the pipe length, also called "major loss": $$ \Delta p_{\mathrm{major}}={fL\over D_H} {\rho U_o^2\over 2}, $$ where $f$ is the <a href="https://en.wikipedia.org/wiki/Darcy%E2%80%93Weisbach_equation" rel="nofollow noreferrer">Darcy-Weisbach friction factor</a>, which depends on the Reynolds number.</p></li> <li><p>Inlet loss, also called "minor loss", $$ \Delta p_{\mathrm{minor}} = \xi{\rho U_o^2\over 2}, $$ where $\xi$ is the minor loss coefficient.</p></li> </ul> <p>Tabulated values for all these losses in various pipe geometries are in <a href="http://pbadupws.nrc.gov/docs/ML1220/ML12209A041.pdf" rel="nofollow noreferrer">Idelchik, Handbook of Hydraulic Resistance</a> (warning: 21 MB download).</p> <p>I don't know what happens inside your vessel, but if you're only interested in the relation between flow rate and pressure, I wouldn't even bother with trying to run a CFD code.</p> <p><img src="https://i.stack.imgur.com/Pgu6q.jpg" alt="flow schematic"></p>

8624

How can I translate a pressure boundary condition to a velocity boundary condition for incompressible, viscous flow?

2016-04-28T04:07:31.670

<p>I am learning by myself structural dynamics but confused about the concept of scaled ground motion data. I can't get the reason why the acceleration data should be scaled. I mean, isn't it better to use the original data "as it is" to compute the response of structures so that the results can reflect actual response of a specific structure? </p> <p><strong>EDIT :</strong> Is it the case that when we design a structure, scaled data should be used, while unscaled data can be utilized when we conduct post-seismic-analysis of a structure?</p> <p>Could someone please summarize the situations when we should use scaled or unscaled data?</p>

|structural-engineering|dynamics|seismic|

<p>A ground motion record (accelerogram) is a measurement with high specificity to the fault that caused the earthquake. The recorded peak acceleration, frequencies, duration etc depend on the fault geometry and movement. Of course there are not two identical faults and every time a fault is activated it doesn't produce the same ground motion. Therefore, the value of designing a structure by one "as recorded" ground motion is small, since it is highly unlikely to encounter it in the future. Typically the design with real ground motions requires a set of them (at least 5). </p> <h2>Why scale a ground motion</h2> <p>In an ideal world a structural engineer would have a vast amount of historical ground motions to select from, for a particular site, when designing a new structure. There would be no need to use any records from other parts of the world. Of course this is not happening because severe earthquakes are not an every-day phenomenon and recording ground accelerations started not to far in the past. There are sites where only one or two good records are available. So the solution is to use available records from all over the world, but then the problem is that the seismic risk is not the same all over the planet. Typically, the seismic code provides us with the expected intensity of the seismic excitation, in the form of peak ground acceleration. A solution is to scale an accelerogram to match the code requirements. A problem arises though, because a distrortion in the frequencies is unavoidable. Therefore, it is better to avoid too much scaling, or alternativly to use artificial accelerograms. </p> <p>Also, scaling of the ground motions is inherently needed for some advanced types of analyses (incremental dynamic analysis). </p>

8627

Why is it useful to scale seismic ground motion data

2016-04-28T18:53:31.517

<p>Some time ago, I was going through some exercises and I came across a odd question, for which my atomic species degrees of freedom did not match my degrees of freedom using molecular species. I showed my lecturer and he couldn't see why this was the case. The (summarized) question is as follows: 2 reactions occur in a reactor: </p> <p>\begin{align} \text{C}_2\text{H}_6\text{O} + \text{H}_2\text{O} &amp;\rightarrow 2\text{CH}_3\text{OH} \\ 2\text{CH}_3\text{OH} &amp;\rightarrow \text{C}_2\text{H}_4 + 2\text{H}_2\text{O} \end{align}</p> <p>The input feed of water and $\text{C}_2\text{H}_6\text{O}$ is specified and the selectivity of the first equation. Reactions do not go to completion. </p> <p>So, there are 4 unknowns ($\text{H}_2\text{O}$, $\text{C}_2\text{H}_4$, $2\text{CH}_3\text{OH}$, and $\text{C}_2\text{H}_6\text{O}$).</p> <p>Doing the atomic analysis: $$\text{DoF} = 4 - 3(\text{C},\text{H},\text{O}) - 1 (\text{Selectivity}) = 0$$</p> <p>Doing the molecular species: $$\text{DoF} = 4 + 2 - 4(\text{H}_2\text{O}, \text{C}_2\text{H}_4, 2\text{CH}_3\text{OH}, \text{C}_2\text{H}_6\text{O}) - 1 (\text{Selectivity}) = 1$$</p> <p>Do you see where the problem is?</p>

|chemical-engineering|mathematics|

<p>The answer to your questions is that you calculate the DOF for a reactor with parallel reactions the same way you would any other process. Your problem is unrelated to the fact that you have parallel reactions.</p> <p>The reason you have a different number of DOF for molecular species vs atomic species is because you have one less equation than you think for the atomic species balance. Remember that the equations resulting from your three atomic balances should be independent. If they are not, you have to count one less equation. The way to check for this (short of trying to solve the problem and failing) is to write the three atomic balance equations and check whether you can get one of them by algebraic manipulations of the other two. Let us label $n_1$ and $n_2$ as the input molar flow rate of diethyl ether (DE) and water (W), respectively; and $n_3$, $n_4$, $n_5$, and $n_6$ as the output molar flow rate of DE, W, ethylene (E), and methanol (M), respectively. Then the atomic balances read: $$ \begin{align} \text{(1) Oxygen: } &amp; n_1+n_2=n_3+n_4+n_6 \\ \text{(2) Hydrogen: } &amp; 6n_1+2n_2=6n_3+2n_2+4n_5+4n_6 \\ \text{(3) Carbon: } &amp; 2n_1=2n_3+2n_5+n_6 \end{align} $$ If you take two times equation (1) and add two times equation (3) you get equation (2) (verify this). This means that the set of equations is not independent and you can only count two atomic species in your balance instead of three. Both approaches then give you 1 DOF.</p> <p>On a side note, the selectivity should be defined as: $$ \frac{\text{moles of desired product}}{\text{moles of undesired product}} $$ which mean that it should not be defined with respect to the first reaction but with respect to the entire reactor.</p>

8637

How does one calculate degrees of freedom in a reactor with parallel reactions?

2016-04-29T04:48:01.353

<p>I am IT engineer not a refrigeration expert, pardon my basic question.</p> <p>Would a NH3 or other absorption cycle work with a generator temperature of say 30C, and condenser temperature of 10C by maintaining correct pressure level. </p> <p>Most of the technical papers I read on the internet, talk about requirement of 80C+ temperature for generator and 25-30C for condenser. This would make sense if one has access to hot water source like geo thermal, or any other free heat source. But if I have access to low temperature cool water of 10C, and relatively stable heat source of 25C+ (say atmospheric temperature), can a absorption refrigeration system be designed to harvest that energy. One of the main objective is to have very low electrical energy input, so compression system may not work out. The second objective is to have a -3C or lower evaporator temperature to make salty ice and store it, and use it when the cold water source is not available. If at all its possible, what major issues to keep in mind.</p> <p>I guess even the standard off the shelf absorption systems can be made to use 30C heat, by first passing it through a heat pump to increase that temperature to 80C+. But that additional stage also need lot of electricity. </p> <p>Any suggestion to effectively harvest this low temperature source would be of great help.</p>

|hvac|refrigeration|

<p>Basically to vaporise the NH3 in generator and pressurise it, you need to add heat that will be greater than the latent heat of ammonia due to fact that water is present it will take some part of heat.</p> <p>To vaporise and seperate water and NH3 at lower temperature you will need to add more amount of heat (red line) with respect to heat you will add to higher temperature (blue line).</p> <p>further more the temperature range you have provided is not practical. I mean the heat added to the aqua ammonia by just atmosphere would be insufficient to vaporise the refrigerant. let alone it to pressurise.</p> <p>to add heat to it you need to give it some external heat.</p> <p>Note:- the diagram below is for ammonia refrigerant I have drawn basic diagram on the left. I didn't find the diagram for aqua ammonia strong solution. Actual curve of the solution may vary</p> <p><a href="https://i.stack.imgur.com/jAQqM.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/jAQqM.png" alt="enter image description here"></a></p>

8647

absorption refrigeration with low generator temperature?

2016-04-29T06:19:07.040

<p>From what I know, for a nice rectangle slab with 4 edges bounded by beams, one can distribute the slab loading to adjacent beam via the <a href="http://www.engr.uky.edu/~gebland/CE%20382/CE%20382%20PDF%20Lecture%20Slides/CE%20382%20L3%20-%20System%20Laoding.pdf" rel="nofollow">Tributary Area method</a>. </p> <p>But what if the slab is irregular, or there are arbitrary line loads on the slab, or the slab is not of uniform method, or not all sides of the slabs are covered by beams? How should the slab's dead load and live load, and any load on slab be distributed to beams? Is there a general way?</p>

|structural-analysis|

<p><em>Extension of my comment to an answer, feel free to edit and expand it!</em></p> <p>The most general method would be to use a numerical model consisting the slab as well, thus the load transferring mechanism is captured by the model. Within the finite element (FEM) paradigm plate elements would suffice. With today's computers and FEM applications it can be easily solved.</p> <p>I do not know about your background and experience but using plate elements with line elements to model slabs with beams is quite widespread. That is why I just mentioned FEM without any further details. If you could provide more details on your experience or formulate more specific questions, e.g. how to model the eccentricity of beams, beam slab connection, then we might be able to give equally specific answers.</p> <p>Here are two practitioners' guides on <strong>modelling of reinforced concrete slabs</strong>:</p> <p><a href="http://publications.lib.chalmers.se/records/fulltext/176734/local_176734.pdf" rel="nofollow">Recommendations for finite element analysis for the design of reinforced concrete slabs</a></p> <p><a href="https://www.scribd.com/doc/19184786/How-to-Design-r-c-Flat-Slabs-Using-Finite-Element-Analysis" rel="nofollow">How to Design r.c. Flat Slabs Using Finite Element Analysis</a></p> <p>Regarding <strong>how the tributary approach compares with FEM plate model</strong> for a simple case:</p> <p><a href="https://s-frame.com/index_files/Files/Approaches%20to%20Area%20Load%20Distribution%20in%20S-FRAME.pdf" rel="nofollow">A Case Study Comparing Two Approaches for Applying Area Loads: Tributary Area Loads vs Shell Pressure Loads</a></p> <p>Based on their simple example for a particular slab-to-beam stiffness ratio: The tributary area method is on the safe side and yields to about 20% overestimation of maximal bending moment and shear force.</p> <p>These are based on a single, simple example thus the results are not conclusive, at best only indicative.</p> <p>Since the structure type was not mentioned and maybe you have other than a building structure in mind, here are two relevant books on <strong>bridge decks</strong>:</p> <p><a href="https://www.scribd.com/doc/72761049/Bridge-Desk-Analysis" rel="nofollow">O'Brien and Keogh: Bridge Deck Analysis</a></p> <p><a href="https://www.scribd.com/doc/76442313/Bridge-Deck-Behaviour-E-C-Hambly" rel="nofollow">Hambly: Bridge Deck Behaviour</a></p>

8648

What is the general way for the slab loading be distributed to adjacent beam?

2016-04-29T09:50:51.820

<p>From <a href="http://www.bgstructuralengineering.com/BGASCE7/BGASCE7005/BGASCE70504.htm">ASCE 7-05 code</a>:</p> <blockquote> <p>ASCE 7-05 Section 4.6 states "The full intensity of the appropriately reduced live load applied only to a portion of a structure or member shall be accounted for if it produces a more unfavorable effect than the same intensity applied over the full structure or member."</p> </blockquote> <p>The article then goes on to demonstrate how can we calculate the pattern of live loading for a few simple, text-book cases. </p> <p>The problem now is, what if the configuration is not as simple? In real life the beam configuration, the support condition can be very different than from textbook examples. </p> <p>How to obtain the most critical pattern of live loading for the most general situation? Is there an algorithm for this?</p>

|structural-engineering|civil-engineering|structural-analysis|

<p>As mentioned in the linked text and in <a href="https://engineering.stackexchange.com/a/8652/3353">@grfrazee's answer</a>, the secret is influence lines. Or, more generically, influence surfaces.</p> <p>For starters, let's stick to influence lines, since they are far easier to describe. An influence line is a diagram for a given point on an object composed of unidimensional beam elements. It describes the internal force that will occur on that point due to a unit load applied at different points along the entire structure.</p> <p>For instance, a simply supported beam has the following bending-moment influence line for the point at a quarter-span (I'm mostly going to talk about bending moment influence lines here, but the general gist of things applies to other forces as well):</p> <p><a href="https://i.stack.imgur.com/Hj6GL.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Hj6GL.png" alt="enter image description here"></a></p> <p>This means that if a concentrated unitary vertical load (say, 1&nbsp;kN) were applied at that point, it would cause a bending moment at that point equal to 0.75&nbsp;kNm (or 7.5&nbsp;kNm if the load were 10&nbsp;kN). If, on the other hand, the unit load were applied at midspan, the moment felt at the quarter-span would be equal to 0.50&nbsp;kNm. And so on.</p> <p>This also tells you that the worst-case scenario for this point is to have the entire structure loaded. This can be seen by the simple fact that all values on the influence line are positive, therefore a load applied at any point on this beam will increase the internal forces occurring at the quarter-span.</p> <p>This, however, is an isostatic structure which can be trivially solved. Once you move into hyperstatic (statically indeterminate) structures, things get messy. For instance, take a look at this relatively simple hyperstatic beam:</p> <p><a href="https://i.stack.imgur.com/dqnry.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/dqnry.png" alt="enter image description here"></a></p> <p>This is a relatively simple structure, but it's already impossible to find a closed-form solution for the precise best location of loads. For non-trivial structures, influence lines are a pain.¹ You can however notice one important thing: at the supports the value is null and shifts from positive on one side to negative on the other. This happens in every structure. If, instead of supports you had columns, then the value at the columns won't actually be equal to zero due to the deformability of the column. That being said, the result is usually very close to zero, so you can usually consider columns as perfectly rigid (i.e. as normal supports) with barely any loss of accuracy (assuming reasonable layouts). </p> <p>So, if you're only dealing with distributed loads (such as in a building), this is the only rule you need to find your solution: if you're looking for the maximum positive (tension on the bottom fiber) bending moment, apply loads on the span in question, don't apply loads on the neighboring spans, apply on the ones neighboring those, etc. In this case, the actual values of the influence line are irrelevant, all that matters is the sign (positive or negative) at each span. Basically, here's the rule in graphic form:</p> <p><a href="https://i.stack.imgur.com/waPiy.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/waPiy.png" alt="enter image description here"></a></p> <p>However, what if you're building a bridge and you need to take into consideration the position of the load train, which is composed of concentrated loads? Complicating matters, the position of the load train usually has a lower (if any) distributed live load, which means there's an interaction between these two parts.</p> <p>So, looking at the second figure, where would you put the load train? It is pretty intuitive that you want to place it near the maximum value (in this case 0.3704). But what if you have an even number of wheels or if your load train is asymmetric? Do you want to put the truck's load center at the maximum? Do you want to certify that the heaviest wheel is at the maximum? Is your uniform load so freakishly high that, actually, you're better off putting the truck far away where it won't reduce the result due to your uniform load?</p> <p>Even worse, what if you're actually looking for your negative bending moment envelope? Then you know you want your truck at the neighboring span, where the influence line sign is negative, but once again, where do you put it? You'll need to derive the equation of that curve in order to find the point of the maximum value (its not at the middle of that span), and then you'll still have the same issues described above.</p> <p>These are all possibilities which can't be reduced to a closed form solution for a generic structure. So you need to rely on software.</p> <p>What most programs actually do is <strong>cheat</strong>. They approximate the solution by doing a moving-load analysis. First they use influence lines as described above to find out where to put the uniform loads. Then, for the load train itself, they simply place it at one spot, calculate the results, move it a certain distance (usually user-defined), calculate the new results, and repeat. It then gets the worst case and adopts that.</p> <p>This method is obviously flawed because, if you use a step size equal to, say, one meter, you don't know if that maximum value found is the <em>true</em> maximum or if there was a certain point between the tested steps which would have given a higher result (there almost certainly is). So it's up to the user to define a step size such that the difference between the actual result and the one obtained is negligible (I usually use a step size <strong><em>at most</em></strong> equal to one tenth of the smallest span, preferably significantly smaller than that).²</p> <p>This entire answer, however, has relied on influence lines. These are useful for linear structures such as simple beam systems and even some bridges. But if you have a truly three-dimensional structure, influence lines don't cut it and must be generalized to influence surfaces. These are no more than the three-dimensional version of influence lines. However, like all such things, influence surfaces are orders of magnitude harder to obtain. Every program I know which can calculate them brute-forces it: they apply a concentrated force at each node, one at a time, and see what happens.</p> <p>That being said, as far as distributed loads go, the same approach suggested above (apply on one span, skip its neighbors, apply on the next ones, etc) can also be applied successfully for influence surfaces as well. In this case it becomes something of an approximation since the boundaries between slabs are usually just beams which are quite flexible for vertical displacements (relative to columns or actual supports). This means that, unlike the case for influence lines, where the influence line value at the supports is equal (or almost) to zero, the value at the slab supports (the beams) is not necessarily so. That being said, the error is usually reasonable (especially considering the low influence values for the slabs other than the one being studied).</p> <p>That being said, it is quite common to simply assume for buildings (<strong><em>not bridges</em></strong>) that the worst case is with the entire structure under load, without considering influence lines. This is assumed knowing it is false and goes against safety (not loading neighboring slabs would result in a larger positive bending moment that that obtained by loading the entire structure), but it is equivalent to assuming that the value of the influence line on neighboring slabs is so small that it can be considered equal to zero. The validity of such an assumption is dependent on each structure's configuration.</p> <p>As mentioned by <a href="https://engineering.stackexchange.com/questions/8650/general-solution-for-the-most-critical-pattern-of-live-loading#comment16068_8666">@Arpi in the comments to this answer</a>, it's also worth mentioning that this all assumes linear behavior. If your analysis is non-linear, then everything falls apart. Non-linearity breaks everything.</p> <p>_{All figures here were created with <a href="https://www.ftool.com.br/Ftool/" rel="nofollow noreferrer">Ftool</a>, a free 2D frame analysis tool.}</p> <hr> <p>¹ It's actually quite easy to determine influence lines yourself if you have an analysis software, even if it doesn't calculate them itself. For bending moments, place a hinge on the desired point and apply equal and opposite bending moments to each side of the hinge such that they create a unit rotation in the deformed configuration. That deformed configuration is your influence line. This same idea (<a href="https://en.wikipedia.org/wiki/M%C3%BCller-Breslau%27s_principle" rel="nofollow noreferrer">the Müller Breslau Principle</a>, which is based on the <a href="https://en.wikipedia.org/wiki/Betti%27s_theorem" rel="nofollow noreferrer">Maxwell-Betti reciprocal work theorem</a>) can be applied to find the other forces' influence lines as well.</p> <p>² The Ftool software used to draw these figures actually uses a genetic algorithm to find the optimal load train position. It isn't analytical and is actually itself something of an approximation, but it is for all intents and purposes exactly correct. The article which developed this method can be found <a href="https://en.wikipedia.org/wiki/Betti%27s_theorem" rel="nofollow noreferrer">here</a> if anyone's interested.</p>

8650

General solution for the most critical pattern of live loading

2016-04-29T21:46:45.403

<p>In a bike frame build, are there any differences between hexagonal cross-section tube instead of the typical round tubes?</p> <p>Assuming that both are made from the same aluminium alloy and have the same wall thickness, are there differences in weight, strength, or durability? Does the hexagonal shape have any particular advantages over round tube?</p>

|structural-engineering|design|stresses|structural-analysis|

<p>An additional factor is that any section which has defined corners will tend to concentrate stress at the corners rather than it being evenly distributed throughout the section. </p> <p>With tubing this can be a double effect that you potentially have work hardening from the manufacturing process concentrated at the corners as well. If the section has a uniform wall thickness then it certainly means that some of the material is effectively wasted as the corners will encounter yield before the flats and at worst it can lead to crack propagation points and fatigue. </p> <p>It is a fairly good rule of thumb in structures that any sort of discontinuity represents at best an inefficiency and at worst a potential failure point. An ideal structure will have a smoothly varying section with section size proportional to the stresses on it, in fact bones are an excellent example of this. Although this is, in most cases, impractical for fabricated frame structures. </p> <p>Of course there are other pragmatic considerations to be taken into account for example square or rectangular tube is much easier to join than round as square tube can just be mitred to the required angle with a saw whereas round tube needs to be notched with a specialist machine or painstakingly hand fitted. </p> <p>It seems likely to me that hex tube would be a fairly poor compromise between square and round/elliptical section tube. </p>

8661

Properties of an hexagonal shape vs round tube shape

2016-04-30T05:13:05.843

<p>Which position of the water pump in the below diagram provide maximum water output to the overhead tank?</p> <p><a href="https://i.stack.imgur.com/5dd2s.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/5dd2s.png" alt="enter image description here"></a></p> <ul> <li>Each floor is 12 feet in height.</li> <li>I already have the pump (Centrifugal Regenerative Self Priming 1HP Peripheral Type, Self Priming - 180sec At 3m STATIC SUCTION LIFT) and would like to know which is the best location to install it for maximum water intake to the overhead tank. The water Supply is only available for a limited period everyday and the goal is to achieve maximum water to the overhead tank.</li> <li>The Water supply line provides water from a bigger (8inch) main line but water pressure from it is not adequate to pump up water on top level. (Water pressure only enough to supply water to first floor and that too with reduced pressure.)</li> <li>The water pipe from main supply to pump and from pump to overhead tank is 1/2 inches wide.</li> </ul>

|pressure|pumps|piping|pipelines|building-design|

<p>Considering this is a <strong>centrifugal water pump</strong>, it can provide very little <strong>air suction</strong>. That means if your pump is airlocked, there's not a shade of a chance it will be able to produce nearly enough suction to pull the water up two floors. And it will be airlocked all the time.</p> <p>With a check valve preventing backflow into the mains, and then filling the water lines and the pump with a bucket, you might have chance to get the 'top pump' config started... until it gets airlocked one way or another again.</p> <p>The "low pump" config will get enough mains pressure to fill the pump with water and then the pump will have no trouble handling the height.</p>

8668

Water pump at top floor or bottom?

2016-04-30T08:44:34.647

<p>I have found that drill bits are always slightly undersized. For example, 10mm bits tend to be more like 9.8mm diameter. And in aluminium, the hole ends up being about 9.9mm. The same seems to be the case for conical stepped drill bits</p> <p>I've found it useful for inserting ball bearings, since they can be hammered in place without being damaged and will stay firmly in place. Also, bolts tend to be slightly undersized so will have a more snug (but still loose) fit.</p> <p>What's the actual reason for drill bits being undersized?</p> <p>Oddly, hole saws tend to be slightly oversized. A 15mm hole saw's teeth outer diameter measures to be 15.05mm and makes a 15.3mm hole. </p> <p>Is there a term for drill bits which are slightly oversized instead? Or hole saws which produce slightly undersized holes? (assuming they exist)</p>

|drilling|

<p>They short answer is that they aren't. Exact sizes will, of course depend on manufacturing tolerances but 0.2mm is a lot undersized by normal standards, although for jobber drills this won't actually matter much. </p> <p>For example metric <a href="http://www.kasthurimmc.com/tap-drill-chart.html" rel="nofollow">tap drill sizes</a> are generally specified to at least 0.1mm </p> <p>One possible explanation for measuring drills as undersized is that the shank often gets worn, especially if it has slipped in a chuck and it is difficult to accurately measure the diameter across the flutes. </p> <p>The actual size of the hole produced by a given drill is a different matter and will depend on the drilling setup and the material being drilled. Twin flute twist drills tend to produce slightly triangular or hexagonal holes, especially when the material being drilled is significantly thinner than the diameter of the drill, but this is more to do with flexing then the actual diameter of the drill bit. This tends to make them functionally undersized in terms of fitting bolts etc and for general fabrication it is normal practice to go 0.5-1.0mm above nominal size for a clearance hole for a metric bolt (notwithstanding any additional issues with fit and clearance between sets of holes etc). </p> <p>With twin flute twist drills in general it tends to be the roundness, and in some cases straightness, of holes which is the limiting factor on accuracy. Better tolerances are usually achieved by using a reamer, a multi-flute drill/mill or by boring with a lathe, mill or specialist drilling machine. There are also specialist drills eg <a href="https://en.wikipedia.org/wiki/Gun_drill" rel="nofollow">gun drills</a> for use where the hole depth is very much larger than the hole diameter. </p>

8672

Why are drill bits always slightly undersized?

2016-05-01T08:15:12.563

<p>On a recent project for a outdoor shed the roof support structure did not look like one of the conventional roof truss designs I have encountered. See snapshot below. (The span is 10 m with the support members 250x125 mm beams)</p> <p><a href="https://i.stack.imgur.com/BHh47.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/BHh47.png" alt="enter image description here"></a></p> <p>Is this still a truss? Or what is this sort of support framework called? </p> <p>It looks like a textbook King Post Truss (to me) but without the vertical member. </p> <p><a href="https://i.stack.imgur.com/Emd0H.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Emd0H.png" alt="enter image description here"></a></p>

|structural-engineering|structural-analysis|

<p>This looks like a <a href="https://en.wikipedia.org/wiki/Collar_beam">Collar Beam Roof</a> (or sometimes a Collar Tie Roof). It is often used in lightly-loaded residential construction. One popular use is to form Cathedral Ceilings.</p> <p>Strictly speaking this is not a truss since the structure relies <em>primarily</em> on bending of the rafters to support the roof loads. By definition <a href="https://en.wikipedia.org/wiki/Truss">trusses</a> are assumed to consist entirely of tension/compression elements and bending of these elements is minimal.</p> <p>In this case the higher up the collar beam is the more bending would be present in the rafters which can result in some outwards movement of the walls. This is not likely to be an issue in a shed but could lead to some drywall cracking in a finished interior.</p>

8679

Steel Roof Truss Shape Question

2016-05-01T13:58:36.513

<p>I am trying to estimate the volume of excavation needed for the foundations of a structure from the drawings provided. </p> <p>The drawings clearly give the dimensions of the RC / cement footing but I suppose the actual excavation will need to be larger than this? How much larger should I plan for in the costing calculations. </p> <p>Any rules of thumb? </p> <p>Another issue that comes to mind is that although the RC / cement footing structures are rectangular or various dimensions, any practical excavation would have a slight slope in the walls that increases the excavation volume? Or is that not so significant?</p>

|structural-engineering|foundations|

<p>I don't know of any code which defines this, but a one meter "working perimeter" around the footing is probably more than enough.</p> <p>Regarding the slope of the excavation walls, that's trickier to define with a blanket statement. The problem is that the slope is necessary to give the soil stability. Unfortunately, the soil's stability is highly dependent on its properties (<a href="https://en.wikipedia.org/wiki/Cohesion_(geology)" rel="nofollow">cohesion</a>, <a href="https://en.wikipedia.org/wiki/Angle_of_repose" rel="nofollow">angle of repose</a>, etc), so the requisite angle to stabilize the excavation is also a function of these properties. Therefore, a blanket statement of "a slope of X is adequate" can't be given without knowledge of the type of soil being discussed.</p> <p>That being said, for non-problematic soils (i.e. non-expansive and non-collapsible soils, with a reasonable SPT, etc), a long-term excavation can usually be given a 2:3 slope (two vertical for three horizontal), so adopting such a slope for a temporary excavation would be very conservative. Depending on the soil, I believe even a 1:1 slope could be possible for a temporary excavation.</p>

8683

Foundation Excavation: How much larger than the RCC footing?

2016-05-02T09:08:33.140

<p>I need to modify a chuck's jaws so they can handle high temperatures, but will not scratch the workpiece. (Glass). The load will be very small. </p> <p>I will probably use aluminium jaws and either coat them with, or attach to them, this material. </p> <p>Example items being held:</p> <ul> <li>Borosilicate glass tube, 30mm long, 6mm diameter, 5g weight</li> <li>Borosilicate glass tube, 50mm long, 30mm diameter, 50g weight</li> </ul> <p>The tube will be held horizontally by the chuck, and should be supported such that the other end will not droop by more than 2mm. That end is held in a propane bunsen flame, on the edge of the hottest part of the flame: a cone 15-20mm height, with a base ~12mm, 1200-1400°c.</p> <p>I do not have figures on what the chuck jaws will have to withstand, but the air passing over the tip of the jaws probably wouldn't be more than about 250°c, and the temperature of the glass at that point will be about the same.</p> <p>Of issue might be the heat radiated from the flame - so a heat reflecting material will help.</p> <p>Also, since the rest of the chuck will be barely above room temperature, it may help to have thermally conductive material to dissipate heat from the tips of the jaws to the chuck.</p> <p>Plastics seemail like an obvious choice for softness, but high temperature plastics seem hard to find. Silicone (260°c) and PTFE (204°c) are barely temperature resistant enough, but so far seem like the best choice, since higher temperature polymers like Vespel (300-400°c) are very hard to come by. </p> <p>What alternatives are there? Metals? Composites?</p> <p>The only other approach I might use is a heat shield directly in front of the chuck, but it would be awkward, as it would have to fit quite snugly around the workpiece, different shields would be needed for different size workpieces, and would be time consuming if batches of workpieces need to be processed. I can't imagine how this could be automated for large batches. Even then, scratching the workpiece could be a problem.</p>

|materials|

<p>Carbon fibre or kevlar textile might work as a soft covering on an aluminium backing plate </p> <p>In terms of metals high purity annealed copper <em>might</em> be soft enough on it's own. You can also get self adhesive copper foil tape.</p> <p>It may also be worth considering using semi-consumable sleeves or collets. For example a hardwood dowel, bored to the appropriate diameter with a couple of slits along its length might do the trick. Even if it chars a bit at the end they might last long enough to be worthwhile. </p> <p>Disposable sleeves also have the advantage that there is less risk of picking up swarf or grit that could scratch the glass. </p> <p>Similarly it might be worth looking at O rings as a cheaper way to protect the rod than manufacturing jaws from stock, or maybe even use PTFE tape. </p> <p>It might also be worth trying some dipping the glass in a natural or synthetic rubber to protect it. </p> <p>In terms of heat shielding maybe look at small, disposable aluminium pie tins you could just punch a hole in the middle and suspend it by the glass rod itself. this might give just enough protection to make an otherwise marginal material viable. Similarly blowing air across the chuck with a portable fan or shop vac on blow mode might do just enough. </p> <p>Another possible option is <a href="http://www.tufnol.com/materials-full/glass-fabric-grades/tufnol-grade-10g-50.aspx" rel="nofollow">Tufnol</a>, this is a phenolic-resin-based composite, laminated with glass, paper, or other textiles and has somewhat better heat resistance than a lot of other composites and is reasonably cheap and available as sheet stock. </p>

8688

Materials which withstand high temperatures, and are relatively soft, for chuck jaws

2016-05-02T12:10:35.960

<p>I am an electrical engineer, making a hobby project on an automated unwanted-material-removal system in grains and pulses (rice, wheat, etc). I am well versed with programming and electrical stuff, but I am stuck at one point with the mechanical design.</p> <p>My system uses a camera located on top of a conveyor belt. The camera scans the belt and identifies various impurities on the belt (using an intelligent machine learning algorithm). A brush is positioned further down the belt that pushes the impurities off the belt.</p> <p>My idea for the system is to simply make the device sit in any container, with the belt part projecting out of the container. A tumbler below the belt will collected the purified grains. Sort of the things that you see in rock crushing plants.</p> <p>So, for this, I need to design a mechanical system that lifts the rice and puts / pours onto the conveyor belt. I am unable to come up with any solid design, given that I am from an electrical engineering background. And the system should be low powered, dust resistant and should work with one slowly rotating motor (or maybe a fast motor with a set of step-down gear system)</p> <p>Can anyone suggest me a good starting point for such systems or what design I can use to achieve the same?</p> <p>Note: The quantity of rice or grains moved in a unit time is very small, say a handful.</p>

|mechanical-engineering|design|structures|

<p>There are several ways to lift grains: </p> <ul> <li>auger/screw conveyor</li> <li>cup elevator - like a vertical conveyor belt with cups mounted to it, cups tilt near top <ul> <li>There are also systems with little plates on a string, beeing pulled through a pipe (i forgot the name)</li> </ul></li> </ul> <p>All of the above are used in large scale grain handling. </p> <p>However, given that the quantitiy is 'a handful', I would consider another approach: Consider a hopper above a <a href="https://en.wikipedia.org/wiki/Rotary_feeder" rel="nofollow">rotary feeder</a> - a bunch of chambers on a wheel. Bulk goods fall into the chamber when it's on top, the wheel turns slowly and the bulk goods fall from the chamber when it's in the bottom position. This allows the dosing of bulk goods. You can use a rotary feeder with a large hopper and forgoe the conveyor.</p>

8691

Mechanism to continously lift grains and put them on a conveyor belt

2016-05-02T12:13:16.833

<p>One of my garage door torsion springs broke again. It makes me wonder why some sort of counterweight system isn't used instead. Torsion springs seem to have a 90 day warranty generally and are "rated" for up to 15,000 cycles. It seems like there is a lot less stress on a system that uses counterweights, <a href="http://hermco.ca/products/2000-series/" rel="noreferrer">this company</a> has a counterweight system for high use places (condo buildings or businesses with large garages or car washes) with warranties for 100,000-500,000 cycles.</p> <p>Is it initial cost? Space required? Just to keep repairmen in business? </p>

|springs|building-design|

<p>The reason counter weights are not used anymore is due to a couple factors. The limited space like they mentioned in the above comments but the main reason for smaller residential garages the added weight of having heavy weights hanging on the side of the door header does nightmares to the structure over the years. I'm dealing with a job right now where the entire front wall has warped due to have heavy counter weights hanging for years. When a spring is used all the tension is on the spring itself and not on the structure.</p>

8692

Why aren't counterweights used more for garage doors?

2016-05-02T18:17:17.700

<p>When building a 12 m tall industrial shed with a peaked roof it is desirable that the sides of the shed should stay open as much as possible for ventilation / safety / access concerns. </p> <p>On the other hand, an entirely open side from top to bottom might cause a lot of rain exposure when the rain comes in angled?</p> <p>Is there a trade-off possible? Perhaps to enclose sides till a certain distance below the roof edge? Any heuristics about how much? (I've shown 4000 mm in the sketch below)</p> <p><a href="https://i.stack.imgur.com/fnqTL.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/fnqTL.png" alt="enter image description here"></a></p> <p>The contents inside the shed are relatively robust / waterproof &amp; hence absolute rain protection is not needed. </p>

|structures|

<p>This will very much depend on the exposure of the building to wind as this is what will drive the rain in at an angle. So you really need to consider the prevailing wind direction and how sheltered the building is by other nearby structures etc. </p> <p>If may be that if the prevailing wind direction is fairly consistent you can have one side more open than another. </p> <p>Another option is to have panels which can be folded down when required eg if you have 4000mm of fixed cladding, you could have the same in fold-down form (this arrangement is fairly common in barns).</p> <p>Also it depends what access you need obviously forklifts will need more height clearance than pedestrians. </p> <p>You could also consider using strip curtains for at least part of the height. </p>

8697

Siding for Industrial Shed: Rain protection vs Ventilation

2016-05-03T08:27:38.007

<p>I've started playing around with brazing pieces of extruded aluminium sections to join them together for strong joins as an alternative to bolting them.</p> <p>The pieces are made of 6082-T6 aluminium alloy. I've found that the heating needed to braze softens the aluminium enormously, to the point where 3mm thick sections can be effortlessly deformed with pliers. </p> <p>I've done a lot of searching and I need to either naturally or artificially age-harden it (precipitation hardening). But how long it takes to naturally harden, or the precise process needed to artificially harden it (temperature &amp; time) escapes me. Also, quenching might help, but I haven't found any definitive answers.</p> <p>How can I re-harden the alloy?</p>

|materials|brazing|

<p>While brazing can be useful for ad-hoc repairs of aluminium AC TIG or MIG welding is vastly preferable for structural applications and in the long run is likely to be less hassle than re heat treating it. </p> <p>The advantage of TIG welding in this context is twofold: a) you can use filler metal which is as strong or stronger than the base metal and with an alloy composition optimized for welding; b) you can keep the heat input into the metal to a minimum as the arc is a much higher temperature than a brazing flame and much more concentrated so the heat goes into he weld pool only and not the bulk material. Also TIG welds are cleaned by the arc itself rather than requiring fluxing, which itself requires more heat put into the base metal. </p> <p>From personal experience you can TIG weld aluminum extrusions with little noticeable loss of tensile strength, this is in the application of structural racking for commercial vehicles. I also have expedience of annealing and cold working extruded aluminium for similar applications. </p> <p>Obviously your mileage may vary depending on application but I would certainly recommend looking seriously at TIG welding for this sort of application </p>

8709

6082 Aluminium softened by brazing. How to harden?

2016-05-04T04:04:37.290

<p>I understand that the concept of line balancing is such that each station takes exactly the same amount of time to process one unit. Clearly, this is an idealization and real manufacturing plants will have bottleneck stations and underutilized stations.</p> <p>I'm wondering if someone has an idea on what numbers are typical in various industries? For instance, if there is a 10% take difference between the slowest and fastest stations, is the line considered balanced enough? </p>

|manufacturing-engineering|engineering-economics|

<p>I can answer for the Chemicals Sector I'm familiar with: Overall line balancing is quite poor between processing operations. 10% difference would be surprisingly good. 50% difference isn't uncommon. </p> <p>Several reasons for this I can think of:</p> <ol> <li>Lot of equipment comes in discrete sizes. e.g. Compressors, Centrifuges etc So often the next size higher means you have a lot of slack built in that equipment from Day-0.</li> <li>Often the exact max capacity of a complex piece of equipment like a distillation column is not known with 100% accuracy. So designers incorporate a factor of safety in mentioning the rated capacity (that will become a contractual obligation). Once plant is operational it can be pushed to max capacity which is often higher than the on-paper capacity the equipment was designed for. Ergo now other pieces of equipment become bottlenecks. And you have an unbalanced line. </li> <li>Some equipment like Heat Exchangers or Plug Flow Reactors will foul over their working life and hence become less effective. There can be a wide margin between their Day-0 throughput and 6 months down the line. As a result, other equipment not suffering from such in-service derating will end up showing excess capacity. </li> <li>Often Capex pricing of specific equipment is highly non-linear with respect to capacity. e.g. You could be adding 50% extra capacity at only 10% extra price. So designers end up having unbalanced lines simply by hedging their risks by over-specifing where the cost penalty isn't high.</li> <li>Over the long life of a plant (e.g. 20 yrs) some innovations come up where with modest investment a equipment can be significantly boosted in throughput. e.g. Better generations of catalysts. The parts of the train that <em>cannot</em> be innovated upon then become bottlenecks. </li> <li>The lead times for new equipment commissioning can be long. e.g. 3 years. So sometimes management will knowingly add larger equipment, consciously realizing that the new equipment will stay underutilized for a while till the other parts of the line can be upgraded.</li> <li><p>Employees &amp; site-engineers come up with operating innovations that debottleneck a specific equipment. Suddenly the rest of the line is underutilized. </p> <p>8.Chemical processing lines can make multiple products. e.g. </p> <p>A-->B-->C-->D-->E</p></li> </ol> <p>where both C &amp; E are salable products. The plant was designed line-balanced assuming you'd sell 500 tons of E &amp; 300 tons of C every month. Unfortunately 5 years down the line the market for C is bad. So parts of the line pre-C become underutilized although the rest of the line is still balanced. </p>

8726

Line balancing - How far do companies go?

2016-05-06T08:11:28.473

<p>For example, metal tape measures can be extended and stay straight when in a "u" orientation, but collapse the other way round.</p> <p>I presume the same phenomenon is why metal shelving has sheet metal on top, and flanges on the bottom rather than the top.</p> <p>Why is this?</p> <p><strong>I'm guessing</strong>: it's because of the direction of compression/tension, and buckling. Since a long piece of material being compressed is more likely to buckle than a short piece. In one orientation, the buckling can only happen when the walls buckle across their height but in the other orientation, the compression acts across the entire length of the channel?</p> <p>Both a common-sense approach, and a mathematical approach are welcome; I assume formulas are well established, using "moment of inertia" values for beam profiles or similar.</p>

|applied-mechanics|

<p>You are correct.</p> <p>A channel section in a "u" or "n" orientation is obviously asymmetric around its transversal x-axis. This means that its centroid is not located at the midpoint of its height. Instead, the centroid will be closer to its web than to the opening.</p> <p>Since the stress at a given fiber at a distance $y$ from the centroid is equal to $$\sigma = \dfrac{My}{I}$$ in bending, this means that the stresses in the web will be lower than the stresses at the ends of the flanges.</p> <p>Using your metal tape measure example, an open tape measure without other supports will behave like a cantilever, with its rotation fixed at the "mouth" of the tape measure. It will therefore be under negative bending moment (compression on bottom).</p> <p>Since the tape measure is very slender, it is more sensitive to buckling due to compression than to simple collapse due to tension. Therefore, in order to avoid buckling, you want to have your tape measure in a "u" orientation, which will reduce the compressive stresses, while increasing the tensile stresses. In an "n" orientation, the stresses are inverted, with low tension and high compression (and therefore lower resistance to buckling).</p>

8756

"u" vs "n" orientation of channel profiles: why difference in strength?

2016-05-06T08:49:27.767

<p>I want to compare strength of a metal before and after heat treatment, to assess to effectiveness of that treatment. (See my previous question, <a href="https://engineering.stackexchange.com/questions/8709/6082-aluminium-softened-by-brazing-how-to-harden">Aluminium softened by brazing. How to harden?</a>).</p> <p>Specifically, I want to find the yield point (the strain a material can take without being deformed, ie the range where it remains elastic).</p> <p>I have two ideas that use a similar setup. Shove a length of material (constant width, height, length, shape) in a vice. Apply a force to the end. Fasten a $1 digital luggage scale to the other end. Hang a container of water or sand, adding a little each time. Deflection could be measured as the distance from that end to some reference point with a digital caliper.</p> <p>Then, either:</p> <ol> <li>Apply an increasing force, until the force is no longer proportional to the deflection (since up till the yield point, materials bend according too hooke's law). Or,</li> <li>Apply a force and release. Repeat with increasing forces until the material no longer returns to its original shape.</li> </ol> <p>The problem is, to achieve any real accuracy, both 1 and 2 above would involve 10's of manual iterations per assessment. </p> <p>Besides buying or hiring a rheometer, are there any practical alternatives?</p>

|measurements|yield-point|strength|

<p>Determining the yield point for aluminium alloys can be difficult as it is not as well defined as steel and the stress/strain curve tends to gradually peel away from the straight elastic line rather than there being a distinct cut-off. </p> <p>For a low tech experiment ultimate tensile strength is easier to measure as you are more likely to observe an obvious failure rather than interpolating from a graph. </p> <p>Similarly using a cantilever loading setup (i.e. bending it with a mass hung off one end) is a bit easier to arrange and should make the failure point a bit more obvious. This makes it more difficult to derive actual numbers for stress but it will certainly make direct comparison easier, especially if you have a known 'good' sample to compare it to. </p> <p>Another possibility for this sort of application is to use a hardness tester, very accurate one's aren't cheap but it's possible that a cheap one may do the trick, depending on the range you are looking at. </p> <p>Also if you are looking at a practical performance test for fabricated assemblies then basic load test which is reasonably close to real world conditions is as good a bet as anything. Bearing in mind also that stress raisers from any joining method may have an effect on practical strength so you may as well check if brazing is any better than bolts for your application. </p>

8757

Low-tech ways of assessing relative strength of a material (yield point)

2016-05-06T14:50:46.373

<p>I am performing a calculation in which I need to plot a generated response spectrum to a seismic tripartite graph. However, I am having a heck of a time finding a blank tripartite graph online. A low-resolution example is shown below.</p> <p><a href="https://i.stack.imgur.com/PhEeg.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/PhEeg.png" alt="tripartite graph" /></a></p> <p>How might I go about generating a blank seismic tripartite graph? As far as I know, Excel does not have this functionality, since you need to be able to plot four separate axes in 2D with two of the axes at an angle. All axes are in logarithmic scale.</p> <p>The unit system should be as such:</p> <ul> <li>Spectral displacement: inches</li> <li>Spectral velocity: in/sec</li> <li>Spectral acceleration: % of <span class="math-container">$g$</span></li> <li>Frequency: Hz</li> </ul> <p>Otherwise, a link to a high-resolution graph would be acceptable.</p> <hr /> <p>Edit January 29, 2021</p> <p>I have figured out how to make one of these graphs using LaTeX. Please see <a href="https://www.glennfrazee.com/eng-tools/tripartite-paper" rel="nofollow noreferrer">this post on my website</a> for details.</p>

|seismic|frequency-response|

<p><a href="https://i.stack.imgur.com/XC36W.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/XC36W.png" alt="Tripartite Paper - Period on X-axis"></a> This is the only place on the internet that I could find that actually gave decent resolution images for tripartite plotting paper. However, it had some limits that affected use in our area: 1) the design earthquakes top out in our area at about 1g and 2) we generally are only concerned with periods between 0.01s and 10s, or 0.1 Hz to 100 Hz. This resulted in plots of the earthquake crammed down in one corner, and the 144-yr operating basis earthquake barely showed up at all. Also, we generally plot with the period, rather than the frequency, on the X-axis.</p> <p>I took a weekend to learn just enough R to modify Biswajit Banerjee's excellent script to come up with a plot that would serve my purposes. I'm going to throw it up in case it's useful for anybody else. If you've not worked with R, you can install the basic R interpreter, cut and paste either Mr. Banerjee's or my code into a script, and run it to get a higher-quality PDF output.</p> <p>It didn't take long to learn enough to actually plot the earthquakes using relatively easy code, but the way I did it is extremely user-unfriendly, if effective.</p> <pre><code>require(ggplot2) require(data.table) # The constant value grid lines periods = unique(c(seq(0.01, 0.02, by = 0.005), seq(0.02, 0.1, by = 0.01), seq(0.1, 0.2, by = 0.05), seq(0.2, 1.0, by = 0.1), seq(1.0, 2.0, by = 0.50), seq(2.0, 10.0, by = 1.0))) vels = unique(c(seq(0.01, 0.02, by = 0.005), seq(0.02, 0.1, by = 0.01), seq(0.1, 0.2, by = 0.05), seq(0.2, 1.0, by = 0.1), seq(1.0, 2.0, by = 0.50), seq(2.0, 10.0, by = 1.0), seq(10.0, 20.0, by = 5.0))) accs = unique(c(seq(0.0001, 0.0002, by = 0.00005), seq(0.0002, 0.001, by = 0.0001), seq(0.001, 0.002, by = 0.0005), seq(0.002, 0.01, by = 0.001), seq(0.01, 0.02, by = 0.005), seq(0.02, 0.1, by = 0.01), seq(0.1, 0.2, by = 0.05), seq(0.2, 1.0, by = 0.1), seq(1.0, 2.0, by = 0.50), seq(2.0, 10.0, by = 1.0), seq(10.0, 20.0, by = 5.0), seq(20.0, 100.0, by = 10.0))) disps = unique(c(seq(0.00001, 0.00002, by = 0.000005), seq(0.00002, 0.0001, by = 0.00001), seq(0.0001, 0.0002, by = 0.00005), seq(0.0002, 0.001, by = 0.0001), seq(0.001, 0.002, by = 0.0005), seq(0.002, 0.01, by = 0.001), seq(0.01, 0.02, by = 0.005), seq(0.02, 0.1, by = 0.01), seq(0.1, 0.2, by = 0.05), seq(0.2, 1.0, by = 0.1), seq(1.0, 2.0, by = 0.50), seq(2.0, 10.0, by = 1.0), seq(10.0, 20.0, by = 5.0))) # Horizontal grid lines (const vel = in/s) vel_grid = data.table() periodmin = min(periods) periodmax = max(periods) for (vel in vels) { xx = c(periodmin, periodmax) yy = c(vel, vel) label = c(paste0("vel",vel), paste0("vel",vel)) color = "vel" local_dt = data.table(x = xx, y = yy, label = label, color = color) vel_grid = rbind(vel_grid, local_dt) } # Vertical grid lines (freq = cycles/sec) period_grid = data.table() velmin = min(vels) velmax = max(vels) for (period in periods) { xx = c(period, period) yy = c(velmin, velmax) label = c(paste0("period",period), paste0("period",period)) color = "period" local_dt = data.table(x = xx, y = yy, label = label, color = color) period_grid = rbind(period_grid, local_dt) } # 45 grid lines (constant acc = omega * vel = (2*pi/period) radians/sec * vel in/sec # = (2*pi/period)*vel in/sec^2 = (2*pi/period)*vel/g Gs) gravity = 32.2*12.0 # in/s^2 acc_grid = data.table() for (acc in accs) { for (period in periods) { Ts = period vel = acc*gravity*period/(2*pi) if (vel &lt; velmin) { vel = velmin Ts = vel*2*pi/(acc*gravity) } if (vel &gt; velmax) { vel = velmax Ts = vel*2*pi/(acc*gravity) } xx = c(Ts) yy = c(vel) label = c(paste0("acc",acc)) color = "acc" local_dt = data.table(x = xx, y = yy, label = label, color = color) acc_grid = rbind(acc_grid, local_dt) } } ## The const acc=1 line is the disp axis acc_1 = acc_grid[acc_grid$label == "acc0.01"] disp_axis = data.table() for (disp in disps) { acc = 0.012 omega = sqrt(acc*gravity/disp) period = (2*pi)/omega vel = omega*disp xx = c(period) yy = c(vel) label = c(paste0("disp_axis",acc)) color = "disp_axis" local_dt = data.table(x = xx, y = yy, z = disp, label = label, color = color) disp_axis = rbind(disp_axis, local_dt) } # -45 grid lines (const disp = vel/omega = vel/(2*pi/period) in) disp_grid = data.table() for (disp in disps) { for (period in periods) { Ts = period vel = disp*(2*pi)/period if (vel &lt; velmin) { vel = velmin Ts = disp*2*pi/vel } if (vel &gt; velmax) { vel = velmax Ts = disp*2*pi/vel } xx = c(Ts) yy = c(vel) label = c(paste0("disp",disp)) color = "disp" local_dt = data.table(x = xx, y = yy, label = label, color = color) disp_grid = rbind(disp_grid, local_dt) } } ## The const disp=0.1 line is the acc axis disp_1 = disp_grid[disp_grid$label == "disp0.1"] acc_axis = data.table() for (acc in accs) { disp = 0.12 omega = sqrt(acc*gravity/disp) period = 2*pi/omega vel = omega*disp xx = c(period) yy = c(vel) label = c(paste0("acc_axis",acc)) color = "acc_axis" local_dt = data.table(x = xx, y = yy, z = acc, label = label, color = color) acc_axis = rbind(acc_axis, local_dt) } # Join vertical and horizontal grid = rbind(vel_grid, period_grid, acc_grid, disp_grid) ### Axis label points disp = 0.08 acc = 0.002 omega = sqrt(acc*gravity/disp) period = 2*pi/omega vel = omega*disp acc_axis_label = data.table(x = c(period), y = c(vel), label="Spectral acceleration (g)") disp = 0.006 acc = 0.008 omega = sqrt(acc*gravity/disp) period = 2*pi/omega vel = omega*disp disp_axis_label = data.table(x = c(period), y = c(vel), label="Spectral displacement (in)") # Plot horizontal grid xticks = periods[c(TRUE, FALSE)] yticks = vels[c(TRUE, FALSE)] plt = ggplot() + geom_line(data = grid, aes(x = x, y = y, group = label, color = color), size=0.25) + geom_line(data = acc_1, aes(x = x, y = y), color = 1, size=0.5) + geom_line(data = disp_1, aes(x = x, y = y), color = 1, size=0.5) + xlab("Period (s)") + ylab("Spectral velocity (in/sec)") + scale_color_discrete(guide = FALSE) + scale_x_log10(limits = c(periodmin, periodmax), expand = c(0,0), breaks=xticks, labels= c("0.01","0.02","0.04","0.06","0.08", xticks[xticks &gt;= 0.1])) + scale_y_log10(limits = c(velmin, velmax), expand = c(0,0), breaks=yticks) + coord_fixed() + theme_bw() + theme(panel.grid.major = element_blank(), panel.grid.minor = element_blank()) + annotate("text", x = acc_axis$x, y = acc_axis$y, label = as.character(acc_axis$z), size = 3, angle = 45) + annotate("text", x = disp_axis$x, y = disp_axis$y, label = as.character(disp_axis$z), size = 3, angle = -45) + annotate("text", x = acc_axis_label$x, y = acc_axis_label$y, label = acc_axis_label$label, size = 4, angle = -45) + annotate("text", x = disp_axis_label$x, y = disp_axis_label$y, label = disp_axis_label$label, size = 4, angle = 45) print(plt) ggsave("TripartitePaper.pdf",width = 10.0, height = 10.0, units = "in") </code></pre>

8760

Generating a Blank Seismic Tripartite Graph

2016-05-06T17:18:34.523

<p>Hello I was watching a video recently about aircraft carriers and how they use compressed air to propel aircraft off the ship. It go me thinking if I had, maybe a 50cm x 50 cm x 50 cm spacecraft that was put into orbit. Would I be able to use compressed air as thrust to orientate the space craft? or even better would I be able to use it to give the space craft a greater distance from the Earth?<img src="https://i.stack.imgur.com/fr0dE.jpg" alt="enter image description here" /><img src="https://i.stack.imgur.com/v93bG.jpg" alt="enter image description here" /></p>

|aerospace-engineering|compressed-air|

<p>The <a href="https://en.wikipedia.org/wiki/Simplified_Aid_For_EVA_Rescue" rel="nofollow">SAFER</a>, used by NASA for EVAs ("Space Walks"), uses compressed <a href="https://www.nasa.gov/missions/shuttle/f_saferspacewalk.html" rel="nofollow">nitrogen</a> for thrust. Air is mostly nitrogen, and the properties are fairly similar. So, essentially it not only could be done, but even has been done.</p>

8763

Can Compressed air be used as a thrust component in a spacecraft to propel the craft through space or to orient the craft?

2016-05-07T05:16:00.917

<p>Is there a way to derive a formula of modulus of rupture (flexural strength) vs compressive strength?</p> <p>I want to know is there any mathematical reason for the $\sqrt{f_{c} ^{'}}$ placed somewhere in the formula?</p>

|structural-engineering|civil-engineering|concrete|

<p>Short answer: No.</p> <p>Longer answer:</p> <p>The code equations (e.g. ACI) relating modulus of rupture with compressive strength cannot be mathematically derived. It is an empirical relationship based on curve fitting to extensive experimental data.</p> <p>The ASTM test for flexural strength (i.e. Modulus of rupture) is <a href="http://www.astm.org/Standards/C78" rel="nofollow noreferrer">ASTM C78/C78M</a>.</p> <p>In essence, the most common test consists of loading an unreinforced simply-supported concrete beam at third points and determining the load at which fracture occurs. From this data, the modulus of rupture can be <a href="https://ceen.et.byu.edu/sites/default/files/snrprojects/26-kirtikumar_k_shah-1969-kwk.pdf" rel="nofollow noreferrer">calculated</a>.</p> <p><a href="https://i.stack.imgur.com/LCGAZ.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/LCGAZ.jpg" alt="Flexural strength test"></a></p> <p>There is generally a large amount of scatter in modulus of rupture data because a number of factors affect $f_r$ (mix design, aggregate size, rate of loading, specimen size, etc.)</p> <p>Many researchers have studied the relationship between modulus of rupture and compressive strength. Relating $f_r$ to $\sqrt f'_c$ seems to provide good correlation. Data from one such researcher is shown below. (Admittedly, it's not the definitive reference, but even Google has its limits when it comes to academic publications and paywalls.) I think some of the more recent research has proposed a 2/3 power relationship, though that may only be for high strength concrete.</p> <p><a href="https://i.stack.imgur.com/2DIXP.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/2DIXP.jpg" alt="Relationship between modulus of rupture and compressive strength"></a></p>

8769

Is there a way to derive a formula of modulus of rupture vs compressive strength?

2016-05-07T10:25:50.643

<p>I know comparatively little about general engineering methodologies, but it strikes me that the (now outdated) "waterfall" method is possibly very apt for mechanical engineering. Perhaps advances in rapidly changing software engineering could also be applied.</p> <p>Are software methodologies influencing other engineering disciplines? Or vice versa? </p> <p>Wikipedia's page on <a href="https://en.m.wikipedia.org/wiki/Engineering" rel="noreferrer">engineering</a> says:</p> <blockquote> <p>Koen argues that the definition of what makes one an engineer should not be based on what he produces, but rather how he goes about it.</p> </blockquote> <p>Which vaguely implies that development processes are to some degree ad-hoc, given that neither the <a href="https://en.m.wikipedia.org/wiki/Engineering#Methodology" rel="noreferrer">methodology</a> section of that page, nor the <a href="https://en.m.wikipedia.org/wiki/Engineering#Problem_solving" rel="noreferrer">problem solving</a> subsection of the engineering page make any references to development stages in the precise ways that is widely encouraged in software engineering.</p> <p>So why the apparent lack of processes?</p>

|project-management|

<p>Software engineering is a relatively newer form of engineering so borrows more than leads - the agile methodologies borrowed heavily from <a href="https://en.wikipedia.org/wiki/Lean_manufacturing" rel="nofollow">lean manufacturing</a>, <a href="https://en.wikipedia.org/wiki/Toyota_Production_System" rel="nofollow">Toyota</a> and <a href="https://en.wikipedia.org/wiki/Kanban" rel="nofollow">Kanban</a>. The Wiki summary of Toyota principles is very close to the <a href="http://www.agilemanifesto.org/" rel="nofollow">agile manifesto</a>.</p>

8771

Are latest software engineering methodologies, such as Agile, being adopted by other engineering disciplines?

2016-05-07T12:35:26.483

<p>What are the differences between the processes of 3D printing with plastics versus metals? Why is metal such a problem? There are plenty of metals with low melting points. And AFAIK metal can be sprayed and directed in a number of ways such as with electric arcs and similar.</p>

|mechanical-engineering|3d-printing|

<p>There are a number of practical processes for 3D printing in metals using various powder sintering techniques as described in other answers. </p> <p>However it is difficult to use the sort of direct extrusion processes that you see with plastics because most metal alloys don't have the same sort of viscous, self adhesive phase that thermoplastics do. Most molten metals are fairly fluid, cast iron, steel, bronze and aluminium have a viscosity not too far off water when melted if there is an intermediate phase it is usually a slush mixture of solid and liquid phases rather than something which can be extruded and adheres to itself.</p> <p>There is also the consideration that consumer 3D printing is designed for relatively inexpensive low volume production, so trying to do the same thing with metal is likely to be self defeating. In most cases it is likely to be easier to print a pattern in wax and plastic and investment cast it if you want it in metal.</p> <p>Similarly in the case of low melting point alloys it is just as cheap and almost easy to make a silicone mould from a master and cast it plus the fact that modern resins may actually be superior to low melt metals in most respects. </p> <p>One example of this is is the models used for table top games. A decade ago these were injection moulded plastic for simple models and white metal for more detailed ones. Now they are almost exclusively cast resin.</p> <p>There is also the fact that metals are inherently a bit more versatile than plastics in terms of rapid prototyping and fabrication. So there are more competing options for low volume manufacturing of metal parts than plastic ones. </p> <p>In terms of direct equivalents to plastic filament extrusion in metal MIG welding is not that far off inn terms of process and a robotic version of the process is certainly reasonably plausible, however as already discussed the flow characteristics of metal mean that MIG certainly doesn't have anything like the resolution of plastic extrusion as there is a minimum voltage/current which is required to strike and arc. </p> <p>There is also the fact that a decent MIG welder which produces anything like a consistent weld bead would start at around £500 with the 3 axis control on top of that, plus the fact that MIG generates a lot of fumes, spatter and UV radiation which needs to be managed. </p> <p>TIG welding offers a bit more precision but is equivalently more expensive for the basic machinery and an automated setup would require additional wire feed equipment. </p>

8773

Why is 3D printing with metals so uncommon?

2016-05-08T10:23:14.167

<p>I'm dealing with hot combustion products flowing through heat exchangers and I'd like to perform some calculations regarding the flow properties (velocity, temperature, pressure and density) as it passes trough the heat exchanger.</p> <p>I've a 1D model already built and I'd like to test it and validate it, but I'm lacking some critical information.</p> <p>Is there any place where I can find the absolute viscosity [Pa*s] as a function of temperature (for pressures of 1atm, or close to it) for some select gases?</p> <p>I need to know the viscosity between 300K and 2500K for gaseous CO2, H2O, N2 and O2. In polynomial form (something similar to the NASA/JANAF polynomials) would be nice, but in table would also be good.</p> <p>I've already tried NIST, but it doesn't have that information for the temperature range I'm interested in.</p> <p>Is it possible to run a macro for FLUENT, Star ccm+ or OpenFOAM and get those values from those programs? I have legal access to all of them.</p>

|fluid-mechanics|modeling|combustion|heat-exchanger|data|

<p>See NASA's computer program <a href="https://www.grc.nasa.gov/WWW/CEAWeb/ceaHome.htm" rel="nofollow noreferrer">CEA</a> (Chemical Equilibrium with Applications). From this site you can download an application to perform calculations. This application contains a text file called <code>trans.inp</code> with coefficients, $A$ through $D$, for an empirical dynamic viscosity equation: $$\ln\eta=A\ln T+\frac{B}{T}+\frac{C}{T^2}+D$$ and for thermal conductivity: $$\ln\lambda=A\ln T+\frac{B}{T}+\frac{C}{T^2}+D$$</p> <p>with $T$ in Kelvin, $\eta$ in $\mu$P (or $10^{-7}$Pa s), and $\lambda$ in $\mu$W/cm K (or $10^{-4}$W/m K).</p> <p>The file contains coefficients for multiple temperature ranges (up to 10,000 K for CO$_2$ and 15,000 K for H$_2$, N$_2$, and O$_2$) and also for binary interaction parameters, e.g., $\eta_{ij}$ for species $i$ and $j$, required for obtaining the viscosity of mixtures.</p> <p>From this site one can download NASA report RP-1311 which describes how to obtain viscosity of mixtures (see section 5.2.1) and RP-1311-P2 describing the file format (see TABLE E.1).</p>

8792

Viscosity of select species at atmospheric pressure for very large temperatures?

2016-05-08T11:19:50.903

<p>I have an thin uniform rectangular steel plate supported on 3 sides and free on one, and supporting a uniform load.</p> <p>I'm trying to determine what thickness of plate I need to keep deflection below a given value, but the deflection may not be "small" compared to the plate thickness, so standard formulae for maximum deflection may not be accurate, and I'm stuck on how to check my answer, or how to calculate it more accurately.</p> <p>Data: </p> <p>The plate is a flat rectangular plate (stainless 316 or mild, undecided) anything from 1.5 -15mm thick with an unsupported area at its edge of 1250x500mm total 0.625m^2. It is simply supported along the 500-1250-500 edges and free on the other 1250 edge. The unsupported area carries a static load of 4800 N/m^2 (approx 360kg spread uniformly over the unsupported area) plus its self weight. </p> <p>The 3 supported edges are unrestrained simple supports that can slide or rotate; they don't resist any movement except in the Z-direction (a bit like it's resting across the 3 edges of a "u" shaped pit).</p> <p>My question is the thickness of steel plate I need, to ensure maximum deflection (at the middle of the free edge?) stays under possible values of 3mm / 5mm / 10mm (the most likely permitted deflections or at least a good selection to choose between)</p> <p>The problem is that I am guessing solutions could be thicknesses around 1.5 - 5mm, which means that the deflection might not be "small" compared to its thickness and the usual simple calculation may not be very accurate or trustworthy. But I'm not sure....</p> <p>Thanks - and any hints how I can work this out myself appreciated but not essential :)</p>

|deflection|numerical-methods|

<p>If you use Mathematica or similar program you can create finite difference solution for bending of a plate. I have done similar calculation based on Timoshenko's work and rather accurate if you use reasonable number of elements. You can vary thickness and dimension and test your formulas, also you can upgrade code to include large deflection.</p> <p>I have tested my code with FEM but is too large to paste it here. It is based on this article: <a href="http://www.teamsociety.eu/team_conferences/team_2010/default.aspx" rel="nofollow noreferrer">Sertic</a></p>

8794

Deflection of thin plate with 1 free edge and deflection > thickness

2016-05-08T15:55:22.193

<p>Do you know if there is some sort of equipment that can measure the distance to a sound? What practical application can such an equipment have? The reason I'm asking is because I just got an idea of how to make such a machine. But it's really not hard to figure out so I guess there already exists such things. I didn't find anything on google though. Thanks.</p>

|acoustics|

<blockquote> <p>Do you know if there is some sort of equipment that can measure the distance to a sound [source]?</p> </blockquote> <p>A familiar example of this is estimating the distance to a lightning strike. The light arrives nearly instantaneously. The sound travels at 340 m/s at sea level (at standard temperature and pressure). By measuring the time between the light and sound arriving we can estimate the distance to the source. It's roughly 3 s/km or 5 s/mile.</p> <p>Without the light source the problem becomes more difficult. </p> <ul> <li>One microphone doesn't give you any location information. </li> <li>Two microphones may tell you the 'line' that the source is on but not the range. e.g. For two microphones 100 m apart a difference in time of 0.1 s (34 m) gives a location of the sound somewhere on an infinite curve that passes between them.</li> <li>A third microphone is required to triangulate the source.</li> </ul> <p>The system will have to deal with background noise, echos, changes is sound <em>ground</em> velocity due to wind, etc.</p>

8797

Measuring the distance to a sound

2016-05-09T00:39:42.670

<p>I was intrigued to read about so-called <a href="https://www.sciencedaily.com/releases/2015/07/150713095732.htm" rel="nofollow">"heat storage ceramic"</a> and was wondering how the energy stored in such way compares to energy stored in a NMC Li-Ion battery, if the latter is used to generate heat (the article lists it a 230kJ L^-1). I know it is not an apples-to-apples comparison, but I am curious about any ballpark assessment, especially in terms of energy stored per kg.</p> <p>The main reason for my interest is potential use in electric vehicles in cold climates. I drive an EV in Quebec, and I know that when the temperature is significantly below freezing, I can expect to spend 5-7KWh per 100km of driving simply to heat the interior. If energy density of "heat-storage ceramic" is significantly higher per weight or volume, it could make sense to install heat storage "batteries" alongside electrical batteries.</p>

|energy-storage|

<p>The energy density of lithium Ion batteries is between 900 kJ/L and 2430 kJ/L according to <a href="https://en.wikipedia.org/wiki/Lithium-ion_battery" rel="nofollow">Wikipedia</a>. This is at least a factor of 4 higher than the energy density you give for the heat storage ceramic so no gains in terms of space. </p> <p>Some <a href="http://global.kyocera.com/prdct/fc/list/tokusei/hiju/index.html" rel="nofollow">example ceramics</a> have a specific gravity ranging from 2.5 kg/L to 6.0 kg/L. At the lower end, the heat storage ceramics will have a specific energy of 92 kJ/kg while the lithium ion batteries range between 100 and 265. So, if the heat storage ceramics have a very low specific gravity, then you might save a little bit of weight over low density lithium ion batteries, but the gains won't be substantial.</p>

8806

Energy density of "heat storage ceramic" vs. NMC li-ion batteries

2016-05-11T01:27:22.703

<p>What are the relative pros and cons between aluminum and steel fasteners when riveting together sheet aluminum (50XX or 60XX) when high amounts of vibration are anticipated? Assuming that the usual guidelines for sheet thickness, hole size and separation are followed, which metal would be the better choice? I don't anticipate atmospheric corrosion to be an issue. </p>

|steel|aluminum|vibration|rivets|

<p>Steel is better than stainless steel, with the same kind of aluminum being even better; as far as corrosion is concerned. The difference between steel and stainless for strength won't matter as much for rivets as the potential for corrosion.</p> <p>See Wikipedia's <a href="https://en.wikipedia.org/wiki/Galvanic_corrosion#Anodic_index" rel="nofollow noreferrer">anodic index</a>:</p> <blockquote> <p>"The compatibility of two different metals may be predicted by consideration of their anodic index. This parameter is a measure of the electrochemical voltage that will be developed between the metal[s] <s>and gold</s>. To find the relative voltage of a pair of metals it is only required to subtract their anodic indices.</p> <p>For normal environments, such as storage in warehouses or non-temperature and humidity controlled environments, there should <strong>not be more than 0.25 V difference</strong> in the anodic index. For controlled environments, in which temperature and humidity are controlled, 0.50 V can be tolerated. For harsh environments, such as outdoors, high humidity, and salt environments, there should <strong>not be more than 0.15 V difference</strong> in the anodic index.".</p> </blockquote> <p>$\begin{array}{lc} \\ \text{Metal} &amp; \text{Index (V)} \\ \hline \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \text{Most Cathodic} &amp; \\ \hline \text{18% chromium type corrosion-resistant steels} &amp; −0.50 \\ \text{Chromium plated; tin plated; 12% chromium type corrosion-resistant steels} &amp; −0.60 \\ \text{Tin-plate; tin-lead solder} &amp; −0.65 \\ \text{Lead, solid or plated; high lead alloys} &amp; −0.70 \\ \text{2000 series wrought aluminum} &amp; −0.75 \\ \text{Iron, wrought, gray or malleable, plain carbon and low alloy steels} &amp; −0.85 \\ \text{Aluminum, wrought other than 2000 series, or cast alloys (silicon type)} &amp; -0.90 \\ \text{Aluminum, cast other than silicon type, cadmium, plated and chromate} &amp; −0.95 \\ \hline \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \text{Most Anodic} &amp; \\ \hline \end{array}$</p> <p>Note: The difference between stainless and aluminum.</p>

8824

Aluminum vs Steel rivets/fasteners in a high vibration environment

2016-05-11T06:54:03.970

<p>I am setting up an experimental station for which I am using a small gear pump to pump water. I have noticed that on the discharge port of the pump, a short row of bubbles comes out at high flows and low pressures. </p> <p>For example, at 650 ml/min and 2 bar of system pressure I might observe a relatively long row of bubbles coming out from the port. By increasing the system pressure up to 7 bar, the row shortens and eventually disappears. The same happens if I decrease the flow. I have checked the feed, and it does not contain any bubbles.</p> <p>This has led me to believe I am having some degree of cavitation in my pump, is this correct?</p>

|mechanical-engineering|pressure|pumps|

<p>Air leak in the suction. No cavitation ; cavitation voids ( almost vacuum ) do not leave the immediate area of the pump.</p>

8825

Small bubbles on a pump discharge

2016-05-11T12:29:14.987

<p>I am reading this story on artist Ugo Rondinone's sculpture in the Nevada desert (<a href="http://www.nytimes.com/2016/05/15/arts/design/building-an-artists-magic-mountains-to-draw-visitors-to-the-desert.html" rel="nofollow noreferrer">Building an Artist’s ‘Magic Mountains’ to Draw Visitors to the Desert</a>) and simply wondering (as an artist myself): how did they make those columns stay up?</p> <p>I didn't see any information on dimensions but from the photo with the artist below, one might guess that these columns are abut 25 feet high?</p> <p>Assuming there is a solid concrete footing for the columns to stand on and some preparation of the "joints" between the stones:</p> <ul> <li>Is the weight of the stones enough to keep them stacked and resistant to wind and weather (as such weather as there might be in the desert)?</li> <li>Is it likely that the stones are drilled to incorporate some steel armature?</li> </ul> <p>I understand that the stones could just be balanced, but given that this is a funded and likely insured project, I find that hard to believe likely.</p> <p><a href="https://i.stack.imgur.com/U9XgI.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/U9XgI.jpg" alt="enter image description here"></a></p> <hr> <p><a href="https://i.stack.imgur.com/DMD1I.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/DMD1I.jpg" alt="enter image description here"></a></p>

|structural-engineering|structures|

<p>I do this for a living and I would certainly say that having a single steel rod running the full height of the sculpture would be the idea way to do it. </p> <p>This method means that all of the major loads on the structure apart form its own weight in compression are supported by steel. </p> <p>If you have individual pins for each joint means that any side loads are transmitted into the stone as shear of bending loads which is far form idea for natural stone (or even concrete for that matter). </p> <p>Another very important consideration is the effect of water ingress into the structure and how this could cause problems with freeze/thaw cycles and corrosion. </p>

8826

How do the columns in the 'magic mountains' sculpture stay up?

2016-05-11T14:28:22.510

<p>I am working on a project for college, and I am stuck on one idea that I don't really know how to represent it. </p> <p>Basically what I want to do is something like a pipe, that if a put a kind of cylinder on inside from the top side, the cylinder will reach the end by force of gravity.</p> <p>So something like this:</p> <p><a href="https://i.stack.imgur.com/KPQfy.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/KPQfy.png" alt="cylinder going down"></a></p> <p>But, if I turn the pipe, so I put the bottom on top and the top on the bottom, that cylinder must be stopped/blocked.</p> <p>So something like this:</p> <p><a href="https://i.stack.imgur.com/9FlQq.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/9FlQq.png" alt="enter image description here"></a></p> <p>So my idea was to put some-kind of corrugated pipe as follows: </p> <p><a href="https://i.stack.imgur.com/LA64U.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/LA64U.png" alt="enter image description here"></a></p> <p>In this way the pipe will be one way only. As creating a pipe like this for a school project will be too hard, my questions is if is there already a pipe like this, if not if you have any other idea to achieve this.</p>

|mechanical-engineering|pipelines|

<p>This is known as a "valve" (duh :-) ). If you have enough room in your pipe, cut a circular piece of similar pipe whose diameter is a bit less than the pipe's diameter. then mount this curved plate with a hinge on one side and mount a small stop on the other side. You want it set up so that when the pipe is in the "go" orientation, the plate falls down due to gravity and since it's curved, it goes flat against the pipe's interior. In the "stop" config, the plate falls against the stop and blocks the pipe. </p> <p>Apologies for poor 3D rendering of the "potato chip" - shaped valve plate. Left picture is in blocking mode; right picture shows the tube with the other end up, in 'passthru' mode.</p> <p><a href="https://i.stack.imgur.com/eW7Ud.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/eW7Ud.jpg" alt="Tube up/down"></a></p>

8828

One way plastic rubber pipe

2016-05-11T15:11:35.843

<p>I am currently MS student in artificial intelligence (AI) and working on my thesis on reinforcement learning, which robot should learn by itself how to control a quadcopter.</p> <p>Long story short, I have come up with a method for such purposes, but before I jump into quadcopter and deal with its non-linear controlling problems, which could take a month to program, I need to test the method to see if it's really a solution for this kind of problem. I've defined a simplified version of my problem, such as following:</p> <blockquote> <p><strong>Simplified scenario:</strong></p> <p>I have a ball <strong>dropped</strong> at an altitude and the program must learn the <strong>force</strong> it needs to apply on the ball to hold it still in the air at a pre-defined altitude.<br> (i.e if target altitude is higher than the altitude that ball's dropped, it should up-force the ball until it reaches the altitude, otherwise, it should moderately stabilize the ball in the target altitude).</p> </blockquote> <p>In this scenario it doesn't matter how the force will apply to the ball, i.e. there is no rope attached or etc. it just gets applied.</p> <p>The last time I read something related to physics was 6-7 years ago, so the motion equation of this problem is beyond my specialty, but it seems to me that the defined scenario is a classic physic problem.</p> <h3>Questions:</h3> <ol> <li>What are the equations to formulate the scenario (falling ball with an external force besides gravity) to be able to write its simulator?</li> <li>What about adding some noise to the problem, like air resistance factor or wind, how would the equations look then?</li> </ol> <p>I would really appreciate if someone help a C.S. fellow here.</p> <p>Thanks in advance.</p> <hr> <p>PS: Please let the equations be simple as they could be, I am not quite good at reading complicated physic equations.</p> <hr> <p><strong>UPDATE:</strong></p> <p>From <a href="http://www.physicsclassroom.com/class/1DKin/Lesson-6/Kinematic-Equations-and-Free-Fall" rel="nofollow">this</a> site and details @Jodes kindly explained, I have come up with following equations:</p> <p>$$ x = {1 \over 2}at^2 + v_0t + x_0 $$ where $a = g + \sum {F_i \over m_{\text{ball}}}$ and $F_i$ are the external forces applied to the ball.</p> <p>If I'm right, the noises could also be combined to the equation as external forces</p> <ol> <li>Does this formulation valid for the defined scenario?</li> <li>If so, how to make it work in 2D and 3D environment(does simply vectorizing the $\vec{x}, \vec{a}, \vec{v}$ will do the job?)?</li> </ol>

|mechanical-engineering|simulation|experimental-physics|

<p>The force equation of an object in a uniform gravitational field with a mysterious compensating force is $$ ma=F_c-mg $$ Where $m$ is the mass of the object, $g$ is the acceleration due to gravity, $F_c$ is the force applied by your controller, and $a$ is the net acceleration. For simulation purposes, you probably want to express this as a differential equation $$ \frac{d^2}{dt^2}y=\frac{F_c}{m} - g. $$ If you want to add a random fluctuating force due to noise, $F_n$ then you can simply add it in on the RHS of the equation $$ \frac{d^2}{dt^2}y=\frac{F_c}{m} - g + \frac{F_n}{m}. $$</p>

8832

Need motion equations for falling ball for simulating

2016-05-13T02:49:53.967

<p>I would like to know how to calculate unit mass of beam having uniform cross section, made by shell, by using Patran or Nastran, not by manual.</p> <p>Would there be any functions to calculate mass or unit mass?</p>

|modeling|simulation|

<p>You can do it in Nastran using the <strong>Grid point weight generator</strong> option. Look up PARAM,GRDPNT in the documentation.</p> <p>Patran probably has something equivalent, but the Nastran output is "guaranteed" to be consistent with the model that you actually analysed.</p>

8855

Calculation of unit mass of beam shell by using patran or nastran

2016-05-13T10:41:53.537

<p>In an answer to <a href="https://engineering.stackexchange.com/questions/8160/how-to-calculate-power-of-motor-when-used-as-a-generator">this question</a>, for a motor used as a generator, the following was said:</p> <blockquote> <p>The maximum current that can be drawn from a generator is generally a fixed value that is not determined by speed.</p> </blockquote> <p>What equation(s) show this? </p>

|electromagnetism|

<p>That statement about the current being independent of speed is just plain wrong. This should be obvious by thinking about the limiting case. When the generator isn't spinning at all, it won't produce any current, but it will produce current at some finite speed.</p> <p>You can easily prove to yourself that the statement is wrong. Take a simple brushed DC motor. These also work backwards as generators. Connect it to a ammeter. That's essentially shorting its output while measuring the current. Now spin the motor (generator). The current will be proportional to the spinning speed.</p> <p>I didn't follow the link to where this quote came from. Perhaps there are various qualifying statements around it. However, taken by itself without some specific context, that statement is wrong.</p>

8858

Why is the current that can be drawn from an electric generator constant?

2016-05-13T12:23:02.843

<p>I'm currently filling out an application for the Massachusetts FE &amp; EIT certification. Section 22 requests information regarding memberships to professional organizations. Unfortunately, I'm not a member of any and I'm afraid of leaving this section blank. Is anyone familiar with how leaving this section blank would impact my chances of getting certified?</p>

|electrical-engineering|

<p>I am not from Mass nor do I plan to go through the EIT process again. Unless they state a requirement for you to complete X task, they should not be able to deny you access to the test based on not having completed the task. Personally, I believe they are more compiling statistics with that information.</p> <p>Joining professional organizations are a great thing and will help your business communications and to develop contacts.</p>

8861

How difficult is attaining EIT certification?

2016-05-13T18:23:49.623

<p>In order to give you a good idea of what I want to accomplish suppose you have a telescope mounted on a single rod extruding from the ground. The telescope is free to move such that you can tilt it to point at a particular altitude but can never point towards the ground. In order to move the telescope you have a wheel which can only be rotated in a single direction (say clockwise). Is there a gear configuration that would make a point alternate rotations every 180 degrees? and also have it configured such that the power loss is consistent throughout the entire cycle so that no half a cycle takes more power than the other? </p> <p>Here is an illustration</p> <p><a href="https://i.stack.imgur.com/Mh2NG.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Mh2NG.png" alt="Img"></a></p>

|mechanical-engineering|gears|applied-mechanics|

<p>A crank (or 4 bar) type linkage can be forced to do this. In a crank mechanism there is a a point at bottom dead centre and top dead centre where the rotation could go in either direction regardless of the direction of the driving wheel/arm. </p> <p>So you could have a stop which prevents the driven arm from rotating past 180 degrees, at which point it will naturally reverse direction. Note that steam trains have opposite wheels driven by cranks 90 degrees out of phase precisely to prevent this. </p> <p>You could also use half gears (ie gears with no teeth on half of their circumference) along with an idler gear on one to achieve a similar effect. </p> <p>There are also lots of different cam and multi-bar linkages which will produce reciprocating motion depending on your exact requirements. </p> <p>The difference between the myriad of possible solutions will depend to a large extent on the function which relates input to output, particularly how linear it is throughout the range of movement. </p> <p>For this sort of application it is also well worth looking through old patents as there are thousands and thousand of mechanical linkage patents out there for almost every imaginable mechanical application, many of which have now expired. </p>

8866

Is there a way rotate a wheel in a single direction but have opposing motions every 180 degrees?

2016-05-14T13:21:42.823

<p>From an Inspection and Test Checklist for flange heaters (the number is the same on both lines):</p> <blockquote> <p><a href="https://i.stack.imgur.com/3Kxp9.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/3Kxp9.png" alt="enter image description here"></a></p> </blockquote> <p>I wonder what the difference between "Design Drawing" (or is it "Drawings"?) and "Manufacturing Drawing(s)" is. Is it that Mfg. Dwg. merely represents some part of Design Dwg. (sheet no.3) that is copied for use in the production shop? Or are there some additions\changes? </p> <p>Sorry if this is a very basic question. I tried to translate this document and dropped the translation because I found it quite hard to understand many of the terms used, especially in the condensed context of a checklist. But I remembered some terms, and it made me curious. </p> <p>P.S. I was told on a translators' site that it could possibly mean "Manufacturer's Drawing Number" in this context. I wonder if that's true. </p>

|drafting|

<p>This describes the difference. <a href="http://www.vista-industrial.com/blog/engineering-drawings-vs-manufacturing-drawings/">http://www.vista-industrial.com/blog/engineering-drawings-vs-manufacturing-drawings/</a></p> <p>To summarize:</p> <p>A <strong>design drawing</strong> or <strong>engineering drawing</strong> is a <em>complete specification</em> of the finished product - in other words, it shows the important properties of each part (dimensions, materials, specifications for electrical components, etc), and how the parts work together.</p> <p>A <strong>manufacturing drawing</strong> has a more limited function. It tells you how to <em>make</em> each a particular part, but nothing else.</p> <p>In your example of a heater, the <em>design</em> drawing would specify the sizes of the screws or bolts that fix different parts of the heater together, but the heater company won't make any <em>manufacturing</em> drawings for the screws or bolts - they will buy those parts ready made. </p> <p>On the other hand, the company that <em>makes</em> the bolts and screws <em>will</em> have detailed manufacturing drawings showing exactly <em>how</em> to make a bolt from a plain metal rod, or whatever type of raw material they use, but they won't have any engineering drawings that show how the bolts are used after they have sold them.</p> <p>Often one "drawing" consists of several pages, or "sheets" (before computers, these were literally several physical sheets of paper). In your example, perhaps the design drawing was the first sheet, and the manufacturing drawing for each component of the heater was on a separate sheet - for example the manufacturing drawing for the "shell" was sheet 3 of the complete set. A complete set of drawings for a complex device may contain hundreds or even thousands of sheets.</p>

8879

Difference between "Design Dwg." and "Mfg.Dwg."

2016-05-14T19:24:55.537

<p>I'm a 3rd-year bachelor student and I may be mistaken but from what I have understood: In control theory we assume that the studied systems have a State Variable Description. Then from there we can derive that for such systems the transfer function must be proper. But, to have a State Variable Description means that the system can be described by a simple differential equation or equivalently by a system of differential equations. I am not really convinced that all physical systems can be described by such equations. Although I don't know what kind of other description there could be, I would like to find some strong argument. I guess it must have to do with the equations of fluid or Newtonian dynamics but if someone could be more specific I would appreciate it.</p>

|control-engineering|control-theory|

<p>A professor of mine, long ago, once quipped:</p> <blockquote> <p>"Give me a word... any word at all... and I'll show you how the root of that word is a PDE!"</p> </blockquote> <p><strong><em>A touch of Philosophy</em></strong><br> All physical systems are modeled by some set of mathematical equations, be they differential, algebraic, integral, stochastic, etc. The simplest models of physical systems only consider variations in the time variable (i.e. ordinary differential equations). While these models tend to be very simplified compared to the "true physics", they are often just "good enough" for specific engineering applications because we, as engineers, often only care about the input-output behavior of the system over time. </p> <p><strong><em>Advantages of ODE Models:</em></strong><br> Ignoring all variations with respect to other variables (especially space) simply makes things easier to both compute and analyze (in comparison to other types of differential equations). You can take advantage of a great deal of analysis tools, including impulse &amp; frequency response, bode plots, etc... If your system of ODE's are linear or otherwise linearizable around a setpoint, you can often use standard control techniques like PID, LQR, and LQG.</p> <p><strong><em>Disadvantages of non-ODE models</em></strong><br> I emphasize that ODE's are simpler to analyze insofar as they require the minimal mathematical complexity required of an engineer. There are, of course, more complicated models of physical systems (e.g. partial differential equations) that are likely more reflective of the true physics in comparison to an ODE model. However, the analysis tools that you learned for ODE's (a.k.a. classical control theory) no longer apply. Instead, you will likely have to use a set of tools that fall under the category of optimal control theory. For most problems of interest to engineers, optimal control requires fairly sophisticated numerical methods (usually adjoint-based optimization). The situation gets even more complicated if you include stochasticity in any of the variables or parameters. </p> <p>In short, control of anything other than an ODE is both very complicated to do in practice and is often unnecessary to begin with. Non-ODE control is a very highly specialized field and is generally not taught as a part of core engineering classes. If you ever get the chance to control anything other than an ODE in practice, it will likely come as a part of state-of-the-art scientific research rather than run-of-the-mill engineering applications.</p>

8884

Control theory - Differential equations and state variable description

2016-05-14T21:58:10.683

<p>I am a complete solidworks novice trying to create the part pictured in the orthographic schematic below but i am really struggling. I hoped someone might be able to guide me in creating this part. </p> <p><a href="https://i.stack.imgur.com/70wxT.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/70wxT.jpg" alt="Sideplate schematic"></a></p> <p>I began by creating a sketch as follows but i get stuck after this. I'm not sure how to align the top two circles in the center of the plate or how to create the rounded edge on the bottom of the plate.</p> <p><a href="https://i.stack.imgur.com/Zv1hq.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Zv1hq.jpg" alt="progress"></a></p> <p>Any help would be greatly appreciated! </p>

|solidworks|

<p>I would highly recommend the first couple of solidworks tutorials. Creating something when you don't understand what you're doing generally leads to underdefined, incorrect parts. </p> <p>But search for centreline (a construction line) or right click on a normal line to give you some geometry to align the top two circles with. A filet is what you want for the rounded edges. Also the green boxes are constraints, and I can see you have some you don't want, like the left and right uprights have been forced to be vertical. To add constraints, right click on the parts you want to constrain and select what you need. To delete them, click on the green squares and press delete </p> <p>Also your sketch is currently floating in space, not fixed to the geometry like it should be. To fix this, select the origin and a circle (hold shift) and select make concentric. Basically you need to do a tutorial. </p> <p>When a line is fully defined, it will turn black. When the whole sketch is fully defined, you can move on, go to the features toolbar, and select extruded boss/Base. If you move on without fully defining the sketch, you can accidently drag lines out of place without knowing. Which is the worst to try to hunt down and fix. </p>

8885

Creating a side plate in Solid Works

2016-05-15T11:53:24.247

<p>it is a simple question but I am new to this software. Both in <em>Postprocessor</em> and within <em>Measure Characteristic</em> options there are only angular velocities and accelerations - and no angle (orientation) itself. I saw <em>Angle</em> and <em>Orientation Measure</em> options but this must be something more advanced, since all I need is generalised coordinate of a part.</p> <p>Also, is it possible to obtain set of generalised coords against time in table form instead of plot? If so, please explain how.</p> <p>Thank you in advance,</p> <p>Maverick</p>

|mechanical-engineering|software|

<ul> <li><p>You can export a Adams/Solver dataset (.adm) that will contain all the variables that were solved in a ascii format. (file export)</p></li> <li><p>you can export a results file.</p></li> <li><p>You can also export measures in the post processor as a ascii tabular data. In the file export dialog in case you want to specify what format the measure is to be had in.</p></li> </ul>

8890

ADAMS/View - how to plot part orientation?

2016-05-16T04:26:40.373

<p>I am checking the leak rate of a carbon seal installed on an aircraft engine bearing. The test cart has a reading of pounds per hour but my manual says the leakage should not be more than 0.08 cubic metres (2.5 cubic feet) per minute of free air.</p> <p>Since pounds are a weight measure and cubic metres/feet are volume measures, I can't compare them directly. How do I convert the volumetric flow rate to a mass flow rate?</p>

|airflow|

<p>There are several pieces of additional information you need to go from units of volume/time to units of force/time.</p> <p>For gasses, temperature and pressure matter. Air is mostly nitrogen and oxygen. These have atomic masses of 14 and 16 respectively. Both form diatomic molecules, so air is a mixture of gasses with molecular masses of 28 and 32. Since there is more nitrogen than oxygen, I'm going to say "air" has a molecular mass of 29. Close enough given your precision of "0.08".</p> <p>At Standard Temperature and Pressure (STP; 1 atm, 0 &deg;C), a mole of ideal gas occupies 22.4 liters. Let's say you're really asking about 21 &deg;C. That's 294 &deg;K as apposed to 273 &deg;K at STP. The volume will therefore be (22.4 l)(294 &deg;K)/(273 &deg;K) = 24.1 l. Since that's one mole, it has a mass of 29 g. Put another way, we've decided the density of your "air" is 29 g/24.1 l.</p> <p>There are 1000 liters in a cubic meter, so your 0.08 cubic meters is 80 liters, which has a mass of 96.2 g. There are 60 minutes in one hour, so we can compute the mass flow rate in your final time units, which is 5.77 kg/h.</p> <p>The remaining problem is to convert kg to pounds. Kg is a unit of mass, and pounds is a unit of weight. The conversion depends on gravity. You didn't say if whether this is on the surface of the earth, on the internal space station, or somewhere else. I'll assume the surface of the earth where 1 kg of mass weighs about 2.2 pounds. (5.77 kg)(2.2 lb/kg) = 13 lb.</p> <p>The final answer is therefore 0.08 cubic meter per minute of air at 1 atmosphere, 21 &deg;C, on the surface of the earth is about 13 pounds/hour.</p>

8898

How do I convert cubic metres/feet per minute to a pounds per hour flow rate?

2016-05-17T07:47:08.880

<p>I have a structure, which contains only frame elements ( beam and column).</p> <p><a href="https://i.stack.imgur.com/8zIej.png" rel="noreferrer"><img src="https://i.stack.imgur.com/8zIej.png" alt="enter image description here"></a></p> <p>From what I know, if I apply point load at the columns on both left and right hand side,and if the beam-column joint is rigid, then the column at the center will experience extra moment, because the beam moment will be transmitted to the center column.</p> <p>In other words, the center column will take moment because of the beam's existence, am I right?</p> <p>In order to eliminate this extra moment taken by the center column, can I then set the beam-column join connection as pin? How would the structural behavior changes?</p>

|structural-engineering|structural-analysis|

<p>If you add hinge to center post you are still going to have loads carried by the center post. The only way to isolate the center post from any load on the lateral posts is to have the 2 side post-beam joints pin connection as well as the center post. </p>

8908

Can I reduce the column moment due to beam, by setting the beam end to pin?

2016-05-18T07:05:56.193

<p>My question might seem a simple one, but it has been troubling me for a while. How many hours in a day do conventional electrical generators work (the ones in large power plants)? </p> <p>Do they by any method conserve or store the power before it reaches the consumer?</p>

|electrical-engineering|energy|power-engineering|energy-storage|

<p><em>How many hours in a day do conventional electrical generators work (the ones in large power plants)?</em></p> <p>Short answer; it depends on the constraints of the generators and the priorities of the system operator.</p> <p>The underlying area is known as <strong><a href="https://en.wikipedia.org/wiki/Unit_Commitment_Problem_in_Electrical_Power_Production" rel="nofollow noreferrer">Unit Commitment</a></strong>:</p> <blockquote> <p>A family of mathematical optimization problems where the production of a set of electrical generators is coordinated in order to achieve some common target, usually either match the energy demand at minimum cost or maximize revenues from energy production.</p> </blockquote> <p>Assuming that you are referring to thermal generators, the constraints <a href="https://en.wikipedia.org/wiki/Unit_Commitment_Problem_in_Electrical_Power_Production#Types_of_production_units" rel="nofollow noreferrer">include</a>:</p> <ul> <li>minimum up/down time </li> <li>ramp up/down rate </li> <li>modulation/stability </li> <li>start-up/shut-down ramp rate</li> </ul> <p><em>Do they by any method conserve or store the power before it reaches the consumer?</em></p> <p>No. At least not over the time frames that unit commitment is considered (of the order of hours/half hours).</p> <p>Over much shorter timeframes (seconds) energy is stored in the rotating mass of the generators. This is described in answers to <a href="https://engineering.stackexchange.com/questions/2245/quantifying-inertia-on-the-electricity-grid">this question</a>.</p>

8920

Working hours of power plant generators

2016-05-18T07:19:07.603

<p>Assume that I am writing a new FEM software for structural analysis, <em>it shouldn't really matter whether this is a frame or frame+shell element analysis tool.</em></p> <p>I understand that <a href="https://www.quora.com/What-is-the-difference-between-primary-secondary-and-tie-beam-and-where-are-located-in-building-and-how-can-we-identify-them-and-what-are-their-purpose/answer/Asafa-Khan-%D8%A7%D8%B3%D8%A7%D9%81%D8%A7-%D8%AE%D8%A7%D9%86?srid=36H4" rel="nofollow noreferrer">primary and secondary beams have their own function</a>, and are designed for different use. </p> <p><a href="https://i.stack.imgur.com/GYtIU.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/GYtIU.jpg" alt="enter image description here"></a></p> <blockquote> <p><strong>Primary Beam</strong>: A horizontal beam connecting columns (simply supported or shear connected.) Function: It will transfer the load from secondary beam(if present) to the columns.</p> <p><strong>Secondary Beam</strong>: A horizontal beam connecting primary beams (simply supported or shear connected.) Function: It will transfer the load to the primary beam and not directly connected to the columns.</p> </blockquote> <p>In FEM, everything is just frame element, and all the beams and columns are equal, so whether one beam should resist more loads ( ie primary beam) than the other is completely up to the outcome of FEM procedure. But as engineers, sometimes we <em>just</em> want <em>this</em> one beam to take more loads than the neighboring ones. </p> <p>How can I modify my FEM engine to do what I want?</p>

|finite-element-method|structural-analysis|

<p>Your description of what primary and secondary beams do is dramatically different to the definition given in the link you present (which is itself correct).</p> <p>To quote the answer in your link:</p> <blockquote> <p><strong>Primary Beam:</strong><br /> A horizontal beam connecting columns (simply supported or shear connected.)<br /> <em>Function</em>: It will transfer the load from secondary beam(if present) to the columns.</p> <p><strong>Secondary Beam:</strong><br /> A horizontal beam connecting primary beams (simply supported or shear connected.)<br /> <em>Function</em>: It will transfer the load to the primary beam and not directly connected to the columns.</p> </blockquote> <p>That is the only difference. Secondary beams take in the loads from the slab and transfer them to the primary beams, which then transfer them to the columns.</p> <p>So all you need to do in your FEA software is allow the user to define how these loads are applied. <em><strong>You don't need to create different categories of beam elements (primary vs. secondary).</strong></em></p> <p>You are calling your software a <em><strong>frame</strong></em> analysis tool. This implies that 2D elements like plates and shells aren't available, only 1D beam elements. This therefore means that the user will have to take some decisions as to how area loads which are applied to the slab (which you can't model, since you don't have 2D elements) are transferred.</p> <ul> <li>If your program is simple, the user will have to apply the linear distributed loads on each beam by hand. In this case, there's no problem: the user applies the relevant loads on the secondary beams which then transfer them to the primary beams.</li> <li>Your program may also include &quot;area loads&quot;, where the user defines an area and a load and the program automatically decides how to distribute this area load to the beams, usually through tributary areas. So long as the user can define the beams that are to be considered for these area loads, that's also not a problem. You just define the area load and then tell the program to only consider the secondary beams (though, depending on the layout, letting some of the load go directly to the primary beams would actually be more correct).</li> <li>If your program is actually a full structural analysis tool with 2D elements (and is therefore more than just a <em><strong>frame</strong></em> analysis tool) as well to model the slab, that's still fine. If the user can tell the slab not to take the specific beams into consideration while meshing (such as your intended primary beams), they won't share any nodes and therefore won't have any direct load transfer. The slab will mesh joining itself only to the remaining beams, transferring loads to them, who will then transfer the loads to the primary beams.</li> </ul> <p>Also, the image in your OP is a poor example of primary and secondary beams because, structurally, they are identical. The secondary ones are a bit shorter, but they share the same top-level with the primaries. This means that when the slab transfers loads, it makes no distinction between them. Look at the tributary areas for each slab, it's clear the vast majority of the loads will naturally go to the secondaries, with only a small triangle of load going to each of the primaries. Therefore, when you model them there should be no distinction between them: the loads should go where they naturally wish to go.</p> <p>However, if you're modelling something like the roof below, then you'll need to control how the loads are distributed. After all, all the loads coming from the roof need to go through the purlins. None of the loads may go directly to the rafters.</p> <p><a href="https://i.stack.imgur.com/tWqF9m.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/tWqF9m.jpg" alt="enter image description here" /></a> ^{<a href="http://www.civilengineeringx.com/building/bce/Steel-roof-truss.jpg" rel="nofollow noreferrer">Source</a>}</p>

8922

How to distinguish primary and secondary beam in FEM formulation?

2016-05-18T07:27:10.757

<p>Assuming that I have a beam that spans across a few supports. I can design the beam as <a href="http://machinedesign.com/whats-difference-between/what-s-difference-between-beam-diagrams">a continuous beam (a beam that is loaded and has more than two supports)</a>, or individual beams that each encompass between two supports. </p> <p>From what I know, the design of continuous beam is always more economical than beam with supports. Why this is the case? I understand that this is because the forces and moment in continuous beam are less than in beams with multiple supports. But I am looking for a deeper, more physical explanation on <em>why</em> the forces are smaller in continuous beam. </p>

|structural-analysis|

<p>I see another answer has given input on the strength side of things, where the economics isn't as clear, so I will add that one of the crucial aspects that usually make continuous beams economical is deflection. </p> <p>For a given load the deflection of a continuous beam is smaller (usually <em>much</em> smaller) than what it would be for a simply supported beam of the same section and span due to the addition of end restraint(s) and their influence on the curvature distribution. For a problem governed by deflection limits in the service state (and many if not most are) this means that a continuous beam can use a smaller section for the same deflection compared with a simply supported beam. Smaller section means less material and less expense.</p>

8923

Why the design of continuous beam is always more economical than beam with supports?

2016-05-18T16:06:30.137

<p>I'm working on a 2D-model (in the <em>xy</em>-plane) in NASTRAN and am getting convergence problems. The main cause of these convergence problems is a "high or negative matrix factor diagonal ratio."</p> <p>I've attached a picture of a setup similar to the one I'm working on. The elements in <em>white</em> are rod elements; the elements in <em>pink</em> are bar elements.</p> <p><a href="https://i.stack.imgur.com/Qt2QL.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Qt2QL.png" alt="enter image description here"></a></p> <p>The loads and BCs are:</p> <ul> <li>Angular velocity of 60 rad/sec about the <em>y</em>-axis</li> <li>Gravitational acceleration of 9.8 m/s$^2$ in the <em>-y</em>-direction</li> <li>$123456$ constraints on the top and bottom nodes to prevent them from moving or rotating</li> <li>$123$ constraints on the inner-most node to prevent it from moving</li> </ul> <p>Here is some text from the f06 file:</p> <p><a href="https://i.stack.imgur.com/yfzmp.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/yfzmp.png" alt="enter image description here"></a></p> <p><a href="https://i.stack.imgur.com/nXkkI.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/nXkkI.png" alt="enter image description here"></a></p> <p>Nodes 13 and 27 are where the rods connect to the bars, and Nodes 20 and 24 are just two random nodes on the right-hand side of the pink section.</p> <p>Under the bar properties, there is an option called "Pinned DOF @ Node 1," which may or may not be helpful here. The options are UX, UY, UZ, RX, RY, RZ.</p> <p>I've tried experimenting with various boundary conditions, but I can't get convergence in either the linear static case or the nonlinear static case.</p> <p>Generally high matrix factor diagonal ratios imply that I have a very stiff element connected to a not-very-stiff element, but Nodes 20 and 24 <strong>aren't</strong> where the rod/bar intersection happens.</p> <p>Does anyone know how I might go about diagnosing the problem? I think I'm missing boundary conditions somewhere, but I'm not entirely sure where that is.</p> <h2>UPDATE</h2> <p>Whoops, I forgot to explain what behavior I <em>expect</em> from this contraption.</p> <p>Since I have a really high angular velocity (and perfectly normal gravity), I would expect the pink nodes to move rightward and downward due to the loads. The only thing stopping them from moving indefinitely are the white rods. Since these aren't beams and don't transmit moment stiffness, all these rods can really do is extend in length and change angle, since they're required to be a straight line.</p> <p>This will result in a shape that looks similar to the pink curve, but probably shifted rightward a bit.</p> <p>I ran the case with PARAM,BAILOUT,-1 (linear static), and I got the following clearly incorrect deflection ($10^7$ m deflection): <strong>image removed since it didn't provide any useful information and it took up too much space</strong>.</p> <h2>UPDATE #$2$</h2> <p>I've pinpointed the problem, I believe.</p> <p><a href="https://i.stack.imgur.com/goCN7.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/goCN7.png" alt="enter image description here"></a><a href="https://i.stack.imgur.com/52vj0.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/52vj0.png" alt="enter image description here"></a></p> <p>Blue elements are rods, pink elements are bars. The first case converged (nonlinear static) and gave results pretty close to what I expected (deflections were slightly high, but my angular velocity is slightly high, too).</p> <p>The second case failed to converge, and it had a problem with my new node - it told me that it's having diagonal ratio problems in the T2 direction.</p> <p>There's something NASTRAN is doing wrong with how these two bar elements are connected at this node. Any ideas?</p>

|structural-engineering|finite-element-method|

<p>After a long discussion with the OP in comments and chat, it emerged that the basic cause of the problem is the geometry of the structure. The curved "beam" has a length/diameter ratio of about 50,000:1. </p> <p>As this answer <a href="https://engineering.stackexchange.com/questions/5311/stiffness-method-fem-slender-members/5321">Stiffness Method FEM - Slender Members</a> points out, the relative size of the axial and bending stiffness of a beam is of the order of $L^2/d^2$, or around $10^9$ for the OP's structure. </p> <p>For practical purposes the <em>elastic</em> bending stiffness of the OP's beam is zero, and it will resist bending only because of the <em>stress</em> stiffness created by the axial tension in it. </p> <p>One approach to modeling this would be to start with a <em>straight</em> configuration of the "beam", modelled by rod elements not beams, to eliminate the vanishingly small elastic bending stiffness from the model. A nonlinear stress analysis could then start by applying a tension force to pre-stress the structure, and then change the loading in increments to the actual loading conditions.</p> <p>Another idea would be to perform a nonlinear <em>dynamic</em> analysis, and let the inertia of the structure constrain its behaviour under the initial loads. You would have to include some fictitious damping in the model to produce a final steady-state solution.</p> <p>A naïve approach to overcome the $10^9$ ratio of axial to bending stiffness would be to use short beam elements so that the stiffness ratios within each element were of similar magnitude. But to achieve that, the model would have to consist of say 10,000 or 100,000 elements, which is not likely to be a sensible way to attack the problem.</p> <p>From my own experience I would not choose Nastran to analyse such extremely nonlinear behaviour. A FEA package that was designed "from the ground up" to deal with nonlinear analysis, such as Abaqus, would be a safer bet.</p> <p>It would be worth searching the literature for similar types of problems - for example the mechanics of cables for mooring floating structures in the offshore oil and gas industry.</p> <p>The OP might get some qualitative insight into the structural behaviour by increasing the diameter of the beam (probably by a few orders of magnitude) to get the model to run.</p>

8931

FEA: What boundary conditions am I missing?

2016-05-18T20:34:22.630

<p>I have a <em>seemingly</em> simple structure and I need to perform a hand calculation to check that the stresses in the members are below the allowable values. In its simplest form, the structure is a rectangular frame made from rectangular tube welded at the corners and suspended from four equal length cables attached at the corners. The cables meet above the center of the rectangular frame and attach to a hook.</p> <p>For loading purposes, each member of the frame can be considered to have the same uniformly distributed load along its entire length.</p> <p>So, it looks something like this: <a href="https://i.stack.imgur.com/TSZp1.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/TSZp1.png" alt="enter image description here"></a> I have been poring over static, mechanics, structural analysis and other books and resources trying to find the methodology for analyzing this but all of these resources seem to stop at planar structures where the loading is in the plane of the structure. Either this is a difficult problem, I am overthinking things, or I have been out of practice with respect to statics for too long.</p> <p>What I have so far is a free body diagram where I have replaced each cable with a spherical moveable support (the 3D equivalent of a roller support). Each corner is free to rotate in all three directions and to translate in the plane of the frame but not out of it. So there is a single vertical reaction force at each corner (which is actually the vertical component of the cable tension). This is an assumption. Please let me know if I am wrong.</p> <p>The degree of static indeterminancy is $(6m + R) - 6j$, where $m$ is the member count, $j$ is the joint count and $R$ is the number of unknown reaction forces. Plugging in the values leads to an answer of 4. However, I think I can actually reduce that to 1 since (due to symmetry) the reaction force at each corner is identical.</p> <p>And this is where I am stuck. What is the next step? Or is this the point at which people say "let the computer solve it".</p>

|structural-analysis|

<p>I am an ex rigger used to slinging heavy loads. You cannot spread the load equally over 4 cables. Three of the cables will support the load - the fourth will provide stability. </p> <p>I know this from practical experience. I was involved in the construction of the Westgate Bridge. The rig we used to lift the box girded which weighed anywhere from 70 to 120 tonnes was a 3 cable arrangement.</p> <p>If I was you I would assume the load is carried over two cables.</p>

8935

Structural Analysis of a Space Frame

2016-05-19T10:09:23.347

<p><a href="http://www.reactivetools.com/datasheets/reflex_lite.pdf" rel="nofollow">This description</a> of the Reflex Lite packer by Reactive states the following among its advantages:</p> <p>"No mechanical delay required"</p> <p>What could that mean? The packer has no built-in system of delayed activation or delayed swelling control? Or "there is no need for complex procedures when installing the packer"?</p>

|drilling|

<p>I've just got a reply from the company:</p> <blockquote> <p>Some of our competitors use an applied outer layer of material to delay their swell curves (mechanical delay). At Reactive, we select the correct polymer for the well, so no fabricated delay system is required.</p> </blockquote> <p>Voila. </p>

8939

Meaning of "no mechanical delay required" in packer description (oil drilling)

2016-05-19T21:15:09.907

<p>I am trying to find the internal forces in the member at point G in the following illustration:</p> <p><a href="https://i.stack.imgur.com/ReGKC.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ReGKC.png" alt=""></a></p> <p>My answer is different from the solution in the textbook and I think it has to do with how I'm looking at the 1500 lb external force acting on point <strong>A</strong>.</p> <p>I've included the 1500 lb force in my free body diagrams for both <strong>AB</strong> and <strong>AE</strong>. I'm getting the feeling that it should only be acting on one of these members; however, I am unsure.</p> <p>Does the load at the joint act on only one member, or on all members in the joint?</p>

|structural-engineering|civil-engineering|statics|

<p>Using the method of virtual work, assume that $AE$ is given a small deflection $d\theta$ downwards and that the unknown force is $Y$.</p> <p>$$ 1500 (6+2+2) d\theta + Y (6+2) d\theta + \left(\int_0^6 \frac{300 l}{6}l \, dl\right)d\theta =0$$</p> <p>$$ \Rightarrow 15000 + 8 Y+ 3600 =0$$</p> <p>$$\Rightarrow Y = -2325 $$</p>

8946

Does an external force at a joint between two members act on one member or on both?

2016-05-20T00:08:06.637

<p>In the electric power equation $P=RI^2$, $P$ represents the product of $RII$.</p> <p>I'm trying to understand the fundamental anatomy of the $RII$ term. More precisely, why does the current, $I$ appear twice as a factor? Does this represent two different dimensions of $I$, in the sense of a measurable, albeit temporal or spatial quantity?</p> <p>For example, is one factor "speed" and the other one "direction?"</p> <p>Update:. This <a href="https://en.wikipedia.org/wiki/Angular_velocity#Particle_in_three_dimensions" rel="nofollow">link</a> about angular velocity seems to support my question regarding direction, speed, and dimensions. </p> <p>Cheers </p>

|electrical-engineering|circuits|current|

<p>Firstly, by temporal I think you mean spatial. Secondly if you consider acceleration, the units are m/s^2. Seconds-squared are not spatial, but they are temporal. But they needn't be either. Consider E=mc^2. c is the speed of light.</p> <p>Your formula comes from these two formulas:</p> <blockquote> <p>P = I V</p> </blockquote> <p>In other words, the power consumed by a device is equal to the voltage across it multiplied by the current flowing through it.</p> <blockquote> <p>V = I R</p> </blockquote> <p>Which means the voltage across a resistive element is equal to the current flowing through it multiplied by its resistance.</p> <p>If you substitute the second into the first, you get:</p> <blockquote> <p>P = I V = I ( I R ) = I^2 R</p> </blockquote> <p>If you look at the equivalent SI units for power, volts, current and resistance, you will find the above to be consistent.</p> <p>If this doesn't answer your question, you really need to be clearer!</p>

8948

In the equation for power as a function of current and resistance, what do the two factors for current represent?

2016-05-20T11:01:47.737

<p>I want to calculate how heavy a mobile hydraulic post table base must be to support an adjustable monitor arm and monitor.</p> <p>Monitor (iMac) Specs:</p> <ul> <li>Height: 20.3 inches (51.6 cm)</li> <li>Width: 25.6 inches (65.0 cm)</li> <li>Weight: 30 pounds (9.54 kg)</li> </ul> <p><a href="http://www.northerntool.com/shop/tools/product_200652091_200652091" rel="nofollow noreferrer">Mobile Hydraulic Post Table Specs</a>:</p> <ul> <li>Ship Weight: 120.0 lbs</li> <li>Load Capacity: 300 lbs</li> <li>Platform Size: 16 x 16 in</li> <li>Raised Height: 49 in</li> <li>Lowered Height: 31 in</li> </ul> <p><a href="http://www.ergotron.com/ProductsDetails/tabid/65/PRDID/787/language/en-US/Default.aspx" rel="nofollow noreferrer">Monitor Arm Specs</a>: </p> <ul> <li>Product Weight: 17.5 lbs (8 kg)</li> <li>20 inches (51 cm) of vertical adjustment</li> <li>62 inches (157 cm) of side-to-side motion</li> <li>Screen ≤ 46"</li> <li>Capacity 14–30 lbs</li> <li>Lift 20"</li> <li>Tilt 80°, Pan 360°, Rotation 90˚</li> </ul> <p>The one capability that I think could cause the post table to collapse is the 62 inches (157 cm) of side-to-side motion. I've included a drawing.</p> <p><a href="https://i.stack.imgur.com/0bTZT.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/0bTZT.jpg" alt="Workstation sketch"></a></p>

|structures|

<p>To calculate this, we need to know what tipping moment your table will suffer. This is pretty simple.</p> <p>With the arm totally extended, it is $31''$ long (half of $62''$). You seem to be clamping the arm to the table, so I'm going to assume that the distance from the screen at the tip of the arm to the center of the table is equal to $31 + \frac{16}{2} = 39''$. The pivot point when tipping, however, will be one of the wheels, so we can subtract the distance from the post to one of the wheels. This isn't given, so I'm going to assume that distance is around $5''$, which gives us a total arm of $34''$.</p> <p>So, now we have to calculate the tipping moment, considering the weight of the screen and of the arm.</p> <p>$$M = 30\cdot34 + 17.5\cdot\left(3 + \dfrac{31}{2}\right) = 1344\text{ lb-in}$$</p> <p>Now, this is already resisted by the weight of the post table (120&nbsp;lb¹), which can be considered concentrated at the post, generating therefore a moment equal to</p> <p>$$M = 120 \cdot 5 = 600\text{ lb-in}$$</p> <p>You therefore have here an excess of $1344-600 = 744\text{ lb-in}$ which you'll need to counter with additional weights. We can find the necessary weight by distributing it uniformly on the base, which gives us:</p> <p>$$W = \dfrac{744}{5} = 148.8\text{ lb}$$</p> <p>This is however only the nominal weight that's necessary. I would put in quite a bit extra weight to be sure the structure will hold in unexpected situations (someone bumping against the screen or some such).</p> <p>The table states it can support 300 lb, so this is still acceptable: $$148.8 + 17.5 + 30 = 196.3 &lt; 300\text{ lb}$$ Hell, I personally don't see why not simply put 250&nbsp;lb of weight, just as a safety precaution.</p> <p>You also need to check the clamping of the arm to the table. For this, however, we need more information regarding the table (thickness, material) and of the clamp.</p> <hr> <p>¹ This is actually the ship weight, which means the table will be lighter. Get the correct weight and repeat these calculations. Make sure to also check the actual distance from the post to the wheels, which I assumed here at $5''$.</p>

8953

Calculate mobile base weight to support adjustable arm

2016-05-20T16:06:21.337

<p>I am having trouble finding a 3.4 µH through hole inductor that is in stock right now. I was wondering if I can just place 2 inductors that add up to 3.4 µH in series to solve this problem. I am not sure if this would introduce any problems, if so what kind of problems and how can I mitigate them.</p>

|electrical-engineering|power-electronics|circuits|circuit-design|

<p>3.4 &micro;H is not a standard value, but 3.3 &micro;H is. There are plenty of those out there, even in thru-hole.</p> <p>Inductors aren't generally accurate enough to justify making a 3.4 &micro;H version when a 3.3 &micro;H version is already in production. The extra 3% more isn't worth another product to manufacture, and for customers isn't worth another part to stock. You better have a really good reason you are committing your company to stock a new inductor with a non-standard value.</p> <p>You need to step back and examine why you need such a unusual inductance value so accurately that 3% less actually matters. Usually you determine the rough inductance you need, find a suitable part, then design the details of the circuit around that. You can get resistors and timing components at much higher accuracies than inductors, for example, so you pick exact values for those last.</p> <p>You should also examine why you want thru-hole. Unless this is a high-current high-power part that is therefore physically heavy, thru-hole makes little sense. Thru-hole will limit you to parts designed 10 years ago or longer. New development is obviously in surface mount parts, except for physically large inductors. You haven't said anything about the saturation current requirement or maximum resistance you can tolerate. "3.4 &micro;H" is only one part of the spec for a inductor.</p>

8960

Circuit Design Inductor Problem

2016-05-22T07:10:37.957

<p>When a material does not show a distinct yield point then 0.2% of strain is considered and a line is drawn parallel to the elastic line and the corresponding stress is called proof stress. My question is why are they taking 0.2% of strain? Is that an assumption and does it differ with materials</p>

|materials|civil-engineering|

<p>Although as mentioned by others this is not universally accepted, in my opinion the reason that the 0.2% strain was used for proof stress, is that <strong>it offers a more straight forward comparison with the yield stress of steel</strong>.</p> <p>Steel has been (and probably still is) the most common material for structural engineering. However, it has distinct differences from other material e.g. aluminium</p> <p><a href="https://i.stack.imgur.com/RybkR.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/RybkR.png" alt="enter image description here" /></a></p> <p>Steel has a distinct yield point, that you can use to set a good safety margin before failure. Most other materials don't have that distinct characteristic. Because most design happens up to yield stress, it was common to try and find similar values for other materials.</p> <p>That is why the 0.002 strain value is used. To offer a better direct comparison of materials to steel. IMHO, it is not a very efficient metric but its one of two values you usually ask when considering a material (how stiff it is and what is its strength).</p>

8977

Why 0.2% of strain is considered while taking proof stress?

2016-05-23T13:29:31.567

<p>I want to know how much I can over-drive a small electric motor. Its speed at the rated voltage is too slow, so I want to find a higher voltage that won't blow it. It will only be used intermittently, but I figure if I run it for 100x the max expected duration with the expected load then it will probably survive repeated intermittent use.</p> <p>So my question is, safety concerns aside, how can I apply a relatively constant load? I can't think of a reliable way of using friction, such as attaching a disk and hold something on it, vaguely resembling a vehicle brake disk. I could attach a weight via a pulley but it would quickly run out of length. I could use a second motor and load that electrically to increase the torque, which is IMO the best idea, but I'm wondering if there are any other mechanical solutions? I think I'm missing something really obvious.</p>

|friction|

<p>Probably the easiest way to add a reasonably constant mechanical load to a motor is a fan. Because aerodynamic forces are exponential in respect to speed this should be consistent enough for evaluation purposes and has the advantage that you don't have the problems of wear and adjustment associated with solid friction brakes. Here the obvious solution for a small motor is to scavenge the fan from a cheap PC cooler. If the fan has an enclosure you can even vary the load by the simple expedient of using duct tape to restrict the flow. </p> <p>For a small motor a fan in air should be fine, for higher power applications a paddle wheel immersed in a fluid is more compact,albeit more complex to set up. </p> <p>If you want to be able to measure of vary the load then a <a href="https://en.wikipedia.org/wiki/Band_brake" rel="nofollow">friction band brake</a> is a good solution as you can measure the tension in the band to give a crude measure of torque and even a simple setup should at least give you a consistent range even if you can't derive precise figures for power. </p> <p>You can also get cheap handled non-contact tachometers to measure rpm </p> <p>If you want more detailed measurements of load then you could always connect the motor to a generator and measure the power generated. This can give fairly reliable direct measures of power output and a simple optical or magnetic tachometer will give you rpm and thus torque. With the right instruments this setup also allows for sampling of variations over time. With some basic programming and electronics knowledge building a data logger to a PC is realistic (especially with a legacy parallel port). </p>

8997

Apply constant load to an electric motor - low tech way

2016-05-23T15:10:42.097

<p>I do not have any engineering background but I am studying some mathematics to try and understand and solve this problem. I have a mesh (M1) in cylindrical coordinates and this deforms into a mesh (M2) (again in cylindrical coordinates). I need to compute strain in radial, longitudinal and circumferential directions.</p> <p>Basically I have done the following computations on each point of the mesh. The deformation gradient F is computed:<br> $$F = x·X^T·(X·X^T)^{-1}$$ where:</p> <ul> <li>$x$: the deformed cylindrical coordinates (3x1 matrix) </li> <li>$X$: the undeformed cylindrical reference coordinates (3x1 matrix) </li> <li>$X^T$: transpose of $X$ </li> <li>$X^{-T}$ : inverse of transpose of $X$ </li> </ul> <p>The Langrangian finite strain tensor $E$ is then computed:<br> $$E = \dfrac{1}{2}(F^T·F-I)$$</p> <p>where: </p> <ul> <li>$F^T$ : transpose of $F$ </li> <li>$I$ : identity matrix (3x3) </li> </ul> <p>I then take the diagonal of $E$ and this gives me the principal strains in the orthogonal cylindrical coordinates.</p> <p>Questions: Am I doing it right? Do I need to do some further computations on the diagonal of $E$ to get the principal strains at each point of this mesh that is deforming?</p>

|structural-engineering|structural-analysis|finite-element-method|kinematics|deformation|

<p>I'm not sure what the interpretation of your deformation gradient is. To my knowledge the deformation gradient gives the local deformation, i.e. the deformation around a certain point. To obtain this deformation, local information (i.e. information in one point only) is not enough, but you also require some information on the surroundings. The formula I know is $$ F_{ij}=\frac{\partial x_i}{\partial X_j} $$ with $F_{ij}$ being the deformation gradient, and $x_i$ and $X_j$ are local coordinates in the deformed and undeformed configuration, respectively. Thus to calculate the continuous $F_{ij}$, you need the differentials of the local coordinates $\partial x_i$ and $\partial X_j$. I'm no expert on discrete meshes, but one possibility is to calculate $F_{ij}$ for each element in your mesh by using finite differences $\Delta x_i$ and $\Delta X_j$. If unsure you should probably ask again more precisely.</p> <p>Your finite strain tensor calculation seems correct to me. Note that when the deformation gradient is calculated for an element, the finite strain tensor is also for that element. Again I'm no expert on the interpretation of the results. The <a href="https://en.wikipedia.org/wiki/Finite_strain_theory#Physical_interpretation_of_the_finite_strain_tensor" rel="nofollow">wikipedia article on finite strain theory</a> seems to have some answers for you.</p> <p>When calculating in curvilinear coordinate systems, things usually become a bit more complicated than in cartesian coordinates. However, since cylindrical coordinates are locally cartesian, your calculation is fine. For more complex curvilinear coordinate systems you would need to evaluate your equations using co- and contravariant bases.</p>

9000

Deformation gradient, strain tensor from cylindrical coordinates

2016-05-25T04:42:05.683

<p>My question has two parts.</p> <ul> <li>What is BSc engineering and Btech engineering ?</li> <li>Which has the highest validity in electronic and computer fields?</li> </ul> <p><br> PS : Specially in North American region.</p>

|electrical-engineering|computer-engineering|

<p>I think Bachelor of Technology (B-Tech) is mostly a designation used by Indian Institute of Technology (IIT) and National Institute of Technology (NIT). <a href="https://www.quora.com/What-is-the-difference-between-a-Bachelor-of-Engineering-and-a-Bachelor-of-Technology" rel="nofollow noreferrer">This</a> write up describes </p> <ul> <li>Bachelor of Engineering as knowledge oriented</li> <li>Bachelor of Technology as skill oriented</li> </ul> <p>In USA some engineering schools offer two engineering tracks per discipline. The traditional engineering track is more theory base content and is referred to as BSc Engineering. On the contrary the technology track offers more hand on and has more laboratory work. </p> <p>It is best you look at syllabus for both traditional and technology track programs from a few recognized programs. Here are a few for you to get started.</p> <ul> <li><a href="https://engineering.purdue.edu/ECE/Academics/Undergraduates/UGO/pdf/BSEE%20Sample%20Plan.pdf" rel="nofollow noreferrer">Purdue BS EE Plan of Study</a></li> <li><a href="https://tech.purdue.edu/sites/default/files/ECET-fall-2011.pdf" rel="nofollow noreferrer">Purdue BS EET Plan of Study</a></li> <li><a href="https://engineering.purdue.edu/ME/Academics/Undergraduate/index.html" rel="nofollow noreferrer">Purdue BS ME Plan of Study</a></li> <li><a href="https://tech.purdue.edu/new-albany/degrees/mechanical-engineering-technology/plan-of-study" rel="nofollow noreferrer">Purdue BS MET Plan of Study</a></li> <li><a href="http://engineering.siu.edu/elec/undergraduate/undergraduate-courses.php" rel="nofollow noreferrer">Southern Illinois University at Carbondale EE courses</a></li> <li><a href="http://engineering.siu.edu/tech/undergraduate/engineering-technology/courses.php" rel="nofollow noreferrer">Southern Illinois University at Carbondale EET courses</a></li> </ul> <p>If you follow an ABET accredited program then you can take both Fundamentals in Engineering exam and Professional Engineer exam. If you pass both the exams per the requirements you can obtain Professional Engineer designation.</p> <p>Both traditional and technology track are equally recognized. In my experience most students with technology based concentration find work in the area of Product Design, Product Development, Product Testing and Product Service in the respective discipline. </p> <p>Below are some references that have a broader and in-depth explanation. </p> <hr> <p><strong>References:</strong> </p> <ul> <li><a href="https://en.wikipedia.org/wiki/Engineer_in_Training" rel="nofollow noreferrer">Engineer in Training</a></li> <li><a href="https://www.quora.com/What-is-the-difference-between-a-BTech-and-a-BSc" rel="nofollow noreferrer">What is the difference between a BTech and a BSc?</a></li> <li><a href="https://www.quora.com/What-is-the-difference-between-a-Bachelor-of-Engineering-and-a-Bachelor-of-Technology" rel="nofollow noreferrer">What is the difference between a Bachelor of Engineering and a Bachelor of Technology?</a></li> <li><a href="https://engineering.stackexchange.com/questions/405/what-are-the-major-differences-between-engineering-degrees-and-engineering-techn?lq=1">What are the major differences between Engineering degrees and Engineering Technology degrees in terms of employability?</a></li> </ul>

9016

BTech or BSc engineering

2016-05-25T12:49:20.793

<p>I am designing a product that requires internal mixing of a sealed container. If you need to visualize the container, imagine some quasi-oil drum that holds about 20&nbsp;L of fluid.</p> <p>I am looking to use as little electrical input as necessary so I will be using a manual hand turner with some sort of paddles attached to mix the liquid appropriately. However, the seal on the joint or input valve connecting the manual crank to the tank <strong>must</strong> be as airtight as possible.</p> <p><strong>How do I make the hinge connecting the crank to the tank as airtight as possible?</strong></p>

|mechanical-engineering|seals|

<p>What you're looking for is a mechanical face seal, which are used to allow rotating shafts to penetrate into a sealed volume such as a pump, mixer, compressor etc. </p> <p>One important thing to remember about these seals is that all such seals leak to some degree. The exact seal you need will be driven by your specific requirements such as:</p> <ul> <li>How much leakage of the process fluids can you tolerate?</li> <li>What is the process fluid?</li> <li>What are the pressures and temperatures?</li> <li>Operating speed</li> <li>Budget</li> <li>etc.</li> </ul> <p>On the higher end are engineered and stock mixer seals and cartridge seals like these from <a href="https://www.flowserve.com/Products/Seals/Mixer" rel="nofollow">Flowserve</a>. Those can easily run into the tens of thousands of dollars depending on your specific requirements.</p> <p>There are also less expensive OEM seals designed for things like swimming pool pumps. Such as <a href="https://www.flowserve.com/Products/Seals/OEM-%26-Special-Duty/ci.PacSeal%2Cen_US.standard--------" rel="nofollow">these, also from Flowserve</a>. Again, depending on the specifics of your application you might get away with a sub $100 solution.</p> <p>At the low end of the scale is simply using one or more radial lip seals that ride on the shaft, usually on either side of the bearing for the shaft.</p> <p>You can also eliminate the need for a seal completely by using a magnetic drive mixer like these from <a href="http://www.magnasafe.com/" rel="nofollow">Magnasafe</a>. I have never seen a hand operated model though.</p> <p>I can give you further guidance, but not without more detail from your end.</p> <p>If your project involves any significant gas pressure, I strongly suggest that you following the ASME code for pressure vessels or contract with someone who is qualified and incorporating safety features to limit the possibility that the vessel could become over pressured and turn into a bomb. People continually underestimate the energy contained in relatively small volumes, like 20l, of pressurized gas and the consequences can be lethal.</p> <p>Disclosure: I used to work as an product development engineer in the Seals division of Flowserve. I cited their literature because I'm more familiar with their products. </p>

9018

How can I make a rotating hinge airtight?

2016-05-25T21:43:58.773

<p>How come axial compressors always narrow and turbines always widen towards the rear? From what I know:</p> <ul> <li>The speed of sound of air increases with temperature, therefore increases (decreases) as it goes through the compressor (turbine).</li> <li>The compressor and turbine blades at each stage should be meeting the air at just below the local speed of sound (more precisely critical Mach number).</li> <li>The rotational speed of all stages of a given spool is constant.</li> <li>The cross-section through which the air is flowing is <span class="math-container">$\pi(r_e^2-r_i^2)$</span>, where <span class="math-container">$r_e$</span> and <span class="math-container">$r_i$</span> are the exterior and interior radii, respectively.</li> </ul> <p>So shouldn't compressors widen and turbines narrow towards the rear, with the interior varying more than the exterior? And shouldn't turbines be much larger (~2x) than the compressors they drive?</p> <p>You can obviously see civil engines don't look like this:</p> <p>RR Trent 1000 (Courtesy of aviationgazette) <img src="https://www.aviationgazette.com/wp-content/uploads/2016/03/4-trent-1000-design.jpg" alt="RR Trent 1000"></p> <p>GE GEnX (Courtesy of www.mikejamesmedia.com) <img src="https://i.stack.imgur.com/886Hb.jpg" alt="GE GEnX"></p> <p>Anything I'm not taking, or taking too much, into account in my analysis?</p>

|geometry|turbomachinery|

<p>Note that there are multiple physical constraints in the design. Unlike in a turboshaft engine, in a turbofan, the torque of the turbine spool has to match the torque of it's associated compressor spool. Also, the axial load on the two spools must largely cancel, because thrust bearings are a pain in this environment. Balancing these requirement puts a severe crimp on the relative size of the two spools.</p> <p>Next, you have the problem of managing tip losses overall. Tip losses account for as much as 1/3 the aerodynamic losses in the machine. Low hub ratio designs minimize tip losses. Better tip design has allowed more efficient cascades and bigger hub ratios, but you still need a clearance that works for the entire spool. This tends to limit how much the hub ratio can vary over the stages of a single spool.</p> <p>And you also want to minimize the surface area of anything that requires cooling. Cooling stuff robs efficiency.</p>

9026

Turbocompressor spool geometry

2016-05-26T10:32:13.610

<p>How is the armature in an electric motor energized (i.e. main conditions on signal as function of position)? Let's consider the following parameters:</p> <ul> <li>$n_a$,$n_f$ the number of armature and field pole pairs, respectively</li> <li>$\gamma_{a,k}={\pi k\over n_a}$, $\gamma_{f,k}={\pi k\over n_f}$ the position of armature and field pole $k$ ($k\in \{1,\ldots,2n\}$)</li> <li>$\gamma$, the angular position of the shaft</li> <li>$I(\gamma)$, the current in armature as a function of position ($\forall\:\theta\in\mathbf{R}\::\:I(\theta+\pi)=-I(\theta)$)</li> <li>$I_k(\gamma)=I(n_f×(\gamma-\gamma_k))$, the current in armature pole $k$.</li> </ul> <p>The most important question is: should $I_k(\gamma_k)=0$ (i.e. should the current commutate whenever a field pole passes an armature pole)?</p> <p>I created the gif below to illustrate. The vectors represent the magnetic moments: blue is armature, yellow is field.</p> <p><a href="https://i.stack.imgur.com/aEoHU.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/aEoHU.gif" alt="Abstract motor model"></a></p>

|electrical-engineering|motors|

<p>Synchronous motors can be operated with a VFD in the same way that induction motors are usually operated. To start the motor, the stator is energized at a very low frequency, 1 Hz or so. The motor is then accelerated by increasing the frequency at a rate that is limited by the torque with which it is desired to accelerate the load or the torque that the system has a the capacity to deliver.</p> <p>The torque can be more tightly controlled by measuring the torque angle and controlling the armature current to provide the desired torque. The torque angle is the angle between the rotor field and the stator field. To control torque, the phase of the armature current would be adjusted so that the moving position of the magnetic poles created by the armature current would lead (for positive torque) the moving position of the field poles as indicated by position encoder feedback. This is called a "flux vector" torque control strategy.</p> <p>In an electric vehicle, for example, the "gas pedal" could provide a torque command. Pressing the pedal down would call for more torque and accelerate the vehicle. Lifting your foot from the pedal would call for less torque (or perhaps braking torque) and decelerate the vehicle.</p> <p>The relationship between the physical stator poles and the rotor poles is not relevant. The relevant thing is the angle between the rotating stator field and the rotor field.</p>

9032

Energizing of armature as function of position in electric motor

2016-05-26T18:31:47.350

<p>I am asked to design a timing/timer pulley based on a timer belt which comes in standard size. How do I do the calculation to find the number of teeth, depth and width of groove etc?</p>

|mechanical-engineering|design|gears|pulleys|

<p>There are two aspects to this problem: mechanical (number of teeth) and geometric (width and depth of groove).<br> Solving the mechanical part requires:</p> <ul> <li>$p$ the pitch (distance from the face of one tooth to the next) (probably mm)</li> <li>$v$ the speed of the belt (m/s)</li> <li>$f_i$ the required rotational speed of an element $i$ driven by the belt (Hz)</li> <li>$n_i$ the number of teeth on element $i$</li> </ul> <p>This gives us the useful rule that, for any element $i$ $$2\pi r_i f_i = v$$ from which we deduce that $$r_i = {v\over2\pi f_i}$$ and since $r_i = p n_i$, we can deduce that $$n_i = {r_i\over p}={v\over2\pi p f_i}$$ so we already have the radius and number of teeth for each element $i$.</p> <p>Solving the geometric part requires knowing the shape of the tooth (round, square, etc.), which depends entirely on the belt.</p>

9037

How to design a timer/timing pulley?

2016-05-26T19:20:03.217

<p>I have a frame, similar to a container frame, with two vessels inside. Each vessel is supported at four points on two beams, as shown in this mockup:</p> <p><a href="https://i.stack.imgur.com/LMlp4.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/LMlp4m.jpg" alt=""></a></p> <p>I am using ANSYS to simulate the loaded frame. How should I represent the impact of the weight of the vessels on the frame so that my structural analysis is correct? Should I divide the total weight of the vessels by four and place that force on each supporting plate or use remote force to represent the weight of each vessel?</p> <p>I know only the mass and center of gravity of vessels. I don't have full models of the vessels, just a model of the frame.</p>

|mechanical-engineering|structural-engineering|structural-analysis|ansys|

<p>If the vessels (and the load distribution of whatever they're containing) are truly symmetric (or close enough), then yes, you can simply replace them with four equal concentrated loads at the points where they are supported.</p> <p>If the vessels aren't symmetric but can be reasonably considered to have a pretty uniform stiffness (they don't have two really strong legs and two weak ones, for example), then you can use the vessels' centers of gravity to calculate the load distribution via this equation: $$R = P\cdot\left(1-\dfrac{c_x}{L_x}\right)\cdot\left(1-\dfrac{c_y}{L_y}\right)$$ where $c_x$ and $c_y$ are the distance in the $x$ and $y$ axes, respectively, from the desired support to the center of gravity, and $L_x$ and $L_y$ are the spans between supports in the respective axes.</p> <p>So, if the center of gravity is perfectly centered between all supports, then $\dfrac{c_x}{L_x}=\dfrac{c_y}{L_y}=\dfrac{1}{2}$ and $R = \dfrac{P}{4}$.</p> <p>If, however, the center of gravity is at coordinate $\left(\frac{L}{3}, \frac{L}{3}\right)$ (so, close to the bottom-left support), then the load will be distributed $\frac{4}{9}$ at the bottom-left support, $\frac{2}{9}$ at the top-left and bottom-right supports, and only $\frac{1}{9}$ at the top-right support.</p> <p>If the vessels are totally asymmetric, then you need to know the actual vessel geometry to calculate the load distributions.</p> <p>This is a simple assumption which is perfectly correct for static analyses. If you need to deal with vibrations or other dynamic behaviors, then there may be fluctuations on how much load goes to each support.</p>

9038

How to represent a load supported at multiple points on multiple beams?

2016-05-26T19:50:57.563

<p>I have heard of lasers being used to cut metal, however all laser cutting machines I've found can't cut metal, I'm guessing because it reflects the light?</p> <p>What is different in the designs of such machines between those that can and can't cut metal? Is it just accepted that energy is lost in reflection, and has a safe way of absorbing reflections?</p>

|lasers|cutting|

<p>I know this is an old question, but I'd like to expand on the general info alephzero stated.</p> <p>In my research for building my own metal cutting CNC laser, I've found that it takes at least a 150 watt CO2 laser to cut through even thin metals. This is also coupled with using a gas (N2 or O2) to either shield the material cut or to help it be cut. Thickness can also be a factor in whether a shield gas or a cutting gas is used.</p> <p><a href="http://am.co.za/laser/thickness" rel="nofollow noreferrer">http://am.co.za/laser/thickness</a></p> <p><a href="http://www.fabricatingandmetalworking.com/2012/11/nitrogen-vs-oxygenwhich-should-you-use-to-cut-steel/" rel="nofollow noreferrer">http://www.fabricatingandmetalworking.com/2012/11/nitrogen-vs-oxygenwhich-should-you-use-to-cut-steel/</a></p> <p>Due to the different reflective nature of different metals, not all metals can be cut at the same power. Thin steel and stainless steel can be cut with a 150 watt laser, but much more powerful lasers are needed for more highly reflective materials, like aluminum and silver.</p> <p>ND or ND:YAG lasers are commonly used for metal cutting, although they are much more expensive laser systems, and therefore much more expensive machines.</p> <p><a href="https://en.wikipedia.org/wiki/Laser_cutting" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Laser_cutting</a></p> <p>For my own machine, I've made it large (4'x8' capacity) and found that a 180 watt CO2 laser tube is between 6.5' and 7' long. Since my CNC frame is so big, I don't have to worry about that, though. The tube, if gotten directly from China can be as "little" as \$900-\$1000, but a more local USA dealer has them for \$1500-\$1600.</p> <p>I've spent around \$8000 on my own machine and still don't have the tube yet, but that's not the £100,000 to £500,000 alephzero mentions. I've seen machines (from China) that were around \$40,000 with power ratings to do metal, but that's still not something you are likely to get with the change in your couch.</p> <p>The thing about the reflections is that they definitely need to be managed. In too small of a laser cutter, the reflections will go back into the tube and basically set up a standing wave that eventually burns out the tube, since not enough of the light is absorbed or dispersed, exceeding the rated power of the tube for too long.</p> <p>In a laser that can cut through metal, the laser can still be reflected into the tube, but only until it pierces the material. At that point, the angle of reflection moves any reflected laser to a different angle. That usually means the laser is reflected into the bottom the of laser cabinet and handled/absorbed there, if not used to heat the metal. By the time the laser hits something at that point, it is usually not focused enough to do any damage.</p> <p>Also, many laser cutters have a full cabinet with windows that absorb the laser wavelength. In the one I have access to, it uses specially coated glass windows to absorb the laser. In the other one, it's Plexiglas/acrylic windows. Acrylic goggles/glasses are used when a laser cutter doesn't have a cabinet.</p>

9039

How to handle reflections when cutting metal with a laser?

2016-05-27T03:28:25.240

<p>About this time of year in my region people often look at the air conditioning compressor pumping heat to the outside air on one side of their house while their swimming pool heater tries to pump heat into the water on the other side, and wonder why they're paying twice to move heat. Inevitably someone thinks of submerging their compressor coils, but then they run the numbers and realize that it won't be too many days before their swimming pool is uncomfortably hot.</p> <p>But this leads to an interesting engineering question: What is the practical breakeven point between air-source and water-source heat pumps (for cooling indoor air)? I.e.: Is it possible, due to its higher thermal conductivity, for a water sink <em>at a higher temperature</em> to be more efficient for cooling than an air sink <em>at a lower temperature</em>?</p> <p>More specifically: Suppose we have an outdoor compressor with surface area <em>x</em> pumping <em>y</em> BTUs of heat. I am assuming that because water has something like 20 times the thermal conductivity of air that it would be a more efficient heat sink even when it's somewhat hotter than the air sink. (Is that correct?) But <em>how much more efficient</em>? I.e., at what temperature differential is air as efficient as water?</p> <p>(Or is this not a practical question? For example, we run fans to force more air through air-cooled compressors. That takes energy, but maybe it raises the effective thermal conductivity to the point that it is breakeven with a convective water sink ... even including the energy to drive the fan?)</p>

|thermodynamics|heat-transfer|hvac|

<p>The hot-side air coil of an AC installation (the part that heats the outdoor air) is a compromise between effectivity of heat transfer from the refrigerant to the air, the size, and the cost. Heat transfer to air typically comes with a significant thermal resistance, for two reasons. Firstly, air is a bad heat conductor. Secondly, air has a low heat capacity, so the air will heat up as it passes through the air coil.</p> <p>A quick Google didn't give me typical values of the effective thermal resistance of the air coil. If you have an IR thermometer, you could measure the temperature of the coil and compare it to the air temperature at the outdoor inlet. It wouldn't surprise me if the coil is 20 K higher in temperature than the outdoor air, assuming that the AC is running at full power.</p> <p>If you put a coil designed for air into water (assuming that corrosion etc. is not a problem), with some means of circulating the water, I think you can safely assume that the coil temperature will be very close to the water temperature. So, the break-even point would be when the water temperature is, say, 19 K higher than the air temperature.</p>

9045

Can it be more efficient to pump heat into warmer water than cooler air?

2016-05-27T19:41:43.003

<p>I have a very large antique mirror with some dense heavy wood framing it. I've stripped out some home electrical wire to use the copper ground wire as the cable that ties to the back of the mirror frame and holds it up on the wall. My question is how can I tell if the copper wire is strong enough to rely on and is it possible it will snap given time?</p> <p>If I search the internet for load bearing on copper wire it refers to the wrong kind of load i.e. electrical which is not what I want to know.</p>

|structural-engineering|

<p>You can find a table of mechanical properties of various gages of copper and aluminum wiring here:</p> <p><a href="http://www.okonite.com/engineering/conductor-properties.html" rel="nofollow">Wire Mechanical Properties</a></p> <p>You just have to Google the right terms.</p>

9058

Is it safe to hold up a large heavy mirror with copper wire?

2016-05-28T03:10:06.480

<p>Inspired by <a href="https://engineering.stackexchange.com/q/8923/3353">this question</a>, and also <a href="https://engineering.stackexchange.com/a/8918/3353">this answer</a> to a separate question. </p> <p>I have two beam models, one with two independent spans (a full hinge), another is a continuous beam over both spans (only column is hinged), the below is the diagram of both beams, and their force/bending moment distribution:</p> <p><a href="https://i.stack.imgur.com/MXqfI.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/MXqfI.png" alt="enter image description here"></a></p> <p>How can I derive from first principles of static analysis or from rules of thumb that continuous beams have a smaller deflection?</p>

|structural-engineering|beam|structural-analysis|

<p>To add to the answers already given, you are only scratching the surface with your 2 span beams. One you get to 3 or more spans, you will see that your bending moment is also reduced. Problems that may arise from having multiple continuous spans include having to restraint your beam's bottom chord, since it will be in compression at the supports. In steel construction, this is usually done by tying the open web steel joists to the column top / beam bottom chord.</p>

9066

Why does a continuous beam have less deflection than a pair of simply supported beams?

2016-05-28T12:19:31.523

<p>I have recently studied that the mmf produced by the stator in an electrical machine varies as a rectangular wave over the air gap periphery of the machine; And to make the wave similar to a sinusoidal wave we use techniques such as distributing the winding,short pitching the coils or using a current sheet.</p> <p>My question is,Why should I go through this process?Why not leave it as a rectangular wave,Since these processes reduce the mmf peak?</p>

|electrical-engineering|motors|

<p>You could actually leave the MMF rectangular (in fact <a href="https://en.wikipedia.org/wiki/Brushless_DC_electric_motor" rel="nofollow">brushless DC motors</a> are synchronous motors with a rectangular MMF) but there is a drawback: the torque induced by a rectangular MMF under steady conditions is much less constant. There is a torque ripple in the form of harmonics (the sinusoidal harmonics that are in the rectangular MMF).</p>

9069

MMF in AC windings

2016-05-28T16:24:08.417

<p>Why does positive phase and gain margins cause any system to be stable whereas negative phase and gain margins cause the system to be unstable ?</p>

|control-engineering|

<p>This is linked to the <a href="https://en.m.wikipedia.org/wiki/Nyquist_stability_criterion" rel="nofollow">Nyquist stability criterion</a>, which comes down to the number of encirclements of the minus one point of the open-loop in the Nyquist diagram. Namely the number of unstable closed-loop poles will be equal to the number open-loop unstable poles minus the number of counterclockwise encirclements of the minus one point.</p> <p>A negative phase margin means that you enter the unit disk, with the minus one point to your right (which is a part of a clockwise encirclement of the minus one point). You can only prevent an encirclement by decreasing in phase and go to the positive real axis outside the unit disk, to close the loop from the Nyquist contour, if you add unstable poles to the open-loop system. Adding this also still makes the closed-loop unstable. However the opposite does not have to be true, that positive phase margin means that the closed-loop will be stable. Namely when the open-loop system is unstable it is possible to have positive phase margin, while the closed-loop system is unstable. So positive phase margin is necessary but not sufficient for closed-loop stability.</p> <p>I am not sure if a negative gain margin has any practical meaning, because multiplying by a negative number also induces 180° of phase shift. So in order to identity such a margin you need to find a crossing of a multiple of 360°. You can get that the closed-loop will become unstable when you multiply the open-loop by a negative number, however the existence of a negative gain margin does not mean that the closed-loop will be unstable.</p>

9073

Condition for system stability

2016-05-29T12:50:13.717

<p>I'm building a APRS tracker for a high altitude balloon and am tracing the voltage as it's not powering up. I'm confused why it seems that this capacitor's ground appears to not be connected to anything. <a href="https://i.stack.imgur.com/2Qdzt.jpg" rel="nofollow noreferrer">topside of board</a></p> <p><a href="https://i.stack.imgur.com/tTwld.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/tTwld.jpg" alt="underside of board"></a></p>

|electrical-engineering|

<p>Actually, your picture clearly shows the pad to be connected to the top plane. Most likely that's ground. With the top plane being ground, how else would you expect pads to be connected to ground?</p> <p>You can clearly see the connections between the pad and the plane at the 12, 3, 6, and 9 hour clock positions. The reason the pad isn't connected to the plane all around is because that would make hand-soldering difficult. With copper connected everywhere, the plane would act like a large heat sink, preventing the pad from getting hot enough for solder to flow properly. These arrangements of isolated connections and gaps are often referred to as <i>thermals</i> in PC board lingo.</p> <p>You can see such thermals on the pad immediately right of the one you marked, on the right-most oval pad at lower right, and on the right pad in the row of 7 at bottom. All these pads are electrically connected to the top layer, which is largely a plane, most likely ground.</p>

9076

Why is capacitor 2 not grounded on this board?

2016-05-29T15:59:54.010

<p>I once saw on TV a factory dipping a product into a tank of paint and it magically emerged with a crisp pattern. (It was an Archery bow, and the pattern was army-style camouflage).</p> <p>Was I deceived or is it actually possible?</p>

|manufacturing-engineering|

<p>This sounds like <a href="https://en.wikipedia.org/wiki/Hydrographics_(printing)" rel="nofollow">hydrographic printing</a>. In short a design is printed onto a water soluble film and floated in a container of water to apply it to a surface. </p> <p>It is generally used to apply complex and detailed patterns to curved and irregularly shaped surfaces as it conforms well to contours, especially as a cheaper and more repeatable alternative to hand painting. </p>

9079

Multicolor dip-painting techniques

2016-05-29T17:45:56.153

<p>I would like to find out if it is possible to reduce the divergence of a red laser pointer beam from 40 degrees to 1 degree. </p> <p>@Ruslan wrote on October 3 2013 that because the active zone of a red diode laser has diameter of order of several micrometers the divergence of a red laser pointer beam is currently 40 degrees.</p> <p>Due to Heisenberg's uncertainty principle $\Delta\times\Delta p \gtrsim \hslash/2$, one can't really make a quantum have zero momentum in any direction. So you can't say that photons go in the same direction - this is just a simplified description of laser operation. In reality, the thinner the beam, the higher the divergence.</p> <p>Compare e.g. a DPSS laser (e.g. green laser pointer) with a diode laser (e.g. a red laser pointer).</p> <p>In a DPSS laser the active material will have diameter of order of hundreds of micrometer, and the exiting beam will start from even smaller diameter for various reasons. The divergence is quite small: if you remove the collimating lens, your light image from a green laser pointer will be several centimeters after the light goes several meters. Divergence angle would be $\lambda/d = 532\ \mathrm{nm}/100\ \mu\textrm{m} \approx 0.3°$.</p> <p>If you try doing the same with a red laser pointer, you'll see that its light diverges quite a lot: after going several centimeters in direction of propagation, it'll already give image of several centimeters. The reason for this is that active zone of diode laser has diameter of order of several micrometers. This makes output beam quite thin, making Δx small and thus Δp high, and this is what leads to high divergence. Divergence angle would be $\lambda/d = 640\ \mathrm{nm}/1\ \mu\textrm{m} \approx 40°$. Actual angle would depend on which transverse direction you select, because active zone is ∼10× longer in one direction than in another.</p> <p>In general, the thicker your starting laser beam, the more collimated it is, so if you manage to make a (visible wavelength) laser with beam starting at 1cm thickness, you'll have almost perfectly collimated laser beam.</p> <p>I would like to know if there is a method to enlarge the active zone of a red diode laser from a diameter of several micrometers to 30 micrometers. Also, could a lens with short focal length be used to reduce the divergence of a red laser pointer beam from 40 degrees to 1 degree?</p>

|optics|lasers|

<p>Of course, a lens with a short focal length can be used. The easiest way is to take the standard lens that comes with most laser pointers and move it a bit closer to the laser crystal. </p> <p>If you want the beam waist of the output beam to be exactly at the lens aperture, then you'll have to do a bit more work and find a suitable microlens. For a beam with 1 deg ($\theta=0.017$ rad) divergence, you'd need the beam to have a diameter $d\approx \lambda/\theta=37~\mathrm{\mu m}$ at the lens. The lens would be at a distance of 52 microns from the laser crystal. So that would be rather difficult to achieve.</p> <p>Update: clarification on beam waist and divergence. Here is what a Gaussian laser beam looks like this (image by DrBob on Wikipedia): </p> <p><img src="https://upload.wikimedia.org/wikipedia/commons/thumb/5/53/GaussianBeamWaist.svg/500px-GaussianBeamWaist.svg.png" alt="Gaussian beam"></p> <p>With a typical laser, the output coupler is at the narrowest part, the beam waist. The divergence angle $\Theta$ is a fixed parameter of the beam, no matter how far you propagate it. If you put a lens in the beam waist or within the Rayleigh length $z_R$, you can't decrease the far-field divergence angle $\Theta$, although you can focus it with a short focal length and achieve an even higher divergence after the focus. If you're not interested in the beam diameter near the beam waist, only in the spot size at 100 m away, then I wouldn't bother with microlenses and just use a regular lens at a few millimeters from the laser aperture.</p> <p>I can't comment on modifying the diameter of the laser crystal itself; if it is possible at all, it will require advanced semiconductor processing techniques.</p> <p>By the way: using Heisenberg to describe beam divergence as a function of beam diameter and wavelength doesn't make much sense, because Planck's constant cancels out once you write out $\Delta p$. The rule of thumb $\theta=\lambda/d$ works just as well for acoustic waves, which are not at all quantum mechanical.</p>

9083

Is it possible to reduce the divergence of a red laser pointer beam from 40 degrees to 1 degree?

2016-05-30T09:49:19.243

<p>Boiling temperature of Acetylene is -84°C and that of Petrol is 95°C (at Atmospheric Pressure). So the spark is efficient to ignite the compressed Acetylene and run the Engine. But why wont that work ? Is any other factor that I miss or misunderstood ?</p>

|automotive-engineering|fuel-economy|

<p>We prepared a bike propelled on acetylene for studying such problems. We found it was possible to do do, but some of the factors that govern the operation of acetylene powered engines are: cooling of the gas, proper mixing of gas and air and satisfactory cooling of the engine.</p>

9092

Why wont Petrol engine run on Acetylene?

2016-05-30T17:50:17.300

<p>I'm trying to understand the implications for buckling and column sizing in the drawings below.</p> <p>A 3000mm 120x60 RHS steel column supports a 350mm horizontal steel beam. The column base and right side of the beam are both fixed but can pivot. They are joined by a continual fillet weld "T" join, so the junction can rotate but will retain its angle. The only load is a vertical static force on the left-most 100mm of the horizontal beam. The column can be placed at any position under the beam, as shown in the 3 drawings (nearer to the right would be better). The thickness of the beam is thick enough that it will not deflect noticeably or materially. The thickness of steel in the RHS column is to be based on calculations.</p> <p>My question is this:</p> <p>The column's required cross-section clearly depends on the axial compressive force, which varies according to the column's position. But the eccentric load is a concern and I can't be quite sure if I am right to ignore it here.</p> <p><strong>Assuming</strong> that the beam is sufficiently rigid and won't noticeably deflect, <strong>then</strong> when the column position changes to the right (as shown), does the column experience only an increased axial compressive force, or should I also consider an additional bending moment imposed on the column as well?</p> <p>My reason for thinking I can ignore it and only consider the axial compressive load on the column, is that wherever the force applies on the beam, the column only experiences it as an axial load. But I'm not 100% sure since if the beam deflects even slightly (as it must), this must cause bending of the column (as they are welded) and therefore a bending moment in the column (however slight).</p> <p>Is this effect significant compared to the increased axial compressive force, or is it trivial and dominated by the increase in axial compression? Do I need to allow for an additional bending moment in my buckling calculations, in this situation, to ensure the column profile is adequate?</p> <p><a href="https://i.stack.imgur.com/xl49Z.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/xl49Z.png" alt="Drawing"></a></p> <p><em>(NOTE - I am based in Europe so any specific calculations would be Eurocode not North America code based. That said, the question isn't really about specific code calculations and it shouldn't matter)</em></p>

|beam|columns|buckling|moments|

<p>A general rule of thumb is that if the center of the vertical load is within the middle 1/3 of the column (kern), the eccentricity of the load can be ignored. You will have to calculate this for each possible position of your load.</p>

9101

Buckling of a cantilever with midway support

2016-05-30T18:42:35.683

<p><strong>How do you design against an earthquake?</strong> Do one initially consider earthquake in early design stages or is earthquake mitigation considered later in the design stage after conceptual stage? </p> <p>What are the advantages and disadvantages of the above-mentioned approach?</p> <p>I am unable to find enough material online on this topic can someone guide me on this so that I can further my research.</p>

|structural-engineering|civil-engineering|structural-analysis|

<p>The general shape of the building will affect earthquake resistance. I think L-shaped buildings are quite bad. You also need to know whether the soil is capable of supporting a heavy, tall building during an earthquake. If you want to keep the cost under control, better make a decision early in the design process on where to put supporting walls and structures so that you get sufficient resistance for a minimum amount of material.</p>

9102

How do you design against an earthquake?

2016-05-30T23:07:27.710

<p>Double diffraction is where a light beam passes through an grating with distance D1 between ridges and then passes through an grating with distance D2 The image after passing through diffraction gratings is a function of the product of two Fourier tranforms in the frequency domain. In other words, the product of two Fourier transforms in the frequency domain is equivalent to the convolution of the two Fourier integrals in the time domain.</p> <p>If we match spatial coherence length for a given wavelength with the spacing between the diffraction grating ridges, what is the intensity of the image after passing through two cascaded diffraction gratings each with distinct spacing between grating ridges?</p> <p>The reason I ask this question is to examine the best linear combination of 2 cascaded diffraction grating where the 2 unique diffraction gratings are centered around 2 of the primary colors, red green and blue. For example, suppose we want to filter an orange laser pointer in the far field(i.e 100 meters) to protect pilot's vision from the orange laser pointer or dual wavelength laser pointer upon takeoff or landing.</p> <p>Each of the diffraction gratings would be centered at a fixed visible light wavelength. Would it be necesssary for one of the cascaded diffraction grating to be centered around a variable visible light wavelength harnessing the Pockels effect where a varying electric field modulates the electro-optic behavior of the diffraction grating aperture radius? </p> <p>Please excuse my novice use of terminology and suggest how I should modify my presentation if you have time.</p> <p>Dr. Chih-Hsien Jason Lin, Ph.D and Dr. Yu-wha Lo , Ph.D of the Cornell Research Foundation filed U.S. Patent 5760960 A, "Cascaded self-induced holography", in June 1998 which harnesses Young's Interference experiment with diffraction gratings. This excellent research may be used to deflect and filter laser pointer beams in the visible light region. Cambridge University Press published a book , Optical coherence and quantum optics, by Professor Leonard Mandel and Professor Emil Wolf of the University of Rochester,around the same time analyzing the complex spatial coherence function from Young's two slit interference experiment using Fourier integrals. Pages 171 to Page 176 of this seminal book directly address the analysis of spatial coherence and optical filters for Young's two slit interference experiment </p> <p>"FIG. 2 illustrates that the size of the central opaque region 40 defines a target exposure area 76 on the substrate 60 by shading the target area 40 from direct illumination by the light source 72. This allows only the interference patterns from the cascaded grating pairs 56 and 58 to reach the target area 76. Preferably, the size of the target area 76 can be from a few millimeters to a centimeter square, and is comparable to the exposure size for an optical stepper."</p> <p>"The linearly polarized monochromatic light from the light source 72 indicated at 80 is first diffracted by the first and third mask gratings 44, 50 on the top surface of the mask 10. Most of the photon energy is contained in the first order diffraction beams labeled as +1 on the right and -1 on the left. Other diffractive beams and the zero order transmitted beams are not shown in FIG. 2 because they do not play a role in the final grating formation. When the +1 and -1 diffraction beams reach the second and fourth mask gratings 48 and 54 on the bottom of the mask 10, two new diffracted beams labeled as +2 and -2 are generated. As shown in FIG. 2, the target area 76 is the area where only the desired +2 and -2 beams can interfere, thereby exposing a grating 90 which has a period of approximately 1/4 of the periods in the mask gratings 44, 48, 50 and 54."</p> <p><a href="https://i.stack.imgur.com/A5qcI.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/A5qcI.png" alt="enter image description here"></a> <a href="https://i.stack.imgur.com/i1aAN.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/i1aAN.png" alt="enter image description here"></a></p> <p><a href="https://i.stack.imgur.com/LYnkg.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/LYnkg.jpg" alt="enter image description here"></a></p> <p>Grover A. Swartzlander wrote in 2000 in the attached article , "Peering into Darkness with a vortex spatial filter" , at the bottom of this question that "It is sometimes desireable to attenuate the intense glare of a bright coherent beam of light to enhance the detection of an incoherent background signal or a weak nearly collinear source". I would like to extend this research to protecting airline pilots from the effects of laser pointers shone from the ground at planes on takeoff and landing. The reason I wish to enhance the detection of a weaker incoherent background signal is to image this signal on a Planar transparent display.</p>

|optics|signal-processing|

<p>The original question was along the lines of: "what happens if light passes through cascaded gratings". Before that question can be answered, you have to understand what a single grating does. See the image below: a single beam of light will be split up into diffraction orders, numbered $n=0, \pm 1, \pm 2, \ldots$. </p> <p><a href="https://i.stack.imgur.com/vsHwo.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vsHwo.png" alt="Generic grating with laser beam"></a></p> <p>If the incident light beam is a mixture of wavelengths, then all diffraction orders other than 0 will have diffraction angles $\theta$ depending on the wavelength and the order, i.e., $\sin\theta=n\lambda/d$, where $d$ is the grating line pitch and $\lambda$ the wavelength. This is for a transmissive grating with normal incidence. If the incidence is close to normal, the formula is approximately correct; for large angles, the formula gets a bit more complicated (<a href="https://en.wikipedia.org/wiki/Diffraction_grating#Theory_of_operation" rel="nofollow noreferrer">Wikipedia</a>). For a reflective grating, the angles are roughly half of this. If there is no solution for $\theta$ for a given value of the $n$, it means that that diffraction order does not exist.</p> <p>There is no such thing as "centered on a particular wavelength" for a grating. At most, you can have "optimized for a particular wavelength" in the sense that the ratio between intensities of the diffraction orders has a particular value for a particular wavelength.</p> <p>A grating can be designed such that the zeroth order is suppressed, like in the picture below. Generally, a high suppression is only possible for a single wavelength-angle combination. In this picture, the incident angle is 0 deg (normal incidence).</p> <p><a href="https://i.stack.imgur.com/UDU23.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/UDU23.png" alt="grating without 0th order"></a></p> <p>You may think that suppressing the zeroth order is a good thing if you want to protect aircraft pilots against laser beams. However, if you consider the scenario where the laser beam can have any angle and may hit the grating at any position, then you'll see that this approach will not really help the pilot. In the following picture, laser beam B1 would have hit the pilot's eye if there hadn't been a grating and laser beam B2 would not have hit the pilot's eye. With the grating in place, the pilot's eye gets hit by the $n=-1$ diffraction from beam B2. For this grating, which has only the orders $n=\pm 1$, you roughly double the chance of a random beam hitting the pilot and any hit will be at half of the original intensity.</p> <p><a href="https://i.stack.imgur.com/fHKQv.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/fHKQv.png" alt="Two beams on a grating"></a></p> <p>Now that we understand what a single grating does, we can look at the case of cascaded gratings, which should speak for itself. Now, each incident beam will result in four new beams -- or even more if the gratings allow other orders than $n=\pm1$. If the gratings have different line pitches $d_1$ and $d_2$, then there will be different diffraction angles $\theta_1$ and $\theta_2$ for the first and the second grating.</p> <p><a href="https://i.stack.imgur.com/vUsDX.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vUsDX.png" alt="One beam, two gratings"></a></p> <p>Now, you can arrange the gratings in such a way that you put some object (maybe a third grating) at the location where beams B1b and B2a cross. Depending on the exact wavelength and path lengths of the two paths, the beams may interfere destructively or constructively. The patent and the paper that you refer to are along those lines. You should bear in mind that in such applications, the incident angle of the beam incident on the first grating and the distances between the gratings and the position of the detector should be exactly right within a fraction of the wavelength; otherwise it won't work. For the application that you have in mind (aircraft cockpits), this is not the way to go. Keep in mind that "background light" and "foreground light" is a subjective distinction. All that matters for the grating is the angle, the wavelength, and the position on the grating where a light ray hits the grating.</p> <p>Moreover, if you replace the cockpit windows by diffraction gratings, the light from the horizon and the landing strip will be diffracted as well and will make it nearly impossible to see objects on the outside. </p> <p>If you want to protect a pilot against the most common high-power laser pointers (532 nm wavelength), you should be using <a href="https://en.wikipedia.org/wiki/Dielectric_mirror" rel="nofollow noreferrer">dielectric mirrors</a> as filters. Such mirrors have a coating that is a stack of thin layers, tuned to reflect light of a certain wavelength. Schematically:</p> <p><a href="https://i.stack.imgur.com/PDRFq.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/PDRFq.png" alt="Dielectric mirror"></a></p> <p>They can be designed to reflect only over a very narrow wavelength range, in which case they are called "notch filter". You can buy these off the shelf, for example <a href="http://www.edmundoptics.eu/optics/optical-filters/notch-filters/od-4-notch-filters/67110/" rel="nofollow noreferrer">Edmunds OD4 notch filter</a>. A disadvantage for pilot protection is that the reflecting wavelength will depend on the angle of incidence $\alpha$. The only way to make them reflect 532 nm wavelength over a wide range of angles is by making them reflect over a large range of wavelengths. If you do that for green light, you will probably end up with something resembling very dark sunglasses, which only transmit deep red and deep blue light and block orange-yellow-green-cyan.</p>

9106

What happens when very bright coherent light and weaker incoherent background light passes through doubly cascaded diffraction gratings?

2016-05-31T09:33:22.507

<p>I want to know what torque a stepper motor can hold when stationary. They seem to only have one figure in specifications for holding torque which, I presume, is while coils are energised. </p> <p>How is this normally handled in usage of stepper motors? Is there a way of calculating holding torque while coils are not energised? I know braking can be done by shorting the coils, but it's obviously not as effective as just holding one coil energised to maintain position. Is one coil usually just kept energised to hold position with maximum torque? Is there a limit to how long one coil can remain energised?</p>

|electrical-engineering|motors|torque|stepper-motor|

<p>The value Detent Torque or <a href="https://en.wikipedia.org/wiki/Cogging_torque" rel="nofollow noreferrer">Cogging Torque</a> is what you should look for in the specification sheet. The detent torque is related to stepper motor power loss. The following is a reference from Minebea' web site.</p> <blockquote> <p><strong>Detent Torque</strong>: amount of torque that the motor produces when it is not energized. No current is flowing through the windings.</p> </blockquote> <p>Here is an excerpt from <a href="https://www.openimpulse.com/blog/wp-content/uploads/wpsc/downloadables/57BYGH420-Stepper-Motor-Datasheet.pdf" rel="nofollow noreferrer">57BYGH420 Stepper Motor Datasheet</a></p> <p><a href="https://i.stack.imgur.com/K3WMU.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/K3WMU.jpg" alt="Detent Torque Specification"></a></p> <hr> <p><strong>Referances:</strong></p> <ul> <li><a href="https://en.wikipedia.org/wiki/Cogging_torque" rel="nofollow noreferrer">Cogging torque</a></li> <li><a href="http://www.nmbtc.com/step-motors/engineering/torque-and-speed-relationship/" rel="nofollow noreferrer">Stepper Motor Speed and Torque Relationship</a></li> <li><a href="http://www.ewh.ieee.org/soc/es/Nov1998/12/BEGIN.HTM" rel="nofollow noreferrer">In-depth Learning of Cogging/Detenting Torque through Experiments and Simulations</a></li> <li><a href="http://www.geckodrive.com/support/step-motor-basics.html" rel="nofollow noreferrer">Step Motor Basics Guide</a></li> </ul>

9111

Stepper motor holding torque without coils energized?

2016-05-31T13:53:02.977

<p>Is there an energy related reason why, in the production of ammonia, air is not introduced in the primary reformer? In this case we don't need the second reformer. I could think of a few reasons like the fact that in the secondary reformer hydrogen gas is needed to react with the oxygen to water, which can react with $\text{CH}_4$. </p> <p>But this is not really energy related?</p>

|chemical-engineering|industrial-engineering|

<p><a href="http://www.owlnet.rice.edu/~ceng403/nh3ref97.html" rel="nofollow">This report from Rice University covers design and economics of synthesis gas reformers.</a></p> <p>Reaction 1: $CH_4+H_2O&lt;-&gt;CO+3H_2$ (Steam reforming)</p> <p>Reaction 2: $CO+H_2O&lt;-&gt;CO_2+H_2$ (Water gas shift)</p> <p>Reaction 3: $CH_4+3/2O_2&lt;-&gt;2H_2O+CO$ (Combustion)</p> <p>Reactions 1 &amp; 2 are endothermic and occur in the primary reformer. Reaction 3, the combustion reaction, is exothermic and occurs along with reactions 1 and 2 in the secondary reformer. </p> <p>Among many other reasons, air is not introduced in the primary reformer so that we get the most methane conversion to hydrogen with the smallest reformer possible. The link above explains sizing considerations between primary and secondary reformers. </p> <p>The secondary reformer is a much simpler vessel and has the primary purpose to burn up extra methane that has slipped out of the primary reformer. The heat from the exothermic combustion reaction is recovered and used for the primary reforming. </p> <p>To answer the air question from the comments, there are two ways to get nitrogen into the process. One method is using an air separation unit. The other is to put air in the secondary reformer, combust the methane and oxygen, and keep the left over nitrogen in the syngas for ammonia conversion. </p>

9115

Why isn't air added to the primary reformer in ammonia production?

2016-05-31T14:02:49.563

<p>This question might sound stupid as I'm more of a software guy and never really had to do control engineering (besides the 101 classes).</p> <p>So I have a second order system I could approximate, with the following denominator:</p> <p>$$s^2 + a_1s + a_0$$</p> <p><strong>Both $a_1$ and $a_0$ are positive, which means, for a 2^nd order system, that it is stable.</strong></p> <p>However this system's input is a motor's command (not a speed but a PWM cycle value), and the output is the robots angle.</p> <p>Technically, my angle will go to $+\infty$ with a constant input, which means my system will never stabilize. <strong>So the system should not be stable.</strong></p> <p>Besides, plotting my transfer function's output to a constant input using scipy gives me the expected result (an angle which ends up growing linearly as the speed stabilizes).</p> <p>So what am I missing here? Is my model completely wrong? Or did I miss something on the definition of "stability" itself? It would make sense if I worked on the speed but the variable I work on is the angle.</p>

|control-engineering|control-theory|stability|

<p>Without any more info, I think your problem arises from the values of <span class="math-container">$a_0$</span> and <span class="math-container">$a_1$</span>. The answer is a little involved, so a bit of systems background is necessary.</p> <h1>The short answer</h1> <p>Your system should be stable, but I don't think you are simulating your system long enough to see it stabilize. Calculate your settling time and simulate it for at least that long to see the complete response transient.</p> <h1>The long answer</h1> <p>The characteristic equation (i.e. denominator of the transfer function) of a general 2nd order system has the following convenient form:</p> <p><span class="math-container">$s^2 + 2 \zeta \omega_n s + w_n^2 $</span></p> <p>In this form, <span class="math-container">$\zeta$</span> is known as the damping ratio and <span class="math-container">$\omega_n$</span> is the natural frequency. As you noted this system is guaranteed to be stable if <span class="math-container">$w_n^2 &gt; 0$</span> and <span class="math-container">$2 \zeta \omega_n &gt; 0$</span>.</p> <p>In systems engineering we have several rules of thumb for determining the transient response of such systems. One of these is called the settling time <span class="math-container">$t_s$</span>. It is a measure of how long it takes for the amplitude of the system's oscillations to decrease below a certain threshold, usually 2% or 5% of the steady state value (i.e. the value at which the system stabilizes). You can calculate your settling time using the following approximation (this one is for the 2% criterion):</p> <p><span class="math-container">$t_s = \frac{4}{\zeta \omega_n}$</span></p> <p>Therefore, if you do not simulate your system until the settling time, you may see a system that does not appear to stabilize. Note that if <span class="math-container">$\omega_n$</span> is very small (therefore your <span class="math-container">$a_0$</span> parameter is very small) the settling time could be very long.</p> <p>For your reference, you can calculate <span class="math-container">$\omega_n$</span> and <span class="math-container">$\zeta$</span> from your parameters via the following:</p> <p><span class="math-container">$\omega_n = \sqrt{a_0} \qquad \zeta = \frac{a_1}{2 \omega_n}$</span></p> <h1>A demonstration</h1> <p>Check out what happens when I simulate a second order system with <span class="math-container">$\omega_n = 0.01$</span> rad/s and <span class="math-container">$\zeta = 0.9$</span> (i.e. <span class="math-container">$a_0 = 0.0001$</span> and <span class="math-container">$a_1 = 0.0180$</span>). Its settling time is approximately 444 seconds. The initial conditions are <span class="math-container">$\phi = 0$</span> and <span class="math-container">$\frac{d\phi}{dt} = 0$</span>. I applied a unit step input (i.e. a constant motor input equal to 1). The following figures show the same system with the exact same conditions, but simulated for different lengths of time. This was all done in MATLAB Simulink, my preferred math simulator.</p> <p><a href="https://i.stack.imgur.com/sD0Lh.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/sD0Lh.png" alt="Simulating for too little time" /></a></p> <p><a href="https://i.stack.imgur.com/KpTGU.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/KpTGU.png" alt="Simulating for an appropriate amount of time" /></a></p> <p>As you can see in the first image it appears that my system is unstable, but in the second image it becomes clear that it is simply taking a very long time to stabilize.</p> <p>I hope this helps.</p>

9116

A stable transfer function which diverges?

2016-05-31T19:09:18.750

<p>Sorry if this is in the wrong place; I haven´t posted in SE Engineering before, and the site said that SEE is also for students.</p> <p>I´m a High School senior and I´m debating between Computer Engineering and Computer Science. Can someone give me practical ¨real world¨ differences between the two? I <em>think</em> I´m more interested in embedded systems, VR, AI, and electronics than just software (though who really knows until I actually study it!) Especially virtual reality and AI because they´re such new, promising fields. If I want to work in those fields would either degree be more applicable? </p> <p>Thanks for your input! I´d really appreciate some replies from outside sources or personal experience.</p>

|computer|

<p>Computer Engineering tend to go more deeper into Electrical engineering topics such as electronics, transistors, micro processor design, embedded systems as well as some computer programming topic. Labs tend to be more hands on working with electronic components, as well as using electronic measurement instruments such as oscilloscopes and logic analyzers </p> <p>Computer Science programs tend go deeper into topic such as programming languages, database design, object oriented programming and so forth. The exposure to electronics are much lighter in a computer science programs. Labs tend to mostly focused on writing software programs. </p> <p>I suggest that you look at computer engineering and computer science curriculum from different universities. This is should help you get better understanding of the differences between programs.</p> <p>Here are the undergraduate curriculum requirement for <a href="http://www.ece.illinois.edu/academics/ugrad/curriculum/ce-curriculum-07.asp" rel="nofollow">Computer Engineering</a> and <a href="https://cs.illinois.edu/current-students/undergraduates/undergraduate-curriculum-requirements#BSENG" rel="nofollow">Computer science</a> at <a href="http://illinois.edu/" rel="nofollow">University of Illinois</a>. I suggest that you do side by side comparison between a both the programs.</p> <p>If you are more interested in Embedded Systems, I suggest that you follow a Computer Engineering program. I also suggest following some of the free online course. These will give you better first hand experiences on what is best for you. </p> <p>Good luck</p>

9119

Computer engineering vs Computer Science?

2016-06-01T10:29:36.307

<p>I'm building something to dry glass tubes. It's raised a question which I'd like some background knowledge on before I draw up requirements for the design.</p> <p>Say I want to dry something with heat alone, then obviously in a room environment (50% humidity, relatively free airflow, atmospheric pressure) heating something above 100°c would quickly evaporate fairly small amounts of water.</p> <p>Similarly, a powerful enough jet of air would blast off most water, quickly evaporating any remains.</p> <p>But say I use a heated airflow. Say the heater power is fixed, what would the relationship be between speed of drying and airflow? I think this is a hard question to answer without specifics, but perhaps one question in isolation could be answered:</p> <p>In principle, for small amounts of water, and a fixed heater power: if airflow is doubled, then the difference in temperature between that airflow and room temperature would be halved. Similarly, if the airflow is halved, the difference in temperature would be doubled. What would the effect on rate of evaporation be?</p> <p>Perhaps this is impossible to answer without more information, in which case, how might this problem be solved? A compromise would have to be established on criteria like cost of higher pressure air flow generation, noise levels, power consumption, energy efficiency, rate of drying, surface area of objects to be dried, etc etc etc. Might this be better off just developed with experimentation?</p> <p>Any level of guidance would be appreciated</p>

|drying|evaporation|

<p>If you are only interested in temperature, flow rate, and drying rate, then you could consider the following quantities for a given air temperature $T$:</p> <ul> <li>How much heat you need for evaporation (based on how much water is on the glassware)?</li> <li>How much heat you need to bring the entire mass of glassware to that temperature?</li> <li>What volume of air do you need to carry the moisture of one batch, assuming saturated vapor and how much heat do you need to heat all that air? </li> </ul> <p>This will give you the the minimum amount of energy needed to dry the glassware. To increase the rate of evaporation, I would put priority on increasing the water-carrying capacity of the resulting air, since evaporation rate is primarily driven by the difference between the actual water content of the air and the maximum water content. Specifically, the rate of evaporation is $$Q = K (c_{\mathrm{max}} - c), $$ where $K$ is a proportionality constant, $c_{\mathrm{max}}$ is the temperature-dependent maximum water content of the air (e.g., in kg/m3 units) and $c$ is the actual water content. For your system, you could start by taking $c$ as the final water content of the air after all the glassware has dried.</p> <p>Next, consider increasing the flow velocity of the air without increasing the volume of air. This will increase the proportionality constant $K$. Turbulent flow, as opposed to laminar flow, is a good thing here.</p>

9125

Optimum airflow for drying with heated air

2016-06-01T21:33:53.423

<p>I asked <a href="https://engineering.stackexchange.com/questions/9125/optimum-airflow-for-drying-with-heated-air">a question</a> where a comment pointed out:</p> <blockquote> <p>Requirements are things that you should be drafting to figure out what you want, before you know exactly how you are going to solve it.</p> </blockquote> <p>I was doing it backwards. I was going to define requirements after I had investigated design ideas to weigh up compromises that might be made. What I called requirements were flexible, depending on the cost vs effectiveness of design ideas.</p> <p>If I do not know what I want before I consider design ideas, if the possible criteria are not requirements, then what are they? </p> <p>The problem is the product fits into a larger system, ultimately the goal is profit from a business perspective, but considering the whole system's economics while designing the one product would be too complicated, so the requirement for "maximum profit" doesn't help!</p>

|project-management|

<p>I think I wrote that comment; I think I also added quite a few examples of what requirements could look like. In that question, you asked something along the lines of "what is the best way to ...". But "best" does not exist if you don't give any guidance on how to evaluate the 'goodness' of a particular solution. What's "best" for me can be something else than "best" for you. That's why I wrote: "to figure out what you want".</p> <p>In such an early stage of the development process, requirements will generally not be final and may not be very detailed. And it's fine if further investigations lead to the conclusion that the requirements were not realistic and need to be modified. </p> <p>Requirements and design can be cascaded:</p> <ol> <li><p>Requirement: make a profit with a factory producing N widgets per year at production cost P per widget, for a capital investment C.</p></li> <li><p>Design: the factory should have places to a) shape the widgets out of raw materials, b) clean them, c) dry them, d) package them. </p></li> <li><p>Requirement: the drying place may cost C/4 and should handle N widgets per year. The widgets come in clean and wet, go out clean and dry. Operation costs of the drying place is max P/10 per widget.</p></li> <li><p>Design: we could do it with a 100 kW heater, blowing 5000 m3/h of hot air over the widgets.</p></li> </ol> <p>Ideally, you break down the engineering tasks such that at the bottom of the chain, different engineers can be working on the different subsystems without having to be aware of all the problems that the other engineers are dealing with. In practice, it will often turn out that some of the choices at steps 2 and 3 were not right and the design (#2) and requirements (#3) have to be tweaked. </p> <p>But in particular if the work is split up over multiple people, having good requirements upfront, even if not final, reduces the probability that someone comes with a design that provokes a reaction along the lines of: "But that's a ridiculous solution; technically it meets the requirements, but that's not what I meant!"</p>

9131

What if project requirements depend on discoveries while designing?

2016-06-02T07:20:25.903

<p>I'm looking for a rubber thread that can be stretched elastically to 10 or 20 times of its original length. Does such material exist? Which is the best one in this respect?</p>

|materials|

<p>This is quite a broad question, but in an engineering context the main properties you are looking for are <a href="https://en.wikipedia.org/wiki/Elastic_modulus" rel="nofollow">elastic modulus</a> which is stress/strain is how much a material stretched under load and <a href="https://en.wikipedia.org/wiki/Ultimate_tensile_strength" rel="nofollow">ultimate tensile strength</a> OR <a href="https://en.wikipedia.org/wiki/Yield_(engineering)" rel="nofollow">yield strength</a> which are both measures of how much load a material will sustain before it fails. </p> <p>Materials with a high modulus of elasticity are broadly classified as 'rubbers', this may include 'natural' rubbers (derived from latex) or various silicone, vinyls, polyurethanes etc etc. This category of materials is usually referred to in materials science as <em>elastomer</em>s, although the term <em>rubber</em> is also widely used </p>

9136

What is the most stretchable rubber thread?

2016-06-02T22:54:38.580

<p>I need to control the temperature with high precision, and for this task would be good a temperature controller with pt-100 sensor, something like this one <a href="http://www.ink-bird.com/product/detail/p/PID_Temperature_Controller_ITC-100/id/7" rel="nofollow">PID Temperature Controller ITC-100</a>. As I see that, it can be very simple: there is an input (pt100 sensor), an output - to the heater element (my heating element needs 12V supply). However, I still have some things which are unclear for me:</p> <ol> <li>Is it generally supposed that there is a power source for the heater element which I also put in the circuit or usually the temperature controller gives the voltage at the output ? Or is that voltages supposed for the external SSR, not my heater?</li> <li>In many sources I have seen circuits with external SSR. However, such temperature controllers already have internal relay. Should it be possible without any problem just wire output to my heater (because in all schemes output goes to the external relays)?</li> <li>Wires soldered to the pt sensor shouldn't influence much to the measurement, right?</li> </ol>

|temperature|heating-systems|

<blockquote> <ol start="3"> <li>Wires soldered to the pt sensor shouldn't influence much to the measurement, right?</li> </ol> </blockquote> <p>A "pt100" sensor has a resistance of about 100 ohms (hence its name!) and changes resistance by about 0.385 ohms per degree C. Your question doesn't say what you mean by "high precision" but those two numbers should give you an idea of what lead resistance is acceptable for your application.</p> <p>If you want to use long leads and still get "high precision", it would be better to use a 3-wire RTD probe, as shown in the ITC-100 documentaton. That configuration uses a Wheatstone bridge type of circuit, and only the <em>difference</em> in resistance between the three leads contributes to the error. Three-wire probes can use leads up to about half a kilometer long. </p>

9149

Controlling the temperature with high precision

2016-06-03T18:08:33.200

<p>What is the name of the angle if I have larger wheels on the rear axles compared to the front, this would make the cars body and chassis not parralel to the road, unlike if the wheels were the same size.</p> <p>In other words the body and chassis would be at an angle. </p> <p>Does this angle have a certain technical name?</p>

|mechanical-engineering|

<p>The fixed (unloaded), or typical angle of the floor of the car is usually referred to as <em>rake angle</em> and can be a combination of suspension geometry (including relative wheel/tyre diameters) and the shape of the floor itself. </p> <p>It is particularly significant in cars which use ground effect to generate down-force where the floor, in conjunction with the front wing and a rear diffuser are designed to accelerate air under the floor of the car to create an area of low pressure by increasing the space under the floor and hence the cross sectional area of the flow from front to rear. Here the space enclosed between the floor and the road surface effectively forms an expanding nozzle although one of the technical challenges which limits this is to seal the gap between the floor and the road at the sides. There are examples where this has been achieved with flexible skirts (especially in early ground effect racing cars) although now it is more often achieved by precise control of airflow, especially vortices, off the front wing and other bodywork. </p> <p>As well as the tyre diameter this can be controlled by the suspension geometry and spring/damper characteristics and it is perfectly possible to have a level, or even actively adjustable rake angle regardless of relative tyre diameters. Indeed, cars with active suspension may adjust this angle in order to control the centre of pressure of downforce generated front to rear and/or to adjust the balance between downforce and drag. Some degree of control can also be achieved by flexible or movable sections of aerodynamic bodywork. </p> <p>This may also be expressed in terms of front to rear <em>ride height</em> or <em>ground clearance</em> (especially in an off-road context).</p>

9170

Entire Car Chassis & Body Angle

2016-06-04T16:22:06.553

<p><a href="https://en.wikipedia.org/wiki/Choked_flow#Mass_flow_rate_of_a_gas_at_choked_conditions" rel="noreferrer">According to compressible flow theory</a>, a downstream pressure of 0.528 times upstream pressure is sufficient to choke a flow device.</p> <p>So, how do industrial grade pressure regulators throttle a wide range of pressures (say 150&nbsp;bar to 20&nbsp;bar)?</p>

|fluid-mechanics|pressure|airflow|gas|compressible-flow|

<p>The short answer is that they don't. Choking is not a bad thing. For a gas flow regulator, it's actually a good thing, because under choking conditions, the mass-flow rate is no longer dependent on the downstream pressure. The mass-flow rate is still dependent on the upstream pressure and the size of the aperture connecting the high and low-pressure sides.</p> <p>In a reduction valve that is designed to keep the downstream pressure constant, rather than the mass flow rate, a mechanism is needed to open and close the aperture depending on whether the downstream pressure is below or above the pressure set point.</p> <p><strong>Update:</strong> if you're worried about energy losses in the gas stream, due to viscous dissipation: don't. If a <a href="https://en.wikipedia.org/wiki/Perfect_gas" rel="nofollow">perfect gas</a> (that's an ideal gas with an additional condition for the specific heat) enters a thermally insulated throttle device at low speed and exits it at low speed (through a pipe with a large diameter), then its temperature will be the same as when it entered. One way is to look at the enthalpy of the gas, $H=m C_V T+ pV$, which must be conserved if no energy is added or removed along the way. Because $pV=nRT$, this condition can only be met if the gas has the same temperature before and after. Note that I mentioned "low speed", which is what you typically want in a pressure regulator. In the nozzle, where the speeds are sonic, the energy balance will also include kinetic energy of the gas jet.</p> <p>Its may take a few moments for you to wrap your head around this (at least, for me it did, years ago).</p>

9183

How do gas flow regulators avoid choking?

2016-06-05T05:39:29.983

<p>I know this might sound ridiculous, but I am attempting to make a fan designed to induce lift of a 200 lbs object. By 'fan', I mean like the fans that lift a drone off the ground. I can only have a maximum of 3 fans.</p> <p>I am trying to lift the object at least 5 feet of the ground.</p> <p>Do you know what type of fans I would need, and/or the RPM of the fans? I am little bit confused on what to ask since there aren't any other questions like this. Also I am a beginner to engineering.</p>

|materials|design|

<p>Start out by doing a momentum ballance. You want to lift 200 pounds, so that means you need to push down on air with 200 pounds (890 N). That means you need to move (890 * Kg * m/s) of air every second.</p> <p>Note that this gives you a tradeoff. You could, for example, push 10 kg of air per second down at 89 m/s, 100 kg per second of air down at 8.9 m/s, etc. In theory, you get the same lift force either way. Basically you can push a little air a lot, or a lot of air a little.</p> <p>However, consider the power requirement. The kinetic energy of a moving mass is (&frac12; * m * V²). Note that annoying V² term. It says that pushing a little air a lot requires more power than pushing a lot of air a little. This is why sail planes have such wide wings. Power usage is a primary design consideration, so the large area being swept across by the wings each time unit allows for pushing a lot of air, which requires imparting less velocity to the air, which saves power since that's proportional to the square of the velocity.</p> <p>To get a start on your numbers, let's pick a air velocity and see what we get. I'm going to pick about half the speed of sound as example, so 165 m/s. This means you have to move 5.4 kg of air per second. At 20 &deg;C, that's about 4500 liters of air per second. The power you'd be imparting to the moving air is 73 kW.</p> <p>The area of that jet (for now assuming constant velocity crossection just to get a rough feel where we're at) is (4.5 m³/s)/(165 m/s) = 0.027 m². That's a circular jet 186 mm in diameter, or about 7.3 inches. Of course the flow crossection won't be constant, so the diameter will need to be bigger, and the power higher because some air will be faster than the average and some slower, but the faster will require disproportionately more power. I don't want to get into those fluid dynamics details here. I'll arbitrarily say the area needs to be about 50% larger, so 9 inches in diameter. I'll also round up the 73 kW to 100 kW.</p> <p>So, we have as a very rough starting point:</p> <p>&nbsp;&nbsp;165 m/s average jet speed<br> &nbsp;&nbsp;9 inch jet diameter<br> &nbsp;&nbsp;100 kW power required</p> <p>From there you can scale things accordingly. The area is inversely proportional to the jet speed, and the diameter goes with the square root of the area. The power goes with the jet speed.</p> <p>For example, let's see what we get with 1/10 the jet speed:</p> <p>&nbsp;&nbsp;16.5 m/s average jet speed<br> &nbsp;&nbsp;28 inch jet diameter<br> &nbsp;&nbsp;10 kW power required</p> <p>Again, these are very rough numbers with a lot of simplifying assumptions. The point is to give you a ballpark idea, and a basis for making relative tradeoffs from.</p>

10188

What type of fan would I need to induce the lift of a 200 lbs object?

2016-06-05T14:10:25.717

<p>It seems like pitch angle could be computed from <a href="http://43oh.com/2010/11/msp430-based-quadcopter-ez430-rf2500/" rel="nofollow">just gyro data</a>, which uses 2 gyros.</p> <p>However, isn't pitch computed from acceleration rather than gyro data?</p> <pre><code>pitch = 180 * arctan (accelerationX/sqrt(accelerationY*accelerationY + accelerationZ*accelerationZ))/PI </code></pre> <p>where no gyro data is needed. So how could a copter be balanced with just gyro data?</p>

|mechanical-engineering|sensors|

<p>It depends on the type of gyro. The cheap and available electronic gyros produce a signal proportional to the rotation speed. You don't get absolute angles. Even if you started stationary and level, then integrated from there, the answer would be useless in a just a few seconds due to offset drift.</p> <p>Some gyros are more accurate, but cost <i>much</i> more. Other gyros use a spinning flywheel as a reference and do produce angle signals, but those are also much more expensive, not to mention heavier and larger. NASA can afford to put them on spacecraft and commercial jetliners used to use them for long over-water hauls where there weren't navigation beacons, but you're not going to put one on your copter. Even then, the mechanical gyros drift over time too. </p> <p>In the end, <i>inertial navigation</i> fundamentally relies on integration to find position, so has a ever-increasing error envelope with time. This is one reason the astronauts on the way to the moon and back had to do regular position fixes from star sightings and the like. The inertial navigation provided the immediate position and orientation, but the accumulated drift had to be occasionally reset from absolute measurements using other methods.</p> <p>In the case of cheap MEMs gyros (the only kind you can reasonably put on your copter) they are really only good to tell you the angular velocity. The absolute vertical direction is determined by measuring the gravity vector, and you can measure the absolute angle about the vertical axis with a Hall effect magnetometer.</p> <p>Nowadays, of course you use GPS for absolute location. It doesn't tell you orientation, but once you're moving you can find heading from multiple GPS readings. GPS is used for the long term data, and inertial navigation only to fill in the short term information that is too detailed for GPS.</p>

10191

Balancing a copter with just gyro data

2016-06-05T19:37:09.907

<p>How can we calculate the minimum bend radius(the radius which it starts to buckle) of a annular cylindrical tube? Intuitively I guess that for two tubes having same wall thickness but different outer diameters, the larger diameter one will have larger bend radius. Is there an analytical way to confirm this?</p>

|mechanical-engineering|

<p>There is also a practical element in that there are quite a few different methods for bending tube ranging from a completely unsupported bend through simple jigs, <a href="http://www.baileigh.com/pipe-bender-rdb-125" rel="nofollow">draw benders</a> (which provide external support) to serious industrial NC <a href="http://www.baileigh.com/mandrel-benders-mb-nce-1-series" rel="nofollow">mandrel benders</a> which support the tube from the inside during bending and can give sophisticated control of strain rate and other parameters. </p> <p>There is also a difference between creating a bend at a point and rolling a continuous radius. </p> <p>In practice the behaviour can vary significantly even from one batch of stock to the next so any calculations need to be treated with some degree of caution. </p>

10198

minimum bend radius

2016-06-06T05:24:31.350

<p>If the angle corresponding to the open loop transfer function of a system at infinite frequency is smaller than that at zero frequency, the polar plot curve in the clockwise direction with increase in frequency. Why is it so ?</p>

|control-engineering|

<p>This is just how the phase component of of a complex value, returned by the transfer function, is defined. Phase is the angle between the positive real axis and the complex value in the counterclockwise direction.</p> <p>So if you start at some phase $\phi_1$ at a frequency of zero and go to phase $\phi_2$ at an infinite frequency, where $\phi_1&gt;\phi_2$, then on average the polar plot should rotate in the clockwise direction (the direction in which the phase decreases).</p>

10203

Polar plot direction with increase in frequency

2016-06-07T13:40:14.700

<p>I'm dealing with an application in which I use small DC brush geared motors to move a robot. The motors are driven by a standard FET driver. I do have the option to coast or brake the motor after every operation. Brake would be preferable since the robot would stop directly after a "forward" or "backward" operation. What I'm wondering is: would the brake operation affect negatively the lifespan of the motor or gearhead? Or is this effect negligible?</p>

|motors|robotics|

<p>The forces of breaking a motor by shorting its electrical connections are the same as starting the motor with full voltage applied. Actually, they'd only be the same in a ideal motor. In any real motor the breaking forces would be a little less due to the less than 100% efficiency of the motor.</p> <p>Motor torque is proportional to the current. During electronic breaking, that current is the generator voltage applied to the coils resistance. During startup, that current is the driving voltage applied to the coils resistance. Unless the mechanical load is providing significant power on its own, the breaking current, and hence the breaking torque, won't exceed the initial stall current when the full voltage is applied.</p> <p>This is all a long way of saying that the maximum mechanical stress is more likely at startup with full voltage, than when electrically breaking by shorting the motor connections.</p>

10222

DC brush geared motor: coast vs brake effects on motor/gearhead lifespan

2016-06-07T13:40:54.763

<p>Would adding a shade (i.e. a piece of plastic) at the back of an A/C increase the efficiency of a window unit A/C? I'm not talking about covering the fins but rather placing a sheet of plastic above to create shade without obstructing the air flow. Or does the forced air easily suck off the heat regardless of how much the sun is shining on the unit?</p>

|heat-transfer|

<p>There will obviously be some performance boost if you shade the fins/housing, but it will be negligible, since the ambient temperature <em>around</em> the a/c unit will not be affected by shading the unit itself. the ambient temperature on the north side of a structure can be considerably lower than the south or west side - my house has a carport on the north side of the house and I frequently see temperatures that are 10 degrees cooler than the south side of my house. </p>

10223

Will a shade hood increase a/c efficiency?

2016-06-07T13:42:18.907

<p>Here's a photo of a bearing plate where a bridge reinforced concrete beam meets the earthfill</p> <p><a href="https://i.stack.imgur.com/ccKdX.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ccKdX.jpg" alt="bearing plate"></a></p> <p>The bridge span is about 20 meters long and consists of two reinforced concrete beams each resting on two bearing plates like the one shown - one plate for each end of each beam, total four plates. The bridge holds a railway track designed for 25 tons per axle cars. The bearing plate is made of cast iron (or maybe steel) and consists of two large parts joined through a hinge.</p> <p>25 tons per axle cars means the bridge bears something like several hundred tons when a train is passing which we can assume causes at least one hundred tons per bearing plate shown. Yes, I just ignored the bridge weight.</p> <p>Not only the plate upper and lower surfaces are rather small but the plate further concentrates the accepted load and transfers it onto the hinge through even smaller surface. Basically this rather small hinge alone accepts more than one hundred tons. And this is designed on purpose.</p> <p>Why is the load deliberately concentrated instead of being distributed or at least transferred through some part with uniform section?</p>

|structural-engineering|civil-engineering|steel|bridges|

<p>If you go back to your basic engineering courses and look at you bending moment diagrams for beams, quite often they will be illustrated with pin roller supports. Pinned at 1 end only allowing rotation, and roller connection at the other allowing for rotation and horizontal translation. This makes the beam statically determinant.</p> <p>When this bridge was originally built, elastomeric/rubber bearing pads and a few others did not exist as an option. This design emulates our formulas that we use for design, or rather the formulas work with this arrangement. So this type of configuration is good. It allows us to use our formulas as intended, keeps the design simple, and used the technology of the time. In addition, as mentioned in other posts, it allows for rotation at the support as a result of live load, variations in dead, or sag after removal of shoring (assumed shored construction. If the girder were lifted into place after casting, the large steel plate embedded into the concrete allows for some inaccuracies in measurement of the span and even placement. It also allows for support should the beam shift slightly due to vibration or earthquake.</p> <p>Note, you will also see a similar configuration with steel girders on various types of bearings. I believe the term shoe or shoe plate will get used, though that may be more for buildings than bridges.</p> <p>On an aside</p> <p>When it comes to &quot;RAIL&quot; bridges, the vast majority in North America designed to AREMA will consist of simple spans whether it be single span or multi-span bridges. I found this statement funny while on my AREMA course as I just finished inspecting about a dozen rail bridges in my city with the vast majority not following this rule. In highway bridges you will tend to see continuous for live load and these bridges as a result are not statically determinant.</p>

10224

Why would a bearing plate that further concentrates load onto small surface be used?

2016-06-07T19:23:17.447

<p>I am trying to measure the height [level] of the water in a box that goes from -3°C (saltwater) with water and 150°C dry. Objects will be placed inside the box but will not take up the whole area. </p> <p>I need a sensor that is robust enough to go through those temperatures and measure accurately. I was thinking of using a float switch on a worm gear, but the motor couldn't stand the heat... </p> <p>What other methods/sensors are there? I also have a clear tube that runs outside of the tank meant for visually inspecting the height of the water.</p> <p><a href="https://engineering.stackexchange.com/questions/2695/can-i-use-a-ultrasound-sensor-to-measure-water-level">Can I use a ultrasound sensor to measure water level?</a> is a closely related question but focuses upon ultrasound. That won't work for my particular case because the contents of the box are effectively unknown. And capacitive techniques won't work either as the box will be grounded.</p>

|sensors|distance-measurement|

<p>I do work for a company that makes sensors that might be applicable. These sensors are used, among other places, to measure levels in ballast tanks of ocean-going ships. The possible gotcha for your case is the size of the tank. These sensors are meant for tanks depths of a few meters at least, up to 30 meters or so. When the tank is too small, the errors in the system usually become too significant. A few mm difference of liquid height in a ballast or fuel tank on a ship is irrelevant, especially considering the ship isn't sitting perfectly still and the liquid always sloshes around a bit anyway.</p> <p>The best description of the technology I found on their web site is at <a href="http://www.jowa-usa.com/jowausaproductslanding.cfm" rel="nofollow noreferrer">Jowa-USA Marine Products</a>. Imagine a resistance wire wrapped around a conducting rod, but with insulators in the right places so that at rest the wire doesn't touch the rod. When external pressure is applied, like from being submersed in a liquid, the wire is squished against the rod and makes electrical contact. The resistance between the rod and the wire at the top of the sensor then tells you the distance from the top to the first contact, from which you can derive the level of the liquid.</p> <p>Of course this is all packaged and encased to be impervious to various fluids. Different materials are used depending on the fluids. The two most common applications are for ballast tanks (seawater) and fuel tanks (diesel fuel). These things last many years on average.</p>

10232

How to measure height of water

2016-06-08T06:40:02.610

<p>Let's say you build a square reinforced concrete deck at height of, say, 3 meters, with the deck supported by its own steel reinforcement and by four reinforced concrete piers (call them, A, B,C, and D) below each corner of the square.</p> <p>If you then build four infill walls, A-B, B-C, C-D, D-A, are those walls under compressive load? From one side, the weight of the concrete slab would say yes, but on the other side the fact that deck was supported prior to the walls, maybe not?</p> <p>Is there a difference, in the context of the loads transmitted to a wall below, between a vertical stack as follows:</p> <p>wall - reinforced concrete ring beam - slab</p> <p>and</p> <p>wall - reinforced concrete deck with reinforcement designed to support the deck by itself and the piers alone.</p> <p>Here's an example, without the ring of beams:</p> <p><img src="https://i.stack.imgur.com/lijJP.png" alt=""></p>

|beam|concrete|

<p>It depends on the order of operations for the construction of the structure.</p> <p>Let's start assuming the structure is as simple as the image you've given us: a single story with a slab supported by columns.</p> <p>If the columns and slab are built, their formwork is removed and only then the wall is built, the wall shouldn't suffer any stresses. This is because the removal of the slab's formwork allowed it to deform itself into a stable configuration. When your wall comes up to the slab, it will simply close the gap, but won't have to support the slab since, well, the slab is already supporting itself. This will only be true in the short term, however. Over time, creep will try to further deflect the slab, and the wall will end up resisting that movement, generating a compressive force. This can be reduced by increasing the time between removal of the formwork and the raising of the wall, thereby increasing the amount of creep which occurs without harming the wall (this effect would be minimal, though).</p> <p>If the columns and slabs are built and the wall is raised before their formwork is removed, then the wall will suffer compression. In this case, when the formwork is removed, the wall will (try to) impede the deflection of the slab and will therefore be compressed.</p> <p>If the slab is actually supported by a ring of beams, the same effects apply as described above. That being said, the beams will dramatically increase the stiffness of the slabs, and will therefore reduce the deflection of the structure above the wall. This will reduce the compression suffered by the wall (either immediately or due to creep), but not eliminate it.</p> <p>If the structure actually has multiple floors, then the entire structure should be built before any walls are fully raised. The walls should also be raised starting from the top floor, and then proceeding downwards. This way, the load due to the wall of an upper floor will already have deformed the structure before a wall is raised. This way we can guarantee that a wall will not participate in supporting the walls above it. If the walls are made of brick, these can be raised along with the structure, so long as they leave a gap between their top course (row of bricks) and the bottom of the slab. Once the entire structure is raised, that gap can be filled (starting from the top floor).</p> <p>As @AndyT well mentioned in a <a href="https://engineering.stackexchange.com/questions/10236/does-a-concrete-slab-deck-cause-compressive-load-in-a-wall-beneath-it#comment18108_10242">comment</a> below, regardless of when the wall is raised, it will participate to some extent in resisting loads due to live/accidental loads.</p>

10236

Does a concrete slab (deck) cause compressive load in a wall beneath it?

2016-06-08T18:14:35.397

<p>I've been translating a truck scale description in a business offer letter, and the specification table there, listing the five parts comprising the prospective order, says on line five:</p> <blockquote> <p>Cable for connecting the load cell to the weighing terminal, <strong>bare end</strong>. Length: 30m.</p> </blockquote> <p>In the Russian original, the phrase "открытый конец" (literally "open end") most likely means that the cable has no connector attached to it. It probably comes out of the load cell, and then is cut at this length, 30m. The Buyer can attach the necessary connector. </p> <p>What would be a naturally-looking phrase for describing the same "bare end" in a similar English document? </p> <p>(<a href="https://ell.stackexchange.com/questions/93187/phrase-for-describing-a-cable-without-a-connector-at-the-end">Also asked on ELL SE</a>) </p>

|terminology|

<p>I'm guessing this is an electrical cable, not a mechanical cable. The "load cell" is the part that actually measures the load, and presumably the "weighing terminal" is the electronics that displays the measurements to the operator - possibly inside a building, to protect the electronics and the operator from the weather etc, while the actual truck weighing mechanism may be some distance away outside. Connecting the load cell to something with 30m of wire rope doesn't seem like a reasonable design for a "truck scale."</p> <p>"Bare end" or "bare ended" is understandable, but a more formal general term would be <strong>unterminated.</strong></p> <p>There are more specific terms for particular forms of "unterminated" wire and cable - see the "wire termination options" and "cable termination options" here: <a href="http://www.dsmt.com/resources/termination-options/" rel="nofollow">http://www.dsmt.com/resources/termination-options/</a> </p> <p>Most of those options only apply to a cable that is already cut to the correct length, and has been prepared ready to connect to something - but that doesn't seem to be the situation that you are describing.</p>

10244

Phrase for describing a cable without a connector at the end

2016-06-09T20:14:45.080

<p>I'd like to drill several very small (~0.015") but relatively long (~1") holes into a bar of stainless steel, as shown in the diagram below.</p> <p><a href="https://i.stack.imgur.com/uhBjJ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/uhBjJ.png" alt=""></a></p> <p>I did a quick check on McMaster and all the drilling lengths are much shorter (&lt;&lt;0.5") than I need. I'm not sure if it's feasible after all. How can I source this part, or machine it myself?</p>

|mechanical-engineering|steel|machining|

<p>It should be technically feasible using EDM (electrical discharge machining). There are purpose-built EDM "hole popper" drilling machines made for this type of work - the actual drilling time would be about 5 seconds or less per hole.</p> <p>Your OP says drilling "into" a bar to a depth of 1in, so it's not clear the holes are completely through 1in thick metal, or 1in deep blind holes in a thicker part. Laser drilling would be possible for through holes, but there might be problems getting an accurately controlled depth for blind holes.</p> <p>Neither of these options is a "do-it-yourself" machining operation, of course.</p>

10255

How can I drill a long, narrow hole into the end of a steel circular bar?

2016-06-10T06:53:08.840

<p>When there is one rectangular beam like below image,</p> <p><a href="https://i.stack.imgur.com/3kce0.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/3kce0.png" alt="enter image description here"></a></p> <p>I would like to know how the beam is deflected according to thicknesses of top plate and bottom plate.</p> <p>My concerning cross section patterns are like below.</p> <p><a href="https://i.stack.imgur.com/L7HG3.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/L7HG3.png" alt="enter image description here"></a></p> <p>The conditions are as below.</p> <p>Thicknesses of both side plates are same.</p> <p>Type 1 : thickness of top &lt; thickness of bottom</p> <p>Type 2 : thickness of top = thickness of bottom</p> <p>Type 3 : thickness of top > thickness of bottom</p> <p>Which type is less deformed? and Why?</p>

|mechanical-engineering|structural-engineering|civil-engineering|beam|

<p>The deflection in type 2 is the least. Type one and type 3 will deform more because of the fact that neutral axis if the beam will move nearer to top or bottom.<br> When this happens the new strain on the narrower flange if the beam will be bigger resulting in more elongation at that flange. Meaning more deflection of the beam.<br> One crude way of looking at it is the new neutral axis will consume some of the web of the beam from the thick side. Hence the beam bends more to stress the left over cross section more to make up for its reduced area. It can easily be seen in drawing of beams cross section stress/ strain digram which i tried to draw on my phone by finger! <a href="https://i.stack.imgur.com/EqHCf.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/EqHCf.jpg" alt="cross section if the beam."></a></p>

10261

Deflection of beam, according to thickness of top plate and bottom plate

2016-06-12T13:19:03.080

<p>How does the Michelson interferometer measure the self-coherence function of coherent light in an incoherent background? <a href="http://discovery.ucl.ac.uk/2670/" rel="nofollow"><em>Detection of coherent light in an incoherent background</em> (Coutinho et al., 1999)</a> is still the best analysis and implementation of this concept despite considerable progress in semiconductor fabrication and photonics since its publication.</p> <p>The reason I ask this question is because airplane pilots need to see the landing strip, ground landmarks and the horizon while at the same time harmful visible light laser rays are filtered.</p>

|electrical-engineering|optics|lasers|

<p>The order of the dark fringe of the coherent source to lock to is the first order because the self coherence function's first minimum corresponding to the sinc function's envelope is located at an interval of lambda\4 multiplied by the distance d where lambda equals wavelength and d is equal to the distance between a fixed mirror's image and an moving mirror's image.</p> <p>The information of the greatest interest is the spectrum in one branch of the interferometer called the interferogram's inverse Fourier transform. As a result, the interferometer's signal as a function of the optical path difference and the spectrum as a function of wavenumber, also known as 1/wavelength,form a Fourier transform pair.</p> <p>If you wish to explore this topic, please read chapter 6 , Fourier Trahsform Spectroscopy , in the book written by the Finnish professors, J. Kauppinen and J. Partanen , titled Fourier Transforms in Spectroscopy.</p> <p>I enjoyed chatting with the New York genius expert, @Chris Mueller , about this topic. </p> <p>Thank you</p>

10277

How does the Michelson interferometer measure the self-coherence function of coherent light in an incoherent background?

2016-06-12T15:56:42.520

<p>I have a hole that contains metal shavings after I have tapped it. Is there a common method to clean out metal shavings from between the threads of the tapped hole?</p>

|fasteners|

<p>There are a few ways to do this obviously it's easier to clean a through hole than a blind one. </p> <ul> <li>Run a clean second or plug tap through the hole. If you keep a tap just for this purpose it also has the benefit of keeping one 'fresh' tap which clean clean up any burrs of defects from worn taps when initially cutting the thread. </li> <li>Flush the hole with cutting fluid, suds or degreaser. </li> <li>Scrub the thread with a cylindrical nylon or brass brush of the appropriate diameter, a brush mounted in a cordless drill can be an effective way to clean multiple threads. </li> <li>Cut a slot down the length of a bolt or stud of the appropriate size and carefully run it into the thread </li> <li>Hold the part inverted and tap with a nylon or hide hammer. </li> </ul> <p>Edit : good point about compressed air. A probe type air duster (either from a compressor or a hand held aerosol) or even a manual blower can be effective, especially where threads are hard to get at by other means. This tends to work best when the hole is reasonably free of oil and grease so flushing with something like a solvent or alkaline degreaser first may help. Obviously care needs to be taken not to blow solvents or debris from the hole into your own (or anybody else's) face and eyes. </p>

10278

Cleaning Shavings from Tapped Hole

2016-06-13T03:20:48.750

<p>In Russia, there's a special term for a pipe that is not circular in its section: "a profiled pipe" (профильная труба). There's even a Wikipedia page for it: <a href="https://ru.wikipedia.org/wiki/%D0%9F%D1%80%D0%BE%D1%84%D0%B8%D0%BB%D1%8C%D0%BD%D0%B0%D1%8F_%D1%82%D1%80%D1%83%D0%B1%D0%B0" rel="nofollow noreferrer">профильная труба</a>. Such pipes come in a wide variety of sizes and usually have an oval, square or rectangular section ("profile" in Russian), but other kinds of sections are possible. </p> <p>I wonder what is the common cover-all technical term describing "any kind of non-circular pipe". </p> <p><a href="https://i.stack.imgur.com/uWE6V.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/uWE6V.jpg" alt="enter image description here"></a></p> <p>Example of usage:</p> <blockquote> <p>Pervouralsk New Pipe Plant produces a range of non-circular pipes, including pipes of variable cross-section. </p> </blockquote> <p>(<a href="https://ell.stackexchange.com/questions/93589/looking-for-a-common-term-for-non-circular-pipe">Also asked on ELL SE</a>)</p>

|terminology|

<p>For structural purposes these are usually called <a href="https://en.wikipedia.org/wiki/Hollow_structural_section" rel="nofollow">Hollow Structural Sections</a> (HSS). For non-structural purposes <em>Mechanical Tubing</em> is often used.</p> <p>In Europe the main body that deals with structural design of these is <a href="http://www.cidect.com/en/" rel="nofollow">CIDECT</a>, who have a series of useful design guides. The equivalent would probably be the <a href="https://steeltubeinstitute.org" rel="nofollow">Steel Tube Institute</a> in the USA who also deal with non-structural components.</p>

10284

Looking for a common term for "non-circular pipe"